A block of gelatin is 120 mm by 120mm by 40mm 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 120x120 comes on the lower and upper surface .find the shearing stress , shearing strain and shear modulus
Answers
Answer: F = 49N ; x = 10mm; L = 40mm; A = 14 400mm²
Substituting, we have that :
Shear Modulus, S = Shear stress/ Shear strain
= FL/Ax
= (49×40)/(14 400× 10)
= 49/3600 N/mm²
= 49/3600 × 10⁶ N/m²
= 13 611.11 kPa
We can also find the shear stress and shear strain respectively.
Shear stress = F/A = 49N/14 400mm²
= 49/14 400 × 10⁶ N/m²
= 3 402. 78 kPa
Shear strain = x/L = 10mm/40mm
= 1/4
= 0.25 ( having no unit).
Explanation:
The shearing stress is 3402.7 N/m²
The shearing strain is 0.25
The shear modulus is 2.304 Pascals.
Shearing stress:
Shearing stress the force applied parallel to the object. It is tangential stress.
It is given by the formula:
τ = F/A
Where,
F = Force = 49 N
A = Area = 120 mm × 120 mm = 0.12 × 0.12 = 0.0144 m²
On substituting the values, we get,
τ = 49/0.0144
∴ τ = 3402.7 N/m²
Shearing strain:
Shearing strain is the deformed height when the shearing stress is applied.
It is given by the formula:
Δl = x/L
Where,
x = Displacement = 10 mm = 0.01 m
L = Heigt of the block = 40 mm = 0.04 m
On substituting the values, we get,
Δl = 0.01/0.04
∴ Δl = 0.25
Shear modulus:
Shear modulus is the elastic coefficient of the material.
It is given by the formula:
G = (F × L)/(A × Δl)
On substituting the values, we get,
G = (49 × 40)/(3402.7 × 0.25)
∴ G = 2.304 Pa