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A 10 cm long steel rod having a diameter of 5 cm fits snugly between two rigid walls that are 10 cm apart, at room temperature. Young's modulus of elasticity and coefficient of linear expansion of steel are 2 x 106 kgf/cm2 and 12 x 10-6/°C, respectively. The stress developed in the rod due to a 100°C rise in temperature will be (in kgf/cm2) ___. (Rounded off to the nearest integer)
Correct answer is '2400'. Can you explain this answer?
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A 10 cm long steel rod having a diameter of 5 cm fits snugly between t...
°C respectively.
(a) What is the force required to pull the rod out of the walls at room temperature?
(b) If the temperature is increased by 50°C, what will be the force required to pull the rod out of the walls?
(c) Will the force required to pull the rod out of the walls increase or decrease with an increase in temperature? Why?
Solution:
Given:
Length of steel rod (L) = 10 cm
Diameter of steel rod (d) = 5 cm
Radius of steel rod (r) = d/2 = 2.5 cm
Distance between walls (d) = 10 cm
Young's modulus of elasticity (Y) = 2 x 106 kgf/cm2
Coefficient of linear expansion (α) = 12 x 10-6/°C
Change in temperature (ΔT) = 50°C
(a) To find the force required to pull the rod out of the walls at room temperature, we need to calculate the stress in the rod.
The cross-sectional area of the rod is given by:
A = πr2 = π(2.5 cm)2 = 19.63 cm2
The original length of the rod at room temperature is equal to the distance between the walls, which is 10 cm.
The stress in the rod is given by:
σ = F/A
where F is the force required to pull the rod out of the walls.
Using Young's modulus of elasticity, we can write:
σ = Y(ΔL/L)
where ΔL is the change in length of the rod and L is the original length.
At room temperature, the change in length of the rod is zero, so we have:
σ = 0
Therefore, the force required to pull the rod out of the walls at room temperature is zero.
(b) To find the force required to pull the rod out of the walls at a temperature of 50°C, we need to calculate the change in length of the rod first.
The change in length of the rod due to temperature change is given by:
ΔL = αLΔT
Substituting the given values, we get:
ΔL = (12 x 10-6/°C)(10 cm)(50°C) = 0.006 cm
The new length of the rod at a temperature of 50°C is:
L' = L + ΔL = 10 cm + 0.006 cm = 10.006 cm
The stress in the rod due to temperature change is given by:
σ = Y(ΔL/L')
Substituting the given values, we get:
σ = (2 x 106 kgf/cm2)(0.006 cm/10.006 cm) = 1.199 kgf/cm2
The force required to pull the rod out of the walls at a temperature of 50°C is:
F = σA = (1.199 kgf/cm2)(19.63 cm2) = 23.55 kgf
Therefore, the force required to pull the rod out of the walls at a temperature of 50°C is 23.55 kgf.
(c) The force required to pull the rod out of the walls will increase with an increase in temperature.
This
(a) What is the force required to pull the rod out of the walls at room temperature?
(b) If the temperature is increased by 50°C, what will be the force required to pull the rod out of the walls?
(c) Will the force required to pull the rod out of the walls increase or decrease with an increase in temperature? Why?
Solution:
Given:
Length of steel rod (L) = 10 cm
Diameter of steel rod (d) = 5 cm
Radius of steel rod (r) = d/2 = 2.5 cm
Distance between walls (d) = 10 cm
Young's modulus of elasticity (Y) = 2 x 106 kgf/cm2
Coefficient of linear expansion (α) = 12 x 10-6/°C
Change in temperature (ΔT) = 50°C
(a) To find the force required to pull the rod out of the walls at room temperature, we need to calculate the stress in the rod.
The cross-sectional area of the rod is given by:
A = πr2 = π(2.5 cm)2 = 19.63 cm2
The original length of the rod at room temperature is equal to the distance between the walls, which is 10 cm.
The stress in the rod is given by:
σ = F/A
where F is the force required to pull the rod out of the walls.
Using Young's modulus of elasticity, we can write:
σ = Y(ΔL/L)
where ΔL is the change in length of the rod and L is the original length.
At room temperature, the change in length of the rod is zero, so we have:
σ = 0
Therefore, the force required to pull the rod out of the walls at room temperature is zero.
(b) To find the force required to pull the rod out of the walls at a temperature of 50°C, we need to calculate the change in length of the rod first.
The change in length of the rod due to temperature change is given by:
ΔL = αLΔT
Substituting the given values, we get:
ΔL = (12 x 10-6/°C)(10 cm)(50°C) = 0.006 cm
The new length of the rod at a temperature of 50°C is:
L' = L + ΔL = 10 cm + 0.006 cm = 10.006 cm
The stress in the rod due to temperature change is given by:
σ = Y(ΔL/L')
Substituting the given values, we get:
σ = (2 x 106 kgf/cm2)(0.006 cm/10.006 cm) = 1.199 kgf/cm2
The force required to pull the rod out of the walls at a temperature of 50°C is:
F = σA = (1.199 kgf/cm2)(19.63 cm2) = 23.55 kgf
Therefore, the force required to pull the rod out of the walls at a temperature of 50°C is 23.55 kgf.
(c) The force required to pull the rod out of the walls will increase with an increase in temperature.
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Community Answer
A 10 cm long steel rod having a diameter of 5 cm fits snugly between t...
l = 10 cm
E = 2 x 106 kgf/cm2
α = 12 x 10-6/oC
ΔT = 100oC
∵ Strain is prevented, stress will be induced in steel rod.
It is statically indeterminate. So, we use one equation on compatibility:
LαΔT = PL/AE
σ = EαΔT
= 12 x 10-6 x 2 x 106 x 100
= 2.4 x 103 kgf/cm2
= 2400
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A 10 cm long steel rod having a diameter of 5 cm fits snugly between two rigid walls that are 10 cm apart, at room temperature. Youngs modulus of elasticity and coefficient of linear expansion of steel are 2 x 106 kgf/cm2 and 12 x 10-6/°C, respectively. The stress developed in the rod due to a 100°C rise in temperature will be (in kgf/cm2) ___. (Rounded off to the nearest integer)Correct answer is '2400'. Can you explain this answer?
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