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Modulus of Rigidity: Definition & Equation

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  • 0:03 The Modulus of Rigidity
  • 1:59 Some Examples
  • 3:34 Shear Modulus from Elasticity
  • 3:59 Lesson Summary
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Lesson Transcript
Instructor: Hassan Alsaud

Earned my B.S. in Civil Engineering back in 2011. Have two years of experience in oil and gas fields and two year as a graduate research assistant. Earned my Master degree in Engineering from Tennessee State University in 2016.

In this lesson, modulus of rigidity is introduced. First, we'll define the term modulus of rigidity, also known as shear modulus, then we 'll look at some equations and examples.

The Modulus of Rigidity

The modulus of rigidity, also known as shear modulus, is defined as a material property with a value equal to the shear stress divided by the shear strain.

The general formula of shear modulus is written as shown:

shear m

where τ is the shear stress in a given member, which has the unit of force divided by area (N/m 2 or lbf/ft2); γ is the shear strain which does not have a unit (strain is the change of length divided by the original length); and G is the shear modulus or the modulus of rigidity, which has the unit of force divided by the area. Similar to the modulus of elasticity, shear modulus is another form of the generalized Hook's Law.

The modulus of rigidity depends on the material. For example, the modulus of rigidity for cast steel is approximately 11 x 106 psi (or 78 GPa), the modulus of rigidity of concrete is 3 x 106 psi (or 21 GPa), and the modulus of rigidity of aluminum is approximately 4 x 106 psi (or 28 GPa).

The modulus of rigidity can be determined by performing a tensile stress test where stress vs strain is plotted. The slope of the line is equal to the modulus of rigidity. Since shear stress is equal to the shear force over area, and strain is equal to the change in length divided by initial length, we get the equation:

mr 1

By substituting the formulas for τ and γ, we get:

mr 2

by rearranging, we get:


mr 3

Some Examples

Example 1

A steel beam is under 50 ksi shear stress. What is its final length if the initial length is 6 in?

We begin with the initial equation:

mr ex1

Remember that G is the sheer modulus, or the modulus of rigidity.

By placing the values into the equation, we get:

mr ex1 a

The value of the strain is γ = 0.0043.

From the obtained strain, we can calculate the final length:

mr ex1 c

Substituting in numbers, we get:

mr ex1 d

Therefore, lfi = 5.97 in

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