Mechanical and structural analysis miscellaneous
- A simply supported beam AB of span, L = 24 m is subjected to two wheel loads acting at a distance, d = 5 m apart as shown in the figure below. Each wheel transmits is a load, P = 3 kN and may occupy any position along the beam. If the beam is an I-section having section modulus, S = 16.2 cm3, the maximum bending stress (in GPa) due to the wheel loads is_______________
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σ = m y I ⇒ σ = m . y = m I z = 28.5 × 106 16.2 × 103
(m = 3 × 9.5 = 28.5)
= 1759.2 GPaCorrect Option: D
σ = m y I ⇒ σ = m . y = m I z = 28.5 × 106 16.2 × 103
(m = 3 × 9.5 = 28.5)
= 1759.2 GPa
- A horizontal beam ABC is loaded as shown in the figure below. The distance of the point of contraflexure from end A (in m) is ____
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Take RC at redundant
By Marshall Reciprocal Theorem
Deflection at C due to load B, ∆BC∆BC = 10 ×(0.75)3 + 10 ×(0.75)3 × 0.25 3EI 2EI = 2.11 ↓ EI
Deflection at C due to redundant Rc, ∆cc∆cc = Rc ×(0.75)3 = 0.141Rc 3EI EI
∴ ∆c = 02.11 - 0.141Rc = 0 EI EI Rc = 2.11 = 15 KN 0.141
Mx = 10x – 15 × (x – 0.25) = 0
⇒ = 10x – 15x + 3.75 = 0
⇒ x = 0.75m
∴ Distanced contraflexure from C = 0.75m
⇒ Distanced contraflexure from A = 1 – 0.75 = 0.25mCorrect Option: A
Take RC at redundant
By Marshall Reciprocal Theorem
Deflection at C due to load B, ∆BC∆BC = 10 ×(0.75)3 + 10 ×(0.75)3 × 0.25 3EI 2EI = 2.11 ↓ EI
Deflection at C due to redundant Rc, ∆cc∆cc = Rc ×(0.75)3 = 0.141Rc 3EI EI
∴ ∆c = 02.11 - 0.141Rc = 0 EI EI Rc = 2.11 = 15 KN 0.141
Mx = 10x – 15 × (x – 0.25) = 0
⇒ = 10x – 15x + 3.75 = 0
⇒ x = 0.75m
∴ Distanced contraflexure from C = 0.75m
⇒ Distanced contraflexure from A = 1 – 0.75 = 0.25m
- For formation of collapse mechanism in the following figure, the minimum value of Pu is cmp /L. mp and 3 mp denote the plastic moment capacities of beam sections as shown in this figure. The value of c is________.
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Mechanism I
3mp .θ + mp .(2 θ ) + mp θP u × L × θ 4 ⇒ 6mp .θ = P u × L × θ 4 ⇒ P u = 24 mp L
Mechanism II
1.Φ = 3θ
⇒ Φ = 3θ= 13.33 mp L
∴ c = 13.33Correct Option: C
Mechanism I
3mp .θ + mp .(2 θ ) + mp θP u × L × θ 4 ⇒ 6mp .θ = P u × L × θ 4 ⇒ P u = 24 mp L
Mechanism II
1.Φ = 3θ
⇒ Φ = 3θ= 13.33 mp L
∴ c = 13.33
- Two beams are connected by a linear spring as shown in the following figure. For a load P as shown in the figure, the percentage of the applied load P carried by the spring is ________.
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∆ = R = R × 2L3 K 3EI ∆ = K = 3EI 2L3 ∆ = 2RL3 3EI ⇒ P = R 3
∴ % of applied load carried by spring= 1 × 100 = = 33.3% 3 Correct Option: A
∆ = R = R × 2L3 K 3EI ∆ = K = 3EI 2L3 ∆ = 2RL3 3EI ⇒ P = R 3
∴ % of applied load carried by spring= 1 × 100 = = 33.3% 3
- For the 2D truss with the applied loads shown below, the strain energy in the member XY is _______ kN-m. For member XY, assume AE=30kN, where A is cross section area and E is the modulus of elasticity.
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RA × 3 + 10 × 9 = 0
⇒ RA = –30 kN
RB = 35 kN
Taking joint A
Taking joint B
Taking joint x
Fx = 10 kNU = F2 × L 2AE = 102 × 3 = 5 kNm 2 × 3
Correct Option: A
RA × 3 + 10 × 9 = 0
⇒ RA = –30 kN
RB = 35 kN
Taking joint A
Taking joint B
Taking joint x
Fx = 10 kNU = F2 × L 2AE = 102 × 3 = 5 kNm 2 × 3