Question

A block 100 mm 100 mm base and 10 mm height. When we apply a tangential force 10 kN to the upper edge it is displaced 1 mm relative to lower face.Then the direct shear stress in the element is:

• A. 0.2
• B. 0.1
• C. 0.02
• D. 0.01

Answer 1

The correct option is B 0.1

In $\Delta ABC$ $\tan r=\frac{1}{10}$

For small angles $\tan r=r$

$\therefore$ Shear strain $r=\frac{1}{10}=0.1$

Shear stress$=\frac{Forceapplied}{Areaonwhichforceisapplied}$

$=\frac{10\times {10}^3}{100\times 100m{m}^2}$

$=1N/m{m}^2=1M{P}_a$

$\therefore$ Stress(Shear)$=1$MPa

Strain(Shear)$=0.1$.