Mechanical and structural analysis miscellaneous Easy Questions and Answers | Page - 16

Mechanical and structural analysis miscellaneous

For the beam shown below, the value of the support moment M is _____ kN-m.

1. 5
1. 6
1. 4
1. 3

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The beam is symmetrical
∴ Take half of beam to left. The same moment M will be there on other end.

Correct Option: A

The beam is symmetrical
∴ Take half of beam to left. The same moment M will be there on other end.

For the state stresses (in MPa) shown in the figure below, the maximum shear stress (in MPa) is _______

1. 3
1. 4
1. 5
1. 6

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Correct Option: C

The axial load (in kN) in the member PQ for the arrangement/assembly shown in the figure given below is _________

1. 50
1. 70
1. 66
1. 77

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By principle of super position
160 × 2³ + 160 × 2² × 2 - V_Q L³ = V_Q L
2EI 2EI 3EI AE

We neglect the annual definition
V_Q L
A E

∴ 160 × 2³ + 160 × 2² × 2 - V_Q L³ = 0
3EI 2EI 3EI

= V_Q 4³ = 160 × 4 × 5
3 3

∴ V_Q = 50 KN

Correct Option: A

By principle of super position
160 × 2³ + 160 × 2² × 2 - V_Q L³ = V_Q L
2EI 2EI 3EI AE

We neglect the annual definition
V_Q L
A E

∴ 160 × 2³ + 160 × 2² × 2 - V_Q L³ = 0
3EI 2EI 3EI

= V_Q 4³ = 160 × 4 × 5
3 3

∴ V_Q = 50 KN

The beam of an overall depth 250 mm (shown below) is used in a building subjected to two different thermal environments. The temperatures at the top and bottom surfaces of the beam are 36°C and 72°C respectively. Considering coefficient of thermal expansion (α) as 1.50 × 10^–5 per °C, the vertical deflection of the beam (in mm) an its mid-span due to temperature gradient is ________

1. 2.38 to 2.4
1. 2.38 to 2.45
1. 2.38 to 3.45
1. 3.38 to 2.45

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Vertical deflection at mid span,
δ = L² ∝ T
8h

T = difference in temperature = (72 – 36) = 36° C
δ = 3² (1.5 × 10^-5) × 36 = 2.43 mm
8 ×(250 × 10^-3)

Correct Option: B

Vertical deflection at mid span,
δ = L² ∝ T
8h

T = difference in temperature = (72 – 36) = 36° C
δ = 3² (1.5 × 10^-5) × 36 = 2.43 mm
8 ×(250 × 10^-3)

The static indeterminacy of the two-span continuous beam with an internal hinge, shown below, is ___________

1. 1
1. 2
1. 3
1. 0

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0 < zero >
Degree of static indeterminacy = D_s = 3_m + r_e – 3_j – r_r
m = no. of members = 4
r_e = no. of external reactions = 4
j = no. of joints = 5
r_r = no. of reactions released = 1 (hinge)
∴ D_s = (3 × 4) + 4 – (3 × 5) –1
= 12 + 4 – 15 – 1 = 0

Correct Option: D

0 < zero >
Degree of static indeterminacy = D_s = 3_m + r_e – 3_j – r_r
m = no. of members = 4
r_e = no. of external reactions = 4
j = no. of joints = 5
r_r = no. of reactions released = 1 (hinge)
∴ D_s = (3 × 4) + 4 – (3 × 5) –1
= 12 + 4 – 15 – 1 = 0