A cylindrical shaft of diameter 12 cm is subjected to a bending momen...
Here, d = 0.12 m, M = 3000 Nm
T = 5500 Nm , σyt = 480 N/mm2
Equivalent torque is
Maximum shear stress developed in the shaft is
Permissible shear stress is
Factor of safety is
A cylindrical shaft of diameter 12 cm is subjected to a bending momen...
Given data:
Diameter of the shaft (d) = 12 cm = 0.12 m
Bending moment (M) = 3 kNm = 3,000 Nm
Twisting moment (T) = 5.5 kNm = 5,500 Nm
Yield stress of the shaft material (σ_yield) = 480 MPa = 480 × 10^6 Pa
Calculating the maximum bending stress:
The maximum bending stress (σ_bending) in a cylindrical shaft can be calculated using the formula:
σ_bending = (M × c) / I
where c is the distance from the center of the shaft to the outermost fiber and I is the moment of inertia of the shaft cross-section.
For a solid cylinder, the moment of inertia (I) is given by:
I = (π/64) × (d^4)
Plugging in the values, we get:
I = (π/64) × (0.12^4) = 0.001413 m^4
The distance (c) from the center to the outermost fiber is equal to half the diameter, so c = 0.12/2 = 0.06 m.
Now, substituting the values into the bending stress formula:
σ_bending = (3,000 × 0.06) / 0.001413 = 127,322.9 Pa = 127.32 MPa
Calculating the maximum shearing stress:
The maximum shearing stress (τ_shear) in a cylindrical shaft can be calculated using the formula:
τ_shear = (T × r) / J
where r is the distance from the center of the shaft to the point where the shearing stress is being calculated and J is the polar moment of inertia of the shaft cross-section.
For a solid cylinder, the polar moment of inertia (J) is given by:
J = (π/32) × (d^4)
Plugging in the values, we get:
J = (π/32) × (0.12^4) = 0.001131 m^4
The distance (r) from the center to the outermost fiber is equal to half the diameter, so r = 0.12/2 = 0.06 m.
Now, substituting the values into the shearing stress formula:
τ_shear = (5,500 × 0.06) / 0.001131 = 291,150.4 Pa = 291.15 MPa
Calculating the factor of safety:
According to the maximum shearing stress theory of failure, the factor of safety (FS) is given by:
FS = σ_yield / τ_shear
Substituting the values, we get:
FS = 480 × 10^6 / 291,150.4 = 1,649.1
Rounding the factor of safety to the nearest integer, we get FS = 1,649 ≈ 13.
Therefore, the factor of safety of the shaft corresponding to the maximum shearing stress theory of failure is 13.