Mechanical and structural analysis miscellaneous Easy Questions and Answers | Page - 23

Mechanical and structural analysis miscellaneous

Considering the symmetry of a rigid frame as shown below, the magnitude of the bending moment (in kNm) at P (preferably using the moment distribution method) is

1. 170
1. 172
1. 176
1. 178

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Distribution Factor

Correct Option: C

Distribution Factor

The values of axial stress (s) in kN/m², bending moment (M) in kNm, and shear force (V) in kN acting at point P for the arrangement shown in the figure are respectively

1. 1000, 75 and 25
1. 1250, 150 and 50
1. 1500, 225 and 75
1. 1750, 300 and 100

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Axial stress
= 50 = 1250 kN/m²
.2 × .2

Bending moment = 50 × 3 = 150 kNm
Distribution Table

Correct Option: B

Axial stress
= 50 = 1250 kN/m²
.2 × .2

Bending moment = 50 × 3 = 150 kNm
Distribution Table

Mathematical idealization of a crane has three bars with their vertices arranged as shown in the figure with a load of 80 kN hanging vertically. The coordinates of the vertices are given in the parentheses. The force in the member QR, F_QR will be

1. 30 kN Compressive
1. 30 kN Tensile
1. 50 kN Compressive
1. 50 kN Tensile

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Take moment about R
∑ M_R = 0
80 × 3 = VQ × 2
∴ V_a = 120 kN
Consider Joint Q

∠ PQR = 104.03°
∴ Q = 104.03° – 90 = 14.03°
Resolve vertically,
∑ F_v = 0
F_QP .cos + V_Q = 0
∴ F_QP = 120
cos14.03°

Resolve horizontally,
∑ F_H = 0
F_QP sin θ = F_QR
120 × sin14.03° = F_QR
cos14.03°

∴ F_QR = 120 × tan 14.03° = 30 kN_M

Correct Option: A

Take moment about R
∑ M_R = 0
80 × 3 = VQ × 2
∴ V_a = 120 kN
Consider Joint Q

∠ PQR = 104.03°
∴ Q = 104.03° – 90 = 14.03°
Resolve vertically,
∑ F_v = 0
F_QP .cos + V_Q = 0
∴ F_QP = 120
cos14.03°

Resolve horizontally,
∑ F_H = 0
F_QP sin θ = F_QR
120 × sin14.03° = F_QR
cos14.03°

∴ F_QR = 120 × tan 14.03° = 30 kN_M

A box of weight 100 kN shown in the figure is to be lifted without swinging. If all forces are coplanar, the magnitude and direction (q) of the force (F) with respect to x-axis should be

1. F = 56.389 kN and θ = 28.28°
1. F = – 56.389 kN and θ = – 28.28°
1. F = 9.055 kN and θ = 1.414°
1. F = – 9.055 kN and θ = – 1.414°

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For no swing, resultant, force along horizontal should be zero
F_H = 90 cos 30° – (40 cos 45° + F cos θ) = 0
∴ F cos θ = 49.658
Find F cos θ values from the given F and θ values in options to find the nearest answer.
In option (a)
F cos θ = 56.389 × cos (28.28°)
= 49.65 kN

Correct Option: A

For no swing, resultant, force along horizontal should be zero
F_H = 90 cos 30° – (40 cos 45° + F cos θ) = 0
∴ F cos θ = 49.658
Find F cos θ values from the given F and θ values in options to find the nearest answer.
In option (a)
F cos θ = 56.389 × cos (28.28°)
= 49.65 kN

If the following equation establishes equilibrium in slightly bent position, the mid-span deflection of a member shown in the figure is
d²y + P y = 0
dx² EI

If a is amplitude constant for y, then

1. y = 1 1 - a cos 2πx
  P L
1. y = 1 1 - a sin 2πx
  P L
1. y = a sin nπx
  L
1. y = a cos nπx
  L

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d²y = - P .y = - m²y
dx² EI

Solution of above equation is:
y = a sin mx + b cos nx
at x = 0, y = 0
0 = a × 0 + b × 1
∴ b = 0
at x = L, y = 0
0 = a sin mL
∴ mL = nπ
∴ m = nπ
L

y = a sin nπx
L

Correct Option: C

d²y = - P .y = - m²y
dx² EI

Solution of above equation is:
y = a sin mx + b cos nx
at x = 0, y = 0
0 = a × 0 + b × 1
∴ b = 0
at x = L, y = 0
0 = a sin mL
∴ mL = nπ
∴ m = nπ
L

y = a sin nπx
L