Mechanical and structural analysis miscellaneous
- In the propped cantilever beam carrying a uniformly distributed loaf of w N/m, shown in the following figure, the reaction at the support B is
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wL4 = RBL3 8EI 3EI ⇒ RB = 3 wL 8 Correct Option: B
wL4 = RBL3 8EI 3EI ⇒ RB = 3 wL 8
- A bar of varying square cross-section is loaded symmetrically as shown in the figure, loads shown are placed on one of the axes of symmetry of cross-section. Ignoring self weight, the maximum tensile stress in N/nm2 anywhere is
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Consider top portion
σ1 = 250 × 103 = 25 MPa 100 × 100
Consider bottom portionσ2 = 50 × 103 = 20 MPa 50 × 50
∴ σmax = σ2 = 25 MPaCorrect Option: C
Consider top portion
σ1 = 250 × 103 = 25 MPa 100 × 100
Consider bottom portionσ2 = 50 × 103 = 20 MPa 50 × 50
∴ σmax = σ2 = 25 MPa
- The stiffness K of a beam deflecting in a symmetric mode, as shown in the figure, is
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θ = ML 2EI ∴ M = 2EI .θ L θ = 1 ⇒ M = 2EI L
Which is the stiffness of the beam.Correct Option: B
θ = ML 2EI ∴ M = 2EI .θ L θ = 1 ⇒ M = 2EI L
Which is the stiffness of the beam.
- A long structural column (length = L) with both ends hinged is acted upon by an axial compressive load, P. The differential equation governing the bending of column is given by
EI d2y = - py dx2
Where y is the structural lateral deflection and EI is the flexural rigidity. The first critical load on column responsible for its buckling is given by
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Effective length, L = l
Critical load, Pc = π2 EI L2 Correct Option: A
Effective length, L = l
Critical load, Pc = π2 EI L2
- In a redundant joint model, three bar members are pin connected at Q as shown in the figure. Under some load placed at Q, the elongation of the members MQ and OQ are found to be 48 mm and 35 mm respectively. Then the horizontal displacement u and the vertical displacement v of the node Q, in mm, will be respectively.
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tanθ1 = 400 = 0.8 500
∴ θ1 = 38.66°tanθ2 = 500 = 1 500
∴ θ2 = 45°
Along MQ,
u sin 38.66° + v cos 38.66° = 48
Along OQ
–u sin 45° + v cos 45° = 35
Solving, for u and v, we get
u = 6.64 mm, v = 56.14 mmCorrect Option: B
tanθ1 = 400 = 0.8 500
∴ θ1 = 38.66°tanθ2 = 500 = 1 500
∴ θ2 = 45°
Along MQ,
u sin 38.66° + v cos 38.66° = 48
Along OQ
–u sin 45° + v cos 45° = 35
Solving, for u and v, we get
u = 6.64 mm, v = 56.14 mm