Elasticity Maharashtra Board Textual Solved Problems

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Example – 01:

  • A mild steel wire of radius 0.5 mm and length 3 m is stretched by a force of 49 N. calculate a) longitudinal stress, b) longitudinal strain c) elongation produced in the body if Y for steel is 2.1 × 1011 N/m².
  • Solution:
  • Given: Initial length of wire = L = 3 m, radius of wire = 0.5 mm = 0.5 × 10-3 m, = 5 × 10-4 m,Force applied =49 N, Young’s modulus for steel = Y = 2.1 × 1011 N/m².
  • To Find: Stress = ?, Strain = ?, elongation = ?

Stress = F / A  = mg /π r²

∴  Stress =  49 /(3.142 ×(5 × 10-4)²)

∴  Stress = 49 /(3.142 × 25 × 10-8)

∴  Stress = 6.238 × 10N/m²



Now, Y = Stress / Strain

∴  Strain = Stress / Y =  (6.238 × 10) / (2.1 × 1011)

Strain = 2.970 × 10-4

Now, Strain = l / L



∴ l = Strain  × L

∴  l = 2.970 × 10-4  × 3

∴  l = 8.91 × 10-4  m = 0. 891 × 10-3  m = 0.891 mm

Ans: Stress = 6.238 × 10N/m², Strain =  2.970 × 10-4, Elongation = 0.891 mm.

Example – 02:

  • 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.
  • Solution:
  • Given: Original Volume = 1 m³, Pressure = dP = 10 atm = 10 × 1.013 × 105 N/m², Bulk modulus of elasticity = K = 8 × 109 N/m².
  • To Find: Change in volume = dV =?, Compressibility = ?

Volumetric Stress = Pressure intensity = dP



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

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

∴  Change in volume = dV  =  (10 × 1.013 × 105 × 1)/ 8 × 109

∴    dV  =  1.27  × 10-4 

Compressibility = 1/K = 1/(8 × 109)



Compressibility =  1.25 × 10-10 m²/N

Ans: Change in volume is  1.27  × 10-4 m³ and compressibility of lead is 1.25 × 10-10 m²/N

Example – 03:

  • 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 – 05:

  • Calculate the strain energy per unit volume in a brass wire of length 2.0 m and cross-sectional area 0.5 mm2, when it is stretched by 2mm and a force of 5 kg-wt is applied to its free end.
  • Solution:
  • Given: Area  = A = 0.5 mm² = 0.5 × 10-6 m² = 5 × 10-7 m², Length of wire = L = 2.0 m, Extension in wire = l = 2 mm = 2 × 10-3 m,  Load applied = F = 5 kg-wt = 5 × 9.8 N
  • To Find: Strain energy per unit volume = dU/V =?

Strain energy per unit volume = dU/V = ½ × Stress × Strain



∴    Strain energy per unit volume = ½ × (F/A) × (l/L)

∴    Strain energy per unit volume = ½ × (Fl/AL)

∴    Strain energy per unit volume = ½ × (5 × 9.8 ×  2 × 10-3)/( 5 × 10-7 × 2)

∴    Strain energy per unit volume = 4.9 × 104  J/m³

Ans: The strain energy per unit volume of the wire  4.9 × 104  J/m³

Example – 06:

  • Calculate the greatest length of a steel wire which when fixed at one end can hang freely without breaking under its own weight? Density of steel = 7800 kg/m³. Breaking stress for steel = 7.2 × 108 N/m².
  • Solution:
  • Given: Density of steel = ρ =  7800 kg/m³. Stress = 7.8 × 108 N/m².
  • To Find: To find whether wire can be stretched by 10 mm.

Stress = F / A   = mg / A =  V ρ g /A



∴ Stress =   A L  ρ g / A

∴ Stress =   L  ρ g

∴ L = Stress /  ρ g

∴ L = 7.2 × 108 /  (7800 × 9.8)



∴ L = 7.2× 108 /  (7800 × 9.8)

∴ L = 9.419 × 103   m

Ans: Maximum length of copper wire is  9.419 × 103   m

Example – 07:

  • A steel wire having cross-sectional area 1 mm² is stretched by 10 N. Find the lateral strain produced in the wire. Young’s modulus for steel is 2 × 1011 N/m² and Poisson’s ratio is 0.291.
  • Solution:
  • Given: Area of cross-section = 1 mm² = 1 × 10-6  m², Stretching Load = 10 N,  Young’s modulus for steel= Y =  2 × 1011 N/m², Poisson’s ratio = σ = 0.291
  • To Find: Lateral strain = ?

Y = Longitudinal Stress / Longitudinal Strain

∴  Y = F  /(A  × Longitudinal  Strain)

∴  Longitudinal strain = F  /(A  × Y)



∴  Longitudinal strain = 10  /(1 × 10-6  × 2 × 1011) = 5 × 10-5

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.291 × 5 × 10-5   = 1.455 × 10-5

Example – 09:

  • Find the maximum load which may be placed on a tungsten wire of diameter 2 mm so that the permitted strain not exceed 1/1000. Young’s modulus fortungsten = Y = 35 × 1010 N/m².
  • Solution:
  • Given:  Strain = 1/1000 =  10-3 , Young’s modulus of elasticity =  Y = 35 × 1010 N/m², Diameter of wire = 2 mm, Radius of wire = 2/2 = 2 mm = 1 × 10-3 m,
  • To Find: Maximum load = F =?

Y = Stress /Strain = (F/A)/Strain

Y = F/(A × strain)

∴  F = π r² × Y× strain

∴  F = 3.142  × (1 × 10-3)² × 35 × 1010× 10-3

∴  F = 3.142  × 1 × 10-6 × 35 × 1010× 10-3

∴  F = 1100 N

Ans: Maximum load can be placed is 1100 N

Problem – 10:

  • A mass of 2kg is hung from a steel wire of radius 0.5 mm and length 3m. Compute the extension produced. What should be the minimum radius of wire so that elastic limit is not exceeded. Elastic limit for steel is 2.4 × 108 N/m², Y for steel = Y = 20 × 1010 N/m²
  • Solution:
  • Given:  Radius of wire = 0.5 mm = 0.5 × 10-3 m  = 5 × 10-4 m. Initial length of wire = L = 3m, Mass attached = m = 2 kg,  Y for steel = Y = 20 × 1010 N/m²
  • To Find: Extension = l =?,
  • Part – I:

Y= F L /A l

∴  l = F L /A Y

∴  l = m g L /π r² Y

∴  l = (2 × 9.8  × 3) /(3.142 1 × (5× 10-4)²× 20 × 1010 )

∴  l = (2 × 9.8  × 3) /(3.142  × 25× 10-8 × 20 × 1010 )

∴  l = 3.743 × 10-4 m = 0.3743 mm

  • Part – II:
  • Given:  Elastic limit for steel = Stress = 2.4 × 108 N/m², Mass attached = m = 2 kg,
  • To Find: Radius of wire at elastic limit = r =?

Stress = F /A =  F /π r²

∴   r² = mg / (π × Stress)

∴   r² = (2 × 9.8)  / (3.142 × 2.4 × 108)

∴   r² = 2.599× 10-8

∴   r = 1.612× 10-4  m = 0.1612× 10-3 m =  0.1612 mm

Ans: Part – I: Change in length of wire is  0.3743 mm

Part – II: Radius of wire at elastic limit = 0.1612 mm

Problem – 11:

  • A compressive force of 4 × 104 N is exerted at the end of a bone of length 30 cm and 4 cm² square cross-sectional area. What will happen to the bone? Calculate the change in length of a bone. Compressive strength of bone is 7.7 × 108 N/m² and Young’s modulus of bone is 1.5 × 1010 N/m²
  • Solution:
  • Given: Initial length of wire = L =  30 cm = 0.30 m, Area of cross-section  = 4 cm² =  4× 10-4 m², Load attached = F = 4 × 104 N . Y = 1.5 × 1010 N/m². Maximum Stress =  7.7 × 108 N/m².
  • To Find: Effect of loading =?Change in length = l = ?,

Applied Stress = Applied force / Area of cross-section

Applied Stress = (4 × 104 )/ (4× 10-4 ) =  4 × 108 N/m²

This stress is less than the maximum allowable stress (7.7 × 108 N/m²)

Hence the bone will not break but will get compressed and its length decreases

Y= F L /A l

∴  l = (4 × 104 × 0.3) /(4× 10-4  × 1.5 × 1010)

∴  l = 2 × 10-3 m = 2 mm

Ans: The length of bone decreases by 2 mm

Science > Physics > ElasticityYou are Here
Physics Chemistry  Biology  Mathematics

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