Test: Solid Mechanics- 4 - Civil Engineering (CE) MCQ

Test: Solid Mechanics- 4 - Civil Engineering (CE) MCQ

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10 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Solid Mechanics- 4

Test: Solid Mechanics- 4 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: Solid Mechanics- 4 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Solid Mechanics- 4 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Solid Mechanics- 4 below.
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Test: Solid Mechanics- 4 - Question 1

In a closed coiled helical spring, the maximum shear stress occurs on the

Detailed Solution for Test: Solid Mechanics- 4 - Question 1

The major stresses in a helical spring are of two types, shear stress due to torsion and direct shear due to applied load. It is observed that for both tensile load as well as compressive load on the spring, maximum shear stress always occurs at the inner side of the spring. Hence, failure of the spring, in the form of crake, is always initiated from the inner radius of the spring.

Test: Solid Mechanics- 4 - Question 2

A rigid bar AB of length ‘L’ is supported by a hinge and two springs of stiffness ‘K’ as shown in figure. The buckling load Pcr for the bar will be-

Detailed Solution for Test: Solid Mechanics- 4 - Question 2

F = Kδ and F1 = K × δ/2

Pδ = F × L + F1 × L/2
⇒ P × δ = Kδ × L + K × δ/2 × L/2
⇒ P = 5KL/4

Test: Solid Mechanics- 4 - Question 3

If 10 coils of spring cut out from a spring of 25 coils, then the stiffness of new spring will be

Detailed Solution for Test: Solid Mechanics- 4 - Question 3

⇒k′ = 2.5 k

Test: Solid Mechanics- 4 - Question 4

A bar of length L is supported by springs of stiffness Coefficient kA, kB, and a Torsional spring of torsional stiffness coefficient kT as shown in the figure below. The critical load is

Detailed Solution for Test: Solid Mechanics- 4 - Question 4

MA = 0

P x Δ = (KAΔ +KBΔ) L + KT θ

P = (KA+KB)L+KT/L

*Answer can only contain numeric values
Test: Solid Mechanics- 4 - Question 5

A column has a rectangular section of 10 mm x 20 mm & effective length of 1 m. The value of buckling stress (in MPa) is, given that E=200 GPa

Detailed Solution for Test: Solid Mechanics- 4 - Question 5

*Answer can only contain numeric values
Test: Solid Mechanics- 4 - Question 6

A weight of 40 N falls from a height of 200 mm on a rigid massless bar supported by two identical springs as shown in the figure. If the maximum deflection is 20 mm then the stiffness of spring (in N/mm) is ………..

Detailed Solution for Test: Solid Mechanics- 4 - Question 6

mg(h+x) = 2 × 1/2 kx2

40 ×(200+x) = kx2

40 ×(200+x) = k ×(20)2

k = 22 N/mm

Test: Solid Mechanics- 4 - Question 7

The euler’s crippling load for a column whose one end is fixed and other free is 60 kN. If one end is made fixed and other hinged, then the crippling load will

Detailed Solution for Test: Solid Mechanics- 4 - Question 7

End Condition - Effective length (Le)
Both end hinged - L
One end fixed other free - 2L
Both end fixed - 1/2
One end fixed and other hinged - L/√2

Calculation:
where Le = effective length of column

Effective length for 1st case = 2L, effective length for 2nd case = L/√2

Test: Solid Mechanics- 4 - Question 8

An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square : (σcirtical)circular’ is

Detailed Solution for Test: Solid Mechanics- 4 - Question 8

*Answer can only contain numeric values
Test: Solid Mechanics- 4 - Question 9

A column has a rectangular cross section of 40 cm × 15 cm and of flexural rigidity 225 Nm2. If the column gets buckle under the axial load of 1110 N. Calculate the slenderness ratio of the column.

Detailed Solution for Test: Solid Mechanics- 4 - Question 9

Given A = 0.15 × 0.4 m2

Flexural rigidity: EI =225 Nm2

Pe = 1110 N

We know
I = 1/12 × 0.153 × 0.4 = 1.125 × 10−4 m4

Slenderness Ratio = S = Le/k = 32.665

*Answer can only contain numeric values
Test: Solid Mechanics- 4 - Question 10

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 _______________.

Detailed Solution for Test: Solid Mechanics- 4 - Question 10

Δspring = Δ− ΔD

Compression of spring

(P – R) – R = 2R

P = 4R

R = P/4

% force carried by spring = 25%

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