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Page - 1 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Structural Calculations High Street, Lewisham GB11/07/082 Date of Issue July 2011
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Page 1: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

Page - 1

JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Structural Calculations High Street, Lewisham GB11/07/082 Date of Issue July 2011

Page 2: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Loading Roof : Tiles = 0.65 kN/m2 Dead Live

Rafters, felt, insulation etc . = 0.30 kN/m2

0.95 kN/m2/Cos 35 = 1.16 kN/m2 Plasterboard = 0.25 kN/m2

TOTAL = 1.41 kN/m2

Attic = 0.25 Roof snow loading = 0.6*((60-35)/30)= 0.50 kN/m2

TOTAL = 0.75 kN/m2 Roof : Tiles = 0.65 kN/m2 Dead Live

Plasterboard = 0.25 kN/m2 Rafters, felt, insulation etc . = 0.30 kN/m2

1.20 kN/m2/Cos 35 = 1.47 kN/m2

TOTAL = 1.47 kN/m2 Roof snow loading = 0.6*((60-35)/30)= 0.50 kN/m2

TOTAL = 0.50 kN/m2 Roof: Joists & boarding , finishes = 0.35 kN/m2

(flat) Plasterboard = 0.25 kN/m2 TOTAL = 0.60 kN/m2

Imposed = 0.75 kN/m2

TOTAL = 0.75 kN/m2 Floor: Joists & boarding = 0.25 kN/m2

Plasterboard = 1.0 kN/m2 TOTAL = 1.25 kN/m2

Imposed = 1.50 kN/m2

TOTAL = 1.50 kN/m2 Walls: 2.4m high, 100mm blockwork = 2.4*1.4 =3.36 kN/m Plasterwork both sides = 2.4*0.25*2 =1.2 kN/m TOTAL = 4.56 kN/m 2.4m high, studwork = 2.4*0.12 =0.29 kN/m Plasterwork both sides = 2.4*0.15*2 =0.72 kN/m TOTAL = 1.01 kN/m 2.7m high, cavity wall blockwork = 2.7*(2.1+1.4) =9.45 kN/m Plasterwork to one side = 2.4*0.25 =0.60 kN/m TOTAL = 10.05 kN/m 2.7m high, external studwork = 2.7*(2.1+1.2) =6.0 kN/m Plasterwork to one side = 2.4*0.25 =0.60 kN/m TOTAL = 6.60 kN/m 2.7m high, solid brickwork = 2.7*(2.1+2.1) =11.34 kN/m Plasterwork to one side = 2.4*0.25 =0.60 kN/m TOTAL = 11.94 kN/m

Page 3: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Existing

Proposed

Page 4: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Beam 1

Beam 5

Beam 4

Post 1

Post 2

Post 3 Post 4

Beam 2 Beam 3

Beam 6

Beam 7

Beam 7 Beam 8

Beam 9

Beam 1.1 Beam 4.1

Page 5: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

MasterFrame : Graphics

Frame Geometry - (Full Frame) - X+030 Y-010 Z+000

Page 6: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Load to Beam 1 Floor (4.4/2+4.6/2)*1.25=5.625 (4.4/2+4.6/2)*1.5=6.75 Partitions 2.0 Total 7.625 kN/m Dead 6.75 kN/m Live

Beam 1 (Beam 1.1) and Beam 4(Beam 4.1) are most dangerous positions

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Beam 1

Member 4 (N.17-N.18) @ Level 1 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 D1 UDLY -007.625 ( kN/m ) L1 UDLY -006.750 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

4 17 0.20C 0.00 51.09 0.00 -0.81 0.00 61.13 0.00 5.24 18 0.20C 0.00 -51.08 0.00 -0.80 0.00 @ 2.425 @ 0.000 @ 2.425

Classification and Effective Area (EN 1993: 2006) Section (48.07 kg/m) 305x127 UB 48 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 4.48, 29.47, 275, 0.2, 61.13, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 0.003 / 474.695 = 0 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 710.7/1 195.443 kN.m Npl.Rd = Ag.fy/ γM0 61.23 x 275/1 = 1683.825 kN n = NEd/Npl.Rd 0.201 / 1683.825 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 710.7, 29.898, 0 710.7 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 710.7 x 275/1 195.443 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (61.128/195.443)²+(0)1= 0.098 OK

Compression Resistance N.b.Rd λy= √A.fy/Ncr √61.23x275/8439.22 0.447 Nb.y.Rd = Area.χ.fy/ γM1 61.23x0.94x275/10/1 = 1582.962 kN Curve a λz= √A.fy/Ncr √61.23x275/407.75 2.031 Nb.z.Rd = Area.χ.fy/ γM1 61.23x0.204x275/10/1 = 343.051 kN Curve b

Equivalent Uniform Moment Factors C1, C.mLT, C.mz, and C.my C1= fn(M1, M2, Mo, ψ,μ) -0.8, -0.7, 61.9, 0.988, -81.958 1.202 Uniform CmLT=0.95+0.05αh Mh= -0.76, Ms= 61.13, ψ = 0.988, αs= -0.012 0.949 Table B.3 Cmz=Max(0.6+0.4ψ, 0.4) M = 0, ψ = 1.000 1 Table B.3 Cmy=0.95+0.05αh Mh= -0.81, Ms= 61.13, ψ = 0.988, αs= -0.013 0.949 Table B.3

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Lateral Buckling Check M.b.Rd Le = 1.2L+2D 1.2 x 4.85 + 2 x 0.311 = 6.442 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.202, 6.442, 462.7, 31.78, 0.1012, 210000 101.263 kN.m λLT= √ W.fy/Mcr √ 710.7 x 275 / 101.263 1.389 χLT= Fn(λLT, λLT5950 ) 1.389, 1.503 0.434 Curve c χLT.mod = Fn(χLT,λLT,kc,f) 0.434, 1.389, 0.912, 0.987 0.440 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.440 x 710.7 x 275 ≤ 195.443 = 85.980 kN.m

Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 0.201 / 1582.962 0.000 OK UN.z = NEd/(χz.NRk/γM1) 0.201 / 343.051 0.001 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 61.128 / 85.98 0.711 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0 / 31.928 0.000 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.949 kzz=Cmz{1+1.4UN.z} 1.001 kyz=0.6 kzz 0.600 kzy= 1- {0.1λz/(CmLT-0.25)}UN.z 1.000 UNy+kyy.UM.y+kyz.UM.z 0.000+0.949x0.711+0.600x0.000 0.675 OK UNz+kzy.UM.y+kzz.UM.z 0.001+1.000x0.711+1.001x0.000 0.711 OK

Deflection Check - Load Case 3 δ ≤ Span/360 5.24 ≤ 4850 / 360 5.24 mm OK

Section (48.07 kg/m) 305x127 UB 48 [Grade 43] Load to Beam 1 Floor (4.4/2+4.6/2)*1.25=5.625 (4.4/2+4.6/2)*1.5=6.75 Partitions 2.0 Total 7.625 kN/m Dead 6.75 kN/m Live +Additional load from roof

Beam 1.1 C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Members 13-14 (N.36-N.37) @ Level 3 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 Part 1 D1 UDLY -007.625 ( kN/m ) L1 UDLY -006.750 ( kN/m ) Part 2 D1 UDLY -007.625 ( kN/m ) L1 UDLY -006.750 ( kN/m )

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3 Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

36 0.19C 0.00 66.14 0.00 -1.66 0.00 102.16 0.00 8.43 38 0.19C 0.00 -77.01 0.00 0.00 0.00 @ 3.108 @ 0.000 @ 2.487

Classification and Effective Area (EN 1993: 2006) Section (48.07 kg/m) 305x127 UB 48 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 4.48, 29.47, 275, 0.19, 102.15, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 40.333 / 474.695 = 0.085 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 710.7/1 195.443 kN.m Vz.Ed/Vpl.z.Rd 0.003 / 557.033 = 0 Low Shear Mc.z.Rd = fy.Wpl.z/ γM0 275 x 116.1/1 31.928 kN.m Npl.Rd = Ag.fy/ γM0 61.23 x 275/1 = 1683.825 kN n = NEd/Npl.Rd 0.187 / 1683.825 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 710.7, 29.898, 0 710.7 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 710.7 x 275/1 195.443 kN.m Wpl.N.z = Fn(Wpl.z, Avz, n) 116.1, 35.084, 0 116.1 cm³ MN.z.Rd = Wpl.N.z.fy/ γM0 116.1 x 275/1 31.928 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (102.153/195.443)²+(0.002/31.928)1= 0.273 OK

Compression Resistance N.b.Rd λy= √A.fy/Ncr √61.23x275/8439.22 0.447 Nb.y.Rd = Area.χ.fy/ γM1 61.23x0.94x275/10/1 = 1582.962 kN Curve a λz= √A.fy/Ncr √61.23x275/407.75 2.031 Nb.z.Rd = Area.χ.fy/ γM1 61.23x0.204x275/10/1 = 343.051 kN Curve b

Equivalent Uniform Moment Factors C1, C.mLT, C.mz, and C.my C1= fn(M1, M2, Mo, ψ,μ) -1.6, 0.1, 97.5, -0.049, -61.186 1.157 Uniform CmLT=0.95+0.05αh(1+2ψ) Mh= -1.59, Ms= 96.77, ψ = -0.049, αs= -0.016 0.949 Table B.3 Cmz=Max(0.6+0.4ψ, 0.4) M = 0, ψ = 0.000 0.6 Table B.3 Cmy=0.95+0.05αh(1+2ψ) Mh= -1.66, Ms= 96.77, ψ = -0.001, αs= -0.017 0.949 Table B.3

Lateral Buckling Check M.b.Rd Le = 1.00 L 1 x 4.85 = 4.85 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.157, 4.850, 462.7, 31.78, 0.1012, 210000 137.356 kN.m λLT= √ W.fy/Mcr √ 710.7 x 275 / 137.356 1.193 χLT= Fn(λLT, λLT5950 ) 1.193, 1.266 0.529 Curve c χLT.mod = Fn(χLT,λLT,kc,f) 0.529, 1.193, 0.930, 0.976 0.542 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.542 x 710.7 x 275 ≤ 195.443 = 105.898 kN.m

Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 0.187 / 1582.962 0.000 OK UN.z = NEd/(χz.NRk/γM1) 0.187 / 343.051 0.001 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 102.153 / 105.898 0.965 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0.002 / 31.928 0.000 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.949 kzz=Cmz{1+1.4UN.z} 0.600 kyz=0.6 kzz 0.360 kzy= 1- {0.1λz/(CmLT-0.25)}UN.z 1.000 UNy+kyy.UM.y+kyz.UM.z 0.000+0.949x0.965+0.360x0.000 0.916 OK UNz+kzy.UM.y+kzz.UM.z 0.001+1.000x0.965+0.600x0.000 0.965 OK

Deflection Check - Load Case 3 δ ≤ Span/360 8.43 ≤ 4850 / 360 8.43 mm OK

Section (48.07 kg/m) 305x127 UB 48 [Grade 43]

Page 9: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Consider Bearings Max Load = 77.01 kN Assuming γm=3.5 fk=3.5 N/mm2 and Bearing Type 1 Required Bearing length = 77.01E3*3.5/(3.5*1.5)/100=513.4mm

Provide 550 long x 102.5 wide 3 Course Eng. Bwk Padstone Load to Beam 4 Floor (4.0/2+3.9/2)*1.25=4.938 (4.0/2+3.9/2)*1.5=5.925 Partitions 2.0 Total 6.938 kN/m Dead 5.925 kN/m Live

Beam 4 C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Beam 4

Member 1 (N.10-N.12) @ Level 1 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 D1 UDLY -006.938 ( kN/m ) L1 UDLY -005.925 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

1 10 0.03C -0.01 54.76 0.00 -0.17 0.01 79.16 9.73 12 0.03C 0.01 -54.86 0.00 -0.46 -0.02 @ 2.900 @ 2.900

Classification and Effective Area (EN 1993: 2006) Section (48.07 kg/m) 305x127 UB 48 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 4.48, 29.47, 275, 0.03, 79.15, 0.01 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 0.053 / 474.695 = 0 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 710.7/1 195.443 kN.m Vz.Ed/Vpl.z.Rd 0.003 / 557.033 = 0 Low Shear Mc.z.Rd = fy.Wpl.z/ γM0 275 x 116.1/1 31.928 kN.m Npl.Rd = Ag.fy/ γM0 61.23 x 275/1 = 1683.825 kN n = NEd/Npl.Rd 0.026 / 1683.825 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 710.7, 29.898, 0 710.7 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 710.7 x 275/1 195.443 kN.m

Page 10: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Wpl.N.z = Fn(Wpl.z, Avz, n) 116.1, 35.084, 0 116.1 cm³ MN.z.Rd = Wpl.N.z.fy/ γM0 116.1 x 275/1 31.928 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (79.149/195.443)²+(0.002/31.928)1= 0.164 OK

Compression Resistance N.b.Rd λy= √A.fy/Ncr √61.23x275/5901.19 0.534 Nb.y.Rd = Area.χ.fy/ γM1 61.23x0.913x275/10/1 = 1537.961 kN Curve a λz= √A.fy/Ncr √61.23x275/285.12 2.429 Nb.z.Rd = Area.χ.fy/ γM1 61.23x0.147x275/10/1 = 247.988 kN Curve b

Equivalent Uniform Moment Factors C1, C.mLT, C.mz, and C.my C1= fn(M1, M2, Mo, ψ,μ) -0.1, -0.4, 79.4, 0.288, -197.043 1.144 Uniform CmLT=0.95+0.05αh Mh= -0.4, Ms= 79.15, ψ = 0.288, αs= -0.005 0.95 Table B.3 Cmz=Max(0.6+0.4ψ, 0.4) M = -0.01, ψ = -0.455 0.418 Table B.3 Cmy=0.95+0.05αh Mh= -0.46, Ms= 79.15, ψ = 0.373, αs= -0.006 0.95 Table B.3

Lateral Buckling Check M.b.Rd Le = 1.00 L 1 x 5.8 = 5.8 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.144, 5.800, 462.7, 31.78, 0.1012, 210000 109.110 kN.m λLT= √ W.fy/Mcr √ 710.7 x 275 / 109.11 1.338 χLT= Fn(λLT, λLT5950 ) 1.338, 1.413 0.457 Curve c χLT.mod = Fn(χLT,λLT,kc,f) 0.457, 1.338, 0.935, 0.986 0.463 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.463 x 710.7 x 275 ≤ 195.443 = 90.492 kN.m

Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 0.026 / 1537.961 0.000 OK UN.z = NEd/(χz.NRk/γM1) 0.026 / 247.988 0.000 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 79.149 / 90.492 0.875 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0.002 / 31.928 0.000 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.950 kzz=Cmz{1+1.4UN.z} 0.418 kyz=0.6 kzz 0.251 kzy= 1- {0.1λz/(CmLT-0.25)}UN.z 1.000 UNy+kyy.UM.y+kyz.UM.z 0.000+0.950x0.875+0.251x0.000 0.831 OK UNz+kzy.UM.y+kzz.UM.z 0.000+1.000x0.875+0.418x0.000 0.875 OK

Deflection Check - Load Case 3 δ ≤ Span/360 9.73 ≤ 5800 / 360 9.73 mm OK

Section (48.07 kg/m) 305x127 UB 48 [Grade 43]

Beam 4.1 Load to Beam 4 Floor (4.0/2+3.9/2)*1.25=4.938 (4.0/2+3.9/2)*1.5=5.925 Partitions 2.0 Total 6.938 kN/m Dead 5.925 kN/m Live +Additional load from roof

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Members 9-10 (N.30-N.32) @ Level 3

Between 0.000 and 2.801 m, in Load Case 1

Page 11: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 Part 1 D1 UDLY -006.938 ( kN/m ) L1 UDLY -005.925 ( kN/m ) Part 2 D1 UDLY -006.938 ( kN/m ) L1 UDLY -005.925 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

30 0.00T 0.00 76.11 0.00 0.00 0.00 139.02 0.00 15.84 33 0.00T -0.01 -74.84 0.00 -0.42 0.00 @ 2.801 @ 0.000 @ 2.891

Classification and Effective Area (EN 1993: 2006) Section (48.07 kg/m) 305x127 UB 48 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 4.48, 29.47, 275, 0, 139.02, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Moment Capacity Check M.c.y.Rd Vy.Ed/Vpl.y.Rd 23.164 / 474.695 = 0.049 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 710.7/1 195.443 kN.m My.Ed/Mc.y.Rd 139.018 / 195.443 = 0.711 OK

Equivalent Uniform Moment Factor C1 C1= fn(M1, M2, Mo, ψ,μ) 0.1, 139.0, -20.9, 0.001, -0.150 2.196 Uniform

Lateral Buckling Check M.b.Rd Le = 1.00 L 1 x 2.801 = 2.801 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 2.196, 2.801, 462.7, 31.78, 0.1012, 210000 555.675 kN.m λLT= √ W.fy/Mcr √ 710.7 x 275 / 555.675 0.593 χLT= Fn(λLT, λLT5950 ) 0.593, 0.867 0.890 Curve c χLT.mod = Fn(χLT,λLT,kc,f) 0.890, 0.593, 0.675, 0.851 1.000 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 1.000 x 710.7 x 275 ≤ 195.443 = 195.443 kN.m My.Ed/Mb.Rd 139.018 / 195.443 0.711 OK

Deflection Check - Load Case 3 δ ≤ Span/360 15.84 ≤ 5800 / 360 15.84 mm OK

Section (48.07 kg/m) 305x127 UB 48 [Grade 43]

This type of beam I will use for the rest of structure to support floor joists.

Consider Bearings Max Load = 74.84 kN Assuming γm=3.5 fk=3.5 N/mm2 and Bearing Type 1 Required Bearing length = 74.84E3*3.5/(3.5*1.5)/100=498.933mm

Provide 550 long x 102.5 wide 3 Course Eng. Bwk Padstone

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Beam 5 Load to Beam 5 Wall 6.60 Total 6.60 kN/m Dead

AXIAL WITH MOMENTS (MEMBER)

Beam 5 Member 31 (N.31-N.37) @ Level 3 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 D1 UDLY -006.600 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

31 31 0.00C 0.00 -21.00 0.00 0.00 0.00 -23.61 0.00 10.55 37 0.00C 0.00 21.11 0.00 0.25 0.00 @ 2.254 @ 0.000 @ 2.254

Classification and Effective Area (EN 1993: 2006) Section (15.95 kg/m) 2 No. 152x89 UB 16 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 5.76, 27.07, 275, 0, 23.61, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 0.057 / 259.658 = 0 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 246.6/1 67.815 kN.m Npl.Rd = Ag.fy/ γM0 40.64 x 275/1 = 1117.6 kN n = NEd/Npl.Rd 0.002 / 1117.6 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 246.6, 16.354, 0 246.6 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 246.6 x 275/1 67.815 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (23.606/67.815)²+(0)1= 0.121 OK

Compression Resistance N.b.Rd λy= √A.fy/Ncr √40.64x275/1702.98 0.81 Nb.y.Rd = Area.χ.fy/ γM1 40.64x0.79x275/10/1 = 882.680 kN Curve a λz= √A.fy/Ncr √40.64x275/184.73 2.461 Nb.z.Rd = Area.χ.fy/ γM1 40.64x0.144x275/10/1 = 160.652 kN Curve b

Equivalent Uniform Moment Factors C1, C.mLT, C.mz, and C.my C1= fn(M1, M2, Mo, ψ,μ) 0.0, -0.2, 23.7, -0.083, -103.542 1.151 Uniform CmLT=0.95+0.05αh(1+2ψ) Mh= -0.23, Ms= 23.61, ψ = -0.083, αs= -0.010 0.95 Table B.3 Cmz=Max(0.6+0.4ψ, 0.4) M = 0, ψ = 1.000 1 Table B.3 Cmy=0.95+0.05αh Mh= -0.25, Ms= 23.61, ψ = 0.000, αs= -0.011 0.949 Table B.3

Lateral Buckling Check M.b.Rd Le = 1.00 L 1 x 4.509 = 4.509 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.151, 4.509, 181.2, 7.121, 0.009376, 210000 40.530 kN.m λLT= √ W.fy/Mcr √ 246.6 x 275 / 40.53 1.294 χLT= Fn(λLT, λLT5950 ) 1.294, 1.353 0.527 Curve b χLT.mod = Fn(χLT,λLT,kc,f) 0.527, 1.294, 0.932, 0.983 0.536 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.536 x 246.6 x 275 ≤ 67.815 = 36.378 kN.m

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Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 0.002 / 882.68 0.000 OK UN.z = NEd/(χz.NRk/γM1) 0.002 / 160.652 0.000 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 23.606 / 36.378 0.649 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0 / 17.16 0.000 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.949 kzz=Cmz{1+1.4UN.z} 1.000 kyz=0.6 kzz 0.600 kzy= 1- {0.1λz/(CmLT-0.25)}UN.z 1.000 UNy+kyy.UM.y+kyz.UM.z 0.000+0.949x0.649+0.600x0.000 0.616 OK UNz+kzy.UM.y+kzz.UM.z 0.000+1.000x0.649+1.000x0.000 0.649 OK

Deflection Check - Load Case 3 δ ≤ Span/360 10.55 ≤ 4509 / 360 10.55 mm OK

Section (15.95 kg/m) 2 No. 152x89 UB 16 [Grade 43]

Consider Bearings Max Load = 21.11 kN Assuming γm=3.5 fk=3.5 N/mm2 and Bearing Type 1 Required Bearing length = 21.11E3*3.5/(3.5*1.5)/100=140.733mm

Provide 330 long x 102.5 wide 2 Course Eng. Bwk Padstone

Beam 6 Load to Beam 6 Wall 6.60 Roof 2.6/2*0.6=0.78 2.6/2*0.75=0.975 (flat) Total 7.38 kN/m Dead 0.975 kN/m Live

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Beam 6

Member 29 (N.28-N.32) @ Level 3 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 D1 UDLY -007.380 ( kN/m ) L1 UDLY -000.975 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

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Member Forces in Load Case 1 and Maximum Deflection from Load Case 3 Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

29 28 0.00C 0.00 -21.31 0.00 0.00 0.00 -19.15 0.00 5.38 32 0.00C 0.00 21.33 0.00 0.04 0.00 @ 1.798 @ 0.000 @ 1.798

Classification and Effective Area (EN 1993: 2006) Section (15.95 kg/m) 2 No. 152x89 UB 16 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 5.76, 27.07, 275, 0, 19.15, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Moment Capacity Check M.c.y.Rd Vy.Ed/Vpl.y.Rd 0.01 / 259.658 = 0 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 246.6/1 67.815 kN.m My.Ed/Mc.y.Rd 19.152 / 67.815 = 0.282 OK

Equivalent Uniform Moment Factor C1 C1= fn(M1, M2, Mo, ψ,μ) 0.0, 0.0, 19.1, -0.526, 300.000 1.127 Uniform

Lateral Buckling Check M.b.Rd Le = 1.00 L 1 x 3.597 = 3.597 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.127, 3.597, 181.2, 7.121, 0.009376, 210000 51.721 kN.m λLT= √ W.fy/Mcr √ 246.6 x 275 / 51.721 1.145 χLT= Fn(λLT, λLT5950 ) 1.145, 1.185 0.611 Curve b χLT.mod = Fn(χLT,λLT,kc,f) 0.611, 1.145, 0.942, 0.978 0.625 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.625 x 246.6 x 275 ≤ 67.815 = 42.404 kN.m My.Ed/Mb.Rd 19.152 / 42.404 0.452 OK

Deflection Check - Load Case 3 δ ≤ Span/360 5.38 ≤ 3597 / 360 5.38 mm OK

Section (15.95 kg/m) 2 No. 152x89 UB 16 [Grade 43]

Consider Bearings Max Load = 21.11 kN Assuming γm=3.5 fk=3.5 N/mm2 and Bearing Type 1 Required Bearing length = 21.11E3*3.5/(3.5*1.5)/100=140.733mm

Provide 330 long x 102.5 wide 2 Course Eng. Bwk Padstone

Beam 7 Load to Beam 7 Wall 2*6.60=13.2 Total 13.2 kN/m Dead Load that distributes from upper beams: 38.0/6.8=5.588kN/m Dead 16.3/6.8=2.397 kN/m Live 19.2/6.8=2.824kN/m Dead 16.3/6.8=2.397 kN/m Live

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

AXIAL WITH MOMENTS (MEMBER) Beam 7

Member 24 (N.16-N.18) @ Level 1 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 D1 UDLY -014.730 ( kN/m ) L1 UDLY -001.913 ( kN/m ) D1 UDLY -005.588 ( kN/m ) L1 UDLY -002.397 ( kN/m ) D1 UDLY -002.824 ( kN/m ) L1 UDLY -002.397 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

24 16 0.01C 0.00 -97.45 0.00 0.00 0.00 -111.71 0.00 3.17 18 0.01C 0.00 98.06 0.00 1.42 0.00 @ 2.300 @ 0.000 @ 2.300

Classification and Effective Area (EN 1993: 2006) Section (88.7 kg/m) 356x171 UB 67 + 275x10 T Plate 88.7 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 5.52, 34.24, 275, 0.01, 111.71, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 0.308 / 567.238 = 0.001 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 1517/1 417.175 kN.m Npl.Rd = Ag.fy/ γM0 112.99 x 275/1 = 3107.225 kN n = NEd/Npl.Rd 0.012 / 3107.225 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 1517, 35.727, 0 1517 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 1517 x 275/1 417.175 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (111.709/417.175)²+(0)1= 0.072 OK

Compression Resistance N.b.Rd λy= √A.fy/Ncr √112.99x275/26167.92 0.345 Nb.y.Rd = Area.χ.fy/ γM1 112.99x0.967x275/10/1 = 3004.223 kN Curve a λz= √A.fy/Ncr √112.99x275/3031.69 1.013 Nb.z.Rd = Area.χ.fy/ γM1 112.99x0.589x275/10/1 = 1829.271 kN Curve b

Equivalent Uniform Moment Factors C1, C.mLT, C.mz, and C.my C1= fn(M1, M2, Mo, ψ,μ) 0.1, -1.3, 112.3, -0.073, -86.327 1.154 Uniform CmLT=0.95+0.05αh(1+2ψ) Mh= -1.3, Ms= 111.71, ψ = -0.073, αs= -0.012 0.95 Table B.3 Cmz=Max(0.6+0.4ψ, 0.4) M = 0, ψ = 1.000 1 Table B.3 Cmy=0.95+0.05αh Mh= -1.42, Ms= 111.71, ψ = 0.000, αs= -0.013 0.949 Table B.3

Lateral Buckling Check M.b.Rd Le = 1.2L+2D 1.2 x 4.6 + 2 x 0.363 = 6.247 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.154, 6.247, 3095, 64.85, 0.6757, 210000 439.400 kN.m λLT= √ W.fy/Mcr √ 1517 x 275 / 439.4 0.974 χLT= Fn(λLT, λLT5950 ) 0.974, 1.261 0.655 Curve c χLT.mod = Fn(χLT,λLT,kc,f) 0.655, 0.974, 0.931, 0.968 0.677 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.677 x 1517 x 275 ≤ 417.175 = 282.285 kN.m

Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 0.012 / 3004.223 0.000 OK UN.z = NEd/(χz.NRk/γM1) 0.012 / 1829.271 0.000 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 111.709 / 282.285 0.396 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0 / 118.8 0.000 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.949 kzz=Cmz{1+1.4UN.z} 1.000

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kyz=0.6 kzz 0.600 kzy= 1- {0.1λz/(CmLT-0.25)}UN.z 1.000 UNy+kyy.UM.y+kyz.UM.z 0.000+0.949x0.396+0.600x0.000 0.376 OK UNz+kzy.UM.y+kzz.UM.z 0.000+1.000x0.396+1.000x0.000 0.396 OK

Deflection Check - Load Case 3 δ ≤ Span/360 3.17 ≤ 4600 / 360 3.17 mm OK

Section (88.7 kg/m) 356x171 UB 67 + 275x10 T Plate 88.7 [Grade 43]

Consider Bearings Max Load = 98.06 kN Assuming γm=3.5 fk=3.5 N/mm2 and Bearing Type 1 Required Bearing length = 98.06E3*3.5/(3.5*1.5)/100=653.733mm

Provide 670 long x 102.5 wide 4 Course Eng. Bwk Padstone

Beam 8 Load to Beam 8 Wall 3*1.73=5.19 Floor 1.5*0.5=0.75 1.5*1.5=2.25 Total 5.94 kN/m Dead 2.25 kN/m Live

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Members 20 and 22 (N.4-N.12) @ Level 1 in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 Part 1 Part 2 D1 UDLY -005.940 ( kN/m ) L1 UDLY -002.253 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

4 0.08C -0.06 -9.98 -0.02 0.27 0.00 -29.25 -0.06 1.20 13 0.08C -0.43 54.26 0.05 1.12 -0.02 @ 3.321 @ 3.287 @ 2.225

Classification and Effective Area (EN 1993: 2006) Section (48.07 kg/m) 305x127 UB 48 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 4.48, 29.47, 275, 0.08, 29.29, 0.06 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 47.036 / 474.695 = 0.099 Low Shear

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Mc.y.Rd = fy.Wpl.y/ γM0 275 x 710.7/1 195.443 kN.m Vz.Ed/Vpl.z.Rd 0.055 / 557.033 = 0 Low Shear Mc.z.Rd = fy.Wpl.z/ γM0 275 x 116.1/1 31.928 kN.m Npl.Rd = Ag.fy/ γM0 61.23 x 275/1 = 1683.825 kN n = NEd/Npl.Rd 0.078 / 1683.825 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 710.7, 29.898, 0 710.7 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 710.7 x 275/1 195.443 kN.m Wpl.N.z = Fn(Wpl.z, Avz, n) 116.1, 35.084, 0 116.1 cm³ MN.z.Rd = Wpl.N.z.fy/ γM0 116.1 x 275/1 31.928 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (29.289/195.443)²+(0.056/31.928)1= 0.024 OK

Compression Resistance N.b.Rd λy= √A.fy/Ncr √61.23x275/12915 0.361 Nb.y.Rd = Area.χ.fy/ γM1 61.23x0.963x275/10/1 = 1621.165 kN Curve a λz= √A.fy/Ncr √61.23x275/624.09 1.642 Nb.z.Rd = Area.χ.fy/ γM1 61.23x0.295x275/10/1 = 496.515 kN Curve b

Equivalent Uniform Moment Factors C1, C.mLT, C.mz, and C.my C1= fn(M1, M2, Mo, ψ,μ) -0.3, -1.1, 9.9, 0.250, -9.348 1.173 Uniform CmLT=0.95+0.05αh Mh= -1.05, Ms= 9.19, ψ = 0.250, αs= -0.115 0.944 Table B.3 Cmz=Max(0.6+0.4ψ, 0.4) M = -0.01, ψ = 0.000 0.6 Table B.3 Cmy=0.95+0.05αh Mh= -1.12, Ms= 18.04, ψ = 0.244, αs= -0.062 0.947 Table B.3

Lateral Buckling Check M.b.Rd Le = 1.2L+2D 1.2 x 3.92 + 2 x 0.311 = 5.326 m Mcr= Fn(C1,Le,Iz,It,Iw,E) 1.173, 5.326, 462.7, 31.78, 0.1012, 210000 123.993 kN.m λLT= √ W.fy/Mcr √ 710.7 x 275 / 123.993 1.255 χLT= Fn(λLT, λLT5950 ) 1.255, 1.342 0.496 Curve c χLT.mod = Fn(χLT,λLT,kc,f) 0.496, 1.255, 0.923, 0.978 0.508 6.3.2.3 Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.508 x 710.7 x 275 ≤ 195.443 = 99.240 kN.m

Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 0.08 / 1621.165 0.000 OK UN.z = NEd/(χz.NRk/γM1) 0.08 / 496.515 0.000 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 29.289 / 99.24 0.295 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0.056 / 31.928 0.002 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.947 kzz=Cmz{1+1.4UN.z} 0.600 kyz=0.6 kzz 0.360 kzy= 1- {0.1λz/(CmLT-0.25)}UN.z 1.000 UNy+kyy.UM.y+kyz.UM.z 0.000+0.947x0.295+0.360x0.002 0.280 OK UNz+kzy.UM.y+kzz.UM.z 0.000+1.000x0.295+0.600x0.002 0.296 OK

Deflection Check - Load Case 3 δ ≤ Span/360 1.2 ≤ 3920 / 360 1.2 mm OK

Section (48.07 kg/m) 305x127 UB 48 [Grade 43]

Consider Bearings Max Load = 10.0 kN Assuming γm=3.5 fk=3.5 N/mm2 and Bearing Type 1 Required Bearing length = 10E3*3.5/(3.5*1.5)/100=66.667mm

Provide 215 long x 102.5 wide 1 Course Eng. Bwk Padstone

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Beam 9 Load to Beam 9 Wall 2*6.60 Total 13.2 kN/m Dead Load that distributes from upper beams: 35.5/6.8=5.22kN/m Dead 19.0/6.8=2.794 kN/m Live 21/6.8=3.09kN/m Dead 17.0/6.8=2.5 kN/m Live

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

AXIAL WITH MOMENTS (MEMBER) Members 19 and 21 (N.2-N.10) @ Level 1

Between 0.000 and 3.900 m, in Load Case 1

Member Loading and Member Forces Loading Combination : 1 UT + 1.35 D1 + 1.5 L1 Part 1 D1 UDLY -013.200 ( kN/m ) D1 UDLY -005.740 ( kN/m ) L1 UDLY -002.790 ( kN/m ) D1 UDLY -002.940 ( kN/m ) L1 UDLY -002.530 ( kN/m ) Part 2 D1 UDLY -013.200 ( kN/m ) D1 UDLY -005.220 ( kN/m ) L1 UDLY -002.794 ( kN/m ) D1 UDLY -003.090 ( kN/m ) L1 UDLY -002.500 ( kN/m )

Member Forces in Load Case 1 and Maximum Deflection from Load Case 3

Mem ber No.

Node End1 End2

Axial Force (kN)

Torque Moment (kN.m)

Shear Force (kN)

Bending Moment (kN.m)

Maximum Moment (kN.m @ m)

MaximumDeflection(mm @ m)y-y z-z y-y z-z y-y z-z

2 1.02C 0.07 -158.99 0.02 3.57 0.00 -322.07 -0.33 18.71 11 1.50C 0.08 168.23 0.11 5.27 0.00 @ 3.900 @ 3.900 @ 3.432

Classification and Effective Area (EN 1993: 2006) Section (88.7 kg/m) 356x171 UB 67 + 275x10 T Plate 88.7 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 5.52, 34.24, 275, 1.5, 322.06, 0.33 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Local Capacity Check Vy.Ed/Vpl.y.Rd 8.004 / 567.238 = 0.014 Low Shear Mc.y.Rd = fy.Wpl.y/ γM0 275 x 1517/1 417.175 kN.m Vz.Ed/Vpl.z.Rd 0.02 / 863.475 = 0 Low Shear Mc.z.Rd = fy.Wpl.z/ γM0 275 x 432/1 118.8 kN.m Npl.Rd = Ag.fy/ γM0 112.99 x 275/1 = 3107.225 kN n = NEd/Npl.Rd 1.021 / 3107.225 = 0.000 OK Wpl.N.y = Fn(Wpl.y, Avy, n) 1517, 35.727, 0 1517 cm³ MN.y.Rd = Wpl.N.y.fy/ γM0 1517 x 275/1 417.175 kN.m Wpl.N.z = Fn(Wpl.z, Avz, n) 432, 54.385, 0 432 cm³ MN.z.Rd = Wpl.N.z.fy/ γM0 432 x 275/1 118.8 kN.m (My.Ed/MN.y.Rd)+(Mz.Ed/MN.z.Rd) (322.051/417.175)²+(0.083/118.8)1= 0.597 OK

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Compression Resistance N.b.Rd λy= √A.fy/Ncr √112.99x275/11974.12 0.509 Nb.y.Rd = Area.χ.fy/ γM1 112.99x0.921x275/10/1 = 2862.859 kN Curve a

Lateral Buckling Check M.b.Rd Mb.Rd = Mc.y.Rd Fully Restrained 417.175 kN.m

Buckling Resistance UN.y = NEd/(χy.NRk/γM1) 1.496 / 2862.859 0.001 OK UN.z = NEd/(χz.NRk/γM1) 1.496 / 3107.225 0.000 OK UM.y = My.Ed/(χLT.My.Rk/γM1) 322.051 / 417.175 0.772 OK UM.z = Mz.Ed/(Mz.Rk/γM1) 0.083 / 118.8 0.001 OK kyy=Cmy{1+(λy-0.2)UN.y} 0.949 kzz=Cmz{1+(2λy-0.6)UN.z} 0.600 kyz=0.6 kzz 0.360 kzy=0.6 kyy 0.570 UNy+kyy.UM.y+kyz.UM.z 0.001+0.949x0.772+0.360x0.001 0.734 OK UNz+kzy.UM.y+kzz.UM.z 0.000+0.570x0.772+0.600x0.001 0.441 OK

Deflection Check - Load Case 3 δ ≤ Span/360 18.71 ≤ 6800 / 360 18.71 mm OK

Section (88.7 kg/m) 356x171 UB 67 + 275x10 T Plate 88.7 [Grade 43]

(Fully Restrained)

Ground Floor Post Design Most Dangerous Case

C:\SIMAS\MASTERSERIES\LEWISHAM\FRAMEREVISED.$5

COLUMNS IN SIMPLE CONSTRUCTION Member 40 (N.5-N.13) @ Level 1 in Load Case 1

Column in Simple Construction check not support by EN 1993:2006. Capacities and components calcualted in accoradance with EN 1993 and design check presented according to BS 5950. Caution

Classification and Effective Area (EN 1993: 2006) Section (36.98 kg/m) 152x152 UC 37 [Grade 43] Class = Fn(b/T,d/t,fy,N,My,Mz) 6.71, 15.45, 275, 478.34, 0, 0 (Axial: Non-Slender) Class 1 Auto Design Load Cases 1

Applied Factored Loads Fc=F+Fy1+Fy2+Fz1+Fz2+Fa 348.707+55.224+0+20.148+54.261+0 478.341 kN Myo=Fy1.ey1 55x181 9.99 kN.m Mzo=Fz1.ez1-Fz2.ez2 20x104-54x104 3.548 kN.m My=Myo.(Iy/L)/(Iy/L+Iy.u/L.u) 9.99x(2211/3.5)/((2211/3.5)+(2211/2.325)) 3.987 kN.m Mz=Mzo.(Iz/L)/(Iz/L+Iz.u/L.u) 3.548x(707/3.5)/((707/3.5)+(707/2.325)) 1.416 kN.m

Page 20: IPXX XXX XX - Structural Calculations · 2011-07-19 · Page - 8 JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011 Member Forces in Load Case 1 and Maximum Deflection from Load

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JOB REF: GB11/07/082 DESIGNED BY: SR DATE: June 2011

Compression Resistance N.b.Rd λy= √A.fy/Ncr √47.11x275/3741.54 0.589 Nb.y.Rd = Area.χ.fy/ γM1 47.11x0.843x275/10/1 = 1091.820 kN Curve b λz= √A.fy/Ncr √47.11x275/1196.37 1.042 Nb.z.Rd = Area.χ.fy/ γM1 47.11x0.516x275/10/1 = 668.621 kN Curve c

Buckling Resistance Moment M.b.Rd λLT = 0.5 L / (iz.π. √ E/fy) 0.5x100x3.5/ (3.87x π x √ 210000/275) 0.52 χLT= Fn(λLT) 0.521 0.952 Curve b Mb.Rd = χ Wpl.y.fy≤ Mc.y.Rd 0.952 x 308.8 x 275 ≤ 84.920 = 80.802 kN.m

Columns in Simple Construction Fc/Nb.Rd+My/Mb.Rd+Mz/fy.Wel.z 478.341/668.621+3.987/80.802+1.416/(275x91.59) 0.821 OK

Section (36.98 kg/m) 152x152 UC 37 [Grade 43]

Use 152x152UC37 For All Posts


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