This document describes the design of a single-floor steel warehouse with specifications including an 18m clear span, 6m eave height, and various loading calculations for dead, live, and wind loads based on structural guidelines. It details the materials used, structural configurations, load combinations, and internal force analysis required to ensure safety and compliance with building codes. The document concludes with a consideration of the frame design and section checking for structural integrity.

1. - 1 -
Steel Warehouse Design Report
1.Design Information
Steel Warehouse : Single floor , 18m clearly span , eave height 6m ,column distance
6m ,totally 12pcs steel frame ,roof slope 1:10 , Seismic intensity of 7 degree , Seismic
acceleration value 0.05g , Steel Frame Layout Plan Figure 1(a) , Steel Frame Type Figure
1(b) , wall and roof panel used corrugated color coated single sheet ,with glass wool
insulation ,Wall Girt and Roof Purlin used Galvanized C section , distance 1.5m , Steel
Materials Q345B ,welding rod : E43

2. - 2 -
2.Loading Calculated
(1) Loading Value :
1)Roof Dead Loading (Horizontal projected area)
YX51-380-760 Corrugated single sheet 0.15 KN/m2
50mmGloass Wool Insulation 0.05 KN/m2
Aluminum Foil and wire mesh 0.02 KN/m2
Purlin and Bracing 0.10 KN/m2
Self-weight of Roof Beam 0.15 KN/m2
Hanging Equipments 0.20 KN/m2
Totally 0.67 KN/m2
2)Roof Live Loading
Roof Live Loading Value : 0.50 KN/m2
。
Snow Loading : Snow PressureS0=0.45 KN/m2
。Double Slope Roof
Roof Slope α=5°42′38,μr=1.0 , Snow Loading : Sk=μrS0=0.45 KN/m2
。
Compared live loading and snow loading used the bigger value : 0.50 KN/m2
Not consider the Dust Load .
3)Light Weight Wall and self-weight of column
(including Column , wall-stud structures ) : 0.50 KN/m2
4)Wind Loading :
According 《Technical Specification for Light-Weighted Steel Portal Frames》
CECS102: 2002
Basic Wind Pressure ω0=1.05×0.45 KN/m2
, Ground Roughness : B,
Height variation factor of Wind Loading according《Building Structural Load Code》
(GB50009-2001), when height less than 10m, according 10m height value, μz=1.0.
Wind Load Shape Factor μs : Windward Column and Roof are respectively＋0.25 and－1.0,
Leeward Column and Roof are respectively ＋0.55 and－0.65(CECS102:2002).

3. - 3 -
3.The Load Standard Calculated Value of Each Part
Roof Loading:
Dead Loading Standard Values : 0.67×6=4.02KN/m
Live Loading Standard Value : 0.50×6=3.00KN/m
Column Loading :
Dead Loading Standard Value : 0.5×6×6+4.02×9=54.18KN
Live Loading Standard Value : 3.00×9=27.00KN
Wind Loading Standard Value :
Windward : Column qw1=0.47×6×0.25=0.71KN/m
Rafter qw2=－0.47×6×1.0=－2.82KN/m
Leeward : Column qw3=－0.47×6×0.55=－1.55KN/m
Rafter qw4=－0.47×6×0.65=－1.83KN/m
4. Inner Force Analysis
(1)Under Dead Loading
λ=l/h=18/6=3
ψ=f/h=0.9/6=0.15
k=h/s=6/9.0449=0.6634
μ=3+k+ψ(3+ψ)=3+0.6634+0.15×(3+0.15)=4.1359
5289.0
1359.44
15.058
4
58







HA=HE=qlλΦ/8=4.02×18×3×0.5289/8=14.35KN
MC=ql2
[1－(1+ψ) Φ]/8=4.02x182
[1－(1+0.15)×0.5289]=63.78KN·m
MB=MD=－ql2
Φ/8=－4.02×182
×0.5289/8=－86.11KN·m
Inner Force of Steel Frame under dead loading as below figure

4. - 4 -

5. - 5 -
(2) Under Live Loading
VA=VE=27.00KN
HA=HE=3.00×18×3×0.5289/8=10.71KN
MC=3.00×182
[1－(1+0.15)×0.5289]/8=47.60KN·m
MB=MD=－3.00×182
×0.5289/8=－64.26KN·m
Inner Force of Steel Frame under live loading as below figure

6. - 6 -
(3) Under Wind Loading
The wind load acting on the roof can be decomposed into the component force qx in the
horizontal direction and the vertical component force qy. Calculated separately, and then
superimposed.
1) The Rafter under the action of wind loading vertical component of forces (qw2y) on
windward side
1322.0)15.058(
1359.416
1
)58(
16
1


 

KN
l
qa
VE 35.6
182
982.2
2
22




VA=2.82×9－6.35=19.03KN
HA=HE=qlλ Φ /4=2.82×18×3×0.1322/4=5.03KN
MB=MD=5.03×6=30.18KN·m
MC= ql2
[α 2
－(1+ψ ) Φ ]/4=2.82×182
×[0.52
－1.15×0.1322]/4=22.38KN·m

7. - 7 -
Steel Frame Internal Force diagram under action of (qw2y)
2) The Rafter under the action of wind loading vertical component of forces (qw2y) on
leeward side
KN
l
qa
VE 12.4
182
983.1
2
22




VA=1.83×9－4.12=12.35KN
HA=HE=qlλ Φ /4=1.83×18×3×0.1322/4=3.27KN
MB=MD=3.27×6=19.62KN·m
MC= ql2
[α 2
－(1+ψ ) Φ ]/4=1.83×182
×[0.52
－1.15×0.1322]/4=14.52KN·m
Steel Frame Internal Force diagram under action of (qw2y)

8. - 8 -
3) The Column under wind loading action qw1 on windward side
α=1,
9803.0]16634.0)6634.015.02(6[
1359.44
1
])2(6[
4
1 22


 

KK
VA=－VB=－qh1
2
/2L=－0.71×62
/(2×18)=－0.71KN
KN
qh
HA 22.3)9803.0
2
1
2(
2
1671.0
)
2
2(
2




HE=0.71×6－3.22=1.04KN
MD=1.04×6=6.24KN·m
mKN
qh
MC 

 81.0)]9803.0)15.01(1[
4
1671.0
])1(1[
4
2222


Steel Frame Internal Force diagram under action of (qw1)
mKN
qh
MB 

 52.6)9803.02(
4
1671.0
)2(
4
2222


9. - 9 -
4) The Column under wind loading action qw3 on windward side
VA=－VB=－qh1
2
/2L=－1.55×62
/(2×18)=－1.55KN
KN
qh
HE 02.7)9803.0
2
1
2(
2
1655.1
)
2
2(
2




HA=1.55×6－7.02=2.28KN
MD=7.02×6－1.55×62
/2=14.22KN·m
MB=2.28×6=13.68KN·m
mKN
qh
MC 

 78.1)]9803.0)15.01(1[
4
1655.1
])1(1[
4
2222


Steel Frame Internal Force diagram under action of (qw3)

10. - 10 -
5) The Beam under wind loading action (qw2x )on windward side
α=1，β=0
0202.0)15.0134(
1359.48
15.0



KNVV EA 91.0)9.062(
182
9.082.2




HA=2.82×0.9(1+0.0202)/2=1.29KN
HE=2.82×0.9－1.29=1.25KN
mKNMC 

 39.0]0202.015.15.015.0[
2
69.082.2
MB=1.29×6=7.74KN·m
MD=1.25×6=7.50KN·m

11. - 11 -
Steel Frame Internal Force diagram under action of (qw2x)
6) The Rafter under the action of wind loading vertical component of forces (qw4x) on
leeward side
KNVV EA 59.0)9.062(
182
9.083.1




HA=1.83×0.9(1+0.0202)/2=0.84KN
HE=1.83×0.9－0.84=0.81KN
mKNMC 

 26.0]0202.015.15.015.0[
2
69.083.1
MB=0.81×6=4.86KN·m
MD=0.84×6=5.04KN·m

12. - 12 -
Steel Frame Internal Force diagram under action of (qw4x)
7) The combined force of steel frame under wind load

13. - 13 -
4.Internal force combination
The frame structure is designed according to the ultimate capacity of the load capacity.
According to the 《Building Structural Load Code》(GB50009－2001), the basic
combination of load effect is used: γ 0S≤R. The safety grade of the structural
components of this project is Grade 2, γ0 = 1.0.
The design value of the load effect combination (S) is determined by taking the most
unfavorable values from the following combination values :
A. 1.2 × constant load standard value calculated load effect + 1.4 × live load standard value
calculated load effect

14. - 14 -
B. 1.0 × constant load load value calculated by the load effect + 1.4 × wind load standard
value calculated load effect
C. 1.2 × constant load standard value calculated load effect + 1.4 × live load standard value
calculated load effect + 0.6 × 1.4 wind load standard value calculated load effect
D. 1.2 × constant load standard value calculated load effect + 1.4 × wind load standard
value calculated load effect + 0.7 × 1.4 × live load standard value calculated load effect
E. 1.35 × Constant load calculated by the load effect of + 0.7 × 1.4 × live load standard
value calculated load effect
The most unfavorable internal force combination of the calculated control section to take
the bottom of the column, the top of the column, end beam and beam cross section, for
the rigid beam, the cross section may be the most unfavorable internal force combination:
Beam end section: (1) Mmax and the corresponding N, V; (2) Mmin and the corresponding
N, V
Beam cross section: (1) Mmax and the corresponding N, V; (2) Mmin and the
corresponding N, V
For the rigid column, the most unfavorable internal forces of the cross section are:
(1) Mmax and the corresponding N, V; (2) Mmin and the corresponding N, V
(3) Nmax and the corresponding ± Mmax, V; (4) Nmin and the corresponding ± Mmax,
Internal Force Combination see below figure

15. - 15 -
Internal Force Combination of Steel Frame Figure 1
Section Internal Force Combination
Loading
combine mode
Loading Combine
portfolio
M
（KN·m）
N
（KN）
V
（KN）
Column
Top Column
（B point）
Mmax, N, V A 1.2×dead +1.4× live 193.30 81.22 －32.21（←）
Mmin, N, V B 1.0×dead +1.4×wind －7.89 1.05 －1.67（←）
Nmax ±Mmax、V A 1.2×dead+1.4×live 193.30 81.22 －32.21（←）
Nmin ±Mmax、V B 1.0×dead+1.4×wind －7.89 1.05 －1.67（←）
Bottom
Column
（A point）
Mmax, N, V
Mmin, N ,V
Nmax ,±Mmax ,V A 1.2×dead+1.4×live 0 102.82 －32.21（→）
Nmin ±Mmax ,V B 1.0×dead+1.4×wind 0 19.05 4.30（←）
Beam
Support
（B point）
Mmax N , V A 1.2×dead+1.4×live 193.30 39.89 77.60（↑）
Mmini N ,V B 1.0×dead+1.4×wind －7.89 1.57 0.89（↑）
Midspan
（C Point）
Mmax, N, V B 1.0×dead+1.4×wind －9.40 －2.03 0.60（↓）
Mmin , N ,V A 1.2×dead+1.4×live －143.18 31.81 －3.21（↑）

17. - 17 -
5.Steel Frame Design
(1) Section design
Column and Beam used welded H section Steel 450×200×8×12，
Section properties：
B=200mm，H=450mm，tw=8.0mm，tf=12.0mm，A=82.1cm2
Ix=28181cm4
，Wx=1252cm3
，ix=18.53cm
Iy=1602cm4
，Wx=160.2cm3
，ix=4.42cm
(2) Section Checking
1) Width-to-thickness ratio checking
Flange Plate : 15235/f15896/12b/t y 
Web Plate : 250235/f50225.356/824/th yw0 
2) Beam Checking
(1)The shearing calculation
Max Shearing of Beam section Vmax=77.60KN
Consider only with support stiffened plate
8.0562.0235/
34.541
/0
 y
w
s f
th

fv=125N/mm2
Vu=hwtwfv=426×8×125=426000N=426.0KN
Vmax=77.60KN<Vu，Ok
(2) Bending moment, Shearing force, Pressure action checking together
Checking and calculated the end of beam
N=39.89KN，V=77.60KN，M=193.30KN·N
Because V<0.5Vu，V=0.5Vu ,According to code , GB70017 .
))(( 22
2
2
1
1
A
N
fhA
h
h
AM fff 

18. - 18 -
)
8210
39890
215)(21912200
219
219
12200(
2

=220.90KN·m>M=193.30KN·m，M=Mf
10)1
5.0
( 2




feu
f
u MM
MM
V
V
, OK
(3) Overall stability checking
N=39.89KN，M=193.30KN·m
A．In-plane Beam Overall stability checking
Calculated Length according beam length : lx=18090mm，
λ x=lx/ix=18090/185.3=97.63<[λ ]=150，b type section ,ψ x=0.570
KN
EA
N e
EX 0.1592
63.971.1
821010206
1.1 2
32
2
0
2
'
0 






，β mx=1.0
)
1592
89.39
570.01(101252
1030.1930.1
8210570.0
39890
)1( 3
6
'
0
1
0 






EX
xe
xmx
ex
N
N
W
M
A
N



=165.15N/mm2
<f=215 N/mm2
, Ok
B．Out-plane Beam Overall stability checking
Calculated length according two purlin distance or flange bracing distance ,ly=3015mm。
As for uniform section γ =0，μ s=μ w=1
λ y=μ sl/iy0=3015/44.2=68.2，b type section , ψ y=0.762
6.0133.1)
4264.4
122.68
(
101252
4268210
2.68
4320 2
32






by
ψ b’=1.07－0.282/ψ by=0.821
975.0)
N
N
(75.0
N
N
-1.0 2
'
EX0
'
EX0
t
22
3
6
10
/215/73.189
101252821.0
1030.193975.0
8210762.0
39890
mmNfmmN
W
M
A
N
bre
t
ey









(4)《Design Code for Steel Structure》(GB50017－2003) Check the beam web to allow high
thickness ratio.
Beam end section:

19. - 19 -
2
6
33
max
min /
24.141
96.150
10.14686.4
1028181
2131030.193
8210
1089.39
mmN







94.1
max
minmax
0 





7.115
235
)2.265.048(25.53 0
0

yw ft
h
 ，OK
Mid-span section :
2
6
33
max
min /
35.104
09.112
22.10887.3
1028181
2131018.143
8210
1081.31
mmN







93.1
max
minmax
0 





3.115
235
)2.265.048(25.53 0
0

yw ft
h
 ，OK
(5) Checking the area compressive bearing capacity under the concentrated load of purlin
The purlin pass the concentrated load to the upper flange of the beam：
F=(1.2×0.27×6+1.4×3.00)×3=18.43KN
Lz=a+5hy+2hR=70+5×12+0=130mm
22
3
/215/72.17
1308
1043.180.1
mmNfmmN
lt
F
zw
c 





Checking the reduce stress at the upper edge of the web
Take the internal force at the end of the beam: M=193.30KN·m，N=39.89KN, V=77.60KN
2
4
3
1 /10.146213
1028181
1030.193
mmNy
I
M
n




σc=17.72N/mm2
2
4
3
/61.34
81028181
419122001060.77
mmN
It
VS
w




222222
61.343)89.3910.146(72.1772.17)89.3910.146(3   cc
=130.65 N/mm2
<1.2f=258 N/mm2
，OK

20. - 20 -
(3)Checking and calculated Column
Shearing Check
Max Shearing for column Vmax=32.21KN
Consider only with supported stiffening rib
8.0562.0235/
34.541
/0
 y
w
s f
th

fv=125N/mm2
Vu=hwtwfv=426×8×125=426000N=426.0KN
Vmax=32.21KN<Vu , OK
(2) Bending moment, Shearing force, Pressure action checking together
Checking and calculated the end of beam
N=81.22KN，V=32.21KN，M=193.30KN·N
Because V<0.5Vu , V=0.5Vu, According to code GB70017
))(( 22
2
2
1
1
A
N
fhA
h
h
AM fff 
)
8210
81220
215)(21912200
219
219
12200(
2

=215.61KN·m>M=193.30KN·m，M=Mf
10)1
5.0
( 2




feu
f
u MM
MM
V
V
, OK
(3) Overall stability checking
Max Internal Force of steel structure : N=102.82KN，M=193.30KN·m
In-plane Steel Frame Overall stability checking
Height of Column H=6000mm，Length of Rafter L=18090mm.
linear rigidity of column K1=Ic1/h=28181×104
/6000=46968.3mm3
linear rigidity of beam K2=Ib0/(2ψS)=28181×104
/(2×9045)=15578.2mm3
K2/K1=0.332, The calculation length coefficient of the column μ=2.934
Calculated Length of Column lx=μh=17604mm
λx=lx/ix=17604/185.3=95。0<[λ]=150，b type section , ψx=0.588

21. - 21 -
KN
EA
N e
EX 4.1681
0.951.1
821010206
1.1 2
32
2
0
2
'
0 






，βmx=1.0
)
4.1681
82.102
588.01(101252
1030.1930.1
8210588.0
1082.102
)1( 3
63
'
0
1
0 







EX
xe
xmx
ex
N
N
W
M
A
N



=181.45N/mm2
<f=215 N/mm2
, OK
Out-plane Overall stability checking of Column
Calculated length according two purlin distance or fly bracing distance ly=3000mm
uniform section structure : γ=0，μs=μw=1
λy=μsl/iy0=3000/44.2=67.9，b type section , ψy=0.764
6.0138.1)
4264.4
129.67
(
101252
4268210
9.67
4320 2
32






by
ψ b’=1.07－0.282/ψ by=0.822
942.0)
N
N
(75.0
N
N
-1.0 2
'
EX0
'
EX0
t
22
3
63
10
/215/32.193
101252822.0
1030.193942.0
8210764.0
1082.102
mmNfmmN
W
M
A
N
bre
t
ey










(4)《Design Code for the Steel Structure》(GB50017－2003) Check the steel column to
allow high thickness ratio
Column Top Section:
2
6
33
max
min /
21.136
99.155
10.14689.9
1028181
2131030.193
8210
1022.81
mmN







87.1
max
minmax
0 





0.111
235
)2.265.048(25.53 0
0

yw ft
h
 ，OK
Column Bottom Section :
00 
5.72
235
)255.016(25.53 0
0

yw ft
h
 , OK

22. - 22 -
4.Check the lateral movement of the frame under wind load , μ
Ic=Ib=28181cm4
，ζt= Ic l/hIb=18000/6000=3.0
The equivalent horizontal force of the column is calculated by the following formula :
H=0.67W=0.67×13.56=9.09KN
W=(ω1+ω4)·h=(0.71+1.55)×6.0=13.56KN
mmhmm
EI
Hh
t
c
40150/][1.14)32(
10281811020612
60001009.9
)2(
12 43
333



 
6.Detail Checking
(1) Column and Beam Connected node
1) Bolt Strength Checking
Column and beam node used 10.9Gr , connected by M22 frictional high-strength bolts,
The contact surface of the component is sandblasted, Friction surface anti-slip coefficient
μ = 0.45, each high-strength bolts of the pre-tension of 190KN, the transmission of
internal force design value: N = 39.89KN, V = 77.60KN , M=193.30KN·m
The tension of each bolt :
KNKN
n
N
y
My
N
i
1521908.065.128
8
89.39
)16.0265.0(4
265.030.193
222
1
1 




KNKN
n
N
y
My
N
i
1521908.070.75
8
89.39
)16.0265.0(4
16.030.193
222
2
2 




Shear force of Bolt Group :
KNVKNpnN f
b
V 60.776.615819045.019.09.0   , OK
Outside row bolt of shear, tension force :
197.0
152
65.128
8/6.615
8/60.77
 b
t
t
b
V
V
N
N
N
N
，OK
2)End Plate thickness Checking
End Plate thickness : t=21mm。
Calculated according double side support end plate :

23. - 23 -
mm
feeebe
Nee
t
wffw
twf
9.20
205)]4640(40220046[
1065.12846406
)](2[
6 3






(3) Calculation of shear stress of beam , column joints
22
6
/125/52.106
10426426
1030.193
mmNfmmN
tdd
M
v
ccb



 , OK
(4)Web Plate strength calculated on bolt area
Nt2=75.70KN<0.4P=0.4×190=76.0KN
22
3
/215/52.206
846
101904.04.0
mmNfmmN
te
P
ww



 ,OK
(2)Mid-span Beam Node
1)Bolt Strength Checking
Mid-span beam node used 10.9Gr , connected by M20 frictional high-strength bolts, The
contact surface of the component is sandblasted, Friction surface anti-slip coefficient
μ = 0.45, each high-strength bolts of the pre-tension of 155KN, the transmission of
internal force design value: N=31.81KN，V=3.21KN，M=143.18KN·m

24. - 24 -
The tension of each bolt :
KNKN
n
N
y
My
N
i
1241558.001.95
8
81.31
)16.0265.0(4
265.018.143
222
1
1 




KNKN
n
N
y
My
N
i
1241558.079.55
8
81.31
)16.0265.0(4
16.018.143
222
2
2 




Shear force of Bolt Group :
KNVKNpnN f
b
V 21.32.502815545.019.09.0   , OK
Outside row bolt of shear, tension force :
177.0
124
01.95
8/2.502
8/21.3
 b
t
t
b
V
V
N
N
N
N
, OK
2)End Plate thickness checking
End Plate thickness t=18mm
Calculated according double side support end plate :
mm
feeebe
Nee
t
wffw
twf
8.17
205)]4640(40220046[
1001.9546406
)](2[
6 3






3) Web Plate strength calculated on bolt area
Nt2=55.79KN<0.4P=0.4×155=62.0KN
22
3
/215/48.168
846
101554.04.0
mmNfmmN
te
P
ww



 , OK

25. - 25 -
Column Foot Design
Column hinge connected with foundation, used plane type hinge connected column foot
1)Column foot internal force design value:
Nmax=102.82KN , V=32.21KN；
Nmin=19.05KN , V=4.30KN。
2)Because column foot shearing force
smaller .
Ｖmax=32.21KN<0.4Nmax=41.13KN，so the
span no need the shear key, but
calculated the column foot which
arrange column bracing need the shear
key.
Nmin>0，Considering the column bracing
pull up force, the bolt is still not bear the
tension, so only consider the stability of column in the installation process o, according to

26. - 26 -
the requirements of the structure can be set to anchor bolt , used 4M24 bolt .
3)Column base plate area and thickness calculated
A. Confirmed the column base plate area
b=b0+2t+2c=200+2×12+2×(20~50)=264~324mm，b=300mm；
h=h0+2t+2c=450+2×12+2×(20~50)=514~574mm，h=550mm；
Checking the concrete under base plate design value of Axial compressive strength:
Foundation used C20 Concrete , fc=9.6N/mm2
22
3
/6.9/62.0
550300
1082.102
mmNfmmN
bh
N
cc 


  , OK
B. Confirmed the thickness of Base plate
According the Column Base Plate area that segmented by column web plate and flange
plate calculate separately the maximum bending moment that base plate bearing, as for
three plate support area : b2/b1=96/426=0.225<0.3 , Calculated by cantilever plate with
length of b2 : mNaM  660814662.0
2
1
2
1 22
4
Cantilever Plate part : mNaM  7755062.0
2
1
2
1 22
4
Base Plate thickness : mmfMt 6.13215/66086/6 max  , t=20mm。
6.Other Structure Calculated
(1) Fly Bracing Design
Fly Bracing design according axle-center pressed structural member, axle center N
calculated as following :

27. - 27 -
KNN
fAf
N
y
16.121009.12
68.44cos60
21512200
235cos60
3



 

Connected bolt used normal C grade M12
Bolt , The calculation length of the corner
supports the distance between the two ends
of the bolt center : l0=633mm , used L50x4 ,
section properties:
A=3.90cm2
，Iu=14.69cm4
，Wu=4.16cm3
，
iu=1.94cm，iv=0.99cm
λ u=l0/ iu=633/19.4=32.6<[λ ]=200，
b type section , ψu=0.927
Strength design value that single side
connected angle multiplied with the
reduction factor , αy：λ=633/9.9=63.94，
αy=0.6+0.0015λ=0.696
22
3
/215/0.48
390927.0696.0
1016.12
mmNfmmN
A
N
uy





 , OK
(2) Purlin Design
Basic Information
Purlin used Galvanized C section steel, design according
single span simple member, roof slope :1/10,purlin
span6m ,set one row sag rod , purlin distance 1.5m ,steel
materials Q235B.
Loading and Internal force
Consider the combination of dead load and roof live load as the control effect ,
Purlin linear load standard value: Pk=(0.27+0.5)×1.5=1.155KN/m
Purlin linear load design value: Pk=(1.2×0.27+1.4×0.5)×1.5=1.536KN/m
Px=Psinα=0.153KN/m，Py=Pcosα=1.528KN/m ;
Design value of bending moment :

28. - 28 -
Mx=Pyl2
/8=1.528×62
/8=6.88KN·m
My=Pxl2
/8=0.153×62
/32=0.17KN·m
Section choose and section properties
Choose : C180×70×20×2.2
Ix=374.90cm4
，Wx=41.66cm3
，ix=7.06cm；
Iy=48.97cm4
，Wymax=23.19cm3
，Wymin=10.02cm3
，iy=2.55cm，χ 0=2.11cm；
Section stress calculated according gross section :
2
3
6
3
6
max
1 /48.172
1019.23
1017.0
1066.41
1088.6
mmN
W
M
W
M
y
y
x
x






 （Press）
2
3
6
3
6
min
2 /18.148
1002.10
1017.0
1066.41
1088.6
mmN
W
M
W
M
y
y
x
x






 （Press）
2
3
6
3
6
max
3 /82.157
1019.23
1017.0
1066.41
1088.6
mmN
W
M
W
M
y
y
x
x






 （Pull）
Stability Coefficient of the stress steel structure
Web Plate
Web Plate is the stiffened Plate,ψ=σmin/σmax=－157.82/172.48=－0.915>－1，
k=7.8－6.29ψ+9.78ψ2
=21.743
Up Flange Plate
Up flange plate is the biggest press act on the support edge of the stiffened plate
ψ=σmin/σmax=148.18/172.48=0.859>－1,
kc=5.89－11.59ψ+6.68ψ2
=0.863
The effective wide of the press plate
Web Plate
k=21.743，kc=0.863，b=180mm，c=70mm，t=2.2mm，σ1=172.48N/mm2
1.1952.1
863.0
743.21
180
70

ck
k
b
c

Coefficient of constraint action between plate components
k1=0.11+0.93/(ξ－0.05)2
=0.367
080.348.172/743.21367.0205/205 11   kk

29. - 29 -
Because ψ=σmin/σmax<0 , α=1.5 .
bc=b/(1－ψ)=180/(1+0.915)=93.99mm
b/t=180/2.2=81.82
18αρ=18×1.15×3.080=63.76，38αρ=38×1.15×3.080=134.60
So 18αρ<b/t<38αρ
So the effective wide of section
mmb
tb
b ce 62.8199.93)1.0
82.81
060.315.18.21
)1.0
/
8.21
( 



be1=0.4be=0.4×81.62=32.65mm，be2=0.6be=0.6×81.62=48.97mm
B. Top flange plate
k=0.863 , kc=21.743, b=70mm, c=180mm , σ1=172.48N/mm2
1.1512.0
743.21
863.0
70
180

ck
k
b
c

Coefficient of constraint action between plate components
398.1512.0/1/11  k
197.148.172/863.0398.1205/205 11   kk
Because ψ=σmin/σmax>0，则 α=1.15－0.15ψ=1.15－0.15×0.859=1.021，
bc=b=70mm，b/t=70/2.2=31.82
18αρ=18×1.021×1.197=22.00，38αρ=38×1.021×1.197=46.44
So 18αρ<b/t<38αρ
So the effective wide of section
mmb
tb
b ce 05.5770)1.0
82.31
197.1021.18.21
)1.0
/
8.21
( 



be1=0.4be=0.4×57.0
5=22.82mm，be2=0.6be=0.6×57.05=34.23mm
Purlin top flange plate and Web plate effective area

30. - 30 -
C. Lower Flange Plate
Lower Flange Plate total cross section under tension and effective .
D. Effective net-section modulus
Top Flange deduct area the wide :70－57.05=12.95mmWeb Plate deduct area the wide :
93.99－81.62=12.37mm, and the calculated section of web plate with aφ 13 sag rod hole
(35mm distance from top flange plate), hole position same as the deduct area, so the
deduct area of web plate calculated accordingφ 13.see figure ,the effective net-section
modulus :
34
224
10813.3
90
)3590(2.213902.295.121090.374
mmWenx 


1.21
)2/2.21.21(2.213)1.2182.222/95.12(2.295.121097.48 224
max

enyW
34
10257.2 mm
1.2170
)2/2.21.21(2.213)1.2182.222/95.12(2.295.121097.48 224
max


enyW
34
10974.0 mm
Wenx/Wx=0.915，Wenymax/Wymax=0.973，Wenymin/Wymin=0.972
Strength Calculated
Consider roof could stop purlin lateral buckling and torsion:
22
4
6
4
6
max
1 /205/97.187
10257.2
1017.0
10813.3
1088.6
mmNfmmN
We
M
W
M
ny
y
enx
x







22
4
6
4
6
min
2 /205/99.162
10974.0
1017.0
10813.3
1088.6
mmNfmmN
We
M
W
M
ny
y
enx
x







Deflection Calculated
mmlmmy 30200/][11.25
109.37410206
6000"38'425cos155.1
384
5
43
4



 

,OK

31. - 31 -
Structural requirement
λx=600/7.06=85.0<[λ]=200 , OK
λy=300/2.55=117.6<[λ]=200 , OK
(3) Wall Girt Design
1)Basic Information
The Building project is single floor workshop, column distance 6m,eave height
6m ,above1.2m is corrugated single color sheet ,wall girt distance 1.5m, one row sag rod,
steel materials Q235B
2)Loading Calculated
Wall Girt used Galvanized C section steel :160x60x20x2.5mm ,wall weight : 0.22KN/m2
Wind Loading
Basic wind loading: ω0=1.05×0.45=0.473KN/m2
, wind loading standard value calculated
according CECS102:2002 ωk=μsμzω0，μs=－1.1（+1.0）
When Calculated the wall girt no need calculated the weight of wall ,the wall sit on ground
qx=1.2×0.07=0.084KN/m，qy=－1.1×0.473×1.5×1.4=－1.093KN/m
3)Internal force calculated
Mx=0.084×62
/8=0.378KN·m , My=1.093×62
/8=4.919KN·m
4)Strength Calculated
Wall Girt :C160×60×20×2.5mm
Wxmax=19.47cm3
，Wmin=8.66cm3
，Wy=36.02cm3
，Iy=288.13cm4
Reference wall purlin calculated result and project experience
Wenx=0.9 Wx , Weny=0.9 Wy
22
3
6
3
6
/205/2.200
1002.369.0
10919.4
1066.89.0
10378.0
mmNfmmN
W
M
W
M
eny
y
enx
x







Under wind suction , the sag rod set at the internal of wall girt ,and arrange the inclined
sag rod at the bottom of column ,and corrugated color sheet strong connected with

32. - 32 -
outside of wall girt ,so don’t need calculated the overall stability of wall girt.
5) Deflection calculated
mmlmm 30200/][3.22
1013.28810206
60005.1473.01.1
384
5
43
4



  , OK
(4) Wind Column Design
Basic information
Wind Column height 6274mm ,distance 6m ,bearing loading including self-weight, wall
girt weight ,and gable wall wind loading ,wind column hinge connected with foundation,
design as compression-bending members. Wind Column used as simple member which
support roof beam and foundation.
Wind Pressure ω0=0.45KN/m2
, Ground roughness category B, Fly Bracing distance
3.0m,Wind Column used Q235B steel .
Loading Calculated
Wind Column used H section steel 300×200×6×10, self-weight:g1=44.6kg/m
Wind Loading calculated according : CECS102:2002
ωk=μsμzω0 ,μs=－1.0(+1.0) ,ω0=1.05×0.45=0.473KN/m2
qz=1.2×(0.07×6×3+44.6×6.274×10-2
)=4.87KN
qy=1.4×1.0×1.0×0.473×6=3.97KN/m
Wind Column eccentricity under wall girt act :1.2×0.07×6×3×0.23=0.35KN·m
Internal Force Calculated
N=4.87KN , M=1/8×3.97×6.2742
+0.35=19.88KN·m
Local Stability Checking of Steel Structure
Width-to-thickness ratio of flange : b/t=96/10=9.6< yf/23513
2
3
63
max
min /
49.30
21.32
101.634
1088.19
56800
1087.4
mmN
W
M
A
N
x
x







947.1
max
minmax
0 




 ，因 1.6<α 0<2.0，
l0=6274mm，λ x= l0/ ix=48.5<[λ ]=150

33. - 33 -
7.46
6
280
5.91
235
)2.265.048( 0
0 
wy t
h
f
 , OK
Strength Calculated
Section Properties : A=56.8cm2
，Ix=9511cm4，Wx=634.1cm3
，ix=12.94cm，
Iy=1334cm4
，Wy=133.4cm3
，iy=4.85cm
22
3
63
/215/7.30
101.63405.1
1088.19
56800
1087.4
mmNfmmN
W
M
A
N
nxx
x
n







Checking the In-plane stability under bending moment
λ=48.5 , b type section , ψx=0.863
KN
EA
NEX 1.4463
5.481.1
568010206
1.1 2
32
2
2
'







，βmx=1.0
)
1.4463
87.4
8.01(101.63405.1
1088.190.1
56800863.0
1087.4
)8.01( 3
63
'1 







EX
xx
xmx
x
N
N
W
M
A
N



=30.85N/mm2
<f=215 N/mm2
, OK
Checking the Out-plane stability under bending moment
Consider the fly bracing as Lateral Support of Out-plane wind column l0y=3000mm，
λy= l0y/ iy=3000/48.5=61.9<[λ]=150，b type section , ψy=0.797
983.0
23544000
07.1
2

yy
b
f
 ，η=1.0，βtx=1.0
3
63
1 101.634983.0
1088.190.10.1
56800797.0
1087.4






xb
xtx
y W
M
A
N




=32.97N/mm2
<f=215 N/mm2
OK
Deflection Calculated
Wind Column under the action of horizontal wind loading could see as single
span ,simply-supported beam and calculated horizontal deflection according below:
mmlmm
EI
l
x
k
7.15400/][1.4
10951110206
627497.3
384
5
384
5
43
44



 



34. - 34 -
Column Foot Design
Because wind column bear small vertical load, so the base plate size used
400x300x20mm,anchor bolt used 2M20.
(5)Column Bracing Design
Column Bracing Layout as Below
Column bracing is diagonal strut, used X cross rod bracing ,strut could also used as purlin,
Column Bracing loading and internal force
Bracing calculated figure as below :
The wind loading at the top of gable wall(gable wall height 7.2m)
μs=0.8+0.5=1.3 ,ω1=1.3×1.0×0.45×18×7.35/2=38.70KN

35. - 35 -
According half wall action on 1/3 wind loading consider the loading standard value:
Fwk=1/3×1/2×38.70=6.45KN
Node loading design value : Fw=1.4×6.45=9.03KN
diagonal strut tension design value N=9.03/cos43.9191°=12.54KN
diagonal strut section design and strength checking
used φ12 rod , A=113.0mm2
Strength Calculated :N/A=12.54×103
/113.0=111.0N/mm2
<f= 215N/mm2
Ridge Calculated : tension rod no need consider the required of slenderness ratio
But Consider the structural used φ16 rod
(6)Roof Horizontal Bracing
Roof Bracing Layout Plan
Purlin distance 1.5m ,horizontal bracing :3m
Roof Bracing loading and internal force
Roof bracing used tension rod, bracing calculated as figure.
On gable wall side wind loading shape coefficient μs=1.0.

36. - 36 -
Node Loading Standard Value: Fwk=0.45×1.0×1.0×3.0×7.35/2=4.96KN ;
Node Loading Design Value : Fw=4.96×1.4=6.94KN；
Diagonal strut tension design value : N=2.5×6.94/cos29.0546°=19.85KN
Diagonal strut section design and strength calculated
Diagonal strut used :φ12 rod , A=113.0mm2
Strength Calculated :N/A=19.85×103
/113.0=175.7N/mm2
<f= 215N/mm2
Ridge Calculated: tension rod no need consider the required of slenderness ratio
But Consider the structural used φ16 rod
(7)Canopy Design
(1) Basic Information
Canopy total length 6000mm,overhang type, overhang length 1500,used Q235B steel,
Purlin used C180×70×20×2.2mm
(2) Loading Calculated
1)Dead Loading
Corrugated single color sheet 0.15KN/m2
Purlin, Canopy Beam and other structures 0.10 KN/m2
Total 0.25 KN/m2
Canopy Beam linear load standard value : 0.25×3=0.75 KN/m2
2)Live Loading
Along the wide of sheet each 1.0m choose one construction and repair loading ,each
concentrated load 1.0Kn, act area at the outside edge of canopy, so the live loading on
canopy is 3.5Kn.
3)Wind Loading
Wind loading shape coefficient of canopy :μs=2.0 ,ω0=0.45KN/m2
ωk=μsμzω0=2.0×1.0×0.45=0.90 KN/m2
Convert the loading standard value act on
canopy beam: 0.90×3=2.70KN/m
(3)Internal calculated and section design

37. - 37 -
Canopy Beam calculated as figure g+q=1.2×0.75+1.4×2.70=4.68KN/m
P=1.4×3.5=4.9KN
Most critical section on the root of beam :
M=12.62KN·m , V=11.92KN
Canopy Beam used variable cross-section built up H beam (200~100)×150×6×8
Beam root section properties :
A=3504m2
，Ix=6×1843
/12+8×150×962
×2=2523×104
mm4
，
Wx=2523×104
/96=26.3×104
mm3
22
3
0
/125/8.10
1846
1092.11
mmNfmmN
th
V
v
w



 , OK
22
4
6
/215/7.45
103.2605.1
1062.12
mmNfmmN
W
M
nxx
x





, OK
Canopy beam connected with column used 4M20 normal C grade bolt
－END－