KUY Limeng,e20190482(I4GCI-B).pdf

Assignment 3
Combined bending and axial compression
Lecturer : Mr. LY Hav
Student : KUY Limeng
ID : e20190482
Group : I4-GCI B
2022-2023
Institut de Technologie
du Cambodge
Faculté de Génie Civil
Département de Génie Civil

Institute de Technologie du Cambodge Construction Métallique II
KUY Limeng Page 1 of 14 Rev 07/07/2023
ASSIGNMENT 3
Uniform members in bending and axial compression [6.3.3]
The stability of uniform members with double symmetric cross sections for sections not
susceptible to distortional deformations should be checked as given in the following clauses, where
a distinction is made for:
Members which are subjected to combined bending and axial compression should satisfy:
NEd
χy × NRk
γM1
+ kyy ×
My,Ed + ΔMy,Ed
χLT ×
My,Rk
γM1
+ kyz ×
Mz,Ed + ΔMz,Ed
Mz,Rk
γM1
≤ 1
NEd
χz × NRk
γM1
+ kzy ×
Mz,Ed + ΔMzEd
χLT ×
Mz,Rk
γM1
+ kzz ×
Mz,Ed + ΔMz,Ed
Mz,Rk
γM1
≤ 1
Where:
NEd, My,Ed, Mz,Ed : are the design values of the compression force and the maximum
moments about the y-y and z-z axis along the member, respectively.
ΔMy,Ed, ΔMz,Ed : are the moments due to the shift of the centroidal axis for class 4
sections. [Table 6.7]
χy, χz : are the reduction factors due to flexural buckling.
χLT : is the reduction factor due to lateral torsional buckling
kyy, kyz, kzy, kzz : are the interaction factors.
The interaction factors kyy, kyz, kzy and kzz have been derived from two alternative
approaches. Values of these factors may be obtained from Annex A (alternative method 1) or from
Annex B (alternative method 2).
Note: For members not susceptible to torsional deformation χLT would be χLT = 1.0
(6.61)
(6.62)

Interaction factors are obtained from one of two methods: Method 1 and Method 2.
Annex A, Method 1: Interaction factors kij for interaction formula in 6.3.3(4)

For, Interaction factor is depended on the classification of section properties.
Wy and Wz is the appropriate section modulus as follow: (6.55)
• Wy = Wpl,y for Class 1 or 2 cross-sections
• Wy = Wel,y for Class 3 cross-sections
• Wy = Weff,y for Class 4 cross-sections
For the exercise we assume, Cross-section IPE 400, strength S235 and length L= 4 m
Section IPE 400 we have, h = 400 mm b = 180 mm tw = 8.6 mm tf = 13.5 mm
r = 21 mm A = 8450 mm2
Iy = 23130 × 104
mm4
Iz = 1318 × 104
mm4
It = 51.1 × 104
mm4
Iw = 4.9 × 109
mm6

• Determine the class of cross-section
- Internal compression parts
Part subject to bending and compression: We assume 𝛼 = 0.2
By 𝛼 ≤ 0.5 𝑐 = ℎ − 2𝑡𝑓 − 2𝑟 = 331 𝑚𝑚 𝑡 = 𝑡𝑤 = 8.6 𝑚
ε = √
235
𝑓𝑦
= √
235
235
= 1
⇒
𝑐
𝑡
= 38.4884 ≤ 36 ×
𝜖
𝛼
= 180
Thus, it is Class 1
- Outstand flanges
Part subject to bending and compression:
𝑐 =
𝑏 − 𝑡𝑤
2
− 𝑟 = 64.7 𝑚𝑚
𝑡 = 𝑡𝑓 = 13.5𝑚𝑚
⇒
𝑐
𝑡
= 4.7925 ≤ 9 ×
𝜖
𝛼
= 45
Thus, it is Class 1
Therefore, Class of section is class 1
For bars with Class 1 and 2 cross-sections, the above expressions reduce to:
NEd
χy × NRk
γM1
+ kyy ×
My,Ed
χLT ×
My,Rk
γM1
+ kyz ×
Mz,Ed + ΔMz,Ed
Mz,Rk
γM1
≤ 1
NEd
χz × NRk
γM1
+ kzy ×
Mz,Ed
χLT ×
Mz,Rk
γM1
+ kzz ×
Mz,Ed + ΔMz,Ed
Mz,Rk
γM1
≤ 1
and the coefficient, kyy = CmyCmLT
μy
1 −
NEd
Ncr,y
1
Cyy
kyz = Cmz
μy
1 −
NEd
Ncr,z
1
Cyz
0.6√
wz
wy
, kzz = Cmz
μz
1 −
NEd
Ncr,z
1
Czz
, kzy = CmyCmLT
μz
1 −
NEd
Ncr,y
1
Czy
0.6√
wy
wz

and,
with,
• Buckling Curve
h
b
=
400
180
= 2.22 > 1.2 and tf = 13.5 mm < 40 mm
Axes Z-Z : curve b
αz = 0.34 (axe faible courbe b)
⇒ lkz = L × kz = 4m , (kz = 1) Articulée 2 axe faible
Ncr,z =
π2
EIz
lkz
2 =
π2
× 21 × 104
MPa × 1.3 × 107
mm4
(4 × 103mm)2
= 426829.3702N
μy =
1 −
NEd
Ncr,y
1 − χy
NEd
Ncr,y
μz =
1 −
NEd
Ncr,z
1 − χz
NEd
Ncr,z
Cyy = 1 + (wy − 1) [(2 −
1.6
wy
Cmy
2
λ
̅max −
1.6
wy
Cmy
2
λ
̅2
max) npl − bLT] ≥
Wel,y
Wpl,y
bLT = 0.5αLTλ
̅0
2
My,Ed
χLTMpl,y,Ed
Mz,Ed
Mpl,z,Ed
Cyz = 1 + (wz − 1) [(2 − 14
Cmy
2
λ
̅2
max
𝑤𝑧
5
) npl − cLT] ≥ 0.6√
𝑤𝑧
𝑤𝑦
Wel,z
Wpl,z
cLT = 10αLT
λ
̅0
2
5 + λ
̅z
4
My,Ed
CmyχLTMpl,y,Ed
Czy = 1 + (wy − 1) [(2 − 14
Cmy
2
λ
̅2
max
𝑤𝑦
5 ) npl − dLT] ≥ 0.6√
𝑤𝑦
𝑤𝑧
Wel,y
Wpl,y
cLT = 2αLT
λ
̅0
2
0.1 + λ
̅z
4
My,Ed
CmyχLTMpl,y,Ed
Mz,Ed
CmzχLTMpl,z,Ed
Czz = 1 + (wz − 1) [(2 −
1.6
wz
Cmz
2
λ
̅max −
1.6
wz
Cmz
2
λ
̅2
max) npl − eLT] ≥
Wel,z
Wpl,z
eLT = 1.7αLT
λ
̅0
2
0.1 + λ
̅z
4
My,Ed
CmyχLTMpl,y,Ed
𝜆̅𝑚𝑎𝑥 = max(𝜆̅𝑦, 𝜆̅𝑧)

λ
̅z = √
Afy
Ncr,z
= √
8450 × 235
426829.3702
= 2.1569
ϕz = 0.5[1 + αz(λ
̅z − 0.2) + λ
̅z
2
] = 0.5[1 + 0.34(2.1569 − 0.2) + 2.15692] = 3.1588
χz =
1
ϕz + √ϕz
2 − λ
̅z
2
=
1
3.1588 + √3.15882 − 2.15692
= 0.1829
Axes-Y-Y : curve a
αy = 0.21(axe faible courbe a)
⇒ lky = L × ky = 8m , (ky = 2) Encastrée libre axe fort
Ncr,y =
π2
EIy
lky
2 =
π2
× 21 × 104
MPa × 2.31 × 108
mm4
(8 × 103mm)2
= 7490563.986N
λ
̅y = √
Afy
Ncr,y
= √
8450 × 235
7490563.986
= 0.5148
⇒ λ
̅max = max(λ
̅z, λ
̅y) = 2.1569
ϕy = 0.5[1 + αy(λ
̅y − 0.2) + λ
̅y
2
] = 0.5[1 + 0.21(0.5148 − 0.2) + 0.51482] = 0.6656
χy =
1
ϕy + √ϕy
2 − λ
̅y
2
=
1
0.6656 + √0.66562 − 0.51482
= 0.9195

• Elastic torsional-flexural buckling force
Ncr,T =
1
i0
2 (GIt +
π2
EIw
Lcr,T
2 )
It =
2
3
(b − 0.63tf)tf
3
+
1
3
(h − 2tf)tw
2
+ 2 (
tw
tf
) (0.145 + 0.1
r
tf
) [
(r +
tw
2
)
2
+ (r + tf)2
− r2
2r + tf
]
4
It = 510754.7221mm4
Iw =
tfb3
24
× (h − tf)2
= 4.9 × 1011
mm6
i0
2
= iy
2
+ iz
2
+ y0
2
+ z0
2
y0 = 0 and z0 = 0 cus it is coordinate of shear center compare to center of gravity and
section is bisymmetry.
Assume iy = 2mm iz = 3mm
⇒ i0
2
= 22
+ 32
= 13mm2
Lcr,T = 2 × 4000mm = 8000mm cus buckling in axe strong
Ncr,T =
1
13
(80769.23077 × 510754.7221 +
π2
× 21 × 104
× 4.9 × 1011
80002
= 4394100697N
β = 1 − (
y0
i0
)
2
= 1
Ncr,TF =
1
2β
[(Ncr,y + Ncr,T) − √(Ncr,y + Ncr,T)
2
− 4βNcr,yNcr,T]
Ncr,TF = 7490563.986N
• Lateral Buckling
h
b
=
400
180
= 2.22 > 1.2 and tf = 13.5 mm < 40 mm

So Lateral Buckling has curved b
⇒ αLT = 0.34
• The value of this coefficient χLT can be obtained by the expression:
χLT =
1
ΦLT + √ΦLT
2
− λ
̅2
LT
where, ΦLT = 0.5[1 + αLT(λ
̅LT − 0.2) + λ
̅2
LT
λ
̅LT = √
Wyfy
Mcr
• Calculation of the elastic Critical moment Mcr
For beams with constant cross-section, the critical buckling moment is given by the
general formula:
Mcr = C1
π2
EIz
(kL)2
× {[(
k
kw
)
2
Iw
Iz
+
(kL)2
× GIt
π2EIz
+ (C2zg − C3zj)
2
]
1
2
− (C2zg − C3zj)}
Where, C1, C2 and C3 factors depending on the load and embedding conditions given
in the tables (Appendix F)
(6.56)

C1 = 2.752 C2 = 0 C3 = 0
k = 1 simple support kw = 1
Coordinate of point which load apply on
za =
h
2
=
400
2
= 200 mm
Coordinate of shear center
zs = 0 m
zg = za − zs = 200 − 0 = 200 mm
zj = 0
⇒ Mcr = 1.163 × 109
Nmm
Wy = Wpl,y = 1307000 mm3
⇒ λ
̅LT = √
Wyfy
Mcr
= √
1307000 × 235
1.163 × 109
= 0.513
⇒ ϕLT = 0.5[1 + 0.34(0.513 − 0.2) + 0.5132] = 0.685
⇒ 𝜒𝐿𝑇 =
1
0.685 + √0.6852 − 0.5132
= 0.878

• Interaction Factors [Annex A, Table A.2]
We assume ψi = −1
Cmy,0 = 0.79 + 0.21ψi + 0.36(ψi − 0.33)
NEd
Ncr,y
= 0.579
Cmz,0 = 0.79 + 0.21ψi + 0.36(ψi − 0.33)
NEd
Ncr,z
= 0.565
λ
̅0 = 0.2
λ
̅0,lim = 0.2√C1 √(1 −
NEd
Ncr,z
) (1 −
NEd
Ncr,TF
)
4
By using Table F.1.2, we get:
C1 = 2.752 C2 = 0 C3 = 0
⇒ λ
̅0,lim = 0.2√2.572√(1 −
13000
426829.3702
) (1 −
13000
7490563.986
)
4
= 0.329
λ
̅0 = 0.2 < λ
̅0,lim = 0.329
⇒ Cmy = Cmy,0 = 0.579
⇒ Cmz = Cmz,0 = 0.565
⇒ CmLT = 1.0

aLT = max (1 −
IT
Iy
, 0) = max (1 −
510754.7221mm4
2.31 × 108mm4
, 0) = 0.9977
γM1 = 1
npl =
NEd
NRk/γM1
=
13000
1985750/1
= 0.006546
Wy = min (
Wpl,y
Wel,y
, 1.5) = min (
1307000
1160000
, 1.5) = 1.1267
Wz = min (
Wpl,z
Wel,z
, 1.5) = min (
229000
146000
, 1.5) = 1.5
μy =
1 −
NEd
Ncr,y
1 − χy.
NEd
Ncr,y
=
1 −
13000
7490563.986
1 − 0.9195 ×
13000
7490563.986
= 0.9998
μz =
1 −
NEd
Ncr,z
1 − χz.
NEd
Ncr,z
=
1 −
13000
426829.3702
1 − 0.1829 ×
13000
426829.3702
= 0.9749
Mpl,y,Rd =
Wpl,y. fy
γM0
=
1307000mm3
× 235MPa
1
= 307145000N. mm
Mpl,z,Rd =
Wpl,z. fy
γM0
=
229000mm3
× 235MPa
1
= 53815000N. mm
χLT = 0.878
bLT = 0.5aLTλ
̅0
2
My,Ed
χLT Mpl,y,Rd
Mz,Ed
Mpl,z,Rd
⇒ bLT = 0.5 × 0.9977 × 0.22
15 × 106
0.878 × 307145000
12 × 106
53815000
= 0.00024
cLT = 10aLT
λ
̅0
2
5 + λ
̅z
4
My,Ed
Cmy. χLT. Mpl,y,Rd
⇒ cLT = 10 × 0.9977
0.22
5 + 2.15694
15 × 106
0.579 × 0.878 × 307145000
= 0.0014
dLT = 2aLT
λ
̅0
0.1 + λ
̅z
4
My,Ed
Cmy. χLT. Mpl,y,Rd
Mz,Ed
Cmz. Mpl,z,Rd

⇒ dLT = 2 × 0.9977
0.2
0.1 + 2.15694
15 × 106
0.579 × 0.878 × 307145000
12 × 106
0.565 × 53815000
dLT = 0.00069
eLT = 1.7aLT
λ
̅0
0.1 + λ
̅z
4
My,Ed
Cmy. χLT. Mpl,y,Rd
⇒ eLT = 1.7 × 0.9977
0.2
0.1 + 2.15694
15 × 106
0.579 × 0.878 × 307145000
= 0.0014
Cyy = max (1 + (Wy − 1) [(2 −
1.6
Wy
Cmy
2
λ
̅max −
1.6
Wy
Cmy
2
λ
̅max
2
) npl − bLT] ,
Wel,y
Wpl,y
)
Cyy = 0.9989
Cyz = max (1 + (Wz − 1) [(2 − 14
Cmz
2
λ
̅max
2
Wz
2
) npl − cLT] , 0.6√
Wz
Wy
Wel,z
Wpl,z
)
Cyz = 0.9968
Czy = max (1 + (Wy − 1) [(2 − 14
Cmy
2
λ
̅max
2
Wy
2
) npl − dLT] , 0.6√
Wy
Wz
Wel,y
Wpl,y
)
Czy = 0.9915
Czz = max (1 + (Wz − 1) [(2 −
1.6
Wz
Cmz
2
λ
̅max −
1.6
Wz
Cmz
2
λ
̅max
2
) npl − eLT] ,
Wel,z
Wpl,z
)
Czz = 0.9981
kyy = CmyCmLT
μy
1 −
NEd
Ncr,y
1
Cyy
⇒ kyy = 0.579 × 1 ×
0.9998
1 −
13000
7490563.986
1
0.9989
= 0.58
kyz = Cmz
μy
1 −
NEd
Ncr,z
1
Cyz
0.6√
wz
wy
⇒ kyz = 0.565 ×
0.9998
1 −
13000
426829.3702
1
0.9968
0.6√
1.5
1.1267
= 0.4

kzy = CmyCmLT
μz
1 −
NEd
Ncr,y
1
Czy
0.6√
wy
wz
⇒ kzy = 0.579 × 1 ×
0.9749
1 −
13000
7490563.986
1
0.9915
0.6√
1.1267
1.5
= 0.296
kzz = Cmz
μz
1 −
NEd
Ncr,z
1
Czz
⇒ kzz = 0.565 ×
0.9998
1 −
13000
426829.3702
1
0.9981
= 0.569
Therefore:
NEd
χy.
NRk
γM1
+ kyy
My,Ed
χLT.
My,Rk
γM1
+ kyz
Mz,Ed
Mz,Rk
γM1
≤ 1
13000
0.9195 ×
1985750
1
+ 0.58 ×
15 × 106
0.136 ×
307145000
1
+ 0.4 ×
12 × 106
53815000
1
= 0.1297 < 1 𝐎𝐤
NEd
χz.
NRk
γM1
+ kzy
My,Ed
χLT.
My,Rk
γM1
+ kzz
Mz,Ed
Mz,Rk
γM1
≤ 1
13000
0.1829 ×
1985750
1
+ 0.296 ×
15 × 106
0.136 ×
307145000
1
+ 0.569 ×
12 × 106
53815000
1
= 0.1793 < 1 𝐎𝐤

REFERENCE:
Eurocode 3: Design of steel structures, Part 1.1: General rules and rules for
buildings, (Together with United Kingdom National Application Document)
Resistance des barres, théorie de stabilité (TOME IV)

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