**Introduction**

Universal beam sections are normally employed in buildings to carry load. Loads on beams may include the load from slab, walls, building services, and their own self weight. It is necessary for structural beams to satisfy ultimate and serviceability limit state requirements.

This post gives a solved design example of a laterally restrained beam according to BS 5950.

**Design example**

A laterally restrained beam 9m long that is simply supported at both ends supports a dead uniformly distributed load of 15 KN/m and an imposed load uniformly distributed load of 5KN/m. It also carries a dead load of 20KN at distance of 2.5m from both ends. Provide a suitable UB to satisfy ultimate and serviceability limit state requirements (P

_{y}= 275 N/mm

^{2}).

**Solution**

At ultimate limit state;

Concentrated dead load = 1.4Gk = 1.4 × 20 = 28 KN

Uniformly distributed load = 1.4Gk + 1.6Qk = 1.4(15) + 1.6(5) = 29 KN/m

**Support Reactions**

Let ∑M

_{B}= 0; anticlockwise negative

(9 × Ay) – (29 × 9 × 4.5) – (28 × 6.5) – (28 × 2.5) = 0

Ay = 158.5 KN

Let ∑M

_{A}= 0; clockwise negative

(9 × By) – (29 × 9 × 4.5) – (28 × 6.5) – (28 × 2.5) = 0

By = 158.5 KN

**Internal Stresses**

**Moment**

M

_{A}= 0 (Hinged support)

M

_{C}= M

_{D}= (158.5 × 2.5) – (29 × 2.5 × 1.25) = 305.625 KNm

M

_{midspan}= (158.5 × 4.5) – (29 × 4.5 × 2.25) – (28 ×2) = 363.625 KNm

**Shear**

Q

_{A}= Ay = 158.5 KN

Q

_{C}

^{L}= 158.5 – (29 × 2.5) = 86 KN

Q

_{C}

^{R}= 158.5 – (29 × 2.5) - 28 = 58 KN

Q

_{D}

^{L}= 158.5 – (29 × 6.5) – 28 = -58 KN

Q

_{D}

^{R}= 158.5 – (29 × 6.5) – 28 - 28 = - 86 KN

Q

_{B}= 158.5 – (29 × 9) – 28 - 28 = - 158.5 KN

**Internal Forces Diagram**

**Structural Design to BS 5950**

P

_{y}= 275 N/mm

^{2}

^{}

**Initial selection of section**

Moment Capacity of section M

_{c}= P

_{y}S ------- (1)

Where S is the plastic modulus of the section

Which implies that S = M

_{c}/P

_{y}= (363.625 × 10

^{6})/275 = 1320963.636 mm

^{3}= 1320.963 cm

^{3}

With this we can go to the steel sections table and select a section that has a plastic modulus that is slightly higher than 1320.963 cm

^{3}

^{}Try section UB 457 × 191 × 67 (S = 1470 cm

^{3})

Properties of the section;

I

_{xx}= 29400 cm

^{4}

Z

_{xx}= 1300 cm

^{3}

Mass per metre = 67.1 kg/m

D = 453.4mm

B = 189.9mm

t = 8.5mm

T = 12.7mm

r = 10.2mm

d = 407.6mm

**Strength classification**

Since T = 12.7mm < 16mm, P

_{y}= 275 N/mm

^{2}

Hence Îµ = √(275/P

_{y}) = √(275/275) = 1.0

**Section classification**

**Flange**

b/T = 7.48 < 9Îµ; Flange is plastic class 1

**Web**

d/t = 48 < 80Îµ; Web is also plastic class 1

**Shear Capacity**

As d/t = 48 < 70Îµ, shear buckling need not be considered (clause 4.4.4)

Shear Capacity P

_{v}= 0.6P

_{y}A

_{v}-----

_{}(2)

P

_{v }= 0.6P

_{y}tD = 0.6 × 275 × 8.5 × 453.4 = 635893.5 N = 635.89 kN

But design shear force F

_{v}= 158.5 KN. Since F

_{v}(158.5) < P

_{v}(635.89), section is ok for shear.

Now, 0.6P

_{v}= 0.6 × 635.89 = 381.54 kN

Since F

_{v}(158.5KN) < 0.6P

_{v}(381.54 KN), we have low shear load.

**Moment Capacity**

Design Moment = 363.625 KNm

Moment capacity of section UB 457 × 191 × 67 (S = 1470 cm

^{3}= 1470 × 10

^{3}mm

^{3})

M

_{c}= P

_{y}S = 275 × 1470 × 10

^{3}= 404.25 × 106 N.mm = 404.25 KNm

1.2P

_{y}Z = 1.2 × 275 × 1300 × 10

^{3}= 429 × 106 N.mm = 429.00 KNm

M

_{c}(404.25 KNm) < 1.2P

_{y}Z (429.00 KNm) Hence section is ok

**Evaluating extra moment due to self weight of the beam**

Self-weight of the beam S

_{w}= 67.1 kg/m = 0.658 KN/m (UDL on the beam)

Moment due to self weight (M

_{sw}) = (ql

^{2})/8 = (0.658 × 9

^{2})/8 = 6.66 KNm

(363.625 + 6.66) < M

_{c}(404.25 KNm) < 1.2P

_{y}Z (429.00 KNm) Hence section is ok for moment resistance.

**Deflection Check**

We check deflection for the unfactored imposed load; E = 205 KN/mm

^{2}= 205 × 10

^{6}KN/m

^{2}; I

_{xx}= 29400 cm

^{4}= 29400 × 10

^{-8}m

^{4}

^{}

The maximum deflection for this structure occurs at the midspan and it is given by;

Î´ = (5ql

^{4})/384EI = (5 × 5 × 9

^{4}) / (384 × 205 × 10

^{6}× 29400 × 10

^{-8}) = 7.087 × 10

^{-3}m = 7.087mm

Permissible deflection; L/360 = 9000/360 = 25mm

7.087mm < 25mm. Hence deflection is satisfactory.

**Web bearing**

According to Clause 4.5.2 of BS 5950-1:2000, the bearing resistance P

_{bw}is given by:

P

_{bw}= (b

_{1}+ nk)tP

_{yw }-----

_{}(3)

Where;

b

_{1}is the stiff bearing length

n = 5 (at the point of concentrated loads) except at the end of a member and n = 2 + 0.6b

_{e}/k ≤ 5 at the end of the member

b

_{e}is the distance to the end of the member from the end of the stiff bearing

k = (T + r) for rolled I- or H-sections

T is the thickness of the flange

t is the web thickness

P

_{yw}is the design strength of the web

_{}

**Web bearing at the supports**

Let us assume that beam sits on 200mm bearing, and b

_{e}= 20mm

k = (T + r) = 12.7 + 10.2 = 22.9mm;

Hence n = 2 + 0.6(20/22.9) = 2.52mm < 5mm.

P

_{bw}= (b

_{1}+ nk)tP

_{yw}

P

_{bw}= [200 + 2.52(22.9)] × 8.5 × 275 = 602392.45 N = 602.392 KN

P

_{bw}(602.392 KN) > F

_{v}(158.5 KN) Hence it is ok

**Contact stress at supports**

P

_{cs}= [b

_{1}× 2(r +T )]P

_{y}= [200 × 2(22.9)] × 275 = 2519000N = 2519 KN

P

_{cs}(2519 KN) > F

_{v}(158.5 KN) Hence contact stress is ok

**Web buckling**

According to clause 4.5.3.1 of BS 5950, provided the distance Î±

_{e}from the concentrated load or reaction to the nearer end of the member is at least 0.7d, and if the flange through which the load or reaction is applied is effectively restrained against both;

(a) rotation relative to the web

(b) lateral movement relative to the other flange

The buckling resistance of an unstiffened web is given by;

P

_{x}= [25Îµt/√(b

_{1}+ nk)d] P

_{bw}-----

_{}(4)

_{}When Î±

_{e}< 0.7d, the buckling resistance of an unstiffened web is given by;

P

_{x}= [(Î±

_{e}+ 0.7d)/1.4d] × [25Îµt/√(b

_{1}+ nk)d)] P

_{bw}------------- (5)

Therefore, Î±

_{e}= 20mm + (200/2) = 120mm

0.7d = 0.7 × 407.6 = 285.32mm

Î±

_{e}(120mm) < 0.7d(285.32mm). Hence equation (5) applies

P

_{x}= [(120 + 285.32)/(1.4 × 407.6)] × [(25 × 1 × 8.5 )/√((200 + 2.52 × 22.9) × 407.6)] × 602.392 = 280.538KN

P

_{x}(280.538 KN) > F

_{v}(158.5 KN) Hence it is ok.

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