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Mechanical Engineering - Practice Test - 28

Q1:

If two springs of stiffness k1  and  k2are connected in series, the stiffness of the equivalent spring is

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IBS
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Q2:

Two closed springs of stiffness "k" and "2k" are arranged in series in one case and in parallel in the other case. The ratio of stiffness of springs connected in series to parallel is -

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Strength of Materials
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Q3:

If three close -coiled and two open - coiled helical springs, each having the stiffness K are connected in series then the overall stiffness is - 

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Q4:

Two co-axial springs are subjected to a force of 1 KN. Springs constant of larger diameter springs is 80 N/mm and that of smaller diameter spring is 120 N/mm. The deformation in the spring combination will be equal to -

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Strength of Materials
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Q5:

What is the equivalent spring stiffness for the system of springs shown in the figure given below?

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Strength of Materials
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Q6:

A close helical spring of 100 mm mean diameter is made of 10 mm diameter rod, and has 20 turns, The spring carries an axial load of 200 KN with G=8.4×104N/mm2.The stiffness of the spring is nearly

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Strength of Materials
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Q7:

A closely coiled helical spring of round steel wire 5 mm in diameter having 12 complete coils of 50 mm mean diameter is subjected to an axial load of 100 N. Modulus of rigidity of the spring is 80 KN/mm2. What is the deflection of the spring?

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Strength of Materials
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Q8:

Two closely coiled helical springs, A and B are equal in all respects but for the number of turns, with A having just half the number of turns of that of B. What is the ratio of deflections in terms of spring A to spring B?

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Strength of Materials
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Q9:

A closed coiled spring is cut into two identical halves. The stiffness of each of the resulting springs will -

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Strength of Materials
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Q10:

Which one of the following pairs is not correctly matched?

Boundary conditions of column

Euler’s buckling load

a) pin – pin

π2EIl2

b) fixed – fixed

4π2EIl2

c) fixed – free

0.25π2EIl2

d)fixed – pin

2π2EIl2

 

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Strength of Materials
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Q11:

For a circular column having its ends hinged, the slenderness ratio is 160. The (l/d) ratio of the column is

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Strength of Materials
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Q12:

Assertion (A) :  The buckling load for a column of specified material, cross – section and end conditions calculated as per euler’s formula varies inversely with the column length.

Reason (R) : Euler’s formula takes into account the end conditions in determining the effective length of column.

Of these statements -

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Strength of Materials
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Q13:

The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends is equal to

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Strength of Materials
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Q14:

A circular column of length 2 m has euler’s crippling load of 1.5 KN. If the diameter of the column is reduced by 10% the reduction in the crippling load will be

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Strength of Materials
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Q15:

The dimensions ration of a compressive member in the context of rankine’s formula is defined as -

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Strength of Materials
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Q16:

If the euler load for a steel column is 1000 KN and crushing load is 1500 KN, the rankine load is equal to

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Strength of Materials
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Q17:

The strain energy per unit volume of a round bar under uniaxial tension with stress σand modulus of elasticity E is

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SOM_CA1_Set1
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Strength of Materials
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Q18:

A cantilever beam, 2m in length, is subjected to a uniformly distributed load of 5 kN/m. If E = 200 GPa and I = 1000 cm4, the strain energy stored in the beam will be

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Strength of Materials
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Q19:

A power transmission solid shaft of diameter d, length l and rigidity modulus G is subjected to a pure torque. The maximum allowable shear stress isτmax. The maximum strain energy unit volume in the shaft is given by:

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Strength of Materials
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Q20:

Match List-I 3with List-II and select the correct answer using the code given below the lists:

List-I

List-II

A. Point of inflection

1. Strain energy

B. Shearing strain

2. Equation of bending

C. Section modulus

3. Equation of torsion

D. Modulus of resilience

4. Bending moment diagram

Code:

ABCD

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3rdyrbaselinenew
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Strength of Materials
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Q21:

A steel specimen 150 mm2 in cross-section stretches by 0.05 mm over a 50 mm gauge length under an axial load of 30 kN. What is the strain energy stored in the specimen?

(Take E = 200 GPa)

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PlaceMe2_SOM1
NEAT_SOM_1
LPU_CA_2020_Set1
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Strength of Materials
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Q22:

The maximum distortion energy theory of failure is suitable to predict the failure of which one of the following types of materials?

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Strength of Materials
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Q23:

A rod of cross- sectional area 100×10−6m2is subjected to a tensile load. Based on the Tresca failure criterion. If the uniaxial yield stress of the material is 200 MPa, the failure load is

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Strength of Materials
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Q24:

A circular solid shaft is subjected to bending moment of 400 kN-m and a twisting moment of 300 kN-m. On the basis of the maximum principal stress theory, the direct stress is σand according to the maximum shear stress theory, the shear is τ. The ratio σ/τis

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Strength of Materials
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Q25:

Permissible bending moment in a circular shaft under pure bending is M according to maximum principal stress theory of failure. The permissible bending moment in the same shaft  as per maximum shear stress of failure is

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Strength of Materials
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