Designing Concrete Elements for Torsion

Torsion in structures is something that is best avoided as far as possible. This is because the twisting of structural elements generates large internal forces which result in significant increase in element sizes and possibly even change in form. In a previous post titled “Avoiding Torsion in Structures“. The factors contributing to torsion in structures were identified, suggestions were also made on how best to avoid or reduce it in structures. However, there are instances where Torsion cannot be avoided and the only course of action is just to allow for it. In such instances, there are two choices available to the designer.

• Provide restraint that prevent the element from being subject to Torsion
• Design the structural elements to resist torsion.

In this post we are concerned witht the latter but with respect to reinforced concrete elements using Eurocode 2.

Torsion in Concrete Elements

Torsion in reinforced concrete elements in most instances will act along with flexure and shear. Thus, when designing concrete elements subjected to a combination of torsion, flexure and shear. The first step is to design separately for flexure, shear and torsion and then combining the area of steel reinforcements in detailing the section.

Torsional moments generate mainly tensile stresses along a member subjected to torsion. Diagonal cracking follows almost immediately if the tensile stresses exceed the concrete’s tensile strength. As a result, the cracks form in the form of spirals around the member. Thus longitudinal and shear reinforcement must be provided to avoid this failure mode.

The torsional design equations of reinforced concrete components are derived from a structural model in which the concrete component behaves like a thin-walled hollow section. The first step is to determine the geometry and wall thickness of this hollow equivalent section. However, when appraising a particular section which geometry has more complexity. For example, a section I, L or T, is divided into rectangular components. Then each part is laid out to carry a proportion of the complete torsion applied.

Designing Torsional Reinforcement

Designing for Torsion and provision of torsional reinforcement is based on the concept of an internal truss within a concrete element. This is the same concept used in designing for shear in reinforced concrete. The steel reinforcement within the concrete act as tension ties while the concrete in compression acts as diagonal struts. Section 6.3 of BS EN 1992-1-1 sets out the analysis and design for Torsion in concrete element. Generally the procedure involves calculating the shear stress due to the applied torsional moment and then providing torsional reinforcement if required.

Procedure for Design

Determine the applied design torsional moment TEd and calculate the shear stress due to torsion using expression 1.

{ \tau }_{ t,i }\cdot { t }_{ ef,i }=\frac { { T }_{ Ed } }{ 2{ A }_{ k } } ————1
Where:
• TEd is the applied design torsional moment
• Ak is the area enclosed by the centrelines of the connecting walls, including inner hollow areas
• t,i is the shear stress in the thin wall due to torsion tef,i is the eﬀective thin wall thickness, which can be assumed to be A/u,
• A is the total area of the cross-section within the outer circumference, including inner areas assumed to be hollow
• u is the outer circumference of the cross-section

Calculate the applied shear force from the shear stress due to torsion using expression 2.

{ V }_{ Ed }={ \tau }_{ t,i }{ t }_{ ef,i }{ z }_{ i }——————2
Where:
• VEd,i is the applied design shear force due to torsion in the thin wall
• zi is the side length of wall i defined by the distance between the intersection points with the adjacent walls.

Determine the areas of longitudinal reinforcement required to resist torsion from. expression 3.

\frac { \sum { { 0.87A }_{ s1 }{ f }_{ yk } } }{ { u }_{ k } } =\frac { { T }_{ Ed } }{ { 2A }_{ k } } cot\theta ————–3
Where
• Asl is the required area of longitudinal steel to resist tension stress due to torsion
• uk is the perimeter of area Ak
• fyd is the design yield stress of the steel reinforcement is the angle of the compression struts in the concrete element, normally taken to be 45

In an instance where a member is subjected to both Torsion in combination with Shear and Flexure. The additive effect of the twisting moment must be taken into cognizance when estimating the area of tension reinforcement. To ensure that these elements comply with this requirement BS EN 1992-1-1 requires that a unit check be carried out. This is defined as expression 4 on this post. Thus, the condition must be satisfied, else the element must be resized

\frac { { T }_{ Ed } }{ { T }_{ Rd,max } } +\frac { { V }_{ Ed } }{ { V }_{ Rd,max } } \le 1.0————–4
Where
• VEd is the applied design transverse force
• TRd,max is the design torsional resistant moment calculated using Equation 5
• VRd,max is the maximum design shear resistance calculated Equation 66
{ T }_{ Rd,max }=2v{ \alpha }_{ cw }{ f }_{ cd }{ A }_{ k }{ t }_{ ef,i }sin\theta cos\theta ———5
{ V }_{ Rd,max }=\frac { { \alpha }_{ cw }{ b }_{ w }zv{ f }_{ cd } }{ \left( cot\theta +tan\theta \right) } ————-6
Where

v is the strength reduction factor for cracked concrete when subjected to shear stress. This is calculated using Expression 7

v=0.6\left( 1-\frac { { f }_{ ck } }{ 250 } \right) —————7

αcw is the coefficient for the state of compression stress in the concrete element, which is taken to be 1.0 for in situ concrete elements.

bw is the overall width of the concrete element in its major axis. z is 0.9d, where d is the eﬀective depth of reinforcement within the element.

Also when considering solid rectangular cross-section, only minimum shear reinforcement is required provided that expression 8 is satisfied.

\frac { { T }_{ Ed } }{ { T }_{ Rd,c } } +\frac { { V }_{ Ed } }{ { V }_{ Rd,c } } —————-8
Where:
• TRd,c is the cracking moment resistance due to torsion, which can be determined by rearranging Eqs. 2 and 3 and setting t,i = fctd
• VRd,c is calculated using Equation 9.
{ V }_{ Rd,c }=0.12k\left( 100\rho { f }_{ ck } \right) ^{ 1/3 }{ b }_{ w }d\ge 0.035{ k }^{ 3/2 }\sqrt { { f }_{ ck } } { b }_{ w }d—9
k=1+\sqrt { \frac { 200 }{ d } } \le 1.5

ρ1 is the ratio of the area of tensile rebar to the area of the section.

Detailing

All detailing requirements of shear reinforcement are valid for Torsion except that closed links must be used. Closed links are required because because all four sides of the link take tension stresses. Thus, all four sides must be anchored very well.

Worked Example

A rectangular floor beam is to support steel railing that is to be installed oﬀ –center from the beam. The resulting design torsion moment from the analysis is 15kNm. The beam is 450mm deep × 225mm wide and is made from C25/30 concrete with a 25mm cover to reinforcement. Calculate the longitudinal tension reinforcement and closed links that are required in the beam to resist the torsion. The applied major axis bending has resulted in a required area of tension steel reinforcement of 805mm2.

Longitudinal Reinforcement Design

Calculate the equivalent wall thickness of the beam

t=\frac { 225\times 450 }{ 2(450+225) } =75mm
{ A }_{ k }=\left( 225-75 \right) \cdot \left( 450-75 \right) =56.25\times { 10 }^{ 3 }{ mm }^{ 2 }
{ u }_{ k }=2\left[ \left( 225-75 \right) +\left( 450-75 \right) \right] =1050mm

Area of longitudinal reinforcement due to torsion

\frac { \sum { { 0.87A }_{ s1 }{ f }_{ yk } } }{ { u }_{ k } } =\frac { { T }_{ Ed } }{ { 2A }_{ k } } cot\theta
{ A }_{ s1 }=\frac { { T }_{ Ed }{ u }_{ k }cot\theta }{ 2{ A }_{ k }\cdot 0.87{ f }_{ yk } } =\frac { 15\times { 10 }^{ 3 }\times 1050cot45 }{ 2\cdot 56.25\times { 10 }^{ 3 }\times 0.87\times 410 } =392.5{ mm }^{ 2 }

Tension reinforcement due to bending – 805mm2 (Bottom). USE- 3Y20 bars (942mm2).

Tension reinforcement due to torsion-392.5mm2 (distributed around the perimeter of the beam). USE- 4Y10 bars.

Shear stress in beam due to torsion

{ \tau }_{ t,i }=\frac { { T }_{ Ed } }{ 2{ A }_{ k }{ t }_{ ef,i } } =\frac { 15\times { 10 }^{ 6 } }{ 2\times 56.25\times { 10 }^{ 3 }\times 75 } =1.78N/{ mm }^{ 2 }
{ V }_{ Ed,i }={ \tau }_{ t,i }{ t }_{ ef,i }{ z }_{ i }=1.78\times 75\times \left( 450-75 \right) \cdot { 10 }^{ -3 }=50.1kN

The area of steel required is determined using the same equation of shear design.

d=h-\left( { c }_{ nom }+\phi /2+links \right) =450-\left( 25+20/2+8 \right) =407mm
\frac { { A }_{ sw } }{ { s }_{ v } } \ge \frac { { V }_{ Ed,i } }{ z{ f }_{ ywd } } =\frac { 50.1\times { 10 }^{ 3 } }{ 0.9\times 407\times 0.8\times 410 } =0.41

Note that the unity check have been avoided, however, for practical designs, this must be verified first before designing the torsional reinforcements.

References

Mosley W.H., Bungey J.H. and Hulse R. (2012) Reinforced concrete design to Eurocode 2 (7th ed.), London: Palgrave Macmillan.

Institution of Structural Engineers (2006) Manual for the design of concrete building structures to Eurocode 2, London: IStructE Ltd.

62 Replies to “Designing Concrete Elements for Torsion”

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