Question

​A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed. A force of .49 N is applied tangentially to the upper surface causing a 10 mm displacement relative to the lower surface. The block is placed such that 120 X 120 comes on the lower and upper surface. Find the shearing stress, shearing strain and shear modulus.

Answer 1

Answer:

Explanation:

Given

Applied force $F=0.49\ N$

Area of displaced Plane $A=120\times 120\ mm^2$

displacement $=10\ mm$

Shearing stress$=\frac{F}{A}$

$\tau _s=\frac{0.49}{0.12\times 0.12}$

$\tau _s=34.02\ N$

Shearing strain $=\tan \theta$

$\varepsilon =\frac{10}{40}=0.25$

Shear modulus $G=\frac{\text{Shearing stress}}{\text{Shearing strain}}$

$G=\frac{34.02}{0.25}=136.11\ N/m^2$