The 16 by 22 by 25-mm rubber blocks shown in Figure P3.10 are used in a double U shear mount to isolate the vibration of a machine from its supports. An applied load of P = 285 N causes the upper frame to be deflected downward by 5 mm. Determine the shear modulus G of the rubber blocks.

Question

Accepted Answer

Consider a FBD of the upper U frame. The downward force P is resisted by two upward shear forces V; therefore, V = 285 N / 2 = 142.5 N.

Next, consider a FBD of one of the rubber blocks. The shear force acting on one rubber block is V = 142.5 N. The area of the rubber block that is parallel to the direction of V is

[latex]A_{V}=(22 mm )(25 mm )=550 mm ^{2}[/latex]

Consequently, the shear stress in one rubber block is

[latex] au=\frac{V}{A_{V}}=\frac{142.5 N }{550 mm ^{2}}=0.259091 MPa[/latex]

The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by:

[latex] an \gamma=\frac{5 mm }{16 mm }=0.312500 \quad herefore \gamma=0.302885 rad[/latex]

From Hooke’s Law for shear stress and shear strain, the shear modulus G can be computed:

[latex] au=G \gamma \quad herefore G=\frac{ au}{\gamma}=\frac{0.259091 MPa }{0.302885 rad }=0.855411 MPa =0.855 MPa[/latex]

Question 3.10: The 16 by 22 by 25-mm rubber blocks shown in Figure P3.10 ar...

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