Question

An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces.

If the top plate displaces 2 mm horizontally when it is subjected to a horizontal force P = 2.3kN , determine the shear modulus of the polymer. The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysis.

Express your answer to three significant figures and include the appropriate units.

An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2 mm horizontally when it is subjected to a horizontal force P = 2.3kN , determine the shear modulus of the polymer. The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysis.

Answer 1

Concepts and reason

First of all calculate the angular deviation of the side of the block. Calculate the area of the surface along which the force is working. Use the expression for the shear stress given by the Hook’s law to calculate the value of the shear modulus for the block.

Hook’s law: this law relates the stress and strain of the object under the elastic limit, it provides a ratio for the stress and the strain which remains constant for the material under the elastic limit.

Fundamentals

The expression for the hook’ s law is given as,

$G = \frac{F}{{A\theta }}$

Here, G is the shear modulus, F is the deforming force, A is the area of the section along which force is acting and $\theta$ is the change in angle of a side from original shape.

Draw the diagram of the problem after block underwent the transverse displacement under the action of applied force,

Calculate the small deviation in the angle.

$\theta = \frac{{\Delta x}}{t}$

Here, $\theta$ is small deviation of the side, t is thickness of the block and $\Delta x$ is small change in the length of the block.

Substitute $2{\rm{ mm}}$ for $\Delta x$ and $200{\rm{ mm}}$ for t.

$\begin{array}{c}\\\theta = \frac{2}{{200}}\\\\ = 0.01\\\end{array}$

Calculate the area of the surface along which the shear force is working.

$A = bl$

Here, $A$ is area of the surface, b is width of the block and l is length of the block.

Substitute $400{\rm{ mm}}$ for $b$ and $100{\rm{ mm}}$ for l.

$\begin{array}{c}\\A = \left( {400{\rm{ mm}}} \right)\left( {100{\rm{ mm}}} \right)\\\\ = 40000{\rm{ m}}{{\rm{m}}^2}\left( {\frac{{{{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}}}{{1{\rm{ m}}{{\rm{m}}^2}}}} \right)\\\\ = 0.04{\rm{ }}{{\rm{m}}^2}\\\end{array}$

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Write the expression for the shear modulus.

$G = \frac{P}{{A \cdot \theta }}$

Substitute $2.3 {\rm{kN}}$ for $P$ , $0.0{\rm{4 }}{{\rm{m}}^2}$ for $A$ and $0.01$ for $\theta$ .

$\begin{array}{l}\\G = \frac{{\left( {2.3 {\rm{kN}}} \right)\left( {\frac{{{{10}^3} {\rm{N}}}}{{1 {\rm{kN}}}}} \right)}}{{\left( {0.0{\rm{4 }}{{\rm{m}}^2}} \right)\left( {0.01} \right)}}\\\\G = 5750000{\rm{ Pa}}\left( {\frac{{{{10}^{ - 6}}{\rm{ MPa}}}}{{1{\rm{ Pa}}}}} \right)\\\\G = 5.75 {\rm{MPa}}\\\end{array}$

Ans:

The shear modulus of the polymer is $5.75 {\rm{MPa}}$ .