Question

The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is . What is the displacement of the upper face in the direction of the applied force

Answer 1

The question is incomplete. The complete question is :

The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is $3.0 \times 10^{10} \ Pa$ . What is the displacement of the upper face in the direction of the applied force?

Solution :

The relation between shear modulus, shear stress and strain,

$\text{Shear modulus, S =} \frac{\text{Shear stress}}{\text{shear strain}}$

Shear stress = shear modulus (S) x shear strain

                     $=3 \times 10^{10} \times 0.0060$

                     $=1.80 \times 10^8$ Pa

                     $=180 \times 10^6$ Pa

                     $=180 \ MPa$

The length represents the distance between the fixed in place portion and where the force is being applied.

Therefore,

$\text{Displacement} = \text{shear strain} \times \text{length}$

                     = 0.006 x 60 cm

                      = 0.360 cm

                      = 3.6 mm

Thus, the displacement of the upper face is 3.6 mm in the direction of the applied force.