Mechanical and structural analysis miscellaneous
- The shear stress at the neutral axis in a beam of triangular section with a base of 40 mm and height 20 mm, subjected to a shear force of 3 kN is
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b = 40 × 2 = 80 mm (by symmetry) 3 3
∴ Area of upper triangle= 1 × 80 × 40 2 3 3 = 1600 mm2 9 y = 40 × 1 = 40 mm 3 3 9 τnetutral axis = V.Ay l.b
= 10 N/mm2 = 10 mPa.Correct Option: C
b = 40 × 2 = 80 mm (by symmetry) 3 3
∴ Area of upper triangle= 1 × 80 × 40 2 3 3 = 1600 mm2 9 y = 40 × 1 = 40 mm 3 3 9 τnetutral axis = V.Ay l.b
= 10 N/mm2 = 10 mPa.
- The maximum and minimum shear stresses in a hollow circular shaft of outer diameter 20 mm and thickness 2 mm, subjected to a torque of 92.7 Nm will be
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τ = T r J τmax = T.rmax J = 92.7 × 10 × 1000 (π/32) × (204 - 104)
≈ 100 N/mm2
τmax is at r = 16/2 = 8 mm
τmax = (100/10) × rmin = 80 N/mm2 = 80 mPaCorrect Option: B
τ = T r J τmax = T.rmax J = 92.7 × 10 × 1000 (π/32) × (204 - 104)
≈ 100 N/mm2
τmax is at r = 16/2 = 8 mm
τmax = (100/10) × rmin = 80 N/mm2 = 80 mPa
- A rigid bar is suspended by three rods made of the same material as shown in the figure. The area and length of the central rod are 3A and L, respectively while that of the two outer rods are 2A and 2L, respectively. If a downward force of 50 kN is applied to the rigid bar, the forces in the central and each of the outer rods will be
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δl is same for 3 bars.
δl = PL AE
∴ PC = 3Ps
∴ 2PC + Ps = 50
⇒ 2PC + 3Ps = 50
∴ Ps = 10 kN
∴ PC = 30 kNCorrect Option: C
δl is same for 3 bars.
δl = PL AE
∴ PC = 3Ps
∴ 2PC + Ps = 50
⇒ 2PC + 3Ps = 50
∴ Ps = 10 kN
∴ PC = 30 kN
- A metal bar of length 100 mm is inserted between two rigid supports and its temperature is increased by 10°C. If the coefficient of thermal expansion is 12 × 10–6 per °C and the young’s modulus is 2 × 105 MPa, the stress in the bar is
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Stress = E × strain
= E × δl l
= Eαt
=2 × 105 × 12 × 10-6 × 10 =24 mPaCorrect Option: C
Stress = E × strain
= E × δl l
= Eαt
=2 × 105 × 12 × 10-6 × 10 =24 mPa
- An axially loaded bar is subjected to a normal stress of 173 MPa. The shear stress in the bar is
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For maximum, θ = 45°
τmax = σx - σy .sin2θ 2 = 173 - 0 × 1 = 86.5 MPa 2 Correct Option: B
For maximum, θ = 45°
τmax = σx - σy .sin2θ 2 = 173 - 0 × 1 = 86.5 MPa 2