Mechanical and structural analysis miscellaneous
- A continuous beam is loaded as shown in the figure below. Assuming a plastic moment capacity equal to MP, the minimum load at which the beam would collapse is
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For point load max moment occurs under the load Consider IJ
Max. moment= PL = 3 MP 4 2 ∴ P = 6MP L
Consider HI,PL = 2MP 4 ∴ P = 8MP L
Collapse load = 6MP/L (Minimum of two.)Correct Option: B
For point load max moment occurs under the load Consider IJ
Max. moment= PL = 3 MP 4 2 ∴ P = 6MP L
Consider HI,PL = 2MP 4 ∴ P = 8MP L
Collapse load = 6MP/L (Minimum of two.)
- A symmetric frame PQR consists of two inclined members PQ and QR, connected at ‘Q’ with a rigid joint, and hinged at ‘P’ and ‘R’. The horizontal length PR is l. If a weight W is suspended at ‘Q’, the bending moment at ‘Q’ is
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VP = VR = ω/2
By symmetry,y = h x l/2 ∴ y = hx l/2
Horizontal reaction for any support,H = ∫myds ∫y2ds ⇒ H = Wl 8h
BM at Q = R1 . x - H.yW × l - Wl . h 2 2 8h = Wl 8 Correct Option: C
VP = VR = ω/2
By symmetry,y = h x l/2 ∴ y = hx l/2
Horizontal reaction for any support,H = ∫myds ∫y2ds ⇒ H = Wl 8h
BM at Q = R1 . x - H.yW × l - Wl . h 2 2 8h = Wl 8
- The sketch shows a column with a pin at the base and rollers at the top. It is subjected to an axial force P and a moment M at mid-height. The reaction(s) at R is/are
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P will cause direct vertical reaction at R. Take moment about Q.
RH .h - M = 0
∴ RH = M/hCorrect Option: C
P will cause direct vertical reaction at R. Take moment about Q.
RH .h - M = 0
∴ RH = M/h
- A simply supported beam is subjected to a uniformly distributed load of intensity w per unit length, on half of the span from the end. The length of the span and the flexural stiffness are denoted as l and EI, respectively. The deflection at mid-span of the beam is
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Apply Maxwell reciprocal theorem.
δc = 5ωl4 384EI ∴ δc1= δc = 5ωl4 2 384EI Correct Option: B
Apply Maxwell reciprocal theorem.
δc = 5ωl4 384EI ∴ δc1= δc = 5ωl4 2 384EI
- The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends, is equal to
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p = π2 EI leff2
For fixed end, leff = l/2
For pinned end, leff = lp ∝ 1 leff2 p1 = l22 = l2 = 4 p2 l12 (l/2)2 Correct Option: D
p = π2 EI leff2
For fixed end, leff = l/2
For pinned end, leff = lp ∝ 1 leff2 p1 = l22 = l2 = 4 p2 l12 (l/2)2