Beam LOADING AND CALCULATION

1/B-C1/C-D1/D-E5/B-C5/C-D

Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Brick Wall : 2.5 kN/m2 x 4 m = 10 kN/m

Total dead load : ( 4.85 x 1.875 ) + 10 = 19.09 kN/m

Live Load (LL)Office : 2.5 x 1.875 = 4.6875kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (19.09) + 1.6 (4.6875)

= 34.226 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 32.226 (8.75)2 / 8 = 327.55 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =327.55x 103

275 = 1191.09cm3

From table properties, try UB 406x178x67Beam self weight = 67.1 kg/mD = 409.4mm, B = 178.8mm, t = 8.8mm, T = 14.3 mm, b/T = 6.25 mm, d/t = 41.0 mm

Sx = 1350 cm3 , Ix = cm4, Zx = cm3 , r = 10.2 mm

New design load = 1.4 (19.09 + 0.671) + 1.6 (4.6875) = 35.17 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 35.17 (8.75)2/ 8 = 336.59 kNmFmax = wL/2 = 35.17 (8.75) / 2 = 153.87 kN

Design strength and determination of section class

Since T = 14.3mm < 16mm,py = 275 N/mm2

ε = √ 275275 = 1.00

b/T=6.25 < 9ε,the flange is plastic (class 1)d/t =41.0 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =41.0 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(409.4)(8.8)= 594.45kN

Beam LOADING AND CALCULATION

2/B-C2/C-D2/D-E

Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Total dead load : ( 4.85 x 3.75 ) = 18.19 kN/m

Live Load (LL)Office : 2.5 x 3.75 = 9.375 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (18.19) + 1.6 (9.375)

= 40.47 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 40.47 (8.75)2 / 8 = 387.31 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =387.31x 103

275 = 1408.4cm3

From table properties, try UB 406x178x74Beam self weight = 74.2 kg/mD = 412.8mm, B = 179.5mm, t = 9.5mm, T = 16 mm, b/T = 5.61 mm, d/t = 37.9 mm

Sx = 1500 cm3 , Ix = 27300 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (18.19 + 0.742) + 1.6 (9.38) = 41.51 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 41.51 (8.75)2/ 8 = 397.26 kNmFmax = wL/2 = 41.51 (8.75) / 2 = 181.61 kN

Design strength and determination of section class

Since T = 16mm ≤ 16mm,py = 275 N/mm2

ε = √ 275275 = 1.00

b/T=5.61 < 9ε,the flange is plastic (class 1)d/t =37.9 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =37.9 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(412.8)(9.5)= 647.064kNFv = 181.61kN < Pv shear is adequate

Beam LOADING AND CALCULATION

3/B-C3/C-D3/D-E

Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Partition : 1 X 4 = 4 kN/m

Total dead load : ( 4.85 x 3.75 ) + 4 = 22.19 kN/m

Live Load (LL)Office : 2.5 x 3.75 = 9.375 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (22.19) + 1.6 (9.375)

= 46.066 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 46.066 (8.75)2 / 8 = 440.87 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =440.87 x103

275 = 1603.16cm3

From table properties, try UB 406x178x85Beam self weight = 85.3 kg/mD = 417.2mm, B = 181.9mm, t = 10.9mm, T = 18.2 mm, b/T = 5 mm, d/t = 33.1 mm

Sx = 1730 cm3 , Ix = 31700 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (22.19 + 0.853) + 1.6 (9.375) = 47.26 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 47.26 (8.75)2/ 8 = 452.3 kNmFmax = wL/2 = 47.26 (8.75) / 2 = 206.76 kN

Design strength and determination of section class

Since T = 18.2mm ≤ 40mm,py = 265 N/mm2

ε = √ 275265 = 1.02

b/T=5 < 9ε,the flange is plastic (class 1)d/t =33.1 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =33.1 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(417.2)(10.9)= 723.05kNFv = 206.76kN < Pv shear is adequate

Beam LOADING AND CALCULATION

3/A-B4/B-C4/C-D

Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Total dead load : ( 4.85 x 3.75 ) = 18.19 kN/m

Live Load (LL)Office : 2.5 x 3.75 = 9.375 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (18.19) + 1.6 (9.375)

= 40.474 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 40.474 (8.75)2 / 8 = 387.35 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =387.35x 103

275 = 1408.55cm3

From table properties, try UB 406x178x74Beam self weight = 85.3 kg/mD = 412.8mm, B = 179.5mm, t = 9.5mm, T = 16 mm, b/T = 5.61 mm, d/t = 37.9 mm

Sx = 1500 cm3 , Ix = 27300 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (18.19 + 0.742) + 1.6 (9.375) = 41.51 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 41.51 (8.75)2/ 8 = 397.26 kNmFmax = wL/2 = 41.51 (8.75) / 2 = 181.61 kN

Design strength and determination of section class

Since T = 16mm ≤ 16mm,py = 275 N/mm2

ε = √ 275275 = 1

b/T=5.61 < 9ε,the flange is plastic (class 1)d/t =37.9 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =37.9 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(412.8)(9.5)= 647.064kNFv = 181.61kN < Pv shear is adequateMoment capacity

Beam LOADING AND CALCULATION

5/D-E Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Brick wall : 2.5 x 4 = 10 kN/m

Total dead load : ( 4.85 x 1.875 ) + 10 = 19.09 kN/m

Live Load (LL)Kitchen / pantry : 3 x 1.875 = 5.625 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (19.09) + 1.6 (5.625)

= 35.726 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 35.726 (8.75)2 / 8 = 341.91 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =341.91x 103

275 = 1243.31cm3

From table properties, try UB 406x178x67Beam self weight = 67.1 kg/mD = 409.4mm, B = 178.8mm, t = 8.8mm, T = 14.3 mm, b/T = 6.25 mm, d/t = 41.0 mm

Sx = 1350 cm3 , Ix = 24300 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (19.09 + 0.671) + 1.6 (5.625) = 36.68 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 36.68 (8.75)2/ 8 = 351.04 kNmFmax = wL/2 = 36.68 (8.75) / 2 = 160.48 kN

Design strength and determination of section class

Since T = 8.8mm ≤ 16mm,py = 275 N/mm2

ε = √ 275275 = 1

b/T=6.25 < 9ε,the flange is plastic (class 1)d/t =41.0 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =41.0 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(409.4)(8.8)= 594.45kNFv = 160.48kN < Pv shear is adequate

Beam LOADING AND CALCULATION

4/D-E Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

partition : 1 x 4 = 4 kN/m

Total dead load : ( 4.85 x 3.75 ) + 4 = 22.19 kN/m

Live Load (LL)Kitchen / pantry : 3 x 3.75 = 10.31 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (22.19) + 1.6 (10.31)

= 47.562 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 473.562 (8.75)2 / 8 = 455.18 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =455.181x 103

275 = 1655.2cm3

From table properties, try UB 406x178x85Beam self weight = 85.3 kg/mD = 417.2mm, B = 181.9mm, t = 10.9mm, T = 18.2 mm, b/T = 5 mm, d/t = 33.1 mm

Sx = 130 cm3 , Ix = 31700 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (22.19 + 0.853) + 1.6 (10.31) = 48.76 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 48.76 (8.75)2/ 8 = 466.65 kNmFmax = wL/2 = 48.76 (8.75) / 2 = 213.33 kN

Design strength and determination of section class

Since T = 18.2 ≤ 40mm,py = 265 N/mm2

ε = √ 275265 = 1

b/T=5 < 9ε,the flange is plastic (class 1)d/t =33.1 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =33.1 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(265)(4417.2)(10.9)= 750.33kNFv = 213.33kN < Pv shear is adequate

Beam LOADING AND CALCULATION

1/A-B Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

Partition = 1.0 kN/m2

---------------- 4.85 kN/m2

Brick wall : 2.5 x 4 = 10 kN/m

Total dead load : ( 5.85 x 1.875 ) = 10.97 kN/m

Live Load (LL)Toilet : 2 x 1.875 = 3.75 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (10 + 10.97) + 1.6 (3.75)

= 35.36 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 35.36 (8.75)2 / 8 = 338.41 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =338.1x 103

275 = 1230.58cm3

From table properties, try UB 406x178x67Beam self weight = 67.1 kg/mD = 409.4mm, B = 178.8mm, t = 8.8mm, T = 14.3 mm, b/T = 6.25 mm, d/t = 41.0 mm

Sx = 1350 cm3 , Ix = 24300 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (10 + 10.97 + 0.671) + 1.6 (3.75) = 36.63 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 36.3 (8.75)2/ 8 = 347.40 kNmFmax = wL/2 = 36.3 (8.75) / 2 = 158.81 kN

Design strength and determination of section class

Since T = 8.8mm ≤ 16mm,py = 275 N/mm2

ε = √ 275275 = 1

b/T=6.25 < 9ε,the flange is plastic (class 1)d/t =41.0 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =41.0 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(409.4)(8.8)= 594.45kN

Beam LOADING AND CALCULATION

2/A-B Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

Partition = 1.0 kN/m2

---------------- 4.85 kN/m2

Brick wall : 2.5 x 4 = 10 kN/m

Total dead load : ( 5.55 x 1.875 ) = 10.41 kN/m

Live Load (LL)office : 2.5 x 3.75 = 8.44 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (10 + 10.41) + 1.6 (8.44)

= 42.08 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 42.08 (8.75)2 / 8 = 402.72 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =402.72 x103

275 = 1464.44cm3

From table properties, try UB 406x152x74Beam self weight = 74.2 kg/mD = 462.0mm, B = 154.5mm, t = 9.6mm, T = 17 mm, b/T = 4.54 mm, d/t = 42.5 mm

Sx = 1630 cm3 , Ix = 32700 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (10 + 1.41 + 0.7) + 1.6 (8.44) = 43.17 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 43.17 (8.75)2/ 8 = 413.15 kNmFmax = wL/2 = 43.17 (8.75) / 2 = 188.887 kN

Design strength and determination of section class

Since T = 17mm ≤ 40mm,py = 265 N/mm2

ε = √ 275265 = 1.02

b/T=4.54 < 9ε,the flange is plastic (class 1)d/t =42.5 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =42.5 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(265)(462)(9.6)= 705.2kN

Beam LOADING AND CALCULATION

4/A-B Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Partition : 1 x 4 = 4 kN/m

Total dead load : ( 4.85 x 3.75 ) + 4 = 22.19 kN/m

Live Load (LL)Stair : 3 x 3.75 = 11.25 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (22.19) + 1.6 (11.25)

= 49.01 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 49.01 (8.75)2 / 8 = 469.04 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =469.04 x 103

275 = 1705.6cm3

From table properties, try UB 406x191x82Beam self weight = 82 kg/mD = 460mm, B = 155.3mm, t = 9.9mm, T = 16 mm, b/T = 5.98 mm, d/t = 41.2 mm

Sx = 1830 cm3 , Ix = 2=37100 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (22.19 x 0.82) + 1.6 (11.25) = 50.214 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 50.214 (8.75)2/ 8 = 480.56 kNmFmax = wL/2 = 50.214(8.75) / 2 = 219.69 kN

Design strength and determination of section class

Since T = 16 mm ≤ 16mm,py = 275 N/mm2

ε = √ 275275 = 1

b/T=5.98< 9ε,the flange is plastic (class 1)d/t =41.2 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =41.2 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(460)(9.9)= 724.09kNFv = 219.69kN < Pv shear is adequate

Beam LOADING AND CALCULATION

5/A-B Using steel grade 275Design load

Dead load (DL)Total dead load on slab : 24 kN/m3 x 0.15 m = 3.6 kN/m2

Ceiling = 0.25 kN/m2

Mechanical and electrical (m & e ) = 0.5 kN/m2

Finishing = 0.5 kN/m2

---------------- 4.85 kN/m2

Partition : 1 x 4 = 4 kN/m

Total dead load : ( 4.85 x 1.875 ) + 4 = 13.09 kN/m

Live Load (LL)Toilet : 3 x 1.875 = 5.625 kN/m

Design Load, Wudl = 1.4DL + 1.6LL = 1.4 (13.09) + 1.6 (5.625)

= 27.326 kN/m

Maximum moment (Trial)

Mmax = wL2 / 8 = 27.326 (8.75)2 / 8 = 261.52 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =261.52x 103

275 = 950.98cm3

From table properties, try UB 356x171x57Beam self weight = 57.0 kg/mD = 358.0mm, B = 172.2mm, t = 8.1mm, T = 13 mm, b/T = 6.62 mm, d/t = 38.5 mm

Sx = 1010 cm3 , Ix = 16000 x 104 cm4, , r = 10.2 mm

New design load = 1.4 (13.09 + 0.57) + 1.6 (5.625) = 28.124 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 = 28.124 (8.75)2/ 8 = 269.16 kNmFmax = wL/2 = 28.124 (8.75) / 2 = 123.04 kN

Design strength and determination of section class

Since T = 13mm ≤ 16mm,py = 275 N/mm2

ε = √ 275275 = 1

b/T=6.62 < 9ε,the flange is plastic (class 1)d/t =38.5 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =38.5 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(275)(358)(8.1)= 478.48kNFv =123.04kN < Pv shear is adequate

Beam LOADING AND CALCULATION

D/1-3C/1-3C/3-5

Using steel grade 275Design load

Dead load (DL)Partition : 1 x 4 = 4 kN/m

Live Load (LL)-Design Load, Wudl = 1.4DL = 1.4 (4)

= 5.6 kN/m Design load from beam 2/C-D, 2/D-E = 41.51 kN/mPoint load acting on the beam = 41.51 x 8.75 = 363.21 kN

Maximum moment (Trial)

Mmax = wL2 / 8 + PL/4 = 5.6 (7.5)2 / 8 + 363.21(7.5) / 4 = 720.4 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =720.4 x103

275 = 2619.64cm3

From table properties, try UB 457x191x133Beam self weight = 133.3 kg/mD = 480.6mm, B = 196.7mm, t = 15.3mm, T = 26.3 mm, b/T = 3.74 mm, d/t = 26.6 mm

Sx =3070 cm3 , , r = 10.2 mm

New design load = 1.4 (4 + 1.33) = 7.562 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 + PL/4 = 7.462 (7.5)2/ 8 + 363.21 (7.5) / 4 = 733.49 kNmFmax = wL/2 + P/2 = 7.462 (7.5) / 2 + 363.21 / 2 = 209.59 kN

Design strength and determination of section class

Since T = 26.3mm ≤ 40mm,py = 265 N/mm2

ε = √ 275265 = 1.02

b/T=3.74 < 9ε,the flange is plastic (class 1)d/t =26.6 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =26.6 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(265)(480.6)(15.3)= 1169.2kNFv < Pv shear is adequateMoment capacity0.6Pv = 0.6 (1169.2) = 701.52 kNFv < Pv ; low shear

Mc= pySx = 265(3070x10−3) = 813.55 kNm

Since Mmax < Mc moment is adequate

AT SUPPORT

BEAM LOADING AND CALCULATIONA/1-3 Using steel grade 275

Design load

Dead load (DL)BRICK WALL : 2.5 x 4 = 10 kN/m

Live Load (LL)-Design Load, Wudl = 1.4DL = 1.4 (10)

= 14 kN/m Design load from beam 2/A-B = 43.17Point load acting on the beam = (43.17 x 4.375) = 188.87 kN

Maximum moment (Trial)

Mmax = wL2 / 8 + PL/4 = 14 (7.5)2 / 8 + 188.87(7.5) / 4 = 452.57 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =452.57 x103

275 = 1645.71cm3

From table properties, try UB 457x152x82Beam self weight = 82.1 kg/mD = 465.8mm, B = 155.3mm, t = 10.5mm, T = 18.9 mm, b/T = 4.11 mm, d/t = 38.8 mm

Sx =1810 cm3 , , r = 10.2 mm

New design load = 1.4 (10 + 0.821) = 15.15 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 + PL/4 = 15.15 (7.5)2/ 8 + 188.87 (7.5) / 4 = 460.65 kNmFmax = wL/2 + P/2 = 15.15 (7.5) / 2 + 188.87 / 2 = 151.25 kN

Design strength and determination of section class

Since T = 26.3mm ≤ 40mm,py = 265 N/mm2

ε = √ 275265 = 1.02

b/T=4.11 < 9ε,the flange is plastic (class 1)d/t =38.8 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =38.8 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(265)(465.8)(10.5)= 777.65kNFv < Pv shear is adequateMoment capacity0.6Pv = 0.6 (777.65) = 466.59 kNFv < Pv ; low shear

Mc= pySx = 265(1810x10−3) = 479.65 kNm

Since Mmax < Mc moment is adequate

AT SUPPORTWeb bearing capacityFv < Pbw b1= t + T + 0.8r – g * g = 10

BEAM LOADING AND CALCULATIONB/3-5 Using steel grade 275

Design load

Dead load (DL)BRICK WALL : 2.5 x 4 = 10 kN/m

Live Load (LL)-Design Load, Wudl = 1.4DL = 1.4 (10)

= 14 kN/m Design load from beam 4/A-B = 50.214, 4/B-C = 41.51Point load acting on the beam = (50.214 x 4.375) + (41.51 X 4.375) = 401.29 kN

Maximum moment (Trial)

Mmax = wL2 / 8 + PL/4 = 14 (7.5)2 / 8 + 401.29(7.5) / 4 = 850.86 kNm

Plastic modulus

Assume the web thickness,T<16mm, py = 275N/mm2

Sx =850.86 x103

275 = 3092.95 cm3

From table properties, try UB 457x19x161Beam self weight = 161.4 kg/mD = 492.0mm, B = 199.4mm, t = 18.0mm, T = 32 mm, b/T = 3.12 mm, d/t = 22.6 mm

Sx =3780 cm3 , r = 10.2 mm

New design load = 1.4 (10 + 1.994) = 16.79 kN/m

Maximun moment and shear (Final)Mmax = wL2/8 + PL/4 = 16.79 (7.5)2/ 8 + 401.29 (7.5) / 4 = 870.47 kNmFmax = wL/2 + P/2 = 16.79 (7.5) / 2 + 401.29 / 2 = 263.61 kN

Design strength and determination of section class

Since T = 32mm ≤ 40mm,py = 265 N/mm2

ε = √ 275265 = 1.02

b/T=3.12 < 9ε,the flange is plastic (class 1)d/t =22.6 < 80ε,the web is plastic (class 1)Since the flange and web are class 1, the cross section is class 1. Plastic design is valid for designing this section.Shear bucklingd/t =26.6 < 70ε, there is no need check need for shear buckling.Shear capacityPv = 0.6pyAv = 0.6(265)(492)(18)= 1408.104kNFv < Pv shear is adequateMoment capacity0.6Pv = 0.6 (1408.104) = 844.86 kNFv < Pv ; low shear

Mc= pySx = 265(3780x10−3) = 1001.7 kNm

Since Mmax < Mc moment is adequate

AT SUPPORTWeb bearing capacityFv < Pbw b1= t + T + 0.8r – g * g = 10