Advance Design 2010 - Validation Guide

Advance Design

Validation Guide

ADVANCE VALIDATION GUIDE

i

Contents

1 INTRODUCTION....................................................................................................................................................................1 1.1 Test description documents coding.....................................................................................................................2 1.2 Test description documents coding example......................................................................................................3 1.3 Results comparison documents coding example ...............................................................................................4 1.4 Margin of error........................................................................................................................................................4 1.5 Synthetic sheet of 2010 version............................................................................................................................5

2 DETAILED TESTS DESCRIPTION .......................................................................................................................................9 2.1 Test No. 01-0001SSLSB_FEM: Cantilever rectangular plate ............................................................................10

2.1.1 Description sheet .....................................................................................................................................10 2.1.2 Overview ..................................................................................................................................................10 2.1.3 Displacement of the model in the linear elastic range ..............................................................................11 2.1.4 Results sheet ...........................................................................................................................................11

2.2 Test No. 01-0002SSLLB_FEM: System of two bars with three hinges ............................................................12 2.2.1 Description sheet .....................................................................................................................................12 2.2.2 Overview ..................................................................................................................................................12 2.2.3 Displacement of the model in C ...............................................................................................................13 2.2.4 Bars stresses ...........................................................................................................................................13 2.2.5 Shape of the stress diagram ....................................................................................................................14 2.2.6 Results sheet ...........................................................................................................................................14

2.3 Test No. 01-0003SSLSB_FEM: Circular plate under uniform load ...................................................................15 2.3.1 Description sheet .....................................................................................................................................15 2.3.2 Overview ..................................................................................................................................................15 2.3.3 Vertical displacement of the model at the center of the plate ...................................................................16 2.3.4 Results sheet ...........................................................................................................................................17

2.4 Test No. 01-0004SDLLB_FEM: Slender beam with variable section (fixed-free) ............................................18 2.4.1 Description sheet .....................................................................................................................................18 2.4.2 Overview ..................................................................................................................................................18 2.4.3 Eigen mode frequencies...........................................................................................................................19 2.4.4 Results sheet ...........................................................................................................................................20

2.5 Test No. 01-0005SSLLB_FEM: Tied (sub-tensioned) beam ..............................................................................21 2.5.1 Description sheet .....................................................................................................................................21 2.5.2 Overview ..................................................................................................................................................21 2.5.3 Compression force in CE bar ...................................................................................................................22 2.5.4 Bending moment at point H......................................................................................................................23 2.5.5 Vertical displacement at point D...............................................................................................................24 2.5.6 Results sheet ...........................................................................................................................................25

2.6 Test No. 01-0006SDLLB_FEM: Thin circular ring fixed in two points..............................................................26 2.6.1 Description sheet .....................................................................................................................................26 2.6.2 Overview ..................................................................................................................................................26 2.6.3 Eigen mode frequencies...........................................................................................................................27 2.6.4 Results sheet ...........................................................................................................................................28

2.7 Test No. 01-0007SDLSB_FEM: Thin lozenge-shaped plate fixed on one side (α = 0 °) .................................29 2.7.1 Description sheet .....................................................................................................................................29 2.7.2 Overview ..................................................................................................................................................29 2.7.3 Eigen mode frequencies relative to the α angle .......................................................................................30 2.7.4 Results sheet ...........................................................................................................................................30

2.8 Test No. 01-0008SDLSB_FEM: Thin lozenge-shaped plate fixed on one side (α = 15 °) ................................31 2.8.1 Description sheet .....................................................................................................................................31

ADVANCE DESIGN VALIDATION GUIDE

ii

2.8.2 Overview.................................................................................................................................................. 31 2.8.3 Eigen modes frequencies function by α angle ......................................................................................... 32 2.8.4 Results sheet ........................................................................................................................................... 32

2.9 Test No. 01-0009SDLSB_FEM: Thin lozenge-shaped plate fixed on one side (α = 30 °)................................ 33 2.9.1 Description sheet ..................................................................................................................................... 33 2.9.2 Overview.................................................................................................................................................. 33 2.9.3 Eigen mode frequencies relative to the α angle....................................................................................... 34 2.9.4 Results sheet ........................................................................................................................................... 34

2.10 Test No. 01-0010SDLSB_FEM: Thin lozenge-shaped plate fixed on one side (α = 45 °)................................ 35 2.10.1 Description sheet ..................................................................................................................................... 35 2.10.2 Overview.................................................................................................................................................. 35 2.10.3 Eigen mode frequencies relative to the α angle....................................................................................... 36 2.10.4 Results sheet ........................................................................................................................................... 36

2.11 Test No. 01-0011SDLLB_FEM: Vibration mode of a thin piping elbow in plane (case 1) .............................. 37 2.11.1 Description sheet ..................................................................................................................................... 37 2.11.2 Overview.................................................................................................................................................. 37 2.11.3 Eigen mode frequencies .......................................................................................................................... 38 2.11.4 Results sheet ........................................................................................................................................... 38



2.14 Test No. 01-0014SDLLB_FEM: Thin circular ring hang from an elastic element............................................ 43 2.14.1 Description sheet ..................................................................................................................................... 43 2.14.2 Overview.................................................................................................................................................. 43 2.14.3 Eigen mode frequencies .......................................................................................................................... 44 2.14.4 Results sheet ........................................................................................................................................... 45

2.15 Test No. 01-0015SSLLB_FEM: Double fixed beam with a spring at mid span................................................ 46 2.15.1 Description sheet ..................................................................................................................................... 46 2.15.2 Overview.................................................................................................................................................. 46 2.15.3 Displacement of the model in the linear elastic range.............................................................................. 47 2.15.4 Results sheet ........................................................................................................................................... 48

2.16 Test No. 01-0016SDLLB_FEM: Double fixed beam ........................................................................................... 49 2.16.1 Description sheet ..................................................................................................................................... 49 2.16.2 Overview.................................................................................................................................................. 49 2.16.3 Displacement of the model in the linear elastic range.............................................................................. 50 2.16.4 Eigen mode frequencies of the model in the linear elastic range............................................................. 50 2.16.5 Results sheet ........................................................................................................................................... 52

2.17 Test No. 01-0017SDLLB_FEM: Short beam on simple supports (on the neutral axis) .................................. 53 2.17.1 Description sheet ..................................................................................................................................... 53 2.17.2 Overview.................................................................................................................................................. 53 2.17.3 Eigen modes frequencies ........................................................................................................................ 54 2.17.4 Results sheet ........................................................................................................................................... 55

2.18 Test No. 01-0018SDLLB_FEM: Short beam on simple supports (eccentric) .................................................. 56 2.18.1 Description sheet ..................................................................................................................................... 56 2.18.2 Overview.................................................................................................................................................. 56 2.18.3 Eigen modes frequencies ........................................................................................................................ 57


iii

2.18.4 Results sheet ...........................................................................................................................................59 2.19 Test No. 01-0019SDLSB_FEM: Thin square plate fixed on one side ...............................................................60

2.19.1 Description sheet .....................................................................................................................................60 2.19.2 Overview ..................................................................................................................................................60 2.19.3 Eigen modes frequencies.........................................................................................................................61 2.19.4 Results sheet ...........................................................................................................................................62

2.20 Test No. 01-0020SDLSB_FEM: Rectangular thin plate simply supported on its perimeter ...........................63 2.20.1 Description sheet .....................................................................................................................................63 2.20.2 Overview ..................................................................................................................................................63 2.20.3 Eigen modes frequencies.........................................................................................................................64 2.20.4 Results sheet ...........................................................................................................................................65

2.21 Test No. 01-0021SFLLB_FEM: Cantilever beam in Eulerian buckling .............................................................66 2.21.1 Description sheet .....................................................................................................................................66 2.21.2 Overview ..................................................................................................................................................66 2.21.3 Critical load on node 5..............................................................................................................................67 2.21.4 Results sheet ...........................................................................................................................................67

2.22 Test No. 01-0022SDLSB_FEM: Annular thin plate fixed on a hub (repetitive circular structure) ..................68 2.22.1 Description sheet .....................................................................................................................................68 2.22.2 Overview ..................................................................................................................................................68 2.22.3 Eigen modes frequencies.........................................................................................................................69 2.22.4 Results sheet ...........................................................................................................................................69

2.23 Test No. 01-0023SDLLB_FEM: Bending effects of a symmetrical portal frame..............................................70 2.23.1 Description sheet .....................................................................................................................................70 2.23.2 Overview ..................................................................................................................................................70 2.23.3 Eigen modes frequencies.........................................................................................................................71 2.23.4 Results sheet ...........................................................................................................................................72

2.24 Test No. 01-0024SSLLB_FEM: Slender beam on two fixed supports ..............................................................73 2.24.1 Description sheet .....................................................................................................................................73 2.24.2 Overview ..................................................................................................................................................73 2.24.3 Shear force at G.......................................................................................................................................75 2.24.4 Bending moment in G...............................................................................................................................76 2.24.5 Vertical displacement at G .......................................................................................................................77 2.24.6 Horizontal reaction at A ............................................................................................................................78 2.24.7 Results sheet ...........................................................................................................................................78

2.25 Test No. 01-0025SSLLB_FEM: Slender beam on three supports.....................................................................79 2.25.1 Description sheet .....................................................................................................................................79 2.25.2 Overview ..................................................................................................................................................79 2.25.3 Bending moment at B...............................................................................................................................80 2.25.4 Reaction in B............................................................................................................................................80 2.25.5 Vertical displacement at B........................................................................................................................81 2.25.6 Results sheet ...........................................................................................................................................81

2.26 Test No. 01-0026SSLLB_FEM: Bimetallic: Fixed beams connected to a stiff element...................................82 2.26.1 Description sheet .....................................................................................................................................82 2.26.2 Overview ..................................................................................................................................................82 2.26.3 Deflection at B and D ...............................................................................................................................83 2.26.4 Vertical reaction at A and C......................................................................................................................84 2.26.5 Bending moment at A and C ....................................................................................................................84 2.26.6 Results sheet ...........................................................................................................................................84

2.27 Test No. 01-0027SSLLB_FEM: Fixed thin arc in planar bending......................................................................85 2.27.1 Description sheet .....................................................................................................................................85 2.27.2 Overview ..................................................................................................................................................85 2.27.3 Displacements at B ..................................................................................................................................86 2.27.4 Results sheet ...........................................................................................................................................87


iv

2.28 Test No. 01-0028SSLLB_FEM: Fixed thin arc in out of plane bending............................................................ 88 2.28.1 Description sheet ..................................................................................................................................... 88 2.28.2 Overview.................................................................................................................................................. 88 2.28.3 Displacements at B.................................................................................................................................. 89 2.28.4 Moments at θ = 15°.................................................................................................................................. 89 2.28.5 Results sheet ........................................................................................................................................... 89

2.29 Test No. 01-0029SSLLB_FEM: Double hinged thin arc in planar bending...................................................... 90 2.29.1 Description sheet ..................................................................................................................................... 90 2.29.2 Overview.................................................................................................................................................. 90 2.29.3 Displacements at A, B and C ................................................................................................................... 91 2.29.4 Results sheet ........................................................................................................................................... 92

2.30 Test No. 01-0030SSLLB_FEM: Portal frame with lateral connections............................................................. 93 2.30.1 Description sheet ..................................................................................................................................... 93 2.30.2 Overview.................................................................................................................................................. 93 2.30.3 Displacements at A.................................................................................................................................. 94 2.30.4 Moments in A........................................................................................................................................... 95 2.30.5 Results sheet ........................................................................................................................................... 95

2.31 Test No. 01-0031SSLLB_FEM: Truss with hinged bars under a punctual load.............................................. 96 2.31.1 Description sheet ..................................................................................................................................... 96 2.31.2 Overview.................................................................................................................................................. 96 2.31.3 Displacements at C and D ....................................................................................................................... 97 2.31.4 Results sheet ........................................................................................................................................... 97

2.32 Test No. 01-0032SSLLB_FEM: Beam on elastic soil, free ends....................................................................... 98 2.32.1 Description sheet ..................................................................................................................................... 98 2.32.2 Overview.................................................................................................................................................. 98 2.32.3 Bending moment and displacement at C ................................................................................................. 99 2.32.4 Displacements at A................................................................................................................................ 100 2.32.5 Results sheet ......................................................................................................................................... 100

2.33 Test No. 01-0033SFLLA_FEM: EDF Pylon ....................................................................................................... 101 2.33.1 Description sheet ................................................................................................................................... 101 2.33.2 Overview................................................................................................................................................ 101 2.33.3 Displacement of the model in the linear elastic range............................................................................ 103 2.33.4 Results sheet ......................................................................................................................................... 105

2.34 Test No. 01-0034SSLLB_FEM: Beam on elastic soil, hinged ends................................................................ 106 2.34.1 Description sheet ................................................................................................................................... 106 2.34.2 Overview................................................................................................................................................ 106 2.34.3 Displacement and support reaction at A ................................................................................................ 107 2.34.4 Displacement and bending moment at D ............................................................................................... 108 2.34.5 Results sheet ......................................................................................................................................... 109

2.35 Test No. 01-0035SSLPB_FEM: Plate with in plane bending and shear ......................................................... 110 2.35.1 Description sheet ................................................................................................................................... 110 2.35.2 Overview................................................................................................................................................ 110 2.35.3 Planes stresses in (x,y).......................................................................................................................... 110 2.35.4 Results sheet ......................................................................................................................................... 111

2.36 Test No. 01-0036SSLSB_FEM: Simply supported square plate..................................................................... 112 2.36.1 Description sheet ................................................................................................................................... 112 2.36.2 Overview................................................................................................................................................ 112 2.36.3 Vertical displacement at O ..................................................................................................................... 113 2.36.4 Results sheet ......................................................................................................................................... 113

2.37 Test No. 01-0037SSLSB_FEM: Caisson beam in torsion ............................................................................... 114 2.37.1 Description sheet ................................................................................................................................... 114 2.37.2 Overview................................................................................................................................................ 114 2.37.3 Displacement and stress at two points................................................................................................... 115


v

2.37.4 Results sheet ......................................................................................................................................... 116 2.38 Test No. 01-0038SSLSB_FEM: Thin cylinder under a uniform radial pressure ............................................ 117

2.38.1 Description sheet ................................................................................................................................... 117 2.38.2 Overview ................................................................................................................................................ 117 2.38.3 Stresses in all points .............................................................................................................................. 118 2.38.4 Cylinder deformation in all points ........................................................................................................... 118 2.38.5 Results sheet ......................................................................................................................................... 118

2.39 Test No. 01-0039SSLSB_FEM: Square plate under planar stresses.............................................................. 119 2.39.1 Description sheet ................................................................................................................................... 119 2.39.2 Overview ................................................................................................................................................ 119 2.39.3 Displacement of the model in the linear elastic range ............................................................................ 120 2.39.4 Results sheet ......................................................................................................................................... 121

2.40 Test No. 01-0040SSLSB_FEM: Stiffen membrane ........................................................................................... 122 2.40.1 Description sheet ................................................................................................................................... 122 2.40.2 Overview ................................................................................................................................................ 122 2.40.3 Results of the model in the linear elastic range...................................................................................... 123 2.40.4 Results sheet ......................................................................................................................................... 124

2.41 Test No. 01-0041SSLLB_FEM: Beam on two supports considering the shear force ................................... 125 2.41.1 Description sheet ................................................................................................................................... 125 2.41.2 Overview ................................................................................................................................................ 125 2.41.3 Vertical displacement of the model in the linear elastic range................................................................ 126 2.41.4 Results sheet ......................................................................................................................................... 126

2.42 Test No. 01-0042SSLSB_FEM: Thin cylinder under a uniform axial load...................................................... 127 2.42.1 Description sheet ................................................................................................................................... 127 2.42.2 Overview ................................................................................................................................................ 127 2.42.3 Stress in all points .................................................................................................................................. 128 2.42.4 Cylinder deformation at the free end ...................................................................................................... 128 2.42.5 Results sheet ......................................................................................................................................... 129

2.43 Test No. 01-0043SSLSB_FEM: Thin cylinder under a hydrostatic pressure................................................. 130 2.43.1 Description sheet ................................................................................................................................... 130 2.43.2 Overview ................................................................................................................................................ 130 2.43.3 Stresses ................................................................................................................................................. 131 2.43.4 Cylinder deformation .............................................................................................................................. 131 2.43.5 Results sheet ......................................................................................................................................... 132

2.44 Test No. 01-0044SSLSB_FEM: Thin cylinder under its self weight................................................................ 133 2.44.1 Description sheet ................................................................................................................................... 133 2.44.2 Overview ................................................................................................................................................ 133 2.44.3 Stresses ................................................................................................................................................. 134 2.44.4 Cylinder deformation .............................................................................................................................. 134 2.44.5 Results sheet ......................................................................................................................................... 134

2.45 Test No. 01-0045SSLSB_FEM: Torus with uniform internal pressure ........................................................... 135 2.45.1 Description sheet ................................................................................................................................... 135 2.45.2 Overview ................................................................................................................................................ 135 2.45.3 Stresses ................................................................................................................................................. 136 2.45.4 Cylinder deformation .............................................................................................................................. 136 2.45.5 Results sheet ......................................................................................................................................... 136

2.46 Test No. 01-0046SSLSB_FEM: Spherical shell under internal pressure ....................................................... 137 2.46.1 Description sheet ................................................................................................................................... 137 2.46.2 Overview ................................................................................................................................................ 137 2.46.3 Stresses ................................................................................................................................................. 138 2.46.4 Cylinder deformation .............................................................................................................................. 138 2.46.5 Results sheet ......................................................................................................................................... 139

2.47 Test No. 01-0047SSLSB_FEM: Spherical shell under its self weight............................................................. 140


vi

2.47.1 Description sheet ................................................................................................................................... 140 2.47.2 Overview................................................................................................................................................ 140 2.47.3 Stresses................................................................................................................................................. 141 2.47.4 Cylinder radial deformation .................................................................................................................... 141 2.47.5 Results sheet ......................................................................................................................................... 142

2.48 Test No. 01-0048SSLSB_FEM: Pinch cylindrical shell ................................................................................... 143 2.48.1 Description sheet ................................................................................................................................... 143 2.48.2 Overview................................................................................................................................................ 143 2.48.3 Vertical displacement at point A............................................................................................................. 144 2.48.4 Results sheet ......................................................................................................................................... 144

2.49 Test No. 01-0049SSLSB_FEM: Spherical shell with holes ............................................................................. 145 2.49.1 Description sheet ................................................................................................................................... 145 2.49.2 Overview................................................................................................................................................ 145 2.49.3 Horizontal displacement at point A ........................................................................................................ 146 2.49.4 Results sheet ......................................................................................................................................... 146

2.50 Test No. 01-0050SSLSB_FEM: Spherical dome under a uniform external pressure ................................... 147 2.50.1 Description sheet ................................................................................................................................... 147 2.50.2 Overview................................................................................................................................................ 147 2.50.3 Horizontal displacement and exterior meridian stress............................................................................ 148 2.50.4 Results sheet ......................................................................................................................................... 149

2.51 Test No. 01-0051SSLSB_FEM: Simply supported square plate under a uniform load ................................ 150 2.51.1 Description sheet ................................................................................................................................... 150 2.51.2 Overview................................................................................................................................................ 150 2.51.3 Vertical displacement and bending moment at the center of the plate................................................... 151 2.51.4 Results sheet ......................................................................................................................................... 151

2.52 Test No. 01-0052SSLSB_FEM: Simply supported rectangular plate under a uniform load......................... 152 2.52.1 Description sheet ................................................................................................................................... 152 2.52.2 Overview................................................................................................................................................ 152 2.52.3 Vertical displacement and bending moment at the center of the plate................................................... 153 2.52.4 Results sheet ......................................................................................................................................... 153

2.53 Test No. 01-0053SSLSB_FEM: Simply supported rectangular plate under a uniform load......................... 154 2.53.1 Description sheet ................................................................................................................................... 154 2.53.2 Overview................................................................................................................................................ 154 2.53.3 Vertical displacement and bending moment at the center of the plate................................................... 155 2.53.4 Results sheet ......................................................................................................................................... 155

2.54 Test No. 01-0054SSLSB_FEM: Simply supported rectangular plate loaded with punctual force and moments............................................................................................................................................................. 156 2.54.1 Description sheet ................................................................................................................................... 156 2.54.2 Overview................................................................................................................................................ 156 2.54.3 Vertical displacement at C ..................................................................................................................... 157 2.54.4 Results sheet ......................................................................................................................................... 157

2.55 Test No. 01-0055SSLSB_FEM: Shear plate perpendicular to the medium surface ...................................... 158 2.55.1 Description sheet ................................................................................................................................... 158 2.55.2 Overview................................................................................................................................................ 158 2.55.3 Vertical displacement at C ..................................................................................................................... 159 2.55.4 Results sheet ......................................................................................................................................... 159

2.56 Test No. 01-0056SSLLB_FEM: Triangulated system with hinged bars ......................................................... 160 2.56.1 Description sheet ................................................................................................................................... 160 2.56.2 Overview................................................................................................................................................ 160 2.56.3 Tension force in BD bar ......................................................................................................................... 161 2.56.4 Vertical displacement at D ..................................................................................................................... 161 2.56.5 Results sheet ......................................................................................................................................... 161


vii

2.57 Test No. 01-0057SSLSB_FEM: 0.01m thick plate fixed on its perimeter, loaded with a uniform pressure .............................................................................................................................................................. 162 2.57.1 Description sheet ................................................................................................................................... 162 2.57.2 Overview ................................................................................................................................................ 162 2.57.3 Vertical displacement at C...................................................................................................................... 163 2.57.4 Results sheet ......................................................................................................................................... 163

2.58 Test No. 01-0058SSLSB_FEM: 0.01333 m thick plate fixed on its perimeter, loaded with a uniform pressure .............................................................................................................................................................. 164 2.58.1 Description sheet ................................................................................................................................... 164 2.58.2 Overview ................................................................................................................................................ 164 2.58.3 Vertical displacement at C...................................................................................................................... 165 2.58.4 Results sheet ......................................................................................................................................... 165

2.59 Test No. 01-0059SSLSB_FEM: 0.02 m thick plate fixed on its perimeter, loaded with a uniform pressure ........ 166 2.59.1 Description sheet ................................................................................................................................... 166 2.59.2 Overview ................................................................................................................................................ 166 2.59.3 Vertical displacement at C...................................................................................................................... 167 2.59.4 Results sheet ......................................................................................................................................... 167



2.62 Test No. 01-0062SSLSB_FEM: 0.01 m thick plate fixed on its perimeter, loaded with a punctual force .... 172 2.62.1 Description sheet ................................................................................................................................... 172 2.62.2 Overview ................................................................................................................................................ 172 2.62.3 Vertical displacement at point C (center of the plate) ............................................................................. 173 2.62.4 Results sheet ......................................................................................................................................... 173

2.63 Test No. 01-0063SSLSB_FEM: 0.01333 m thick plate fixed on its perimeter, loaded with a punctual force ....... 174 2.63.1 Description sheet ................................................................................................................................... 174 2.63.2 Overview ................................................................................................................................................ 174 2.63.3 Vertical displacement at point C (the center of the plate) ....................................................................... 175 2.63.4 Results sheet ......................................................................................................................................... 175

2.64 Test No. 01-0064SSLSB_FEM: 0.02 m thick plate fixed on its perimeter, loaded with a punctual force.................................................................................................................................................................... 176 2.64.1 Description sheet ................................................................................................................................... 176 2.64.2 Overview ................................................................................................................................................ 176 2.64.3 Vertical displacement at point C (the center of the plate) ....................................................................... 177 2.64.4 Results sheet ......................................................................................................................................... 177

2.65 Test No. 01-0065SSLSB_FEM: 0.05 m thick plate fixed on its perimeter, loaded with a punctual force .... 178 2.65.1 Description sheet ................................................................................................................................... 178 2.65.2 Overview ................................................................................................................................................ 178 2.65.3 Vertical displacement at point C center of the plate) .............................................................................. 179 2.65.4 Results sheet ......................................................................................................................................... 179

2.66 Test No. 01-0066SSLSB_FEM: 0.1 m thick plate fixed on its perimeter, loaded with a punctual force ...... 180 2.66.1 Description sheet ................................................................................................................................... 180 2.66.2 Overview ................................................................................................................................................ 180 2.66.3 Vertical displacement at point C (center of the plate) ............................................................................. 181


viii

2.66.4 Results sheet ......................................................................................................................................... 181 2.67 Test No. 01-0067SDLLB_FEM: Vibration mode of a thin piping elbow in space (case 1)............................ 182

2.67.1 Description sheet ................................................................................................................................... 182 2.67.2 Overview................................................................................................................................................ 182 2.67.3 Eigen modes frequencies ...................................................................................................................... 183 2.67.4 Results sheet ......................................................................................................................................... 183

2.68 Test No. 01-0068SDLLB_FEM: Vibration mode of a thin piping elbow in space (case 2)............................ 184 2.68.1 Description sheet ................................................................................................................................... 184 2.68.2 Overview................................................................................................................................................ 184 2.68.3 Eigen modes frequencies ...................................................................................................................... 185 2.68.4 Results sheet ......................................................................................................................................... 185

2.69 Test No. 01-0069SDLLB_FEM: Vibration mode of a thin piping elbow in space (case 3)............................ 186 2.69.1 Description sheet ................................................................................................................................... 186 2.69.2 Overview................................................................................................................................................ 186 2.69.3 Eigen modes frequencies ...................................................................................................................... 187 2.69.4 Results sheet ......................................................................................................................................... 187

2.70 Test No. 01-0077SSLPB_FEM: Reactions on supports and bending moments on a 2D portal frame (Rafters) .............................................................................................................................................................. 188 2.70.1 Description sheet ................................................................................................................................... 188 2.70.2 Overview................................................................................................................................................ 188 2.70.3 Moments and actions on supports M.R. calculation on a 2D portal frame. ............................................ 189 2.70.4 Results sheet ......................................................................................................................................... 189

2.71 Test No. 01-0078SSLPB_FEM: Reactions on supports and bending moments on a 2D portal frame (Columns) ........................................................................................................................................................... 190 2.71.1 Description sheet ................................................................................................................................... 190 2.71.2 Overview................................................................................................................................................ 190 2.71.3 Moments and reactions on supports M.R. calculation on a 2D portal frame. ......................................... 191 2.71.4 Results sheet ......................................................................................................................................... 191

2.72 Test No. 01-0084SSLLB_FEM: Short beam on two hinged supports ............................................................ 192 2.72.1 Description sheet ................................................................................................................................... 192 2.72.2 Overview................................................................................................................................................ 192 2.72.3 Reference results................................................................................................................................... 192 2.72.4 Results sheet ......................................................................................................................................... 193

2.73 Test No. 01-0085SDLLB_FEM: Slender beam of variable rectangular section with fixed-free ends (β=5) .................................................................................................................................................................... 194 2.73.1 Description sheet ................................................................................................................................... 194 2.73.2 Overview................................................................................................................................................ 194 2.73.3 Reference results................................................................................................................................... 195 2.73.4 Results sheet ......................................................................................................................................... 197

2.74 Test No. 01-0086SDLLB_FEM: Slender beam of variable rectangular section (fixed-fixed)........................ 198 2.74.1 Description sheet ................................................................................................................................... 198 2.74.2 Overview................................................................................................................................................ 198 2.74.3 Reference results................................................................................................................................... 198 2.74.4 Results sheet ......................................................................................................................................... 199

2.75 Test No. 01-0089SSLLB_FEM: Plane portal frame with hinged supports..................................................... 200 2.75.1 Description sheet ................................................................................................................................... 200 2.75.2 Overview................................................................................................................................................ 200 2.75.3 Calculation method used to obtain the reference solution ..................................................................... 201 2.75.4 Reference values................................................................................................................................... 201 2.75.5 Results sheet ......................................................................................................................................... 201

2.76 Test No. 01-0090HFLSB_FEM: Simply supported beam in Eulerian buckling with a thermal load ............ 202 2.76.1 Description sheet ................................................................................................................................... 202 2.76.2 Overview................................................................................................................................................ 202 2.76.3 Displacement of the model in the linear elastic range............................................................................ 203


ix

2.76.4 Results sheet ......................................................................................................................................... 204 2.77 Test No. 01-0091HFLLB_FEM: Double fixed beam in Eulerian buckling with a thermal load...................... 205

2.77.1 Description sheet ................................................................................................................................... 205 2.77.2 Overview ................................................................................................................................................ 205 2.77.3 Displacement of the model in the linear elastic range ............................................................................ 206 2.77.4 Results sheet ......................................................................................................................................... 206

2.78 Test No. 01-0092HFLLB_FEM: Cantilever beam in Eulerian buckling with thermal load............................. 207 2.78.1 Description sheet ................................................................................................................................... 207 2.78.2 Overview ................................................................................................................................................ 207 2.78.3 Displacement of the model in the linear elastic range ............................................................................ 207 2.78.4 Results sheet ......................................................................................................................................... 208

2.79 Test No. 01-0094SSLLB_FEM: 3D bar structure with elastic support ........................................................... 209 2.79.1 Description sheet ................................................................................................................................... 209 2.79.2 Overview ................................................................................................................................................ 209 2.79.3 Results ................................................................................................................................................... 210 2.79.4 Results sheet ......................................................................................................................................... 214

2.80 Test No. 01-0095SDLLB_FEM: Fixed/free slender beam with centered mass .............................................. 215 2.80.1 Description sheet ................................................................................................................................... 215 2.80.2 Test data ................................................................................................................................................ 215 2.80.3 Reference results ................................................................................................................................... 216 2.80.4 Results sheet ......................................................................................................................................... 219

2.81 Test No. 01-0096SDLLB_FEM: Fixed/free slender beam with eccentric mass or inertia ............................. 220 2.81.1 Description sheet ................................................................................................................................... 220 2.81.2 Problem data.......................................................................................................................................... 220 2.81.3 Reference frequencies ........................................................................................................................... 221 2.81.4 Results sheet ......................................................................................................................................... 222

2.82 Test No. 01-0097SDLLB_FEM: Double cross with hinged ends..................................................................... 223 2.82.1 Description sheet ................................................................................................................................... 223 2.82.2 Problem data.......................................................................................................................................... 223 2.82.3 Reference frequencies ........................................................................................................................... 224 2.82.4 Results sheet ......................................................................................................................................... 225

2.83 Test No. 01-0098SDLLB_FEM: Simple supported beam in free vibration ..................................................... 226 2.83.1 Description sheet ................................................................................................................................... 226 2.83.2 Problem data.......................................................................................................................................... 226 2.83.3 Reference frequencies ........................................................................................................................... 227 2.83.4 Results sheet ......................................................................................................................................... 228

2.84 Test No. 01-0099HSLSB_FEM: Membrane with hot point............................................................................... 229 2.84.1 Description sheet ................................................................................................................................... 229 2.84.2 Problem data.......................................................................................................................................... 229 2.84.3 σyy stress at point A: ............................................................................................................................... 231 2.84.4 Results sheet ......................................................................................................................................... 231

2.85 Test No. 01-0100SSNLB_FEM: Beam on 3 supports with T/C (k = 0)............................................................. 232 2.85.1 Description sheet ................................................................................................................................... 232 2.85.2 Overview ................................................................................................................................................ 232 2.85.3 References solutions.............................................................................................................................. 233 2.85.4 Results sheet ......................................................................................................................................... 234

2.86 Test No. 01-0101SSNLB_FEM: Beam on 3 supports with T/C (k → ∞)........................................................... 235 2.86.1 Description sheet ................................................................................................................................... 235 2.86.2 Overview ................................................................................................................................................ 235 2.86.3 References solutions.............................................................................................................................. 236 2.86.4 Results sheet ......................................................................................................................................... 237

2.87 Test No. 01-0102SSNLB_FEM: Beam on 3 supports with T/C (k = -10000 N/m)............................................ 238 2.87.1 Description sheet ................................................................................................................................... 238


x

2.87.2 Overview................................................................................................................................................ 238 2.87.3 References solutions ............................................................................................................................. 239 2.87.4 Results sheet ......................................................................................................................................... 240

2.88 Test No. 01-0103SSLLB_FEM: Linear system of truss beams....................................................................... 241 2.88.1 Description sheet ................................................................................................................................... 241 2.88.2 Overview................................................................................................................................................ 241 2.88.3 References solutions ............................................................................................................................. 242 2.88.4 Results sheet ......................................................................................................................................... 243

2.89 Test No. 01-0104SSNLB_FEM: Non linear system of truss beams................................................................ 244 2.89.1 Description sheet ................................................................................................................................... 244 2.89.2 Overview................................................................................................................................................ 244 2.89.3 References solutions ............................................................................................................................. 245 2.89.4 Results sheet ......................................................................................................................................... 246

2.90 Test No. 02-0112SMLLB_P92: Study of a mast subjected to an earthquake................................................ 247 2.90.1 Description sheet ................................................................................................................................... 247 2.90.2 Model overview...................................................................................................................................... 247 2.90.3 Material strength model ......................................................................................................................... 247 2.90.4 Seismic hypothesis in conformity with PS92 regulation ......................................................................... 248 2.90.5 Modal analysis ....................................................................................................................................... 248 2.90.6 Spectral study ........................................................................................................................................ 249 2.90.7 Results sheet ......................................................................................................................................... 251

2.91 Test No. 02-0158SSLLB_B91: Linear element in combined bending/tension - without compressed reinforcements - Partially tensioned section................................................................................................... 252 2.91.1 Description sheet ................................................................................................................................... 252 2.91.2 Overview................................................................................................................................................ 252 2.91.3 Reinforcement calculation...................................................................................................................... 253 2.91.4 Results sheet ......................................................................................................................................... 255

2.92 Test No. 02-0162SSLLB_B91: Linear element in simple bending - without compressed reinforcement... 256 2.92.1 Description sheet ................................................................................................................................... 256 2.92.2 Overview................................................................................................................................................ 256 2.92.3 Reinforcement calculation...................................................................................................................... 257 2.92.4 Results sheet ......................................................................................................................................... 258

2.93 Test No. 03-0206SSLLG_CM66: Design of a Steel Structure according to CM66. ....................................... 259 2.93.1 Data ....................................................................................................................................................... 259 2.93.2 Effel Structure results............................................................................................................................. 261 2.93.3 CM66 Effel Expertise results.................................................................................................................. 262 2.93.4 Effel Structure / Advance Design comparison........................................................................................ 267

2.94 Test No. 03-0207SSLLG_CM66: Design of a 2D portal frame......................................................................... 269 2.94.1 Data ....................................................................................................................................................... 269 2.94.2 Effel Structure Results ........................................................................................................................... 270 2.94.3 Effel Expert CM results .......................................................................................................................... 271 2.94.4 Effel Structure / Advance Design comparison........................................................................................ 274

2.95 Test No. 03-0208SSLLG_BAEL91: Design of a concrete floor with an opening........................................... 276 2.95.1 Data ....................................................................................................................................................... 276 2.95.2 Effel Structure Results ........................................................................................................................... 279 2.95.3 Effel RC Expert Results ......................................................................................................................... 281 2.95.4 Effel Structure / Advance Design comparison........................................................................................ 283

3 EUROCODES 1 TESTS DESCRIPTION........................................................................................................................... 285 Wind calculation .............................................................................................................................................................. 286 3.1 Portal frame with 11.31° angle – Example A.fto .............................................................................................. 286

3.1.1 Wind pressure calculation...................................................................................................................... 286 3.1.2 Cpe and Wind force calculation for front wind (Y).................................................................................. 294


xi

3.1.3 Cpe and Wind force calculation for parallel wind (X) .............................................................................. 306 3.2 Monopitch frame with 15 ° angle – Example B.fto........................................................................................... 313

3.2.1 Lateral Wind (X direction): Cpe calculation on rooftop ........................................................................... 313 3.3 Portal frame with 10° angle – Example C.fto.................................................................................................... 316

3.3.1 Lateral Wind (X): Cpe calculation and Wind force calculation ................................................................ 316 Snow calculation .............................................................................................................................................................. 318 3.4 Portal frame with 11.31° angle – Example A.fto............................................................................................... 318

4 EUROCODES 2 TESTS DESCRIPTION ........................................................................................................................... 321 4.1 Test No. 01 – EC2: Minimum reinforcement of a beam................................................................................... 322

4.1.1 Overview ................................................................................................................................................ 322 4.1.2 Reference results ................................................................................................................................... 323 4.1.3 Results sheet ......................................................................................................................................... 323

4.2 Test No. 02 – EC2: Longitudinal reinforcement of a beam under a linear load - horizontal level behavior law ....................................................................................................................................................... 324 4.2.1 Overview ................................................................................................................................................ 324 4.2.2 Reference results ................................................................................................................................... 325 4.2.3 Results sheet ......................................................................................................................................... 325

4.3 Test No. 03 – EC2: Longitudinal reinforcement of a beam under a linear load - inclined stress strain behavior law ....................................................................................................................................................... 326 4.3.1 Overview ................................................................................................................................................ 326 4.3.2 Reference results ................................................................................................................................... 327 4.3.3 Results sheet ......................................................................................................................................... 328

4.4 Test No. 04 – EC2: Longitudinal reinforcement of a beam under a concentrated load - horizontal level behavior law ....................................................................................................................................................... 329 4.4.1 Overview ................................................................................................................................................ 329 4.4.2 Reference results ................................................................................................................................... 330 4.4.3 Results sheet ......................................................................................................................................... 330


4.6 Test No. 06 – EC2: Transverse reinforcement of a beam subjected to a linear load.................................... 333 4.6.1 Overview ................................................................................................................................................ 333 4.6.2 Reference results ................................................................................................................................... 334 4.6.3 Results sheet ......................................................................................................................................... 335


5 EUROCODES 3 TESTS DESCRIPTION ........................................................................................................................... 339 5.1 Test No. 01 – EC3: 2D frame Design................................................................................................................. 340

5.1.1 Data........................................................................................................................................................ 340 5.1.2 Axial forces – N ...................................................................................................................................... 342 5.1.3 Shear forces – T..................................................................................................................................... 342 5.1.4 Bending moments – M ........................................................................................................................... 342 5.1.5 Classification of the beam cross section (IPE 240) ................................................................................ 343 5.1.6 Resistance of the beam cross section (IPE 240).................................................................................... 344 5.1.7 Classification of the column cross section (HEA 200) ............................................................................ 351 5.1.8 Resistance of the column cross section (HEA 200) ...............................................................................352


xii

5.2 Example 1 - Class of cross section (EC3)........................................................................................................ 359 5.2.1 Overview................................................................................................................................................ 359



5.5 Example 4: Tension column – design value of the resistance (EC3) ............................................................ 367 5.6 Example 5: Tension column – design value of the resistance (EC3) ............................................................ 368

1 Introduction

Before official releases, each version of GRAITEC software, includingAdvance Design, undergoes a series of validation tests from the"standard tests".

This validation is performed in parallel and in addition to the Beta-Test in order to obtain the "operational version" status.

At the moment, the database for automatic test includes 96 testswhich are codified and stored in a precise manner.

Each test is the subject of several reference documents:

Test description document.

Result sheet.

Information sheet corresponding to the tested model.

The test description documents coding is detailed below.


2

1.1 Test description documents coding

The test description documents coding is summarized in the following table:

Year Regulations

Test number

Application field Analysis type

Behavior type Model type Results comparison

O1 OOO1 Static Linear Linear O2 OOO2

Structure mechanics S L L

To theoretical reference

Finite elements

. OOO3 S Dynamic B FEM

. . D Non-linear Planar Eurocodes X

. . N S

To standard mesh

ECX . . Thermomechanical

Eulerian buckling N BAEL

H F Plane (2D) B91 Spectral P

To a different software CM66

M A C66

Transitory Discrete Blind CB71 Thermal T D G C71

T Stationary . P .

The first column of the table corresponds to the last two digits of the year of creation of the test.

The second column of the table contains the number of the test (4 digits). The first test is numbered 0001. (9999 tests can be created each year.)

Columns 3 to 6 of the table are from the Structure Calculation Software Validation Guide of AFNOR. They regard all GRAITEC software according to their application fields.

Regarding the selection of the analysis type, if two types of analysis are used for the same test, the test of the highest level is accepted for the codification (the test from the lowest level in the corresponding column); for example: a test defined in static and in dynamic will get the dynamic corresponding short name (D)

Regarding the choice of model type, the surface denomination includes models consisting of planar elements only, as well as models consisting of planar and linear elements.

The results comparison column presents the reference taken into account to validate the results obtained with GRAITEC software.

The last column contains the regulation used for the test:


3

Type Regulations Used code

Climatic NV65-84 N65

DIN-1055 D55 NBE-EA-95-EHE NBE NBN-B03 NBN RSA-98 R98 STAS 10101/21-92/20-90 S01

Seismic NCSE94 N94

P100-92 P00 PS69 P69 PS92 P92 RPA88 R88 RPA99 R99 RSA-98 S98 SI413 S13

Reinforced concrete ACI ACI

BAEL B91 DIN DIN EC2 EC2 EHE EHE STAS 10107/0-90 S07

Steelwork CM66 C66

EA-95 E95 EC3 EC3

Timber construction CB71 C71

Finite elements FEM FEM

The test description documents coding consists of the 8 columns. For the results comparison documents, the version number used is added next (without dot).

1.2 Test description documents coding example

SSLL: Analysis FEM: Regulation

01: Date 01-0001SSLLA_FEM

0001: Test number A: Results comparison


4

1.3 Results comparison documents coding example

01-0001SSLLA_FEM-092N

92N: Version used for the test calculation

01-0001SSLLA_FEM-101M 101M: Version used for the test calculation

1.4 Margin of error

The acceptable margin of error for a test validation is:

Static 2% Dynamic 5%

Eulerian buckling 5% Spectral 5%

Stationary 5% Transitory 5% Climatic 5% Seismic 5%

Reinforced concrete 10% Steelwork 10%

Timber construction 10%


5

1.5 Synthetic sheet of 2010 version

Software Solver Code Title Reference value Calculated value Deviation

Advance Design 2010 CM2 01-0001SSLSB_FEM Cantilever Rectangular plate -9.71 -9.59 -1.24%

Advance Design 2010 CM2 01-0002SSLLB_FEM System of two bars with three hinges -0.30 -0.30 0.00%

Advance Design 2010 CM2 01-0003SSLSB_FEM Circular plate under uniform load -6.50 -6.47 -0.46%

Advance Design 2010 CM2 01-0004SDLLB_FEM Slender beam with variable section (fixed-free) 1112.28 1075.70 -3.29%

Advance Design 2010 CM2 01-0005SSLLB_FEM Tied (sub-tensioned) beam 584584 584584 0.00%

Advance Design 2010 CM2 01-0006SDLLB_FEM Thin circular ring fixed in two points. 2801.5 2777.43 -0.86%

Advance Design 2010 CM2 01-0007SDLSB_FEM Thin lozenge-shaped plate fixed on one side (a = 0°) 8.7266 8.67 -0.65%



Advance Design 2010 CM2 01-0010SDLSB_FEM Thin lozenge-shaped plate fixed on one side (a = 45°) 26.3897 28.08 6.41%

Advance Design 2010 CM2 01-0011SDLLB_FEM Vibration mode of a thin piping elbow in plane (case 1) 119 120.09 0.92%

Advance Design 2010 CM2 01-0012SDLLB_FEM Vibration mode of a thin piping elbow in plane (case 2) 180 184.68 2.60%

Advance Design 2010 CM2 01-0013SDLLB_FEM Vibration mode of a thin piping elbow in plane (case 3) 25.300 24.961 -1.34%

Advance Design 2010 CM2 01-0014SDLLB_FEM Thin circular ring suspended by an elastic leg. 682.00 683.9 0.28%

Advance Design 2010 CM2 01-0015SSLLB_FEM Double fixed beam with a spring in the middle -0.11905 -0.11905 0.00%

Advance Design 2010 CM2 01-0014SDLLB_FEM Double fixed beam 26.228 25.758 -1.79%

Advance Design 2010 CM2 01-0017SDLLB_FEM Short beam on simple supports (on the neutral axis) 1498.295 1537.156 2.59%

Advance Design 2010 CM2 01-0018SDLLB_FEM Short beam on simple supports (eccentric) 902.2 945.35 4.78%

Advance Design 2010 CM2 01-0019SDLSB_FEM Thin square plate fixed on one side 136.0471 134.655 -1.02%

Advance Design 2010 CM2 01-0020SDLSB_FEM Thin rectangular plate simply supported on the perimeter 197.32 195.545 -0.90%

Advance Design 2010 CM2 01-0021SFLLB_FEM Cantilever beam in Eulerian buckling -98696 -98699.278 0.00%

Advance Design 2010 CM2 01-0022SDLSB_FEM Thin ring plate fixed on a hub 609.7 593.83 -2.60%


6

Advance Design 2010 CM2 01-0023SDLLB_FEM Bending effects of a symmetrical portal frame 206 206.2 0.10%

Advance Design 2010 CM2 01-0024SSLLB_FEM Slender beam on two fixed supports -540 -540 0.00%

Advance Design 2010 CM2 01-0025SSLLB_FEM Slender beam on three supports 63000 63000 0.00%

Advance Design 2010 CM2 01-0026SSLLB_FEM Bimetallic: Fixed beams connected to a stiff element -0.125 -0.125 0.00%

Advance Design 2010 CM2 01-0027SSLLB_FEM Fixed thin arc in planar bending 0.3791 0.37891 -0.05%

Advance Design 2010 CM2 01-0028SSLLB_FEM Fixed thin arc in out of plane bending -96.5925 -96.5779 -0.02%

Advance Design 2010 CM2 01-0029SSLLB_FEM Double hinged thin arc in planar bending 5.3912 5.386 -0.10%

Advance Design 2010 CM2 01-0030SSLLB_FEM Portal frame with lateral connections 113.559 113.7044 0.13%

Advance Design 2010 CM2 01-0031SSLLB_FEM Truss with hinged bars under a punctual load 0.08839 0.08817 -0.25%

Advance Design 2010 CM2 01-0032SSLLB_FEM Beam on elastic soil, free ends 5759 5772.05 0.23%

Advance Design 2010 CM2 01-0033SFLLA_FEM EDF Pylon 2.77 2.830 2.17%

Advance Design 2010 CM2 01-0034SSLLB_FEM Beam on elastic soil, hinged ends 11674 11644.36 -0.25%

Advance Design 2010 CM2 01-0035SSLPB_FEM Plate with in plane bending and shear -80 -79.46 -0.68%

Advance Design 2010 CM2 01-0036SSLSB_FEM Simply supported square plate -0.158 -0.16491 4.37%

Advance Design 2010 CM2 01-0037SSLSB_FEM Caisson beam in torsion -0.11 -0.10 -9.09%

Advance Design 2010 CM2 01-0038SSLSB_FEM Thin cylinder under uniform radial pressure -2.38 x 10-6 -2.39 x 10-6 -0.42%

Advance Design 2010 CM2 01-0039SSLSB_FEM Square plate under planar stresses -14.66 -14.61 -0.34%

Advance Design 2010 CM2 01-0040SSLSB_FEM Stiffen membrane 11.55 11.55 0.00%

Advance Design 2010 CM2 01-0041SSLLB_FEM Beam on two supports considering the shear force -1.017 -1.017 0.00%

Advance Design 2010 CM2 01-0042SSLSB_FEM Thin cylinder under uniform axial load -7.14 x 10-7 -7,109 x 10-7 0.43%

Advance Design 2010 CM2 01-0043SSLSB_FEM Thin cylinder under a hydrostatic pressure -2.86 x 10-6 -2,854 x 10-6 -0.20%

Advance Design 2010 CM2 01-0044SSLSB_FEM Thin cylinder under its self weight 3.14 x 105 3.11 x 105 -0.95%

Advance Design 2010 CM2 01-0045SSLSB_FEM Torus with a uniform internal pressure 7.5 x 105 7.43 x 105 -0.94%

Advance Design 2010 CM2 01-0046SSLSB_FEM Spherical shell under internal pressure 8.33 x 10-7 8.34 x 10-7 0.12%

Advance Design 2010 CM2 01-0047SSLSB_FEM Spherical shell under its self weight -7.85 x 104 -7.90 x 104 -0.63%

Advance Design 2010 CM2 01-0048SSLSB_FEM Pinch cylindrical shell -113.9 x 10-3 -113.29 x 10-3 -0.53%


7

Advance Design 2010 CM2 01-0049SSLSB_FEM Spherical shell with holes 94.0 92.13 -1.99%

Advance Design 2010 CM2 01-0050SSLSB_FEM Spherical dome under uniform external pressure -1.73 x 10-3 -2.14 x 10-3 23%

Advance Design 2010 CM2 01-0051SSLSB_FEM Simply supported square plate under a uniform load 0.0479 0.0471 -1.67%

Advance Design 2010 CM2 01-0052SSLSB_FEM Simply supported rectangular plate under a uniform load 1.106 x 10-2 1.102 x 10-2 -0.36%

Advance Design 2010 CM2 01-0053SSLSB_FEM Simply supported rectangular plate under a uniform load 1.416 x 10-2 1.401 x 10-2 -1.06%

Advance Design 2010 CM2 01-0054SSLSB_FEM Simply supported rectangular plate with punctual forces and moments -12.480 -12.667 1.50%

Advance Design 2010 CM2 01-0055SSLSB_FEM Shear plate perpendicular to the medium surface 35.37 x 10-3 35.67 x 10-3 0.85%

Advance Design 2010 CM2 01-0056SSLLB_FEM Triangulated system with hinged bars 43633 43688 0.13%

Advance Design 2010 CM2 01-0057SSLSB_FEM 0.01m thick plate fixed on its perimeter, loaded with a uniform pressure 0.66390 x 10-2 0.65879 x 10-2 -0.77%

Advance Design 2010 CM2 01-0058SSLSB_FEM 0.01333 m thick plate fixed on its perimeter, loaded with a uniform pressure 0.28053 x 10-2 0.28045 x 10-2 -0.03%




Advance Design 2010 CM2 01-0062SSLSB_FEM 0.01 m thick plate fixed on its perimeter, loaded with a punctual force 0.29579 0.29215 -1.23%

Advance Design 2010 CM2 01-0063SSLSB_FEM 0.01333 m thick plate fixed on its perimeter, loaded with a punctual force 0.11837 0.12458 5.25%

Advance Design 2010 CM2 01-0064SSLSB_FEM 0.02 m thick plate fixed on its perimeter, loaded with a punctual force 0.037454 0.036980 -1.27%

Advance Design 2010 CM2 01-0065SSLSB_FEM 0.05 m thick plate fixed on its perimeter, loaded with a punctual force 0.25946 x 10-2 0.25723 x 10-2 -0.86%

Advance Design 2010 CM2 01-0066SSLSB_FEM 0.1m thick plate fixed on its perimeter, loaded with a punctual force 0.42995 x 10-2 0.41209 x 10-2 -4.15%

Advance Design 2010 CM2 01-0067SDLLB_FEM Vibration mode of a thin piping elbow in space (case 1) 125 120.09 -3.93%

Advance Design 2010 CM2 01-0068SDLLB_FEM Vibration mode of a thin piping elbow in space (case 2) 100 94.62 -5.38%

Advance Design 2010 CM2 01-0069SDLLB_FEM Vibration mode of a thin piping elbow in space (case 3) 24.800 24.43 -1.49%

Advance Design 2010 CM2 01-0077SSLPB_FEM Reactions on supports and bending moments on a 2D portal frame (Rafters) 1671 1673.35 0.14%

Advance Design 2010 CM2 01-0078SSLPB_FEM Reactions on supports and bending moments on a 2D portal frame (Columns) -302.7 -302.06 -0.21%

Advance Design 2010 CM2 01-0084SSLLB_FEM Short beam on two hinged supports -1.25926 -1.25926 0.00%

Advance Design 2010 CM2 01-0085SDLLB_FEM Slender beam of variable rectangular section with fixed-free ends (beta = 5) 56.55 58.49 3.43%

Advance Design 2010 CM2 01-0086SDLLB_FEM Slender beam Slender beam of variable rectangular section fixed-fixed 143.303 145.88 1.80%


8

Advance Design 2010 CM2 01-0089SSLLB_FEM 2D portal frame with hinged supports -0.03072 -0.03072 0.00%

Advance Design 2010 CM2 01-0090HFLSB-FEM Simply supported beam in Eulerian buckling with a thermal load -40774 -41051.95 0.68%

Advance Design 2010 CM2 01-0091HFLLB_FEM Double fixed beam in Eulerian buckling with thermal load -117.724 -118.08 0.30%

Advance Design 2010 CM2 01-0092HFLLB _FEM Cantilever beam in Eulerian buckling with thermal load -0.5 -0.5 0.00%

Advance Design 2010 CM2 01-0094SSLLB_FEM 3D bar structure with elastic support -1436 -1436,55 -3.83%

Advance Design 2010 CM2 01-0095SDLLB_FEM Fixed/free slender beam with a centered mass 16.07 16.06 -0.06%

Advance Design 2010 CM2 01-0096SDLLB_FEM Fixed/free slender beam with eccentric mass or inertia 61.61 63.091 2.40%

Advance Design 2010 CM2 01-0097SDLLB_FEM Double cross with hinged ends 57.39 56.06 -2.32%

Advance Design 2010 CM2 01-0098SDLLB_FEM Simply supported beam in free vibration 42.649 43.11 1.08%

Advance Design 2010 CM2 01-0099HSLSB_FEM Membrane with hot point 50 50.87 1.74%

Advance Design 2010 CM2 01-0100SSNLB_FEM Beam on 3 supports with T/C (k = 0) -0.000153 -0.000153 0.00%

Advance Design 2010 CM2 01-0101SSNLB_FEM Beam on 3 supports with T/C (k tends to infinite) -93.75 -93.649 -0.11%

Advance Design 2010 CM2 01-0102SSNLB_FEM Beam on 3 supports with T/C (k = -10000 N/m) -58.15 -58.117 -0.06%

Advance Design 2010 CM2 01-0103SSLLB_FEM Linear system of truss beams 0.000649 0.000649 0.00%

Advance Design 2010 CM2 01-0104SSNLB_FEM Non-linear system of truss beams 0.001195 0.001190 -0.42%

Advance Design 2010 CM2 02-0112SMLLB_P92 Study of a mast subjected to an earthquake 2.929 2.927 -0.07%

Advance Design 2010 CM2 02-158SSLLB_BAEL91 Linear element in combined bending/tension - without compressed reinforcements - partially tensioned section -- -- Ok

Advance Design 2010 CM2 02-162SSLLB_BAEL91 Linear element in simple bending - without compressed reinforcement -- -- Ok

Advance Design 2010 CM2 03-0206SSLLG_CM66 Optimization of a Steel Structure according to CM66 339.74 347.46 2.20%

Advance Design 2010 CM2 03-0207SSLLG_CM66 Design of a 2D portal frame 230.34 230.34 0.00%

Advance Design 2010 CM2 03-0208SSLLG_BAEL91 Design of a concrete floor with an opening 0.18 0.175 2.85%

2 Detailed tests description

Test 01-0001SSLSB_FEM

10

2.1 Test No. 01-0001SSLSB_FEM: Cantilever rectangular plate

2.1.1 Description sheet

Reference: Structure Calculation Software Validation Guide, test SSLS 01/89.

Analysis type: linear static.

Element type: planar.

2.1.2 Overview The 1 m long plate is fixed at one end and has a "p" planar load.

Cantilever rectangular plate Scale =1/4 01-0001SSLSB_FEM

Units

S.I.

Geometry

Thickness: e = 0.005 m,

Length: l = 1 m,

Width: b = 0.1 m.

Materials properties

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

Poisson's ratio: ν = 0.3.


11

Boundary conditions

Outer: Fixed at end x = 0,

Inner: None.

Loadings

External: Uniform load p = -1700 Pa on the upper surface,

Internal: None.

2.1.3 Displacement of the model in the linear elastic range Reference solution

The reference displacement is calculated for the unsupported end located at x = 1m.

u = bl4p8EIz =

0.1 x 14 x 1700

8 x 2.1 x 1011 x 0.1 x 0.0053

12 = -9.71 cm

Finite elements modeling

Planar element: plate, imposed mesh,

1100 nodes,

990 surface quadrangles.

Deformed shape

Deformed cantilever rectangular plate Scale =1/4 01-0001SSLSB_FEM

2.1.4 Results sheet

Results comparison: vertical displacement

Solver Positioning Units Reference AD 2010 Deviation CM2 Free extremity cm -9.71 -9.59 -1.24%

Test 01-0002SSLLB_FEM

12

2.2 Test No. 01-0002SSLLB_FEM: System of two bars with three hinges


Reference: Structure Calculation Software Validation Guide, test SSLL 09/89;

Analysis type: linear static;

Element type: linear.

2.2.2 Overview A punctual load, "F", is hanged from two connected hinged bars, also hinged at extremities.

System of two bars with three hinges Scale =1/33 0002SSLLB_FEM

4.500 m

30° 30°

4.500 m

AA BB

CC

FF

X

Y

Z X

Y

Z

Units

I. S.

Geometry

Bars angle relative to horizontal: θ = 30°,

Bars length: l = 4.5 m,

Bar section: A = 3 x 10-4 m2.


Longitudinal elastic modulus: E = 2.1 x 1011 Pa.

Boundary conditions

Outer: Hinged in A and B,

Inner: Hinge on C


13

Loading

External: Punctual load in C: F = -21 x 103 N.

Internal: None.

2.2.3 Displacement of the model in C

Reference solution

uc = -3 x 10-3 m


Linear element: beam, imposed mesh,

21 nodes,

20 linear elements.

Displacement shape

System of two bars with three hinges Scale =1/33 Displacement in C 0002SSLLB_FEM

2.2.4 Bars stresses

Reference solutions

σAC bar = 70 MPa

σBC bar = 70 MPa



21 nodes,

20 linear elements.


14

2.2.5 Shape of the stress diagram System of two bars with three hinges Scale =1/34

Bars stresses 0002SSLLB_FEM

2.2.6 Results sheet


Solver Positioning Units Reference AD 2010 Deviation CM2 In C cm -0.30 -0.30 0.00%

Results comparison: tensile stress

Solver Positioning Units Reference AD 2010 Deviation CM2 AC bar MPa 70 70 0.00% CM2 BC bar MPa 70 70 0.00%


15

2.3 Test No. 01-0003SSLSB_FEM: Circular plate under uniform load


Reference: Structure Calculation Software Validation Guide, test SSLS 03/89;



2.3.2 Overview A circular plate of 5 mm thickness and 2 m diameter with an uniform load perpendicular on the plan of the plate.

Circular plate under uniform load Scale =1/10 01-0003SSLSB_FEM

Units

I. S.

Geometry

Circular plate radius: r = 1m,

Circular plate thickness: h = 0.005 m.





16

Boundary conditions

Outer: Plate fixed on the side (in all points of its perimeter), For the modeling, we consider only a quarter of the plate and we impose symmetry conditions on some nodes (see the following model; yz plane symmetry condition):translation restrained nodes along x and rotation restrained nodes along y and z: translation restrained nodes along x and rotation restrained nodes along y and z:

Inner: None.

Loading

External: Uniform loads perpendicular on the plate: pZ = -1000 Pa,

Internal: None.

2.3.3 Vertical displacement of the model at the center of the plate

Reference solution

Circular plates form:

u = pr4

64D = -1000 x 14

64 x 2404 = - 6.50 x 10-3 m

with the plate radius coefficient: D = Eh3

12(1-ν2) = 2.1 x 1011 x 0.0053

12(1-0.32)

D = 2404



70 nodes,

58 planar elements. Circular plate under uniform load Scale =1.5

Meshing 01-0003SSLSB_FEM


17

Deformed shape

Circular plate under uniform load Scale =1.5 Deformed 01-0003SSLSB_FEM

2.3.4 Results sheet


Solver Positioning Units Reference AD 2010 Deviation CM2 Plate center mm -6.50 -6.47 -0.46%

Test 01-0004SDLLB_FEM

18

2.4 Test No. 01-0004SDLLB_FEM: Slender beam with variable section (fixed-free)


Reference: Structure Calculation Software Validation Guide, test SDLL 09/89;

Analysis type: modal analysis;


2.4.2 Overview Find the first eigen mode frequencies for a beam of variable section, subjected to its own weight only.

Slender beam with variable section (fixed-free) Scale =1/4 01-0004SDLLB_FEM

Units

I. S.

Geometry

Beam length: l = 1 m,

Initial section (in A): o Height: h1 = 0.04 m, o Width: b1 = 0.04 m, o Section: A1 = 1.6 x 10-3 m2, o Flexure moment of inertia relative to z-axis: Iz1 = 2.1333 x 10-7 m4,


19

Final section (in B): o Height: h2 = 0.01 m, o Width: b2 = 0.01 m, o Section: A2 = 10-4 m2, o Flexure moment of inertia relative to z-axis: Iz2 = 8.3333 x 10-10 m4.


Longitudinal elastic modulus: E = 2 x 1011 Pa,

Density: 7800 kg/m3.

Boundary conditions

Outer: Fixed in A,

Inner: None.

Loading

External: None,

Internal: None.

2.4.3 Eigen mode frequencies

Reference solutions

Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):

∂2

∂x2 (EIz ∂2v∂x2 ) = -ρA

∂2v∂x2 where Iz and A vary with the abscissa.

The result is: fi = 12π λi

h2l2

E12ρ

λ1 λ2 λ3 λ4 λ5 23.289 73.9 165.23 299.7 478.1


Linear element: variable beam, imposed mesh,

31 nodes,

30 linear elements.


20

Eigen mode shapes

2.4.4 Results sheet

Results comparison: eigen modes frequencies

Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Mode 1 Hz 54.18 54.01 -0.31% CM2 Mode 2 Hz 171.94 170.58 -0.79% CM2 Mode 3 Hz 384.4 378.87 -1.44% CM2 Mode 4 Hz 697.24 681.31 -2.28% CM2 Mode 5 Hz 1112.28 1075.70 -3.29%


21

2.5 Test No. 01-0005SSLLB_FEM: Tied (sub-tensioned) beam



Analysis type: static, thermoelastic (plane problem);


2.5.2 Overview Beam reinforced by a system of hinged bars with a uniform linear load, p.

Tied (sub-tensioned) beam Scale =1/37 01-0005SSLLB_FEM

Units

I. S.

Geometry

Length: o AD = FB = a = 2 m, o DF = CE = b = 4 m, o CD = EF = c = 0.6 m, o AC = EB = d = 2.088 m, o Total length: L = 8 m,

AD, DF, FB Beams: o Section: A = 0.01516 m2, o Shear area: Ar = A / 2.5, o Inertia moment: I = 2.174 x 10-4 m4,


22

CE Bar: o Section: A1 = 4.5 x 10-3 m2,

AC, EB bar: o Section: A2 = 4.5 x 10-3 m2,

CD, EF bars: o Section: A3 = 3.48 x 10-3 m2.


Isotropic linear elastic material,


Shearing module: G = 0.4x E.

Boundary conditions

Outer: Hinged in A, support connection in B (blocked vertical translation),

Inner: Hinged at bar ends: AC, CD, EF, EB.

Loading

External: Uniform linear load p = -50000 N/ml,

Internal: Shortening of the CE tie of δ = 6.52 x 10-3 m (dilatation coefficient: αCE = 1 x 10-5 /°C and temperature variation ΔT = -163°C).

2.5.3 Compression force in CE bar

Reference solution

The solution is established by considering the deformation effects due to the shear force and normal force:

μ = 1 - 43 x

aL

k = AAr

= 2.5

t = IA

γ = (L/c)2 x (1+ (A/A1) x (b/L) + 2 x (A/A2) x (d/a)2 x (d/L) + 2 x (A/A3) (c/a)2 x (c/L)

τ = k x [(2Et2) / (GaL)]

ρ = μ + γ + τ

μ0 = 1 – (a/L)2 x (2 – a/L)

τ0 = 6k x (E/G) x (t/L)2 x (1 + b/L)

ρ0 = μ0 + τ0

NCE = - (1/12) x (pL2/c) x (ρ0 /ρ) + (EI/(Lc2)) x (δ/ρ) = 584584 N


Linear element: without meshing,

AD, DF, FB: S beam (considering the shear force deformations),

AC, CD, EF, EB: bar,

CE: beam, 6 nodes.


23

Force diagrams

Tied (sub-tensioned) beam Scale =1/31 Compression force in CE bar

2.5.4 Bending moment at point H

Reference solution

MH = - (1/8) x pL2 x [1- (2/3) x (ρ0/ρ)] – (EI/(Lc)) x (δ/p) = 49249.5 N


Linear element: without meshing,

AD, DF, FB: S beam (considering the shear force deformations),

AC, CD, EF, EB: bar,

CE: beam, 6 nodes.


24

Shape of the bending moment diagram

Tied (sub-tensioned) beam Scale =1/31 Mz bending moment

2.5.5 Vertical displacement at point D

Reference solution

The reference displacement vD provided by AFNOR is determined by averaging the results of several software with implemented finite elements method.

vD = -0.5428 x 10-3 m


Linear element: without meshing, o AD, DF, FB: S beam (considering the shear force deformations), o AC, CD, EF, EB: bar, o CE: beam,

6 nodes.


25

Deformed shape

Tied (sub-tensioned) beam Scale =1/31 Deformed

2.5.6 Results sheet

Results comparison: Tension force

Solver Positioning Units Reference AD 2010 Deviation CM2 CE Bar N 584584 584584 0.00%


26

2.6 Test No. 01-0006SDLLB_FEM: Thin circular ring fixed in two points



Analysis type: modal analysis, plane problem;


2.6.2 Overview Research of the first eigen modes frequencies for a circular ring fixed in two points, subjected to its own weight only.

Thin circular ring fixed in two points Scale =1/2 01-0006SDLLB_FEM

Units I. S.

Geometry

Average radius of curvature: OA = OB = R = 0.1 m,

Angular spacing between points A and B: 120° ;

Rectangular straight section: oThickness: h = 0.005 m, oWidth: b = 0.010 m, oSection: A = 5 x 10-5 m2, oFlexure moment of inertia relative to the vertical axis: I = 1.042 x 10-10 m4,

Point coordinates: o O (0 ;0),

o A (-0.05 3 ; -0.05),

o B (0.05 3 ; -0.05).


27


Longitudinal elastic modulus: E = 7.2 x 1010 Pa Poisson's ratio: ν = 0.3, Density: ρ = 2700 kg/m3.

Boundary conditions

Outer: Fixed at A and B, Inner: None.

Loading

External: None, Internal: None.

2.6.3 Eigen mode frequencies Reference solutions The deformation of the fixed ring is calculated from the deformations of the free-free thin ring

Symmetrical mode: o u’i = i cos(iθ) o v’i = sin (iθ)

o θ’i = 1-i2R sin (iθ)

Antisymmetrical mode: o u’i = i sin(iθ) o v’i = -cos (iθ)

o θ’i = 1-i2R cos (iθ)

From Green’s method results:

fj = π21

λj ⋅2R

h 12E

ρ with a support angle of 120°.

i 1 2 3 4 Symmetrical mode 4.8497 14.7614 23.6157

Antisymmetrical mode 1.9832 9.3204 11.8490 21.5545


Linear element: beam, without meshing, 32 nodes, 32 linear elements.

Eigen mode shapes


28

2.6.4 Results sheet

Results comparison: eigen mode frequencies

Solver

Eigen mode type Eigen mode order j i

Units Reference AD 2010 Deviation

CM2 Mode 1 1 antisymmetric 1 Hz 235.3 236.32 0.43% CM2 Mode 2 2 symmetric 1 Hz 575.3 578.52 0.56% CM2 Mode 3 3 antisymmetric 2 Hz 1105.7 1112.54 0.62% CM2 Mode 4 4 antisymmetric 3 Hz 1405.6 1414.22 0.61% CM2 Mode 5 5 symmetric 2 Hz 1751.1 1760 0.51% CM2 Mode 6 6 antisymmetric 4 Hz 2557 2569.97 0.51% CM2 Mode 7 7 symmetric 3 Hz 2801.5 2777.43 -0.86%

Test 01-0007SDLSB_FEM

29

2.7 Test No. 01-0007SDLSB_FEM: Thin lozenge-shaped plate fixed on one side (α = 0 °)


Reference: Structure Calculation Software Validation Guide, test SDLS 02/89;



2.7.2 Overview The 10 mm thick plate is fixed on one side and is with its self weight only.

Thin lozenge-shaped plate fixed on one side Scale =1/10 01-0007SDLSB_FEM

Units

I. S.

Geometry

Thickness: t = 0.01 m, Side: a = 1 m, α = 0° Points coordinates:

o A ( 0 ; 0 ; 0 ) o B ( a ; 0 ; 0 ) o C ( 0 ; a ; 0 ) o D ( a ; a ; 0 )


Longitudinal elastic modulus: E = 2.1 x 1011 Pa, Poisson's ratio: ν = 0.3, Density: ρ = 7800 kg/m3.


30

Boundary conditions

Outer: AB side fixed,

Inner: None.

Loading

External: None,

Internal: None.

2.7.3 Eigen mode frequencies relative to the α angle

Reference solution M. V. Barton formula for a side "a" lozenge, leads to the frequencies:

fj = ⋅⋅π 2a21

λi2

)1(12Et

2

2

ν−ρ where i = 1,2, and λi

2 = g(α).

α = 0 λ1

2 3.492 λ2

2 8.525

M.V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method.



61 nodes,


Eigen mode shapes

2.7.4 Results sheet


Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Mode 1 Hz 8.7266 8.67 -0.65% CM2 Mode 2 Hz 21.3042 21.21 -0.44%


31






2.8.2 Overview The 10 mm thick plate is fixed on one side and is loaded with its self weight only.


Units I. S.

Geometry

Thickness: t = 0.01 m,

Side: a = 1 m,

α = 15°

Points coordinates: o A ( 0 ; 0 ; 0 ) o B ( a ; 0 ; 0 ) o C ( 0.259a ; 0.966a ; 0 ) o D ( 1.259a ; 0.966a ; 0 )



Poisson's ratio: ν = 0.3,

Density: ρ = 7800 kg/m3.


32

Boundary conditions


Inner: None.

Loading

External: None,

Internal: None.

2.8.3 Eigen modes frequencies function by α angle

Reference solution

M. V. Barton formula for a lozenge of side "a" leads to the frequencies:

fj = ⋅⋅π 2a21

λi2

)1(12Et

2

2

ν−ρ where i = 1,2, or λi

2 = g(α).

α = 15° λ1

2 3.601 λ2

2 8.872

M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method.



961 nodes,


Eigen mode shapes

2.8.4 Results sheet


Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Mode 1 Hz 8.999 8.95 -0.54% CM2 Mode 2 Hz 22.1714 21.69 -2.17%


33


2.9.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar.



Units I. S.

Geometry

Thickness: t = 0.01 m, Side: a = 1 m, α = 30° Points coordinates:

o A ( 0 ; 0 ; 0 ) o B ( a ; 0 ; 0 )

o C ( 0.5a ; 3 2 a ; 0 )

o D ( 1.5a ; 3 2 a ; 0 )

Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa, Poisson's ratio: ν = 0.3, Density: ρ = 7800 kg/m3.


34

Boundary conditions


Inner: None.

Loading

External: None,

Internal: None.


Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies:

fj = ⋅⋅π 2a21

λi2

)1(12Et

2

2


2 = g(α).

α = 30° λ1

2 3.961 λ2

2 10.19




961 nodes,


Eigen mode shapes

2.9.4 Results sheet

Results comparison: eigen modes frequencies Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Mode 1 Hz 9.8987 9.82 -0.80% CM2 Mode 2 Hz 25.4651 23.44 -7.95%


35








Units

I. S.

Geometry


Side: a = 1 m,

α = 45°

Points coordinates: o A ( 0 ; 0 ; 0 ) o B ( a ; 0 ; 0 )

o C ( 22

a ; 22

a ; 0 )

o D (2

22 +a ;

22

a ; 0 )


36





Boundary conditions


Inner: None.

Loading

External: None,

Internal: None.


Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies:

fj = ⋅⋅π 2a21

λi2

)1(12Et

2

2


2 = g(α).

α = 45° λ1

2 4.4502 λ2

2 10.56




961 nodes,


Eigen mode shapes

2.10.4 Results sheet

Results comparison: eigen mode frequencies Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Mode 1 Hz 11.1212 11.28 1.43% CM2 Mode 2 Hz 26.3897 28.08 6.41%


37

2.11 Test No. 01-0011SDLLB_FEM: Vibration mode of a thin piping elbow in plane (case 1)



Analysis type: modal analysis (plane problem);


2.11.2 Overview A thin piping elbow with a radius of 1m has fixed ends and is loaded with its self weight only.

Vibration mode of a thin piping elbow in plane Scale = 1/7 Case 1 01-0011SDLLB_FEM

Units

I. S.

Geometry

Average radius of curvature: OA = R = 1 m,

Straight circular hollow section:

Outer diameter: de = 0.020 m,

Inner diameter: di = 0.016 m,

Section: A = 1.131 x 10-4 m2,

Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4,

Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4,

Polar inertia: Ip = 9.274 x 10-9 m4.

Points coordinates (in m): o O ( 0 ; 0 ; 0 ) o A ( 0 ; R ; 0 ) o B ( R ; 0 ; 0 )


38





Boundary conditions

Outer: Fixed at points A and B ,

Inner: None.

Loading

External: None,

Internal: None.


Reference solution

The Rayleigh method applied to a thin curved beam is used to determine parameters such as:

in plane bending:

fj = 2

2i

R2 ⋅π

λ

AEIz

ρ where i = 1,2,


Linear element: beam,

11 nodes,

10 linear elements.

Eigen mode shapes



Solver Eigen mode type Units Reference AD 2010 Deviation CM2 In plane 1 Hz 119 120.09 0.92% CM2 In plane 2 Hz 227 227.10 0.04%


39



Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear.

2.12.2 Overview A thin piping elbow with a radius of 1m is extended by two straight elements of length L is loaded with its self weight only.

Vibration mode of a thin piping elbow Scale = 1/11 Case 2 01-0012SDLLB_FEM

Units I. S.

Geometry

Average radius of curvature: OA = R = 1 m, L = 0.6 m, Straight circular hollow section: Outer diameter de = 0.020 m, Inner diameter di = 0.016 m, Section: A = 1.131 x 10-4 m2, Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4, Polar inertia: Ip = 9.274 x 10-9 m4. Points coordinates (in m):

o O ( 0 ; 0 ; 0 ) o A ( 0 ; R ; 0 ) o B ( R ; 0 ; 0 ) o C ( -L ; R ; 0 ) o D ( R ; -L ; 0 )


40





Boundary conditions

Outer: o Fixed at points C and D o At A: translation restraint along y and z, o At B: translation restraint along x and z,

Inner: None.

Loading

External: None,

Internal: None.

2.12.3 Eigen mode frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as:

in plane bending:

fj = 2

2i

R2 ⋅π

λ

AEIz

ρ where i = 1,2,



23 nodes,

22 linear elements.

Eigen mode shapes

2.12.4 Results sheet Results comparison: eigen modes frequencies

Solver Eigen mode type Units Reference AD 2010 Deviation CM2 In plane 1 Hz 94 94.62 0.66% CM2 In plane 2 Hz 180 184.68 2.60%


41






2.13.2 Overview A thin piping elbow with a radius of 1m is extended by two straight elements of length L is loaded with its self weight only.

Vibration mode of a thin piping elbow Scale = 1/12 Case 3 01-0013SDLLB_FEM

Units I. S.

Geometry





Section: A = 1.131 x 10-4 m2,




Points coordinates (in m): o O ( 0 ; 0 ; 0 ) o A ( 0 ; R ; 0 ) o B ( R ; 0 ; 0 ) o C ( -L ; R ; 0 ) o D ( R ; -L ; 0 )


42





Boundary conditions

Outer: o Fixed at points C and Ds, o At A: translation restraint along y and z, o At B: translation restraint along x and z,

Inner: None.

Loading

External: None,

Internal: None.


Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as:

in plane bending:

fj = 2

2i

R2 ⋅π

λ

AEIz

ρ where i = 1,2,



41 nodes,

40 linear elements.

Eigen mode shapes



Solver Eigen mode type Units Reference AD 2010 Deviation CM2 In plane 1 Hz 25.300 24.961 -1.34% CM2 In plane 2 Hz 27.000 26.710 -1.07%


43

2.14 Test No. 01-0014SDLLB_FEM: Thin circular ring hang from an elastic element



Analysis type: modal analysis, plane problem;


2.14.2 Overview Search for the first eigen modes frequencies of a circular ring hang from an elastic element and loaded with its own weight only.

Thin circular ring hang from an elastic element Scale = 1/1 01-0014SDLLB_FEM

Units

I. S.

Geometry

Average radius of curvature: OB = R = 0.1 m,

Length of elastic element: AB = 0.0275 m ;


44

Straight rectangular section: o Ring Thickness: h = 0.005 m, Width: b = 0.010 m, Section: A = 5 x 10-5 m2, Flexure moment of relative to the vertical axis: I = 1.042 x 10-10 m4,

o Elastic element Thickness: h = 0.003 m, Width: b = 0.010 m, Section: A = 3 x 10-5 m2, Flexure moment of inertia relative to the vertical axis: I = 2.25 x 10-11 m4,

Points coordinates:

o O ( 0 ; 0 ), o A ( 0 ; -0.0725 ), o B ( 0 ; -0.1 ).





Boundary conditions

Outer: Fixed in A,

Inner: None.

Loading

External: None,

Internal: None.


Reference solutions

The reference solution was established from experimental results of a mass manufactured aluminum ring.



43 nodes,

43 linear elements.


45

Eigen mode shapes



Solver Eigen mode type Eigen mode order Units Reference AD 2010 Deviati

on CM2 Mode 1 Asymmetrical Hz 28.80 28.81 0.03% CM2 Mode 2 Symmetrical Hz 189.30 189.69 0.21% CM2 Mode 3 Asymmetrical Hz 268.80 269.38 0.22% CM2 Mode 4 Asymmetrical Hz 641.00 642.15 0.18% CM2 Mode 5 Symmetrical Hz 682.00 683.9 0.28% CM2 Mode 6 Asymmetrical Hz 1063.00 1065.73 0.26%


46

2.15 Test No. 01-0015SSLLB_FEM: Double fixed beam with a spring at mid span


Reference: internal GRAITEC test;



2.15.2 Overview Consider the double fixed beam described below. This beam consists of four elements of the length l having identical characteristics.

Units

I. S.

Geometry

l = 1 m

S=0.01 m2

I = 0.0001 m4




Boundary conditions

Outer:

Fixed at ends x = 0 and x = 4 m,

Elastic support with k = EI/l rigidity

Inner: None.

Loading

External: Punctual load P = -10000 N at x = 2m,

Internal: None.


47

2.15.3 Displacement of the model in the linear elastic range

Reference solution

The reference vertical displacement v3, is calculated at the middle of the beam at x = 2 m.

Rigidity matrix of a plane beam:

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−

−

−

−

=

llll

llll

ll

lll

llll

ll

EIEIEIEI

EIEIEIEI

EIEIEIEI

EIEIEIEI

460260

61206120

00l

ES00ES

260460

61206120

00ES-00ES

K

22

2323

22

2323

e

Given the symmetry / X and load of the structure, it is unnecessary to consider the degrees of freedom associated with normal work (u2, u3, u4).

The same symmetry allows the deduction of:

v2 = v4

β2 = -β4

β3 = 0

( )( )( )( )( )( )654321

000

00

4626

612612

268026

612024612

268026

6120124612

268026

612024612

2646

612612

5

5

1

1

5

5

4

4

3

3

2

2

1

1

22

22

22

22

22

22

22

22

22

22

⎪⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−

−

−−−

−

−⎟⎠

⎞⎜⎝

⎛ +−−

−

−−−

−

−

MR

P

MR

v

v

v

v

v

EI

β

β

β

β

β

llll

llll

lllll

lllll

lllll

llllll

lllll

lllll

llll

llll

33

333

333

333

33


48

The elementary rigidity matrix of the spring in its local axis system, [ ])()(

1111

6

35 U

UEIk ⎥⎦

⎤⎢⎣

⎡−

−=

l, must be expressed in the

global axis system by means of the rotation matrix (90° rotation):

[ ]

( )( )( )( )( )( )6

6

6

3

3

3

5

000000010010000000000000010010000000

β

β

vu

vu

EIK

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=l

→ 344332 43 0826 vvllll

−=⇒=++ βββ

→ 344332332 024612 vvvv =⇒=+−−

lllβ

→ y)unnecessar(usually 026826244423222 vvvv =⇒=+−++ βββ

lllll

(3) → ( ) m 10 11905.03

612124612 032

3

34243332223−−=

+−=⇒−=−−⎟

⎠

⎞⎜⎝

⎛ ++−−EIl

PvEIPvvv

llllllββ



6 nodes,

4 linear elements + 1 spring,

Deformed shape

Double fixed beam with a spring at mid span Deformed

Note: the displacement is expressed here in μm



Solver Positioning Units Reference AD 2010 Deviation CM2 Middle of the beam Mm -0.11905 -0.11905 0.00%


49

2.16 Test No. 01-0016SDLLB_FEM: Double fixed beam


Reference: internal GRAITEC test (beams theory);

Analysis type: static linear, modal analysis;


2.16.2 Overview Consider the double fixed beam described below. This beam consists of eight elements of the length l having identical characteristics.

Units

I. S.

Geometry

Length: l = 16 m,

Axial section: S=0.06 m2

Inertia I = 0.0001 m4


Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,


Density: ρ = 7850 kg/m3

Boundary conditions

Outer: Fixed at both ends x = 0 and x = 8 m,

Inner: None.

Loading

External: Punctual load P = -50000 N at x = 4m,

Internal: None.


50


Reference solution

The reference vertical displacement v5, is calculated at the middle of the beam at x = 2 m.

m 05079.00001.0111.2192

1650000192

33

5 =××

×==

EEIPlv



9 nodes,

8 elements.

Deformed shape

Double fixed beam Deformed

2.16.4 Eigen mode frequencies of the model in the linear elastic range

Reference solution

Knowing that the first four eigen mode frequencies of a double fixed beam are given by the following formula:

SIE

Lf nn .

...2 2

2

ρπχ

= where for the first 4 eigen modes frequencies

⎪⎪

⎩

⎪⎪

⎨

⎧

→=

→=

→=

→=

Hz 26.228=f 8.199

Hz 15.871=f 9.120

Hz 8.095=f 67.61

Hz 2.937=f 37.22

424

323

222

121

χ

χ

χ

χ



9 nodes,

8 elements.


51

Modal deformations

Double fixed beam Mode 1




52

Double fixed beam

Mode 4


1 Results comparison: vertical displacement

Solver Positioning Units Reference AD 2010 Deviation

CM2 Middle of the beam m -0.05079 -0.05079 0.00%

2 Results comparison: eigen mode frequencies

Solver Eigen modes Units Reference AD 2010 Deviation

CM2 1 Hz 2.937 2.937 0.00%

CM2 2 Hz 8.095 8.087 -0.10%

CM2 3 Hz 15.870 15.789 -0.51%

CM2 4 Hz 26.228 25.758 -1.79%


53

2.17 Test No. 01-0017SDLLB_FEM: Short beam on simple supports (on the neutral axis)





2.17.2 Overview Search for the first eigen mode frequencies of a beam on simple supports (the supports are located on the neutral axis).

Short beam on simple supports on the neutral axis Scale = 1/6 01-0017SDLLB_FEM

Units

I. S.

Geometry

Height: h = 0.2 m,

Length: l = 1 m,

Width: b = 0.1 m,

Section: A = 2 x 10-2 m4,

Flexure moment of inertia relative to z-axis: Iz = 6.667 x 10-5 m4.


54





Boundary conditions

Outer: o Hinged at A (null horizontal and vertical displacements), o Simple support in B.

Inner: None.

Loading

External: None.

Internal: None.

2.17.3 Eigen modes frequencies

Reference solution

The bending beams equation gives, when superimposing, the effects of simple bending, shear force deformations and rotation inertia, Timoshenko formula.

The reference eigen modes frequencies are determined by a numerical simulation of this equation, independent of any software.

The eigen frequencies in tension-compression are given by:

fi = ⋅π

λl2

i ρE

where λi = 2

)1i2( −


Linear element: S beam, imposed mesh,

10 nodes,

9 linear elements.

Eigen mode shapes


55



Solver Eigen mode type Units Reference AD 2010 Deviation

CM2 Mode 1 Hz 431.555 437.118 1.29% CM2 Mode 2 Hz 1265.924 1264.319 -0.13% CM2 Mode 3 Hz 1498.295 1537.156 2.59% CM2 Mode 4 Hz 2870.661 2911.456 1.42% CM2 Mode 5 Hz 3797.773 3754.542 -1.14% CM2 Mode 6 Hz 4377.837 4281.235 -2.21%


56

2.18 Test No. 01-0018SDLLB_FEM: Short beam on simple supports (eccentric)



Analysis type: modal analysis, (plane problem);


2.18.2 Overview Search for the first eigen mode frequencies of a beam on simple supports (the supports are eccentric relative to the neutral axis).

Short beam on simple supports (eccentric) Scale = 1/5 01-0018SDLLB_FEM

Units

I. S.

Geometry

Height: h = 0.2m,

Length: l = 1 m,

Width: b = 0.1 m,

Section: A = 2 x 10-2 m4,

Flexure moment of inertia relative to z-axis: Iz = 6.667 x 10-5 m4.


57





Boundary conditions

Outer: o Hinged at A (null horizontal and vertical displacements), o Simple support at B.

Inner: None.

Loading

External: None.

Internal: None.


Reference solution

The problem has no analytical solution, the solution is determined by averaging several software: Timoshenko model with shear force deformation effects and rotation inertia. The bending modes and the traction-compression are coupled.


Linear element: S beam, imposed mesh,

10 nodes,

9 linear elements.

Eigen modes shape

Short beam on simple supports (eccentric) Mode 1


58

Short beam on simple supports (eccentric)

Mode 2




59

Short beam on simple supports (eccentric)

Mode 5



Solver Eigen mode type Units Reference Reference

margin AD 2010 Deviati

on CM2 Mode 1 Hz 392.8 ± 2.5% 393.70 0.23% CM2 Mode 2 Hz 902.2 ± 5% 945.35 4.78% CM2 Mode 3 Hz 1591.9 ± 3% 1595.94 0.25% CM2 Mode 4 Hz 2629.2 ± 5% 2526.22 -3.92%CM2 Mode 5 Hz 3126.2 ± 4% 3118.91 -0.23%


60

2.19 Test No. 01-0019SDLSB_FEM: Thin square plate fixed on one side





2.19.2 Overview Research of the first eigen modes frequencies of a thin square plate fixed on one side.

Thin square plate fixed on one side Scale = 1/6 01-0019SDLSB_FEM

Units I. S.

Geometry

Side: a = 1 m,

Thickness: t = 1 m,

Points coordinates in m: o A (0 ;0 ;0) o B (1 ;0 ;0) o C (1 ;1 ;0) o D (0 ;1 ;0)






61

Boundary conditions

Outer: Edge AD fixed.

Inner: None.

Loading

External: None.

Internal: None.


Reference solution

M. V. Barton formula for a square plate with side "a", leads to:

fj = 2a2

1⋅π

λi2

)1(12Et

2

2

ν−ρ where i = 1,2, . . .

i 1 2 3 4 5 6 λi 3.492 8.525 21.43 27.33 31.11 54.44


Planar element: shell,

959 nodes,

900 planar elements.

Eigen mode shapes

Thin square plate fixed on one side Mode 1


62

Thin square plate fixed on one side

Mode 2

Thin square plate fixed on one side Mode 3




CM2 Mode 1 Hz 8.7266 8.671 -0.64% CM2 Mode 2 Hz 21.3042 21.218 -0.40% CM2 Mode 3 Hz 53.5542 53.130 -0.79% CM2 Mode 4 Hz 68.2984 67.737 -0.82% CM2 Mode 5 Hz 77.7448 77.147 -0.77% CM2 Mode 6 Hz 136.0471 134.655 -1.02%


63

2.20 Test No. 01-0020SDLSB_FEM: Rectangular thin plate simply supported on its perimeter





2.20.2 Overview Search for the first eigen mode frequencies of a thin rectangular plate fixed simply supported on its perimeter.

Rectangular thin plate simply supported on its perimeter Scale = 1/8 01-0020SDLSB_FEM

Units

I. S.

Geometry

Length: a = 1.5 m,

Width: b = 1 m,



64

Points coordinates in m: o A (0 ;0 ;0) o B (0 ;1.5 ;0) o C (1 ;1.5 ;0) o D (1 ;0 ;0)





Boundary conditions

Outer: o Simple support on all sides, o For the modeling: hinged at A, B and D.

Inner: None.

Loading

External: None.

Internal: None.


Reference solution

M. V. Barton formula for a rectangular plate with supports on all four sides, leads to:

fij = 2π

[ ( ai

)2 + ( bj

)2] )1(12

Et2

2

ν−ρ

where: i = number of half-length of wave along y ( dimension a)

j = number of half-length of wave along x ( dimension b)



496 nodes,



65

Eigen mode shapes



Solver Eigen mode type i j


CM2 1 1 Hz 35.63 35.580 -0.14% CM2 2 1 Hz 68.51 68.292 -0.32% CM2 1 2 Hz 109.62 109.982 0.33% CM2 3 1 Hz 123.32 123.021 -0.24% CM2 2 2 Hz 142.51 141.976 -0.37% CM2 3 2 Hz 197.32 195.545 -0.90%

Test 01-0021SFLLB_FEM

66

2.21 Test No. 01-0021SFLLB_FEM: Cantilever beam in Eulerian buckling


Reference: internal GRAITEC test (Euler theory);

Analysis type: Eulerian buckling;


2.21.2 Overview

Units

I. S.

Geometry

L = 10 m

S=0.01 m2

I = 0.0002 m4




Boundary conditions


Inner: None.

Loading

External: Punctual load P = -100000 N at x = L,

Internal: None.

Test 01-0021SFLLB_FEM

67

2.21.3 Critical load on node 5

Reference solution

The reference critical load established by Euler is:

98696.010000098696N 98696

L4EIP 2

2

critique ==λ⇒=π

=


Planar element: beam, imposed mesh,

5 nodes,

4 elements.

Deformed shape


Results comparison: critical load

Solver Positioning Units Reference AD 2010 Deviation CM2 On node 5 N -98696 -98699.278 0.00%


68

2.22 Test No. 01-0022SDLSB_FEM: Annular thin plate fixed on a hub (repetitive circular structure)




Element type: planar element.

2.22.2 Overview Search for the eigen mode frequencies of a thin annular plate fixed on a hub.

Annular thin plate fixed on a hub (repetitive circular structure) Scale = 1/3 01-0022SDLSB_FEM

Units

I. S.

Geometry

Inner radius: Ri = 0.1 m,

Outer radius: Re = 0.2 m,

Thickness: t = 0.001 m.

Material properties





69

Boundary conditions

Outer: Fixed on a hub at any point r = Ri.

Inner: None.

Loading

External: None.

Internal: None.


Reference solution

The solution of determining the frequency based on Bessel functions leads to the following formula:

fij = 1

2πRe2 λij

2 Et2

12ρ(1-ν2)

where: i = the number of nodal diameters

j = the number of nodal circles

and λij2 such as:

j \ i 0 1 2 3 0 13.0 13.3 14.7 18.5 1 85.1 86.7 91.7 100


Planar element: plate,

360 nodes,




Solver Eigen mode type i j

Corresponding eigen mode in AD 2010


CM2 0 0 1 Hz 79.26 79.05 -0.26% CM2 0 1 18 Hz 518.85 521.84 0.58% CM2 1 0 2 Hz 81.09 80.52 -0.70% CM2 1 1 20 Hz 528.61 529.49 0.17% CM2 2 0 4 Hz 89.63 88.43 -1.34% CM2 2 1 22 Hz 559.09 552.43 -1.19% CM2 3 0 7 Hz 112.79 110.27 -2.23% CM2 3 1 25 Hz 609.70 593.83 -2.60%


70

2.23 Test No. 01-0023SDLLB_FEM: Bending effects of a symmetrical portal frame





2.23.2 Overview Search for the first eigen mode frequencies of a symmetrical portal frame with fixed supports.

Bending effects of a symmetrical portal frame Scale = 1/5 01-0023SDLLB_FEM


71

Units

I. S.

Geometry

Straight rectangular sections for beams and columns:

Thickness: h = 0.0048 m,

Width: b = 0.029 m,

Section: A = 1.392 x 10-4 m2,


Points coordinates in m: A B C D E F x -0.30 0.30 -0.30 0.30 -0.30 0.30 y 0 0 0.36 0.36 0.81 0.81





Boundary conditions

Outer: Fixed at A and B,

Inner: None.

Loading

External: None.

Internal: None.


Reference solution

Dynamic radius method (slender beams theory).



60 nodes,

60 linear elements.


72

Deformed shape

Bending effects of a symmetrical portal frame Scale = 1/7 Mode 13


Results comparison: eigen mode frequencies


CM2 1 antisymmetric Hz 8.8 8.8 0.00% CM2 2 antisymmetric Hz 29.4 29.4 0.00% CM2 3 symmetric Hz 43.8 43.8 0.00% CM2 4 symmetric Hz 56.3 56.3 0.00% CM2 5 antisymmetric Hz 96.2 96.1 -0.10% CM2 6 symmetric Hz 102.6 102.7 0.10% CM2 7 antisymmetric Hz 147.1 147.1 0.00% CM2 8 symmetric Hz 174.8 174.9 0.06% CM2 9 antisymmetric Hz 178.8 178.9 0.06% CM2 10 antisymmetric Hz 206 206.2 0.10% CM2 11 symmetric Hz 266.4 266.6 0.08% CM2 12 antisymmetric Hz 320 319.95 -0.02% CM2 13 symmetric Hz 335 334.96 -0.01%


73

2.24 Test No. 01-0024SSLLB_FEM: Slender beam on two fixed supports





2.24.2 Overview A straight slender beam with fixed ends is loaded with a uniform load, several punctual loads and a torque.

Slender beam on two fixed supports Scale = 1/4 01-0024SSLLB_FEM


74

Units

I. S.

Geometry

Length: L = 1 m,

Beam inertia: I = 1.7 x 10-8 m4.


Longitudinal elastic modulus: E = 2.1 x 1011 Pa.

Boundary conditions

Outer: Fixed at A and B,

Inner: None.

Loading

External: o Uniformly distributed load from A to B: py = p = -24000 N/m, o Punctual load at D: Fx = F1 = 30000 N, o Torque at D: Cz = C = -3000 Nm, o Punctual load at E: Fx = F2 = 10000 N, o Punctual load at E: Fy = F = -20000 N.

Internal: None.


75

2.24.3 Shear force at G

Reference solution

Analytical solution:

Shear force at G: VG

VG = 0.216F – 1.26 LC



5 nodes,

4 linear elements.

Results shape

Slender beam on two fixed supports Scale = 1/5

Shear force


76

2.24.4 Bending moment in G

Reference solution


Bending moment at G: MG

MG = pL2

24 - 0.045LF – 0.3C



5 nodes,

4 linear elements.

Results shape


Bending moment


77

2.24.5 Vertical displacement at G

Reference solution


Vertical displacement at G: vG

vG = pl4

384EI + 0.003375FL3

EI + 0.015CL2

EI



5 nodes,

4 linear elements.

Results shape


Deformed


78

2.24.6 Horizontal reaction at A

Reference solution


Horizontal reaction at A: HA HA = -0.7F1 –0.3F2



5 nodes,

4 linear elements.


1 Results comparison: shear force

Solver Positioning Units Reference AD 2010 Deviation CM2 In G N -540 -540 0.00%

2 Results comparison: bending moment

Solver Positioning Units Reference AD 2010 Deviation CM2 In G Nm -2800 -2800 0.00%


Solver Positioning Units Reference AD 2010 Deviation CM2 In G cm -4.90 -4.90 0.00%

4 Results comparison: horizontal reaction

Solver Positioning Units Reference AD 2010 Deviation CM2 In A N 24000 24000 0.00%


79

2.25 Test No. 01-0025SSLLB_FEM: Slender beam on three supports



Analysis type: static (plane problem);


2.25.2 Overview A straight slender beam on three supports is loaded with two punctual loads.

Slender beam on three supports Scale = 1/49 01-0025SSLLB_FEM

Units I. S.

Geometry Length: L = 3 m,

Beam inertia: I = 6.3 x 10-4 m4.

Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa.

Boundary conditions Outer:

o Hinged at A, o Elastic support at B (Ky = 2.1 x 106 N/m), o Simple support at C.

Inner: None.


80

Loading

External: 2 punctual loads F = Fy = -42000N.

Internal: None.

2.25.3 Bending moment at B Reference solution The resolution of the hyperstatic system of the slender beam leads to:

k = Ky3LEI6

Bending moment at B: MB

MB = ± 2L

)k8(F)k26(

++−



5 nodes,

4 linear elements.

Results shape

Slender beam on three supports Scale = 1/49

Bending moment

2.25.4 Reaction in B Reference solution

Compression force in the spring: VB

VB = -11F8 + k



5 nodes,

4 linear elements.


81

2.25.5 Vertical displacement at B

Reference solution

Deflection at the spring location: vB

vB = 11F

Ky(8 + k)



5 nodes,

4 linear elements.

Results shape

Slender beam on three supports Deformed



Solver Positioning Units Reference AD 2010 Deviation CM2 In B Nm -63000 -63000 0.00%


Solver Positioning Units Reference AD 2010 Deviation CM2 In B cm -1.00 -1.00 0.00%

3 Results comparison: reaction

Solver Positioning Units Reference AD 2010 Deviation CM2 In B N -21000 -21000 0.00%


82

2.26 Test No. 01-0026SSLLB_FEM: Bimetallic: Fixed beams connected to a stiff element





2.26.2 Overview Two beams fixed at one end and rigidly connected to an undeformable beam is loaded with a punctual load.

Fixed beams connected to a stiff element Scale = 1/10 01-0026SSLLB_FEM

Units

I. S.

Geometry

Lengths: o L = 2 m, o l = 0.2 m,

Beams inertia moment: I = (4/3) x 10-8 m4,

The beam sections are squared, of side: 2 x 10-2 m.


83


Longitudinal elastic modulus: E = 2 x 1011 Pa.

Boundary conditions

Outer: Fixed in A and C,

Inner: The tangents to the deflection of beams AB and CD at B and D remain horizontal; practically, we restraint translations along x and z at nodes B and D.

Loading

External: In D: punctual load F = Fy = -1000N.

Internal: None.

2.26.3 Deflection at B and D

Reference solution

The theory of slender beams bending (Euler-Bernouilli formula) leads to a deflection at B and D:

The resolution of the hyperstatic system of the slender beam leads to:

vB = vD = FL3

24EI



4 nodes,

3 linear elements.

Results shape

Fixed beams connected to a stiff element Scale = 1/10 Deformed


84

2.26.4 Vertical reaction at A and C

Reference solution

Analytical solution.



4 nodes,

3 linear elements.

2.26.5 Bending moment at A and C

Reference solution




4 nodes,

3 linear elements


1 Results comparison: deflection

Solver Positioning Units Reference AD 2010 Deviation CM2 In B m 0.125 0.125 0.00% CM2 In D m 0.125 0.125 0.00%

2 Results comparison: vertical reaction

Solver Positioning Units Reference AD 2010 Deviation CM2 In A N -500 -500 0.00% CM2 In C N -500 -500 0.00%

3 Results comparison: Bending moment

Solver Positioning Units Reference AD 2010 Deviation CM2 In A Nm 500 500.01 0.00% CM2 In C Nm 500 500.01 0.00%


85

2.27 Test No. 01-0027SSLLB_FEM: Fixed thin arc in planar bending



Analysis type: static linear (plane problem);


2.27.2 Overview Arc of a circle fixed at one end, subjected to two punctual loads and a torque at its free end.

Fixed thin arc in planar bending Scale = 1/24

01-0027SSLLB_FEM


86

Units

I. S.

Geometry

Medium radius: R = 3 m ,

Circular hollow section: o de = 0.02 m, o di = 0.016 m, o A = 1.131 x 10-4 m2, o Ix = 4.637 x 10-9 m4.


Longitudinal elastic modulus: E = 2 x 1011 Pa.

Boundary conditions

Outer: Fixed in A.

Inner: None.

Loading

External: At B:

o punctual load F1 = Fx = 10 N, o punctual load F2 = Fy = 5 N, o bending moment about Oz, Mz = 8 Nm.

Internal: None.

2.27.3 Displacements at B

Reference solution

At point B:

displacement parallel to Ox: u = R2

4EI [F1πR + 2F2R + 4Mz]

displacement parallel to Oy: v = R2

4EI [2F1πR + (3π - 8)F2R + 2(π - 2)Mz]

rotation around Oz: θ = R

4EI [4F1R + 2(π - 2)F2R + 2πMz]



31 nodes,

30 linear elements.


87

Results shape

Fixed thin arc in planar bending Scale = 1/19 Deformed


1 Results comparison: horizontal displacement

Solver Positioning Units Reference AD 2010 Deviation CM2 In B m 0.3791 0.37891 -0.05%


Solver Positioning Units Reference AD 2010 Deviation CM2 In B m 0.2417 0.2417 0.00%

3 Results comparison: rotation about Z-axis

Solver Positioning Units Reference AD 2010 Deviation CM2 In B rad -0.1654 -0.1654 0.00%


88

2.28 Test No. 01-0028SSLLB_FEM: Fixed thin arc in out of plane bending



Analysis type: static linear;


2.28.2 Overview Arc of a circle fixed at one end, loaded with a punctual force at its free end perpendicular to the plane.

Fixed thin arc in out of plane bending Scale = 1/6 01-0028SSLLB_FEM

Units I. S.

Geometry







89

Boundary conditions

Outer: Fixed at A.

Inner: None.

Loading

External: Punctual force in B perpendicular on the plane: Fz = F = 100 N.

Internal: None.

2.28.3 Displacements at B

Reference solution Displacement out of plane at point B:

uB = FR3

EIx [ π4 +

EIx KT

(3π4 - 2)]

where KT is the torsional rigidity for a circular section (torsion constant is 2Ix).

KT = 2GIx = EIx

1 + ν ⇒ uB = FR3

EIx [ π4 + (1 + ν) (

3π4 - 2)]



46 nodes,

45 linear elements.

2.28.4 Moments at θ = 15°

Reference solution

Torsion moment: Mx’ = Mt = FR(1 - sinθ)

Bending moment: Mz’ = Mf = -FRcosθ



46 nodes,

45 linear elements.


1 Results comparison: displacement out of plane

Solver Positioning Units Reference AD 2010 Deviation CM2 In B m 0.13462 0.13516 0.40%

2 Results comparison: torsion moment

Solver Positioning Units Reference AD 2010 Deviation CM2 In θ = 15° Nm 74.1180 74.1220 0.01%


Solver Positioning Units Reference AD 2010 Deviation CM2 In θ = 15° Nm -96.5925 -96.5779 -0.02%


90

2.29 Test No. 01-0029SSLLB_FEM: Double hinged thin arc in planar bending





2.29.2 Overview Double hinged thin arc in planar bending Scale = 1/8

01-0029SSLLB_FEM


91

Units

I. S.

Geometry






Boundary conditions

Outer: o Hinge at A, o At B: allowed rotation along z, vertical displacement restrained along y.

Inner: None.

Loading

External: Punctual load at C: Fy = F = - 100 N.

Internal: None.

2.29.3 Displacements at A, B and C

Reference solution

Rotation about z-axis

θA = - θB = ( π2 - 1)

FR22EI

Displacement;

Vertical at C: vC = π8

FREA + (

3π4 - 2)

FR3

2EI

Horizontal at B: uB = FR

2EA - FR3

2EI



37 nodes,

36 linear elements.


92

Displacements shape

Fixed thin arc in planar bending Scale = 1/11 Deformed



Solver Positioning Units Reference AD 2010 Deviation CM2 In A rad 0.030774 0.030778 0.01% CM2 In B rad -0.030774 -0.030778 0.01%


Solver Positioning Units Reference AD 2010 Deviation CM2 In C cm -1.9206 -1.9202 -0.02%


Solver Model Units Reference AD 2010 Deviation CM2 In B cm 5.3912 5.386 -0.10%


93

2.30 Test No. 01-0030SSLLB_FEM: Portal frame with lateral connections





2.30.2 Overview Portal frame with lateral connections Scale = 1/21

01-0030SSLLB_FEM

Units

I. S.

Geometry

Beam Length Moment of inertia AB lAB = 4 m IAB =

643 x 10-8 m4

AC lAC = 1 m IAC = 112 x 10-8 m4

AD lAD = 1 m IAD = 112 x 10-8 m4

AE lAE = 2 m IAE = 43 x 10-8 m4


94

G is in the middle of DA.

The beams have square sections: o AAB = 16 x 10-4 m o AAD = 1 x 10-4 m o AAC = 1 x 10-4 m o AAE = 4 x 10-4 m



Boundary conditions

Outer: o Fixed at B, D and E, o Hinge at C,

Inner: None.

Loading

External: o Punctual force at G: Fy = F = - 105 N, o Distributed load on beam AD: p = - 103 N/m.

Internal: None.

2.30.3 Displacements at A

Reference solution

Rotation at A about z-axis:

We say: kAn = EIAnlAn

where n = B, C, D or E

K = kAB + kAD + kAE + 34 kAC

rAn = kAnK

C1 = FlAD

8 - plAB

2

12

θ = C14K



6 nodes,

5 linear elements.


95

Displacements shape

Portal frame with lateral connections Deformed

2.30.4 Moments in A

Reference solution

MAB = plAB

2

12 + rAB x C1

MAD = - FlAD

8 + rAD x C1

MAE = rAE x C1

MAC = rAC x C1



6 nodes,

5 linear elements


Displacement results comparison: rotation θ about z-axis

Solver Positioning Units Reference AD 2010 Deviation CM2 In A rad -0.227118 -0.227401 0.12%

Moments results comparison: bending moment

Solver Positioning Units Reference AD 2010 Deviation CM2 In A (MAB) Nm 11023.72 11020.998 -0.02% CM2 In A (MAC) Nm 113.559 113.7044 0.13% CM2 In A (MAD) Nm 12348.588 12347.476 -0.01% CM2 In A (MAE) Nm 1211.2994 1212.7730 0.12%


96

2.31 Test No. 01-0031SSLLB_FEM: Truss with hinged bars under a punctual load





2.31.2 Overview Truss with hinged bars under a punctual load Scale = 1/10

01-0031SSLLB_FEM

Units I. S.

Geometry Elements Length (m) Area (m2)

AC 0.5 2 2 x 10-4 CB 0.5 2 2 x 10-4 CD 2.5 1 x 10-4 BD 2 1 x 10-4


Boundary conditions

Outer: Hinge at A and B,

Inner: None.


97

Loading

External: Punctual force at D: Fy = F = - 9.81 x 103 N.

Internal: None.

2.31.3 Displacements at C and D

Reference solution Displacement method.



4 nodes,

4 linear elements.

Displacements shape Truss with hinged bars under a punctual load Scale = 1/9

Deformed

2.31.4 Results sheet Results comparison: horizontal displacement

Solver Positioning Units Reference AD 2010 Deviation CM2 In C mm 0.26517 0.26469 -0.18% CM2 In D mm 3.47902 3.47531 -0.11%

Results comparison: vertical displacement Solver Positioning Units Reference AD 2010 Deviation CM2 In C mm 0.08839 0.08817 -0.25% CM2 In D mm -5.60084 -5.5950 -0.10%


98

2.32 Test No. 01-0032SSLLB_FEM: Beam on elastic soil, free ends





2.32.2 Overview A beam under 3 punctual loads lays on a soil of constant linear stiffness.

Beam on elastic soil, free ends Scale = 1/21 01-0032SSLLB_FEM

Units I. S.

Geometry

L = (π 10 )/2,

I = 10-4 m4.


Boundary conditions

Outer: o Free A and B extremities, o Constant linear stiffness of soil ky = K = 840000 N/m2.

Inner: None.


99

Loading

External: Punctual load at A, C and B: Fy = F = - 10000 N.

Internal: None.

2.32.3 Bending moment and displacement at C

Reference solution

β = 4

K/(4EI)

ϕ = βL/2

λ = sh (2ϕ) + sin (2ϕ)

Bending moment: MC = (F/(4β))(ch(2ϕ) - cos (2ϕ) – 8sh(ϕ)sin(ϕ))/λ

Vertical displacement: vC = - (Fβ/(2K))( ch(2ϕ) + cos (2ϕ) + 8ch(ϕ)cos(ϕ) + 2)/λ



72 nodes,

71 linear elements.

Bending moment diagram

Beam on elastic soil, free ends Scale = 1/20 Bending moment


100

2.32.4 Displacements at A

Reference solution

Vertical displacement: vA = (2Fβ/K)( ch(ϕ)cos(ϕ) + ch(2ϕ) + cos(2ϕ))/λ

Rotation about z-axis

θA = (-2Fβ2/K)( sh(ϕ)cos(ϕ) - sin(ϕ)ch(ϕ) + sh(2ϕ) - sin(2ϕ))/λ



72 nodes,

71 linear elements



Solver Positioning Units Reference AD 2010 Deviation CM2 In C Nm 5759 5772.05 0.23%


Solver Positioning Units Reference AD 2010 Deviation CM2 In C m -0.006844 -0.006844 0.00%


Solver Positioning Units Reference AD 2010 Deviation CM2 In A m -0.007854 -0.00786073 0.09%

4 Results comparison: rotation θ about z-axis

Solver Positioning Units Reference AD 2010 Deviation CM2 In A rad -0.000706 -0.000707 0.14%

Test 01-0033SFLLA_FEM

101

2.33 Test No. 01-0033SFLLA_FEM: EDF Pylon


Reference: Internal GRAITEC test;

Analysis type: static linear, Eulerian buckling;

Element type: linear

2.33.2 Overview

Units

I. S.

Geometry


102




Boundary conditions

Outer: o Hinged support, o For the modeling, a fixed restraint and 4 beams were added at the pylon supports level.

Inner: None.

Loading

External: Punctual loads corresponding to a wind load.

o FX = 165550 N, FY = - 1240 N, FZ = - 58720 N on the main arms ?, o FX = 50250 N, FY = - 1080 N, FZ = - 12780 N on the upper arm, o FX = 11760 N, FY = 0 N, FZ = 0 N on the lower horizontal frames

Internal: None.


103


Reference solution

Software ANSYS 5.3 NE/NASTRAN 7.0 Max deflection (m) 0.714 0.714 λ dominating mode 2.77 2.77



402 nodes,

1034 elements.


104

Deformed shape

Buckling modal deformation (dominating mode)


105


1 Results comparison: displacement

Solver Positioning Units Reference AD 2010 Deviation CM2 Top of pylon m 0.714 0.71254 -0.20%

2 Results comparison: dominating buckling mode

Solver Mode Units Reference AD 2010 Deviation CM2 λcritique - 2.77 2.830 2.17%


106

2.34 Test No. 01-0034SSLLB_FEM: Beam on elastic soil, hinged ends

2.34.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLL 16/89;



2.34.2 Overview A beam under a punctual load, a distributed load and two torques lays on a soil of constant linear stiffness.

Beam on elastic soil, hinged ends Scale = 1/27 01-0034SSLLB_FEM

Units I. S.

Geometry

L = (π 10 )/2,

I = 10-4 m4.


Boundary conditions

Outer: o Free A and B ends, o Soil with a constant linear stiffness ky = K = 840000 N/m2.

Inner: None.


107

Loading

External: o Punctual force at D: Fy = F = - 10000 N, o Uniformly distributed force from A to B: fy = p = - 5000 N/m, o Torque at A: Cz = -C = -15000 Nm, o Torque at B: Cz = C = 15000 Nm.

Internal: None.

2.34.3 Displacement and support reaction at A

Reference solution

β = 4

K/(4EI)

ϕ = βL/2

λ = ch(2ϕ) + cos(2ϕ)

Vertical support reaction:

VA = -p(sh(2ϕ) + sin(2ϕ)) - 2βFch(ϕ)cos(ϕ) + 2β2C(sh(2ϕ) - sin(2ϕ)) x 1

2βλ

Rotation about z-axis:

θA = p(sh(2ϕ) – sin(2ϕ)) + 2βFsh(ϕ)sin(ϕ) - 2β2C(sh(2ϕ) + sin(2ϕ)) x 1

(K/β)λ



50 nodes,

49 linear elements.

Deformed shape Beam on elastic soil, hinged ends Scale = 1/20

Deformed


108

2.34.4 Displacement and bending moment at D

Reference solution

Vertical displacement:

vD = 2p(λ - 2ch(ϕ)cos(ϕ)) + βF(sh(2ϕ) – sin(2ϕ)) - 8β2Csh(ϕ)sin(ϕ) x 1

2Kλ

Bending moment:

MD = 4psh(ϕ)sin(ϕ) + βF(sh(2ϕ) + sin(2ϕ)) - 8β2Cch(ϕ)cos(ϕ) x 1

4β2λ



50 nodes,

49 linear elements.

Bending moment diagram

Beam on elastic soil, hinged ends Scale = 1/20 Bending moment


109


1 Results comparison: rotation around z

Solver Positioning Units Reference AD 2010 Deviation CM2 In A rad 0.003045 0.003043 -0.07%

2 Results comparison: vertical reaction

Solver Positioning Units Reference AD 2010 Deviation CM2 In A N -11674 -11644.36 -0.25%


Solver Positioning Units Reference AD 2010 Deviation CM2 In D cm -0.423326 -0.42330 -0.01%


Solver Positioning Units Reference AD 2010 Deviation CM2 In D Nm -33840 -33835.87 -0.01%

Test 01-0035SSLPB_FEM

110

2.35 Test No. 01-0035SSLPB_FEM: Plate with in plane bending and shear

2.35.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLP 01/89;



CAD tolerance 0.1 mm

2.35.2 Overview

Units

I. S.

Geometry

Thickness: h = 1 mm,

Length: L = 48 mm,

Height: H = 12 mm.




Boundary conditions

Outer: fixed in any point on the edge x = 0.

Inner: None.

Loading

External: Uniformly distributed force at any point x = 48 mm: fy = p = - 3333.33 N/m.

Internal: None.

2.35.3 Planes stresses in (x,y)

Reference solution

Analytical method by Airy function:

σxx = 12Py(x - L)

H3

σyy = 0

σxy = 6P(

H2

4 - y2)

H3

Test 01-0035SSLPB_FEM

111



784 nodes,



1 Results comparison: σxx plane stress

Solver Positioning Units Reference AD 2010 Deviation CM2 At (0, H/2) MPa -80 -79.46 -0.68%

2 Results comparison: σxx plane stress

Solver Positioning Units Reference AD 2010 Deviation CM2 At (0, -H/2) MPa -80 -79.46 -0.68%

3 Results comparison: σxy plane stress

Solver Positioning Units Reference AD 2010 Deviation CM2 At any point y = 0 MPa -5 -4.97 -0.60%


112

2.36 Test No. 01-0036SSLSB_FEM: Simply supported square plate





2.36.2 Overview A plate simply supported on its perimeter and loaded with its self weight only.

Simply supported square plate Scale = 1/9 01-0036SSLSB_FEM

Units I. S.

Geometry

Side = 1 m,

Thickness h = 0.01m.





Boundary conditions

Outer: o Simple support on the plate perimeter, o For the modeling, we add a fixed support at B.

Inner: None.


113

Loading

External: Self weight (gravity = 9.81 m/s2).

Internal: None.

2.36.3 Vertical displacement at O

Reference solution

According to Love- Kirchhoff hypothesis, the displacement w at a point (x,y):

w(x,y) = Σ wmnsinmπxsinnπy

where wmn = 192ρg(1 - ν2)

mn(m2 + n2)π6Eh2



441 nodes,


Deformed shape

Simply supported square plate Scale = 1/6 Deformed



Solver Positioning Units Reference AD 2010 Deviation CM2 In O μm -0.158 -0.16491 4.37%


114

2.37 Test No. 01-0037SSLSB_FEM: Caisson beam in torsion





2.37.2 Overview A caisson beam fixed at one end is loaded with torsion.

Caisson beam in torsion Scale = 1/4 01-0037SSLSB_FEM

Units I. S.

Geometry

Length; L = 1m,

Square section of side: b = 0.1 m,

Thickness = 0.005 m.




Boundary conditions

Outer: Beam fixed at end x = 0;

Inner: None.


115

Loading

External: Torsion moment M = 10N.m applied to the free end (for modeling, 4 forces of 50 N).

Internal: None.

2.37.3 Displacement and stress at two points

Reference solution

The reference solution is determined by averaging the results of several calculation software with implemented finite elements method.

Points coordinates:

A (0,0.05,0.5)

B (-0.05,0,0.8)

Note: point O is the origin of the coordinate system (x,y,z).



90 nodes,

88 planar elements.

Deformed shape

Caisson beam in torsion Scale = 1/4 Deformed


116


1 Results comparison: displacement

Solver Positioning Units Reference AD 2010 Deviation CM2 In A m -0.617 x 10-6 -0.6159 x 10-6 -0.18% CM2 In B m -0.987 x 10-6 -0.9868 x 10-6 -0.02%


Solver Positioning Units Reference AD 2010 Deviation CM2 In A rad 0.123 x 10-4 0.123 x 10-4 0.00% CM2 In B rad 0.197 x 10-4 0.197 x 10-4 0.00%

3 Results comparison: σxy stress

Solver Positioning Units Reference AD 2010 Deviation CM2 In A MPa -0.11 -0.10 -9.09% CM2 In B MPa -0.11 -0.10 -9.09%


117

2.38 Test No. 01-0038SSLSB_FEM: Thin cylinder under a uniform radial pressure

2.38.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLS 06/89;

Analysis type: static elastic;


2.38.2 Overview A cylinder of length L and radius R loaded with an uniform internal pressure.

Thin cylinder under a uniform radial pressure Scale = 1/18 01-0038SSLSB_FEM

Units I. S.

Geometry

Length: L = 4 m,

Radius: R = 1 m,

Thickness: h = 0.02 m.




Boundary conditions

Outer: o Free conditions o For the modeling, only ¼ of the cylinder is considered and the symmetry conditions are applied. On the

other side, we restrained the displacements at a few nodes in order to make the model stable.

Inner: None.


118

Loading

External: Uniform internal pressure: p = 10000 Pa,

Internal: None.

2.38.3 Stresses in all points

Reference solution

Stresses in the planar elements coordinate system (x axis is parallel with the length of the cylinder):

σxx = 0

σyy = pRh



209 nodes,


2.38.4 Cylinder deformation in all points

Radial deformation:

δR = pR2

Eh

Longitudinal deformation:

δL = -pRνL

Eh


1 Results comparison: σxx stress

Solver Positioning Units Reference AD 2010 Deviation CM2 At all points Pa 0 0 -

2 Results comparison: σyy stress

Solver Positioning Units Reference AD 2010 Deviation CM2 At all points Pa 5 x 105 5 x 105 0.00%

3 Results comparison: δL radial deformation of the cylinder

Solver Positioning Units Reference AD 2010 Deviation CM2 At all points m 2.38 x 10-6 2.39 x 10-6 0.42%

4 Results comparison: δL longitudinal deformation of the cylinder

Solver Positioning Units Reference AD 2010 Deviation CM2 At all points m -2.86 x 10-6 -2.85 x 10-6 0.34%


119

2.39 Test No. 01-0039SSLSB_FEM: Square plate under planar stresses

2.39.1 Description sheet Reference: Internal GRAITEC test;


Element type: planar (membrane).

2.39.2 Overview A square plate of 2 x 2 m is fixed on 3 sides and has a surface load "p" on it’s upper face.

Square plate under planar stresses Scale = 1/19 Modeling

[ ]1;1, −∈ηξ

Units I. S.

Geometry

Thickness: e = 1 m,

4 square elements of side h = 1 m.




Boundary conditions

Outer: Fixed on 3 sides,

Inner: None.

Loading

External: Uniform load p = -1. 108 N/ml on the upper surface,

Internal: None.


120

2.39.3 Displacement of the model in the linear elastic range Reference solution The reference displacements are calculated on nodes 7 and 9.

v9 = -6ph(3 + ν)(1 - ν2)

E(8(3 - ν)2 - (3 + ν)2) = -0.1809 x 10-3 m,

v7 = 4(3 - ν)3 + ν v9 = -0.592 x 10-3 m,

For element 1.4: (For the stresses calculated above, the abscissa point (x = 0; y = 0) corresponds to node 8.)

σyy = E

1 - ν2 (v9 - v7)

2h (1 + ξ) for

σxx = νσyy for

σxy = E

1 + ν (v9 + v7) + η(v9 - v7)

4h (1 + ξ) for


Planar element: membrane, imposed mesh,

9 nodes,


Deformed shape

ξ = -1 ; σxx = 0 ξ = 0 ; σxx = -14.23 MPa ξ = 1 ; σxx = -28.46 MPa

η = -1 ; ξ = 0 ; σxy = -47.82 MPa η = 0 ; ξ = 0 ; σxy = -31.21 MPa η= 1 ; ξ = 0 ; σxy = -14.61 MPa

ξ = -1 ; σyy = 0 ξ = 0 ; σyy = -47.44 MPa ξ = 1 ; σyy = -94.88 MPa


121



Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1.4 node 7 mm -0.592 -0.592 0.00% CM2 Element 1.4 node 9 mm -0.1809 -0.1809 0.00%

2 Results comparison: σxx stresses

Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1.4 in x = 0 m MPa 0 0 - CM2 Element 1.4 in x = 0.5 m MPa -14.23 -14.23 0.00% CM2 Element 1.4 in x = 1 m MPa -28.46 -28.46 0.00%

3 Results comparison: σyy stresses

Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1.4 in x = 0 m MPa 0 0 - CM2 Element 1.4 in x = 0.5 m MPa -47.44 -47.44 0.00% CM2 Element 1.4 in x = 1 m MPa -94.88 -94.88 0.00%

4 Results comparison: σxy stresses

Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1.4 in x = 0 m MPa -14.66 -14.61 -0.34% CM2 Element 1.4 in x = 0.5 m MPa -31.21 -31.21 0.00% CM2 Element 1.4 in x = 1 m MPa -47.82 -47.82 0.00%


122

2.40 Test No. 01-0040SSLSB_FEM: Stiffen membrane


Reference: Klaus-Jürgen Bathe - Finite Element Procedures in Engineering Analysis, Example 5.13;


Element type: planar (membrane).

2.40.2 Overview The 8 x 12 cm plate is fixed in the middle on 3 supports and has a " punctual load "P at its free node A.

[ ]1;1, −∈ηξ

Units I. S.

Geometry

Thickness: e = 0.1 cm,

Length: l = 8 cm,

Width: B = 12 cm.


Longitudinal elastic modulus: E = 30 x 106 N/cm2,


Boundary conditions

Outer: Fixed on 3 sides,

Inner: None.

Loading

External: Uniform load Fx = F = 6000 N at A,

Internal: None.


123

2.40.3 Results of the model in the linear elastic range

Reference solution

Point B is the origin of the coordinate system used for the results positions.

( ) ( )

( )

( )⎪⎩

⎪⎨

⎧

−=−==−=−==

=−=+

+−=

⎪⎩

⎪⎨

⎧

==−====

===

⎪⎩

⎪⎨

⎧

==−====

==−

−=

=+

=+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

+−

== −

MPa 96.17N/cm 1796 ;1MPa 98.8N/cm 898 ;0

0 ;1for 1

81

MPa 55.11N/cm 1155 ;1MPa 77.5N/cm 577 ;0

0 ;1for

MPa 49.38N/cm 3849 ;1MPa 24.19N/cm 1924 ;0

0 ;1for 1

21

3410.97510.367410.2

6000

211

12

3

2xy1

2xy1

xy1

1

2yy1

2yy1

yy1

11

2xx1

2xx1

xx1

21

466

222

σξσξ

σξξ

νσ

σηση

σηνσσ

σηση

σηη

νσ

νν

buE

auE

cm

aES

baeabE

FKFu

Axy

xxyy

Axx

A


Planar element: membrane, imposed mesh,

6 nodes,

2 quadrangle planar elements and 1 bar.

Deformed shape


124



Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1 in A cm -9.34 10-4 -9.34 x 10-4 0.00%


Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1 in y = 0 cm MPa 38.49 38.49 0.00% CM2 Element 1 in y = 3 cm MPa 19.24 19.24 0.00% CM2 Element 1 in y = 6 cm MPa 0 0 -


Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1 in y = 0 cm MPa 11.55 11.55 0.00% CM2 Element 1 in y = 3 cm MPa 5.77 5.77 0.00% CM2 Element 1 in y = 6 cm MPa 0 0 -

4 Results comparison: σxy stress

Solver Positioning Units Reference AD 2010 Deviation CM2 Element 1 in x = 0 cm MPa 0 0 - CM2 Element 1 in x = 4 cm MPa -8.98 -8.98 0.00% CM2 Element 1 in x = 8 cm MPa -17.96 -17.96 0.00%


125

2.41 Test No. 01-0041SSLLB_FEM: Beam on two supports considering the shear force 2.41.1 Description sheet

Reference: Internal GRAITEC test;



2.41.2 Overview The 300 cm long beam consists of an I shaped profile of a total height of 20.04 cm, a 0.96 cm thick web and 20.04 cm wide flanges of 1.46 cm thick.

Units I. S.

Geometry l = 300 cm h = 20.04 cm b= 20.04 cm tw = 1.46 cm tf = 0.96 cm Sx= 74.95 cm2 Iz = 5462 cm4 Sy = 16.43 cm2


Longitudinal elastic modulus: E = 2285938 daN/cm2,

Transverse elastic modulus G = 879207 daN/cm2


Boundary conditions

Outer: o Simple support on node 11, o For the modeling, put an hinge at node 1 (instead of a simple support).

Inner: None.


126

Loading

External: Vertical punctual load P = -20246 daN at node 6,

Internal: None.

2.41.3 Vertical displacement of the model in the linear elastic range

Reference solution

The reference displacement is calculated in the middle of the beam, at node 6.

( )cm 017.1105.0912.0

43.163.012

22859384

300202465462228593848

30020246448

33

6 −=−−=

+

−+

−=+=

xx

xxx

xGSPl

EIPlv

shear

y

flexion

z

876876


Planar element: S beam, imposed mesh,

11 nodes,

10 linear elements.

Deformed shape



Solver Model Units Reference AD 2010 DeviationCM2 At node 6 cm -1.017 -1.017 0.00%


127

2.42 Test No. 01-0042SSLSB_FEM: Thin cylinder under a uniform axial load 2.42.1 Description sheet


Analysis type: static elastic;


2.42.2 Overview A cylinder of radius R and length L under a uniform axial load.

Thin cylinder under a uniform axial load Scale = 1/19 01-0042SSLSB_FEM

Units I. S.

Geometry


Length: L = 4 m,

Radius: R = 1 m.




Boundary conditions

Outer: o Null axial displacement at the left end: vz = 0, o For the modeling, only a ¼ of the cylinder is considered.

Inner: None.

Loading

External: Uniform axial load q = 10000 N/m

Inner: None.


128

2.42.3 Stress in all points

Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis.

σxx = qh

σyy = 0


Planar element: shell, imposed mesh,

697 nodes,


2.42.4 Cylinder deformation at the free end

Reference solution

δL longitudinal deformation of the cylinder:

δL = qLEh

δR radial deformation of the cylinder:

δR = -qνREh



697 nodes,


Deformation shape Thin cylinder under a uniform axial load Scale = 1/22 Deformation shape


129



Solver Positioning Units Reference AD 2010 Deviation CM2 At all points Pa 5 x 105 5 x 105 0.00%



3 Results comparison: δL longitudinal deformation

Solver Positioning Units Reference AD 2010 Deviation CM2 At the free end m 9.52 x 10-6 9.52 x 10-6 0.00%

4 Results comparison: δR radial deformation

Solver Positioning Units Reference AD 2010 Deviation CM2 At the free end m -7.14 x 10-7 -7.109 x 10-7 -0.43%


130

2.43 Test No. 01-0043SSLSB_FEM: Thin cylinder under a hydrostatic pressure



Analysis type: static, linear elastic;


2.43.2 Overview A cylinder of radius R and length L under a hydrostatic pressure.

Thin cylinder under a hydrostatic pressure Scale = 1/25 01-0043SSLSB_FEM

Units I. S.

Geometry


Length: L = 4 m,

Radius: R = 1 m.




Boundary conditions

Outer: For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the nodes that are parallel with the cylinder’s axis.

Inner: None.

Loading

External: Radial internal pressure varies linearly with the "p" height, p = p0 zL ,

Internal: None.


131

2.43.3 Stresses

Reference solution

x axis of the local coordinate system of planar elements is parallel to the cylinders axis.

σxx = 0

σyy = p0RzLh



209 nodes,


2.43.4 Cylinder deformation

Reference solution


δL = -p0Rνz2

2ELh

δL radial deformation of the cylinder:

δR = p0R2zELh



209 nodes,


Deformation shape

Thin cylinder under a hydrostatic pressure Deformed


132





Solver Positioning Units Reference AD 2010 Deviation CM2 In z = L/2 Pa 5 x 105 5 x 105 0.00%

3 Results comparison: δL longitudinal deformation of the cylinder

Solver Positioning Units Reference AD 2010 Deviation CM2 Inferior extremity m -2.86 x 10-6 -2.854 x 10-6 -0.21%

4 Results comparison: δL radial deformation of the cylinder

Solver Positioning Units Reference AD 2010 Deviation CM2 In z = L/2 m 2.38 x 10-6 2.38 x 10-6 0.00%


133

2.44 Test No. 01-0044SSLSB_FEM: Thin cylinder under its self weight





2.44.2 Overview A cylinder of R radius and L length subject of self weight only.

Thin cylinder under its self weight Scale = 1/24 01-0044SSLSB_FEM

Units I. S.

Geometry


Length: L = 4 m,

Radius: R = 1 m.




Density: γ = 7.85 x 104 N/m3.

Boundary conditions

Outer: o Null axial displacement at z = 0, o For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the

nodes that are parallel with the cylinder’s axis.

Inner: None.


134

Loading

External: Cylinder self weight,

Internal: None.

2.44.3 Stresses

Reference solution

x axis of the local coordinate system of planar elements is parallel to the cylinders axis.

σxx = γz

σyy = 0



697 nodes,



Reference solution


δL = γz2

2E

δR radial deformation of the cylinder:

δR = -γνRz

E




Note: To obtain this result, you must generate a calculation note “Planar elements stresses by load case in neutral fiber" with results on center.


Solver Model Units Reference AD 2010 Deviation CM2 for z = L Pa 3.14 x 105 3.11 x 105 -0.95%

3 Results comparison: δL longitudinal deformation

Solver Model Units Reference AD 2010 Deviation CM2 for z = L m x 10-6 2.99 2.99 0.00%


Solver Model Units Reference AD 2010 Deviation CM2 for z = L m x 10-6 -0.44 -0.44 0.00%


135

2.45 Test No. 01-0045SSLSB_FEM: Torus with uniform internal pressure





2.45.2 Overview A torus with radius "a", and in transverse section with a radius "b", loaded with an uniform internal pressure.

Torus with uniform internal pressure

01-0045SSLSB_FEM

Units I. S.

Geometry


Transverse section radius: b = 1 m,

Average radius of curvature: a = 2 m.




Boundary conditions

Outer: For the modeling, only 1/8 of the cylinder is considered, so the symmetry conditions are imposed to end nodes.

Inner: None.


136

Loading

External: Uniform internal pressure p = 10000 Pa

Internal: None.

2.45.3 Stresses

Reference solution

(See stresses description on the first scheme of the overview)

If a – b ≤ r ≤ a + b

σ11 = pb2h

r + ar

σ22 = pb2h



361 nodes,



Reference solution

δR radial deformation of the torus:

δR = pb

2Eh (r - ν(r + a))



361 nodes,



1 Results comparison: σ11 stresses

Solver Positioning Units Reference AD 2010 Deviation CM2 for r = a - b Pa 7.5 x 105 7.43 x 105 -0.94% CM2 for r = a + b Pa 4.17 x 105 4.15 x 105 -0.48%

2 Results comparison: σ22 stress

Solver Positioning Units Reference AD 2010 Deviation CM2 for all r Pa 2.50 x 105 2.49 x 105

3 Results comparison: δL radial deformations of the torus

Solver Model Units Reference AD 2010 Deviation CM2 for r = a - b m 1.19 x 10-7 1.18 x 10-7 0.84% CM2 for r = a + b m 1.79 x 10-6 1.79 x 10-6 0.00%


137

2.46 Test No. 01-0046SSLSB_FEM: Spherical shell under internal pressure





2.46.2 Overview A spherical shell of radius R2 is subjected to an internal pressure.

Spherical shell under internal pressure 01-0046SSLSB_FEM


138

Units I. S.

Geometry


Radius: R2 = 1 m,

θ = 90° (hemisphere).




Boundary conditions

Outer: Simple support (null displacement along vertical displacement) on the shell perimeter.

For modeling, we consider only half of the hemisphere, so we impose symmetry conditions (DOF restrains placed in the vertical plane xy in translation along z and in rotation along x and y). In addition, the node at the top of the shell is restrained in translation along x to assure the stability of the structure during calculation).

Inner: None.

Loading

External: Uniform internal pressure p = 10000 Pa

Internal: None.

2.46.3 Stresses

Reference solution (See stresses description on the first scheme of the overview)

If 0° ≤ θ ≤ 90°

σ11 = σ22 = pR2

2

2h



343 nodes,



Reference solution

δR radial deformation of the calotte:

δR = pR2

2 (1 - ν) sin θ 2Eh



343 nodes,



139

Deformed shape

Spherical shell under internal pressure Scale = 1/11 Deformed



Solver Positioning Units Reference AD 2010 Deviation CM2 for all θ Pa 2.50 x 105 2.50 x 105 0.00%


Solver Positioning Units Reference AD 2010 Deviation CM2 for all θ Pa 2.50 x 105 2.50 x 105 0.00%

2 Results comparison: δR radial deformations

Solver Model Units Reference AD 2010 Deviation CM2 for θ = 90° m 8.33 x 10-7 8.34 x 10-7 0.12%


140

2.47 Test No. 01-0047SSLSB_FEM: Spherical shell under its self weight





CAD tolerance 0.5 mm

2.47.2 Overview A spherical shell of radius R2 is subjected to its own weight.

Spherical shell under its self weight 01-0047SSLSB_FEM


141

Units I. S.

Geometry


Radius: R2 = 1 m,

θ = 90° (hemisphere).




Density: γ = 7.85 x 104 N/m3.

Boundary conditions

Outer: Simple support (null displacement along vertical displacement) on the shell perimeter.

For modeling, we consider only a quarter of the hemisphere, so we impose symmetry conditions (DOF restrains placed in the vertical yz plane in translation along x and in rotation along y and z. Nodes placed on the vertical xy plane are restrained in translation along z and in rotation along x and y),

Inner: None.

Loading

External: Self weight, the vertical axis is y-axis,

Internal: None.

2.47.3 Stresses

Reference solution (See stresses description on the first scheme of the overview)

σ11 = γR2

1 + cosθ

σ22 = -γR2 ( 1

1 + cosθ - cosθ)



2071 nodes,


CAD tolerance. = 0.0005 m

2.47.4 Cylinder radial deformation

Reference solution

δR = -γR2

2 sinθ E (

1 + ν1 + cosθ - cosθ)



2071 nodes,



142



Solver Positioning Units Reference AD 2010 Deviation CM2 for all θ = 90° Pa 7.85 x 104 7.83 x 104 -0.25%


Solver Positioning Units Reference AD 2010 Deviation CM2 for all θ = 90° Pa -7.85 x 104 -7.90 x 104 -0.63%


Solver Positioning Units Reference AD 2010 Deviation CM2 for all θ = 90° m 4.86 x 10-7 4.85 x 10-7 -0.20%


143

2.48 Test No. 01-0048SSLSB_FEM: Pinch cylindrical shell





2.48.2 Overview A cylinder of length L is pinched by 2 diametrically opposite forces (F).

Pinch cylindrical shell 01-0048SSLSB_FEM


144

Units

I. S.

Geometry

Length: L = 10.35 m (total length),

Radius: R = 4.953 m,

Thickness: h = 0.094 m.




Boundary conditions

Outer: For the modeling, we consider only half of the cylinder, so we impose symmetry conditions (nodes in the horizontal xz plane are restrained in translation along y and in rotation along x and z),

Inner: None.

Loading

External: 2 punctual loads F = 100 N,

Internal: None.

2.48.3 Vertical displacement at point A

Reference solution

The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution.



777 nodes,




Solver Positioning Units Reference Reference margin AD 2010 Deviation CM2 At point A m -113.9 x 10-3 ± 2% -113.29 x 10-3 -0.53%


145

2.49 Test No. 01-0049SSLSB_FEM: Spherical shell with holes





2.49.2 Overview A spherical shell with holes is subjected to 4 forces, opposite 2 by 2.

Spherical shell with holes

01-0049SSLSB_FEM

Units

I. S.

Geometry

Radius: R = 10 m Thickness: h = 0.04 m, Opening angle of the hole: ϕ0 = 18°.


Longitudinal elastic modulus: E = 6.285 x 107 Pa, Poisson's ratio: ν = 0.3.


146

Boundary conditions

Outer: For modeling, we consider only a quarter of the shell, so we impose symmetry conditions (nodes in the vertical yz plane are restrained in translation along x and in rotation along y and z. Nodes on the vertical xy plane are restrained in translation along z and in rotation along x and y),

Inner: None.

Loading

External: Punctual loads F = 1 N, according to the diagram,

Internal: None.

2.49.3 Horizontal displacement at point A

Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution.



99 nodes,


Deformed shape Spherical shell with holes Scale = 1/79

Deformed


Results comparison: horizontal displacement Solver Positioning Units Reference Reference margin AD 2010 Deviation CM2 At point A(R,0,0) mm 94.0 ± 2% 92.13 -1.99%


147

2.50 Test No. 01-0050SSLSB_FEM: Spherical dome under a uniform external pressure





2.50.2 Overview A spherical dome of radius (a) is subjected to an uniform external pressure.

Spherical dome under a uniform external pressure

01-0050SSLSB_FEM

Units

I. S.

Geometry

Radius: a = 2.54 m,


Angle: θ = 75°.


148




Boundary conditions

Outer: Fixed on the dome perimeter,

Inner: None.

Loading

External: Uniform pressure p = 0.6897 x 106 Pa,

Internal: None.

2.50.3 Horizontal displacement and exterior meridian stress

Reference solution

The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution.



401 nodes,


Deformed shape


149


1 Results comparison: horizontal displacements

Solver Positioning Units Reference Reference margin AD 2010 DeviationCM2 In ψ = 15° m 1.73 x 10-3 ± 2% 1.7306 x 10-3 0.04% CM2 In ψ = 45° m -1.02 x 10-3 ± 2% -1.01 x 10-3

2 Results comparison: σXX external meridian stresses

Solver Model Units Reference Reference margin AD 2010 Deviation CM2 In ψ = 15° MPa -74 ± 2% -72.26 -2.35% CM2 In ψ = 45° MPa -68 ± 2% -68.99

(7.5° between two nodes)


150

2.51 Test No. 01-0051SSLSB_FEM: Simply supported square plate under a uniform load





2.51.2 Overview A square plate simply supported is subjected to an uniform load.

Simply supported square plate under a uniform load Scale = 1/9 01-0051SSLSB_FEM

Units

I. S.

Geometry

Side: a =b = 1 m,





Boundary conditions

Outer: Simple support on the plate perimeter (null displacement along z-axis),

Inner: None


151

Loading

External: Normal pressure of plate p pZ = -1.0 Pa,

Internal: None.

2.51.3 Vertical displacement and bending moment at the center of the plate

Reference solution

Love-Kirchhoff thin plates theory.



361 nodes,




Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center m -4.43 x 10-3 -4.358 x 10-3 -1.61%

2 Results comparison: MX bending moment

Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center Nm 0.0479 0.0471 -1.67%

3 Results comparison: MY bending moment



152

2.52 Test No. 01-0052SSLSB_FEM: Simply supported rectangular plate under a uniform load





2.52.2 Overview A rectangular plate simply supported is subjected to an uniform load.

Simply supported rectangular plate under a uniform load Scale = 1/11 01-0052SSLSB_FEM

Units

I. S.

Geometry

Width: a = 1 m,

Length: b = 2 m,





Boundary conditions


Inner: None.


153

Loading

External: Normal pressure of plate p = pZ = -1.0 Pa,

Internal: None.


Reference solution




435 nodes,




Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center m -1.1060 x 10-2 -1.1023 x 10-2 -0.33%


Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center Nm -0.1017 -0.1017 0.00%


Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center Nm -0.0464 -0.0465 0.22%


154

2.53 Test No. 01-0053SSLSB_FEM: Simply supported rectangular plate under a uniform load





2.53.2 Overview A simply supported rectangular plate is subjected to an uniform load.

Simply supported rectangular plate under a uniform load Scale = 1/25 01-0053SSLSB_FEM

Units

I. S.

Geometry

Width: a = 1 m,

Length: b = 5 m,





Boundary conditions


Inner: None.


155

Loading

External: Normal pressure of plate p = pZ = -1.0 Pa,

Internal: None.


Reference solution




793 nodes,




Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center m 1.416 x 10-2 1.401 x 10-2 -1.06%




Solver Positioning Units Reference AD 2010 Deviation CM2 At plate center Nm 0.0375 0.0376 0.27%


156

2.54 Test No. 01-0054SSLSB_FEM: Simply supported rectangular plate loaded with punctual force and moments

2.54.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLS 26/89;



2.54.2 Overview A rectangular plate simply supported is subjected to a force and to punctual moments.

Simply supported rectangular plate loaded with punctual force and moments 01-0054SSLSB_FEM

Units I. S.

Geometry

Width: DA = CB = 20 m,

Length: AB = DC = 5 m,

Thickness: h = 1 m,


Longitudinal elastic modulus: E =1000 Pa,


Boundary conditions

Outer: Punctual support at A, B and D (null displacement along z-axis),

Inner: None.

Loading

External: o In A: MX = 20 Nm, MY = -10 Nm, o In B: MX = 20 Nm, MY = 10 Nm, o In C: FZ = -2 N, MX = -20 Nm, MY = 10 Nm, o In D: MX = -20 Nm, MY = -10 Nm,

Internal: None.


157

2.54.3 Vertical displacement at C

Reference solution




867 nodes,


Deformed shape

Simply supported rectangular plate loaded with punctual force and moments Deformed



Solver Positioning Units Reference AD 2010 Deviation CM2 At point C m -12.480 -12.667 1.50%


158

2.55 Test No. 01-0055SSLSB_FEM: Shear plate perpendicular to the medium surface



Analysis type: static;


2.55.2 Overview Rectangular plate fixed at one end, loaded with two forces.

Shear plate Scale = 1/50 01-0055SSLSB_FEM

Units

I. S.

Geometry

Length: L = 12 m,

Width: l = 1 m,



Longitudinal elastic modulus: E =1.0 x 107 Pa,


Boundary conditions

Outer: Fixed AD edge,

Inner: None.


159

Loading

External: o At B: Fz = -1.0 N, o At C: FZ = 1.0 N,

Internal: None.


Reference solution




497 nodes,


Deformed shape

Shear plate Scale = 1/35 Deformed



Solver Positioning Units Reference Reference margin AD 2010 Deviation CM2 At point C m 35.37 x 10-3 ± 3% 35.67 x 10-3 0.85%


160

2.56 Test No. 01-0056SSLLB_FEM: Triangulated system with hinged bars

2.56.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLL 12/89;

Analysis type: static (plane problem);


2.56.2 Overview A truss with hinged bars is placed on 3 punctual supports (subjected to imposed displacements) and is loaded with two 2 punctual forces. A thermal load is applied to all the bars.

Units

I. S.

Geometry

θ = 30°,

Section A1 = 1.41 x 10-3 m2,

Section A2 = 2.82 x 10-3 m2.


Longitudinal elastic modulus: E =2.1 x 1011 Pa,

Coefficient of linear expansion: α = 10-5 °C-1.

Boundary conditions

Outer: o Hinge at A (uA = vA = 0), o Roller supports at B and C ( uB = v’C = 0),

Inner: None.

Loading

External: o Support displacement: vA = -0.02 m ; vB = -0.03 m ; v’C = -0.015 m , o Punctual loads: FE = -150 KN ; FF = -100 KN, o Expansion effect on all bars for a temperature variation of 150° in relation with the assembly temperature

(specified geometry),

Internal: None.


161

2.56.3 Tension force in BD bar

Reference solution Determining the hyperstatic unknown with the section cut method.


Linear element: S beam, automatic mesh,

11 nodes,

17 S beams + 1 rigid S beam for the modeling of the simple support at C.

2.56.4 Vertical displacement at D

Reference solution vD displacement was determined by several software with implemented finite elements method.



11 nodes,

17 S beams + 1 rigid S beam for the modeling of simple support at C.

Deformed shape Triangulated system with hinged bars 01-0056SSLLB_FEM


1 Results comparison: FX traction force

Solver Positioning Units Reference AD 2010 Deviation CM2 BD bar N 43633 43688 0.13%


Solver Positioning Units Reference AD 2010 Deviation CM2 At point D m -0.01618 -0.01616 -0.12%


162

2.57 Test No. 01-0057SSLSB_FEM: 0.01m thick plate fixed on its perimeter, loaded with a uniform pressure


Reference: Structure Calculation Software Validation Guide, test SSLV 09/89;



2.57.2 Overview Square plate of side "a". For the modeling, only a quarter of the plate is considered.

Units

I. S.

Geometry

Side: a = 1 m,


Slenderness: λ = ah = 100.


Reinforcement,



Boundary conditions

Outer: Fixed sides: AB and BD,

For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),

Inner: None.

Loading

External: 1 MPa uniform pressure,

Internal: None.


163


Reference solution

This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.



289 nodes,




Solver Positioning Units Reference AD 2010 Deviation CM2 At point C m -6.639 x 10-2 -6.565 x 10-2 -1.11%


164

2.58 Test No. 01-0058SSLSB_FEM: 0.01333 m thick plate fixed on its perimeter, loaded with a uniform pressure





2.58.2 Overview Square plate of side "a", for the modeling, only a quarter of the plate is considered.

Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,



Boundary conditions

Outer: Fixed sides: AB and BD,


Inner: None.

Loading


Internal: None.


165


Reference solution




289 nodes,






166







Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,



Boundary conditions

Outer: Fixed edges: AB and BD,


Inner: None.

Loading


Internal: None.


167


Reference solution




289 nodes,






168


2.60.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLV 09/89;




Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,



Boundary conditions



Inner: None.

Loading


Internal: None.


169


Reference solution




289 nodes,






170







Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,



Boundary conditions



Inner: None.

Loading


Internal: None.

Test 01-0061SSLSB_MEF

171


Reference solution




289 nodes,






172

2.62 Test No. 01-0062SSLSB_FEM: 0.01 m thick plate fixed on its perimeter, loaded with a punctual force





2.62.2 Overview Square plate of side "a".

0.01 m thick plate fixed on its perimeter Scale = 1/5 01-0062SSLSB_FEM

Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,



Boundary conditions

Outer: Fixed edges,

Inner: None.


173

Loading

External: Punctual force applied on the center of the plate: FZ = -106 N,

Internal: None.

2.62.3 Vertical displacement at point C (center of the plate)

Reference solution




961 nodes,




Solver Positioning Units Reference AD 2010 Deviation CM2 At point C m -0.29579 -0.292146 -1.23%


174


2.63.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLV 09/89;



2.63.2 Overview Square plate of side a.


Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,



Boundary conditions

Outer: Fixed sides,

Inner: None.


175

Loading

External: Punctual force applied on the center of the plate: FZ = -106 N,

Internal: None.

2.63.3 Vertical displacement at point C (the center of the plate)

Reference solution




961 nodes,






176






2.64.2 Overview Square plate of side "a".


Units

I. S.

Geometry

Side: a = 1 m,


Slenderness: λ = 50.


Reinforcement,



Boundary conditions

Outer: Fixed edges,

Inner: None.


177

Loading

External: punctual force applied in the center of the plate: FZ = -106 N,

Internal: None.

2.64.3 Vertical displacement at point C (the center of the plate)

Reference solution




961 nodes,






178








Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,




179

Boundary conditions

Outer: Fixed sides,

Inner: None.

Loading

External: Punctual force applied at the center of the plate: FZ = -106 N,

Internal: None.

2.65.3 Vertical displacement at point C center of the plate)

Reference solution




961 nodes,






180








Units

I. S.

Geometry

Side: a = 1 m,




Reinforcement,




181

Boundary conditions

Outer: Fixed edges,

Inner: None.

Loading

External: punctual force applied in the center of the plate: FZ = -106 N,

Internal: None.

2.66.3 Vertical displacement at point C (center of the plate)

Reference solution




961 nodes,




Solver Positioning Units Reference AD 2010 Deviation CM2 At point C m -0.42995 x 10-3 -0.41209 x 10-3 -4.15

Test 01-0067SDLLB_MEF

182

2.67 Test No. 01-0067SDLLB_FEM: Vibration mode of a thin piping elbow in space (case 1) 2.67.1 Description sheet


Analysis type: modal analysis (space problem);


2.67.2 Overview A thin piping elbow with a radius of 1m is fixed on its ends and loaded with its self weight only.

Vibration mode of a thin piping elbow Scale = 1/7 01-0067SDLLB_FEM

Units

I. S.

Geometry





Section: A = 1.131 x 10-4 m2,




Points coordinates (in m): o O ( 0 ; 0 ; 0 ) o A ( 0 ; R ; 0 ) o B ( R ; 0 ; 0 )


183





Boundary conditions

Outer: Fixed at points A and B,

Inner: None.

Loading

External: None,

Internal: None.


Reference solution


transverse bending:

fj = μi

2

2π R2 GIpρA where i = 1,2.



11 nodes,

10 linear elements.

Eigen mode shapes



Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Transverse 1 Hz 44.23 44.12 -0.25% CM2 Transverse 2 Hz 125 120.09 -3.93%


184



Analysis type: modal analysis (in space);


2.68.2 Overview A thin piping elbow with a radius of 1m is extended with two straight elements of length "L" and is loaded with its self weight only.


Units I. S.

Geometry


L = 0.6 m,




Section: A = 1.131 x 10-4 m2,






185





Boundary conditions

Outer: o Fixed at points C and D o In A: translation restraint along y and z, o In B: translation restraint along x and z,

Inner: None.

Loading

External: None,

Internal: None.

2.68.3 Eigen modes frequencies Reference solution


transverse bending:

fj = μi

2

2π R2 GIpρA where i = 1,2.



23 nodes,

22 linear elements.

Eigen mode shapes



Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Transverse 1 Hz 33.4 33.19 -0.63% CM2 Transverse 2 Hz 100 94.62 -5.38%


186



Analysis type: modal analysis (space problem);


2.69.2 Overview A thin piping elbow with a radius of 1m is extended by two straight elements of length "L" and is loaded with its self weight only.


Units

I. S.

Geometry


L = 2 m,




Section: A = 1.131 x 10-4 m2,






187





Boundary conditions

Outer: o Fixed at points C and D o At A: translation restraint along y and z, o At B: translation restraint along x and z,

Inner: None.

Loading

External: None,

Internal: None.

2.69.3 Eigen modes frequencies Reference solution


transverse bending:

fj = μi

2

2π R2 GIpρA where i = 1,2 with i = 1,2:



41 nodes,

40 linear elements.

Eigen mode shapes

2.69.4 Results sheet Results comparison: eigen modes frequencies

Solver Eigen mode type Units Reference AD 2010 Deviation CM2 Transverse 1 Hz 17.900 17.65 -1.40% CM2 Transverse 2 Hz 24.800 24.43 -1.49%

Test 01-0077SSLPB_MEF

188

2.70 Test No. 01-0077SSLPB_FEM: Reactions on supports and bending moments on a 2D portal frame (Rafters)


Reference: Design and calculation of metal structures.



2.70.2 Overview Moments and actions on supports calculation on a 2D portal frame.

The purpose of this test is to verify the results of Advance Design for the M. R. study of a 2D portal frame.

Portal frame geometry

Suppose there is a symmetric portal frame with a range of 20 meters, with columns (of the same inertia as the rafters), that are hinged on 7.5 meters high base plates, and the ridge is at 10 meters altitude. The column and rafters sections are identical.

Portal frame solicitation

The portal frame is successively subjected to:

A linear load of q=100 daN/ml on the rafters, perpendicular to them.


189

2.70.3 Moments and actions on supports M.R. calculation on a 2D portal frame.

RDM results, for the linear load perpendicular on the rafters, are:

2qLVV EA == ( ) ( ) H

fh3f3k²hf5h8

32²qLHH EA =

++++

==

HhMM DB −== ( )fhH8

²qLMC +−=


Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords

Solver Forces Units Reference AD 2010 Deviation CM2 Vertical reaction VA = VE DaN -1000 -1000 0.00%

Horizontal reaction HA = HE DaN -332.9 -332.67 -0.07% MB = MD DaN.m -2496.8 -2494.99 -0.07% MC DaN.m -1671 -1673.35 0.14%


190

2.71 Test No. 01-0078SSLPB_FEM: Reactions on supports and bending moments on a 2D portal frame (Columns)


Reference: Design and calculation of metal structures.



2.71.2 Overview Moments and actions on supports calculation on a 2D portal frame.

The purpose of this test is to verify the results of Advance Design for the M. R. study of a 2D portal frame.

Portal frame geometry

Suppose there is a symmetric portal frame with a range of 20 meters, with columns (of the same inertia as the rafters), that are hinged on 7.5 meters high base plates, and the ridge is at 10 meters altitude. The profiles are identical for the column and the rafters.

Portal frame solicitation

The portal frame is successively subjected to:

A linear load of q=100 daN/ml on the column, perpendicular on it.


191

2.71.3 Moments and reactions on supports M.R. calculation on a 2D portal frame. RDM results, for the linear load perpendicular on the column, are:

L2²qhVV EA −=−=

( )( ) ( )fh3f3k²h

fh26kh516

²qhHE +++++

= qhHH EA −=

hHqhM EB −=2

² ( )fhH

4²qhM EC +−= hHM ED −=


Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the column

Solver Forces Units Reference AD 2010 Deviation CM2 Vertical reaction VA = -VE DaN -140.6 -140.63 0.02%

Horizontal reaction HA DaN 579.1 579.17 0.01% Horizontal reaction HE DaN 170.9 170.83 -0.04% MB DaN.m 1530.8 1531.26 0.03% MC DaN.m -302.7 -302.06 -0.21% MD DaN.m -1281.7 -1281.23 -0.04%

Test 01-0084SSLLB_MEF

192

2.72 Test No. 01-0084SSLLB_FEM: Short beam on two hinged supports

2.72.1 Description sheet Reference: Structure Calculation Software Validation Guide, test SSLL 02/89



2.72.2 Overview Short beam on two hinged supports

Units

I. S.

Geometry

Length: L = 1.44 m,

Area: A = 31 x 10-4 m²

Inertia: I = 2810 x 10-8 m4

Shearing coefficient: az = 2.42 = A/Ar


E = 2 x 1011 Pa

ν = 0.3

Boundary conditions

Hinge at end x = 0,

Hinge at end x = 1.44 m.

Loading

Uniformly distributed force of p = -1. X 105 N/m on beam AB.

2.72.3 Reference results Calculation method used to obtain the reference solution

The deflection on the middle of a non-slender beam considering the shear force deformations given by the Timoshenko function:

GA8pl

EIpl

3845v

r

24

+=

where ( )υ+=

12EG and

zr a

AA =

where "Ar" is the reduced area and "az" the shear coefficient calculated on the transverse section.

Uncertainty about the reference: analytical solution:


193

Reference values

Point Magnitudes and units Value C V, deflection (m) -1.25926 x 10-3


Results comparison

Solver Point Magnitudes and units Reference value AD 2010 Deviation CM2 C V, deflection (m) -1.25926 -1.25926 0.00%


194

2.73 Test No. 01-0085SDLLB_FEM: Slender beam of variable rectangular section with fixed-free ends (β=5)





2.73.2 Overview Slender beam with variable rectangular section (fixed-free)

Units

I. S.

Geometry

Length: L = 1 m,

Straight initial section: o h0 = 0.04 m o b0 = 0.05 m o A0 = 2 x 10-3 m²

Straight final section o h1 = 0.01 m o b1 = 0.01 m o A1 = 10-4 m²


E = 2 x 1011 Pa

ρ = 7800 kg/m3

Boundary conditions

Outer: o Fixed at end x = 0, o Free at end x = 1

Inner: None.

Loading

External: None,

Internal: None.


195

2.73.3 Reference results Calculation method used to obtain the reference solution

Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):

²t²A

²x²EIz

x2

2

δνδ

ρ−=⎟⎠

⎞⎜⎝

⎛δ

νδδδ

where Iz and A vary with the abscissa.

The result is:

( ) ρβαλπ

= 12E

²l1h,i

21fi with

⎪⎪⎩

⎪⎪⎨

⎧

=ο

=β

=ο

=α

51b

b

41h

h

λ1 λ2 λ3 λ4 λ5

β = 5 24.308 75.56 167.21 301.9 480.4


Reference values

Eigen mode type Frequency (Hz) 1 56.55 2 175.79 3 389.01 4 702.36

Flexion

5 1117.63

MODE 1 Scale = 1/4


196

MODE 2 Scale = 1/4

MODE 3 Scale = 1/4


197

MODE 4 Scale = 1/4

MODE 5 Scale = 1/4


Results comparison

Eigen mode type Frequency Theoretical (Hz)

AD 2010 (Hz)

Deviation

1 56.55 58.49 3.43% 2 175.79 177.67 1.07% 3 389.01 388.85 -0.04% 4 702.36 697.38 -0.71%

Bending

5 1117.63 1106.31 -1.01%


198

2.74 Test No. 01-0086SDLLB_FEM: Slender beam of variable rectangular section (fixed-fixed) 2.74.1 Description sheet




2.74.2 Overview Slender beam with variable rectangular section (fixed-fixed)

Units I. S.

Geometry

Length: L = 0.6 m,

Constant thickness: h = 0.01 m

Initial section: o b0 = 0.03 m o A0 = 3 x 10-4 m²

Section variation: o with (α = 1) o b = b0e-2αx o A = A0e-2αx


E = 2 x 1011 Pa

ν = 0.3

ρ = 7800 kg/m3

Boundary conditions

Outer: o Fixed at end x = 0, o Fixed at end x = 0.6 m.

Inner: None.

Loading

External: None,

Internal: None.

2.74.3 Reference results

Calculation method used to obtain the reference solution

ωi pulsation is given by the roots of the equation:

( ) ( ) ( ) ( ) 0rlsinslshrs2

²r²sslchrlcos1 =−

+−


199

with

( ) 0²²s;²r;EIA 2

isi2

i2i

zo

2i04

i >α−λ⎯→⎯α−λ=λ+α=ωρ

=λ

Therefore, the ν translation components of φi(x) mode, are:

( ) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡−

−−

+−=Φ α ))sx(rsh)rxsin(s()rlsin(s)sl(rsh

)sl(ch)rlcos(sxchrxcosex xi


Reference values

Eigen mode φi(x)* Eigen mode order

Frequency (Hz) x = 0 0.1 0.2 0.3 0.4 0.5 0.6

1 143.303 0 0.237 0.703 1 0.859 0.354 0 2 396.821 0 -0.504 -0.818 0 0.943 0.752 0 3 779.425 0 0.670 0.210 -0.831 0.257 1 0 4 1289.577 0 -0.670 0.486 0 -0.594 1 0

* φi(x) eigen modes* standardized to 1 at the point of maximum amplitude.

Eigen modes



Solver Eigen mode type: Flexion Units Reference AD 2010 Deviation CM2 1 Hz 143.303 145.88 1.80% CM2 2 Hz 396.821 400.26 0.87% CM2 3 Hz 779.425 783.15 0.48% CM2 4 Hz 1289.577 1293.42 0.30%


200

2.75 Test No. 01-0089SSLLB_FEM: Plane portal frame with hinged supports





2.75.2 Overview Calculation of support reactions of a 2D portal frame.

Units

I. S.

Geometry

Length: L = 20 m,

I1 = 5.0 x 10-4 m4

a = 4 m

h = 8 m

b = 10.77 m

I2 = 2.5 x 10-4 m4


Isotropic linear elastic material.

E = 2.1 x 1011 Pa

Boundary conditions

Hinged base plates A and B (uA = vA = 0 ; uB = vB = 0).


201

Loading

p = -3 000 N/m

F1 = -20 000 N

F2 = -10 000 N

M = -100 000 Nm

2.75.3 Calculation method used to obtain the reference solution

K = (I2/b)(h/I1)

p = a/h

m = 1 + p

B = 2(K + 1) + m

C = 1 + 2m

N = B + mC

VA = 3pl/8 + F1/2 – M/l + F2h/l

HA = pl²(3 + 5m)/(32Nh) + (F1l/(4h))(C/N) + F2(1-(B + C)/(2N)) + (3M/h)((1 + m)/(2N))

2.75.4 Reference values Point Magnitudes and units Value

A V, vertical reaction (N) 31 500.0 A H, horizontal reaction (N) 20 239.4 C vc (m) -0.03072


Results comparison: force

Solver Point Magnitude Units Reference AD 2010 Deviation CM2 A Vertical reaction V N -31500 -31500 0.00%

A Horizontal reaction H N -20239.4 -20239.4 0.00%

Results comparison: displacement

Solver Point Magnitude Units Reference AD 2010 Deviation CM2 C vc displacement m -0.03072 -0.03072 0.00%

Test 01-0090HFLSB_MEF

202

2.76 Test No. 01-0090HFLSB_FEM: Simply supported beam in Eulerian buckling with a thermal load


Reference: Euler theory;

Analysis type: Eulerian buckling;


2.76.2 Overview The 300 cm long beam is a I shaped profile of 20.04 cm height, 0.96 cm thick web and 20.04 cm wide flanges of 1.46 cm thick.

Units

I. S.

Geometry

L = 300 cm h = 20.04 cm b= 20.04 cm tw = 0.96 cm tf = 1.46 cm

Cross Section Sx cm² Sy cm² Sz cm² Ix cm4 Iy cm4 Iz cm4

PRS 74.952 (20.04-0.46) x 0.96=18.58

2x20.04x1.46=58.52

46.54 1959.63 5462.06


203


Longitudinal elastic modulus: E = 2285938 daN/cm2,


Transverse elastic modulus: G = 879207 daN/cm2,

Coefficient of thermal expansion: α = 0.00001. Boundary conditions

Outer: Webs simply supported at Z = 0 and Z = 300 cm,

Inner: None. Loading

External: None,

Internal: Thermal load ΔT = 100°C on the center corresponding to a normal compression force of: ( ) daN 4077410000001.096.046.104.202285938TESN =×××−×=Δα=


Reference solution


daN 491243300

632.19592285938L

EIN 2

2

2

2

k =××π

=π

=

To take into account the effect of the shear force, we correct this load by:

93.1140774

486597daN 486597N9905.0

GSN

1

NN k

z

k

kk ==λ⇒=×=

+=′


Planar element: Thick shell, imposed mesh,

2025 nodes,

1920 elements.

Thermal efforts


204

Deformed shape of mode 1


Results comparison

Solver CASE Magnitude Units Reference AD 2010 Deviation CM2 101 Total normal effort* daN -40774 -41051.95 0.68%

102 Critical coefficient - 11.93 11.873 -0.48% * To obtain this force, select the support from one of the column extremities and display the actions on supports on this selection. This value is displayed in the document along with the values on the selected support. The value is represented by the Fz sum of loads on linear supports per load case - global coordinate system (if the z-axis of the global coordinate system corresponds to the x-axis in the local coordinate system of the bar).

Test 01-0091HFLLB_MEF

205

2.77 Test No. 01-0091HFLLB_FEM: Double fixed beam in Eulerian buckling with a thermal load


Reference: Euler theory; Analysis type: Eulerian buckling; Element type: linear.

2.77.2 Overview

Units

I. S.

Geometry

L = 10 m

Cross Section Sx m² Sy m² Sz m² Ix m4 Iy m4 Iz m4 Vx m3 V1y m3 V1z m3 V2y m3 V2z m3

IPE200 0.002850 0.001400 0.001799 0.0000000646 0.0000014200 0.0000194300 0.00000000 0.00002850 0.00019400 0.00002850 0.00019400


Longitudinal elastic modulus: E = 2.1 x 1011 N/m2, Poisson's ratio: ν = 0.3. Coefficient of thermal expansion: α = 0.00001

Boundary conditions


Inner: None.


206

Loading

External: Punctual load FZ = 1 N at = L/2 (load that initializes the deformed shape),

Internal: ΔT = 5°C corresponding to a compression force of:

kN 925.29500001.000285.011E1.2TESN =×××=Δα=


Reference solution


93.3724.117925.29k 724.117

2

P 2

2

critical ==⇒=

⎟⎠⎞

⎜⎝⎛

= λπ NLEI

Observation: in this case, the thermal load has no effect over the critical coefficient



1 1 nodes,

10 elements.

Deformed shape of mode 1


Results comparison: normal force

Solver CASE Node Magnitude Units Reference AD 2010 Deviation CM2 101 6 Normal Force kN -29.925 -29.90 -0.08% CM2 102 6 Normal Force kN -117.724 -118.08 0.30%


207

2.78 Test No. 01-0092HFLLB_FEM: Cantilever beam in Eulerian buckling with thermal load


Reference: Euler theory; Analysis type: Eulerian buckling; Element type: linear.

2.78.2 Overview

Units

I. S.

Geometry

L = 10 m S=0.01 m2 I = 0.0002 m4


Longitudinal elastic modulus: E = 2.0 x 1010 N/m2, Poisson's ratio: ν = 0.1. Coefficient of thermal expansion: α = 0.00001

Boundary conditions

Outer: Fixed at end x = 0, Inner: None.

Loading

External: Punctual load P = -100000 N at x = L, Internal: T = -50°C (Contraction equivalent to the compression force)

( 5000001.0T0005.001.010.2

100000ESN

100 −×=Δα==×

−==ε )


Reference solution


98696.010000098696 98696

4P 2

2

critical ==⇒== λπ NLEI

Observation: in this case, the thermal load has no effect over the critical coefficient


208



5 nodes,

4 elements.

Deformed shape


Results comparison: normal force

Solver CASE Node Magnitude Units Reference AD 2010 Deviation CM2 101 5 Vertical displacement v5 cm -1.0 -1.0 0.00%

101 A Normal Force N -100000 -100000 0.00% 102 A Normal Force N -98696 -98696 0.00%


209

2.79 Test No. 01-0094SSLLB_FEM: 3D bar structure with elastic support 2.79.1 Description sheet

Reference: Internal GRAITEC; Analysis type: static linear; Element type: linear.

2.79.2 Overview Suppose we have the following system of bars:

Units

I. S.

Geometry

For all bars:

H = 3 m B = 3 m S = 0.02 m2

Element Node i Node j 1 (bar) 1 5 2 (bar) 2 5 3 (bar) 3 5 4 (bar) 4 5

5 (spring) 5 6


Isotropic linear elastic materials Longitudinal elastic modulus: E = 2.1 E8 N/m2,

Boundary conditions

Outer: At node 5: K = 50000 kN/m ; Inner: None.


210

Loading

External: Vertical load at node: P = -100 kN, Internal: None.

2.79.3 Results

System solution

2

22 BHL += . Also, U1 = V1 = U5 = U6 = V6 = 0

Stiffness matrix of bar 1

( ) ( )

1L2x= where

.121.1

21)(.).1()(

−

++−=⇔+−=

ξ

ξξξ jiji uuuuLxu

Lxxu

in the local coordinate system:

[ ] [ ] [ ][ ] [ ] [ ]

)()()()(

0000010100000101

)()(

1111

41

41

41

41

2=

221

21

2121

1

1

1

101

j

j

i

i

j

i

L Te

T

v

vuvu

LES

uu

LESd

LES

dL

ESdxBBESdVBHBke

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

⎭⎬⎫

⎩⎨⎧−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−===

∫

∫∫∫

−

−

ξ

ξ

where [ ]

)v()u()v()u(

0000010100000101

LESk

5

5

1

1

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=


211

The elementary matrix [ ]ek expressed in the global coordinate system XY is the following: (θ angle allowing the transition from the global base to the local base):

[ ] [ ] [ ][ ] [ ]

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

θθθθ−θθ−θθθθθ−θ−

θ−θθ−θθθθθ−θ−θθθ

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

θθ−θθ

θθ−θθ

==−

22

22

22

22

e

eeeT

ee

sinsincossinsincossincoscossincoscos

sinsincossinsincossincoscossincoscos

LESK

cossin00sincos00

00cossin00sincos

Ravec RkRK

Knowing that LHsin and

2cos == θθ

LB

then:

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=θθ

=θ

=θ

2

2

22

2

22

L2HBcossin

LHsin

L2Bcos

[ ]

)()()()(

22

2222

22

2222

:)DH(arctan =5,1 nodes 1element for

5

5

1

1

22

22

22

22

31

VUVU

HHBHHB

HBBHBB

HHBHHB

HBBHBB

LESK

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−−

=→ θ

Stiffness matrix of spring support 5

[ ]

)()()()(

0000010100000101

)()(

1111

:system coordinate local in the

4KKsay We

5

j

j

i

i

j

i

vuvu

Kuu

Kk

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

′=⎥⎦

⎤⎢⎣

⎡−

−′=

=′

[ ]

)()()()(

101000001010

0000

':90=6,5 nodes 5element for

6

6

5

5

5

VUVU

KK

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=°→ θ


212

System [ ]{ } { }FQK =

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

′′−

′−′+−−

−−

−−

−−

6Y

6X

5X

1Y

1X

6

6

5

5

1

1

233

233

3

2

33

2

3

233

233

3

2

33

2

3

RR

PRRR

VUVUVU

K0K000000000

K0KHLES

2HB

LESH

LES

2HB

LES

002

HBLES

2B

LES

2HB

LES

2B

LES

00HLES

2HB

LESH

LES

2HB

LES

002

HBLES

2B

LES

2HB

LES

2B

LES

If U1 = V1 = U5 = U6 = V6 = 0, then:

m 001885.0

4KH

LES

4P

KHLES

4P

V2

32

3

5 −=+

−=

′+

−=

And N 23563V

4KRN 1436VH

LESR

0RN 1015V2

HBLESRN 1015V

2HB

LESR

56Y52

31Y

6X535X531X

=−==−=

=−===−=

Note:

The values on supports specified by AD 2010 correspond to the actions,

RY6 calculated value must be multiplied by 4 in relation to the double symmetry,

x1 value is similar to the one found by AD 2010 by dividing this by 2

Effort in bar 1:

⎭⎬⎫

⎩⎨⎧

−=

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

17591759

1111

and

200

200

002

002

5

1

5

1

5

5

1

1

5

5

1

1

NN

uu

LES

VUVU

LB

LH

LH

LB

LB

LH

LH

LB

vuvu

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

5

1

5

1

5

5

1

1

5

5

1

1

1111

and

cossin00sincos00

00cossin00sincos

NN

uu

LES

VUVU

vuvu

θθθθ

θθθθ


213

Reference values

Point Magnitude Units Value 5 V2 m -1.885 10-3

All bars Normal force N -1759 Supports 1, 3, 4 and 5 Fz action N -1436 Supports 1, 3, 4 and 5 Action Fx=Fy N 7182/1015 ±=

Support 6 Fz action N 23563 x 4=94253


Linear element: beam, automatic mesh,

5 nodes,

4 linear elements.


214

Deformed shape

Normal forces diagram


Results comparison

Solver Node Magnitude Units Reference AD 2010 Deviation 5 V2 m -1.885 10-3 -1.885 10-3 0.00%

All bars Normal force N -1759 -1759 0.00% Supports 1, 3, 4 and 5 Fz reaction N -1436 -1436,55 -3.83% Supports 1, 3, 4 and 5 Action Fx=Fy N 7182/1015 ±= 718.27 0.04%

CM2

Support 6 Fz reaction N 23563 x 4=94253 94253.81 0.001%


215

2.80 Test No. 01-0095SDLLB_FEM: Fixed/free slender beam with centered mass





Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending, transverse bending, punctual mass.

2.80.2 Test data

Units

I. S.

Geometry

Outer diameter de= 0.35 m,

Inner diameter: di= 0.32 m,

Beam length: l = 10 m,

Area: A= 1.57865 x 10-2 m2

Polar inertia: IP= 4.43798 x 10-4m4

Inertia: Iy= Iz = 2.21899 x 10-4m4

Punctual mass: mc= 1000 kg

Beam self-weight: M


216


Longitudinal elastic modulus: E = 2.1 x 1011 Pa, Density: ρ = 7800 kg/m3

Poisson's ratio: ν=0.3 (this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more appropriate to obtain the correct frequency value of mode No. 8 with NE/NASTRAN)

Boundary conditions

Outer: Fixed at point A, x = 0, Inner: none

Loading

None for the modal analysis


Reference frequency

For the first mode, the Rayleigh method gives the approximation formula

)M24.0m(IEI3

x2/1fc

3z

1 +Π=

Mode Shape Units Reference 1 Flexion Hz 1.65 2 Flexion Hz 1.65 3 Flexion Hz 16.07 4 Flexion Hz 16.07 5 Flexion Hz 50.02 6 Flexion Hz 50.02 7 Traction Hz 76.47 8 Torsion Hz 80.47 9 Flexion Hz 103.2 10 Flexion Hz 103.2

Comment: The mass matrix associated with the beam torsion on two nodes, is expressed as:

⎥⎦

⎤⎢⎣

⎡×

××ρ12/12/11

3Il P

And to the extent that Advance Design 2010 uses a condensed mass matrix, the value of the torsion mass inertia

introduced in the model is set to: 3

Il p××ρ

Uncertainty about the reference frequencies

Analytical solution mode 1 Other modes: ± 1%


Linear element AB: Beam Beam meshing: 20 elements.


217

Modal deformations


218

Observation: the deformed shape of mode No. 8 that does not really correspond to a torsion deformation, is actually the display result of the translations and not of the rotations. This is confirmed by the rotation values of the corresponding mode.

Eigen modes vector 8

Node DX DY DZ RX RY RZ 1 -3.336e-033 6.479e-031 -6.316e-031 1.055e-022 5.770e-028 5.980e-028 2 -5.030e-013 1.575e-008 -1.520e-008 1.472e-002 6.022e-008 6.243e-008 3 -1.005e-012 6.185e-008 -5.966e-008 2.944e-002 1.171e-007 1.214e-007 4 -1.505e-012 1.365e-007 -1.317e-007 4.416e-002 1.705e-007 1.769e-007 5 -2.002e-012 2.381e-007 -2.296e-007 5.887e-002 2.206e-007 2.289e-007 6 -2.495e-012 3.648e-007 -3.517e-007 7.359e-002 2.673e-007 2.774e-007 7 -2.983e-012 5.149e-007 -4.963e-007 8.831e-002 3.106e-007 3.225e-007 8 -3.464e-012 6.867e-007 -6.618e-007 1.030e-001 3.506e-007 3.641e-007 9 -3.939e-012 8.785e-007 -8.464e-007 1.177e-001 3.873e-007 4.023e-007 10 -4.406e-012 1.088e-006 -1.049e-006 1.325e-001 4.207e-007 4.371e-007 11 -4.863e-012 1.315e-006 -1.267e-006 1.472e-001 4.508e-007 4.684e-007 12 -5.310e-012 1.556e-006 -1.499e-006 1.619e-001 4.777e-007 4.964e-007 13 -5.746e-012 1.811e-006 -1.744e-006 1.766e-001 5.015e-007 5.210e-007 14 -6.169e-012 2.077e-006 -2.000e-006 1.913e-001 5.221e-007 5.423e-007 15 -6.580e-012 2.353e-006 -2.265e-006 2.061e-001 5.396e-007 5.605e-007 16 -6.976e-012 2.637e-006 -2.539e-006 2.208e-001 5.541e-007 5.755e-007 17 -7.357e-012 2.928e-006 -2.819e-006 2.355e-001 5.658e-007 5.874e-007 18 -7.723e-012 3.224e-006 -3.104e-006 2.502e-001 5.746e-007 5.965e-007 19 -8.072e-012 3.524e-006 -3.393e-006 2.649e-001 5.808e-007 6.028e-007 20 -8.403e-012 3.826e-006 -3.685e-006 2.797e-001 5.844e-007 6.065e-007 21 -8.717e-012 4.130e-006 -3.977e-006 2.944e-001 5.856e-007 6.077e-007

With NE/NASTRAN, the results associated with mode No. 8, are:


219


Results comparison

Solver Mode Shape Units Reference AD 2010 Deviation (%)

1 Flexion Hz 1.65 1.65 0.00% 2 Flexion Hz 1.65 1.65 0.00% 3 Flexion Hz 16.07 16.06 -0.06% 4 Flexion Hz 16.07 16.06 -0.06%

CM2 5 Flexion Hz 50.02 50.00 -0.04% 6 Flexion Hz 50.02 50.00 -0.04% 7 Traction Hz 76.47 76.46 -0.01% 8 Torsion Hz 80.47 9 Flexion Hz 103.20 103.14 -0.06% 10 Flexion Hz 103.20 103.14 -0.06%

Comment: The difference between the reference frequency of torsion mode (mode No. 8) and the one found by AD 2010 may be explained by the fact that AD 2010 is using a lumped mass matrix (see the corresponding description sheet).


220

2.81 Test No. 01-0096SDLLB_FEM: Fixed/free slender beam with eccentric mass or inertia


Reference: Structure Calculation Software Validation Guide, test SDLL 15/89; Analysis type: modal analysis; Element type: linear. Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending,

transverse bending, punctual mass..

2.81.2 Problem data

Units I. S.

Geometry

Outer diameter: de= 0.35 m, Inner diameter: di = 0.32 m, Beam length: l = 10 m, Distance BC: lBC = 1 m Area: A =1.57865 x 10-2 m2

Inertia: Iy = Iz = 2.21899 x 10-4m4 Polar inertia: Ip = 4.43798 x 10-4m4 Punctual mass: mc = 1000 kg


Longitudinal elasticity modulus of AB element: E = 2.1 x 1011 Pa, Density of the linear element AB: ρ = 7800 kg/m3 Poisson's ratio ν=0.3(this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more

appropriate to obtain the correct frequency value of modes No. 4 and 5 with NE/NASTRAN: Elastic modulus of BC element: E = 1021 Pa Density of the linear element BC: ρ = 0 kg/m3


221

Boundary conditions

Fixed at point A, x = 0,

Loading None for the modal analysis

2.81.3 Reference frequencies

Reference solutions

The different eigen frequencies are determined using a finite elements model of Euler beam (slender beam).

fz + t0 = flexion x,z + torsion

fy + tr = flexion x,y + traction

Mode Units Reference 1 (fz + t0) Hz 1.636 2 (fy + tr) Hz 1.642 3 (fy + tr) Hz 13.460 4 (fz + t0) Hz 13.590 5 (fz + t0) Hz 28.900 6 (fy + tr) Hz 31.960 7 (fz + t0) Hz 61.610 1 (fz + t0) Hz 63.930

Uncertainty about the reference solutions The uncertainty about the reference solutions: ± 1%


Linear element AB: Beam

Imposed mesh: 50 elements.

Linear element BC: Beam

Without meshing

Modal deformations


222


Results comparison

Solver Mode Units Reference AD 2010 Deviation (%)

1 (fz + t0) Hz 1.636 1.636 0.00% 2 (fy + tr) Hz 1.642 1.649 0.43% 3 (fy + tr) Hz 13.46 13.455 -0.04% 4 (fz + t0) Hz 13.59 13.647 0.42%

CM2 5 (fz + t0) Hz 28.90 29.72 2.84% 6 (fy + tr) Hz 31.96 31.96 0.00% 7 (fz + t0) Hz 61.61 63.091 2.40% 8 (fy + tr) Hz 63.93 63.93 0.00%

Note:

fz + t0 = flexion x,z + torsion

fy + tr = flexion x,y + traction

Observation: because the mass matrix of AD 2010 is condensed and not consistent, the torsion modes obtained

are not taking into account the self rotation mass inertia of the beam.


223

2.82 Test No. 01-0097SDLLB_FEM: Double cross with hinged ends 2.82.1 Description sheet

Reference: NAFEMS, FV2 test Analysis type: modal analysis; Tested functions: Eigen frequencies, Crossed beams, In plane bending.

2.82.2 Problem data

Units I. S.

Geometry

Full square section:

Arm length: L = 5 m Dimensions: a x b = 0.125 x 0.125 Area: A = 1.563 10-2 m2 Inertia: IP = 3.433 x 10-5m4

Iy = Iz = 2.035 x 10-5m4


Longitudinal elastic modulus: E = 2 x 1011 Pa, Density: ρ = 8000 kg/m3

Boundary conditions

Outer: A, B, C, D, E, F, G, H points restraint along x and y; Inner: None.

Loading None for the modal analysis


224

2.82.3 Reference frequencies Mode Units Reference

1 Hz 11.336 2,3 Hz 17.709

4 to 8 Hz 17.709 9 Hz 45.345

10,11 Hz 57.390 12 to 16 Hz 57.390


Linear elements type: Beam

Imposed mesh: 4 Elements / Arms

Modal deformations


225


Results comparison

Solver Mode Units Reference AD 2010 Deviation (%)

1 Hz 11.336 11.333 -0.03% 2, 3 Hz 17.709 17.662 -0.27%

CM2 4 to 8 Hz 17.709 17.691 -0.10% 9 Hz 45.345 45.016 -0.73% 10, 11 Hz 57.390 56.059 -2.32% 12 to 16 Hz 57.390 56.344 -1.82%


226

2.83 Test No. 01-0098SDLLB_FEM: Simple supported beam in free vibration 2.83.1 Description sheet

Reference: NAFEMS, FV5 Analysis type: modal analysis; Tested functions: Shear force, eigen frequencies.

2.83.2 Problem data

Units

I. S.

Geometry

Full square section:

Dimensions: a x b = 2m x 2 m Area: A = 4 m2 Inertia: IP = 2.25 m4

Iy = Iz = 1.333 m4


Longitudinal elastic modulus: E = 2 x 1011 Pa, Poisson's ratio: ν = 0.3. Density: ρ = 8000 kg/m3

Boundary conditions

Outer: o x = y = z = Rx = 0 at A ; o y = z =0 at B ;

Inner: None.


227

Loading

None for the modal analysis

2.83.3 Reference frequencies Mode Shape Units Reference

1 Flexion Hz 42.649 2 Flexion Hz 42.649 3 Torsion Hz 77.542 4 Traction Hz 125.00 5 Flexion Hz 148.31 6 Flexion Hz 148.31 7 Torsion Hz 233.10 8 Flexion Hz 284.55 9 Flexion Hz 284.55

Comment: Due to the condensed (lumped) nature of the mass matrix of AD 2010, the frequencies values of 3 and 7 modes cannot be found by this software. The same modeling done with NE/NASTRAN gave respectively for mode 3 and 7: 77.2 and 224.1 Hz.


Straight elements: linear element

Imposed mesh: 5 meshes

Modal deformations


228


Results comparison

Solver Mode Shape Units Reference AD 2010 Deviation (%) 1 Flexion Hz 42.649 43.11 1.08% 2 Flexion Hz 42.649 43.11 1.08% 3 Torsion Hz 77.542 4 Tension Hz 125.00 124.49 -0.41%

CM2 5 Flexion Hz 148.31 149.39 0.73% 6 Flexion Hz 148.31 149.39 0.73% 7 Torsion Hz 233.10 8 Flexion Hz 284.55 269.55 -5.27% 9 Flexion Hz 284.55 269.55 -5.27%

Comment: The torsion modes No. 3 and 7 that are calculated with NASTRAN cannot be calculated with Advance

Design CM2 solver and therefore the mode No. 3 of the Advance Design analysis corresponds to mode No. 4 of the reference. The same problem in the case of No. 7 - Advance Design, that corresponds to mode No. 8 of the reference.

Test 01-0099HSLSB_MEF

229

2.84 Test No. 01-0099HSLSB_FEM: Membrane with hot point


Reference: NAFEMS, Test T1

Analysis type: static, thermo-elastic;

Tested functions: Stresses.

2.84.2 Problem data

Observation: the units system of the initial NAFEMS test, defined in mm, was transposed in m for practical reasons. However, this has no influence on the results values.

Units

I. S.


230

Geometry / meshing

A quarter of the structure is modeled by incorporating the terms of symmetries.

Thickness: 1 m




Elongation coefficient α = 0.00001.

Boundary conditions

Outer: o For all nodes in y = 0, uy =0; o For all nodes in x = 0, ux =0;

Inner:None.

Loading

External: None,

Internal: Hot point, thermal load ΔT = 100°C;


231

2.84.3 σyy stress at point A:

Reference solution:

Reference value: σyy = 50 MPa in A


Planar elements: membranes,

28 planar elements,

39 nodes.


Results comparison

Solver Quantity Units Reference AD 2010 Deviation (%) CM2 σyy in A MPa 50 50.87 1.74%

Note: This value (50.87) is obtained with a vertical cross section through point A. The value represents σyy at the

left end of the diagram.

With CM2, it is essential to display the results with the “Smooth results on planar elements” option deactivated.

A – hot point limit

Test 01-0100SSNLB_MEF

232

2.85 Test No. 01-0100SSNLB_FEM: Beam on 3 supports with T/C (k = 0)


Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.

2.85.2 Overview Consider the beam on 3 supports as described below. This beam consists of two elements of the same length and identical characteristics.

Units

I. S.

Geometry

L = 10 m Section: IPE 200, Iz = 0.00001943 m4


Longitudinal elastic modulus: E = 2.1 x 1011 N/m2, Poisson's ratio: ν = 0.3.

Boundary conditions

Outer: o Support at node 1 restrained along x and y (x = 0), o Support at node 2 restrained along y (x = 10 m), o T/C stiffness ky = 0,

Inner: None.

Loading

External: Vertical punctual load P = -100 N at x = 5 m, Internal: None.


233

2.85.3 References solutions ky being null, the non linear model behaves the same way as the structure without support 3.

Displacements

( )( )

( )( )

( )( )

( ) rad 000153.0Lk2EI3EI32

LkEI6PL

m 00153.0Lk2EI316

PL3v

rad 000153.0Lk2EI3EI16

LkEI3PL


LkEI2PL3

3yzz

3yz

2

3

3yz

3

3

3yzz

3yz

2

2

3yzz

3yz

2

1

=+

+−=β

=+

−=

=+

+−=β

−=+

+=β

Mz Moments

( )( )

N.m 2502

MM4

PL)m5x(M

0Lk2EI316

PLk3M

0M

1z2zz

3yz

4y

2z

1z

−=−

+==

=+

=

=



3 nodes,

2 linear elements + 1 T/C.

Deformed shape


234

Moment diagrams


Results comparison

Solver Nature Positioning Units Reference AD 2010 Deviation

β1 rotation Node 1 rad -0.000153 -0.000153 0.00% β2 rotation Node 2 rad 0.000153 0.000153 0.00% v3 displacement Node 3 m 0.00153 0.00153 0.00%

CM2 β3 rotation Node 3 rad 0.000153 0.000153 0.00% Mz1 moment Node 1 N.m 0 0 - Mz2 moment Node 2 N.m 0 0 - Mz moment middle span 1 N.m -250 -250 0.00%


235

2.86 Test No. 01-0101SSNLB_FEM: Beam on 3 supports with T/C (k → ∞)


Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.

2.86.2 Overview Consider the beam on 3 supports described below. This beam consists of two elements of the same length and identical characteristics.

Units

I. S.

Geometry

L = 10 m Section: IPE 200, Iz = 0.00001943 m4


Longitudinal elastic modulus: E = 2.1 x 1011 N/m2, Poisson's ratio: ν = 0.3.

Boundary conditions

Outer: o Support at node 1 restrained along x and y (x = 0), o Support at node 2 restrained along y (x = 10 m), o T/C stiffness ky → ∞ (1.1030N/m),

Inner: None.

Loading

External: Vertical punctual load P = -100 N at x = 5 m, Internal: None.


236

2.86.3 References solutions ky being infinite, the non linear model behaves the same way as a beam on 3 supports.

Displacements

( )( )

( )( )

( )( )

( ) rad 000038.0Lk2EI3EI32

LkEI6PL

0Lk2EI316

PL3v


LkEI3PL


LkEI2PL3

3yzz

3yz

2

3

3yz

3

3

3yzz

3yz

2

2

3yzz

3yz

2

1

−=+

+−=β

=+

−=

=+

+−=β

−=+

+=β

Mz Moments

( )( )

N.m 13.2032

MM4

PL)m5x(M

N.m 75.93Lk2EI316

PLk3M

0M

1z2zz

3yz

4y

2z

1z

−=−

+==

−=+

=

=



3 nodes,


Deformed shape


237

Moment diagram


Results comparison


β1 rotation Node 1 rad -0.000115 -0.000115 0.00% β2 rotation Node 2 rad 0.000077 0.000077 0.00% v3 displacement Node 3 m 0 0 -

CM2 β3 rotation Node 3 rad -0.000038 -0.000038 0.00% Mz1 moment Node 1 N.m 0 0 - Mz2 moment Node 2 N.m -93.75 -93.649 -0.11% Mz moment middle span 1 N.m -203.13 -203.176 0.02%


238

2.87 Test No. 01-0102SSNLB_FEM: Beam on 3 supports with T/C (k = -10000 N/m)


Reference: internal GRAITEC test;

Analysis type: static non linear;

Element type: linear, T/C.

2.87.2 Overview Consider the beam on 3 supports, as described below. This beam consists of two elements of the same length and identical characteristics.

Units

I. S.

Geometry

L = 10 m

Section: IPE 200, Iz = 0.00001943 m4




Boundary conditions

Outer: o Support at node 1 restrained along x and y (x = 0), o Support at node 2 restrained along y (x = 10 m), o T/C ky Rigidity = -10000 N/m (the – sign corresponds to an upwards restraint),

Inner: None.

Loading

External: Vertical punctual load P = -100 N at x = 5 m,

Internal: None.


239

2.87.3 References solutions

Displacements

( )( )

( )( )

( )( )

( ) rad 000034.0Lk2EI3EI32

LkEI6PL

m 00058.0Lk2EI316

PL3v


LkEI3PL


LkEI2PL3

3yzz

3yz

2

3

3yz

3

3

3yzz

3yz

2

2

3yzz

3yz

2

1

=+

+−=β

=+

−=

=+

+−=β

−=+

+=β

Mz Moments

( )( )

N.m 9.2202

MM4

PL)m5x(M

N.m 15.58Lk2EI316

PLk3M

0M

1z2zz

3yz

4y

2z

1z

−=−

+==

−=+

=

=



3 nodes,


Deformed shape


240

Moment diagram


Results comparison


β1 rotation Node 1 rad -0.000129 -0.000129 0.00% β2 rotation Node 2 rad 0.000106 0.000106 0.00% v3 displacement Node 3 m 0.00058 0.00058 0.00%

CM2 β3 rotation Node 3 rad 0.000034 0.000034 0.00% Mz1 moment Node 1 N.m 0 0 - Mz2 moment Node 2 N.m -58.15 -58.117 -0.06% Mz moment middle span 1 N.m -220.9 -220.9 0.00%


241

2.88 Test No. 01-0103SSLLB_FEM: Linear system of truss beams


Reference: internal GRAITEC test; Analysis type: static linear; Element type: linear, bar.

2.88.2 Overview Consider the bar system described below. This system contains 4 elements of the same length and S sections (bars

1 to 4) and 2 diagonals of 2L length and 2

S section (bars 5 and 6).

Units

I. S.

Geometry

L = 5 m Section S = 0.005 m2


Longitudinal elastic modulus: E = 2.1 x 1011 N/m2.

Boundary conditions

Outer: o Support at node 1 restrained along x and y, o Support at node 2 restrained along x and y,

Inner: None.

Loading

External: Horizontal punctual load P = 50000 N at node 3, Internal: None.


242

2.88.3 References solutions

Displacements

m 000108.0ES11PL5v

m 000541.0ES11PL25u

m 000129.0ES11PL6v

m 000649.0ES11PL30u

4

4

3

3

==

==

−=−

=

==

N normal forces

N 32141P11

25N N 22727P115N

N 38569P11

26N N 27272P116N

N 22727P115N 0N

4243

1323

1412

−=−===

==−=−=

===


Linear element: bar, without meshing,

4 nodes,

6 linear elements.

Deformed shape


243

Normal forces


Results comparison


u3 displacement Node 3 m 0.000649 0.000649 0.00% v3 displacement Node 3 m -0.000129 -0.000129 0.00% u4 displacement Node 4 m 0.000541 0.000541 0.00% v4 displacement Node 4 m 0.000108 0.000108 0.00%

CM2 N12 normal force Element 1 N 0 0 - N23 normal force Element 2 N -27272 -27272 0.00% N43 normal force Element 3 N 22727 22727 0.00% N14 normal force Element 4 N 22727 22727 0.00% N13 normal effort Element 5 N 38569 38569 0.00% N42 normal force Element 6 N -32141 -32141 0.00%


244

2.89 Test No. 01-0104SSNLB_FEM: Non linear system of truss beams


Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, bar, tie.

2.89.2 Overview Consider the bar system described below. This system contains 4 elements of the same length and S sections (bars

1 to 4) and 2 diagonals of 2L length and 2

S section (ties 5 and 6).

Units I. S.

Geometry

L = 5 m Section S = 0.005 m2


Longitudinal elastic modulus: E = 2.1 x 1011 N/m2.

Boundary conditions

Outer: o Support at node 1 restrained along x and y, o Support at node 2 restrained along x and y,

Inner: None.

Loading

External: Horizontal punctual load P = 50000 N at node 3, Internal: None.


245

2.89.3 References solutions In non linear analysis without large displacement, the introduction of ties for the diagonal bars removes bar 5 (test No. 0103SSLLB_FEM allows finding an compression force in this bar at the linear calculation).

Displacements

0v

m 000238.0ESPLv

m 001195.0ES11PL5uu

4

3

43

=

−=−=

===

N normal forces

0N 0NN 70711P2N N 50000PN

0N 0N

4243

1323

1412

==

==−=−=

==


Linear element: bar, without meshing, 4 nodes, 6 linear elements.

Deformed shape


246

Normal forces


Results comparison


u3 displacement Node 3 m 0.001195 0.001190 -0.42% v3 displacement Node 3 m -0.000238 -0.000238 0.00% u4 displacement Node 4 m 0.001195 0.001190 -0.42% v4 displacement Node 4 m 0 0 -

CM2 N12 normal force Element 1 N 0 0 - N23 normal force Element 2 N -50000 -50000 0.00% N43 normal force Element 3 N 0 0 - N14 normal force Element 4 N 0 0 - N13 normal force Element 5 N 70711 70711 0.00% N42 normal force Element 6 N 0 0 -

Test 02-0112SMLLB_P92

247

2.90 Test No. 02-0112SMLLB_P92: Study of a mast subjected to an earthquake


Reference: internal GRAITEC test; Analysis type: modal and spectral analyses; Element type: linear, mass.

2.90.2 Model overview The structure below consists of 2 beams and 2 punctual masses, subject to a lateral earthquake along X.

2.90.3 Material strength model

Units

I. S.

Geometry

Length: L = 35 m,

Outer radius: Rext = 3.00 m

Inner radius: Rint = 2.80 m

Axial section: S= 3.644 m2

Polar inertia: Ip = 30.68 m4

Bending inertias: Ix =15.34 m4 Iy = 15.34 m4

Masses

M1 =203873.6 kg

M2 =101936.8 kg




Density: ρ = 25 kN/m3


248

Boundary conditions

Outer: Fixed in X = 0, Y = 0 m,

Loading

External: Seismic excitation on X direction


Linear element: beam, automatic mesh,

2.90.4 Seismic hypothesis in conformity with PS92 regulation

Zone: Nice Sophia Antipolis (Zone II).

Site: S1 (Medium soil, 10m thickness).

Construction type class: B

Behavior coefficient: 3

Material damping: 4% (Reinforced concrete).

2.90.5 Modal analysis

Eigen periods reference solution

Substract the value of structure’s specific horizontal periods by solving the following equation:

( ) 0MKdet 2 =ω−

⎟⎟⎠

⎞⎜⎜⎝

⎛≡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−≡

2

1

3

M00M

M

25516

L7EI48K

Eigen modes Units Reference 1 Hz 2.085 2 Hz 10.742

Modal vectors

For ω1:

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛χ⇒=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

ωω

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

−055.31

UU

0UU

M00M

25516

L7EI48

2

11

2

1

12

2

211

3

For ω2:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛χ

655.01

UU

2

12

Normalizing relative to the mass

⎟⎟⎠

⎞⎜⎜⎝

⎛

××

χ−

−

3

4

1 10842.210305.9

; ⎟⎟⎠

⎞⎜⎜⎝

⎛

×−×

χ−

−

3

3

2 10316.11001.2


249

Modal deformations

2.90.6 Spectral study

Design spectrum

Nominal acceleration:

2n1 sm5411.5a 2.085Hzf =⇒=

2n2 sm25.6aHz742.01f =⇒=

Observation: the gap between pulses is greater than 10%, so the modal responses can be regarded as independent.


250

Reference participation factors

Δχ=γ Mii

séismedudirectionladedirecteurVecteur: ⋅⋅⋅⋅⋅⋅Δ

Eigen modes Reference 1 479.427 2 275.609

Pseudo-acceleration

iiii a χ×γ×ξ×=Γ in (m/s2)

4.0%5

⎟⎟⎠

⎞⎜⎜⎝

⎛η

=ξ : Damping correction factor.

η: Structure damping.

⎟⎟⎠

⎞⎜⎜⎝

⎛=Γ

8.25562.7026

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛=Γ

2.4783-3.7852

1

Reference modal displacement

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=ψ024.814E021.576E

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=ψ

045.446E-048.318E

2

Equivalent static forces

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=058.415E055.510E

F1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

052.526E-057.717E

F2

Displacement at the top of the mast

( ) ( )( )221 04E446.502E81.4U −−+−=

Units Reference m 4.814 E-02

Shear force at the top of the mast

( ) ( )( )3

05E526.205E415.8T

22

1+−++

=

3: Being the behavior coefficient of forces

Units Reference N 2.929 E+05

Moment at the base

( ) ( )( ) ( ) ( )( )3

05E717.705E510.55.1705E526.205E415.835M

2222 +++×++++×=

Units Reference N.m 1.578 E+07


251


Results comparison: Frequencies

Solver Modes Units Reference AD 2010 Deviation CM2 1 Hz 2.085 2.085 0.00% CM2 2 Hz 10.742 10.737 -0.05%

Results comparison: Eigen vectors

Solver Modes Node Reference AD 2010 Deviation CM2 1 9.305 E-04 9.305 E-04 0.00% CM2 1 2 2.842 E-03 2.842 E-03 0.00% CM2 1 2.010 E-03 2.010 E-03 0.00% CM2 2 2 -1.316 E-03 -1.316 E-03 0.00%

Results comparison: Participation factors

Solver Modes Reference AD 2010 Deviation CM2 1 479.43 479.43 0.00% CM2 2 275.61 275.61 0.00%

Results comparison: Displacement at the top of the mast

Solver Units Reference AD 2010 Deviation CM2 cm 4.814 4.812 -0.04%

Results comparison: Forces at the top of the mast

Solver Units Reference AD 2010 Deviation CM2 N 2.929E+05 2.927E+5 -0.07%

Test 02-0158SSLLB_B91

252

2.91 Test No. 02-0158SSLLB_B91: Linear element in combined bending/tension - without compressed reinforcements - Partially tensioned section


Reference: J. Perchat (CHEC) reinforced concrete course



2.91.2 Overview Beam with 8 isostatic spans subjects to uniform loads and compression normal forces.

Units

Forces: kN

Moment: kN.m

Stresses: MPa

Reinforcement density: cm²

Geometry

Beam dimensions: 0.2 x 0.5 ht

Length: l = 48 m in 8 spans of 6m,


Longitudinal elastic modulus: E = 20000 MPa,

Poisson's ratio: ν = 0.

Boundary conditions

Outer: o Hinged at end x = 0, o Vertical support at the same level with all other supports

Inner: Hinged at each beam end (isostatic)

Loading

External: o Case 1 (DL): uniform linear load g= -5kN/m (on all spans except 8)

Fx = 10 kN at x = 42m: Ng = -10 kN for spans from 6 to 7 Fx = 140 kN at x = 32m: Ng = -150 kN for span 5 Fx = -50 kN at x = 24m: Ng = -100 kN for span 4 Fx = 50 kN at x = 18m: Ng = -50 kN for span 3 Fx = 50 kN at x = 12m: Ng = -100 kN for span 2 Fx = -70 kN at x = 6m: Ng = -30 kN for span 1

o Case 10 (DL): uniform linear load g = -5 kN/m (span 8) Fx = 10 kN at x = 48m: Ng = -10 kN Fx = -10 kN at x = 42m


253

o Case 2 to 8 (LL): uniform linear load q = -9 kN/m (on spans 1, 3 to 7) uniform linear load q = -15 kN/m (on span 2) Fx = 30 kN at x = 6m (case 2 span 1) Fx = -50 kN at x = 6m (case 3 span 2) Fx = 50 kN at x = 12m (case 3 span 2) Fx = -40 kN at x = 12m (case 4 span 3) Fx = 40 kN at x = 18m (case 4 span 3) Fx = -100 kN at x = 18m (case 5 span 4) Fx = 100 kN at x = 24m (case 5 span 4) Fx = -150 kN at x = 24m (case 6 span 5) Fx = 150 kN at x = 30m (case 6 span 5) Fx = -8 kN at x = 30m (case 7 span 6) Fx = 8 kN at x = 36m (case 7 span 6) Fx = -8 kN at x = 36m (case 8 span 7) Fx = 8 kN at x = 42m (case 8 span 7)

o Case 9 (ACC): uniform linear load a = -25 kN/m (on 8th span) Fx = 8 kN at x = 36m (case 9 span 8) Fx = -8 kN at x = 42m (case 9 span 8) Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107) Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102) Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103) Comb BAELS: 1xDL + 1*LL (comb 108 to 114) Comb BAELA: 1xDL + 1xACC with duration of less than 1h (comb 115)

Internal: None.

Reinforced concrete calculation hypothesis: All concrete covers are set to 5 cm

BAEL 91 calculation (according to 99 revised version)

Span Concrete Reinforcement Application Concrete Cracking 1 B20 HA fe500 D>24h No Non

prejudicial 2 B35 Adx fe235 1h<D<24h No Non

prejudicial 3 B50 HA fe 400 D<1h Yes Non

prejudicial 4 B25 HA fe500 D>24h Yes Prejudicial 5 B25 HA fe500 D>24h No Very

prejudicial 6 B30 Adx fe235 D>24h Yes Prejudicial 7 B40 HA fe500 D>24h Yes 160 MPa 8 B45 HA fe500 D<1h Yes Non

prejudicial

2.91.3 Reinforcement calculation

Reference solution

Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 fc28 20 35 50 25 25 30 40 45 ft28 1.8 2.7 3.6 2.1 2.1 2.4 3 3.3 fe 500 235 400 500 500 235 500 500

teta 1 0.9 0.85 1 1 1 1 0.85 gamb 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.15 gams 1.15 1.15 1.15 1.15 1.15 1.15 1.15 1

h 1.6 1 1.6 1.6 1.6 1 1.6 1.6

fbu 11.33 22.04 33.33 14.17 14.17 17.00 22.67 39.13 fed 434.78 204.35 347.83 434.78 434.78 204.35 434.78 500.00

sigpreju 250.00 156.67 264.00 250.00 250.00 156.67 160.00 252.76 sigtpreju 200.00 125.33 211.20 200.00 200.00 125.33 160.00 202.21


254

Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 g 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00 q 9.00 15.00 9.00 9.00 9.00 9.00 9.00 25.00 pu 20.25 29.25 20.25 20.25 20.25 20.25 20.25 30.00

pser 14.00 20.00 14.00 14.00 14.00 14.00 14.00 G -30.00 -100.00 -50.00 -100.00 -150.00 -10.00 -10.00 -10.00 Q -30.00 -50.00 -40.00 -100.00 -100.00 -8.00 -8.00 -8.00 l 6.00 6.00 6.00 6.00 6.00 6.00 6.00 6.00

Mu 91.13 131.63 91.13 91.13 91.13 91.13 91.13 135.00 Nu -85.50 -210.00 -127.50 -285.00 -352.50 -25.50 -25.50 -18.00

Mser 63.00 90.00 63.00 63.00 63.00 63.00 63.00 Nser -60.00 -150.00 -90.00 -200.00 -250.00 -18.00 -18.00 Vu 60.75 87.75 60.75 60.75 60.75 60.75 60.75 90.00

Main reinforcement calculation according to ULS Mu/A 74.03 89.63 65.63 34.13 20.63 86.03 86.03 131.40 ubu 0.161 0.100 0.049 0.059 0.036 0.125 0.094 0.083 a 0.221 0.133 0.062 0.077 0.046 0.167 0.123 0.108 z 0.410 0.426 0.439 0.436 0.442 0.420 0.428 0.430

Au 6.12 20.57 7.97 8.35 9.18 11.27 5.21 6.46 Main reinforcement calculation with prejudicial cracking according to SLS

Mser/A 51.000 60.000 45.000 23.000 13.000 59.400 59.400 0.000 a 0.4186 0.6678 0.6303 0.4737 0.4737 0.6328 0.6923 0.6157

Mrb 87.53 220.78 302.44 121.16 121.16 182.01 258.82 267.55 A 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 B -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 C -0.4533 -0.8511 -0.3788 -0.2044 -0.1156 -0.8426 -0.8250 0.0000 D 0.4533 0.8511 0.3788 0.2044 0.1156 0.8426 0.8250 0.0000

alpha1 0.238 0.432 0.428 z 0.414 0.385 0.386

Aserp 10.22 10.99 10.75 Main reinforcement calculation with very prejudicial cracking according to SLS

Mser/A 51.00 60.00 45.00 23.00 13.00 59.40 59.40 0.00 a 0.47 0.72 0.68 0.53 0.53 0.68 0.69 0.67

Mrb 96.93 231.67 319.66 132.43 132.43 192.27 258.82 283.60 A 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 B -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 C -0.5667 -1.0638 -0.4735 -0.2556 -0.1444 -1.0532 -0.8250 0.0000 D 0.5667 1.0638 0.4735 0.2556 0.1444 1.0532 0.8250 0.0000

alpha1 0.203 z 0.420

Asertp 14.049 Transverse reinforcement calculation

tu 0.68 0.98 0.68 0.68 0.68 0.68 0.68 1.00 k 0.57 0.40 0.00 -0.14 -0.41 0.00 0.00 0.00

At/st 1.87 7.08 4.31 3.90 4.77 7.34 3.45 4.44 Recapitulation

Aflex 6.12 20.57 7.97 10.22 14.05 11.27 10.75 6.46 e0 -0.95 -1.67 -1.43 -3.17 -3.97 -0.29 -0.29 -0.13

Aminfsimp 0.75 2.38 1.86 0.87 0.87 2.11 1.24 1.37 Aminfcomp 0.83 2.54 2.01 0.90 0.90 2.80 1.65 0.30

At 1.87 7.08 4.31 3.90 4.77 7.34 3.45 4.44 Atmin 1.60 3.40 2.00 1.60 1.60 3.40 1.60 1.60


255


Linear elements: beams with imposed mesh

29 nodes,

28 linear elements.


Results comparison

Solver Model Units Reference RC Expert Deviation

Inf. main reinf. T1 cm2 6.12 6.12 0.00% Sup. main reinf. T1 cm2 0 0 Min. main reinf. T1 cm2 0.75 0.75 0.00% Trans. reinf. T1 cm2 1.87 1.87 0.00% Inf. main reinf. T2 cm2 20.57 20.57 0.00% Sup. main reinf. T2 cm2 0 0 Min. main reinf. T2 cm2 2.38 2.38 0.00% Trans. reinf. T2 cm2 7.08 7.08 0.00% Inf. main reinf. T3 cm2 7.97 7.97 0.00% Sup. main reinf. T3 cm2 0 0 Min. main reinf. T3 cm2 1.86 1.86 0.00% Trans. reinf. T3 cm2 4.31 4.31 0.00% Inf. main reinf. T4 cm2 10.22 10.23 0.10% Sup. main reinf. T4 cm2 0 0

CM2 Min. main reinf. T4 cm2 0.87 0.87 0.00% Trans. reinf. T4 cm2 3.90 3.90 0.00% Inf. main reinf. T5 cm2 14.05 14.05 0.00% Sup. main reinf. T5 cm2 0 0.94 Min. main reinf. T5 cm2 4.20 4.20 Trans. reinf. T5 cm2 4.77 4.77 0.00% Inf. main reinf. T6 cm2 11.27 11.27 0.00% Sup. main reinf. T6 cm2 0 0 Min. main reinf. T6 cm2 2.11 2.11 0.00% Trans. reinf. T6 cm2 7.34 7.34 0.00% Inf. main reinf. T7 cm2 10.75 10.76 0.09% Sup. main reinf. T7 cm2 0 0 Min. main. reinf. T7 cm2 1.24 1.24 0.00% Trans. reinf. T7 cm2 3.45 3.45 0.00% Inf. main reinf. T8 cm2 6.46 6.46 0.00% Sup. main reinf. T8 cm2 0 0 Min. main reinf. T8 cm2 1.37 1.37 0.00% Trans. reinf. T8 cm2 4.44 4.44 0.00%

The "Mu limit" method must be applied in order to achieve the same results.


256

2.92 Test No. 02-0162SSLLB_B91: Linear element in simple bending - without compressed reinforcement


Reference: J. Perchat (CHEC) reinforced concrete course



2.92.2 Overview Beam with 8 isostatic spans subjected to uniform loads.

Units

Forces: kN

Moment: kN.m

Stresses: MPa

Reinforcement density: cm2

Geometry

Beam dimensions: 0.2 x 0.5 ht

Length: l = 42 m in 7 spans of 6m,


Longitudinal elastic modulus: E = 20000 MPa,

Poisson's ratio: ν = 0.

Boundary conditions

Outer: o Hinged at end x = 0, o Vertical support at the same level with all other supports

Inner: Hinge z at each beam end (isostatic)

Loading

External: o Case 1 (DL): uniform linear load g = -5 kN/m (on all spans except 8) o Case 2 to 8 (LL): uniform linear load q = -9 kN/m (on spans 1, 3 to 7)

uniform linear load q = -15 kN/m (on span 2) o Case 9 (ACC): uniform linear load a = -25 kN/m (on 8th span) o Case 10 (DL): uniform linear load g = -5 kN/m (on 8th span)

Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107) Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102) Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103) Comb BAELS: 1xDL + 1*LL (comb 108 to 114) Comb BAELUA: 1xDL + 1xACC (comb 115)

Internal: None.


257

Reinforced concrete calculation hypothesis:

All concrete covers are set to 5 cm

BAEL 91 calculation with the revised version 99 Span Concrete Reinforcement Application Concrete Cracking

1 B20 HA fe500 D>24h No Non prejudicial

2 B35 Adx fe235 1h<D<24h No Non prejudicial

3 B50 HA fe 400 D<1h Yes Non prejudicial

4 B25 HA fe500 D>24h Yes Prejudicial 5 B60 HA fe500 D>24h No Very

prejudicial 6 B30 Adx fe235 D>24h Yes Prejudicial 7 B40 HA fe500 D>24h Yes 160 MPa 8 B45 HA fe500 D<1h Yes Non

prejudicial

2.92.3 Reinforcement calculation

Reference solution

Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 fc28 20 35 50 25 60 30 40 45 ft28 1.8 2.7 3.6 2.1 4.2 2.4 3 3.3 fe 500 235 400 500 500 235 500 500

teta 1 0.9 0.85 1 1 1 1 0.85 gamb 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.15 gams 1.15 1.15 1.15 1.15 1.15 1.15 1.15 1

h 1.6 1 1.6 1.6 1.6 1 1.6 1.6

fbu 11.33 22.04 33.33 14.17 34.00 17.00 22.67 39.13 fed 434.78 204.35 347.83 434.78 434.78 204.35 434.78 500.00

sigpreju 250.00 156.67 264.00 250.00 285.15 156.67 160.00 252.76 sigtpreju 200.00 125.33 211.20 200.00 228.12 125.33 160.00 202.21

g 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00 q 9.00 15.00 9.00 9.00 9.00 9.00 9.00 25.00

pu 20.25 29.25 20.25 20.25 20.25 20.25 20.25 30.00 pser 14.00 20.00 14.00 14.00 14.00 14.00 14.00

l 6.00 6.00 6.00 6.00 6.00 6.00 6.00 6.00 Mu 91.13 131.63 91.13 91.13 91.13 91.13 91.13 135.00

Mser 63.00 90.00 63.00 63.00 63.00 63.00 63.00 Vu 60.75 87.75 60.75 60.75 60.75 60.75 60.75 90.00

Longitudinal reinforcement calculation according to ELU ubu 0.199 0.147 0.068 0.159 0.066 0.132 0.099 0.085 a 0.279 0.200 0.087 0.217 0.086 0.178 0.131 0.111 z 0.400 0.414 0.434 0.411 0.435 0.418 0.426 0.430

Au 5.24 15.56 6.03 5.10 4.82 10.67 4.91 6.28 Main reinforcement calculation with prejudicial cracking according to SLS

A 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 B -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 C -0.56000 -0.89362 -0.87500 D 0.56000 0.89362 0.87500

alpha1 0.367 0.442 0.438 z 0.395 0.384 0.384

Aserp 6.38 10.48 10.25 Main reinforcement calculation with very prejudicial cracking according to SLS

A 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 B -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 C -0.70 -1.60 -0.66 -0.70 -0.61371 -1.12 -0.88 0.00 D 0.70 1.60 0.66 0.70 0.61371 1.12 0.88 0.00


258

Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 alpha1 0.381

z 0.393 Asertp 7.030

Transversal reinforcement calculation tu 0.68 0.98 0.68 0.68 0.68 0.68 0.68 1.00 k 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00

At/st 0.69 1.79 4.31 3.45 3.45 7.34 3.45 4.44 Recapitulation

Aflex 5.24 15.56 6.03 6.38 7.03 10.67 10.25 6.28 Aminflex 0.75 2.38 1.86 0.87 1.74 2.11 1.24 1.37

At 0.69 1.79 4.31 3.45 3.45 7.34 3.45 4.44 Atmin 1.60 3.40 2.00 1.60 1.60 3.40 1.60 1.60


Linear elements: beams with imposed mesh

29 nodes,

28 linear elements.


Results comparison

Solver Model Units Reference AD 2010 RC Expert Deviation

Inf. main reinf. T1 cm2 5.24 5.24 0.00% Sup. main reinf. T1 cm2 0 0 Min. main reinf. T1 cm2 0.75 0.75 0.00% Trans. reinf. T1 cm2 0.69 0.69 0.00% Inf. main reinf. T2 cm2 15.56 15.56 0.00% Sup. main reinf. T2 cm2 0 0 --- Min. main reinf. T2 cm2 2.38 2.38 0.00% Trans. reinf. T2 cm2 1.79 1.79 0.00% Inf. main reinf. T3 cm2 6.03 6.03 0.00% Sup. main reinf. T3 cm2 0 0 --- Min. main reinf. T3 cm2 1.86 1.86 0.00% Trans. reinf. T3 cm2 4.31 4.31 0.00% Inf. main reinf. T4 cm2 6.38 6.38 0.00% Sup. main. reinf. T4 cm2 0 0 ---

CM2 Min. main reinf. T4 cm2 0.87 0.87 0.00% Trans. reinf. T4 cm2 3.45 3.45 0.00% Inf. main reinf. T5 cm2 7.03 7.04 0.14% Sup. main reinf. T5 cm2 0 0 --- Min. main reinf. T5 cm2 1.74 1.74 0.00% Trans. reinf. T5 cm2 3.45 3.45 0.00% Inf. main reinf. T6 cm2 10.67 10.67 0.00% Sup. main reinf. T6 cm2 0 0 --- Min. main reinf. T6 cm2 2.11 2.11 0.00% Trans. reinf. T6 cm2 7.34 7.34 0.00% Inf. main reinf. T7 cm2 10.25 10.27 0.20% Sup. main reinf. T7 cm2 0 0 --- Min. main reinf. T7 cm2 1.24 1.24 0.00% Trans. reinf. T7 cm2 3.45 3.45 0.00% Inf. main reinf. T8 cm2 6.28 6.28 0.00% Sup. main. reinf. T8 cm2 0 0 --- Min. main reinf. T8 cm2 1.37 1.37 0.00% Trans. reinf. T8 cm2 4.44 4.44 0.00%

The "Mu limit" method must be applied to attain the same results.

Test 03-0206SSLLG_CM66

259

2.93 Test No. 03-0206SSLLG_CM66: Design of a Steel Structure according to CM66.

2.93.1 Data

Calculation model: Simple metallic framework with a concrete floor.

Load case: o Permanent loads: 150 kg/m² for the floor and 25kg/m² for the roof. o Overloads: 250 kg/m² on the floor. o Wind loads on region II for a normal location o Snow loads on region 2B at an altitude of 750m.

CM66 Combinations

Model preview


260

Structure’s load case Code No. Type Title CMP 1 Static SW + Dead loads CMS 2 Static Overloads for usage CMV 3 Static Wind overloads along +X in overpressure CMV 4 Static Wind overloads along +X in depression CMV 5 Static Wind overloads along -X in overpressure CMV 6 Static Wind overloads along -X in depression CMV 7 Static Wind overloads along +Z in overpressure CMV 8 Static Wind overloads along +Z in depression CMV 9 Static Wind overloads along -Z in overpressure CMV 10 Static Wind overloads along -Z in depression CMN 11 Static Normal snow overloads


261

2.93.2 Effel Structure results Displacement Envelope (“CMCD" load combinations)

Envelope of linear element forces D DX DY DZ Env. Case No. Max.

location (cm) (cm) (cm) (cm) Max(D) 213 148 CENTER 12.115 0.037 12.035 -1.393 Min(D) 188 1.1 START 0.000 0.000 0.000 0.000

Max(DX) 204 72.1 START 3.138 3.099 0.434 0.244 Min(DX) 204 313 END 2.872 -1.872 -0.129 -2.174 Max(DY) 213 148 CENTER 12.115 0.037 12.035 -1.393 Min(DY) 213 61.5 END 9.986 -0.118 -9.985 0.046 Max(DZ) 201 371 CENTER 4.149 -0.006 -0.188 4.145 Min(DZ) 203 370 CENTER 4.124 -0.006 -0.240 -4.118

Envelope of forces on linear elements (“CMCFN” load combinations)

Envelope of linear element forces Fx Fy Fz Mx My Mz Env. Cas

e No. MaxSite (T) (T) (T) (T*m) (T*m) (T*m) Max (Fx) 120 4.1 START 19.423 -4.108 -1.384 -0.003 1.505 7.551 Min (Fx) 138 98 START -41.618 -0.962 -0.192 0.000 0.000 0.000 Max(Fy) 120 57 END -13.473 16.349 -0.016 -0.003 0.002 55.744 Min(Fy) 120 60 START -15.994 -16.112 -0.006 -3E-004 6E-006 53.096 Max(Fz) 177 371 START -3.486 -0.118 2.655 0.000 0.000 0.000 Min(Fz) 187 370 START -3.666 -0.147 -2.658 0.000 0.000 0.000 Max(Mx) 120 111 END 3.933 4.840 0.278 0.028 -4E-005 11.531 Min(Mx) 120 21 END -22.324 13.785 -0.191 -0.028 -0.004 42.562 Max(My) 177 371 CENTER -3.099 -0.118 -0.323 0.000 4.403 -0.500 Min(My) 179 370 CENTER -3.283 -0.155 0.321 0.000 -4.373 -0.660 Max (Mz) 120 57 END -13.473 16.349 -0.016 -0.003 0.002 55.744 Min (Mz) 120 59.2 END -19.455 -8.969 -0.702 -0.003 -0.001 -57.105 Envelope of linear element stresses (“CMCFN” load combinations)

Envelope of linear element stresses sxxMax sxyMax sxzMax sFxx sMxxMax Env. Cas

e No. MaxSite (MPa) (MPa) (MPa) (MPa) (MPa) Max(sxxMax) 120 59.2 END 273.860 -14.696 -1.024 -16.453 290.312 Min(sxxMax) 120 292 START -150.743 0.000 0.000 -150.743 0.000 Max(sxyMax) 120 57 START 262.954 37.139 -0.030 -15.609 278.562 Min(sxyMax) 120 60 END 241.643 -36.595 -0.011 -18.536 260.179 Max(sxzMax) 185 371 START -2.949 -0.183 3.876 -2.949 0.000 Min(sxzMax) 179 370 START -3.104 -0.255 -3.882 -3.104 0.000 Max(sFxx) 120 293 END 161.095 9E-005 -0.002 161.095 0.000 Min(sFxx) 120 292 START -150.743 0.000 0.000 -150.743 0.000

Max(sMxxMax) 120 59.2 END 273.860 -14.696 -1.024 -16.453 290.312 Min(sMxxMax) 1 1.1 START -4.511 3.155 -0.646 -4.511 0.000


262

2.93.3 CM66 Effel Expertise results

Hypotheses

For columns

Deflections: 1/150 Envelopes deflections calculation.

Buckling XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on fixed nodes

Lateral-torsional buckling: Ldi automatic calculation: hinged restraint Lds automatic calculation: hinged restraint

For the rafters


Buckling: XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on fixed nodes

Lateral-torsional buckling: Ldi automatic calculation: no restraint Lds automatic calculation: hinged restraint

For the columns


Buckling: XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on displaceable nodes

Lateral-torsional buckling: Ldi automatic calculation: hinged restraint Lds automatic calculation: hinged restraint

Optimization parameters

Work ratio optimization between 90 and 100%

All the sections from the library are available.

Labels optimization. The results of the optimization given below correspond to an iteration of the finite elements calculation.


263

Deflection verification

Ratio

Max values on the element

Columns: L / 168

Rafter: L / 96

Column: L / 924


264

CM Stress diagrams

Work ratio

Stresses

Max values on the element

Columns: 375.16 MPa

Rafter: 339.79 MPa

Column: 180.98 MPa


265

Buckling lengths

Lfy

Lfz


266

Lateral-torsional buckling lengths

Ldi

Lds


267

Optimization

2.93.4 Effel Structure / Advance Design comparison

Maximum displacement (CMCD) Effel value

(cm) Advance Design 2010 value

(cm) Deviation

(%) 12.115 12.126 0.09%

Envelope normal force (CMCFN)

Values Effel value (T)

AD 2010 value (T)

Deviation (%)

Min (Fx) -41.618 -41.629 0.03% Max (Fx) 19.423 19.466 0.2%

Envelope bending moment (CMCFN)

Element Effel value (T.m)

AD 2010 value (T.m)

Deviation (%)

Min (Mz) -57.105 -57.094 0.02% Max (Mz) 55.744 55.745 0.00%

Warning, the Mz bending moment of Effel Structure corresponds to the My bending moment of Advance Design.


268

CM deflections Element Effel value

(Ratio) AD 2010 value

(Ratio) Deviation

(%) Columns L / 168 (89%) L / 167.92 (89%) 0.05% Rafters L / 96 (208%) L / 96.13 (208%) -0.13%

Columns L / 924 (16%) L / 925.30 (16.2%) 0.14%

CM stresses Element Effel value

(MPa) AD 2010 value

(MPa) Deviation

(%) Columns 374.67 374.59 0.02% Rafters 339.74 347.46 2.20%

Columns 180.98 180.71 0.15%

Buckling lengths Element Value Effel

(m) AD 2010

(m) Deviation

(%) Columns Lfy 24.07 24.07 0.00%

Lfz 8.02 8.00 -0.25% Rafters Lfy 20.25 20.25 0.00%

Lfz 1.72 1.72 0.00% Columns Lfy 5.67 5.67 0.00%

Lfz 4.20 4.196 0.09%

Warning, the local axes in Effel Structure are opposite to those in Advance Design.

Lateral-torsional buckling lengths Element Value Effel

(m) AD 2010

(m) Deviation

(%) Columns Ldi 8.5 8.5 0.00%

Lds 8.5 8.5 0.00% Rafters Ldi 8.61 8.61 0.00%

Lds 1.72 1.72 0.00% Columns Ldi 2 2 0.00%

Lds 2 2 0.00%

Optimization Element Effel optimization AD 2010 optimization

Initial section

Rate (%)

Final section

Rate (%)

Initial section

Rate (%)

Final section

Rate (%)

Columns IPE500 159 IPE600 92 IPE500 159.4% IPE600 92.2% Rafters IPE400 145 IPE500 88 IPE400 147.9% IPE500 88.2%

Columns IPE400 77 IPE360 92 IPE400 76.9% IPE360 92.4%


269

2.94 Test No. 03-0207SSLLG_CM66: Design of a 2D portal frame

2.94.1 Data

Calculation model: 2D metallic portal frame. o Column section: IPE500 o Rafter section: IPE400 o Base plates: hinged. o Portal frame width: 20m o Columns height: 6m o Portal frame height at the ridge: 7.5m

Load case: o Permanent loads: 150 kg/m on the roof + elements self weight. o Usage overloads: 800 kg/ml on the roof

Mesh density: 1m

Model preview

Combinations

Code Numbers Type Title CMP 1 Static Permanent load + self weight CMS 2 Static Usage overloads

CMCFN 101 Comb_Lin 1.333P CMCFN 102 Comb_Lin 1.333P+1.5S CMCFN 103 Comb_Lin P+1.5S CMCD 104 Comb_Lin P+S


270

2.94.2 Effel Structure Results

Ridge displacements (combination 104)

Diagram of normal force envelope


271

Envelope of bending moments diagram

2.94.3 Effel Expert CM results

Main hypotheses

For columns


Buckling: XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes

Ka-Kb Method Lateral-torsional buckling: Ldi automatic calculation: no restraints

Lds imposed value: 2 m

For the rafters


Buckling: XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes Ka-Kb Method

Lateral-torsional buckling: Ldi automatic calculation: No restraints Lds imposed value: 1.5m

Optimization criteria

Work ratio optimization between 90 and 100%

Labels optimization (on Advance Design templates)


272

Deflection verification

Ratio

CM Stress diagrams

Work ratio

Stresses


273

Buckling lengths Lfy

Lfz

Lateral-torsional buckling lengths Ldi


274

Lds

Optimization


Displacement at the ridge (load combination 104) Effel value


(cm) Deviation

(%) 9.36 9.36 0.00%

Envelope normal forces

Element Effel value (T)

AD 2010 value (T)

Deviation (%)

Columns (min) -15.77 -15.77 0.00% Rafters (max) -1.02 -1.02 0.00%

Envelope bending moments

Element Effel value (T.m)

AD 2010 value (T.m)

Deviation (%)

Columns (min) -42.41 -42.41 0.00% Rafters (max) 42.41 42.41 0.00%


275

CM deflections Element Effel value

(Ratio) AD 2010 value

(Ratio) Deviation

(%) Columns L / 438 (34%) L / 438 (34%) 0.00% Rafters L / 111 (180%) L / 111 (180%) 0.00%

CM stresses

Element Effel value (MPa)

AD 2010 value (MPa)

Deviation (%)

Columns 230.34 230.34 0.00% Rafters 458.38 458.38 0.00%

Buckling lengths

Element Value Effel (m)

AD 2010 (m)

Deviation (%)

Columns Lfy 5.84 5.84 0.00% Lfz 6 6 0.00%

Rafters Lfy 7.08 7.08 0.00% Lfz 10.11 10.11 0.00%

Warning, the local axes in Effel Structure have different orientation in Advance Design.

Lateral-torsional buckling lengths Element Value Effel

(m) AD 2010

(m) Deviation

(%) Columns Ldi 6 6 0.00%

Lds 2 2 0.00% Rafters Ldi 10.11 10.11 0.00%

Lds 1.5 1.5 0.00%

Optimization Element Effel optimization AD 2010 optimization

Initial section

Rate (%)

Final section

Rate (%)

Initial section

Rate (%)

Final Section

Rate (%)

Columns IPE500 98 - 98 IPE500 98 - 98 Rafters IPE400 195 IPE550 77 IPE400 195.1 IPE550 76.8

Test 03-0208SSLLG_BAEL91

276

2.95 Test No. 03-0208SSLLG_BAEL91: Design of a concrete floor with an opening

2.95.1 Data

Calculation model: 2D concrete slab. o Slab thickness: 20 cm o Slab length: 20m o Slab width: 10m o The supports (punctual and linear) are considered as hinged. o Supports positioning (see scheme below) o 1,50m*2,50m opening => see positioning on the following scheme

Materials: o Concrete B25 o Young module: E= 36000 MPa

Load case: o Permanent loads: 100 kg/m2 o Permanent loads: 200 kg/ml around the opening o Punctual loads of 2T in permanent loads (see the following definition) o Usage overloads: 250 kg/m2

Mesh density: 0.5 m

Slab geometry


277

Support positions

Positions of punctual loads


278

Global loading overview

Load Combinations

Code Numbers Type Title BAGMAX 1 Static Permanent loads + self weight

BAQ 2 Static Usage overloads BAELS 101 Comb_Lin Gmax+Q BAELU 102 Comb_Lin 1.35Gmax+1.5Q


279

2.95.2 Effel Structure Results

SLS max displacements (load combination 101)

Mx bending moment for ULS load combination


280

My bending moment for ULS load combination

Mxy bending moment for ULS load combination


281

2.95.3 Effel RC Expert Results

Main hypothesis

Top and bottom concrete covers: 3 cm Slightly dangerous cracking Concrete B25 => Fc28= 25 MPa Reinforcement calculation according to Wood method. Calculation starting from non averaged forces.

Axi reinforcements

Ayi reinforcements


282

Axs reinforcements

Ays reinforcements


283


Max displacement for SLS (load combination 101) Effel value


(cm) Deviation

(%) 0.176 0.175 2.85%

Mx and My bending moments for ULS (load combination 102) Magnitude Effel value

(kN.m) AD 2010 value

(kN.m) Deviation

(%) Max(Mx) 25.20 25.25 0.20% Min(Mx) -15.71 -15.68 0.19% Max(My) 31.17 31.24 0.22% Min(My) -18.79 -18.77 0.11%

Max (Mxy) 10.26 10.25 0.10% Min (Mxy) -10.14 -10.15 0.10%

Theoretic reinforcements

Magnitude Effel value (cm²)

AD 2010 value (cm²)

Deviation (%)

Axi 3.84 3.83 -0.24% Axs 3.55 3.628 2.23% Ayi 3.75 3.728 -0.57% Ays 4.53 4.619 1.97%

These values are obtained from the maximum values from the mesh.

3 Eurocodes 1 Tests description

The objective of this chapter is to validate the results from ADVANCEDesign 2010 according to Eurocodes 1.

Eurocodes 1 Tests description

286

Wind calculation

3.1 Portal frame with 11.31° angle – Example A.fto

3.1.1 Wind pressure calculation

Calculation according to French Appendix

Wind speed region: 1

Type of site: 0

Topographic factor: 1

Turbulence factor: 1


287

ADVANCE DESIGN 2010 is set on:

Wind load cases are set on:


288

a) Wind speed (Vb) calculation:

Vb = Cdirection * Cseason * Vb0 = 1 * 1 * Vb0 = 22m/s

b) Site factor (Kr) calculation:

Kr = 0.19 * (z0/z0,II)^(0.07) = 0.19*(0.005/0.05)^(0.07) = 0.162

Note (see 4.3.2 (1) – French Appendix):

Terrain category Zo Zmin

0 Sea or coastal sea winds area; lakes and water course with wind over a distance of at least 5 m

0.005 1

II Open field, with or without a few isolated obstacles (trees, buildings, etc.) separated from each other by more than 40 times their height

05 2

IIIa Countryside with hedgerows, vineyards, hedged farmland, dispersed habitat

0.20 5

IIIB Urban or industrial zones, densely hedged farmland, orchards

0.5 9

IV Urban areas with at least 15% of the surface covered with buildings whose average height is greater than 15 m, forests

1 15

ADVANCE DESIGN 2010 returns: Kr = 0.16.


289

c) Roughness coefficient Cr (z) calculation:

h = 5m

b = 15m

So: h < 2b

Cr (z) = Kr*ln(z/z0) = 0.162 * ln (z / 0.005) = 1.119


290

French Appendix will give the same value:

ADVANCE DESIGN 2010 returns: 1.12.


291

d) Average Wind Vm(z) speed calculation:

Vm(z) = Cr(z)*C0(z)*Vb = Cr(z) * 22 = 1.119 * 22 = 24.618 m/s

ADVANCE DESIGN 2010 returns: 24.58 m/s

e) Turbulence Intensity Iv(z) calculation:

kl: turbulence factor = 1

Co(z): topographic factor = 1

Iv(z) = 1/ln(z/z0) = 0.145

AD 2010 returns: 0.14


292

f) Basic velocity pressure Qb calculation:

Qb = 0.5 * Ro_air * Vb² = 0.5 * 1.225 * 22² = 296.45 N/m²

Advance Design 2010 returns:

g) Peak velocity pressure Qp(z) calculation

Qp = (1+7Iv(z)) * 0.5*Ro_air * Vm² = (1+7*0.145) * 0.5*1.225 * 24.618² = 748 N/m²

Advance Design 2010 returns:744.82 N/m²


293

h) Exposure factor Ce(z) calculation:

Ce (z) = Qp (z) / Qb = 748 / 296.45 = 2.52

French Appendix will give the same value:

ADVANCE DESIGN 2010 returns: 2.51


294

3.1.2Cpe and Wind force calculation for front wind (Y)

Building characteristics:

h: building height = 5m

b: crosswind direction = 10m

d: parallel wind direction = 30m

e = min (b ; 2h) = 10m

h/d = 5/30 = 0.1667

Note: h, b, d and e are given by the picture below:

For each windwall: Cpe = Cpe10

Table 7.1 Recommended values of external pressure coefficients for the vertical walls of buildings on a rectangular plane:

Note: For buildings with h / d> 5, the total wind load may be based on the information from Sections 7.6 to 7.8 and 7.9.2.


295

a) Windwall n° 1

Cpe calculation:

Windwall n°1: D area

h/d = 5/30 = 0.1667

So: h/d < 0.25



Cpe = Cpe10 = +0.7

ADVANCE DESIGN 2010 returns +0.7:

D


296

Wind force calculation:

Wind force: F = Qp(z)*(Ce-Ci).

Note: in this example, Qp = 748 N/m².

F = 748 * 0.7 = 523.6 N/m²


b) Windwall n° 4

Windwall n°4: E area

E


297

h/d = 5/30 = 0.1667

h/d < 0.25



Cpe = Cpe10 = -0.3

ADVANCE DESIGN 2010 returns -0.3:

F = 748 * (-0.3) = -224.4 N/m²



298

c) Windwalls n° 2 and n°3

e<d:

A area

B area

C area


299

h/d = 4/30 = 0.133

So: h/d < 0.25




Wind force for area A:

F = 748 * (-1.2) = -898 N/m²



300

Wind force for area B:

F = 748 * (-0.8) = -598 N/m²


Wind force for area C:

F = 748 * (-0.5) = -374 N/m²



301

d) Windwalls n°5 and 6

There are 4 areas: F,G,H et I:

Note: e = 10m

e/4 = 2.5m

e/10 = 1m

e/2 = 5m

Cpe depends on the angle: 11.31°.


302

For 11.31°, the values are between those two lines: 5° et 15°:

Table 7.4b - External pressure coefficient for double-sided sloped roofs

Between 5° and 15°:

11.31° -1.41 -2.07 -1.3 -2 -0.64 -1.2 -0.54

F area:

Cpe10 = -1.41

Advance Design 2010 returns:-1.411


303

G area:

Cpe10 = -1.3

Advance Design 2010 returns -1.3

H area:

Cpe = Cpe10 = – 0.64

Advance Design 2010 returns -0.637

I area:

Cpe = Cpe10 = -0.54

Advance Design 2010 returns -0.537:


304

Wind force for area F:

F = Qp(z)*(Ce-Ci) = 748 * (-1.41) = -1055 N/m²

Fx = F * sin(alpha) = 207 N/m²

Fy = F * cos(alpha) = 1035 N/m²


Wind force for area G:

F = 748 * (-1.30) = -972 N/m²

Fx = 191 N/m²

Fy = 953 N/m²



305

Wind force for area H:

F = 748 * (-0.637) = -476 N/m²

Fx = 93.35 N/m²

Fy = 467 N/m²


Wind force for area I:

F = 748 * (-0.537) = -402 N/m²

Fx = 79 N/m²

Fy = 394 N/m²



306

3.1.3Cpe and Wind force calculation for parallel wind (X)


h: building height = 5m (or 4m for windwalls n° 2 and 3)


d: parallel wind direction = 10m

e = min (b ; 2h) = 10m (or 8m for windwalls n° 2 and 3)

h/d = 5/10 = 0.5

Note: h, b, d and e are given by the picture below:





307

a) Windwall n° 3

Cpe calculation:

Windwall n°3: D area

h/d = 4/10 = 0.4

So: 0.25 < h/d < 0.5

The values will be between those two lines:



0.5 -1.2 -1.4 -0.8 -1.1 -0.5 +0.72 +1.0 -0.34

Cpe = Cpe10 = +0.720

D


308



Wind force: F = Qp(z)*(Ce-Ci) = 748 * 0.720 = 538 N/m²

Advance Design 2010 returns: F = 511 N/m².


309

b) Windwall n° 2

0.5 -1.2 -1.4 -0.8 -1.1 -0.5 +0.72 +1.0 -0.34


Cpe = Cpe10 = -0.34


E


310


Wind force: F = Qp(z)*(Ce-Ci) = 748 * (-0.34) = -250 N/m²

Advance Design 2010 returns: F = 241 N/m².

c) Windwalls n° 1 and n°4

e = 5m

d = 10m

So: e<d:

There will be 2 areas: A and B:


311

h/d = 5/10 = 0.5

For each area, Cpe = Cpe10.

Table 7.1 Recommended values of external pressure coefficients for the vertical walls of buildings on a rectangular plane

0.5 -1.2 -1.4 -0.8 -1.1 -0.5 +0.73 +1.0 -0.367



A area

B area


312

Wind force for area A:

F = 748 * (-1.2) = -898 N/m²


Wind force for area B:

F = 748 * (-0.8) = -598 N/m²



313

3.2 Monopitch frame with 15° angle – Example B.fto 3.2.1Lateral Wind (X direction): Cpe calculation on rooftop


h: building height = 6.68m


e = min (b ; 2h) = 13.358m

h/d = 5/10 = 0.5

Roof angle = 15°

Loads areas must be: F, G and H.


314



Table 7.3a - External pressure coefficients applied on single-sided sloped roofs

Note: for F, G and H areas, the table will give two values for Cpe10, which means that for each area, there will be a positive and a negative to take into account.

Note: When θ = 0° (see Table a), the pressure varies rapidly between positive and negative values for an angle α ranging from + 5° to + 45°; this is the reason the positive and negative values are specified for these slopes. For these roofs, it is necessary to consider two cases: a case for all the positive values, and a case for all negative values. A mixture of positive and negative values on one side is not allowed.

F

G

F

H


315

F area: Cpe neg = -0.9

Cpe pos = = +0.2

G area: Cpe neg = -0.8

Cpe pos = +0.2

H area: Cpe neg = -0.3

Cpe pos = +0.2



316

3.3 Portal frame with 10° angle – Example C.fto

3.3.1Lateral Wind (X): Cpe calculation and Wind force calculation


317

The values for a 10° angle must be calculated:

Note 1: For each area, the interpolation must be done between values with same sign.

• For I area, the first value will be an interpolation between -0.6 and -0.4. The second value will be 0, because there is no other positive value to take into account.

• For J area, the first value will be an interpolation between -0.6 and -1.0. The second value will be an interpolation between +0.2 and +0.0.

Note 2: It is not allowed to have negative and positive values for the same pitch.

Advance Design 2010 returns: 3 problems


318

Snow calculation

3.4 Portal frame with 11.31° angle – Example A.fto

Region of structure: A2

Altitude of site: 350m

Exposure factor: 1

Thermal factor: 1

Snow load cases are set on:


319

Normal snow:s = µi • Ce • Ct • sk( 5.1 )

µi is given by:

Table 5.2 - Shape coefficients

sk,0 is given by:

Regions:

A1 A2 B1 B2 C1 C2 D E

Sk characteristic snow load value on soil at an altitude less than 200 m

0.45 0.45 0.55 0.55 0.65 0.65 0.90 1.40

SAd calculation value of exceptional snow load on soil

- 1.00 1.00 1.35 - 1.35 1.80 -

Variation law of the characteristic load for an altitude greater than 200 m

ΔS1 ΔS2

Which means that sk = 0.45 + 350/1000 – 0.20 = 0.60 kN/m²

So: Normal snow: s = µi • Ce • Ct • sk = 0.8 * 1 * 1 * 0.60 kN/m² = 0.48 kN/m²


320


Accidental snow:s = µi • sAd = 0.8 * 1 = 0.80 kN/m²( 5.1 )

Regions:

A1 A2 B1 B2 C1 C2 D E

Sk characteristic snow load value on soil at an altitude less than 200 m

0.45 0.45 0.55 0.55 0.65 0.65 0.90 1.40

SAd calculation value of exceptional snow load on soil

- 1.00 1.00 1.35 - 1.35 1.80 -





322

4.1 Test No. 01 – EC2: Minimum reinforcement of a beam

4.1.1 Overview Search of the minimum longitudinal reinforcement to be implemented on a beam. The calculation is performed on EC2, French DAN.

Units

I. S.

Geometry

Length: l = 5.50 m (5.90m between axes)

Section: b = 0.4m and h = 0.8 m.

Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²

Concrete density: 25 kN/m3

Boundary conditions

Hinge at end x = 0,

Simple support at extremity x = 5.50 m (horizontal restraint).

Loading

Self weight


323


Reference solution

Clause 7.3.2

s

cteffctcs

AFkkAσ

×××= .min,

14,0 ≤=ck (in simple bending)

65,0=k (rectangular beam with h ≥ 80 cm)

Fct.eff = Fctm = 2.56MPa

²16.02

8.04.0 mAct =×

=

50016.056.265.04.0min, ×××=sA ²13.2 cm=

Clause 9.2.1.1

dbffA tyk

ctms .26.0min, =

ctmf =2.56MPa

ykf =500MPa

tb = 0.4m

d = 0.72m

=××== 72.04.0500

56.226.0.26.0min, dbffA tyk

ctms ²84.3 cm

Therefore: ²84.3min, cmAs =

4.1.3 Results sheet

ADVANCE Design: See test 01.zip


324

4.2 Test No. 02 – EC2: Longitudinal reinforcement of a beam under a linear load - horizontal level behavior law

4.2.1 Overview Research of the longitudinal reinforcement to implement on a beam. The calculation is performed on EC2, French DAN.

Units

I. S.

Geometry

Length: l = 5.50 m (5.90m between axes),


Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²


Boundary conditions

Hinge at end x = 0,


Loading

No self weight

G = 100 kN/ml


325


Reference solution

C25/30 concrete strength: MPaffc

ckcccd 67,16

5,125. ===

γα

ELU maximum bending moment calculation:

o mKNlPMEd .5878

²50,5)10035,1(8

².=

××==

Calculation of the reduced moment:

170,067,16²72,040,0

587,0²

=××

==cd

Ed

FbdMμ

ξ calculation: [ ] [ ] 234,0)170,021(125,1)21(125,1 =×−−=−−= μξ

zb leverage calculation: mdzb 653,0)234,04,01(72,0)4,01( =×−=−= ξ

Reinforcement bars cross section calculation: ²68,2078,434653,0

587,0.

cmzMA

sb

Edu =

×==

σ

4.2.3 Results sheet



326

4.3 Test No. 03 – EC2: Longitudinal reinforcement of a beam under a linear load - inclined stress strain behavior law


Units

I. S.

Geometry



Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²


Steel ductility: Class A

Boundary conditions

Hinge at end x = 0,


Loading

No self weight

G = 100 kN/ml


327

4.3.2 Reference results Reference solution


ckcccd 67,16

5,125. ===

γα


o mKNlPMEd .5878

²50,5)10035,1(8

².=

××==

Calculation of the reduced moment: mKNlPMEd .5878

²50,5)10035,1(8

².=

××==

ξ calculation: [ ] [ ] 234,0)170,021(125,1)21(125,1 =×−−=−−= μξ


To determine the ultimate stress sσ of tensile reinforcement, the elongation of the tensile reinforcement must be determined.

For this, a concrete shortening of 3,5‰ is taken into account.

Therefore, we have ‰46.11234,0

234,015,315,35,3 =−

=−

=−

=ξ

ξε

xxd

S

We verify that this value remains less than ukε9,0 , or 0,9*25=22,5‰ for an A class steel.

Table 2.1 Reinforcement classes

Then we determine if it is in the plastic range:

11.46‰ ‰ 175,2200000

78,4340 <===

S

yds E

fε : is in the plastic range.


328

Thus, we obtain:

05,1=k from the 2.1 table of the C annex of EC2 (class A steel)

Table 2.1 Reinforcement classes

( )( )

( ) MPakff

suk

SS

S

ykyd 63,443

)175,225()175,246.11(05,01

15,150011

0

0 =⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−+=

εεεε

γ

We deduce then the section of the corresponding reinforcement:

²26,2063,443653,0

587,0.

cmzMA

sb

Edu =

×==

σ

4.3.3 Results sheet



329

4.4 Test No. 04 – EC2: Longitudinal reinforcement of a beam under a concentrated load - horizontal level behavior law


Units

I. S.

Geometry



Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²


Boundary conditions

Hinge at end x = 0,


Loading

No self weight

G = 150 kN and Q = 250 kN at mid-span


330



ckcccd 67,16

5,125. ===

γα


o mKNlPM Ed .8524

90,5)2505,115035,1(4.

=××+×

==

Calculation of the reduced moment: 246,067,16²72,040,0

852,0²

=××

==cd

Ed

FbdMμ

ξ calculation: [ ] [ ] 359,0)246,021(125,1)21(125,1 =×−−=−−= μξ


Reinforcement bars cross section calculation:

²76,31²10.76,3178,434617,0

852,0.

4 cmmzMA

sb

Edu ==

×== −

σ

4.4.3 Results sheet



331


4.5.1 Overview Search of reinforcement to be implemented on a beam. The calculation is performed on EC2, French DAN.

Units

I. S.

Geometry



Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²


Boundary conditions

Hinge at end x = 0,


Loading

No self weight

G = 25 kN/ml and Q = 30 kN/ml


332



ckcccd 67,16

5,125. ===

γα


o mKNlPMEd .4048

²40,6)305,12535,1(8

².=

××+×==


404,0²

=××

==cd

Ed

FbdMμ

ξ calculation: [ ] [ ] 356,0)244,021(125,1)21(125,1 =×−−=−−= μξ



404,0.

cmzMA

sb

Edu =

×==

σ

4.5.3 Results sheet

Thonier: See test 05.XLS



333

4.6 Test No. 06 – EC2: Transverse reinforcement of a beam subjected to a linear load

4.6.1 Overview Search of reinforcement to be implemented on a beam. The calculation is performed on EC2, French DAN.

Units

I. S.

Geometry



Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²


Boundary conditions

Hinge at end x = 0,


Loading

No self weight



334


Shear force calculation: EdV

kNVEd 2522

305.12535.1=

×+×=

Shear force calculation with reduction: redEdV , : non considered by Advance Design

Concrete shear resistance: max,RdV

θθα

gtgfvzbV cdwcw

Rd cot.... 1

max, +=

o θ: connecting rod slope angle = 45° o z: internal forces lever arms. It can be z=0,9d = 0.567m

o 1=cwα in simple bending without prestress.

o 1v : cracked concrete resistance reduction coefficient of the shear force:

54.02502516,0

25016,01 =⎥⎦

⎤⎢⎣⎡ −=⎥⎦

⎤⎢⎣⎡ −= ckfv

MNgtgfvzbV cdwcw

Rd 630.011

67.1654.0567.025.01cot

.... 1max, =

+××××

=+

=θθ

α

max,RdEd VV < : no crushing of concrete struts. The section is properly sized.

Shear resistance calculation in the absence of transverse reinforcement: cRdV ,

[ ]{ } dbkvfkCMaxV wcpckLcRdcRd ..;)..100(. 1min3/1

,, σρ +=

o 12.05.118.018.0

, ===c

cRdC γ

o 0,2563.163020012001

)lim(

≤=+=+=ètresmild

k

o 02,0.

≤=db

A

w

sLLρ

²19.16 cmAsL = (see test 05)

02,001028.063.025.0

1019.16.

4

≤=××

==−

dbA

w

sLLρ

o 177.0255.1

053,0.053,0 2/12/1min =×== ck

c

fVγ

[ ]{ } MNMaxV cRd 0866.063.025.055.063.025.0177.0;)2501028.0100(563.112.0 3/1, =××=×××××=

cRdEd VV ,> : transverse reinforcement implementation


335

Shear force on transverse vertical reinforcement:

mlcmfzV

sAVfz

sAV

ywd

EdswEdywd

swsRd /²22.10

178.434567.0252.0

cot..cot..., =

××=>⇒>=

θθ

4.6.3 Results sheet



336



Units

I. S.

Geometry



Material properties

Concrete: C25/30

Fck= 25 N/mm²

Fyk= 500 N/mm²


Boundary conditions

Hinge at end x = 0,


Loading

Self weight



337



ckcccd 67,16

5,125. ===

γα

Self weight: mlkNPp /75.32560.025.0 =××=


o mKNlPMEd .5238

²80,5)755,1)575.3(35,1(8

².=

××++×==


523,0²

=××

==cd

Ed

FbdM

μ

Limited reduced moment: 372.0lim =iteμ (for Fe500)

itelimμμ > : compressed reinforcement bars implementation (or resize of the concrete section)

Tensile reinforcement calculation (section A1):

The tensile reinforcement calculation must be associated with a moment corresponding to μlim.

buFbdM

²lim

lim =μ => mMNFbdM bu .452,067,16²54,025,0372,0²limlim =×××== μ

αlim calculation: [ ] 617,0)372,021(125,1lim =×−−=α

zb leverage calculation:


452,0lim1 cm

FzM

Aedb

=×

==

Compressed reinforcement calculation (section A’)

Compressed reinforcement stress calculation:

o ( ) ( ) 00000

000

018.3'03.054.0617.0

54.0617.05.3'5.3

=−××

=−= ddd lu

lusce α

αε

o MPafEf

cdsces

cdsce 78.43418.2

20000078.434

000 ==⇒==> σε

Compressed reinforcement calculation: ²20,378.434)03,054,0(

452,0523,0)'(

' cmddMM

Asc

ulu =−

−=

−−

=σ

A2 reinforcement bars calculation section to balance A’: ²20,378,43478.43403,2'2 cm

fAA

cd

sce ===σ


338

Total section to implement:

The total section to implement is:

A=A1+A2=28,74cm² in the inferior part (tensile reinforcement)

A’=3,20cm² in the superior part (compressed reinforcement)

4.7.3 Results sheet





340

5.1 Test No. 01 – EC3: 2D frame Design

5.1.1 Data

Calculation model: 2D metallic portal frame o Column section: HEA200 o Beam section: IPE240 o Base plates: fixed. o Portal frame width: 10m o Columns height: 3m o Portal frame height at the ridge: 4m

Load case: o Gravitational load: 15kN

Model preview

Material properties

Steel S355: fy = 355N/mm2, ε = (235/fy)0.5 = 0.81


341

Cross sections

1. Beam

EC3 Iy Iz Iw It A Sy Sz Wel,y Wel,z U.M. cm4 cm4 cm6 cm4 cm2 cm2 cm2 cm3 cm3

3892.00 283.6 37.39 12.88 39.12 24.83 19.14 324.00 47.30

2. Column

EC3 Iy Iz Iw It A Sy Sz Wel,y Wel,z U.M. cm4 cm4 cm6 cm4 cm2 cm2 cm2 cm3 cm3

3692.0 1336.0 108.0 20.98 53.88 41.59 18.08 388.6 133.6


342

5.1.2 Axial forces – N

5.1.3 Shear forces – T

5.1.4 Bending moments – M


343

5.1.5 Classification of the beam cross section (IPE 240)

Flange:

( )⇒=⋅>=

− 29.7981.58.9

2/2.6120 ε class 1

Web:

5.047.000.190.012<=−>−=−

⋅⋅

= αψ forfAN

y

Ed

⇒⋅

=<=α

ε3604.6270.30wtc

class 1

1. Reference – the class of cross section = max(class of flange; class of web) = CLASS 1.

2. AD 2010 – CLASS 1.


344

5.1.6 Resistance of the beam cross section (IPE 240)

Compression

The design value of the compression force NEd, should satisfy:

kNNwhereNN

EdRdc

Ed 55.67,00.1,

=≤

The design resistance of cross section for uniform compression Nc,Rd, should be:

00.105.076.1388

55.67

76.138800.1

103551012.39

sec32,1

,

34

0,

0,

≤==⇒

=⋅⋅⋅

=⋅

=

⋅=⇒−

−

Rdc

Ed

M

yRdc

M

yRdc

NN

kNfA

N

fANtioncrossorclassfor

γ

γ

AD 2010:

Shear

The design value of shear force VEd, should satisfy:

kNVwhereVV

EdRdc

Ed 49.64,00.1,

=≤

where Vc,Rc is the design shear resistance:

( )

( )

00.116.029.392

49.64

29.39200.11

3103551014.193/

14.19

40.1614.1922

,

34

0,

2

22

≤==⇒

=⋅⋅⋅⋅

=⋅

=

=

>⇒⋅⋅>⋅⋅++⋅⋅−=

−

Rdpl

Ed

M

yvRdpl

v

wwfwfv

VV

kNfA

V

cmA

cmcmthtrttbAA

γ

η

AD 2010:


345

Bending moment

The design value of the bending moment MEd, should satisfy:

kNmMwhereMM

EdRdc

Ed 91.92,00.1,

=≥

Mc,Rd – the design resistance for bending moment about one principal axis is:

00.171.014.13091.92

14.1301000.1

355060.366

60.366,sec21

,

4,

3

0,

<==⇒

=⋅⋅

=

=−⋅

=

−

Rdc

Ed

Rdc

plM

yplRdc

MM

kNmM

cmWwheretioncrossorclassforfW

Mγ

AD 2010:

Lateral torsional buckling

( )( ) ( ) ( )

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⋅−⋅+

⋅⋅⋅⋅⋅

+⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⋅⋅⋅

⋅= ggz

t

z

w

w

zcr zCzC

IEIGLk

II

kk

LkIECM 2

222

22

21 ππ

( )

( )

( )

01324.060.283

7559.3

00.0

107559.34

046051.0/1008.8

88.12

9877.225

10.500.1

60.283/101.2

642

22

2

27

4

2

24

28

==

=

⋅=−⋅

=

=⋅⋅

⋅⋅⋅⇒

⎪⎭

⎪⎬⎫

⋅=

=

=⋅

⋅⋅⇒

⎪⎪⎭

⎪⎪⎬

⎫

===

=

⋅=

z

w

g

fzw

z

tt

z

w

z

II

mz

cmthI

I

mIEIGLk

mkNG

cmI

kNLkIE

mLkk

cmImkNE

π

π


346

C1 and C2 coefficients:

[ ] kNmNmM cr 07.1545.154075046051.001324.017.22598780.2 ==+⋅⋅⋅=

Slenderness for lateral torsional buckling, LTλ :

( ) ( )acurveforx

MfW

LTLT

cr

yyplLT

72.0919.0

919.01007.154

35506.3664

,

==⇒

=⋅

⋅=

⋅=

λ

λ

Design buckling resistance moment:

00.1981.070.9392.91

70.931000.1

35506.36672.0

,

4

1,,

≤==⇒

=⋅⎟⎠⎞

⎜⎝⎛ ⋅⋅

=⋅⋅= −

Rdb

Ed

M

yyplLTRdb

MM

kNmf

WxMγ

AD 2010:

MM =1

MM ⋅=ψ22010

,51.080.2

2009:

53.065.2

524.091.928

10.5158

521.091.9244.48

44.4891.92

2

1

2

1

22

2

1

ADwithobtainedvaluesthewithcontinue

willncalculatiotheCC

ADObs

CC

MLq

kNmMMkNmMM

⎩⎨⎧

==

⇒

⎩⎨⎧

==

⇒

−=⋅⋅

=⋅⋅

=

−=−=⇒⎭⎬⎫

=⋅=−==

μ

ψψ


347

Buckling resistance:

( )

( )

( )

( )⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≤⋅

⋅+⋅

≤⋅

⋅+⋅

==Δ=Δ⇒==

≤Δ+

⋅+⋅

Δ+⋅+

⋅

≤Δ+

⋅+⋅

Δ+⋅+

⋅

62.600.1

61.600.1

:

00.000.000.0

62.600.1

61.600.1

1

,

,

1

1

,

,

1

,,,

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

EdzRdzRdyNzNy

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

Mx

MkNx

N

Mx

MkNx

N

becomeonsverificatithe

MsiMMeefor

MMM

kMx

MMkNx

N

MMM

kMx

MMkNx

N

γγ

γγ

γγγ

γγγ

Characteristic resistance of the critical cross section

kNmWfM

kNAfN

yplyRky

yRk

14.1301060.366355

76.13881012.393553

,,

2

=⋅⋅=⋅=

=⋅⋅=⋅=

Buckling about axis y-y:

( ) ( )( )

( )

kNN

xN

bcurvexNfA

kNN

cmImkNE

L

IEN

mL

M

RkyRdby

yycr

yy

ycr

y

ycr

yycr

ycr

01.111100.1

76.138880.0

80.0672.005.3074

103551012.39

05.307410.5

1000.3892101.214.3

00.3892/101.2

10.5

1,

34

,

2

882

,

4

28

2,

2

,

,

=⋅=⋅=

=⇒=⋅⋅⋅

=⋅

=

=⋅⋅⋅⋅

=⇒

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=

⋅=

⋅⋅=

=

−

−

γ

λ

π

Buckling about axis z-z:

( ) ( )( )

( )

kNNxN

ccurvexNfA

kNN

cmImkNE

LIEN

mL

M

RkzRdbz

yycr

yy

zcr

z

zcr

zzcr

zcr

314.20800.1

76.138815.0

15.0488.2314.224

103551012.39

314.22410.5

10284101.214.3

00.284/101.2

10.5

1,

34

,

2

882

,

4

28

2,

2

,

,

=⋅=⋅=

=⇒=⋅⋅⋅

=⋅

=

=⋅⋅⋅⋅

=⇒

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=

⋅=

⋅⋅=

=

−

−

γ

λ

π


348

The kyy and kzy coefficients – are calculated with “Alternative method 1”:

`60.01

1

1

1

,

,

z

y

zy

ycr

Ed

zmLTmyzy

yy

ycr

Ed

ymLTmyyy

ww

CNN

CCk

CNN

CCk

⋅⋅⋅−

⋅⋅=

⋅−

⋅⋅=

μ

μ

50.150.1562.130.4790.7350.1131.1

00.3246.366

,

,

,

, =⇒≥===≤=== zzel

zplz

yel

yply w

WW

wandWW

w

732.030114.015.01

30114.01

314.22455.6715.01

314.22455.671

1

1

995.002197.080.01

02197.01

05.307455.6780.01

05.307455.671

1

1

,

,

,

,

=⋅−

−=

⋅−

−=

⋅−

−=

=⋅−

−=

⋅−

−=

⋅−

−=

zcr

Edz

zcr

Ed

z

ycr

Edy

ycr

Ed

y

NNx

NN

NNx

NN

μ

μ

Ncr,T and Ncr,TF :

( ) ( ) ( ) ( )

kNN

NNIII

NNNNII

IN

kNN

mI

mI

mA

mzAIII

LIEIG

IA

N

TFcr

Tcrzcrzy

TcrzcrTcrzcrzy

TFcr

Tcr

w

t

g

szy

Tcr

wt

gTcr

25.199

42

64.1108

107559.3

1088.12

101.39

104.9806

,

,,0

2,,,,

0,

,

68

48

24

4820

,

2

0,

=⇒

⎥⎦

⎤⎢⎣

⎡⋅⋅

+⋅−+−+⋅

+⋅=

=⇒

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⋅=

⋅=

⋅=

⋅=⋅++=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅⋅+⋅⋅=

−

−

−

−

π


349

Calculation of 0λ and zyyymLTmy CCCC ,,, coefficients:

( )

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

≥

⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⋅=

=

⋅+

⋅⋅−+=

⇒

>=⇒

⎟⎠⎞

⎜⎝⎛ −⋅⎟

⎠⎞

⎜⎝⎛ −⋅⋅=⎟

⎟⎠

⎞⎜⎜⎝

⎛−⋅⎟

⎟⎠

⎞⎜⎜⎝

⎛−⋅⋅

00.1

11

11

20.0276.0

25.19955.671

314.22455.67180.220.01120.0

,,

2

0

00

0

44

,,1

TFcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

ε

ε

λ

Cmy,0 :

MM =1

MM ⋅=ψ2

( )

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

==

==

====

≥=−==⋅=

⇒

=⋅⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−

⋅

⋅⋅⋅+=⇒

⋅−=−=

=

−=−=⇒⎭⎬⎫

=⋅=−==

−

714.0;9345.0

;224.1;003.1

00.0;00.0;0486.0/

0997.01;05.4

013.111

1081.481.4

91.92

521.091.9244.48

44.4891.92

1

,

,

,,2

2

0,

2

2

1

zyyy

mLTmy

LTLTMRk

Edpl

y

tLT

yelEd

Edyy

ycr

Ed

Edy

xymy

x

x

CC

CC

dbNNn

IIa

WA

NM

NN

xMLIE

C

mcm

kNmMkNmMM

kNmMM

γ

ε

δπ

δ

ψψ


350

Calculation of kyy and kzy:

67.050.113.160.0

714.01

05.307455.671

732.0224.1003.160.01

1

336.19345.01

05.307455.671

995.0224.1003.11

1

,

,

=⋅⋅⋅−

⋅⋅=⋅⋅⋅−

⋅⋅=

=⋅−

⋅⋅=⋅−

⋅⋅=

z

y

zy

ycr

Ed

zmLTmyzy

yy

ycr

Ed

ymLTmyyy

ww

CNNCCk

CNNCCk

μ

μ


( )

( ) 00.198.0

00.114.13072.0

91.9267.0

00.176.138815.0

55.67:61.6

00.137.1

00.114.13072.0

91.92336.1

00.176.138880.0

55.67:61.6

1

,

,

1

1

,

,

1

<=⋅

⋅+⋅

=⋅

⋅+⋅

>=⋅

⋅+⋅

=⋅

⋅+⋅

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

Mx

MkNx

N

Mx

MkNx

N

γγ

γγ

AD 2010:


351

5.1.7 Classification of the column cross section (HEA 200)

Flange:

( )⇒=⋅>=

− 29.79510

2/5.6200 ε class 1

Web:

5.045.000.192.012

<=−>−=−⋅

⋅= αψ for

fAN

y

Ed

⇒⋅

=<=α

ε368.6415.26wtc

class 1

1. Reference – the class of cross section = max(class of flange; class of web) = CLASS 1.

2. AD 2010 – CLASS 3.


352

5.1.8 Resistance of the column cross section (HEA 200)

Compression

The design value of compression force NEd, should satisfy:

kNNwhereNN

EdRdc

Ed 49.76,00.1,

=≤

The design resistance of cross section for uniform compression Nc,Rd, should be:

00.104.0965.191049.76

965.191000.1

103551083.53

sec32,1

,

34

0,

0,

≤==⇒

=⋅⋅⋅

=⋅

=

⋅=⇒−

−

Rdc

Ed

M

yRdc

M

yRdc

NN

kNfA

N

fANtioncrossorclassfor

γ

γ

AD 2010:

Shear

The design value of shear force VEd, should satisfy:

kNVwhereVV

EdRdc

Ed 59.53,00.1,

=≤

where Vc,Rc is the design shear resistance:

( )( )

00.1145.057.37059.53

57.37000.11

3103551008.183/

08.1822

,

34

0,

2

≤==⇒

=⋅⋅⋅⋅

=⋅

=

=⋅⋅++⋅⋅−=−

Rdpl

Ed

M

yvRdpl

fwfv

VV

kNfA

V

cmtrttbAA

γ

AD 2010:


353

Bending moment

The design value of the bending moment MEd, should satisfy:

kNmMwhereMM

EdRdc

Ed 91.92,00.1,

=≥

Mc,Rd – the design resistance for bending moment about one principal axis is determined as follows:

00.161.047.15291.92

47.1521000.1

355050.429

50.429,1

,

4,

3

0,

<==⇒

=⋅⋅

=

=−⋅

=

−

Rdc

Ed

Rdc

plM

yplRdc

MM

kNmM

cmWwhereclassforfW

Mγ

AD 2010:

Lateral torsional buckling

( )( ) ( ) ( )

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⋅−⋅+

⋅⋅⋅⋅⋅

+⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⋅⋅⋅

⋅= ggz

t

z

w

w

zcr zCzC

IEIGLk

II

kk

LkIECM 2

222

22

21 ππ

( )

( )

( )

2424

642

22

2

27

4

2

24

28

1000.8100.8100.1336108216.10

00.0

108216.104

005510.0/1008.8

98.20

685.3076

00.300.100.1336

/101.2

mcmII

mz

cmthI

I

mIEIGLk

mkNG

cmI

kNLkIE

mLkk

cmImkNE

z

w

g

fzw

z

tt

z

w

z

−⋅==⋅

=

=

⋅=−⋅

=

=⋅⋅

⋅⋅⋅⇒

⎪⎭

⎪⎬⎫

⋅=

=

=⋅

⋅⋅⇒

⎪⎪⎭

⎪⎪⎬

⎫

===

=

⋅=

π

π


354

C1 and C2 coefficients

[ ] kNmM cr 455.992005510.00081.01685.307657.2 =+⋅⋅⋅=

Slenderness for lateral torsional buckling, LTλ :

( ) ( )acurveforx

MfW

LTLT

cr

yyplLT

951.0406.0

406.010455.922

355050.4294

,

==⇒

=⋅

⋅=

⋅=

λ

λ

Design buckling resistance moment:

00.164.084.14491.92

84.1441000.1

35505.42995.0

,

4

1,,

≥==⇒

=⋅⎟⎠⎞

⎜⎝⎛ ⋅⋅

=⋅⋅= −

Rdb

Ed

M

yyplLTRdb

MM

kNmf

WxMγ

AD 2010:

MM =1

MM ⋅= ψ2

⎩⎨⎧

==

⇒

⎩⎨⎧

==

⇒

−=−=⇒⎭⎬⎫

=⋅=−==

00.057.2

2010:

00.057.2

730.091.9287.67

87.6791.92

2

1

2

1

2

1

CC

ADObs

CC

kNmMMkNmMM

ψψ


355


( )

( )

( )

( )⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≤⋅

⋅+⋅

≤⋅

⋅+⋅

==Δ=Δ⇒==

≤Δ+

⋅+⋅

Δ+⋅+

⋅

≤Δ+

⋅+⋅

Δ+⋅+

⋅

62.600.1

61.600.1

:

00.000.000.0

62.600.1

61.600.1

1

,

,

1

1

,

,

1

,,,

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

EdzRdzRdyNzNy

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

Mx

MkNx

N

Mx

MkNx

N

becomeonsverificatithe

MsiMMeefor

MMM

kMx

MMkNx

N

MMM

kMx

MMkNx

N

γγ

γγ

γγγ

γγγ

Characteristic resistance of the critical cross section:

kNmWfM

kNAfN

yplyRky

yRk

4725.152105.42910355

74.19121088.531035563

,,

43

=⋅⋅⋅=⋅=

=⋅⋅⋅=⋅=−

−

Buckling about axis y-y:

( ) ( )( )

( )

kNNxN

ccurvexNfA

kNN

cmImkNE

LIE

N

mL

M

RkyRdby

yycr

yy

ycr

y

ycr

yycr

ycr

96.164400.1

74.191286.0

86.0474.033.8502

103551088.53

33.850200.3

1000.3692101.214.3

00.3692/101.2

00.3

1,

34

,

2

882

,

4

28

2,

2

,

,

=⋅=⋅=

=⇒=⋅⋅⋅

=⋅

=

=⋅⋅⋅⋅

=⇒

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=

⋅=

⋅⋅=

=

−

−

γ

λ

π


356

Buckling about axis z-z:

( ) ( )( )

( )

kNNxN

bcurvexNfA

kNN

cmImkNE

LIEN

mL

M

RkzRdbz

yycr

yy

zcr

z

zcr

zzcr

zcr

3.139600.1

74.191273.0

73.0789.0566.3073

103551088.53

566.307300.3

101336101.214.3

00.1336/101.2

00.3

1,

34

,

2

882

,

4

28

2,

2

,

,

=⋅=⋅=

=⇒=⋅⋅⋅

=⋅

=

=⋅⋅⋅⋅

=⇒

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=

⋅=

⋅⋅=

=

−

−

γ

λ

π

The kyy and kzy coefficients– are calculated with “Alternative method 1”:

z

y

zy

ycr

Ed

zmLTmyzy

yy

ycr

Ed

ymLTmyyy

ww

CNNCCk

CNNCCk

⋅⋅⋅−

⋅⋅=

⋅−

⋅⋅=

60.01

1

1

1

,

,

μ

μ

50.150.1525.160.13380.20350.110.1

60.3885.429

,

,

,

, =⇒≥===≤=== zzel

zplz

yel

yply w

WW

wsiWW

w

975.0024886.073.01

024886.01

566.307349.7673.01

566.307349.761

1

1

998.0008996.086.01

008996.01

33.850249.7686.01

33.850249.761

1

1

,

,

,

,

=⋅−

−=

⋅−

−=

⋅−

−=

=⋅−

−=

⋅−

−=

⋅−

−=

zcr

Edz

zcr

Ed

z

ycr

Edy

ycr

Ed

y

NNx

NN

NNx

NN

μ

μ

Ncr,T:

kNN

mI

mI

mA

mzAIII

LIEIG

IA

N

Tcr

w

t

g

szy

Tcr

wt

gTcr

68.2279

108216.10

1098.20

1088.53

1067.9890

,

68

48

24

4820

,2

2

0,

=⇒

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⋅=

⋅=

⋅=

⋅=⋅++=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅⋅+⋅⋅=

−

−

−

−

π


357

Calculation of 0λ and zyyymLTmy CCCC ,,, coefficients

( )

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

≥

⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅⎟

⎟⎠

⎞⎜⎜⎝

⎛−

⋅=

=

⋅+

⋅⋅−+=

⇒

>=⇒

⎟⎠⎞

⎜⎝⎛ −⋅⎟

⎠⎞

⎜⎝⎛ −⋅⋅=⎟

⎟⎠

⎞⎜⎜⎝

⎛−⋅⎟

⎟⎠

⎞⎜⎜⎝

⎛−⋅⋅

00.1

11

11

20.0316.0

68.227949.761

566.307349.76157.220.01120.0

,,

2

0

00

0

44

,,1

TFcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

Tcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NN

C

ε

ε

λ

Cmy,0 :

MM =1

MM ⋅= ψ2

( )

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

==

==

≥=−===

===⋅=

⇒

=⋅−⋅+⋅+=⇒

−=−=⇒⎭⎬⎫

=⋅=−==

987.0;00.1

;001.1;989.0

0994.01;04.0/

0.0;0.0;104.4

945.033.036.021.079.0

73.091.9285.67

85.6791.92

1

,

,

,0,

2

1

zyyy

mLTmy

y

tLT

MRk

Edpl

LTLTyelEd

Edyy

ycr

Ediimy

CC

CC

II

aNN

n

dbWA

NM

NN

C

kNmMMkNmMM

γ

ε

ψψ

ψψ


358

Calculation of kyy and kzy:

507.05.11.16.0

987.01

33.850249.761

975.0001.1989.06.01

1

997.000.11

33.850249.761

998.0001.1989.01

1

,

,

=⋅⋅⋅−

⋅⋅=⋅⋅⋅−

⋅⋅=

=⋅−

⋅⋅=⋅−

⋅⋅=

z

y

zy

ycr

Ed

zmLTmyzy

yy

ycr

Ed

ymLTmyyy

ww

CNNCCk

CNNCCk

μ

μ


( )

( ) 0.138.0

00.14725.15295.0

91.92507.0

00.174.191273.0

49.76:61.6

0.168.0

00.14725.15295.0

91.92997.0

00.174.191286.0

49.76:61.6

1

,

,

1

1

,

,

1

<=⋅

⋅+⋅

=⋅

⋅+⋅

<=⋅

⋅+⋅

=⋅

⋅+⋅

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

Mx

MkNx

N

Mx

MkNx

N

γγ

γγ

AD 2010:


359

5.2 Example 1 - Class of cross section (EC3)

5.2.1 Overview Determinate the class of the bellow cross section subjected to uniform compression. The calculation is performed on EC3.

Material properties

Steel S355: fy = 355N/mm2, ε = (235/fy)0.5 = 0.81

Cross section properties

A = 174cm2;

Iy = 1.179 · 105 cm4;

Iz = 2.145 · 104 cm4;

Iw = 8.93 · 106 cm6

Wy,el = 2885cm3;

Wy,el = 5321cm3

Cross section stresses


360

Top flange

The compressed top flange is composed by two plates perpendicular on the web, with uniform compression stresses.

Class 4

Bottom flange

Class 4

34.111467.1318

2/)8500(=⋅>=

−= εtc

my case

34.111417.1212

2/)8300(=⋅>=

−= εtc


361

Web

The Web is a single plate, with uniform compression stresses.

Class 4

The class of cross section = max (class top flange; class bottom flange; class web) = CLASS 4.

In AD : The class of cross section -- CLASS 4.

02.3442758

600=⋅>== ε

tc

my case


362


5.3.1 Overview Determinate the class of the bellow cross section subjected to strong axis bending. The calculation is performed on EC3.

Material properties

Steel S355: fy = 355N/mm2, ε = (235/fy)0.5 = 0.81

Cross section properties

A = 174cm2;

Iy = 1.179 · 105 cm4;

Iz = 2.145 · 104 cm4;

Iw = 8.93 · 106 cm6

Wy,el = 2885cm3;

Wy,el = 5321cm3



363

Top flange

The compressed top flange is composed by two plates perpendicular on the web, with uniform compression stresses.

Class 4

my case

34.111467.1318

2/)8500(=⋅>=

−= εtc


364

Web

The class of cross section = max (class top flange; class web) = CLASS 4. In AD : The class of cross section -- CLASS 4.

Class 1

my case

7.8536758

6005.034.0

188.188.1 12

=⋅

<==<=

−<−=⇒⋅−=

αεα

ψσσ

tc


365


5.4.1 Overview Determinate the class of the bellow cross section subjected to strong axis bending. The calculation is performed on EC3.

If the cross section is a symmetric I:


Top flange

Class 4

34.111467.13

182/)8500(

=⋅>=−

= εtc


366

Web

Web it’s a single plate, with bending stresses.

OBS: The class of cross section = max (class top flange; class web) = CLASS 4.

In AD : The class of cross section -- CLASS 4.

44.100124758

60023.67835.0

100.1 12

=⋅<==<=⋅=

−=⇒⋅−=

εεα

ψσσ

tc

my case

Class 3


367

5.5 Example 4: Tension column – design value of the resistance (EC3)

Determine the design value of the resistance to tension force, knowing the following:

Material properties

Steel S235: fy = 235N/mm2; fu = 360N/mm2;

Cross section properties (IPE 200)

A = 28.48cm2;

Iy = 1943 cm4;

Iz = 142.40 cm4

Design value of resistance to normal force:

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

==⋅⋅

=⋅⋅

==⋅

=⋅

= kNkNdaNfA

kNdaNfA

N

M

unet

M

y

Rdt 28.66920.73873820

25.1360048.289.09.0

28.6696692800.1

235048.28

min

2

0,

γ

γ

In AD 2010 => Nt,Rd = 669.28kN


368

5.6 Example 5: Tension column – design value of the resistance (EC3)

Determine the design value of the resistance to tension force, knowing the following:

Material properties

Steel S235: fy = 235N/mm2; fu = 360N/mm2;

Cross section properties (D20/1.5)

A = 87.18cm2;

Iy = 3754.15 cm4;

Iz = 3754.15 cm4;

Design value of resistance to normal force:

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

==⋅⋅

=⋅⋅

==⋅

=⋅

= kNkNdaNfA

kNdaNfA

N

M

unet

M

y

Rdt 73.20487.2259225970

25.1360018.879.09.0

73.204820487300.1

235018.87

min

2

0,

γ

γ

In AD 2010 => Nt,Rd = 2048.71kN

031009-0409-0914

Canada GRAITEC Inc. 183, St. Charles St. W. Suite 300 Longueuil (Québec) Canada J4H1C8 Tel. (450) 674-0657 Fax (450) 674-0665 Hotline (450) 674-0657 Toll free 1-800-724-5678 Web www.graitec.com/en Email [email protected]

France GRAITEC France Sarl 17 Burospace 91573 Bièvres Cedex Tel. 33 (0)1 69 85 56 22 Fax 33 (0)1 69 85 33 70 Web http://www.graitec.com/Fr/ Email [email protected]

Germany, Switzerland, Austria GRAITEC GmbH Centroallee 263a D-46047 Oberhausen Germany Tel. +49-(0) 208 / 62188-0 Fax +49-(0) 208 / 62188-29 Web http://www.graitec.com/Ge/ Email [email protected]

Czech Republic and Slovakia AB Studio spol. s r.o. Jeremenkova 90a 140 00 PRAHA 4 Tel. +420/244 016 055 Fax +420/244 016 088 Hotline +420/244 016 050 Web http://www.abstudio.cz/ Email [email protected]

United Kingdom GRAITEC UK Ltd. The Old Forge Suth Road Weybridge Surrey KT13 9DZ Tel. +44 (0)1932 858516 Fax +44 (0)1932 859099 Email [email protected]

Russia GRAITEC CJSC Locomotivny Proezd 21, Build. 5, Office 503 Moscow 127238 - Russia Tel. +7(495) 225-13-65 Fax. +7(495) 488-67-81 Email [email protected]

Romania GRAITEC Roumanie SRL Str. Samuil Vulcan, Nr. 10 Sector 5 Bucureşti, Romania Tel. +40 (21) 410.01.19 Fax +40 (21) 410.01.24 Mobile 0729 002 107 Web http://www.graitec.com/Ro/ Email [email protected]

Advance Design 2010 - Validation Guide

Documents