# Elasticity Maharashtra Board Textual Unsolved Problems

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#### Problem – 01:

• A wire of length 2 m and cross-sectional area 10-4 m² is stretched by a load 102 kg. The wire is stretched by 0.1 cm. Calculate longitudinal stress, longitudinal strain and Youn’s modulus of the material of wire.
• Solution:
• Given: Initial length of wire = L =  2 m, Cross-sectional area = A =  10-4 m, Stretching weight = 102 kg wt = 102 × 9.8 N, Increase in length = l = 0.1 cm = 0.1 × 10-2 m = 1 × 10-3 m, g = 9.8 m/s² .
• To Find: Stress = ?, Strain = ?, Young’s modulus of material = Y = ?

Stress = F / A  = mg /A

∴  Stress =  ( 102 × 9.8) /10-4

∴  Stress =  1 × 10N/m²

Strain = l / L =   1 × 10-3 / 2

Strain = 0.5 × 10-3 = 5 × 10-4

Now, Young’s modulus of elasticity= Y = Stress / Strain = ( 1 × 107) / ( 5 × 10-4)

Y = 2 × 1010 N/m²

Ans : Stress =  1 × 10N/m², Strain = 5 × 10-4 , Young’s modulus of elasticity= Y = 2 × 1010 N/m²

#### Example – 02:

• Find the increase in the pressure required to decrease volume of mercury by 0.001%. Bulk modulus of mercury = 2.8 × 1010 N/m².
• Solution:
• Given: Volumetric strain = 0.001% = 0.001 × 10-2 = 10-5, Bulk modulus of elasticity = 2.8 × 1010 N/m².
• To Find: Pressure intensity =?

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴  Volumetric stress = K ×Volumetric strain

∴  Pressure intensity = K ×Volumetric strain

∴  Pressure intensity = 2.8 × 1010 × 10-5

∴  Pressure intensity = 2.8 × 105 N/m²

Ans: Pressure intensity is 2.8 × 105 N/m²

#### Example – 03:

• 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 – 05:

• Calculate the work done in stretching a wire of length 2 m and cross-sectional area 0.0225 mm² when a load of 100 N is applied slowly to its free end. Young’s modulus of elasticity = 20 × 1010 N/m².
• Solution:
• Given: Area  = A =0.0225 mm² =0.0225 × 10-6 m² = 2.25 × 10-8 m², Length of wire = L = 2 m, Load applied = F = 100 N, Young’s modulus of elasticity = Y = 20 × 1010 N/m².
• To Find: Work done = W =?

Young’s modulus of elasticity = Y = FL/Al

∴  l = FL/AY

∴  l = (100 × 2)/( 2.25 × 10-8 × 20 × 1010)

∴  l = 4.444 × 10-2 m

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 100 × 4.444 × 10-2

∴  Work done = 2.222 J

Ans: Work done in stretching wire is 2.222 J

#### Example – 06:

• 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?
• Solution:
• Given: Original Volume = 0.305 m³, Pressure = dP = 2 × 107 N/m²², Bulk modulus of elasticity = K =6.1 × 1010 N/m²
• To Find: Change in volume = dV =?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴  Change in volume = dV  =  (dP × V)/ K

∴  Change in volume = dV  =  (2 × 107 × 0.305)/ (6.1 × 1010)

∴    dV  =  10-4

Ans: Change in volume = 10-4

#### Example – 13:

• A uniform steel wire of length 3 m and area of cross-section 2 mm² is extended through 3mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. Young’s modulus of elasticity = Y = 20 × 1010
• Solution:
• Given: Area  = A =2 mm² =2 × 10-6 m² , Length of wire = L = 3 m, Extension = l = 3 mm = 3 × 10-3 m, Young’s modulus of elasticity = Y = 20 × 1010 N/m².
• To Find: Energy stored  = dU =?

Young’s modulus of elasticity = Y = FL/Al

∴  F = YAl/L

∴   F = ( 20  × 1010 × 2 × 10-6 × 3 × 10-3 )/3

∴   F = 400 N

Now Work done in stretching wire = ½ Load  ×Extension

∴  Work done = ½ × 400 × 3 × 10-3

∴  Work done = 0.6 J

Energy stored = work done in stretching wire = 0.6 J

Ans: Energy stored is 0.6 J

#### Example – 14:

• 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

#### Example – 15:

• A load 1 kg produces a certain extension in the wire of length 3 m and radius 5 × 10-4  m. How much will be the lateral strain produced in the wire? Given Y = 7.48 × 1010 N/m², σ = 0.291.
• Solution:
• Given: Load attached = F = 1 kg = 1 × 9.8 NLength of wire = L = 3 m, Radius of cross-section = r = 5 × 10-4  m cross-section = 1 mm² = 1 × 10-6  m², Stretching Load = 10 N,  Young’s modulus = Y =  7.48 × 1010  N/m²,  Poisson’s ratio = σ = 0.291
• To Find: Lateral strain = ?

Y = Longitudinal Stress / Longitudinal Strain

∴  Y = F  /(A  × Longitudinal  Strain)

∴  Y = F  /(π r²  × Longitudinal  Strain)

∴  Longitudinal strain = (1 × 9.8)  /(3.142 × (5 × 10-4)²  ×   7.48 × 1010)

∴  Longitudinal strain = (1 × 9.8)  /(3.142 × 25 × 10-8 ×   7.48 × 1010)

∴  Longitudinal strain = 1.67   × 10-4

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.291 × 1.67   × 10-4   = 4.86 × 10-5

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