Universal beam sections are normally employed in buildings to carry floor and wall 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.

The structural design of steel beams to BS 5950 involves following specific guidelines and principles outlined in the British Standard. BS 5950 is a widely used code of practice for the design of steel structures in the United Kingdom, but has been replaced by Eurocode 3 (EN 1993-1-1).

When designing steel beams, several factors are considered to ensure structural integrity and safety. These factors include determining the appropriate loadings, selecting the appropriate section shape and size, analyzing the beam’s resistance to bending, shear, and deflection, and ensuring proper connection details.

The code specifies various load combinations, such as dead loads, live loads, wind loads, and imposed loads, that need to be considered during the design process. In terms of section selection, the code provides tables and charts to determine the suitable steel section based on the applied loads and required span. These sections include universal beams (UB), universal columns (UC), and parallel flange channels (PFC), among others. The appropriate section is chosen based on its moment resistance, shear capacity, and deflection limits.

Design calculations involve checking the beam’s capacity to resist bending, shear, and deflection. These calculations consider the applied loads, section properties, and material properties of the steel. The code provides formulas and design charts to assess these aspects.

**Steel Beam Design Example**

A laterally restrained beam 9m long that is simply supported at both ends support a dead uniformly distributed load of 15 kN/m and an imposed load uniformly distributed load of 5 kN/m. It also carries a dead load of 20 kN at a 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 = 1.4G_{k} = 1.4 × 20 = 28 kN

Uniformly distributed load = 1.4G_{k} + 1.6Q_{k} = 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 ε = √(275/P_{y}) = √(275/275) = 1.0

**Section classification**

**Flange**

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

**Web**

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

**Shear Capacity**

As d/t = 48 < 70ε, 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}(158.5 kN) < 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;

δ = (5ql^{4})/384EI = (5 × 5 × 9^{4}) / (384 × 205 × 10^{6} × 29400 × 10^{-8}) = 7.087 × 10^{-3} m = 7.087 mm

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 200 mm 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 α_{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} = [25εt/√(b_{1} + nk)d] P_{bw} —– (4)

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

P_{x} = [(α_{e} + 0.7d)/1.4d] × [25εt/√(b_{1} + nk)d)] P_{bw} ————- (5)

Therefore, α_{e} = 20mm + (200/2) = 120mm

0.7d = 0.7 × 407.6 = 285.32mm

α_{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 = 280.538 kN

P_{x}(280.538 kN) > F_{v}(158.5 kN) Hence it is ok.

Thank you for reading, and feel free to share.

Like Our Facebook page

www.facebook.com/structville

• It’s in reality a great and useful piece of information. I am happy that you simply shared this helpful info with us. Please keep us informed like this. Thank you for sharing.

CoBie adaptation in USA

Interference Analysis in USA

MEP F modelling in USA

Revit Modeling in USA

Fabrication Drawings preparation in USA

Shop Drawings Preparation in USA

Structural Analysis in USA

Point Cloud to BIM conversion in USA

ANIMATION SERVICES in USA

BIM Implementation in USA

Thank you Ubani Obinna Uzodimma for presenting such a clearly detailed design procedure

Simple and straight detailed example

Thank you so much

Am really appreciate for this easy way of learning.

Thank Engr for this solution.

But the check for lateral torsional buckling was not presented

Good stuff!

Lateral torsional buckling is a crucial check….

Stirling presentation!

Elaborate and simple!.. appreciated