Steel Beam Design Worked Example [Universal Beam]

The steel beam design worked example elaborates the design of a simply supported beam having a uniformly distributed load. The beam is considered as simply supported and the design data for calculating the bending moment and shear forces are given below.

Further, the section properties to be considered are also given in each stage fo checking the section.

Theoretical aspects and code design procedures are discussed in the article design of the steel beam as per BS 5950.

Design Data

• Load UDL 20 kN/m
• Span of the beam 6m
• Beam is simply supported
• Desing strength of steel, Py = 275 N/mm²

Maximum Bending Moment

= wl²/8 = 20 x 6² / 8 = 90 kNm

Maximum shear force

= wl/2 = 20 x 6 / 2 = 60 kN

Consider universal beam 500x200x89.7 kg/m (P_y = 275 N/mm² )

Section Data

• D = 500 mm
• T = 16 mm
• t = 10 mm
• B = 200 mm
• b = 100 mm
• r₁ = 20 mm
• d = 500 – 16 x 2 – 2 x 20 = 428 mm

• S_x = 2175×10³ mm³
• Z_x = 1914×10³ mm³
• r_y = 43.3 mm

Let’s start the steel beam design calculation. Under this steel beam design calculation, the following checks are done.

1. Section Classification
2. Shear Design
3. Design for bending
4. Check for lateral torsional buckling 
5. Deflection check
6. Web bearing check
7. Web buckling check

Classification of Section

The first step of the steel beam design is the classification of the section to know whether it is plastic, semi-plastic, compact, slender.

T = 16 mm, P_y = 275 N/mm²

ε = (275/P_y)^0.5 = 1

Check Flange

b/T = 100 / 16 = 6.25 < 9ε = 9 – Flange is Plastic

Check Web

d/t = 428 / 10 = 42.8 < 80ε = 80 – Web is Plastic

Further, d/t < 70ε = 70 – Therefore, no need to check for shear buckling

Therefore, section is Plastic

Design for Shear

Design shear force, F_v = 60 kN

P_v = 0.6 P_y A_v = 0.6 P_y tD = 0.6 x 275 x 10 x 500 x 10^-3 = 825 kN

F_v < P_v Shear capacity is OK

Depending on the shear force, it is decided whether section is subjected to low or high shear in steel beam design.

Design for Bending

Check whether section is subjected to low shear or high shear

60% x P_v = 0.6 x 825 = 495 kN

F_v < 0.6 P_v Section subjected to low shear

M_c should be less than 1.2P_yZ_x or P_y S_x as per Cl. 4.2.5.1 and Cl. 4.2.5.2

M_c ≤ 1.2P_yZ_x = 1.2 x 275 x 1914 x10³ x 10^-6 = 613.62 kNm

M_c = P_y S_x = 275 x 2175 x 10³ x 10^-6 = 598.125 kNm

Therefore,

M_c = 598.125 kNm > 90 kNm

Bending ok

Check Lateral Torsional Buckling

M_x < M_b / m_LT

In this example, no intermediate restrains were considered

m_LT = 0.925, Table 18, BS 5950

M_b = P_b S_x Cl. 4.3.6.4

There are two methods to check lateral-torsional buckling as discussed in the article steel beam design as per BS 5950. They are the rigorous method and the simplified method.

In this steel beam design example, we discussed both the methods to elaborate on the procedures that need to be followed using either method.

Further, the main difference between these two methods is evaluating bending strength.

Rigorous Method

M_b = P_b S_x

P_b is depending on the λ_LT and P_y

λ_LT = uvλ√(β_w)

λ = L_E / r_y

L_E – to be taken from Table 13 as per the Cl. 4.3.5.1 and consider L_LT = L – span

L_E = 1.0 L_LT = 1 x 6 = 6 m

λ = L_E / r_y = 6000 / 43.3 = 138.568

For rolled I and H sections, Cl. 4.3.6.8

x = D / T used with u = 0.9

x = D / T = 500 / 10 = 50

β_w to be obtained from Cl 4.3.6.9

β_w = 1 for Class 1 plastic or Class 2 Compact sections

v – slenderness factor to be obtained from Table 19 based on the λ / x and η

λ / x = 138.568 / 50 = 2.771

η = 0.5 for equal flanges

v = 0.919 from Table 19

λ_LT = uvλ√(β_w) = 0.9 x 0.919 x 138.568 x √(1) = 114.6

λ_LO can be obtained from Table 16 (refer bottom of the table)

If λ_LO ≥ λ_LT ; P_b = P_y or Othewise P_b shall be taken from Table 16 for rolled sections.

If λ_LO ≥ λ_LT no allowance needs to be made for lateral-torsional buckling and otherwise check for lateral-torsional buckling.

P_y = 275 N/mm² ; λ_LO = 37.3

λ_LO < λ_LT Therefore check for lateral torsional buckling

From Table 16, for λ_LT = 114.6 ; P_b = 102 N/mm²

M_b = P_b S_x = 102 x 2175 x 10³ x 10^-6 = 221.85 kNm

M_b / m_LT = 221.85 / 0.925 = 239.838 kNm

Therefore, M_x = 90 kNm < M_b / m_LT = 239.838 kNm

Section is adquate.

Simplified Method

Note: Both methods need not be done when designing a beam. Following either the simplified method or rigorous method is adequate.

M_b = P_b S_x : Cl. 4.3.7

In this method, the determination of the P_b is different when compared with the previous method. This method provides conservative answeres. P_b can be obtained from Table 20 of BS 5950 based on √(β_w) (L_E / r_y ) and D / T

β_w = 1 ; previous calculation

L_E / r_y = 138.568 ; previous calculation

√(β_w) (L_E / r_y ) = 1^0.5 x 135.568 = 138.568

D / T = 500 / 16 = 31.25

P_b = 116.646 N/mm² From Table 20

M_b = P_b S_x = 116.646 x 2175 x 10³ x 10^-6 = 253.705 kNm

M_b / m_LT = 253.705 / 0.925 = 274.3 kNm

Therefore, M_x = 90 kNm < M_b / m_LT = 274.3 kNm

Section is adequate

Defelection

Maximum defelection ( δ ) for simply supported beam having uniformly distributed can be evaluated from following equation.

δ = 5W_eL⁴ / (384EI)

This equation can be further simplifed as follows.

δ = 0.104M_maxL² / (EI)

Since we evaluate the deflection due to the imposed loads, consider imposed load as 10 kN/m in this calculation.

δ = 0.104 x 90 x 10⁶ 6000² / (205 x 10³ x 478 x 10⁶ ) = 1.7 mm

Assuming brittle finishes to be used

Span / 360 = 6000 / 360 = 16.7 mm

δ < Span / 360, Hence deflection is Ok

Different methods of calculating the defelections are discussed in the Wikipedia article deflection (engineering).

Web Bearing Capacity

It is a must to check the capacity of the web in steel beam design. Web bearing and buckling is done in this catogory. 

P_bw = (b₁ + nk) tP_yw

t = 10 mm

T = 10 mm

r = 10 mm

g = 5 mm

b_e = 5 mm

b₁ = t + T + 0.8r – g = 10 + 10 + 0.8 x 10 – 5 = 23 mm

k = T + r = 16 + 20 = 36 mm (for rolled section)

at end,

n = (2 + 0.6b_e/k) but ≤ 5

n = (2 + 0.6 x 5 / 36) = 2.083 < 5 P_yw = 275 N/mm² from Table 9

P_bw = (b₁ + nk) tP_yw = (23 +2.083 x 36 ) x 10 x 275 x 10^-3 = 269.5 kN

Support reation is 60 kN.

Web bearing is Ok and no stiffeners are required.

When F_x > P_bw , we need to provide stiffeners to cater to balance force ( F_x – P_bw ). The capacity of the stiffeners shall be obtained from P_s = A_s.net P_y. Where A_s.net is the cross-sectional area of the stiffness. If the web and stiffness have different strengths, the smaller value should be used to calculate both P_s and P_bw.

Web Buckling

When a_e ≥ 0.7d

P_x = 25εt P_bw /√[ ( b₁ + nk ) d ]

When a_e < 0.7d

P_x = [ (a_e +0.7d)/1.4d] {25εt P_bw /√[ ( b₁ + nk ) d ] }

Where

a_e = 0.7d = 23 / 2 = 11.5 mm < 0.7 x 428 = 300 mm

P_x = [ (a_e +0.7d)/1.4d] {25εt P_bw /√[ ( b₁ + nk ) d ] }

P_x = [ (11.5 +0.7×428)/1.4×428] {25x1x10x275 /√[ ( 23 + 2.083×36 ) 428 ] } = 174.3 kN

F_x < P_x

No stiffeners are required.

Beam satisfies all the checks.