Topic 6: Torsion, Rivets & Welds

• 6.1 Torsion: Transverse Shear Stress
  • 6.1a Shear Stress - Example 1
  • 6.1b Shear Stress - Example 2
• 6.2 Torsion: Deformation - Angle of Twist
• 5.3 Torsion: Power Transmission
  • 6.3a Power Transmission - Example 1
  • 6.3b Power Transmission - Example 2
  • 6.3c Power Transmission - Example 3
  • 6.3d Compound Shaft - Example 4
  • Additional General Torsion Examples: #1, #2, #3, #4, #5, #6 [Previous test problems]
• 6.4a Torsion - Problem Assignment 1 (required)
• 6.4b Torsion - Problem Assignment 2 (supplementary)
• 6.4c Torsion - Problem Assignment 3 (supplementary)
• 6.4d Torsion - Problem Assignment 4 (required)
• 6.5 Rivets & Welds - Riveted Joints
  • 6.5a Riveted Joints - Example 1
  • 6.5b Riveted Joints - Example 2
• 6.6 Rivets & Welds - Riveted Joint Selection
  • 6.6a Riveted Joint Selection - Example1
  • Additional Rivet Examples: #2, #3, #4, #5, #6, #7 [Previous test problems]
• 6.7 Rivets & Welds - Welded Joints
  • 6.7a Welded Joints - Example 1
  • Additional Weld Examples: #2, #3, #4, #5 [Previous test problems]
• 6.8a Rivets & Welds - Problem Assignment 1 (required)
• 6.8b Rivets & Welds - Problem Assignment 2 (supplementary)
• 6.9 Torsion, Rivets & Welds - Topic Examination

Topic 6.1: Torsion: Transverse Shear Stress

In this topic we wish to consider shafts, to which an external torque is applied (such as in power transmission), and an internal torque, shear stress, and deformation (twist) develops in response to the externally applied torque. We will consider solid and hollow circular shafts, in which we assume the material is homogeneous and isotropic (that is properties of the materials are the same in all directions in the material), that the stress which develop remain within the elastic limits, and that plane sections of the shaft remain plane under the applied torque.

In Diagram 1 we have shown a section of a solid shaft. An external torque T is applied to the left end of the shaft, and an equal internal torque T develops inside the shaft.

Additionally there is a corresponding deformation (angle of twist) which results from the applied torque and the resisting internal torque causing the shaft to twist through an angle, theta, shown in Diagram 1. There is also an internal shear stress which develops inside the shaft. This may be thought of as being due to the adjacent cross sectional areas of the shaft trying to twist passed each other. (As a practical example of this, take a long stick of chalk, grasp each end and twist strongly and quickly in opposite directions at each end. The result will be a shearing of the chalk into two pieces. If it is really a torsion failure the sheared end of the chalk will have a conical twisting fracture at the end.) This shearing stress on the cross sectional area varies from zero at the center of the shaft linearly to a maximum at the outer edge as also shown in Diagram 1.

We will now go through a relative brief derivation to develop a torsion formula to determine transverse shear stress.

In Diagram 2, we have shown the cross sectional area of the shaft of radius R, and indicated a small area A, a radial distance r from the center.

The shear stress at that location may be written in terms of the maximum shear stress as:
= (r/R) .
Then the shear force acting on the small area may be written as: .
We next may write the small amount of torque due to the force as:.
We may now write the total internal torque as the sum of the small torque acting on the areas A's. (that is, we now mentally divide the cross sectional area into a large number of small areas, A's, and sum the torque on these areas.):
T = .
We may rewrite the last summation by taking radius R and the maximum shear stress to the outside of the summation sign, since these values are constant (they are the same for each A). And then we can write:
T =
The summation of r²A, is the quantity known as the polar moment of inertia, J. (For a brief review of the moment of inertia, select Centroids/Moment of Inertia: ) And now we solve for
= T R/J;
where T is the internal torque in that section of the shaft, R = maximum radius of the shaft, and J = polar moment of inertia.
We may also extend this formula for the shear stress at any arbitrary radius as follows:
= T r / J;
where as above T is the internal torque in that section of the shaft, r = the radial distance from the center of the shaft to the point where we wish to find the shear stress, and J = polar moment of inertia. [ J = (/32) d⁴ for a solid shaft, and J = (/32)(d_o⁴ -d_i⁴) for a hollow shaft, where d_o = outer diameter and d_i = inner diameter.]

The formula, = T r / J, is the relationship for the transverse shearing stress in a shaft under torsion. There is also a longitudinal shear stress in the shaft and, without going through the derivation, we will simply state that at any point in the shaft, the value of the transverse shear stress and the longitudinal shear stress are equal. And often in wood shafts, the shaft will fail in longitudinal shear stress, along the shaft, along the grain of the wood, where the failure stress in lower than across the grain (transverse).

We will now look at two examples of determining the shear stress in a shaft.
Please select :
Topic 6.1a: Shear Stress - Example 1
Topic 6.1b: Shear Stress - Example 2

When finished with Torsion Shear Stress examples, Continue to:
Topic 6.2: Torsion:  Deformation - Angle of Twist
or Select:
Topic 6: Torsion, Rivets & Welds - Table of Contents
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