Beam Design for Torsion | Worked Example

The method of beam design for torsion with worked example is debussed in this article. There are basic steps to be followed same as the beam longitudinal reinforcement design.

In the beam design for torsion, torsional link requirements and an additional area of reinforcement to be added to the longitudinal reinforcement are calculated.

More importantly, torsional reinforcement links are not the same as the usual shear links provided for the beams. The torsional links terminate at two ends not like the conversational link.

Worked Example Beam Design for Torsion

• Span of the beam = 5m
• Span of cantilever = 2m
• Height of the beam = 600mm
• Width of the beam = 300mm
• Shear link diameter = 10mm
• Cover to reinforcement = 25mm
• Design load on beam including all load factors = 4 kN/m²
• Grade of concrete C30
• Characteristic yield strength = 460 kN/m²

Calculation of Torsional Bending Moment

Bending Moment = 4 x 2 x 5 = 40 kNm

Bending moment per support = 40 / 2 = 20 kNm

Calculation of Torsional Shear Stress, v_t

v_t  = 2T / [ h_min² (h_max – h_min/3)]

T = 20 kNm, h_min = 300, h_max = 600

v_t  = 2 x 20 x 10⁶ / [ 300² (600 – 300/3)] = 1.24 N/mm²

From Table 2.3 of BS 8110 Part 2,

v_{t min} = 0.36 N/mm²  and v_tu = 4.38 N/mm²

v_{t min} < v_tu and v_t >  v_{t min}, Therefore, torsional reinforcement needs to be provided.

In this worked example, we are not going to design the beam for shear and bending. However, we need to know the shear stress and shear capacity of the beam for this design. Thus, let’s assume some values for shear stress.

Say, v = 0.25 N/mm² and v < v_c 

There are two segments in the beam based on the shear and torsional reinforcement requirement.

• v_t ≤ v_{t min} Nominal shear reinforcements are adequate and no beam design for torsion required.
• v_t > v_{t min} Design for torsional shear links.

We can easily calculate the segment that we need only the nominal shear links.

First, let’s calculate the torsional moment that specified the minimum requirement for torsional links.

T = v_{t min} h_min² [(h_max – h_min/3)]/2

T = 0.37x 300² [(600 – 300/3)]/2 = 8.33 kNm

Now let’s calculate the distance to this moment

Say it is x,

x = (8.33/20) x 2.5 = 1.0 m

Thus, we need to provide torsional links up to 1.5m distance from each support.

Calculation of Nominal Shear Links

A_sv / S_v ≥ 0.4b / 0.95f_yv

Consider T10 bars = 78 x 2 = 157 mm²  * two legs, therefore, rebar area multiped by 2; ( 78 x 2).

S_v ≤ 157 x 0.95 x 460 / (0.4 x 300) = 575 mm

Provide nominal shear links T10@250mm spacing. This satisfies the minimum shear links requirement also.

Calculate Design Torsional Reinforcement

A_sv / S_v = T / [ 0.8x₁y₁(0.95f_yv)]

x₁ = Smallest center-to-center distance of shear links

x₁ = 300 – 25 – 10/5 = 230 mm

x₁ = Largest center-to-center distance of shear links

x₁ = 600 – 25 – 10/5 = 530 mm

S_v ≤ 157 x 0.8 x 230 x 530 x 0.95 x 460 = 334 mm

Provide torsional reinforcement, torsional shear links T10 @ 200 spacing

Calculation of Additional Longitudinal Reinforcement

The following equation given in the BS 8110 Part 2 can be used.

A_s = A_sv f_yv [ x₁ +y₁)] / S_v f_y

A_s = 157 x 460[ 230 +530)] / (334 x 530) = 358 mm²

This area of reinforcement needs to be distributed at the top, middle and bottom.

Since the width of the beam is 300mm, we can provide three bars at the top, two bars in the middle and three bars at the bottom. Therefore, there will be 8 bars.

Area of a bar = 358 / 8 = 44.7 mm²

We may add this area of reinforcement to each top three bars and bottom three bars. In the middle we may provide 2T12.

In addition to this, we have to check the maximum spacing of the torsional links which is 300mm and make sure to provide torsional reinforcement at four corners. 

Also, we have to keep in mind that the torsional link is not the same as the normal shear link.