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Table of ContentsPart A
1. INTRODUCTION TO PLATE AND SHELL ELEMENT 22. ASSIGNMENT 1 SIMPLY SUPPORTED ONE-WAY SLABS 63. ASSIGNMENT 2 TWO-WAY SLABS 114. ASSIGNMENT 3 CANTILEVER BEAM 155. ASSIGNMENT 4 WATER TANKS 17
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global or local directions.
Linearly varying pressure on element surface in local directions.
Temperature load due to uniform increase or decrease of temperature.
Temperature load due to difference in temperature between top and
bottom surfaces of the element.
The distinguishing features of this finite element are:
Displacement compatibility between the plane stress component of one
element and the plate bending component of an adjacent element which is
at an angle to the first is achieved by the elements. This compatibility
requirement is usually ignored in most flat shell/plate elements.
The out of plane rotational stiffness from the plane stress portion of each
element is usefully incorporated and not treated as a dummy as is usually
done in most commonly available commercial software.
Despite the incorporation of the rotational stiffness mentioned previously,
the elements satisfy the patch test absolutely.
These elements are available as triangles and quadrilaterals, with corner
nodes only, with each node having six degrees of freedom.
These elements are the simplest forms of flat shell/plate elements
possible with corner nodes only and six degrees of freedom per node. Yet
solutions to sample problems converge rapidly to accurate answers even
with a large mesh size.
These elements may be connected to plane/space frame members with
full displacement compatibility. No additional restraints/releases are
required.
Out of plane shear strain energy is incorporated in the formulation of the
plate bending component. As a result, the elements respond to Poisson
boundary conditions which are considered to be more accurate than the
customary Kirchoff boundary conditions.
The plate bending portion can handle thick and thin plates, thus extending
the usefulness of the plate elements into a multiplicity of problems. In
addition, the thickness of the plate is taken into consideration in
calculating the out of plane shear.
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The plane stress triangle behaves almost on par with the well known linear
stress triangle. The triangles of most similar flat shell elements incorporate
the constant stress triangle which has very slow rates of convergence.
Thus the triangular shell element is very useful in problems with double
curvature where the quadrilateral element may not be suitable.
Stress retrieval at nodes and at any point within the element.
Plate Element Local Coordinate System
The orientation of local coordinates is determined as follows:
The vector pointing from I to J is defined to be parallel to the local x- axis.
The cross-product of vectors IJ and IK defines a vector parallel to the local
z-axis, i.e., z = IJ x IK.
The cross-product of vectors z and x defines a vector parallel to the local
y- axis, i.e., y = z x x.
The origin of the axes is at the center (average) of the 4 joint locations (3
joint locations for a triangle).
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Results:
Elementsizemxm
No. ofElements
BMkN.m
%error
inBM
Thicknessm
1.00 18 10.20-
10.29 0.150.50 72 10.93 -2.93 0.150.30 200 11.40 1.32 0.150.25 288 11.50 2.17 0.150.20 450 11.60 3.02 0.150.15 800 11.60 3.02 0.150.10 1800 11.60 3.02 0.15
Convergence
-12.00
-10.00
-8.00
-6.00
-4.00
-2.00
0.00
2.00
4.00
18 72 200 288 450 800 1800
No. of eleme nts
%Error
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Exact solution = w*l^2/8 = 11.25 kN.m
Conclusion :
From the above results we can conclude that the bending moments in the
slab converges to exact solution by mesh refinement .
For the element size 0.5m x 0.5m we get minimum +positive error and for
element size 0.3m x 0.3m we get minimum negative error.
Refinement of mesh further does help, and the negative error goes on
increasing till the element size of 0.1m x 0.1m.
PART 2Here we will model the supporting edge with beam elements instead of pinned
supports of different sizes and get the results of bending moments for both slab
and beam and check with exact classical solution.
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Results:
Beamsizemmxmm
BM inslabkN.m
BM inbeamkN.m
%errorin BM
300x300 13.40 40.88 -65.12300x450 12.00 55.45 -21.73300x600 11.61 60.81 -11.00300x750 11.46 63.00 -7.14300x900 11.39 64.03 -5.42
300x1200 11.32 64.88 -4.04
Convergence
-70.00
-60.00
-50.00
-40.00
-30.00
-20.00
-10.00
0.00
300x300 300x450 300x600 300x750 300x900 300x1200
Size of Beam
%Error
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Conclusion :
From the above results it can be concluded that, increasing the size ofedge beams the result converges to exact solution.
Hence relative stiffness of beams and slab is governing criteria forconvergence of the bending moments for both, beam and slab.
If the sizes of supporting beams are lesser than required, there is no slab-beam effect and act together as slab elements with just edged havingmore stiffness.
Again if the sizes of edge beams are different in sizes, the results arehigher for stiffer beam and carry more bending moments than the onewhich is comparatively weaker.
300x900
BM = 69.5kN.m
300x600
BM = 55.6kN.m
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3. AASSSSIIGGNNMMEENNTT 22 TTWWOO--WWAAYY SSLLAABBSS
Problem Statement :
1. Comparative study of two-way slabs bending moments using IS-456guidelines (Table 26) and finite element analysis.
2. Convergence of results for slab with line loads.3. Effect on panel under consideration due to change in size of adjacent
panel.
Data :
Slab Panel : 5m x 6m
Loading : 10kN/m2
Slab : 150mm
Boundary condition : One short edge continuous (case 8)
For the given problem we will start the meshing of panel from 1m x 1m and use
mesh refinement for convergence.
Analysis will be done using staad-prov8i software.
Results
Elementsizemxm
No. ofElements
BM MX+ve
BM MY+ve
BM MX-ve
BM MY-ve
% errorBM
X+ve
% errorBM
Y+ve
% errorBM Y-
ve
1.00 30 11.20 10.20 0.00 12.10 -31.70 -5.39 -17.77
0.50 120 11.30 10.30 0.00 17.40 -30.53 -4.37 18.10
0.25 480 11.40 10.40 0.00 20.90 -29.39 -3.37 31.82
0.20 750 11.50 10.40 0.00 21.70 -28.26 -3.37 34.33
0.10 3000 11.50 10.45 0.00 23.20 -28.26 -2.87 38.58
IS-456 14.75 10.75 0.00 14.25
Convergence
-60.00
-40.00
-20.00
0.00
20.00
40.00
60.00
0 500 1000 1500 2000 2500 3000 3500
No. of Elements
%Erro MX+ve
My +ve
MY -ve
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Conclusion :
For the given case, the mesh refinement converges the positive X and Y
moments for size 0.25m x 0.25m.
From the results it can be concluded that the positive moments from IS-
456 table is higher than results from finite elements analysis after
convergence.
For negative Y moments the results crosses the exact solution for element
size between 1.0m x 1.0m to 0.5m x 0.5m.
So for negative moments further mesh refinement add to positive error
and the results of finite element analysis are more than the one calculated
from IS-456, which is exactly opposite of the case with positive moments.
Also on changing the size of adjacent panel by half of the original, there is
decrease in negative moments at support and increase in mid span
moments.
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4. AASSSSIIGGNNMMEENNTT 33 CCAANNTTIILLEEVVEERR BBEEAAMM
Problem Statement :
Convergence of results for cantilever beam with concentrated load at end.
Data :
Beam span : 3m
Loading : 10kN at free end
Size : 300x600mm
For the given problem we will start meshing size from 0.2m x 0.2m and furtherrefinement will be done to get the convergence.
Results :
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Element sizemxm
No. ofElements
BMkN.m
%errorBM
0.2mx0.2m 45 25.8 -16.500.15mx0.15m 80 27.4 -9.550.10mx0.1m 180 28.6 -4.730..05mx0.5m 720 29.511 -1.66
Convergence
-20.00
-15.00
-10.00
-5.00
0.00
0 100 200 300 400 500 600 700 800
No. of Elements
%Erro
BM
Conclusion :
For the given problem we can conclude that the result for bendingmoments converges to exact solution for the plate size of 0.05m x 0.05m.
Also the results vary for different width of beam, here we tried to convergethe results for beam size of 300x600mm
The results for width other than 300mm, the bending moments aredifferent for same number of elements.
Hence unlike classical solution, where we dont take the effect of width incalculation of bending moments, in finite element analysis the width of
beam influences the bending moments.
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5. AASSSSIIGGNNMMEENNTT 44 WWAATTEERR TTAANNKKSS
Problem Statement :
Comparison of .moments, shear and hoop forces for water tanks using FEMand IS-3370.a) Rectangular tanksb) Cylindrical tanks
Data :
a) Rectangular tanks
Tank size : 3.75m x 5.0m hieght
Loading : 50kN/m
Wall: 150mm
For the given problem we will use the meshing of panel 0.25m x 0.25m as we
have concluded this size from above examples
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Vertical Table 1
y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)
b/H x/H MY staad %error MY staad %error MY staad %error
0.75 0 0.00 0.49 100.00 0.00 0.08 100.00 0.00 -0.36 100.00
1/4 1.25 1.65 24.24 0.00 0.42 100.00 -2.50 -1.84 -35.87
1/2 6.25 4.80 -30.21 2.50 1.68 -48.81 -5.00 -3.59 -39.31
3/4 11.25 10.00 -12.50 6.25 4.75 -31.58 -6.25 -3.90 -60.46
1 0.00 3.55 100.00 0.00 2.12 100.00 0.00 -0.72 100.00
Horizontal Table 1
y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)
b/H x/H MX staad %error MX staad %error MX staad %error
0.75 0 0.00 1.02 100.00 0.00 0.15 100.00 0.00 -1.23 100.00
1/4 7.50 6.68 -12.28 2.50 1.06 -135.85 -15.00 -10.91 -37.45
1/2 13.75 12.90 -6.59 3.75 2.48 -51.27 -27.50 -21.53 -27.76
3/4 13.75 14.02 1.93 6.25 3.93 -59.03 -31.25 -24.42 -27.951 0.00 2.54 100.00 0.00 1.09 100.00 0.00 -3.52 100.00
Vertical Table 2
y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)
b/H x/H MY staad %error MY staad %error MY staad %error
0.75 0 0.00 -0.15 100.00 0.00 -0.11 100.00 0.00 -0.59 100.00
1/4 1.25 1.00 -25.00 0.00 -0.27 100.00 -3.75 -2.21 -69.68
1/2 6.25 4.34 -44.11 2.50 1.42 -75.56 -5.00 -3.64 -37.36
3/4 12.50 14.02 10.83 7.50 4.71 -59.40 -6.25 -3.89 -60.59
1 0.00 3.55 100.00 0.00 2.12 100.00 0.00 -0.72 100.00
Horizontal Table 2y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)
b/H x/H MX staad %error MX staad %error MX staad %error
0.75 0 6.25 5.35 -16.82 1.25 0.75 -66.67 -10.00 -6.42 -55.69
1/4 10.00 8.11 -23.30 2.50 0.89 -179.64 -16.25 -12.86 -26.38
1/2 13.75 13.13 -4.72 5.00 2.39 -109.21 -27.50 -21.76 -26.36
3/4 15.00 9.89 -51.67 5.00 3.91 -27.91 -32.50 -24.45 -32.92
1 0.00 2.54 100.00 0.00 1.09 100.00 0.00 -3.52 100.00
Vertical Table 3
y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)
b/H x/H MY staad %error MY staad %error MY staad %error
0.75 0 0.00 -0.12 100.00 0.00 -0.09 100.00 0.00 -0.61 100.00
1/4 1.25 0.91 -36.91 0.00 -0.05 100.00 -2.50 -2.51 0.40
1/2 6.25 5.13 -21.83 2.50 2.00 -25.00 -3.75 -3.26 -15.03
3/4 8.75 7.76 -12.76 3.75 4.00 6.25 -3.75 -2.77 -35.38
1 -30.00 -20.67 -45.14 -18.75 -11.91 -57.43 0.00 -1.07 100.00
Horizontal Table 3
y = 0 (Centre) y = b/4 (Quarter ) y = b/2 (at wall)
b/H x/H MX staad %error MX staad %error MX staad %error
0.75 0 5.00 5.37 6.82 1.25 0.75 -66.67 -8.75 -6.57 -33.18
1/4 10.00 7.87 -27.10 2.50 0.97 -157.73 -13.75 -12.54 -9.65
1/2 12.50 11.74 -6.48 3.75 2.47 -51.82 -21.25 -19.79 -7.39
3/4 8.75 9.72 9.98 3.75 3.19 -17.55 -16.25 -17.81 8.75
1 -6.25 -3.18 -96.36 -3.75 -1.85 -103.14 0.00 -1.07 100.00
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Conclusion :
For the water tank wall panels, with the given panel ratio and wallthickness the results are having 50% error
This error may be due to the fact that IS-3370 coefficients does not takeinto effect the thickness of wall panel of tank apart from the width andheight.
b) Cylindrical tanks
Tank size : 5m (Height) x 6m (Radius)
Loading : 50kN/m
Wall: 150mm
Boundary conditions : Top of tank Free and bottom hinged
IS 3370 : Table 12,13 and 14.
Results
Hoop Tension IS 3370 Moment My IS 3370
7.95 -0.03 -0.01 0.00
37.05 29.38 -0.02 0.00
66.60 59.10 -0.06 0.00
97.65 89.75 -0.13 -0.13
131.40 122.55 -0.18 -0.13
168.60 159.47 -0.04 -0.12
206.10 197.78 0.56 0.03
230.40 228.12 1.89 1.53229.35 225.07 2.78 3.29
119.85 153.30 3.66 4.17
42.75 0.00 1.57 0.00
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Vertical Moments
0
1
2
3
4
5
6
-1.00 0.00 1.00 2.00 3.00 4.00 5.00
kNm
mt.
IS 3370
Staad
Hoop Tension
0
1
2
3
4
5
6
-100.00 0.00 100.00 200.00 300.00
kN
mt.
IS 3370
Staad
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Conclusion:
For the given case, the results of staad pro and IS 3370 coefficients
are nearly same with negligible, 1% error in case of hoop tension and
about 13% error in moments.