Strength Of Materials Miscellaneous
- A machine element XY, fixed at end X, is subjected to an axial load P, transverse load F, and a twisting moment T at its free end Y. The most critical point from the strength point of view is
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At centre σb = 0 and torsional shear stress are zero.Correct Option: C
At centre σb = 0 and torsional shear stress are zero.
- A shaft with a circular cross-section is subjected to pure twisting moment. The ratio of the maximum shear stress to the largest principal stress is
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τmax = τxy
σ1 = τxy∴ τmax = 1 σ1 Correct Option: B
τmax = τxy
σ1 = τxy∴ τmax = 1 σ1
- A solid circular shaft of diameter d is subjected to a combined bending moment M and torque, T. The material property to be used for designing the shaft using the relation
16 √M² + T² is πd³
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For a circular shaft of diameter d,
T ' = τ J (d/2)
where, T ' = net torsional moment
τ = shear stress (torsional shear strength)
J = polar moment of inertia∴ t = T '(d/2) J
But for a solid circular shaftJ = π d4 32 ∴ τ = 16T ' 32
But T ' = √M² + T²∴ τ = 16T ' √M² + T² πd3 Correct Option: C
For a circular shaft of diameter d,
T ' = τ J (d/2)
where, T ' = net torsional moment
τ = shear stress (torsional shear strength)
J = polar moment of inertia∴ t = T '(d/2) J
But for a solid circular shaftJ = π d4 32 ∴ τ = 16T ' 32
But T ' = √M² + T²∴ τ = 16T ' √M² + T² πd3
- A solid shaft of diameter d and length L is fixed at both the ends. A torque, T0 is applied at a distance, L/4 from the left end as shown in the figure given below.
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τmax = 16 [√M² + T²] πd³
But M = 0∴ τmax = 16T0 πd³
AlternatelyWe know, T = τ J r Where, J = πd4 32 ∴ τ = Tr = T0 × R × 32 = 16T0 J πd4 πd3 Correct Option: A
τmax = 16 [√M² + T²] πd³
But M = 0∴ τmax = 16T0 πd³
AlternatelyWe know, T = τ J r Where, J = πd4 32 ∴ τ = Tr = T0 × R × 32 = 16T0 J πd4 πd3
Direction: A triangular-shaped cantilever beam of uniformthickness is shown in the figure. The Young's modulus of the material of the beam is E. A concentrated load P is applied at the free end of the beam.
- The maximum deflection of the beam is
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Maximum deflection of the beam
ymax = Pl³ = 6Pl³ 3EI Ebt³ where, I = bt³ 18l Correct Option: D
Maximum deflection of the beam
ymax = Pl³ = 6Pl³ 3EI Ebt³ where, I = bt³ 18l