Strength Of Materials Miscellaneous
- Two solid circular shafts of radii R1 and R2 are subjected to same torque. The maximum shear stresses developed in the two shafts are τ1 and τ2. If R1/R2 = 2, then τ2/τ1 is ______.
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R1 = 2 R2
T1 = T2.= τJ + τJ r 1 r 2 τ1d41 = τ2d42 d1 d2 τ1 = d³2 = 1 τ1 d³1 8 d1 d2 τ2 = 8 τ1 Correct Option: A
R1 = 2 R2
T1 = T2.= τJ + τJ r 1 r 2 τ1d41 = τ2d42 d1 d2 τ1 = d³2 = 1 τ1 d³1 8 d1 d2 τ2 = 8 τ1
- Consider a stepped shaft subjected to a twisting moment applied at B as shown in the figure. Assume shear modulus, G = 77 GPa. The angle of twist at C (in degree) is _______.
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Angle of twist at (C) = Angle of twist at (B)
⇒ θ = TL GJ ⇒ 10 × 0.5 × 32 77 × 109 × π × .024
⇒ 0.236050Correct Option: A
Angle of twist at (C) = Angle of twist at (B)
⇒ θ = TL GJ ⇒ 10 × 0.5 × 32 77 × 109 × π × .024
⇒ 0.236050
- A hollow shaft (d0 = 2di where d0 and di are the outer and inner diameters respectively) needs to transmit 20 kW power at 3000 RPM. If the maximum permissible shear stress is 30 MPa, d0 is
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P = Tω
20 × 10³ = T × 2π × 3000 60
∴ T = 63.662 N– mNow T = τ Ip r So, 63.662 = 30 × 106 (r0 = d), π/32(15d41) d1
∴ d1 = 11.295 mm
∴ d0 = 2d1 = 22.59 mmCorrect Option: B
P = Tω
20 × 10³ = T × 2π × 3000 60
∴ T = 63.662 N– mNow T = τ Ip r So, 63.662 = 30 × 106 (r0 = d), π/32(15d41) d1
∴ d1 = 11.295 mm
∴ d0 = 2d1 = 22.59 mm
- A frame is subjected to a load P as shown in the figure. The frame has a constant flexural rigidity EI. The effect of axial load is neglected. The deflection at point A due to the applied load P is
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Strain energy (t) = L∫0 M²xdx + L∫0 M²ydx 2EI 2EI
Mx = Px MyPy1 Pδ = L∫0 (px)²dx + L∫0 (px)²dx 2 2EI 2EI = 2 P²L² = 1 P × 4 PL³ 3 EI 2 3 EI δ = 4 PL³ 3 EI Correct Option: D
Strain energy (t) = L∫0 M²xdx + L∫0 M²ydx 2EI 2EI
Mx = Px MyPy1 Pδ = L∫0 (px)²dx + L∫0 (px)²dx 2 2EI 2EI = 2 P²L² = 1 P × 4 PL³ 3 EI 2 3 EI δ = 4 PL³ 3 EI
- A simply supported beam of length 2L is subjected to a moment M at the mid-point x = 0 as shown in the figure. The deflection in the domain 0 < x < L is given by
w = - Mx (L - x)(x + c) 12EIL
where E is the Young's modulus, I is the area moment of inertia and C is a constant (to be determined).
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ΔBA = (first moment area APB) EI = -1 M × 1 × L × 2 L M EI 2 2 3 2EI = ML² 6EI Slope at A = ΔBA = ML L 6EI Correct Option: C
ΔBA = (first moment area APB) EI = -1 M × 1 × L × 2 L M EI 2 2 3 2EI = ML² 6EI Slope at A = ΔBA = ML L 6EI