Volumetric stress:

  • When the deforming forces are such that there is a change in the volume of the body, then the stress produced in the body is called volume stress.

Volumetric Stress = Load / Area = Pressure Intensity = dP

  •  S.I. Unit of stress is N m-2  or Pa (pascal) and its dimensions are [L-1M1T-2]. Units and dimensions of stress are the same as that of pressure.

Volumetric strain:

  • When the deforming forces are such that there is a change in the volume of the body, then the strain produced in the body is called volume strain. The volumetric strain is unitless and dimensionless quantity.

Volumetric strain = – Change in volume (dV)/ Original Volume (V)

Negative sign indicates the decrease in the volume

Bulk Modulus of Elasticity:

  • Within elastic limit, the ratio of volumetric stress to the corresponding volumetric strain in a body is always constant, which is called as Bulk modulus of elasticity.
  • It is denoted by letter ‘K’. Its S.I. Unit of stress is N m-2  or Pa (pascal) and its dimensions are [L-1M1T-2].

Strain Elasticity 15

Compressibility:

  • The reciprocal of bulk modulus of elasticity is called as compressibility.
  • Its S.I. unit is m2 N-1 or Pa-1 and its dimensions are  [L-1M-1T2].


Shear Stress:

  • When the deforming forces are such that there is a change in the shape of the body, then the stress produced is called shearing stress. Shear stress is also called as tangential stress.

Elasticity 16

Shear stress = Shearing force (F) / Area under shear

  • Its S.I. Unit of stress is N m-2  or Pa (pascal) and its dimensions are [L-1M1T-2].

Shear Strain:

  • When the deforming forces are such that there is a change in the shape of the body, then the strain produced in the body is called shear strain.
  • Shearing strain is defined as the ratio of relative displacement of any layer to its perpendicular distance from fixed layer.

tan θ = x/h

Modulus of Rigidity:

  • Within elastic limit, the ratio of the shear stress to the corresponding shear strain in the body is always constant, which is called as modulus of rigidity.
  • It is denoted by letter ‘η’. Its S.I. Unit of stress is N m-2  or Pa (pascal) and its dimensions are [L-1M1T-2].

Bulk Modulus



Concept of Thermal Stress:

  • If a bar which is heated and prevented from expansion or heated rod is prevented from contraction as it cools, then stress is produced in the bar. This stress is called a thermal stress. If bar is prevented from expansion or contraction, the change in length is given by

Δl = l ∝ Δθ

Where,  Δl = change in length prevented.

l = original length of the rod

Δθ = Change in temperature.

 ∝ = coefficient of linear expansion of the material of the rod.

Shear Stress, Shear Strain and Modulus of Rigidity

Where Y = Young’s modulus of elasticity.

If A is the area of cross-section of the bar then the force acting on the rod is

F = Thermal stress x area of cross-section

Force = Y α θ A

Problems on Volumetric Stress and Volumetric Strain, Bulk Modulus

  1. A solid rubber ball has its volume reduced by 14.5% when subjected to a uniform stress of 1.45 × 104 N/m². Find the bulk modulus for rubber. Ans: (105 N/m²)
  2. What pressure should be applied to a lead block to reduce its volume by 10% Bulk modulus for lead = 6 × 109 N/m². Ans: (6 × 108 N/m²)
  3. A volume of 5 litres of water is compressed by a pressure of 20 atmospheres. If the bulk modulus of water is 20 × 108 N/m². , find the change produced in the volume of water. Density of Mercury = 13,600 kg/m³; g = 9.8 m/s². Normal atmospheric pressure  = 75 cm of mercury. Ans: (5 cc).
  4. A volume of 10-3 m³ of water is subjected to a pressure of 10 atmospheres. The change in volume is 10-6 m³. Find the bulk modulus of water. Atm. pressure = 10 N/m². Ans: (1.01 × 109 N/m²)
  5. Two litres of water, when subjected to a pressure of 10 atmospheres, are compressed by 1.013 cc. Find the compressibility of water. Ans: (5 × 10-10 m²/N).
  6. The bulk modulus of water is 2.05 × 109 N/m². What change of pressure will compress a given quantity of water by 0.5%? Ans: (1.025 × 107  N/m²)
  7. Calculate the change in volume of a lead block of volume 1 m³ subjected to pressure of 10 atmospheres. Also calculate compressibility of lead. 1 atm = 1.013 × 105 N/m², K = 8 × 105 N/m². Ans: (1.27  × 10-4 m³  1.25 × 10-10 m²/N)
  8. Find the increase in the pressure required to decrease volume of mercury by 0.001%. Bulk modulus of mercury = 2.8 × 1010 N/m². Ans: (2.8 × 105 N/m²).
  9. A solid brass sphere of volume 0.305 m³ is dropped in an ocean. where water pressure is 2 × 107 N/m². The bulk modulus of water is 6.1 × 1010 N/m². What is the change in volume of the sphere? Ans: (10-4 m³)


For Solution of Above Problems Click Here

Problems on Shear Stress, Shear Strain and Modulus of Rigidity

  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. Ans:  (1.5 × 10-3,  6.75 × 107  N/m², 1.69 × 10N)
  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. Ans: (3.92 × 107 N,  6 × 10-4, 6.53 × 1010   N/m²)
  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. Ans: (769  N/m²)
  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. Ans:  (1010  N/m²)
  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. Ans: (5 × 10-3, 4.2 × 108  N/m², 4.2 × 10 N)
  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. Ans: (10-3, 4.2 × 103  N/m²)
  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. Ans: (3mm)
  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. Ans: (1.792 × 109 N)

For Solution of Above Problems Click Here

 

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