Torsion

Torsion

Learning Objectives

Understand the basic principles and definition of Torsion and torsional stress in circular sections.
Calculate maximum shear stress and angle of twist using torsion formulas.
Comprehend elastic-plastic torsion and determine fully plastic torque capacity.
Recognize the effect of stress concentrations in shafts.
Analyze statically indeterminate torques and calculate power transmission via rotating shafts.
Analyze torque transmission through flanged bolt couplings and helical springs.
Apply Saint-Venant's coefficients for torsion in non-circular (solid rectangular) sections.
Utilize Bredt's formula to analyze shear flow and shear stress in thin-walled closed tubes.

Torsion is an important structural concept dealing with the twisting of members under applied torque.

Torsion is the twisting of an object due to an applied torque. In engineering, it primarily refers to the twisting of a straight member (like a cylindrical shaft) when it is subjected to a moment that tends to produce a rotation about its longitudinal axis. This action induces shearing stresses that vary linearly from zero at the central neutral axis to a maximum value at the outer surface of the member.

Fundamental Assumptions in Linear Elastic Torsion

The standard derivation for torsion of circular shafts relies on several critical assumptions:

Plane sections remain plane: Cross-sections that are flat before twisting remain flat after twisting (they do not warp). This is strictly true only for circular solid and hollow sections.
Radii remain straight: A straight line drawn along the radius before twisting remains straight after twisting.
Small deformations: The angle of twist is very small, meaning the length and radius of the shaft remain unchanged.
Linear elastic material: The material obeys Hooke's Law, meaning shear stress is directly proportional to shear strain ( $\tau = G\gamma$ ).
Homogeneous and isotropic material: The material properties are the same everywhere and in all directions.

Torsion of Circular Sections

Torsional Stress (\tau)

For a circular shaft (solid or hollow) subjected to a torque $T$ , the shear stress $\tau$ at any radial distance $\rho$ from the center is given by the Torsion Formula.

Torsion Formula

Calculates the shear stress at any radial distance from the center.

\tau = \frac{T\rho}{J}

Variables

Symbol	Description	Unit
$T$	the applied torque (N $\cdot$ m, kN $\cdot$ m, lb $\cdot$ in)	-
$\rho$	the radial distance from the center to the point of interest (m, mm, in)	-
$J$	the Polar Moment of Inertia of the cross-sectional area (m $^4$ , mm $^4$ , in $^4$ )	-

Maximum Shear Stress (\tau_{max})

Occurs at the outer surface where $\rho = c$ (radius of the shaft).

Maximum Shear Stress

Calculates the maximum shear stress at the outer surface of a circular shaft.

\tau_{max} = \frac{Tc}{J}

Variables

Symbol	Description	Unit
$T$	the applied torque	-
$c$	the radius of the shaft	-
$J$	the Polar Moment of Inertia	-

Polar Moment of Inertia (J)

Represents a cross-section's resistance to twisting deformation. For a solid circular shaft, $J = \frac{\pi c^4}{2} = \frac{\pi d^4}{32}$ . For a hollow shaft, $J = \frac{\pi (c_o^4 - c_i^4)}{2} = \frac{\pi (d_o^4 - d_i^4)}{32}$ .

Angle of Twist (\theta)

The angle of twist measures the rotation of one end of the shaft relative to the other due to the applied torque. It is a measure of the shaft's stiffness.

Sign Convention Use the right-hand rule to determine the direction of the torque vector. Positive torque causes rotation in a specific direction (usually counter-clockwise is positive).

Angle of Twist

Measures the relative rotation of a shaft subjected to torque.

\theta = \frac{TL}{JG}

Variables

Symbol	Description	Unit
$T$	the applied torque	-
$L$	the length of the shaft	-
$J$	the Polar Moment of Inertia	-
$G$	the Shear Modulus of Elasticity (Modulus of Rigidity)	-
$\theta$	the angle of twist in radians	-

Elastic-Plastic Torsion and Stress Concentrations

Fully Plastic Torque (T_P)

The maximum possible torque a shaft can carry, reached when the entire cross-section has yielded and flows plastically at a constant yield shear stress ( $\tau_Y$ ). Unlike elastic torsion, the stress distribution is uniform across the entire radius.

Elastic-Plastic Torsion

When the torque applied to a solid circular shaft increases beyond the proportional limit, the extreme outer fibers will begin to yield first. In ductile materials (like mild steel), instead of fracturing, the material flows plastically while maintaining a constant yield shear stress ( $\tau_Y$ ). As the torque continues to increase, this zone of yielding progresses inward toward the center.

Comparing this fully plastic state to the maximum elastic torque ( $T_Y = \frac{\pi}{2} \tau_Y c^3$ ), the plastic torque is exactly $4/3$ (or 1.33 times) greater. This extra capacity is critical in limit state design and crashworthiness analysis.

Fully Plastic Torque for Solid Circular Shaft

The maximum torque capacity when the entire section has yielded.

T_P = \frac{2\pi}{3} \tau_Y c^3

Variables

Symbol	Description	Unit
$T_P$	the Fully Plastic Torque	-
$\tau_Y$	the yield shear stress	-
$c$	the radius of the solid circular shaft	-

Torsional Failure Modes

When a shaft fails under torsion, the fracture pattern depends heavily on whether the material is ductile or brittle. This provides a critical forensic clue in failure analysis.

Ductile Materials (e.g., structural steel, aluminum): Fail primarily in shear. The maximum shear stresses occur on transverse (cross-sectional) and longitudinal planes. A ductile shaft will typically fail with a clean, flat, transverse break perpendicular to the shaft's axis.
Brittle Materials (e.g., cast iron, chalk): Are weaker in tension than in shear. The maximum principal tensile stress in a twisted shaft occurs on a plane oriented at 45 degrees to the longitudinal axis. Consequently, a brittle shaft will fail with a characteristic helical, 45-degree fracture surface.

Torsional Stress Concentration Factor (K_t)

A factor used to calculate the amplified maximum stress caused by sudden geometric changes in a shaft's cross-section, such as fillets, keyways, or grooves.

Stress Concentrations in Torsion

Shafts used in power transmission often feature sudden changes in cross-section to mount gears, pulleys, or bearings. These discontinuities cause the otherwise uniform torsional shear stress to amplify significantly.

For instance, the sharp internal corner of a keyway (a rectangular groove cut along the shaft) causes extremely high stress concentrations, often being the initiation point for fatigue failure in rotating machinery.

Amplified Maximum Shear Stress

Calculates the maximum stress considering stress concentrations.

\tau_{max} = K_t \frac{T c}{J}

Variables

Symbol	Description	Unit
$\tau_{max}$	the maximum amplified shear stress	-
$K_t$	the Torsional Stress Concentration Factor	-
$T$	the applied torque	-
$c$	the radius of the shaft at the discontinuity	-
$J$	the Polar Moment of Inertia at the discontinuity	-

Statically Indeterminate Torques & Power

Statically Indeterminate Torques

Similar to axially loaded members, a shaft can be statically indeterminate if it has redundant supports (e.g., fixed at both ends) or if it is a composite shaft made of two different materials securely bonded together.

To solve these problems, the equilibrium equations ( $\sum T = 0$ ) must be supplemented by compatibility equations based on geometry.

Compatibility for Shaft Fixed at Both Ends

The total angle of twist must equal zero.

\theta_{total} = \sum \frac{T_i L_i}{J_i G_i} = 0

Variables

Symbol	Description	Unit
$T_i$	the internal torque in segment i	-
$L_i$	the length of segment i	-
$J_i$	the Polar Moment of Inertia of segment i	-
$G_i$	the Shear Modulus of segment i	-

Compatibility for Composite Shafts

Since the materials are bonded, they must twist by the same amount.

\theta_1 = \theta_2 \quad \Rightarrow \quad \left(\frac{TL}{JG}\right)_1 = \left(\frac{TL}{JG}\right)_2

Variables

Symbol	Description	Unit
$\theta_1, \theta_2$	the angle of twist for material 1 and material 2	-
$T$	the torque carried by the respective material (Note: Total applied torque $T = T_1 + T_2$ )	-

Power Transmission

Shafts are often used to transmit power in machines.

Unit Conversions

1 hp = 746 Watts (approx) or 550 ft $\cdot$ lb/s.
Ensure units are consistent (N, m, s).

Power, Torque, and Angular Velocity Relationship

Relates mechanical power transmitted by a shaft to torque and rotational speed.

P = T\omega = 2\pi f T

Variables

Symbol	Description	Unit
$P$	power in Watts (W) or Horsepower (hp)	-
$T$	torque in N $\cdot$ m	-
$\omega$	angular velocity in rad/s	-
$f$	frequency in Hz (revolutions per second)	-

Couplings and Springs

Flanged Bolt Couplings

Flanged couplings are used to connect two shafts end-to-end to transmit power. The torque is transmitted through the shearing force in the bolts connecting the two flanges.

For couplings with multiple concentric rings of bolts, the shear stress in any bolt ring is assumed to be proportional to its radial distance from the center, provided the flanges are rigid.

Torque Transmitted by a Bolt Ring

The torque transmitted by a single ring of bolts.

T = P R = (A_{bolt} \tau_{bolt} n) R = \left(\frac{\pi d^2}{4}\right) \tau n R

Variables

Symbol	Description	Unit
$T$	the applied torque	-
$n$	the number of bolts	-
$R$	the radius of the bolt circle	-
$d$	the diameter of each bolt	-
$\tau$	the shear stress in the bolts	-

Wahl Correction Factor (K_W)

A factor used in calculating the maximum shear stress in helical springs, which accounts for both direct shear stress and the curvature effect of the tightly coiled spring.

Helical Springs

Closely coiled helical springs subjected to an axial load $P$ experience both torsional shear stress and direct shear stress. The maximum shear stress is on the inner face of the coils and is given by Wahl's Formula.

Wahl's Formula for Maximum Shear Stress

Calculates the maximum shear stress in a closely coiled helical spring.

\tau_{max} = \frac{16 P R}{\pi d^3} \left( \frac{4m - 1}{4m - 4} + \frac{0.615}{m} \right)

Variables

Symbol	Description	Unit
$P$	the axial load	-
$R$	the mean radius of the spring coil	-
$d$	the diameter of the spring wire	-
$m$	the spring index ( $2R/d$ )	-

Shear Stress Without Curvature Effect

Calculates stress using only static torsion and direct shear, without the curvature effect.

\tau = \frac{16PR}{\pi d^3} \left( 1 + \frac{d}{4R} \right)

Variables

Symbol	Description	Unit
$P$	the axial load	-
$R$	the mean radius of the spring coil	-
$d$	the diameter of the spring wire	-

Torsion of Non-Circular Sections

Torsion of Non-Circular Sections

The standard torsion formula ( $\tau = T\rho/J$ ) relies on the assumption that plane cross-sections remain plane and undeformed during twisting. This is ONLY true for circular cross-sections (solid or hollow). For non-circular shapes (like rectangles, squares, or triangles), the cross-section warps (bulges in and out).

For a solid rectangular shaft with width $b$ and thickness $c$ (where $b \ge c$ ), the maximum shear stress and angle of twist are given by Saint-Venant's coefficients. The maximum shear stress occurs at the middle of the longest side ( $b$ ).

Maximum Shear Stress for Solid Rectangular Sections

Calculates the max shear stress using Saint-Venant's coefficient \alpha.

\tau_{max} = \frac{T}{\alpha b c^2}

Variables

Symbol	Description	Unit
$\tau_{max}$	the maximum shear stress	-
$T$	the applied torque	-
$b$	the width of the rectangular shaft	-
$c$	the thickness of the rectangular shaft	-
$\alpha$	Saint-Venant's coefficient depending on the ratio $b/c$	-

Angle of Twist for Solid Rectangular Sections

Calculates the angle of twist using Saint-Venant's coefficient \beta.

\theta = \frac{TL}{\beta b c^3 G}

Variables

Symbol	Description	Unit
$\theta$	the angle of twist	-
$T$	the applied torque	-
$L$	the length of the shaft	-
$b, c$	width and thickness	-
$G$	the Shear Modulus	-
$\beta$	Saint-Venant's coefficient depending on the ratio $b/c$	-

Torsion of Thin-Walled Tubes (Bredt's Formula)

Shear Flow (q)

In a thin-walled closed tube, the product of shear stress and thickness ( $\tau t$ ) is constant around the entire perimeter. This product is called Shear Flow.

Torsion of Thin-Walled Tubes (Bredt's Formula)

For closed thin-walled cross-sections (like hollow box beams or aircraft wings), we use Bredt's Formula. The shear stress ( $\tau$ ) is assumed to be constant across the thickness ( $t$ ) of the wall.

Design Note: Because shear flow $q$ is constant, the maximum shear stress ( $\tau_{max}$ ) occurs where the wall thickness ( $t$ ) is the minimum.

Shear Flow

The product of shear stress and thickness, constant around the perimeter.

q = \tau t = \frac{T}{2A_m}

Variables

Symbol	Description	Unit
$q$	the Shear Flow	-
$\tau$	the shear stress	-
$t$	the thickness of the wall	-
$T$	the applied torque	-
$A_m$	the mean area enclosed by the centerline of the tube wall	-

Bredt's Formula for Average Shear Stress

Calculates the average shear stress in the wall of a closed thin-walled tube.

\tau = \frac{T}{2tA_m}

Variables

Symbol	Description	Unit
$\tau$	the average shear stress	-
$T$	the applied torque	-
$t$	the thickness of the wall at the point of interest	-
$A_m$	the mean area enclosed by the centerline	-

Angle of Twist for Thin-Walled Tubes

The angle of twist ( $\theta$ ) for a thin-walled closed tube can be calculated using a line integral around the centerline perimeter.

Angle of Twist for Thin-Walled Closed Tubes

Calculates the angle of twist over length L.

\theta = \frac{TL}{4A_m^2 G} \oint \frac{ds}{t}

Variables

Symbol	Description	Unit
$\theta$	the angle of twist	-
$T$	the applied torque	-
$L$	the length of the tube	-
$A_m$	the mean area	-
$G$	the shear modulus	-
$\oint \frac{ds}{t}$	the line integral around the centerline perimeter ( $s$ ) divided by the thickness ( $t$ ) at each segment	-

Interactive Simulation

Adjust the parameters below to see how torque, length, and diameter affect the shear stress and angle of twist.

Torsion Calculator & Visualizer

\text{Torque } (T)

2000 Nm

\text{Shaft Length } (L)

2 m

\text{Diameter } (d)

50 mm

\text{Shear Modulus } (G)

Max Shear Stress (

\tau

):81.49 MPa

Angle of Twist (

\theta

):4.67°

Radians:0.0815 rad

Visual exaggerates angle for clarity.
Red gradient indicates shear stress intensity.

Key Takeaways

Torsional Shear Stress ( $\tau = Tc/J$ ): Varies linearly with radius. Maximum at the surface, zero at the center. Valid ONLY for circular sections.
Fundamental Assumptions: The torsion formula requires that plane sections remain plane and undeformed (true only for circular sections) and that the material is linear elastic.
Angle of Twist ( $\theta = TL/JG$ ): Measures stiffness. Depends on length ( $L$ ), Polar Moment of Inertia ( $J$ ), and Shear Modulus ( $G$ ).
Polar Moment of Inertia ( $J$ ): Represents the geometric resistance to twisting. Hollow shafts are efficient because material is located far from the center where stress is highest.
Elastic-Plastic Torsion ( $T_P$ ): Occurs when the entire cross-section yields, reaching the absolute maximum torque capacity (for a solid shaft, 4/3 of the yield torque).
Failure Modes: Ductile materials fail on transverse shear planes (flat breaks), while brittle materials fail on 45-degree principal tension planes (helical breaks).
Stress Concentrations in Torsion ( $K_t$ ): Sudden geometric changes (keyways, fillets) heavily amplify torsional shear stress, risking fatigue failure.
Indeterminate Torsion: Similar to axial loading, use compatibility equations (sum of twists equals zero or a specific value) to solve for unknown torques.
Power Transmission ( $P = T\omega$ ): Relates mechanical power to torque and rotational speed.
Flanged Couplings transmit torque via shear stress in the connecting bolts.
Helical Springs experience a combination of torsional and direct shear stress, requiring the Wahl Correction Factor for accurate maximum stress calculation.
Non-Circular Sections warp under torsion. The maximum shear stress occurs at the middle of the longest side, not at the corners.
Bredt's Formula: Calculates average shear stress ( $\tau = T / 2tA_m$ ) in closed thin-walled tubes. The highest shear stress occurs where the wall is thinnest.
Shear flow ( $q$ ) is constant around the perimeter of a closed thin-walled tube.

Previous TopicStrain and Hooke's Law - Examples & Applications

Quiz Me

Next TopicShear and Moment in Beams - Theory & Concepts

Prev Next

Quiz Me