Shear Stress

An introduction Shear Stress, Modulus of Rigidity and Strain Energy.

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Contents

Shear Stress
Complementary Shear Stress.
Shear Strain
The Modulus Of Rigidity
Strain Energy
Page Comments

Shear Stress

If the applied load consists of two equal and opposite parallel forces which do not share the same line of action, then there will be a tendency for one part of the body to slide over, or shear from the other part.

In the figure below, if the section LM is parallel to the forces and has an area A, then the average Shear Stress $\displaystyle S = \frac{F}{A}$ . If the Shear Force varies then at a point $\displaystyle S = \frac{\delta F}{\delta A}$

Note : The Shear Stress is tangential to the area over which it acts, and is expressed in the same units as Direct Stress, i.e. Load per unit Area.

See Also the section on Shear Force and Bending Moment

Complementary Shear Stress.

A,B,C,D is a small rectangular element whose sides are x, y and z which are perpendicular to the figure. A Shear Stress S acts on the planes A,B and C,D

It can be seen that these Stresses form a Couple $\displaystyle (s,xz)y$ which can only be balanced by tangential Stresses on the planes A,D and B,C. These are known as Complementary Shear Stresses. Let S' be the Complementary Shear Stress induced on planes A,D and B,C. Then as the element is in equilibrium:

$(S.xz)\;y = (S'.yz)\;x$

(1)

$\therefore\;\;\;\;\;\;S = S'$

(2)

This shows that every Shear Stress is accompanied by an equal Complementary Shear Stress on planes at right angles. To produce balancing Couples, the direction of the Shear Stress on the element are either both towards, or both away from a corner.

The existence of Complementary Shear Forces may be an important factor in the failure of anisotropic materials such as timber, which is weak in Shear along the grain.

Near to a Free Boundary there are no externally applied Forces and it can be seen that The Shear Stress on any cross section must act in a direction parallel to the Boundary. This is because if there were a component in a direction at right angles to boundary it would require a Complementary Shear Stress on the Boundary wall. For Example, the Shear Stress distribution over a section of a rivet must be as in the left-hand diagram and NOT as on the right.

This produces complications in that the Shear Stress varies in both magnitude and direction, although in the particular case of rivets, it is not normal to make any allowance for this in the design.

Example:

[imperial]

Example - Shear stress on shaft bolts

Problem

A Flange Coupling which joins two sections of a Shaft is required to transmit 250 h.p. at 1000 r.p.m. If six bolts are used on a pitch circle of 6 ins. find the diameter of the bolts. The allowable mean Shear Stress is 5 tons/in².

Workings

Torque to be transmitted s given by:

$T\;=\;\frac{h.p.\times 33,000}{2\;\pi \;N}\;\;\;\;\;\;\;\;\;\;Note\;1\;h.p.\;\equiv 550\;ft.lbs\;per\;sec.\;\;\;0r\;\; \;33,000\;ft.lbs\;per\;min.$

(1)

$i.e.\;\;\;\;\;T\;=\;\frac{250\times 33,000}{2\;\pi \times1000}\;=\;1313\:lb.ft.\;\approx \;15800\;lb.ft.$

(2)

If d is the diameter of a bolt the load carried by one bolt is:

$5\times 2240\times \frac{\pi \;d^2}{4}\;lb.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(Note\;1\;ton\;=\;2240\;lbs.)$

(3)

Multiplying by the number of bolts and their radius arm, the torque carried is:

$5\times 2240\times \frac{\pi \;d^2}{4}\times 6\times \frac{6}{2}\;=\;15,800\;lb.in.\;\;\;the\;Torque\;to\;be\;carried$

(4)

$\therefore\;\;\;\;\;\;d^2\;=\;\frac{15,800\times 4}{5\times 2240\times \pi \times 18}\;=\;0.10$

(5)

$Hence\;\;\;\;\;\;d\;=\;0.316\;in.\;\;\;say\;\;\;\frac{5}{16}\;in.$

(6)

Solution

Diameter of bolt, $d = 0.316 in$

Shear Strain

The distortion produced by Shear Stress on an element or Rectangular Block is shown in the following diagram.

The Shear Strain or "Slide" is $\phi$ and can be defined as the change in the right angle. It is measured in Radians and is dimensionless.

The Modulus Of Rigidity

For Elastic materials it is found that within certain limits, Shear Strain is proportional to the Shear Stress producing it.

The Ratio $\displaystyle \frac{Shear\;Stress}{Shear\;Strain}$ is called the Modulus of Rigidity and is denoted by C.

$\displaystyle C = \frac{S}{\phi }$ and in the Imperial system is in $\displaystyle Tons\;in^{-2}\;\;\;\;or\;\;\;lbs.in^{-2}$

Strain Energy

Within the limits of proportionality Strain is proportional to Stress and:

Strain Energy (U) = Work done in Straining.

$i.e.\;\;\;\;\;\;\;\;U = \frac{1}{2}(Final\;Couple)\times (Angle\;turned\;through)$

(3)

For a gradually applied stress the work done is shown on the diagram as the shaded area.

$U = \frac{1}{2}(Syz\;.\;x)\;\phi = \frac{1}{2}\times Sxyz\times \frac{S}{C}$

(4)

$\therefore\;\;\;\;\;\;U = \left(\frac{S^2}{2\;C} \right)\times \;Volume$

(5)