Mechanical and structural analysis miscellaneous
- Carry-over factor CAB for the beam shown in the figure below is
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There is no role for internal hinge in carry over factor
Correct Option: D
There is no role for internal hinge in carry over factor
- Consider the beam ABCD and the influence line as shown below. The influence it pertains to
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By muller Breslaw principle.
Correct Option: B
By muller Breslaw principle.
- Vertical reaction developed at B in the frame below due to the applied load of 100 kN (with 150,000 mm2 cross-sectional area and 3.125 × 109 mm2 moment of inertia for both members) is
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Due to load P, the member AC undergo compression
Let the reaction at C = R
Thus the remaining vertical force is 100 – R. It is taken up by AB at B
Deflection at A along AC = deflection at A for cantilever AB.⇒ RL = (100 - R)L3 AE 3EI R × 1000 = (100 - R)10003 150000 × E 3 × 3.125 × 109 100 - R = 1000 × 3 × 3.125 × 109 R 10003 × 150000
∴ R = 94.1 kN
∴ Reaction at B = 100 – R = 5.9 kNCorrect Option: A
Due to load P, the member AC undergo compression
Let the reaction at C = R
Thus the remaining vertical force is 100 – R. It is taken up by AB at B
Deflection at A along AC = deflection at A for cantilever AB.⇒ RL = (100 - R)L3 AE 3EI R × 1000 = (100 - R)10003 150000 × E 3 × 3.125 × 109 100 - R = 1000 × 3 × 3.125 × 109 R 10003 × 150000
∴ R = 94.1 kN
∴ Reaction at B = 100 – R = 5.9 kN
- A thin-walled long cylindrical tank of inside radius r is subjected simultaneously to internal gas pressure p and axial compressive force F at its ends. In order to produce ‘pure shear’ state of stress in the wall of the cylinder,F should be equal to
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Consider a thin long cylindrical tank
2σ × t × l = p.2r.l∴ σ0 = pr (tensile) t
Due to pressure at longitudinal section
πr2 × p = 2πrt × σzσz = pr 2t
Due to compressive force, FLongitudinal force = F 2πrt
∴ Total longitudinal stress= pr - F (compressive) 2t 2πrt τ = σ1 - σ2 2 ⇒ pr = - pr - F t 2t 2πrt ⇒ F = 3pr 2πrt 2t
⇒ F = 3πpr2Correct Option: C
Consider a thin long cylindrical tank
2σ × t × l = p.2r.l∴ σ0 = pr (tensile) t
Due to pressure at longitudinal section
πr2 × p = 2πrt × σzσz = pr 2t
Due to compressive force, FLongitudinal force = F 2πrt
∴ Total longitudinal stress= pr - F (compressive) 2t 2πrt τ = σ1 - σ2 2 ⇒ pr = - pr - F t 2t 2πrt ⇒ F = 3pr 2πrt 2t
⇒ F = 3πpr2
- If a beam of rectangular cross-section is subjected to a vertical shear force V, the shear force carried by the upper one-third of the cross-section is
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Variation of shear stress,
τ = V.Ay lb
Taking a height y from N. A.Correct Option: B
Variation of shear stress,
τ = V.Ay lb
Taking a height y from N. A.