Torsion

Calculation assumptions

The behavior of a reinforced concrete section subjected to torsion can be divided into two categories - before and after the time when the cracks may first be expected to occur. Before a crack the cross-section behaves as an elastic material. Torsion stress can be expressed by formula   

 $$\tau =~\frac{{{T}_{Ed}}}{{{W}_{t}}}$$ 

where W_t je sectional module in torsion.

Cracks in the unreinforced member due to principal tensile torsion stress are also ultimate limit state. The behavior of a reinforced concrete section subjected to torsion can be described on the basis of a thin-walled closed section, see Fig. below. 

$$\textsf{\textit{\footnotesize{\qquad Equivalent thin-walled cross-section.}}}$$

Calculation procedure

The process of a reinforced concrete check for torsion is very similar to the check for shear. First of all, we check the concrete resistance. If the concrete check is satisfied, the reinforcement can be designed using the detailing rules. Otherwise, we need to verify the reinforcement and compressive diagonal resistance by calculation.

$$\textsf{\textit{\footnotesize{\qquad Process diagram for torsion check.}}}$$

Resistance

Shear flow in a wall of a thin-walled cross-section under torsion can be expressed as:

$${{\tau }_{t}}{{t}_{ef}}=~\frac{{{T}_{Ed}}}{2{{A}_{k}}}$$

Shear force in a wall of a thin-walled cross-section can be expressed as:

$$V={{\tau }_{t}}{{t}_{ef}}z$$

Where 

τ          Shear flow in wall,

t_ef         is the effective wall thickness,

z           is the side length of wall,

T_Ed       is the torsion moment,

A_k        is the area enclosed by the center-lines of the connecting walls, including inner hollow areas.

Torsion cracking moment, which may be determined by setting f_ctd to previous expression. Thus we get expression for the resistance in torsion without torsion reinforcement.

$${{T}_{Rd,c}}=2{{A}_{k}}{{t}_{ef}}{{f}_{ctd}}$$

where  f_ctd       design axial tensile strength of concrete

$$\textsf{\textit{\footnotesize{\qquad Principles of Truss analogy for member under torsion moment.}}}$$

The member resistance with torsion reinforcement is composed from the resistance of compressive concrete diagonals which is based again on the truss analogy method. Compressive stress in the diagonal can be expressed with help of shear force in the wall of a thin-walled cross-section on the wall surface which is in consideration, i.e.

$${{\sigma }_{c}}=\frac{\frac{{{T}_{Ed}}z}{2{{A}_{k}}\sin \theta }}{z~{{t}_{ef}}\cos \theta }=\frac{{{T}_{Ed}}}{2{{A}_{k}}{{t}_{ef}}\sin \theta \cos \theta }$$

Substitution of  σ_c=σ_cwf_cd and T_Ed=T_Rd,max and expressing of T_Rd,max we get an equation for the compressive diagonal resistance

$${{T}_{Rd,max}}=2~\nu ~{{\alpha }_{cw}}~{{f}_{cd}}~{{A}_{k}}~{{t}_{ef~\sin \theta ~\cos \theta }}$$

where  

ν          = 0,6 pro f_ck ≤ 60MPa or  for f_ck > 60MPa

α_cw       coefficient which takes account the state of compressive stress in compression chord

f_cd        design value of concrete compressive strength

the shear reinforcement resistance subjected to torsion is again based on stress in the compression diagonal. The stirrup force is equal to the stress in the compressed diagonal on the area which corresponds to the particular stirrup line, i.e.

$${{A}_{sw}}{{f}_{ywd}}=\frac{{{T}_{Ed}}}{2{{A}_{k}}{{t}_{ef}}\sin \theta \cos \theta }~{{t}_{ef}}~s{{\sin }^{2}}\theta =\frac{{{T}_{Ed}}~s}{2{{A}_{k}}\cot \theta }~$$

Substituting  T_Ed=T_Rd,s and expressing T_Rd,s  we get equation:

 $${{T}_{Rd,s}}=2{{A}_{k}}\frac{{{A}_{sw}}{{f}_{ywd}}}{s}~\cot \theta$$ 

If the longitudinal and shear reinforcement quantity is known, we can define angle θ by the expression

$${{\tan }^{2}}\theta =\frac{\frac{{{A}_{sw}}{{f}_{ywd}}}{s}}{\frac{{{A}_{sl}}{{f}_{yd}}}{{{u}_{k}}}}$$ 

Substitution for T_Rd,s we get

$${{T}_{Rd,s}}=2{{A}_{k}}\sqrt{\frac{{{A}_{sw}}}{s}{{f}_{ywd~}}\frac{{{A}_{sl}}}{{{u}_{k}}}~{{f}_{yd}}}$$  

Where

A_sw      shear reinforcement area

s           is the radial spacing of stirrups of shear reinforcement

f_ywd      is the effective design strength of the shear reinforcement

A_sl       longitudinal reinforcement area

u_k         is the outer circumference of the cross-section

f_ywd      is the effective design strength of the longitudinal reinforcement

The force in the longitudinal reinforcement can be deducted from The shear force in a wall of a section subject to a pure torsional moment, which is give as:

$$V=\frac{{{T}_{Ed}}}{2{{A}_{k}}}{{u}_{k}}$$

That force is transformed to longitudinal direction and we get:

$${{F}_{l}}=\frac{{{T}_{Ed}}{{u}_{k}}}{2{{A}_{k}}~\tan \theta }$$

The permitted range of the values for angle θ is similar to the shear check, i.e 1 < cot θ < 2,5. Dependency between resistances can be seen in Fig. below. The diagram shows that with increasing the angle θ  the resistance T_Rd,max is growing,  the resistance T_Rd.s is decreasing and resistance T_Rd,c is constant, since it is not based on the truss analogy method.

$$\textsf{\textit{\footnotesize{\qquad Závislost únosnosti průřezu v kroucení na úhlu θ.}}}$$

Calculation of cross-section characteristics for torsion

To check the cross-section for torsion it is necessary to establish a so-called equivalent thin-walled closed section. In determining the dimensions of the equivalent thin-walled cross-section by assuming a rectangular shape. For the true area of rectangular states A = b×h and for the circumference of rectangle u =2 (b +h). Using these two equations can provide alternative thin rectangle-shaped area and periphery of the original cross-section. Solving two equations with two unknowns we get:

$$b=\frac{-u\pm \sqrt{{{u}^{2}}-16A}}{-4}\text{ }\!\!~\!\!\text{ }$$

$$h=\frac{\left( u-2\text{b} \right)}{2}$$

The wall thickness of the effective cross-section can be define from periphery and section area as:

$$t=\text{A}/\text{u}$$

Then the area and periphery defined by center line of the effective cross-section:

$${{A}_{k}}=\left( \text{h}-\text{t} \right)\text{ }\!\!~\!\!\text{ }\left( \text{b}-\text{t} \right)\text{ }\!\!~\!\!\text{ }$$

$${{u}_{k}}=2\left( \left( \text{h}-\text{t} \right)+\text{ }\!\!~\!\!\text{ }\left( \text{b}-\text{t} \right) \right)$$

The problem with this method is for cross-section of type T with a wide plate when the overall area and periphery is taken to calculate the dimensions (including this plate). In the future versions of IDEA RCS program, the selection of the most massive cross-section part will be enabled, which will be used to check the torsion.