PROJECT: DESIGN FOR A BUS PARKING WITH AN OPERATION OFFICE IN UTHM

JOB: SIMPLY SUPPORTED UNRESTRAINED BEAM (3/A-B)

REF CALCULATION OUTPUT

BS EN 1990NA.2.2.3.2TableNA.A1.2(B)

BS EN 1990Eq. (6.10b)

Beam Span Length: 6mAssumed that the floor slab does not offer lateral restraint

Characteristic actions – 1st Floors

selfweight floor 3.0 kN/m2

Self weight of ceilings and furnishing 0.2kN/m2

Brickwall 2.4kN/m2

Selfweight of beam 59.8kg x 9.81 x 10-3 0.57kN/m2

Total permanent load 6.17 kN/m2

Variable actionsImposed floor load for offices 3.0kN/m2

Ultimate limit state (ULS)

Partial factors for actions

Partial factor for permanent actions γG= 1.35Partial factor for variable actions γQ = 1.5Reduction factor ξ = 0.925

Design load combine action : ξ γGgk + ξ γQqk

= 0.925 x 1.35 x 6.17 + 1.5 x 3.0= 12.2kN/m2

w = 12.2 x 6 = 73.0kN/m

Design moment and shear force

Maximum design moment, My,Ed, occurs at mid-span, and for bending about the major (y-y) axis is:

My,ed = w l2

8 = 73.2.0 x62

8 = 329.4kNm

gk : 6.17kN/m2

qk : 3.0kN/m2

w = 69.0kN/m

Design moment = 329.4kNm

Design shear

Table section of propertiesBS EN 1993-1-1

Maximum design shear force, VEd, occurs at the end supports is:

Ved = wl2 = 73.2 x 6

2 = 219.6kN

Partial factors for resistanceγMo = 1.0γM1 = 1.0

Trial section

457 x 152 x 60 UKB, S275

Depth of cross-section h = 454.6 mmWidth of cross-section b = 152.9 mmWeb depth between fillets d = 407.6 mmWeb thickness tw = 8.1 mmFlange thickness tf = 13.3 mmRoot radius r = 10.2 mmSection area A = 76.2 cm2

Second moment, y-y Iy = 25500 cm4

Second moment, z-z Iz = 795cm4

Radius of gyration, z-z iz= 3.23 cmWarping constant Iw = 387000cm6

Torsion constant It = 33.8 cm4

Elastic section modulus, y-y Wel,y = 1120 cm3

Plastic section modulus, y-y Wpl,y = 1290cm3

Ratio for local buckling cf/tf = 4.68 cw/tw = 50.3

force = 219.6kN

EN 1993-1-15.5.2TABLE 5.2

EN 1993-1-16.2.5

EN 1993-1-16.3.2.2(1)

Nominal yeild strength, fy

tw = 8.1 mmtf = 13.3 mm < 40mm

fy for S275 ( EN10025-2) IS 275N/mm2

Section Classification

For section classification the coefficient ε is:

ε=√ 235275

= 0.92

Outstand flange: flange under uniform compression

cf/tf < 9ε4.68< 9(0.92) = 8.28 -------- class 1

Internal compression part subject to bending

cw/tw < 72ε50.3< 72(0.92) = 66.24 -------- class 1

Therefore the section is Class 1

Bending Resistance of the cross-section

The design resistance of the cross-section for bending about themajor axis (y-y) for a class 1 section is:

Mc,y,Rd = Wpl , yfyγ Mo

Mc,y,Rd = 1290 x 103 x 2751.0

x10−6 = 354.8kNm

329.4kNm < 354.8kNm

Med < Mc, rd ok!

Shear resisteance of

fy = 275N/mm2

Overall classification is class 1

EN 1993-1-16.3.2.3 (1)

EN 1993-1-1Table 6.3Table 6.4

Lateral torsional buckling resistance

Elastic critical moment

The critical moment may be calculated from the following expression

Mcr = C1π 2 E I z

Lcr2 ( I w

I z+

Lcr2G I T

π2 E I z)

0.5

where:

E is Young’s modulus: E = 210000 N/mm2G is the shear modulus: G = 81000 N/mm2L is the span: Lcr = 6.00 m

C1 = 1.132 (table 6.11)

Mcr =1.132

π 2210000 x79.5 x106

60002 ¿( 387 x 109

79.5 x 106 +60002 x81000 x338 x103

π2 x210000 x79.5 x106 )0.5

Mcr = 540kNm

Non-dimensional slendernessThe non-dimensional slenderness is obtained from:

λ¿= √ Wy fyMcr

λ¿= √ 1290 x103 x 275540 x106

= 0.81

χ¿ = 1

ϕ¿+√ϕ¿2+λ¿

2

ϕ¿ = 0.5( 1 +α ¿ ( λ¿−0.2 )+λ¿2

For h/b = 465.8 / 155.2 = 2.97 > 2

Mcr = 540kNm

Using buckling curves b:α ¿=0.34(table 6.3) < 1.0 ok!

ϕ¿ = 0.5( 1 +0.34 (0.81−0.2 )+0.812 ¿ =0.93

χ¿ = 1

0.93+√0.932+0.812 = 0.96 < 1.0 ok!

Mb,rd = χ¿Wy fyγ M 1

Mb,rd = 0.9 6 x1290 x103 x 2751.0 = 340.5kNm

Med < Mb,rd

329.4kNm < 340.5kNm ok!

So the section is acceptable : 457 x 152 x 60 UKB, S275

Mb,rd = 340.5