Mechanical Engineering

Section I.2 Torsion of Non-Circular Bars

The observations made for torsion of members with circular cross sections do not hold for those with non-circular cross sections. Consider the following facts for members with non-circular cross sections:

The shear stress is not constant at a given distance from the axis of rotation. As a result sections perpendicular to the axis of member warp, indicating out-of-plane displacement.
The theory of elasticity shows that the shear stress at the corners is zero.
Maximum shear strain and stress are not at the farthest distance from the rotational axis of a homogeneous non-circular member.

Members with rectangular cross sections:

For a rectangular member under torsion the corners do not distort; the corner square angles remain square after torque is applied. This indicates that shear strain is zero at the corners since there is no distorsion. This fact is illustrated in this figure.

St. Venant was the first to accurately describe the shear stress distribution on the cross section of a non-circular member using the Theory of Elasticity. For more information click here.

Theory of Elasticity shows that:

the maximum shear strain and stress occur at the centerline of the long sides of the rectangular cross section
the shear strain and stress at the corners and center of the rectangular cross section are zero.
the strain and stress variations on the cross section are primarily nonlinear.
the preceding statements are demonstrated in the following figure.

The theory of Elasticity has been applied to find analytical solutions for the torsion of rectangular elastic members. The resulting equations for shear stress and angle of twist are as follows:

Here is Table A6.1

NOTE: The equations described in this section can only be used for members having OPEN cross-sectional shapes. For open cross sections composed of multiple thin plates refer to Section I.5.

Section I.3 Elastic Membrane Analogy

Prandtl showed that the Laplace equation describing the torsion of an elastic member is identical to that used to describe the deflection of an elastic membrane subjected to a uniform pressure.

The elastic membrane analogy allows the solution of a torsion problem to be determined in a simpler way than that found by the theory of elasticity which requires the availability of the warping function.

Elastic Membrane and Twisted Bar Relationships

Consider a tube which has the same cross-sectional boundary as the bar. For example, if the bar has a solid square cross section of side dimension b, then the tube will have a hollow square cross section of side dimension b as well. Next we stretch an elastic membrane over the tube's cross section and apply internal pressure. The deflected shape of the membrane helps us visualize the stress pattern in the bar under torsion.

The analogy can be viewed as follows:

Lines of equal deflection on the membrane (contour lines) correspond to shearing stress lines of the twisted bar.
The direction of a particular shear stress resultant at a point is at right angle to the maximum slope of the membrane at the same point.
The slope of the deflected membrane at any point, with respect to the edge support plane is proportional in magnitude to the shear stress at the corresponding point on the bar's cross section.
The applied torsion on the twisted bar is proportional to twice the volume included between the deflected membrane and plane through the supporting edges.

Examples

Keep in mind that the slope at a point on the deflected membrane, and not the displacement from the base, is the parameter that is related to the shear stress in the bar. In all of the following examples, the slope is zero at the very top of the membrane, therefore the stress is zero, not the maximum, at the same location on bar's cross section.

Figure 1
Figure 2
Figure 3 Notice that the stress is maximum at the inside corner and zero at the outside corners.

Section I.4 Torsion of Thin - Wall Open Sections

The sections shown below are extruded sections

Example

EXAMPLE PROBLEMS

Example Problem 1 Maximum shear stress and angle of twist of a FORMED section
Example Problem 2 Maximum shear stress and angle of twist of an EXTRUDED section

Formulae for Torsional Deflection and Stress^*

Angle of twist per unit length

T = Torque

G = Modulus of rigidity

K = Apparent polar moment of inertia (given in table below)

Solid Section	K	Max. Shear Stress
Ellipse with major axis = 2a and minor axis = 2b		(at ends of minor axis)
Square of side = a		(at midpoint of each side)
Rectangle of long side = 2a and short side = 2b		(at midpoint of long side)
Equilateral triangle of side = a		(at midpoint of each side)