**Introduction**

This post deals with the design of simply supported I-beam section subjected to permanent and variable loads according to Eurocode 3. The design involves selecting the appropriate section that will satisfy limit state requirements.

**Problem Statement**

It is desired to select an appropriate section to satisfy ultimate and serviceability limit state requirements for a laterally restrained simply supported beam that is subjected to the following loads;

Permanent Load Gk = 38 kN/m

Variable Load Qk = 12 kN/m

The length of the beam = 7.5m

**Solution**

P

_{Ed}= 1.35Gk + 1.5Qk

P

_{Ed}= 1.35(38) + 1.5(12) = 69.3 kN/m

An advanced UK beam S275 is to be used for this design.

F

_{y}= 275 N/mm

^{2}

Î³

_{m0 }= 1.0 (Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005)

The required section is supposed to have a plastic modulus about the y-y axis that is greater than;

W

_{pl,y}= M

_{y,Ed}Î³

_{m0}/F

_{y}

W

_{pl,y}= (487.265 × 10

^{3}× 1.0)/275 = 1771.872 cm

^{3}

From steel tables, try section 457 x 191 x 82 W

_{pl,y}= 1830 cm

^{3}

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**Properties**

h = 460.0mm; b = 191.3mm; d = 407.6mm; t

_{w}= 9.9mm; t

_{f}; r = 10.2mm; A = 104 cm

^{4}; I

_{y}= 37100 cm

^{4}; I

_{z}= 1870 cm

^{4}; W

_{el,y}= 1610 cm

^{3}; W

_{el,y}= 1830 cm

^{3}

h

_{w}= h - 2t

_{f}= 428.0mm

E (Modulus of elasticity) = 210000 N/mm

^{2 }(Clause 3.2.6(1))

**Classification of section**

Îµ = √(235/F

_{y}) = √(235/275) = 0.92 (Table 5.2 BS EN 1993-1- 1:2005)

Outstand flange: flange under uniform compression c = (b - t

_{w}- 2r)/2 = [191.3 - 9.9 - 2(10.2)]/2 = 80.5mm

c/t

_{f}= 80.5/16.0 = 5.03

The limiting value for class 1 is c/t

_{f}≤ 9Îµ = 9 × 0.92

5.03 < 8.28

Therefore, outstand flange in compression is class 1

**Internal Compression Part (Web under pure bending)**

c = d = 407.6mm

c/t

_{w}= 407.9/9.9 = 41.17

The limiting value for class 1 is c/t

_{w}≤ 72Îµ = 72 × 0.92 = 66.24

41.17 < 66.24

Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.

**Member Resistance Verification**

__Moment Resistance__For the structure under consideration, the maximum bending moment occurs where the shear force is zeo. Therefore, the bending moment does not need to be reduced for the presence of shear force (clause 6.2.8(2))

M

_{Ed}/M

_{c,Rd}≤ 1.0 (clause 6.2.5(1))

M

_{c,Rd}= M

_{pl,Rd }= (M

_{pl,y }× F

_{y})/Î³

_{m0}

M

_{c,Rd}= M

_{pl,Rd }= [(1830 × 275)/1.0] × 10

^{-3}= 503 kNm

M

_{Ed}/M

_{c,Rd}= 487.265/503 = 0.9687 < 1.0 Ok

__Shear Resistance (clause 6.6.2)__The basic design requirement is;

V

_{Ed}/V

_{c,Rd}≤ 1.0

V

_{c,Rd}= V

_{pl,Rd }= A

_{v}(F

_{y }/ √3)/Î³

_{m0 }(for class 1 sections)

For rolled I-section with shear parallel to the web, the shear area is;

A

_{v}= A - 2bt

_{f}+ (t

_{w}+ 2r)t

_{f}(for class 1 sections) but not less than Î·h

_{w}t

_{w}

A

_{v}= (104 × 10

^{2}- (2 × 191.3 × 16) + [9.9 + 2(10.2)] × 16 = 4763 mm

^{2}

Î· = 1.0 (conservative)

Î·h

_{w}t

_{w }= (1.0 × 428 × 9.9) = 4237 mm

^{2}

4763 > 4237

Therefore, A

_{v}= 4763 mm

^{2}

The shear resistance is therefore;

V

_{c,Rd}= V

_{pl,Rd }= [4763 × (275

_{ }/ √3)/1.0] × 10

^{-3}= 756 kN

V

_{Ed}/V

_{c,Rd}= 259.875/756 = 0.343 < 1.0 Ok

**Shear Buckling**

Shear buckling of the unstiffnened web will not need to be considered if;

h

_{w}/t

_{w }≤ 72Îµ/Î·

h

_{w}/t

_{w }= 428.0/9.9 = 43

72Îµ/Î· = (72 × 0.92)/1.0 = 66

43 < 66 Therefore shear buckling need not be considered.

**Serviceability limit state**

Vertical deflections are computed based on variable loads. Permanent loads need not be considered.(BS EN 1993-1-1 NA 2.23)

Q

_{k}= 12 kN/m

w = 5ql

^{4}/384EI

w = (5 × 12 × 7500

^{4})/(384 × 210000 × 37100 × 10

^{4}) = 6.345mm

Span/360 = 7500/360 = 20.833mm (BS EN 1993-1-1 NA 2.23)

6.345mm < 20.833mm. Therefore, deflection is satisfactory

Thank you for visiting Structville Today.

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