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NDECK

This document contains analysis of the structural capacity of steel joist elements. It describes the geometry and material properties of standard joist sections, and performs checks for several limit states including bending capacity, shear strength, and vibration criteria. The calculations reference design standards AISC 360 and Design Guide 11. Three joist elements are analyzed, with the nominal strengths for bending and shear calculated and compared to applied demands to check for compliance.

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Nhân Lê Trọng
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© © All Rights Reserved
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59 views12 pages

NDECK

This document contains analysis of the structural capacity of steel joist elements. It describes the geometry and material properties of standard joist sections, and performs checks for several limit states including bending capacity, shear strength, and vibration criteria. The calculations reference design standards AISC 360 and Design Guide 11. Three joist elements are analyzed, with the nominal strengths for bending and shear calculated and compared to applied demands to check for compliance.

Uploaded by

Nhân Lê Trọng
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as TXT, PDF, TXT or read online on Scribd
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____ _____ _____ _____

| \| _ | __| __|
| | | | __|__ |
|____/|__|__|__| |_____|

Version : v1.0

Standard : AISC 360 and Design Guide 11 (Published by AISC)

DAFS CODE

# Model
1 S=0,0 E=0,6 RS=M RE=M
T 350 350 8 200 10 200 10
Bracing 0T 1.5T 3T 4.5T 6T

# Loading
1 DLPRE PEGG 14ssđc3.85 -0.00
1 DLPRE PEGG 96.97 -0.25
1 DLPRE PEGG 96.97 -0.50
1 DLPRE PEGG 96.97 -0.75
1 DLPRE PEGG 143.85 -1.00
1 LLPRE PEGG 5.06 -0.00
1 LLPRE PEGG 3.d37 -0.25
1 LLPRE PEGG 3.38 -0.50
1 LLPRE PEGG 3.3d6 -1.00
1 DLCOM PEGG 164.d47 -0.25
1 DLCOM PEGG 110.đá7 -0.50
1 DLCOM PEGG 110.47 -0.75
1 DLCOM PEGG 164.1 -1.00
1 LLCOM PEGG 16.87 -0.00
1 LLCOM PEGG 11.25 -0.25
1 LLCOM PEGG 11.25 -0.50
1 LLCOM PEGG 11.25 -0.75
1 LLCOM PEGG 16.87 -1.00

CHECKING VIBRATION FLOOR

Criterion : WALKING
Occupancy : PAPER OFFICE
Accelarate Ratio Limit : [a/g] = 0.50 %
The Effecitive Panel W : W = 746.48 kN
The frequency of system: fn = 4.64 Hz
Calculation for a/g : a/dedeg = 0.29*exp(-0.35*fn)/(0.03*W) = 0.26 % =>
Satisfield

CHECKING CAPACITY

***********************************************************************************
*****
Shear Stud Checking
***********************************************************************************
*****
Factor : Rg = 1.00
Factor : Rp = 0.60
Diameter : d = 19.00 mm²
Modulus of Concrete : Ec = 4700 x fc^0.5 = 15938.48 MPa
Tensile strength of stud : Fu = 500.00 MPa
Maximum shear force for stud : V' = ekN dededeededAISC 360)
Capacity per stud : Qn = min(Rg*Rp*Asc*Fu,0.5*Asc*sqrt(fc*Ec)) =
60.69 kN
Number of stud : n = V'/Qn = 78 pcs
Stud Spacing : d = L_joist/(n-1) = 77.92 mm < 250mm => Non-
Satisfiled

***********************************************************************************
*****
Element ID : 1 Type Element : Joist
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm
ee
- Section Properties (Referred To Compree cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.d83 cm^4
Section Modulus : edSxc = 855.ed23 e³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.d.00 cm
de
II.2- Allowable Stress in Bending Mn:
Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00

+ Shear force : 60.06 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.11 (Satisfied)

***********************************************************************************
*****
Element ID : 2 Type Element : Joist
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm

- Section Properties (Referred To Compression Flange):


Section Area : Ax = 68.00 cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.67 cm^4
Moment Inertia : Iy = 1334.83 cm^4
Section Modulus : Sxc = 855.23 cm³
Section Modulus : Sxt = 855.23 cm³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.00 cm
Shear Center : y0 = 0.00 cm

II.2- Allowable Stress in Bending Mn:


Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00
=> Flange is Non-Compact
- Calculate the lateral-torsional buckling modification factor Cb:
Rm = 1.0 (single curvature bending)
Cb = (12.5*Mmax)/(2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 1.00 ≤ 3.00
- The nominal flexural strength :
---------------------------------------------
1- Compression Flange Yielding:
Mn = Fy*Zp = 332.92 kN.m

2- Lateral-Torsional Buckling:
Lp = 1.76*ry*sqrt(E/Fy) = 187.75 cm
Lr =
1.95*rts*(E/(0.7*Fy))*sqrt(J/(Sxc*h0))*sqrt(1+sqrt(1+6.76*((0.7*Fy)*Sxc*h0/
(E*J))²));
Lr = 529.94 cm
Lb = 0.00 cm
Due to Lb ≤ Lp => The limit state of lateral-torsional buckling does not
apply

3- Compression Flange Local Buckling:


λf = bfc/(2*tfc) = 10.00
λpf = λ_pf = 9.15
λrf = λ_pf = 21.26
For Section with Non-Compact Flanges
Mn = Mp-(Mp-0.7*Fy*Sxc)*(λ_f - λ_pf)/(λ_rf - λ_pf) = 324.05 kN.m

Then Mn (Control) = 324.05 kN.m

II.3- Unity Check For Axial and Bending:


------------------------------------
Mr = 0.00 kN.m
Mct = Mnt * Ω = 291.64 kN.m
Ratio check = M/Mct = 0.00 ≤ 1 => Satisfield

III - Check for Shear:


+ Shear force : 121.72 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.22 (Satisfied)

***********************************************************************************
*****
Element ID : 3 Type Element : Joist
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm

- Section Properties (Referred To Compression Flange):


Section Area : Ax = 68.00 cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.67 cm^4
Moment Inertia : Iy = 1334.83 cm^4
Section Modulus : Sxc = 855.23 cm³
Section Modulus : Sxt = 855.23 cm³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.00 cm
Shear Center : y0 = 0.00 cm

II.2- Allowable Stress in Bending Mn:


Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00
=> Flange is Non-Compact
- Calculate the lateral-torsional buckling modification factor Cb:
Rm = 1.0 (single curvature bending)
Cb = (12.5*Mmax)/(2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 1.00 ≤ 3.00
- The nominal flexural strength :
---------------------------------------------
1- Compression Flange Yielding:
Mn = Fy*Zp = 332.92 kN.m

2- Lateral-Torsional Buckling:
Lp = 1.76*ry*sqrt(E/Fy) = 187.75 cm
Lr =
1.95*rts*(E/(0.7*Fy))*sqrt(J/(Sxc*h0))*sqrt(1+sqrt(1+6.76*((0.7*Fy)*Sxc*h0/
(E*J))²));
Lr = 529.94 cm
Lb = 0.00 cm
Due to Lb ≤ Lp => The limit state of lateral-torsional buckling does not
apply

3- Compression Flange Local Buckling:


λf = bfc/(2*tfc) = 10.00
λpf = λ_pf = 9.15
λrf = λ_pf = 21.26
For Section with Non-Compact Flanges
Mn = Mp-(Mp-0.7*Fy*Sxc)*(λ_f - λ_pf)/(λ_rf - λ_pf) = 324.05 kN.m

Then Mn (Control) = 324.05 kN.m

II.3- Unity Check For Axial and Bending:


------------------------------------
Mr = 0.00 kN.m
Mct = Mnt * Ω = 291.64 kN.m
Ratio check = M/Mct = 0.00 ≤ 1 => Satisfield

III - Check for Shear:


+ Shear force : 121.72 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.22 (Satisfied)

***********************************************************************************
*****
Element ID : 4 Type Element : Joist
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm

- Section Properties (Referred To Compression Flange):


Section Area : Ax = 68.00 cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.67 cm^4
Moment Inertia : Iy = 1334.83 cm^4
Section Modulus : Sxc = 855.23 cm³
Section Modulus : Sxt = 855.23 cm³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.00 cm
Shear Center : y0 = 0.00 cm

II.2- Allowable Stress in Bending Mn:


Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00
=> Flange is Non-Compact
- Calculate the lateral-torsional buckling modification factor Cb:
Rm = 1.0 (single curvature bending)
Cb = (12.5*Mmax)/(2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 1.00 ≤ 3.00
- The nominal flexural strength :
---------------------------------------------
1- Compression Flange Yielding:
Mn = Fy*Zp = 332.92 kN.m

2- Lateral-Torsional Buckling:
Lp = 1.76*ry*sqrt(E/Fy) = 187.75 cm
Lr =
1.95*rts*(E/(0.7*Fy))*sqrt(J/(Sxc*h0))*sqrt(1+sqrt(1+6.76*((0.7*Fy)*Sxc*h0/
(E*J))²));
Lr = 529.94 cm
Lb = 0.00 cm
Due to Lb ≤ Lp => The limit state of lateral-torsional buckling does not
apply

3- Compression Flange Local Buckling:


λf = bfc/(2*tfc) = 10.00
λpf = λ_pf = 9.15
λrf = λ_pf = 21.26
For Section with Non-Compact Flanges
Mn = Mp-(Mp-0.7*Fy*Sxc)*(λ_f - λ_pf)/(λ_rf - λ_pf) = 324.05 kN.m

Then Mn (Control) = 324.05 kN.m

II.3- Unity Check For Axial and Bending:


------------------------------------
Mr = 0.00 kN.m
Mct = Mnt * Ω = 291.64 kN.m
Ratio check = M/Mct = 0.00 ≤ 1 => Satisfield

III - Check for Shear:


+ Shear force : 121.72 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.22 (Satisfied)

***********************************************************************************
*****
Element ID : 5 Type Element : Joist
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm

- Section Properties (Referred To Compression Flange):


Section Area : Ax = 68.00 cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.67 cm^4
Moment Inertia : Iy = 1334.83 cm^4
Section Modulus : Sxc = 855.23 cm³
Section Modulus : Sxt = 855.23 cm³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.00 cm
Shear Center : y0 = 0.00 cm

II.2- Allowable Stress in Bending Mn:


Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00
=> Flange is Non-Compact
- Calculate the lateral-torsional buckling modification factor Cb:
Rm = 1.0 (single curvature bending)
Cb = (12.5*Mmax)/(2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 1.00 ≤ 3.00
- The nominal flexural strength :
---------------------------------------------
1- Compression Flange Yielding:
Mn = Fy*Zp = 332.92 kN.m

2- Lateral-Torsional Buckling:
Lp = 1.76*ry*sqrt(E/Fy) = 187.75 cm
Lr =
1.95*rts*(E/(0.7*Fy))*sqrt(J/(Sxc*h0))*sqrt(1+sqrt(1+6.76*((0.7*Fy)*Sxc*h0/
(E*J))²));
Lr = 529.94 cm
Lb = 0.00 cm
Due to Lb ≤ Lp => The limit state of lateral-torsional buckling does not
apply

3- Compression Flange Local Buckling:


λf = bfc/(2*tfc) = 10.00
λpf = λ_pf = 9.15
λrf = λ_pf = 21.26
For Section with Non-Compact Flanges
Mn = Mp-(Mp-0.7*Fy*Sxc)*(λ_f - λ_pf)/(λ_rf - λ_pf) = 324.05 kN.m

Then Mn (Control) = 324.05 kN.m

II.3- Unity Check For Axial and Bending:


------------------------------------
Mr = 0.00 kN.m
Mct = Mnt * Ω = 291.64 kN.m
Ratio check = M/Mct = 0.00 ≤ 1 => Satisfield

III - Check for Shear:


+ Shear force : 60.06 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.11 (Satisfied)

***********************************************************************************
*****
Element ID : 6 Type Element : Girder
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm

- Section Properties (Referred To Compression Flange):


Section Area : Ax = 68.00 cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.67 cm^4
Moment Inertia : Iy = 1334.83 cm^4
Section Modulus : Sxc = 855.23 cm³
Section Modulus : Sxt = 855.23 cm³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.00 cm
Shear Center : y0 = 0.00 cm

II.2- Allowable Stress in Bending Mn:


Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00
=> Flange is Non-Compact
- Calculate the lateral-torsional buckling modification factor Cb:
Rm = 1.0 (single curvature bending)
Cb = (12.5*Mmax)/(2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 1.92 ≤ 3.00
- The nominal flexural strength :
---------------------------------------------
1- Compression Flange Yielding:
Mn = Fy*Zp = 332.92 kN.m

2- Lateral-Torsional Buckling:
Lp = 1.76*ry*sqrt(E/Fy) = 187.75 cm
Lr =
1.95*rts*(E/(0.7*Fy))*sqrt(J/(Sxc*h0))*sqrt(1+sqrt(1+6.76*((0.7*Fy)*Sxc*h0/
(E*J))²));
Lr = 529.94 cm
Lb = 150.00 cm
Due to Lb ≤ Lp => The limit state of lateral-torsional buckling does not
apply

3- Compression Flange Local Buckling:


λf = bfc/(2*tfc) = 10.00
λpf = λ_pf = 9.15
λrf = λ_pf = 21.26
For Section with Non-Compact Flanges
Mn = Mp-(Mp-0.7*Fy*Sxc)*(λ_f - λ_pf)/(λ_rf - λ_pf) = 324.05 kN.m

Then Mn (Control) = 324.05 kN.m

II.3- Unity Check For Axial and Bending:


------------------------------------
Mr = 226.62 kN.m
Mct = Mnt * Ω = 291.64 kN.m
Ratio check = M/Mct = 81596.25 ≤ 1 => Satisfield

III - Check for Shear:


+ Shear force : 181.78 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.33 (Satisfied)

***********************************************************************************
*****
Element ID : 7 Type Element : Girder
***********************************************************************************
*****
I - Section Information :
---------------------------------
- Section Dimensions :
Section Type = Major
Web Dimension = 350 mm x 8 mm
Top Flange = 200 mm x 10 mm
Bottom Flange = 200 mm x 10 mm
========
Compression Flange = 200 mm x 10 mm
Tension Flange = 200 mm x 10 mm

- Section Properties (Referred To Compression Flange):


Section Area : Ax = 68.00 cm²
Gravity Center : YG = 18.50 cm ( From Bottom Flange )
Moment Inertia : Ix = 15821.67 cm^4
Moment Inertia : Iy = 1334.83 cm^4
Section Modulus : Sxc = 855.23 cm³
Section Modulus : Sxt = 855.23 cm³
Plastic Modulus : Zx = 965.00 cm³
Warping Contants : Cw = 432000.00 cm^6
Shear Center : x0 = 0.00 cm
Shear Center : y0 = 0.00 cm

II.2- Allowable Stress in Bending Mn:


Cm = 1.0 From Code
- Check Of Web
hc = 35.00 cm , hp = 35.00 cm
Mp = 332.92 kN.m, My = Sx(Min) * Fy = 295.05 kN.m
Compact Limit = [(hc/hp)*sqrt(E/Fy)]/[0.54*(Mp/My)-0.09]² = 89.28
NonCompact Limit = 5.70*sqrt(E/Fy) = 137.24
hw/tw = 43.75
hw/tw ≤ λ_pw => Web is Compact
- Check Of Compression Flange
kc = 4.0 / sqrt(hw/tw)= 0.60
Compact Limit = 0.38*sqrt(E/Fy) = 9.15
NonCompact Limit = 0.95*sqrt(Kc*E/FL) = 21.26
λ = (b/t)_c = 10.00
=> Flange is Non-Compact
- Calculate the lateral-torsional buckling modification factor Cb:
Rm = 1.0 (single curvature bending)
Cb = (12.5*Mmax)/(2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 1.92 ≤ 3.00
- The nominal flexural strength :
---------------------------------------------
1- Compression Flange Yielding:
Mn = Fy*Zp = 332.92 kN.m

2- Lateral-Torsional Buckling:
Lp = 1.76*ry*sqrt(E/Fy) = 187.75 cm
Lr =
1.95*rts*(E/(0.7*Fy))*sqrt(J/(Sxc*h0))*sqrt(1+sqrt(1+6.76*((0.7*Fy)*Sxc*h0/
(E*J))²));
Lr = 529.94 cm
Lb = 150.00 cm
Due to Lb ≤ Lp => The limit state of lateral-torsional buckling does not
apply

3- Compression Flange Local Buckling:


λf = bfc/(2*tfc) = 10.00
λpf = λ_pf = 9.15
λrf = λ_pf = 21.26
For Section with Non-Compact Flanges
Mn = Mp-(Mp-0.7*Fy*Sxc)*(λ_f - λ_pf)/(λ_rf - λ_pf) = 324.05 kN.m

Then Mn (Control) = 324.05 kN.m

II.3- Unity Check For Axial and Bending:


------------------------------------
Mr = 226.62 kN.m
Mct = Mnt * Ω = 291.64 kN.m
Ratio check = M/Mct = 81596.25 ≤ 1 => Satisfield

III - Check for Shear:


+ Shear force : 181.78 kN

- Allowable Stress In Shear Fv:


Where a/h > 3.00 or a/h > [260/(h/tw)]² = 35.32
Then kv = 5.00
Due to hw/tw = 43.75 ≤ 1.1*sqrt(kv*E/Fy) = 59.22
=> Cv = 1.0
Then Vn = 0.6 * Fy * Aw * Cv = 612.72 kN
Take Ω = 0.9
Vr/(Vn * Ω) = 0.33 (Satisfied)

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