Design of two-span trapezoidal roof sheeting - Steel-stainless.org

Job No. Sheet 1 of 6 Rev A
CALCULATION SHEET
Job Title
Subject
Client
RFCS Stainless Steel Valorisation Project
Design Example 11 – Design of a two-span trapezoidal
roof sheeting
Made by JG/AO Date Feb 2006
RFCS
Checked by GZ Date March 2006
DESIGN EXAMPLE 11 – DESIGN OF A TWO-SPAN TRAPEZOIDAL ROOF SHEETING
This example deals with a two-span trapezoidal roof sheeting with a thickness of 0,6
mm from stainless steel grade 1.4401 CP500, i.e. cold worked to a yield strength of
500 N/mm 2 . Comparisons will be made to a similar sheeting of grade 1.4401 in the
annealed condition, i.e. fy = 240 N/mm 2 (see also Design Example 3).
If the nominal yield strength in all directions of the sheet is not guaranteed by the
producer it should be reduced to 80% of its value. In this example it is assumed that
the strength was not guaranteed in order to demonstrate this.
Section 3.2.4
The dimensions of the roof sheeting are shown below.
212,5 212,5 212,5 212,5
70
65 57
850
A detailed sketch of the roof sheeting is given in the figure below. The lower flange
will be in compression over the mid support and therefore this case will be checked in
this example.
b u0 /2
h 0
b r /2
h r
b r0 /2
Mid line dimensions:
h0 = 70 mm
w0 = 212,5 mm
bu0 = 57 mm
br = 20 mm
hr = 6 mm
br0 = 8 mm
bl0 = 65 mm
h r
b r0 /2
b r /2
r
radius, r = 3 mm
angle, ϕ = 57,1°
b l0 /2
w 0 /2
177

Job No. Sheet 2 of 6 Rev A
CALCULATION SHEET
Job Title
Subject
Client
RFCS Stainless Steel Valorisation Project
Design Example 11 – Design of a two-span trapezoidal
roof sheeting
Made by JG/AO Date Feb 2006
RFCS
Checked by GZ Date March 2006
Data
Span length
L = 3,5 m
Load q = 1,4 kN/m 2
Self weight g = 0,07 kN/m 2
Sheeting thickness
t = 0,6 mm
Width of support
ss = 100 mm
Yield strength fy = 0 ,8× 500 = 400 N/mm 2
Modulus of elasticity E = 200 000 N/mm 2
Partial factor γM0 = 1,1
Partial factor γM1 = 1,1
Load factor
γG = 1,35 (permanent loads)
Load factor γQ = 1,5 (variable loads)
Table 3.5
Table 2.1
Table 2.1
Section 2.3.2
Section 2.3.2
Effective section properties
Maximum width-to-thickness ratios
Table 4.1
max( bl0 / t, bu0 / t) = bl0
/ t = 108 < 400
h / t = 117 < 400
0
Location of the centroidal axis when the web is fully effective
Effective width of the compression flange Section 4.4.1
bl0
− br
235 E
bp
= = 22,5 mm ε = = 0,75
Table 4.2
2
fy
210000
bp
/ t
Table 4.3
k
σ
= 4
λp
= = 0,883
Eq. 4.2
28, 4ε
kσ
0,772 0,125
ρ = 0,714
2
λ
− λ
= b = ρb
= 16,1 mm
Eq. 4.1a
eff,l p Table 4.3
p
p
Reduced thickness of the flange stiffener: Section 4.5.3
The lower compressed flange is shown in detail below.
b p b eff,l /2
t sl
Intermediate stiffener
e s
A s , I s
b s
178

Job No. Sheet 3 of 6 Rev A
CALCULATION SHEET
Job Title
Subject
Client
RFCS Stainless Steel Valorisation Project
Design Example 11 – Design of a two-span trapezoidal
roof sheeting
Made by JG/AO Date Feb 2006
Effective thickness of the inclined part of the stiffener
⎛
2
br
b
⎞
r0 2
⎜
⎛ − ⎞
+ h ⎟
r
t
⎜
⎜
2
⎟
⎝ ⎠ ⎟
trl
=
⎝
⎠
= 0,85 mm
h
A b b t ht
r
RFCS
Checked by GZ Date March 2006
2
s
= (
eff,l
+
r0) + 2
r rl
= 24,62 mm
Figure 4.3
hr
bht
r0 r
+ 2hr trl
e
2
s
= = 2, 41 mm
A
s
The second moment of area for the stiffener is calculated with two strips of width 15t
adjacent to the stiffener (smaller terms neglected)
2 3
2 2 2 ⎛hr ⎞ trlhr
4
Is = 2× 15 t es + br0t( hr − es) + 2ht r rl⎜
− es⎟
+ 2 = 159,1 mm
⎝ 2 ⎠ 12
2
2 ⎛br − br0
⎞
s r r0
b = 2 h + ⎜ + b = 24,97 mm
2
⎟
⎝ ⎠
2
4 s p p s
b 3
l
s
Figure 4.3
Ib (2b + 3 b)
= 3,07 = 251,0 mm
Eq. 4.9
t
2
⎛w0 −bu0 −bl0
⎞ 2
w
= + h0 = 83,4 mm
⎜
⎝
d p s
2
⎟
⎠
b = 2b + b = 70,0 mm
Eq. 4.11
k
s
+ 2b
w d
wo
= =
sw
+ 0,5bd
1, 37
Eq. 4.10
lb s
w
= 3, 01> 2 kw = kwo = 1,37 Eq. 4.7
4, 2k E I t
σ = = 557,5 N/mm
4 (2 3 )
λ
3
w
s
cr,s 2
As bp bp + bs
f
y
d
= = 0,85
d d
σ
cr,s
2
Eq. 4.3
χ = 1, 47 − 0, 723λ
= 0,86
Eq. 4.16
tred = χdt
= 0,51 mm
Optionally iterate to refine the value of the reduction factor for buckling of the
stiffener.
prEN 1993-
1-3, clause
5.5.3.3 (3)
179

Job No. Sheet 4 of 6 Rev A
CALCULATION SHEET
Job Title
Subject
Client
RFCS Stainless Steel Valorisation Project
Design Example 11 – Design of a two-span trapezoidal
roof sheeting
Made by JG/AO Date Feb 2006
RFCS
Checked by GZ Date March 2006
Distance to the neutral axis from the compressed flange (fully effective web)
2
A = A i
= 84,0 mm
e
c
tot
∑
∑
Ae
i i
= = 36,3 mm
A
tot
Effective cross-section of the web Section 4.4.1
h0 − ec
2
ψ =− =− 0,929
k σ
= 7,81− 6,29ψ + 9,78ψ
= 22,1
e
Table 4.3
c
bp,w = sw = 83,4 mm
0,772 0,125
ρ = 0, 490
λ
− λ
=
s
p
2
p
λ
b
/ t
p,w
p
= = 1,391
Eq. 4.2
28,4ε
kσ
b
b
1−ψ
p,w
eff,w
= ρ =
eff,1
= 0, 4beff,w
= 8, 47 mm seff,2 beff,w
21,2 mm
Eq. 4.1a,
Table 4.3
= 0,6 = 12,7 mm
Table 4.3
Effective cross section properties per half corrugation
∑
∑
Aeff,tot = Aeff, i
= 70,8 mm
e
A
e
eff, i eff, i
eff,c
= =
Aeff,tot
∑ ∑
2
40,0 mm
( ) 2 4
Itot = Ieff,i + Aeff,i ec − eeff,i = 51710 mm
Bending resistance per unit width (1m) Section 4.7.4
1000 mm
I = Itot
= 486685 mm
0,5w
W
eff,l
I
= = 12165 mm
e
c
0
3
4
W
eff,u
= I
16227 mm
h − e
=
0 c
3
Weff,l < Weff,u Weff,min = Weff,l
Mc,Rd = Weff,min fy γ
M0
= 4, 42 kNm
Eq. 4.29
180

Job No. Sheet 5 of 6 Rev A
CALCULATION SHEET
Job Title
Subject
Client
RFCS Stainless Steel Valorisation Project
Design Example 11 – Design of a two-span trapezoidal
roof sheeting
Made by JG/AO Date Feb 2006
RFCS
Checked by GZ Date March 2006
Resistance to local transverse forces at intermediate support
Resistance to local transverse forces per unit width (1 m)
α = 0,15 (for sheeting profiles) and la = ss
1000 mm
R = αt f E − r t + l t + ϕ γ
2
( 1 0,1 / ) ⎡0,5 0,02 / ⎤( 2,4 ( /90)
)
2
w,Rd y ⎣
a ⎦
M1
0,5w0
Rw,Rd = 20,9 kN
prEN 1993-
1-3, Eq.
6.20c, 6.19b
and 6.18
Interaction between bending moment and transverse force
The maximum bending moment will appear at the intermediate support where it will
interact with the support reaction and therefore the following checks must be
performed.
M
Ed
1
M ≤ FEd
MEd
FEd
≤ 1
+ ≤ 1, 25
R
M R
c,Rd
w,Rd
c,Rd
w,Rd
prEN 1993-
1-3, Eqs.
6.28a-c
Design load per unit width (1 m) Section 2.3.2
qd = γ
Gg+ γ
Qq= 2,20 kN/m
Eq. 2.3
The design load, qd, gives the following bending moment and support reaction at the
intermediate support.
2
qL
5
M
Ed
= = 3,37 kNm FEd
= qL=
9,63 kN
8
4
M
Ed
0,76
M = FEd
MEd
FEd
= 0, 46
+
R
M R
= 1, 22 OK
c,Rd
w,Rd
c,Rd
w,Rd
Deflection at serviceability limit state
For verification in the serviceability limit state the effective width of compression
elements should be based on the compressive stress in the element under serviceability
limit state loading. The maximum compression stress is calculated as follows. A
conservative approximation is made based on Weff,min from ultimate limit state.
( q+
g) L
2
prEN 1993-
1-3, clause
5.5.1(4)
M
Ed,ser
= = 2, 25 kNm
Section 2.3.4
8
M
Ed,ser 2
σ
com,Ed,ser
= = 186 N/mm
W
eff,min
Now, the effective section properties are determined as before but with fy replaced by
σcom,Ed,ser. The calculations will not be shown here but the interesting results are:
I = 573 150 mm 4
Wu = 15 866 mm 3
Wl = 16 919 mm 3
181

Job No. Sheet 6 of 6 Rev A
CALCULATION SHEET
Job Title
Subject
Client
RFCS Stainless Steel Valorisation Project
Design Example 11 – Design of a two-span trapezoidal
roof sheeting
Made by JG/AO Date Feb 2006
RFCS
Checked by GZ Date March 2006
Determination of the deflection:
Secant modulus corresponding to the stresses in the tension and compression flange
respectively.
M
Ed,ser 2
σ
1,Ed,ser
= = 142 N/mm
W
u
M
σ
2,Ed,ser
= = 133 N/mm
W
E
E
Ed,ser 2
l
E
= = 199 604 N/mm
s,1 n−1
E ⎛σ
⎞
1,Ed,ser
1+ 0,002
f ⎜
y
f ⎟
⎝ y ⎠
E
= = 199 730 N/mm
n−
E ⎛σ
⎞
2,Ed,ser
1+ 0,002
f ⎜
y
f ⎟
⎝ y ⎠
s,2 1
Es,1 + Es,2 2
Es
= = 199 667 N/mm
2
2
2
n = 7,0
As a simplification, the variation of Es along the length of the member may be
neglected and the minimum value of Es of that member may conservatively be used
throughout its length, i.e.
Es = Es,1 = 199 603 N/mm 2
The permitted deflection is L/300 = 11,7 mm
1+
33
x= L= 1, 47 m (location of maximum deflection)
16
( ) 4 3 4
g+ q L ⎛ x x x ⎞
δ = ⎜ − 3 + 2 10,4 mm
3 4 ⎟=
48Es,1I ⎝L L L ⎠
OK
Appendix C
Appendix C
Table C.1
Appendix C
Comparison with sheeting in grade 1.4401 in the annealed condition
The bending resistance per unit width of identical sheeting in grade 1.4401 in the
annealed condition (fy = 240 N/mm 2 ) is:
Mc,Rd = 3,22 kNm
and the resistance to local transverse forces is:
Rw,Rd = 16,2 kN
With sheeting made from grade 1.4401 in the annealed condition, the span must be
reduced to 2,9 m compared to 3,5 m for material in the cold worked strength
condition. Hence, sheeting made from cold worked material enables the span to be
increased, meaning that the number of secondary beams or purlins could be reduced,
leading to cost reductions.
182