Metal Structures Design Project I Steel truss – examples of calculation
Metal Structures
Design Project I
Steel truss – examples of calculation
Today will be presented 9 examples of calculations various sub-parts of steel truss:
Ist example: roofing selection;
IInd example: purlin and wall girt in bi-axial bending;
IIIrd example: roofing column;
IVth example: truss member;
Vth example: welds;
VIth example: resistance of node;
VIIth example: field splice;
VIIIth example: support on masonry structure;
IXth example: roof bracing.
Theoretical explanations of above problems will be presented during lectures, here are
way of calculation only.
• tmin = 11 cm
• Loads I (dead-weight + snow + wind
pressure + imposed load) = 5,40 kN / m2
• Loads II (wind suction) = 0,50 kN / m2
• Distance between purlins = 2,10 m
• Fire resistance EI30
Ist example of calculations – roofing selection
Photo: steelprofil.eu
Tables propose 10 and 12 cm thick sandwich panels. With the minimum required thickness
of 11 cm, the second should be chosen.
Photo: pruszyński.com.pl
But such type of panels not satisfied condition for loads:
Epressure / Rpressure = 5,40 / 5,27 = 1,025 > 1,0
Esuction / Rsuction = 0,50 / 4,22 = 0,118 < 1,0
Next thickness, 16 / 20 cm: Photo: pruszyński.com.pl
Epressure / Rpressure = 5,40 / 5,55 = 0,973 < 1,0
Esuction / Rsuction = 0,50 / 4,23 = 0,118 < 1,0
Photo: pruszyński.com.pl
According to the manufacturer's declaration,
fire resistance is sufficient (min EI 30):
Sandwich panel PWD PIR 160 (14,20 kg / m2)
is applied to structure. Dead weight:
gk = 0,139 kN / m2
g = 0,188 kN / m2
s = 4,00 [kN / m2];
g = 0,40 [kN / m2];
w = 0,50 [kN / m2];
IPE 300
S 235
a = 5o
d = 2,50 m → d1 = d / cos a = 2,510 m
Photo: M. Łubiński, W. Żółtowski, Konstrukcje
Metalowe t. II, Arkady, Warszawa 2004
IInd example of calculations – purlin in bi-axial bending
d
d1
L
d1
y
z
qz
qy d1 = 2,510 m
L = 6,000 m
Photo: Author
IPE 300:
Jy = 8356,0 cm4
Jz = 603,8 cm4
Wpl, y = 628,4 cm3
Wpl, z = 125,2 cm3
h = 300 mm
d = 150 mm
tw = 7,1 mm
tf = 10,7 mm
AV, y = 32,1 cm2
AV, z = 21,3 cm2
Photo: Author
qH
Photo: Author
qV
s → qH, qV ; l
g → qH, qV ; l1
w → qy, qz ; l1
d1
d
qH(s) = 0,000 kN / m
qV(s) = s d = 10,000 kN / m
qH(g) = 0,000 kN / m
qV(g) = g d1 = 1,004 kN / m
qy(w) = 0,000 kN / m
qz(w) = w d1 = 1,255 kN / m
d1
d
l1 = 2 d1 / 2 = d1
l = 2 d / 2 = d
qH
Photo: Author
qV
qH → qy , qz
qV → qy , qz
qy(qH) = qH cos a
qz(qH) = - qH sin a
qy(qV) = qV sin a
qz(qV) = qV cos a
qy[qH(s)] = qH(s) cos a = 0,000 kN / m
qz[qH(s)] = - qH(s) sin a = 0,000 kN / m
qy[qV(s)] = qV(s) sin a = 0,872 kN / m
qz[qV(s)] = qV(s) cos a = 9,962 kN / m
qy[qH(g)] = qH(g) cos a = 0,000 kN / m
qz[qH(g)] = - qH(g) sin a = 0,000 kN / m
qy[qV(g)] = qV(g) sin a = 0,088 kN / m
qz[qV(g)] = qV(g) cos a = 1,000 kN / m
y Photo: Author
z
qz = 12,217 kN/m
qy = 0,960 kN/m
qy = qy[qH(s)] + qy[qV(s)] + qy[qH(g)] + qy[qV(g)] + qy(w) =
= 0,000 + 0,872 + 0,000 + 0,088 + 0,000 = 0,960 kN / m
qz = qz[qH(s)] + qz[qV(s)] + qz[qH(g)] + qz[qV(g)] + qz(w) =
= 0,000 + 9,962 + 0,000 + 1,000 + 1,255 = 12,217 kN / m
MEd, y = qz L2 / 8 = 54,977 kNm
MEd, z = qy L2 / 8 = 4,320 kNm
VEd, y = qy L / 2 = 2,880 kN
VEd, z = qz L / 2 = 36,651 kN
Resistances of cross-section:
MRd, y = fy Wpl, y / gM0 = 235 [MPa] ∙ 628,4 [cm3] / 1,00 = 147,674 kNm
MRd, z = fy Wpl, z / gM0 = 235 [MPa] ∙ 125,2 [cm3] / 1,00 = 29,422 kNm
VRd, y = fy AV, y / (gM0 √3) = 235 [MPa] ∙ 32,1 [cm2] / (1,732 ∙ 1,000) = 435,524 kN
VRd, z = fy AV, z / (gM0 √3) = 235 [MPa] ∙ 21,3 [cm2] / (1,732 ∙ 1,000) = 288,993 kN
Checking resistances
E / R ≤ 1,0
MEd, y / MRd, y ≤ 1,0 → 54,977 / 147,674 = 0,372
MEd, z / MRd, z ≤ 1,0 → 4,320 / 29,422 = 0,147
VEd, y / VRd, y ≤ 1,0 → 2,880 / 435,524 = 0,007
VEd, z / VRd, z ≤ 1,0 → 36,651 / 288,993 = 0,127
Bi-axial bending of I-beam (EN 1993-1-1 (6.41)):
(MEd, y / MRd, y)2 + MEd, z / MRd, z ≤ 1,0 → 0,3722 + 0,147 = 0,285
Checking stability:
Lateral buckling can occurs as effect of bending. For Your range of design project you
can asume (not completely correct), that sanwich panels protect purlins from instability.
Checking deflections:
Dy = 5 qy L4 / (384 E Jz) =
= 5 ∙ 0,960 [kN/m] ∙ 6,04 [m4] / (384 ∙ 210 [GPa] ∙ 603,8 [cm4]) = 12,8 mm
Dz = 5 qz L4 / (584 E Jy) =
= 5 ∙ 12,217 [kN/m] ∙ 6,04 [m4] / (384 ∙ 210 [GPa] ∙ 8356 [cm4]) = 11,7 mm
D = √(Dz2 + Dz
2) = 17,3 mm
Dacc = EN 1993-1-1 N.A. 22 = L / 200 = 30 mm
D / Dacc = 0,578
Photo: Author
Dz
Dy
D
Conclusions:
E MEd, y MEd, z VEd, y VEd, z MEd, y + MEd, z LT D
E / R 0,372 0,147 0,007 0,127 0,285 0,578
1. For one-span beams longer than ~ 5,0 m the most important requirement concerns
deformations, not cross-sectional forces, nearly always.
2. Each results satisfy fundamental requirement E / R ≤ 1,0. But the highest value for analysed
case (0,578) is far from upper limit (1,000) – this means, cross-section is too big (too expensive)
for situation. It’s somethink like situation: total cost of structure = 10 M€, but only 5,780 M€ is
justified, the rest (4,220 M€) is the result of the designer's inability.
You should check smaller one I-beam (IPE 270? IPE 240?). You no need recalculated values of
loads: influence of change dead-weight of purlin for total loads (snow, wind…) is neglegible.
3. This purlin will be recalculated after analysis of roof bracings. Axial force in purlin occurs as
the effect of coopreation bracings-purlins. Max effort could be bigger that now. Because of this,
efforts for smaller one purlin (IPE 270? IPE 240?) should be:
• E / R ≤ 1,0 for D;
• E / R ≤ 0,8 all the rest conditions.
Wind action
on fron wall
Photo: Author
Roofing Front wall girts
Roofing
column
Purlin Top chord
Horizontal bracings
IIIrd example of calculations – roofing column
1,0 m
Photo: Author
1,0 m
1,0 m
1,0 m
0,80 kN
0,80 kN
0,80 kN
4,80 kN
4,80 kN
4,80 kN
2,40 kN
7,20 kN
7,20 kN
7,20 kNm
7,20 kNm
9,60 kNm MEd
VEd
NEd 7,20 kN
7,20 kN
2,40 kN
2,40 kN
0,80 kN
1,60 kN
2,40 kN
Hot-rolled RHS 80x8 (80 mm side, 8 mm thickness):
A = 22,4 cm2
Jy = Jz = 189,0 cm4
Wel, y = Wel, z = 47,3 cm3
Wpl, y = Wpl, z = 59,5 cm3
AV, y = AV, z = 12,8 cm2
iy = iz = 2,91 cm
S 235 → fy = 235 MPa
e = √ (235 / fy) = 1,0
l cr =
l =
4,0
m
S 235 + RHS hot rolled → buckling curve a → a = 0,21
(EN 1993-1-1 tab. 6.1, 6.2)
4,80 kN – from wind action
0,80 kN – from deadweigth
Resistances of cross-section:
NRd = fy A / gM0 = 235 [MPa] ∙ 22,4 [cm2] / 1,00 = 526,400 kN
MRd, y = fy Wpl, y / gM0 = 235 [MPa] ∙ 59,5 [cm3] / 1,00 = 13,983 kNm
VRd, y = fy AV, y / (gM0 √3) = 235 [MPa] ∙ 12,8 [cm2] / (1,732 ∙ 1,000) = 173,672 kN
Checking resistances
E / R ≤ 1,0
NEd / NRd ≤ 1,0 → 2,400 / 526,400 = 0,005
MEd, y / MRd, y ≤ 1,0 → 9,600 / 13,983 = 0,687
VEd, y / VRd, y ≤ 1,0 → 7,200 / 173,672 = 0,041
Checking interaction NEd ↔ MEd, y for RHS (EN 1993-1-1 (6.39)):
n = NEd / NRd = 0,005
aw = min [0,5 ; (A – 2 b t) / A] = min [0,5 ; (22,4 – 2 ∙ 8,0 ∙ 0,8) / 22,4] = 0,429
MN, Rd, y = min [MRd, y ; MRd, y (1 – n) / (1 – 0,5 aw)] = min[13,983 ; 17,712] = 13,983 kNm
MEd, y / MN, Rd, y ≤ 1,0 → 9,600 / 13,983 = 0,687
Checking stability (EN 1993-1-1 6.3.1.2):
l = (lcr / i) / (93,9 e) = (4,0 [m] / 2,91 [cm]) / 93,3 = 1,463
F = [1 + a (l - 0,2) + l2] / 2 = [1 + 0,21 (1,463 - 0,2) + 1,4632] / 2 = 1,703
c = min {1 / [F + √ (F2 - l2)] ; 1,0} = min {1 / [1,703 + √ (1,7032 – 1,4632)] ; 1,0} =
= min {0,388 ; 1,0} = 0,388
NEd / (c NRd) = 0,013
Checking stability with interaction NEd ↔ MEd, y for RHS (EN 1993-1-1 NA.20):
NEd / (c NRd) + Cmy MEd, y / (cLT MRd, y) ≤ 1,0 - D0
Latreral buckling is not danger for SHS → cLT = 1,0
Cmy could be taken into consideration as 1,0 (rought approximation)
RHS 80x8 → Ist class of cross-section (Lec. #4) → D0 = 0,1 + 0,2 [ (Wpl, y / Wel, y) - 1) ] =
= 0,1 + 0,2 [(59,5 [cm3] / 47,3 [cm3]) – 1,0 ] = 0,152
NEd / (c NRd) + Cmy MEd, y / (cLT MRd, y) = 0,013 + 0,687
1,0 - D0 = 1,0 – 0,152 = 0,848
0,700 < 0,848
(0,943 < 1,000)
Checking deflections:
Rought approximation: deflection under 3 forces in point ≈ deflection under equivalent
continous load
qeq = 3 F / l = 3 ∙ 4,80 [kN] / 4 [m] = 3,60 [kN / m]
D = 5 qeq L4 / (384 E Jy) =
= 5 ∙ 3,6000 [kN/m] ∙ 4,04 [m4] / (384 ∙ 210 [GPa] ∙ 189 [cm4]) = 31 mm
Dacc = EN 1993-1-1 N.A. 22 = L / 150 = 27 mm
D / Dacc = 1,148
Conclusion:
E NEd MEd, y VEd, y MEd, y + Ned NEd stab. MEd, y + Ned stab. D
E / R 0,005 0,687 0,041 0,687 0,013 0,943 1,148
1. Deflection are bigger than accepted limit. You should analised bigger one cross-section
(RHS 80x10?), or check more accurate value of deflection from FEM calculations.
IVth example of calculations – truss members
Photo: Author
Chord Max tension [kN] Max compression [kN] Cross-section
Top -45,987 228,322 Hot-rolled
CHS 57x10 Bottom -235,621 43,181
Connections with roof bracings S 235 → fy = 235 MPa
e = √ (235 / fy) = 1,0
Hot-rolled CHS 57x10 (57 mm diameter, 10 mm thickness):
A = 14,8 cm2
Jy = Jz = 42,6 cm4
iy = iz = 1,70 cm
S 235 + CHS tor tolled→
buckling curve a → a = 0,21
(EN 1993-1-1 tab. 6.1, 6.2)
L = 24,000 m
Resistances of cross-section:
NRd = fy A / gM0 = 235 [MPa] ∙ 14,8 [cm2] / 1,00 = 347,800 kN
Checking resistances
E / R ≤ 1,0
NEd, top, comp / NRd ≤ 1,0 → 228,322 / 347,800 = 0,656
NEd, top, tens / NRd ≤ 1,0 → 45,957 / 347,800 = 0,132
NEd, bott, comp / NRd ≤ 1,0 → 43,181 / 347,800 = 0,124
NEd, bott, tens / NRd ≤ 1,0 → 235,621 / 347,800 = 0,677
Flexural buckling of chords:
Top chords in compression; buckling in plane: critical length = distance between nodes
Photo: Author
Critical length for compressed chord of truss is depend on plane of analysis and part of
structure. For web members, critical length = distance between nodes.
→ Des #1 / 48
Flexural buckling of chords:
Top chords in compression; buckling out of plane: critical length = distance between
horizontal bracings
Photo: Author
→ Des #1 / 49
Flexural buckling of chords:
Bottom chords in compression; buckling in plane: critical length = distance between
nodes
Photo: Author
→ Des #1 / 50
Flexural buckling of chords:
Bottom chords in compression; buckling out of plane: critical length = distance between
vertical bracings
Photo: Author
→ Des #1 / 51
Checking stability for CHS (EN 1993-1-1 6.3.1.2):
l = (lcr / i) / (93,9 e)
F = [1 + a (l - 0,2) + l2] / 2
Chord lcr in plane [m] lcr out of plane [m]
Top 3,047 2 x 3,047 = 6,094
Bottom 3,047 4 x 3,047 = 12,188
Chord l in plane l out of plane
Top 1,909 3,818
Bottom 1,909 7,635
Photo: Author
Chord F in plane F out of plane
Top 2,502 16,337
Bottom 2,502 30,427
Checking stability for CHS (EN 1993-1-1 6.3.1.2):
c = min {1 / [F + √ (F2 - l2)] ; 1,0}
NEd / (c NRd)
Chord c in plane c out of plane
Top 0,243 0,031
Bottom 0,243 0,017
Chord E / R in plane E / R out of plane
Top 2,700 21,161
Bottom 0,510 7,291
Checking deflection:
Max vertical deformation under loads (FEM calculation): 52 mm
Dacc = EN 1993-1-1 N.A. 22 = L / 250 = 96 mm
D / Dacc = 0,542
Conclusions:
1. Stiffness of bars is too small and make instability. Cross-section should be much more
massive (CHS 219,1 x 6,3 ?).
2. Smaller distances between bracings should be considered (→ smaller lcr → bigger c).
Chord E / R tension E / R compression
Top 0,132 0,656
Bottom 0,677 0,124
Resistance:
Stability:
Chord E / R in plane E / R out of plane
Top 2,700 21,161
Bottom 0,510 7,291
Deflection: 0,542
Vth example of calculations – welds
Photo: Author
S235 → fu = 360 MPa, βw = 0,80
(EN 1993-1-8 tab 4.1)
CHS diameter 51 mm
CHS thickness of flange 3,2 mm
Thickness of weld 3 mm
A = p · [(51 / 2 + 3)2 - (51 / 2)2] = 509 mm2
NEd = 64,73 kN
s = NEd / A = 127,171 MPa
t = 0 MPa
σ┴ = t┴ = σ / √2 = 89,923 MPa
t║ = t = 0,000 MPa
γM2 = 1,25
EN 1993-1-8 (4.1):
Condition 1: √[(σ┴)2 + 3(t║2 + t┴
2)] = 254,342 MPa < fu / (βw γM2) = 360,000 MPa
Condition 2: σ┴ = 89,923 MPa < 0,9 fu / γM2 = 259,200 MPa
Photo: Author
Two types of joints: T and KT will be presented in calculations.
Photo: EN 1993-1-8 fig. 7.1
VIth example of calculations – resistance of node
Photo: Author
Truss
Model 3
Model 2
Model 1a 1b 1c
e = 0
e inside
limits
e outside
limits
satisfy
satisfy
Generally, five various models of truss
structure must be taken into
consideration. The choice of model is
determined by the satisfy or dissatisfy of
many conditions. You don’t check these
conditions for your range of project. You
take into consideration model 1a (ideal
truss).
Six various models of joint’s destruction
must be analised. Not all are equally
important for various shapes of truss
joints. Satisfaction of conditions
presented in EN 1993-1-8 tab. 7.1 allows
taken into consideration 1-2 mechanism
only of destruction for each type of joint.
You don’t check these conditions for
your range of project. You make initial
assumptions that these conditions are
satisfied.
Photo: scielo.br
Photo: offshoremechanics.asmedigitalcollection.asme.org
Photo: EN 1993-1-8 fig 7.3, 7.4
General formula of resistance for joints in truss chord – web members as CHS-CHS
(EN 1993-1-8 (7.3)):
Ni, Ed / Ni, Rd + (Mip, i, Ed / Mip, i, Rd)2 + Mop, i, Ed / Mop, i, Rd ≤ 1,0
Ni, Ed – force in web member;
Ni, Rd – resistance of joint under force;
Mip, i, Ed , Mop, i, Rd - local bending moments in joints, in plane (ip) and out of plane (op);
effects of eccenticities e 0 (→ #t / 34). In your range of project, Mip, i, Ed = Mop, i, Rd = 0.
Formulas for Ni, Rd for truss CHS-CHS are presented in EN 1993-1-8 tab. 7.2, 7.3, 7.4, 7.5,
7.6, 7,7. Joints T and KT are listed in:
• 7.2;
• 7.5 – but this table is important for joints under bending, not important for your range of
project;
• 7.6.
Photo: EN 1993-1-8 tab. 7.2
Photo: EN 1993-1-8 tab. 7.2
Photo: EN 1993-1-8 tab. 7.6
T joint, condition 1
T joint, condition 2
KT joint, condition 1
KT joint, condition 2
T joint, condition 1:
Photo: Author
76,1 x 3,2
NP, Ed = more important (Nleft ; Nright)
A0 = 7,33 cm2
t0 = 3,2 mm
d0 = 76,1 mm
51 x 3,2
N1, Ed
d1 = 51 mm
EN 1993-1-8 tab. 7.2:
N1, Rd = g0,2 kp fy0 t02 (2,8 + 14,2 b) / (gM5 sin q)
kp =
for compressive NP, Ed :
min[ 1,0 ; 1,0 – 0,3 np (1 + np)]
for tensile NP, Ed : 1,0
EN 1993-1-8 1.5:
g = d0 / 2 t0
b = d1 / d0
np = NP, Ed / (A0 fy0 gM5)
Case NP, Ed [kN] N1, Ed [kN] np kp g b N1, Rd [kN] N1, Ed / N1, Rd
Comp. 37,584 -10,752 0,218 0,920 11,891 1,492 87,129 0,123
Tens. -192,754 58,235 -1,119 1,000 94,706 0,615
S235
q = 90o
T joint, condition 2:
Condition 2 is important for initial assumption only:
d1 ≤ d0 – 2 t0
d1 = 51 mm
t0 = 3,2 mm
d0 = 76,1 mm
d0 – 2 t0 = 69,7 mm
Condition important for checking
N1, Rd = fy0 t0 p d1 (1 + sin q) / [ 2 (√3) gM5 (sin q)2] = 69,563 kN
Case NP, Ed [kN] N1, Ed [kN] N1, Rd [kN] N1, Ed / N1, Rd
Comp. 37,584 -10,752 69,563 0,155
Tens. -192,754 58,235 0,837
KT joint, condition 1:
Condition 1 is important for initial assumption only:
di ≤ d0 – 2 t0
d1 = d2 = d3 = 51 mm
t0 = 3,2 mm
d0 = 76,1 mm
d0 – 2 t0 = 69,7 mm
Condition important for checking
76,1 x 3,2
NP, Ed = more important (Nleft ; Nright)
A0 = 7,33 cm2
t0 = 3,2 mm
d0 = 76,1 mm
51 x 3,2
N1, Ed
d1 = 51 mm
51 x 3,2
N3, Ed
d1 = 51 mm
51 x 3,2
N2, Ed
d1 = 51 mm
Photo: Author q1 = 45o
q2 = 90o
q3 = 45o
Case NP, Ed [kN] N1, Ed [kN] N2, Ed [kN] N3, Ed [kN]
Comp. 42,322 14,008 -9,221 -17,955
Tens. -207,155 -78,252 71,072 81,321
N1, Rd = fy0 t0 p d1 (1 + sin q) / [ 2 (√3) gM5 (sin q)2] = 356,254 kN
N2, Rd = fy0 t0 p d2 (1 + sin q) / [ 2 (√3) gM5 (sin q)2] = 69,563 kN
N3, Rd = fy0 t0 p d3 (1 + sin q) / [ 2 (√3) gM5 (sin q)2] = 356,254 kN
Member Ni, Ed Ni, Ed / Ni, Rd
1 14,008 0,039
-78,252 0,220
2 -9,221 0,133
71,072 1,022
3 -17,955 0,050
81,321 0,228
KT joint, condition 2: Condition important for situation:
N1, Ed compressive (> 0);
N2, Ed compressive (> 0);
N3, Ed tensile (< 0);
Photo: EN 1993-1-8 tab. 7.6
Case NP, Ed [kN] N1, Ed [kN] N2, Ed [kN] N3, Ed [kN]
Comp. 42,322 14,008 -9,221 -17,955
Tens. -207,155 -78,252 71,072 81,321
Such one situation doesn’t exist in analised case
Conclusion:
1. Joint KT, member 2, condition 1: exceeding resistance. Two parameters are important
for resistance it this case: t0 and di . Resistance and stability for member 2 and chord is
satisfied, but, because of problem in joint, one of them / both must have more massive
cross-section.
Algorithm
Photo: Author
→ Lec #3 / 96
Photo: Author
Max compressive force: not important for such one joint.
Max tensile force: NEd = -297,345 kN
Bolts: M24, class 8.8
M24 → dbolt = 24 mm
Class: X.Y
X = fub / 100 fub = 100 X = 800 MPa
Y = 10 fyb / fub fyb = 10 X Y = 640 MPa
VIIth example of calculations – field splice
tp tp
Photo: Author
76,1x3,2 → D = 76,1 mm, t0 = 3,2 mm
a = 5 mm
d = 24 mm
d0 = 24 + 2 = 26 mm
x = 40 mm
y = 35 mm
tp = 30 mm
p2
e1 e2
r0
ri
re
rp e1 = y = 35 mm
e2 = x + a √2 = 47 mm
r0 = D / 2 + e2 = 85 mm
p2 = L
ri = D / 2 – t0 = 35 mm
re = D / 2 = 38
rp = r0 + e1 = 120 mm
S235
A new Eurocode (working name EN 1993-1-801) is being developed; will contain
information about the calculation of joints for CHS / RHS. But at now calculation
procedure is presented based on literature, not on Eurocode.
k1 = ln (r0 / ri) = 0,887
k2 = k1 + 2 = 2,887
k = [k1 + √(k22 - k1
2)] / (2 k1) = 2,049
eeff = min (e2 ; 1,25 e1) = 44 mm
reff = re + e2 + eeff = 129 mm
k5 = ln (reff / r0) = 0,482
k3 = 1 - 1 / k + 1 / (k k5) = 0,417
Circumference along a line of bolts: c = 2 p r0 = 534 mm
Max distance between bolts p2 = L = 200 mm
Number of bolts: n = c / L ≈ 4 p2 = L = 134 mm
d0 = d + 2 mm (d ≤ 24 mm)
d0 = d + 3 mm (d > 24 mm)
Initial assumptions: d0 = d ; recommended value d0 = d + 2 or 3 mm is consistent with
EN 1090-2, technical requirements for steel structures. Value d0 = 26 mm is accepted.
2,2 d0 ≤ p2 ≤ min (14 tp ; 200 mm)
2,2 d0 = 57 mm < 134 mm < min (420 mm ; 200 mm) ; OK
1,2 d0 ≤ e2 ≤ (1,5 - 2,0) d
1,2 d0 = 31 mm < 47 mm < 48 mm ; OK.
1,2 d0 ≤ e1
1,2 d0 = 31 mm < 35 mm ; OK.
Verification of geometrical parameters, recommended in literature:
Resistance of joint is the effect of plate resistance (NRd1) and bolts resistance (NRd2):
NRd = min (NRd1 ; NRd2)
NRd1 = tp2 fyp p k / (2 gM0) = 394,349 kN
NRd2 = n Ft, Rd / k3
EN 1993-1-8 tab. 3.4
Ft,Rd = k2 fub As / gM2
k2 = 0,90
As - area of threaded portion of bolt; M24 → As = 3,53 cm2
Ft,Rd = 162,662 kN
NRd2 = 1560,307 kN
NRd = 394,349 kN
NEd / NRd = 0,754
Photo: Author
Top plate
Rocker
Bottom plate
VIIIth example of calculations – support on masonry structure
Top and bottom plate: tp = 30 mm, dp x dp = 400x400 mm
Rocker: tr = 30 mm, br x dr = 200x400 mm, rr = 500 mm
Masonry structure: fd = 4,0 MPa
Masonry wall thickness: t = 400 mm
Vertical distance floor – support: hc = 4,70 m
S235
Photo: Author
Vertical action from the first truss (vertical reaction
from red area) REdc, 1 = 295,328 kN
Vertical action from the second truss (vertical reaction
from blue area) REdc, 2 = 590,696 kN
Photo: EN 1996-1-1 fig 6.2
Resistance depends on
distane between support and
end of wall:
a1 = 0,00 for first truss,
a1 = 5,80 for the second one.
first truss second truss not important case
not important
case
NEdc / NRdc ≤ 1,0
NRdc = b Ab fd
b = min { 1,25 + a1 / (2 hc) ; 1,5 ; max [(1 + 0,3 a1 / hc) ∙ (1,5 - 1,1 Ab / Aef) ; 0 ] }
Aef = lefm t
lefm, 1 = dp + (hc / 2) / tg 60o = 1,757 m
lefm, 2 = dp + 2 (hc / 2) / tg 60o = 3,114 m
Truss Aef [m2] Ab [m
2] b NRdc [kN] NEdc / NRdc
1 0,703 0,160 min(1,250 ; 1,500 ; 1,250) = 1,250 800,000 0,396
2 1,246 min(2,484 ; 1,500 ; 1,862) = 1,500 960,000 0,584
Photo: Author
tp tr
br
rr br
dr
dp
dp dp
dp
Rocker-beam compression:
EN 1337-6, Structural bearings; rocker
bearings
Plate in contact with round rocker:
(NEd / bp) / [23 rr fu2 / (E γ
M)] ≤ 1,0
γM
= 1,0
1476,740 [kN/m] / 7097,143 [kN/m] =
= 0,208
Rocker in contact with plate:
(NEd / bp) / [fy (2 tr + bp) / γM ] ≤ 1,0
γM
= 1,1
1476,740 [kN/m] / 98272,727 [kN/m] =
= 0,015
Three various types of actions must be analised.
Photo: Author
1. Wind action on front wall. Load depends on area of front wall and wind pressure.
Additionally friction of wind on roof could be taken into consideration.
IXth example of calculations – roof bracing
Photo: Author
2. Instability of roof girders.
Bracing prevent roof girder
from instability. Witout them,
buckling of girder occurs on
total length. Bracings act as
complex of forces, make girder
more straight. These forces – as
reaction – must be applied to
bracings. Walue of these forces
is equivalent load, in proportion
to compressive force in girder.
Girders lose stability randolmy,
in left or right. Total effect for
entire roof is not multiplied
[(single effect) by (number of
girders m)]
but by (reduction factor am).
Photo: Author
3. Imperfections of roof girders.
Roof girders are not perfectly
straight. Bracings act as complex
of forces, make girder more
straight. These forces – as
reaction – must be applied to
bracings. Walue of these forces
depends on initial bow
imperfection of girder and
compresive force in girder.
Girders are imperfected
randolmly, in left or right. Total
effect for entire roof is not
multiplied
[(single effect) by (number of
girders m)]
but by (reduction factor am).
Way of calculation - 2D vs. 3D - is very important for algorithm.
Photo: mesilo.pl
3D model in FEM calculations: we have full information about cooperation between trusses,
purlins and bracing bars right away. Calculation is made in two steps:
• Initial calculation: dead weight, climatic actions, live loads etc.
• Equivalent forces: for bracing bars additionally important are equivalent forces from
imperfections of trusses and from prevention of instability of trusses. Values of these
forces are calculated based on forces in trusses after initial calculations. Cumulative
effect: loads from initial calculations and equivalent forces is final result of static
calculations.
Rys: Autor
Purlin: bi-axial bending
Flat frame
Roof bracings
Loads as in initial calculations
Equivalent forces
Cooperation frame-bracings:
additional forces in purlins,
additional forces in frame.
Purlin: bi-axial bending +
compressive axial force
(recalculation)
Frame: additional axial forces
from cooperation with
bracings (recalculation)
2D model (method of equivalent flat frames) is much more complicated:
Photo: Author
6 x 3,70 [m]
12 x
6,0
0 [
m]
1,50 [m]
2,15 [m]
Example is made for 2D model.
Wind pressure and winf suction od
both front walls are beared by roof
bracings.
Roof bracings prevent trusses
chords from buckling → buckling
equivalent forces from chords act
on bracings.
Trusses chords are not ideal
straight (imperfection) →
imperfection equivalent forces act
on roof bracings.
Photo: Author
Chord Max tension [kN] Max compression [kN] Cross-section
Top -55,112 302,157 Hot-rolled
CHS 108x10 Bottom -304,853 52,789
Connections with roof bracings S 235 → fy = 235 MPa
e = √ (235 / fy) = 1,0
Hot-rolled CHS 108x10 (108 mm diameter, 10 mm thickness):
A = 30,8 cm2
Jy = Jz = 373,0 cm4
iy = iz = 3,48 cm
S 235 + CHS hot rolled→
buckling curve a → a = 0,21
(EN 1993-1-1 tab. 6.1, 6.2)
L = 22,200 m
Chord E / R - tension E / R - compression in plane E / R - compression out of plane
Top 0,076 0,503 0,851
Bottom 0,421 0,081 0,412
Purlin: IPE 240
A = 37,22 cm2
Jy = 3675,10 cm4
Jz = 282,68 cm4
Wel, y = 306,26 cm3
Wel, z = 47,11 cm3
Wpl, y = 346,39 cm3
Wpl, z = 72,69 cm3
iy = 9,94 cm
iz = 2,76 cm
Photo: Author
y
z
qy = 0,942 kN/m
qz = 4,842 kN/m
E MEd, y MEd, z MEd, y + MEd, z D
E / R 0,268 0,248 0,320 0,960
Photo: Author
Roofing columns collect wind action and transport them to roof bracings and to columns’s
supports on masonry wall. Only half of action from each column acts on roof bracings: wind
action from blue area of fron wall.
Photo: Author
Purlins and roof bracings are in different parallel planes. Cooperation occurs for olny such
purlins, which have common nodes with bracings.
Bracing bars ( 25) are very slender and flexible. They lose stability under very small
compressive force. After buckling they not cooperate with resyt of structure. Only tensed
bars are taken into calculations.
Photo: Author
Photo: Author
Because of shape of truss, areas for each column are different. But it could be approximated
as equvalent areas: awerage height of columns x distance between columns.
Acol = [(1,50 + 1,50 + 2,15) /2 ] x 3,70 = 11,932 m2
Atotal = 6 Acol = 71,595 m2
Wind pressure + wind suction at the both ends of roof: qw = 0,8 kN / m2
Additionally important is wind friction on roof. It depends on wind action on roof,
qw1 = 0,5 kN / m2
Photo: Author
L = 12 x 6,0 m h = 1,50 m + 2,15 m (truss only; masonry structure is not taken into consideration) A (for roof bracings) = 1533,390 m2
h
d
h / 2
h / 2
L1
L1 = L – min (2d ; 4h)
EN 1991-1-4 tab 7.10
Surface cfr
steel, smooth concrete 0,01
rough concrete, tar-boards 0,02
ripples, ribs, folds 0,04
Friction of wind (EN 1991-1-4 7.5)
Wind total: Ftotal = Atotal qw + A cfr qw1 = 88,235 kN qw-t = Ftotal / d = 3,975 kN / m
Photo: Author
12 bands between trusses → 13 trusses. 2 braced bands in roof → m = 6,5 trusses per braced
band.
12 x
6,0
0 [
m]
Reduction factor for effects from instability
and imperfections (EN 1993-1-1 (5.12)):
am = √[ 0,5 (1 + 1 / m)]
m = 6,5
am = 0,760
Max compressive force in top chord of
vertical truss: NEd, c = 302, 157 kN
Buckling equivalent force from truss chord to
bracings:
Fbuck = am NEd, c / 100 = 2,295 kN
Initial horizontal bow inperfection for truss
chord: e0 = am d / 500 = 33,744 mm.
d = 6 x 3,70 = 22,20 [m]
Photo: Author
Two alternative schemes of action are taken into consideration:
• wind and equivalent effects from buckling;
• wind and equivalent effects from imperfections.
For final analysis more important will be bigger value of both above.
qw-t
Fbuck Fbuck Fbuck Fbuck Fbuck Fbuck Fbuck
qimperf
qw-t
Calculation of equivalet loads from imperfection qimperf for roof bracings is complicated.
According to EN 1993-1-1 (5.13), total eqiuvalent load (from wind and imperfection) is
calculated as:
qimperf = qd = S [8 NEd, c (e0 + dq) / L2]
dq – deformation comes from load qd
Total load qd depends on dq ,but dq is the effect of qd . Such problem can be solved by iteration
procedure:
dq0 = d(qw-t) = dw-t
qd(0) = qd
(0)(e0 + dw-t)
dq(1) = dq
(1)(qd(0))
qd(1) = qd
(1)(e0 + dq(1) + dw-t)
dq(2) = dq
(2)(qd(1))
qd(2) = qd
(2)(e0 + dq(2) + dw-t)
…
Static calculations of bracings must be made few times for iteration procedure.
First of all, deformation under wind action dq0 = d(qw-t) = dw-t must be calculated
Photo: Author
Fw-t = 3,975 [kN / m] ∙ 3,70 [m]
0,5 Fw-t 0,5 Fw-t
Steps of iteration:
dq0 = d(qw-t) = dw-t = 8,31 mm
qd(0) = S [8 NEd, c (e0 + dw-t) / L
2] =
= 6,5 ∙ 8 ∙ 302,157 [kN] ∙ (33,74 [mm] + 8,31 [mm]) / (22,2 [m])2 = 1,332 kN / m
Static callculation of deformation
dq(1) = dq
(1)(qd(0)) = 2,78 mm
qd(1) = S [8 NEd, c (e0 + dq
(1) + dw-t) / L2] =
= 6,5 ∙ 8 ∙ 302,157 [kN] ∙ (33,74 [mm] + 8,31 [mm] + 2,78 [mm]) / (22,2 [m])2 = 1,429 kN / m
Static callculation of deformation
dq(2) = dq
(2)(qd(1)) = 2,99 mm
qd(2) = S [8 NEd, c (e0 + dq
(2) + dw-t) / L2] = 1,435 kN / m
Static callculation of deformation
dq(3) = dq
(3)(qd(2)) = 3,00 mm
qd(3) = S [8 NEd, c (e0 + dq
(3) + dw-t) / L2] = 1,436 kN / m
Very small difference, end of iteration.
Photo: Author
Photo: Author
Photo: Author
Two situations must be taken into consideration: wind + equivalent effects of buckling…
Photo: Author
9,649 kN 9,649 kN
17,003 kN
17,003 kN
17,003 kN
17,003 kN
17,003 kN
…and wind + equivalent effects of imperfections.
Photo: Author
10,011 kN 10,011 kN
20,021 kN
20,021 kN
20,021 kN
20,021 kN
20,021 kN
Forces for the second one are bigger the second is important for calculations of
horizontal bracing..
Results (signs of forces in opposite way to Eurocode):
Photo: Author
Max tensile force in bracing bar: NEd, t, bb = 58,8 kN
Max compressive force in purlin: NEd, c, p = 60,1 kN
Max compressive force in chord: NEd, c, ch = 55,6 kN
Max tensile force in chord: NEd, t, ch = 49,4 kN
Checking of resistance: bracing bar
NRd, bb = fy A / gM0 = 115,355 kN
NEd, c, bb / NEd, bb = 0,510
Purlin: bi-axial bendint and axial force is calculated according to the same way as roofing
column, IInd example.
Checking resistance of cross-section:
NRd, p = fy A / gM0 = 874,670 kN
NEd, c, p / NRd, p = 0,069
Checking interaction NEd ↔ MEd, y for I-beam (EN 1993-1-1 (6.33) – (6.38)):
Influence of axial force for bending resistance of cross-section:
n = NEd / Npl, Rd
a = min [ 0,5 ; (A - 2 b tf) / A]
NEd ≤ min ( 0,25 Npl, Rd ;
0,5 hw tw fy / gM0)
NEd > min (0,25 Npl, Rd ;
0,5 hw tw fy / gM0)
MN, y, Rd Mpl, y, Rd min [ Mpl, y, Rd ; Mpl, y, Rd (1 - n) / (1 - 0,5 a) ]
NEd ≤ hw tw fy / gM0 NEd > hw tw fy / gM0
MN, z, Rd
Mpl, z, Rd
n ≤ a n > a
Mpl, z, Rd Mpl, z, Rd {1 - [(n - a) / (1 - a)]2}
min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0) = 218,668 kN
hw tw fy / gM0 = 490,680 kN
Limits for both directions are bigger than axial force no influence of axial force for resistance of cross-section.
Checking stability for NEd and for interaction NEd ↔ MEd, y
For Your range of design project you can asume (not completely correct), that sanwich panels
protect purlins from instability.
Chords of roof bracings.
Both chords of roof bracings are simultaneously top chords of vertical trusses. New values of
forces in trusses must be taken into consideration as additional forces applied to trusses.
Max compressive force from bracings together with combination of loads which gives max
compression in top chord.
Max tensile force from bracings together with combination of loads which gives max tension
(or min compression) in top chord.
Photo: Author
Chord Max tension [kN] Max compression [kN] Cross-section
Top -103,362 356,222 Hot-rolled
CHS 108x10 Bottom -315,973 57,729
New forces in chords
Chord E / R - tension E / R - compression in plane E / R - compression out of plane
Top 0,143 0,591 1,003
Bottom 0,436 0,089 0,451
More massive cross-section must be applied for top chord. For such a litte difference change
from CHS 108x10 to CHS 108x11 is enough. Bottom chord should be the same more
massive to avoid mistakes durnig manufacturing proces (two very similar cross-sections:
CHS 108x10 and 108x11 could be difficult to distinguish).
Of course, recalculation must be made for each truss members, nort only for chords.