A solid circular beam with radius of 0.25 m and length of 2 m is subje...
Problem:
A solid circular beam with radius of 0.25 m and length of 2 m is subjected to a twisting moment of 20 kNm about the z-axis at the free end, which is the only load acting as shown in the figure. The stress component at Point M in the cross-section of the beam at a distance of 1m from the fixed end is? Explain in details.
Solution:
Step 1: Calculation of Polar Moment of Inertia
Polar Moment of Inertia (J) = πr⁴/2
where, r = radius of the beam = 0.25 m
Therefore, J = (3.14 x 0.25⁴)/2 = 0.00098275 m⁴
Step 2: Calculation of Shear Stress
The formula for shear stress (τ) due to torsion is given as:
τ = T*r/J
where, T = twisting moment = 20 kNm, r = distance of point M from the axis of rotation = 1m
Therefore, τ = (20 x 10³ x 1)/(0.00098275) = 20,352.94 kPa
Step 3: Calculation of Shear Strain
Shear strain (γ) is defined as the ratio of angular displacement (θ) to the length of the beam (L).
γ = θ/L
Since the beam is subjected to pure torsion, the angle of twist (θ) can be calculated using the formula:
θ = TL/GJ
where, G = Shear Modulus of Elasticity
For steel, the value of G is around 80 GPa
Therefore, θ = (20 x 10³ x 1)/(80 x 10⁹ x 0.00098275) = 0.002551 rad
Thus, γ = 0.002551/2 = 0.001275
Step 4: Calculation of Shear Strain
The shear stress (τ) and shear strain (γ) are related by the equation:
τ = Gγ
where, G = Shear Modulus of Elasticity
For steel, the value of G is around 80 GPa
Therefore, τ = 80 x 10⁹ x 0.001275 = 102,000 kPa
Step 5: Conclusion
The stress component at Point M in the cross-section of the beam at a distance of 1m from the fixed end is 102,000 kPa.