Restrained Steel Beam (Pure...

Restrained Steel Beam (Pure Bending)

Start

Estimate support condition of the steel beam

Estimate all the load subjected onto the beam

Determine the maximum design of shear force, VEd and moment, MEd

Chose steel grade and cross section size that suitable for the design (EN 1993-1-1:2005, Table 3.1)

Classify the cross section of the steel beam (EN 1993-1-1:2005, Sheet 1, 2 and 3)

Determine the maximum shear force of the section, Vc, Rd (EN 1993-1-1:2005, Clause 6.2.6)

VEd ≤ Vc, Rd

Determine the maximum moment resistance, Mc, Rd (EN 1993-1-1:2005, Clause 6.2.5)

MEd ≤ Mc, Rd

No

Yes

B

Is shear buckling resistance checking for web need? (EN 1993-1-1:2005, Clause 6.2.6)

Check for shear buckling resistance of web (EN 1993-1-5, Section 5)

Yes

No

A

Yes

No

168

Check for combined bending and shear force resistance (EN 1993-1-1:2005, Clause 6.2.8)

Determine value of permanent load that gives effect to the beam deflection

B A

Adopt section

End

No

Is the combined bending and shear checking need?

Yes

No

Determine the allowable deflection (EN 1990:2002, Clause A1.4.3)

Actual deflection ≤ Allowable deflection

Yes

MEd ≤ MV, Rd No

Yes

169

Worked Example 1: Restrained Steel Beam (Pure Bending)

Universiti Teknologi Malaysia

Example 1: Fully Restrained Beam (Pure Bending) Designed by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Checked by: Prof Ir. Dr Mahmood Md. Tahir Page : 1

Reference Calculation Output

Example 1: Fully Restrained Beam (Pure Bending) The beam shown in Figure E1 is fully restrained along its length. Design the beam in Fe 430 steel (S 275). From analysis, VEd = 289.9 kN and MEd = 241.6 kNm.

Figure E1 Fully restrained beam Unfactored load Permanent load; UDL including selfweight of beam, gk = 17.5 kN/m Point load, Gk = 30 kN Variable load; UDL, qk = 35 kN/m Point load, Qk = 55 kN Safety factor for load Permanent load, γG = 1.35 Variable load, γQ = 1.50 Factored load Point load = 1.35Gk + 1.50Qk = 1.35(30) + 1.50(55) = 123 kN UDL = 1.35gk + 1.50qk = 1.35(17.5) + 1.50(35) = 76.13 kN/m VEd = 123/2 + 76.13x6/2 = 289.9 kN and MEd = (PL/4) + (WL/8) = (123x6/4) + (76.13x6/8) = 241.6 kNm.

6000 mm

3000 mm

170

Universiti Teknologi

Malaysia



Table 3.1

Table 5.2

Try section 610 x 305 x 149 UB Fe 430 (S275) h = 609.60 mm d = 537.20 mm Iz = 9300 cm4

b = 304.80 mm Av = 78.80 cm2 Wpl, y = 4570 cm3

tw = 11.9 mm c/tf = 7.74 Wel, y = 4090 cm3

tf = 19.7 mm d/tw = 45.1 A = 190 cm2

r = 16.5 mm Iy = 125000 cm4 Material properties Steel grade = Fe 430 (S275) t < 40 mm ; fy = 275 N/mm2 fu = 430 N/mm2 Classification of section Flange subject to compression c/tf = 7.74 < 9ε ; ε = √(235/fy) = 0.92 Flange is in Class 1 Web subject to bending d/tw = 45.1 < 72ε Web is in Class 1

Class 1

171



Reference Calculation Output Clause 6.2.6

Clause 6.2.6 number (6)

Shear force resistance checking The checking must satisfy

VEd / Vc, Rd ≤ 1.0 For Class 1 cross section Vc, Rd = Vpl, Rd Vpl, Rd = [Av( fy/√3)] / γM0 = [(78.80 x 102)(275/√3)] / (1 x 103) = 1251.12 kN

VEd = 289.9 kN < Vc, Rd = 1251.12 kN Shear buckling resistance checking for web Shear buckling resistance checking for web is not required if hw / tw < 72(ε/η) if hw / tw > 72(ε/η) the checking should be according to EN 1993-1-5, Section 5 Therefore; hw / tw = [609.60 – 2(19.70)] / 11.9 = 47.92 72(ε/η) = 72 x (0.92/1.0) ; n = 1.0 = 66.24 Since hw / tw < 72(ε/η) the shear buckling resistance checking for web is Not Required

OK

172




Clause 6.2.8

Bending moment resistance checking The checking must satisfy

MEd / Mc, Rd ≤ 1.0 For Class 1 cross section Mc, Rd = Mpl, Rd Mpl, Rd = Wplfy / γM0 = (4570 x 103)(275) / (1 x 106) = 1256.75 kNm MEd = 241.6 kNm < Mc, Rd = 1256.75 kNm Combined bending and shear resistance checking The checking must satisfy

MEd / MV, Rd ≤ 1.0 0.5Vc, Rd = 0.5 x 1251.12 = 625.56 kN Maximum shear force applied at maximum moment is equal to

VEd at max. moment = 289.9 – (76.13x3) = 61.51 kN Since VEd < 0.5Vc, Rd the combined bending and shear resistance is Not Required

OK

173




Clause A1.4.3 EN1990:2002

Lateral torsional buckling (LTB) checking Beam is fully restrained. Therefore, it will not be susceptible to lateral torsional buckling (LTB). Serviceability limits check Vertical deflection checking: The checking must satisfy

Wmax ≤ Wallow Wmax = W1 + W2 - WC W1 = (5/384)[(∑gk)(L4)/EI] = (5/384)[(17.5)(60004) / (210000)(125000 x 104)] = 1.13 mm W2 = (1/48)[(∑Gk x 103)(L3)/EI] = (1/48)[(30000)(60003) / (210000)(125000 x 104)] = 0.51 mm Wc = 0 mm Thus, Wmax = 1.13 + 0.51 – 0 = 1.64 mm Wallow = L / 200 = 6000 / 200 = 30 mm Wmax = 1.64 mm < Wallow = 30 mm ADOPT SECTION 610 x 305 x 149 UB, Fe 430 (S275)

OK

174

Restrained Steel Member (Pure Compression)

Start

Estimate support condition of the steel member

Estimate all the load subjected onto the member

Determine the maximum design of compression force, NEd


Classify the cross section of the steel member (EN 1993-1-1:2005, Sheet 1, 2 and 3)

Determine the maximum compression resistance, Nc, Rd (EN 1993-1-1:2005, Clause 6.2.4)

NEd ≤ Nc, Rd

Yes

No

Adopt section

End

175

Worked Example 2: Restrained Steel Member (Pure Compression)


Example 2: Steel Member Subject to Pure Compression Designed by: Dee Aguindrew Gundeh Checked By: Prof. Dr. Shahrin Mohammad Checked by: Prof Ir. Dr Mahmood Md. Tahir Page : 1


Example 2: Member subject to pure compression The beam shown in Figure E2 is restrained at the ends and at the point loads. Design the beam in Fe 430 (S355). From analysis, NEd = 350 kN.

Figure E2 Member subject to pure compression Safety factor for load: Permanent load, γG = 1.35 Variable load, γQ = 1.50 Try section 254 x 254 x 73 UC Fe 430 (S355) h = 254.0 mm d = 200.3 mm Iz = 3870 cm4

b = 254.0 mm Av = 25.6 cm2 Wpl, y = 989cm3

tw = 8.6 mm c/tf = 8.94 Wel, y = 894 cm3

tf = 14.2 mm d/tw = 23.3 A = 92.9 cm2

r = 12.7 mm Iy = 11400 cm4

NEd

8000 mm

176




Table 3.1

Table 5.2

Material properties: Steel grade = Fe 430 (S355) t < 40 mm ; fy = 355 N/mm2 fu = 430 N/mm2 Classification of section: Flange subject to compression c/tf = 8.94 < 14ε ; ε = √(235/fy) Flange is in Class 3 Web subject to compression d/tw = 23.3 < 33ε Web is in Class 1

Class 3

177




Clause 6.3.1

Compression resistance checking The checking must satisfy

NEd / Nc, Rd ≤ 1.0 For Class 3 cross section Nc, Rd = Npl, Rd Npl, Rd = (Afy) / γM0 = [(92.9 x 102)(355)] / (1 x 103) = 3297.95 kN NEd = 350 kN < Nc, Rd = 3297.95 kN Buckling resistance checking Beam is fully restrained. Therefore, it will not be susceptible to buckling resistance.

ADOPT SECTION 254 x 254 x 73 UB, Fe 430 (S355)

OK

178

Restrained Steel Beam (Combined Bending & Compression)

Start







VEd ≤ Vc, Rd


MEd ≤ Mc, Rd

No

Yes

B



Yes

No

A

Yes

No

179



B A

Adopt section

No


Yes

No



Yes

MEd ≤ MV, Rd No

Yes

End

Check for combined bending and compression resistance (EN 1993-1-1:2005, Clause 6.2.9)

MEd ≤ MN, Rd No

Yes

Check for compression resistance (EN 1993-1-1:2005, Clause 6.2.4)

NEd ≤ Nc, Rd No

Yes

180

Example 3: Restrained Steel Beam (Combined Bending & Compression)


Example 3: Steel Beam Subject to Combined Bending and Compression Designed by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Checked by: Prof Ir. Dr Mahmood Md. Tahir Page : 1


Example 3: Steel Beam Subjected to Combined Bending and Compression The beam shown in Figure E3 is fully restrained along its length. Design the beam in Fe 430 steel (S 275). From analysis, VEd = 276.57 kN, MEd = 554.22 kNm and NEd = 1987 kN.

Figure E3 Beam subjected to combined bending and compression

Unfactored load Permanent load; UDL including selfweight of beam, gk = 15 kN/m Point load, Gk = 40 kN Variable load; UDL, qk = 30 kN/m Point load, Qk = 50 kN Safety factor for load Permanent load, γG = 1.35 Variable load, γQ = 1.50 Factored load Point load = 1.35Gk + 1.50Qk = 1.35(40) + 1.50(50) = 129 kN UDL = 1.35gk + 1.50qk = 1.35(15) + 1.50(30) = 65.25 kN/m

6500 mm

NEd

3250 mm

181




Table 3.1

Table 5.2


b = 304.80 mm Av = 78.80 cm2 Wpl, y = 4570 cm3

tw = 11.9 mm c/tf = 7.74 Wel, y = 4090 cm3

tf = 19.7 mm d/tw = 45.1 A = 190 cm2

r = 16.5 mm Iy = 125000 cm4 Material properties Steel grade = Fe 430 (S275) t < 40 mm ; fy = 275 N/mm2 fu = 430 N/mm2 Classification of section Flange subject to compression c/tf = 7.74 < 9ε ; ε = √(235/fy) Flange is in Class 1 Web subject to bending d/tw = 45.1 < 72ε Web is in Class 1

Class 1

182


Example 3: Steel Beam Subject to Combined Bending and Compression Design by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Page : 3





VEd = 276.57 kN < Vc, Rd = 1251.12 kN Shear buckling resistance checking for web Shear buckling resistance checking for web is not required if hw / tw < 72(ε/η) if hw / tw > 72(ε/η) the checking should be according to EN 1993-1-5, Section 5 Therefore; hw / tw = [609.60 – 2(19.70)] / 11.9 = 47.92 72(ε/η) = 72 x (0.92/1,0) ; n = 1.0 = 66.24 Since hw / tw < 72(ε/η) the shear buckling resistance checking for web is Not Required

OK

183




Clause 6.2.8


MEd / Mc, Rd ≤ 1.0 For Class 1 cross section Mc, Rd = Mpl, Rd Mpl, Rd = Wplfy / γM0 = (4570 x 103)(275) / (1 x 106) = 1256.75 kNm

MEd = 554.22 kNm < Mc, Rd = 1256.75 kNm Combined bending and shear resistance checking The checking must satisfy

MEd / MV, Rd ≤ 1.0 0.5Vc, Rd = 0.5 x 1251.12 = 625.56 kN Maximum shear force applied at maximum moment is equal to

VEd at max. moment = 276.57 – (65.25x3.25) = 64.51 kN Since VEd < 0.5Vc, Rd the combined bending and shear resistance is Not Required

OK

184




Clause 6.2.9


NEd / Nc, Rd ≤ 1.0 For Class 1 cross section Nc, Rd = Npl, Rd Npl, Rd = (Afy) / γM0 = [(190 x 102)(275)] / (1 x 103) = 5225 kN NEd = 1987.00 kN < Nc, Rd = 5225 kN Combined bending and compression resistance checking The checking must satisfy

MEd / MN, Rd ≤ 1.0

No reduction to the plastic resistance moment due to the effect of axial force (compression) is required when both of the following criteria are satisfied.

NEd ≤ 0.25Npl, Rd and NEd ≤ 0.5hwtwfy / γM0 From analysis, NEd = 1987 kN Thus, 0.25Npl, Rd = 0.25 x 5225 = 1306.25 kN ; Not satisfied NEd ≤ 0.5hwtwfy / γM0 ≤ [0.5 x (609.60 – 2(19.70)) x 11.9 x 275] / 1 x103 ≤ 932.99 kN ; Not satisfied Therefore, allowance for the effect of axial force on the plastic moment resistance of the cross section must be made.

OK

185




Thus, the reduced plastic moment resistance is (for I cross section) MN, y, Rd = Mpl, y, Rd [(1 – n) / (1 – 0.5a)] ; MN, y, Rd ≤ Mpl, y, Rd n = NEd / Npl, Rd = 1987 / 5225 = 0.38 a = A – 2btf / A = [ (190 x 102) – 2(304.8)(19.7) ] / (190 x 102) = 0.37 Thus, MN, y, Rd = 1256.75 x [ (1 – 0.38) / 1 – 0.5(0.37)) ] = 956. 06 kNm < Mpl, y, Rd

MEd = 554.22 kNm < MN, y, Rd = 956. 06 kNm

OK

186




Clause 6.3.2 Clause A1.4.3 EN1990:2002

Lateral torsional buckling (LTB) checking Beam is fully restrained. Therefore, it will not be susceptible to lateral torsional buckling (LTB). Serviceability limits check Vertical deflection checking: The checking must satisfy

Wmax ≤ Wallow Wmax = W1 + W2 - WC W1 = 5/384[(∑gk)(L4)/EI] = 5/384[(15)(65004) / (210000)(125000 x 104)] = 1.33 mm W2 = 1/48[(∑Gk x 103)(L3)/EI] = 1/48[(40000)(65003) / (210000)(125000 x 104)] = 0.87 mm Wc = 0 mm Thus, Wmax = 1.33 + 0.87 – 0 = 2.20 mm Wallow = L / 200 = 6500 / 200 = 32.50 mm

Wmax = 2.20 mm < Wallow = 32.50 mm


OK

187

Unrestrained Steel Beam (Uniform Bending)

Start







VEd ≤ Vc, Rd


MEd ≤ Mc, Rd

No

Yes

B



Yes

No

A

Yes

No

188


Checking for lateral torsional buckling resistance (EN 1993-1-1:2005, Clause 6.3.2)

B A

Adopt section

End

No


Yes

No



Yes

MEd ≤ MV, Rd No

Yes


MEd ≤ Mb, Rd No

Yes

189

Worked Exampled 4: Unrestrained Steel Beam (Uniform Bending)


Example 4: Unrestrained Steel Beam with Uniform Bending Designed by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Checked by: Prof Ir. Dr Mahmood Md. Tahir Page : 1


Example 4: Unrestrained beam (uniform bending) The beam shown in Figure E4 is unrestrained beam with fixed end. Design the beam in Fe 430 steel (S 275). From analysis, VEd = 83.84 kN and MEd = 197.57 kNm. Figure E4 Unstrained beam with fixed end. Unfactored load: Permanent load; Selfweight of beam, gk = 3 kN/m Point load, Gk = 30 kN Variable load; Point Load, Qk = 60 kN Safety factor for load: Permanent load, γG = 1.35 Variable load, γQ = 1.50 Factored load: Point load = 1.35Gk + 1.50Qk = 1.35(30) + 1.50(60) = 130.5 kN UDL = 1.35gk = 1.35(3) = 4.05 kN/m

5100 mm

2500 mm

190




Table 3.1

Table 5.2


b = 266.70 mm Av = 115 cm2 Wpl, y = 6200 cm3

tw = 14.3 mm c/tf = 6.17 Wel, y = 5390 cm3

tf = 21.6 mm d/tw = 48 A = 220 cm2 r = 16.5 mm Iy = 205000 cm4 It = 267 cm4 Iw = 9.38 dm6 E = 210000 N/mm2 G = 81000 N/mm2 Material properties: Steel grade = Fe 430 (S275) t < 40 mm ; fy = 275 N/mm2 fu = 430 N/mm2 Classification of section: Flange subject to compression c/tf = 6.17 < 9ε ; ε = √(235/fy) Flange is in Class 1 Web subject to bending d/tw = 48 < 72ε Web is in Class 1

Class 1

191





Shear force resistance checking: The checking must satisfy

VEd / Vc, Rd ≤ 1.0 For Class 1 cross section Vc, Rd = Vpl, Rd

Vpl, Rd = [Av( fy/√3)] / γM0

= [(115 x 102)(275/√3)] / (1 x 103) = 1825.87 kN

VEd = 83.84 kN < Vc, Rd = 1825.87 kN

Shear buckling resistance checking for web Shear buckling resistance checking for web is not required if hw / tw < 72(ε/η) if hw / tw > 72(ε/η) the checking should be checked in accordance to EN 1993-1-5, Section 5 Therefore; hw / tw = [762 – 2(21.6)] / 14.3 = 50.27 72(ε/η) = 72 x (0.92/1,0) η ; = 1,0 = 66.24 Since hw / tw < 72(ε/η) the shear buckling resistance checking for web is Not Required

OK

192




Clause 6.2.8

Bending moment resistance checking: The checking must satisfy

MEd / Mc, Rd ≤ 1.0 For Class 1 cross section Mc, Rd = Mpl, Rd Mpl, Rd = Wplfy / γM0 = (6200 x 103)(275) / (1 x 106) = 1705 kNm

MEd = 197.57 kNm < Mc, Rd = 1705 kNm Combined bending and shear resistance checking The checking must satisfy

MEd / MV, Rd ≤ 1.0 0.5Vc, Rd = 0.5 x 1825.87 = 912.94 kN Maximum shear force applied

VEd = 83.84 kN Since VEd < 0.5Vc, Rd the combined bending and shear resistance is Not Required

OK

193




Designers’ Guide to EN

1993-1-1 Eurocode 3: Design of

steel structures

general rules and rules for

buildings (L Gardner

and D A

Nethercot)

Lateral torsional buckling (LTB) checking: The checking must satisfy MEd / Mb, Rd ≤ 1.0 Mb, Rd = χLT Wy fy/ γM1 Where; Wy = Wpl, y (Class 1 and Class 2) Wy = Wel, y (Class 3) Therefore; C1= 1.88 – 1.40Ψ+ 0.52Ψ2 but C1 ≤ 2.70 where Ψ is the ratio of the end moments equal to 1.285 for restrained ends. λLT = √ [(Wy)(fy)/Mcr] = √ [(6200 x 103)(275)/( 1994 x 106)] = 0.92 χLT = 1 / [ФLT + √ (ФLT

2 - λLT2)]

Where; ФLT = 0.5[1 + αLT( λLT – 0.2) + λLT

2] ; h/B < 2, αLT = 0.21 h/B > 2, αLT = 0.34 = 0.5[1 + 0.34( 0.92 – 0.2) + 0.922] = 1.04 Thus, χLT = 1 / [1.04 + √ (1.042 – 0.922)] = 0.66 Mb, Rd = (0.66)(6200 x 103)(275) / (1.0 x 106) = 1125.3 kNm MEd = 1619.75 kNm > Mb, Rd = 1125.3 kNm Not Satisfactory

OK

( )

kNm199410685021000010267810003570

1068501038.9

51007.0106850210000285.1

5.0

42

42

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12

2

42

5.0

2

2

2

2

1

=

⎥⎦

⎤⎢⎣

⎡××××××+

××

×××=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

cr

cr

z

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z

w

cr

zcr

M

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EIGIL

II

LEICM

ππ

ππ

194


Example 4: Unrestrained Steel Beam with Uniform Bending Design by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Page : 6

Reference Calculation Output Clause A1.4.3 EN1990:2002

Serviceability limits check Vertical deflection checking: The checking must satisfy




OK

195

Unrestrained Steel Member (Uniform Compression)

Start

Estimate support condition of the steel member

Estimate all the load subjected on to the member

Determine the maximum design of compression force, NEd


Classify the cross section of the steel member (EN 1993-1-1:2005, Sheet 1, 2 and 3)

Determine the maximum compression resistance, Nc, Rd (EN 1993-1-1:2005, Clause 6.2.4)

NEd ≤ Nc, Rd

B A

Yes

No

196

Checking for buckling resistance (EN 1993-1-1:2005, Clause 6.3.1)

A

Adopt section

End

Yes

NEd ≤ Nb, Rd No

Yes

B

197

Worked Exampled 5: Unrestrained Steel Member (Uniform Compression)


Example 5: Unrestrained Steel Member with Uniform Compression Designed by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Checked by: Prof Ir. Dr Mahmood Md. Tahir Page : 1


Table 3.1

Example 5: Unrestrained member (uniform compression) For the loading shown in Figure E5, design the member in Fe 430 (S275) steel grade using EC3. From analysis, NEd = 252 kN.

Figure E5 Unrestained member at uniform compression load. Safety factor for load: Permanent load, γG = 1.35 Variable load, γQ = 1.50 Try section 350 x 350 x 154 H Fe 430 (S275) h = 350 mm d = 286 mm Iz = 14428 cm4

b = 357 mm Av = 69.29 cm2 Wpl, y = 2730 cm3

tw = 19 mm c/tf = 9.39 Wel, y = 218 cm3

tf = 19 mm d/tw = 15.1 A = 196.4 cm2

r = 13 mm Iy = 42316 cm4 E = 210000 N/mm2 G = 81000 N/mm2

Material properties: Steel grade = Fe 430 (S275) t < 40 mm ; fy = 275 N/mm2 fu = 430 N/mm2

7200 mm

NEd

198

University of Technology Malaysia


Reference Calculation Output Table 5.2 Clause 6.2.4

Classification of section: Flange subject to compression c/tf = 9.39 < 14ε ; ε = √(235/fy) Flange is in Class 3 Web subject to compression d/tw = 15.1 < 33ε Web is in Class 1 Compression resistance checking The checking must satisfy

NEd / Nc, Rd ≤ 1.0 For Class 3 cross section Nc, Rd = Npl, Rd Npl, Rd = (Afy) / γM0 = [(196.4 x 102)(275)] / (1 x 103) = 5401 kN

NEd = 252.00 kN < Nc, Rd = 5401 kN

Class 3 OK

199




Designers’ Guide to EN

1993-1-1 Eurocode 3: Design of

steel structures

general rules and rules for

buildings (L Gardner

and D A

Nethercot)

Buckling resistance checking: The checking must satisfy

NEd / Nb, Rd ≤ 1.0 For Class 3 cross section Nb, Rd = χAfy / γM1 Therefore; Ncr = [ ((π2)(E)(Iy)) / (Lcr

2) ] (1 x 103) = [ ((π2)(E)(42316 x 104)) / (0.7x72002) ) / (1 x 103) = 34,527 kN λ = √ [(A)(fy)/Ncr] = √ [(196.4 x 102)(275)/(34,527 x 103)] = 0.40 χ = 1 / [Ф + √ (Ф2 - λ2)] Where; Ф = 0.5[1 + α( λ – 0.2) + λ2] = 0.5[1 + 0.34( 0.40 – 0.2) + 0.402] = 0.61 Thus, χ = 1 / [0.61 + √ (0.612 – 0.402)] = 0.93 Nb, Rd = (0.93)(196.4 x 102)(275) / (1.0 x 103) = 5045.1 kN

NEd = 252.00 kN < Nb, Rd = 5045.1kN

ADOPT SECTION 350 x 350 x 154 H, Fe 430 (S275)

OK

200

Unrestrained Steel Beam (Combined Bending & Compression)

Start







VEd ≤ Vc, Rd


MEd ≤ Mc, Rd

No

Yes

B



Yes

No

A

Yes

No

201


Checking for compression resistance (EN 1993-1-1:2005, Clause 6.2.4)

B A

No


Yes

No

Checking for lateral buckling resistance (EN 1993-1-1:2005, Clause 6.3.2)

Yes

MEd ≤ MV, Rd No

Yes

Checking for buckling resistance (EN 1993-1-1:2005, Clause 6.3.1)

NEd ≤ Nc, Rd No

Yes

NEd ≤ Nb, Rd

Yes

No

MEd ≤ Mb, Rd

C D

202

Checking for buckling resistance in combined bending and compression (EN 1993-1-1:2005, Clause 6.3.3)

C

Yes

No

Yes

D

[NEd/((χyNRk)/γM1)] + kyy[My, Ed/((χLTMy,

Rk)/γM1)] + kyz[Mz, Ed/(Mz, Rk/γM1)] ≤ 1

and

[NEd/((χyNRk)/γM1)] + kzy[My, Ed/((χLTMy,

Rk)/γM1)] + kzz[Mz, Ed/(Mz, Rk/γM1)] ≤ 1



Adopt section


End

No

203

Worked Exampled 6: Unrestrained Steel Beam (Combined Bending & Compression)


Example 6: Unrestrained Steel Beam with Combined Bending and Compression Designed by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Checked by Prof. Ir. Dr Mahmood Md. Tahir Page : 1


Example 6: Unrestrained Steel Beam with Combined Bending and Compression The beam shown in Figure E6 is fully restrained along its length. Design the beam in Fe 430 steel (S 275). From analysis, VEd = 338.90 kN, MEd = 625.44 kNm and NEd = 121 kN. Figure E6 Unrestrained steel beam with combined bending and compression Unfactored load Permanent load; UDL including selfweight of beam, gk = 15 kN/m Point load 1, Gk = 60 kN Variable load; UDL, qk = 30 kN/m Point load 1, Qk = 50 kN Safety factor for load Permanent load, γG = 1.35 Variable load, γQ = 1.50 Factored load Point load = 1.35Gk + 1.50Qk = 1.35(60) + 1.50(50) = 156 kN UDL = 1.35gk + 1.50qk = 1.35(15) + 1.50(30) = 65.25 kN/m

7200 mm

NEd

2400 mm

204


Example 6: Unrestrained Steel Beam with Combined Bending and Compression Designed by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Checked by Prof. Ir. Dr Mahmood Md. Tahir

Page : 2


Table 3.1

Table 5.2


b = 304.80 mm Av = 78.80 cm2 Wpl, y = 4570 cm3

tw = 11.9 mm c/tf = 7.74 Wel, y = 4090 cm3

tf = 19.7 mm d/tw = 45.1 A = 190 cm2

r = 16.5 mm Iy = 125000 cm4 E = 210000 N/mm2 G = 81000 N/mm2

Material properties Steel grade = Fe 430 (S275) t < 40 mm ; fy = 275 N/mm2 fu = 430 N/mm2 Classification of section Flange subject to compression c/tf = 7.74 < 9ε ; ε = √(235/fy) Flange is in Class 1 Web subject to bending d/tw = 45.1 < 72ε Web is in Class 1

Class 1

205




Clause 6.2.6




VEd = 338.90 kN < Vc, Rd = 1251.12 kN Shear buckling resistance checking for web Shear buckling resistance checking for web is not required if hw / tw < 72(ε/η) if hw / tw > 72(ε/η) the checking should be according to EN 1993-1-5, Section 5 Therefore; hw / tw = [609.60 – 2(19.70)] / 11.9 = 47.92 72(ε/η) = 72 x (0.92/1,2) ; n = 1.2 = 55.20 Since hw / tw < 72(ε/η) the shear buckling resistance checking for web

is Not Required

OK

206




Clause 6.2.5

Clause 6.2.8


MEd / Mc, Rd ≤ 1.0 For Class 1 cross section Mc, Rd = Mpl, Rd Mpl, Rd = Wplfy / γM0 = (4570 x 103)(275) / (1 x 106) = 1256.75 kNm

MEd = 625.44 kNm < Mc, Rd = 1256.75 kNm Combined bending and shear resistance checking The checking must satisfy

MEd / MV, Rd ≤ 1.0 0.5Vc, Rd = 0.5 x 1251.12 = 625.56 kN Maximum shear force applied

VEd = 338.90 kN

Since VEd < 0.5Vc, Rd the combined bending and shear resistance is Not Required

OK

207




Clause 6.2.4

Clause 6.3.1

Designers’ Guide to EN 1993-1-1

Eurocode 3: Design of steel structures general

rules and rules for buildings

(L Gardner and D A Nethercot)


NEd / Nc, Rd ≤ 1.0 For Class 1 cross section Nc, Rd = Npl, Rd Npl, Rd = (Afy) / γM0 = [(190 x 102)(275)] / (1 x 103) = 5225 kN

NEd = 121.00 kN < Nc, Rd = 5225 kN Buckling resistance checking The checking must satisfy

NEd / Nb, Rd ≤ 1.0 For Class 1 cross section Nb, Rd = χAfy / γM1 Therefore; Ncr, y = [ ((π2)(E)(Iy)) / (Lcr

2) ] (1 x 103) = [ ((π2)(E)(125000 x 104)) / (0.7x72002) ) / (1 x 103) = 71,395kN λy = √ [(A)(fy)/Ncr] = √ [(190x 102)(275)/(71395 x 103)] = 0.27

OK

208




Ncr, z = [ ((π2)(E)(Iz)) / (Lcr2) ]

(1 x 103) = [ ((π2)(E)(9300 x 104)) / (48002) ) / (1 x 103) = 8366.03 kN λz = √ [(A)(fy)/Ncr] = √ [(190 x 102)(275)/(8366.03 x 103)] = 0.79 Buckling curves, major (y-y) axis: χ y = 1 / [Фy + √ (Фy

2 - λy2)]

Where; Фy = 0.5[1 + α( λy – 0.2) + λy

2] = 0.5[1 + 0.21( 0.32 – 0.2) + 0.322] = 0.56 Thus, χ y = 1 / [0.56 + √ (0.562 – 0.322)] = 0.94 Nb, y, Rd = (0.94)(190 x 102)(275) / (1.0 x 103) = 4911.5 kN

NEd = 121.00 kN < Nb, y, Rd = 4911.5 kN

OK

209




Buckling curves, minor (z-z) axis: χ z = 1 / [Фz + √ (Фz

2 – λz2)]

Where; Фz = 0.5[1 + α( λz – 0.2) + λz

2] = 0.5[1 + 0.34( 0.79 – 0.2) + 0.792] = 0.91 Thus, χ z = 1 / [0.91 + √ (0.912 – 0.792)] = 0.73 Nb, z, Rd = (0.73)(190 x 102)(275) / (1.0 x 103) = 3814.25 kN

NEd = 121.00 kN < Nb, z, Rd = 3814.25 kN

OK

210


Example 6: Unrestrained Steel Beam with Combined Bending and Compression Design by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Page : 8


Clause 6.3.2

Designers’ Guide to EN 1993-1-1

Eurocode 3: Design of steel structures general

rules and rules for buildings

(L Gardner and D A Nethercot)

Lateral torsional buckling (LTB) checking The checking must satisfy

MEd / Mb, Rd ≤ 1.0 Mb, Rd = χLT Wy fy/ γM1 Where; Wy = Wpl, y (Class 1 and Class 2) Wy = Wel, y (Class 3) Therefore; Mcr = [ ( ((C1)(π2)(E)(Iz)) / (Lcr

2) ) ( (Iw/Iz) + ((Lcr2)(G)(It) /

(π2)(E)(Iz)) )0.5 ](1 x 106) = [ ( ((1.565)(π2)(E)(9300 x 104)) / (48002) ) ( (8.10 x 1012)/(9300 x 104)+ ((48002)(81000)(201 x 104) / (π2)(E)( 9300 x 104)) )0.5 ] / (1 x 106) = 6161 kNm λLT = √ [(Wy)(fy)/Mcr] = √ [(4570 x 103)(275)/( 6161 x 106)] = 0.45 χLT = 1 / [ФLT + √ (ФLT

2 - λLT2)]

Where; ФLT = 0.5[1 + αLT( λLT – 0.2) + λLT

2] ; h/B < 2, αLT = 0.21 h/B > 2, αLT = 0.34 = 0.5[1 + 0.34(0.45 – 0.2) + 0.452] = 0.64 Thus, χLT = 1 / [0.64 + √ (0.642 – 0.452)] = 0.91 Mb, Rd = (0.91)(4570x 103)(275) / (1.0 x 106) = 1143.6 kNm

MEd = 625.44 kNm < Mb, Rd = 1143.6 kNm

OK

211




Clause 6.3.3

Buckling resistance checking for combined bending and compression (Using Annex A) The checking must satisfy [NEd/((χyNRk)/γM1)] + kyy[My, Ed/((χLTMy, Rk)/γM1)] + kyz[Mz, Ed/(Mz,

Rk/γM1)] ≤1

and

[NEd/((χzNRk)/γM1)] + kzy[My, Ed/((χLTMy, Rk)/γM1)] + kzz[Mz, Ed/(Mz,

Rk/γM1)] ≤ 1

Determination of interaction factors kij (Annex A) For class 1 and 2 cross section kyy = CmyCmLT [ μy / (1 – (NEd / Ncr, y)) ] (1 / Cyy) kzy = CmyCmLT [ μy / (1 – (NEd / Ncr, y)) ] (1 / Czy) (0.6√(wy / wz)) Non dimensional slendernesses From the buckling resistance check λy = 0.32, λz = 0.79 therefore, λmax = 0.79 From the lateral torsional buckling check λLT = 0.49 therefore λo = 0.49 Equivalent uniform moment factors Cmi Torsional deformation is possible (λo > 0). From the bending moment diagram ψy = 0. Therefore, from Table A.2, Annex A Cmy, 0 = 1 + 0.03 (NEd / Ncr, y) = 1 + 0.03 (121 / 49976.30) = 1.0 Cmz, 0 = Cmz (no need to consider since Mz, Ed = 0)

212




εy = (My, Ed / NEd) (A / Wel, y) = [(625.44 x 106 / 121 x 103)] (19000 / 4090000) = 24.01 aLT = 1 – IT / Iy = 1 – [(201 x 104) / (125000 x 104)] = 0.998 The elastic torsional buckling force Ncr, T = 1 / i0

2 [ GIT + (π2EIw / LT2) ]

iy = (Iy / A)0.5 = [ (125000 x 104) / 19000 ]0.5 = 256.49 mm iz = (Iz / A)0.5 = [ (9300 x 104) / 19000 ]0.5 = 69.96 mm yo = zo =0 (since the shear centre and centroid of gross section coincide) Thus, io

2 = iy2 + iz

2 + yo2 + zo

2 = 256.492 + 69.962 = 70681.52 mm2 Therefore, Ncr, T = (1 / 70681.52) [ ((81000)(201 x 104)) + (π2E(8.1 x 106) / 72002) ] = 2303435.52 N = 2303.44 kN Cmy = Cmy, 0 + (1 – Cmy, 0) [ (√(εy) aLT) / (1 + √(εy) aLT )] = 1 + (1 – 1) [ (√(24.01) x 0.998) / (1 + (√(24.01) x 0.998)) ] = 1.0 CmLT = Cmy

2 [aLT / √( (1 – (NEd/Ncr, z)) (1 – (NEd/Ncr, T)) ) ] = 1.02 [ 0.998 / √( (1 – (121/8366.03)) (1 – (121/ 2303.44)) ) ] = 1.03 ≥ 1.0 therefore, CmLT = 1.0

213




Other auxiliary terms μy = [ 1 – (NEd/ Ncr, y) ] / [ 1 – χy(NEd/Ncr, y) ] = [ 1 – (121/49976.30) ] / [ 1 – 0.96(121/49976.30) ] = 1.0 μz = [ 1 – (NEd/ Ncr, z) ] / [ 1 – χz(NEd/Ncr, z) ] = [ 1 – (121/8366.03) ] / [ 1 – 0.73(121/8366.03) ] = 0.996 wy = Wpl, y / Wel, y = (4570 x 103) / (4090 x 103) = 1.12 < 1.5 ; ok wz = Wpl, z / Wel, z = (938 x 103) / (611 x 103) = 1.54 > 1.5, therefore wz = 1.5 For class 1 cross section NRk = Afy = (190 x 102)(275) = 5225000 N = 5225 kN Thus, npl = NEd / (NRk/γM1) = 121 / (5225/1.0) = 0.02 bLT = 0 (since Mz, Ed = 0) dLT = 0 (since Mz, Ed = 0) Cij factors Cyy = 1 + (wy – 1) { [(2 – (1.6/wy)Cmy

2 λmax) – ((1.6/wy)Cmy2 λmax

2)] npl -bLT } = 1 + (1.12 - 1) { [(2 – (1.6/1.12)(12)(0.79)) – ((1.6/1.12) )(12)(0.792))] 0.02 – 0 } = 1.0 > (Wel, y / Wpl, y = 0.89) ; ok Czy = 1 + (wy – 1) { [2 – 14(Cmy

2λmax2/wy

5)]npl – dLT } = 1 + (1.12 – 1) { [2 – 14((12)(0.792)/1.125)] 0.02 – 0 } = 0.99 > { 0.6[ (√(wy/wz)) (Wel, y/Wpl, y) ] = 0.46 } ; ok

214




Interaction factors kij kyy = CmyCmLT [ μy / (1 – (NEd / Ncr, y)) ] (1 / Cyy) = (1) (1) [ 1 / (1 – (121/49976.30)) ] (1 / 1) = 1.0 kzy = CmyCmLT [ μy / (1 – (NEd / Ncr, y)) ] (1 / Czy) (0.6√(wy / wz)) = (1) (1) [ 1 / (1 – (121 / 49976.30)) ] (1 / 0.99) (0.6√(1.12 / 1.5)) = 0.52 My, Rk = Mc, y, Rd NRk = Nc, Rd Therefore, [NEd/((χyNRk)/γM1)] + kyy[My, Ed/((χLTMy, Rk)/γM1)] + kyz[Mz, Ed/(Mz,

Rk/γM1)] ≤1

= [121/((0.94 x 5225)/1)] + 1[625.44/((0.89 x 1256.75)/1)] + 0 = 0.02 + 0.56 + 0 = 0.58 < 1 [NEd/((χzNRk)/γM1)] + kzy[My, Ed/((χLTMy, Rk)/γM1)] + kzz[Mz, Ed/(Mz,

Rk/γM1)] ≤ 1 = [121/((0.73 x 5225)/1)] + 0.52[625.44/((0.89 x 1256.75)/1)] + 0 = 0.03 + 0.29 + 0 = 0.32 < 1

OK

OK

215


Example 6: Unrestrained Steel Beam with Combined Bending and Compression Design by: Dee Aguindrew Gundeh Checked by: Prof. Dr. Shahrin Mohammad Page : 13


Clause A1.4.3 EN1990:2002

Serviceability limits check Vertical deflection checking: The checking must satisfy




OK

216

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