👋 Introduction

The calculation of torsion links with a given section and torsion moment is not straight-forward. For some basic idea, you can refer to this article. For quick steps guide, below is an example, hope you find it helpful.

📐 Design Data

Below are the design information of the torsion design of the section.

Geometry
Section
	Width	500mm
	Height	600mm
Reinforcement
	Tension Area	1470 mm²

Design Action
	Torsion Tx	100kNm
	Shear Fy	50kN

Material
Concrete
	Strength	30 MPa
Rebar
	Strength	500 MPa

Assumption
$cover, cc$	=	$50mm$
$tension \ reinforcement \ dia., \phi_{tension}$	=	$25mm$
$shear \ links \ dia., \phi_{shear}$	=	$8mm$

🔢 Calculation Steps

*Unless otherwise specified, all clauses and equations quoted herein are based on BS EN 1992-1-1:2004.

1. Determine the equivalent hollow thin-wall section for torsional analysis

Effective thickness of walls(Fig. 6.11)
$t_{ef,i}$	=	$\frac{b \cdot h}{2 \cdot (b + h)}$
	=	$\frac{500 \cdot 600}{2 \cdot (500 + 600)}$
	=	$136 mm$

Area enclosed by line of wall(Fig. 6.11)
$A_{k}$	=	$(b – t_{ef,i})(h – t_{ef,i})$
	=	$(500 – 136)(h – 136)$
	=	$1.69 x 10^{5} mm^{2}$

Perimeter for line of wall(Fig. 6.11)
$U_{k}$	=	$2 \cdot ((b – t_{ef,i}) + (h – t_{ef,i}))$
	=	$2 \cdot ((500 – 136) + (600 – 136))$
	=	$1191 mm$

2. Determine if concrete is sufficient for compression

2.1Determine compression strut angle

Strength Reduction Factor of Concrete(Eq. 6.6N)
$v$	=	$0.6 \cdot (1 - \frac{f_{ck}}{250})$
	=	$0.6 \cdot (1 - \frac{30 MPa}{250})$
	=	$0.53$

Effective Depth
$d$	=	$h - cc - \phi_{shear} - \phi_{tension}/2$
	=	$600 mm - 50 mm - 8 mm - 25 mm /2$
	=	$530 mm$

Design Shear Stress
$v_{t,Ed}$	=	$V_{Ed,max} / (b \cdot d)$
	=	$50kN / (500mm \cdot 530mm)$
	=	$0.19 MPa$

Torsion Shear Stress
$\tau_{t,i}$	=	$\frac{T_{Ed,max}}{2 \cdot A_{k} \cdot t_{ef,i}}$
	=	$\frac{100 kNm}{2 \cdot 1.69 x 10^{5} mm^{2} \cdot 136 mm}$
	=	$2.17 MPa$

Inclined angle of compression strut
$\theta$	=	$0.5 \cdot sin^{-1}(\frac{2 \cdot (v_{t,Ed} / 0.9 + \tau_{t,i})}{\alpha_{cw} \cdot f_{cd} \cdot v})$
	=	$0.5 \cdot sin^{-1}(\frac{2 \cdot (0.19 MPa / 0.9 + 2.17 MPa)}{1 \cdot 17 MPa \cdot 0.53})$
	=	$15.8° < 21.8°, take 21.8° where 21.8° < \theta < 45°(Eq. 6.7N)$

2.2Determine the max. tosional resistance and max. shear resistance

Maximum Torsional Resistance(Eq. 6.30)
$T_{Rd,max}$	=	$2 \cdot v_{1} \cdot \alpha_{cw} \cdot f_{cd} \cdot A_{k} \cdot t_{ef,i} \cdot sin(θ) \cdot cos(θ)$
	=	$2 \cdot 0.53 \cdot 1 \cdot 17 MPa \cdot 1.69 x 10^{5} mm^{2} \cdot 136 mm \cdot sin(21.8°) \cdot cos(21.8°)$
	=	$143 kNm$

Maximum Shear Resistance(Eq. 6.9)
$V_{Rd,max}$	=	$\frac{\alpha_{cw} \cdot b \cdot 0.9 \cdot d \cdot v_{1} \cdot f_{cd}}{cot(θ) + tan(θ)}$
	=	$\frac{1 \cdot 500 mm \cdot 0.9 \cdot 530 mm \cdot 0.53 \cdot 17 MPa}{cot(21.8°) + tan(21.8°)}$
	=	$740 kN$

2.3Check if concrete section in compression is adequate

Combine Shear and Torsion Check(Eq. 6.29)
$\frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}}$	=	$\frac{T_{Ed}}{T_{Rd,max}} + \frac{V_{Ed}}{V_{Rd,max}}$
	=	$\frac{T_{100kNm}}{143kNm} + \frac{50kN}{740kN}$
	=	$0.77 ≤ 1, pass$

∴ Current concrete section is sufficient for compression.

3. Determine if concrete alone w/o links/reinforcement is sufficient for tension

3.1Determine torsional resistance w/o links

Torsional Resistance w/o links(Eq. 6.26)
$T_{Rd,c}$	=	$2 \cdot A_{k} \cdot f_{ct,d} \cdot t_{ef,i}$
	=	$2 \cdot 1.69 x 10^{5} mm^{2} \cdot (2 MPa / 1.5) \cdot 136mm$
	=	$62 kNm$

3.2Determine shear resistance w/o links

Concrete Factor(N.A. Cl.6.2.2(1))
$C_{Rd,c}$	=	$\frac{0.18}{\gamma_{c}}$
	=	$\frac{0.18}{1.5}$
	=	$0.12$

Reinforcement Ratio(Eq. 6.6N)
$\rho_{1}$	=	$min(\frac{A_{sl}}{d * b}, 0.02)$
	=	$min(\frac{1470 mm²}{529 mm * 500 mm}, 0.02)$
	=	$0.0056$

Depth factor(Cl.6.2.2(1))
$k$	=	$1 + \sqrt{\frac{200}{d}}$
	=	$1 + \sqrt{\frac{200}{529}}$
	=	$1.61$

Minimum Shear Resistance
$V_{Rd,min}$	=	$0.035k^{1.5} \cdot \sqrt{f_{ck}} \cdot b_{w} \cdot d$
	=	$0.035k^{1.5} \cdot \sqrt{30 MPa} \cdot 500 mm \cdot 530 mm$
	=	$0.39 MPa$

Design Shear Resistance(Eq. 6.2a)
$V_{Rd,c}$	=	$C_{Rd,c} \cdot k \cdot (100 \cdot \rho_{1} \cdot f_{ck})^{1/3} \cdot b \cdot d$
	=	$C_{Rd,c} \cdot 1.61 \cdot (100 \cdot 0.0056 \cdot 30 MPa)^{1/3} \cdot 500 mm \cdot 530 mm$
	=	$131 kN$

3.3Check if concrete section in tension is adequate

Combine Shear and Torsion Check(Eq. 6.31)
$\frac{T_{Ed}}{T_{Rd,c}} + \frac{V_{Ed}}{V_{Rd,c}}$	=	$\frac{T_{Ed}}{T_{Rd,c}} + \frac{V_{Ed}}{V_{Rd,c}}$
	=	$\frac{T_{100kNm}}{62kNm} + \frac{50kN}{131kN}$
	=	$1.99 > 1, not sufficient$

∴ Solely concrete resistance is not sufficient for tension, torsional links required.

4. Determine the required torsional reinforcement

Required Area for Torsional Reinforcement
$\frac{A_{sw,req}}{s}$	=	$\frac{T_{Ed}}{2 \cdot A_{k} \cdot f_{yd} \cdot cot(\theta)}$
	=	$\frac{100 kNm}{2 \cdot 1.69 x 10^{5} mm^{2} \cdot 435 MPa \cdot cot(21.8°)}$
	=	$0.27 mm^2/2$

Maximum Vertical Rebar Spacing(Eq. 9.2.3(3))
$s_{w,max}$	=	$min(\frac{U_k}{8} , b , d)$
	=	$min(\frac{1191mm}{8} , 500mm , 530mm)$
	=	$149mm$

Suggested Torsion Link Arrangement

Bar Dia.	spacing	As/s(mm2/m)	notation
8	125	0.4021	R8@125
12	125	0.9048	R12@125
16	125	1.6085	R16@125

5. Determine the required longitudinal reinforcement

Additional Long. Reinforcement for Torsion(Eq. 6.8)
$A_{sl,req}$	=	$\frac{T_{Ed} \cdot U_{k} \cdot cot(\theta)}{2 \cdot A_{k} \cdot f_{yd}}$
	=	$\frac{100 kNm \cdot 1191mm \cdot cot(21.8°)}{2 \cdot 1.69 x 10^5 \cdot 435MPa}$
	=	$2031 mm²$

Suggested Longitudinal Arrangement

Potential Arrangement	area (mm²)	utilization (%)
8T20	2513.27	80.8
12T16	2412.74	84.2
6T25	2945.24	69

∴ The Minimum Shear Rebar Area is 0.84 mm2/m and suggested shaer rebar arrangement is R8@100 2legs.

📄 Need a Full Design Check Report?

If you want similiar calculation analysis with your own design data, feel free to try our application, what you need is just the design data value, and a ready-to-submit design PDF report will be generated in a minute.

The ready-to-submit PDF Report is not generated by AI but programmed by chartered engineer, which is accurate, no hallucination and same input same output.

#rc design #torsion design #formula #structural

CivilSimple Team

The CivilSimple Team writes practical engineering guides for the profession and the curious. All articles are reviewed for technical accuracy before publication.