Torsion - Theory & Concepts

Torsion

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.

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:

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

where:

$T$ is the applied torque (N $\cdot$ m, kN $\cdot$ m, lb $\cdot$ in).
$\rho$ is the radial distance from the center to the point of interest (m, mm, in).
$J$ is 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):

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

Polar Moment of Inertia (J)

Polar Moment of Inertia (J)

Solid Shaft: $J = \frac{\pi c^4}{2} = \frac{\pi d^4}{32}$
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.

\theta = \frac{TL}{JG}

where:

$L$ is the length of the shaft.
$G$ is the Shear Modulus of Elasticity (Modulus of Rigidity).
$\theta$ is in radians.

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).

Elastic-Plastic Torsion and Stress Concentrations

Elastic-Plastic Torsion (Fully Plastic Torque)

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.

When the entire cross-section has yielded, the shaft carries the maximum possible torque, called the Fully Plastic Torque ( $T_P$ ). Unlike elastic torsion (where stress varies linearly from center to surface), the stress distribution in a fully plastic state is uniform across the entire radius at

\tau_Y

For a solid circular shaft of radius

c

, the Fully Plastic Torque is:

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

Comparing this 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.

Stress Concentrations in Torsion

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

Similar to axial and bending loads, the maximum amplified stress is calculated using a Torsional Stress Concentration Factor ( $K_t$ ):

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

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.

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.

For a shaft fixed at both ends:

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

For a composite shaft (e.g., steel core in a brass tube):

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

The total applied torque is shared between them:

T = T_1 + T_2

Power Transmission

Shafts are often used to transmit power in machines. The relationship between power (

P

), torque (

T

), and angular velocity (

\omega

) is:

P = T\omega = 2\pi f T

where:

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

Unit Conversions

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

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.

The torque transmitted by a single ring of bolts is:

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

where:

$T$ is the applied torque.
$n$ is the number of bolts.
$R$ is the radius of the bolt circle.
$d$ is the diameter of each bolt.
$\tau$ is the shear stress in the bolts.

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.

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:

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

where:

$P$ is the axial load.
$R$ is the mean radius of the spring coil.
$d$ is the diameter of the spring wire.
$m = 2R/d$ is the spring index.

The term in the parenthesis is the Wahl Correction Factor ( $K_W$ ), which accounts for both direct shear stress and curvature effect. Without the curvature effect (using only static torsion and direct shear), the stress is:

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

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).

Solid Rectangular Sections

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:

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

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

where

\alpha

and

\beta

are coefficients that depend on the ratio

b/c

. The maximum shear stress occurs at the middle of the longest side (

b

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.

Shear Flow and Bredt's Formula

In a thin-walled closed tube, the shear stress (

\tau

) is assumed to be constant across the thickness (

t

) of the wall. The product of shear stress and thickness is called Shear Flow ( $q$ ), which is constant around the entire perimeter.

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

From this, the average shear stress in the wall is:

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

where:

$T$ is the applied torque.
$t$ is the thickness of the wall at the point of interest.
$A_m$ is the mean area enclosed by the centerline of the tube wall (NOT the cross-sectional area of the material).

Design Note

Because shear flow

q

is constant, the maximum shear stress (

\tau_{max}

) occurs where the wall thickness (

t

) is the minimum.

Angle of Twist for Thin-Walled Tubes

The angle of twist (

\theta

) for a thin-walled closed tube of length

L

and shear modulus

G

is:

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

where

\oint \frac{ds}{t}

is the line integral around the centerline perimeter (

s

) divided by the thickness (

t

) at each segment.

Interactive Tool: Torsion Simulator

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

Torsion of a Circular Shaft

T = 1.5 kN·m

The red dashed line represents a longitudinal line on the surface of the shaft before twisting. When torque is applied, the line twists, demonstrating angular deformation.

Max Shear Stress (

\tau_{max}

)

61.1 MPa

Angle of Twist (

\theta

)

3.50°

(0.0611 rad)

Torque (

T

): 1.5 kN·m

Diameter (

d

): 50 mm

Length (

L

): 2 m

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.
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).
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.

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