Problems on Shear Stress, Shear Strain and Modulus of Rigidity

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Example – 1:

  • The area of the upper face of a rectangular block is 0.5 m x 0.5 m and the lower face is fixed. The height of the block is 1 cm. a shearing force applied to the top face produces a displacement of 0.015 mm. Find the strain, stress and the shearing force. Modulus of rigidity = η = 4.5 × 1010 N/m².
  • Solution:
  • Given: Area under shear = A = 0.5 m x 0.5 m  = 0.25 m², Height of the block = h = 1 cm = 1 × 10-2 m, Displacement of top face = x = 0.015 mm = 0.015 × 10-3 m =  1.5 × 10-5 m, Modulus of rigidity = η = 4.5 × 1010 N/m².
  • To Find: Shear strain =? Shear stress =? Shearing force = F =?

Shear strain = tanθ = x/h = (1.5 × 10-5 ) / (1 × 10-2 ) = 1.5 × 10-3

Mdulus of rigidity = η = Shear stress / Shear strain

∴  Shear stress = η × Shear strain =  4.5 × 1010 × 1.5 × 10-3

∴  Shear stress = 6.75 × 107 N/m².



Shear stress =  F/A

∴  F = Shear stress × Area

∴  F = 6.75 × 107 × 0.25

∴  F = 1.69 × 10N



Ans: Shear strain = 1.5 × 10-3, Shear stress = 6.75 × 107  N/m², Shearing force = 1.69 × 10N.

Example – 2:

  • A metallic cube of side 5 cm, has its lower surface fixed rigidly. When a tangential force of 10kg. wt. is applied to the upper surface, it is displaced through 0.03 mm. Calculate (1) the shearing stress (2) the shearing strain and (3) the modulus of rigidity of the metal.
  • Solution:
  • Given: Area under shear = A = 5 cm x 5 cm  = 25 cm² = 25 × 10-4 m², Height of the block = h = 5 cm = 5 × 10-2 m, Displacement of top face = x = 0.03 mm = 0.03 × 10-3 m =  3 × 10-5 m, Shearing force  = 10kg-wt = 104 × 9.8 N.
  • To Find: Shear strain =? Shear stress =? Modulus of rigidity = η =?

Shear stress =  F/A

∴   Shear stress =  (104 × 9.8)/( 25 × 10-4)

∴   Shear stress = 3.92 × 107 N

Shear strain = tanθ = x/h = (3 × 10-5 ) / (5 × 10-2 ) = 6 × 10-4



Modulus of rigidity = η = Shear stress / Shear strain

η = (3.92 × 107) / (6 × 10-4) = 6.53 × 1010   N/m²

Ans: Shear stress = 3.92 × 107 N Shear strain = 6 × 10-4, Modulus of rigidity = 6.53 × 1010   N/m²

Example – 3:

  • A 5 cm cube of substance has its upper face displaced by 0.65 cm by a tangential force of 0.25 N. Calculate the modulus of rigidity of the substance.
  • Solution:
  • Given: Area under shear = A = 5 cm x 5 cm  = 25 cm² = 25 × 10-4 m², Height of the block = h = 5 cm = 5 × 10-2 m, Displacement of top face = x = 0.65 cm = 0.65 × 10-2 m = 6.5 × 10-3 m, Shearing force  = 0.25 N.
  • To Find: Modulus of rigidity = η =?

Modulus of rigidity = η = Fh/Ax

∴  η = (0.25 × 5 × 10-2)/(25 × 10-4  × 6.5 × 10-3)



∴  η =769  N/m²

Ans: Modulus of rigidity = 769  N/m²

Example – 4:

  • A tangential force of 2100 N is applied on a surface area 3 × 10-6 m² which is 0.1 m from a fixed face of a block of material. The force produces a shift of 7 mm of the upper surface with respect to the bottom. Calculate the modulus of rigidity of the material.
  • Solution:
  • Given: Area under shear = A = 3 × 10-6 m², Height of the block = h = 0.1 m, Displacement of top face = x = 7 mm = 7 × 10-3 m, Shearing force  = 2100 N.
  • To Find: Modulus of rigidity = η =?

Modulus of rigidity = γ = Fh/Ax

∴  η = (2100 × 0.1)/(3 × 10-6  × 7 × 10-3)

∴  η =1010  N/m²



Ans: Modulus of rigidity = 1010  N/m²

Example – 5:

  • A metal plate has an area of face 1m x 1m and thickness of 1 cm. One face of a larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced thereby is 0.005 cm. Find the shearing stress, strain and magnitude of the tangential force applied. Modulus of rigidity of metal is ϒ = 8.4 × 1010  N/m²
  • Solution:
  • Given: Area under shear = A = 1 m x 1 cm  = 1 m²,Thickness of plate = h = 1 cm = 1 × 10-2 m, Displacement of top face = x = 0.005 cm = 0.005 × 10-2 m =  5 × 10-5 m, Modulus of rigidity = η = 8.4 × 1010  N/m²
  • To Find: Shear strain =? Shear stress =? Shearing force = F =?

Shear strain = tanθ = x/h = (5 × 10-5 ) / (1 × 10-2 ) = 5 × 10-3

Modulus of rigidity = η = Shear stress / Shear strain

∴  Shear stress = η × Shear strain =  8.4 × 1010 × 5 × 10-3



∴  Shear stress = 4.2 × 108 N/m².

Shear stress =  F/A

∴  F = Shear stress × Area

∴  F = 4.2 × 108 ×1

∴  F =  4.2 × 10 N

Ans: Shear strain = 5 × 10-3, Shear stress = 4.2 × 108  N/m², Shearing force = 4.2 × 10 N.





Example – 6:

  • A metal plate has an area of face 1m x 1m and thickness of 5 cm. One face of a larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced thereby is 0.005 cm. Find the shearing stress and shear strain. Modulus of rigidity of metal is η = 4.2 × 106  N/m²
  • Solution:
  • Given: Area under shear = A = 1 m x 1 cm  = 1 m²,Thickness of plate = h = 5 cm = 5 × 10-2 m, Displacement of top face = x = 0.005 cm = 0.005 × 10-2 m =  5 × 10-5 m, Modulus of rigidity = η = 4.2 × 106  N/m²
  • To Find: Shear strain =? Shear stress =? Shearing force = F =?

Shear strain = tanθ = x/h = (5 × 10-5 ) / (5 × 10-2 ) =  10-3

Modulus of rigidity = η = Shear stress / Shear strain

∴  Shear stress = η × Shear strain =  4.2 × 106  × 10-3

∴  Shear stress = 4.2 × 103 N/m².



 

Ans: Shear strain =  10-3, Shear stress = 4.2 × 103  N/m².

Example – 7:

  • A copper metal cube has each side of length 1m. The bottom edge of a cube is fixed and tangential force of 4.2 × 108 N is applied to the top surface. Calculate the lateral displacement of the of the surface, if the modulus of rigidity of copper is 14 × 1010  N/m².
  • Solution:
  • Given: Area under shear = A = 1 m x 1 cm  = 1 m²,Height of cube = h =1 m,  Modulus of rigidity = η = 14 × 1010  N/m², Shearing force = F = 4.2 × 108 N
  • To Find: Displacement of top face = x =?

Modulus of rigidity = η = Fh/Ax

∴  x = Fh/Aη

∴  x =  ( 4.2 × 10× 1)/(1 ×14 × 1010 )

∴  x =  ( 4.2 × 10× 1)/(1 ×14 × 1010 )



∴  x = 3 × 10-3 m = 3mm

Ans: Displacement of top face is  3mm

Example – 8:

  • The frame of a brass plate of an outer door design has area 1.60 m² and thickness 1cm. The brass plate experiences a shear force due to the earthquake.  How large parallel force must be exerted on each of the edges if the lateral displacement is 0.32 mm. Modulus of rigidity for brass is 3.5 × 1010  N/m².
  • Solution:
  • Given: Area under shear = A = 1.60 m²,Thickness = h =1 cm =1 × 10-2 m,  Modulus of rigidity = η = 3.5 × 1010  N/m², Displacement of top face = x = 0.32 mm = 0.32 × 10-3 m =3.2 × 10-4 m
  • To Find: Shearing force = F = ?

Modulus of rigidity = η = Fh/Ax

∴ F =  Aηx /h

∴ F =  Aηx /h

∴  x =  ( 1.60 × 3.5 × 1010  × 3.2 × 10-4)/(1 × 10-2 )

∴  x = 1.792 × 109 N

Ans: Shearing force is 1.792 × 109 N

Science > Physics > Elasticity > You are Here
Physics Chemistry Biology Mathematics

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