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Strength of Materials - 1 - Mechanical Engineering MCQ


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20 Questions MCQ Test - Strength of Materials - 1

Strength of Materials - 1 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Strength of Materials - 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Strength of Materials - 1 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Strength of Materials - 1 below.
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Strength of Materials - 1 - Question 1

For the state of plane stress shown in the figure, the maximum and minimum principal stresses are:

Detailed Solution for Strength of Materials - 1 - Question 1

Maximum and minimum principal stresses:

Strength of Materials - 1 - Question 2

Bending moment M and torque T is applied on a solid circular shaft. If the maximum bending stress equals to maximum shear stress developed, then M is equal to

Detailed Solution for Strength of Materials - 1 - Question 2

Bending stress due to bending moment:

Shear stress due to twisting moment/ torque:

If the maximum bending stress equals to maximum shear stress developed:

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Strength of Materials - 1 - Question 3

When bending moment M and torque T is applied on a shaft then equivalent torque is

Detailed Solution for Strength of Materials - 1 - Question 3

In some applications the shaft are simultaneously subjected to bending moment M and Torque T.

From the simple bending theory equation:

If σb is the maximum bending stresses due to bending moment M on shaft:

Torsion equation:

The maximum shear stress developed on the surface of the shaft due to twisting moment T:

Equivalent Bending Moment:

Equivalent Torque:

Strength of Materials - 1 - Question 4

The bending moment diagram for the case shown below in Figure will be:

Detailed Solution for Strength of Materials - 1 - Question 4

Here RA = RB = W

Between A to C:

Mx=W×x⇒Ma=0,Mc=WaMx=W×x⇒Ma=0,Mc=Wa

Between C to D:

Mx=W×x−W(x−a)⇒MD=W×4a−W(4a−a)=WaMx=W×x−W(x−a)⇒MD=W×4a−W(4a−a)=Wa

Between D to B:

Mx=W×x−W(x−a)−W(x−4a)⇒MB=W×5a−W(5a−a)−W(5a−4a)=0

Strength of Materials - 1 - Question 5

If a prismatic bar be subjected to an axial tensile stress σ, then shear stress induced on a plane inclined at θ with the axis will be

Detailed Solution for Strength of Materials - 1 - Question 5

Pn = The component of force P, normal to section FG = P cos θ

Pt = The component of force P, along the surface of the section FG (tangential to the surface FG) = P sin θ

σn = Normal Stress across the section FG

σn=σcos2θσn=σcos2θ

σt = Tangential/Shear Stress across the section FG

Strength of Materials - 1 - Question 6

Select the proper sequence

1. Proportional Limit       2. Elastic limit

3. Yielding                      4. Failure

Detailed Solution for Strength of Materials - 1 - Question 6
  • So it is evident form the graph that the strain is proportional to stress or elongation is proportional to the load giving a straight line relationship. This law of proportionality is valid upto a point A. Point A is known as the limit of proportionality or the proportionality limit.
  • For a short period beyond the point A, the material may still be elastic in the sense that the deformations are completely recovered when the load is removed. The limiting point B is termed as Elastic Limit.
  • Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus permanent deformation or permanent set when load is removed. These two points are termed as upper and lower yield points respectively. The stress at the yield point is called the yield strength.
  • A further increase in the load will cause marked deformation in the whole volume of the metal. The maximum load which the specimen can with stand without failure is called the load at the ultimate strength. The highest point ‘E' of the diagram corresponds to the ultimate strength of a material.
  • Beyond point E, the bar begins to form neck. The load falling from the maximum until fracture occurs at F.
Strength of Materials - 1 - Question 7

In the case of bi-axial state of normal stresses, the normal stress on 45° plane is equal to

Detailed Solution for Strength of Materials - 1 - Question 7

In the case of bi-axial state of normal stresses:

Normal stress on an inclined plane:

Shear Stress on an inclined plane:

Normal stress on 45° plane:

Strength of Materials - 1 - Question 8

The temperature stress is a function of

1. Coefficient of linear expansion

2. Temperature rise

3. Modulus of elasticity 

Detailed Solution for Strength of Materials - 1 - Question 8

Thermal strain ϵis proportional to the temperature change ΔT

ϵT=αΔTϵT=αΔT

α is coefficient of thermal expansion.

When there is some restriction to the bar to expand, thermal stress will generate in the material:

σT=strain×E=α.ΔT.E

Strength of Materials - 1 - Question 9

A steel rod of 1 sq. cm. cross sectional area is 100 cm long and has a Young’s modulus of elasticity 2 × 106 kgf/cm2. It is subjected to an axial pull of 2000 kgf. The elongation of the rod will be

Detailed Solution for Strength of Materials - 1 - Question 9

Change in length:

Strength of Materials - 1 - Question 10

Assume that young's modulus of a wire of length L and radius 'r' is Y. If length is reduced to L/2 and radius r/2 then what will be its young's modulus?

Detailed Solution for Strength of Materials - 1 - Question 10

Young's is a property of the material of construction of a wire. It depends only on the nature of the material used for making the wire.

Young's modulus does not depend on the physical dimensions such as the length and diameter of wire. Thus there will be no change in Young's modulus.

Strength of Materials - 1 - Question 11

Circumferential and longitudinal strains in cylindrical boiler under internal steam pressure, are ε1 and ε2 respectively. Change in volume of the boiler cylinder per unit volume will be

Detailed Solution for Strength of Materials - 1 - Question 11

Strength of Materials - 1 - Question 12

A metal pipe of 1 m diameter contains a fluid having a pressure of 10 kgf/cm2. If the permissible tensile stress in the metal is 200 kgf/cm2, then the thickness of the metal required for making the pipe would be

Detailed Solution for Strength of Materials - 1 - Question 12

As in thin cylinder shell:

Longitudinal stress = Half of circumferential stress

∴ circumferential stress ≤ Permissible stress

Strength of Materials - 1 - Question 13

Hoops stress and longitudinal stress in a boiler shell under internal pressure are 100 MN/m2 and 50 MN/m2 respectively. Young’s modulus of elasticity and Poisson’s ratio of the shell material are 200 GN/m2 and 0.3 respectively. The hoop strain in boiler shell is

Detailed Solution for Strength of Materials - 1 - Question 13

Strength of Materials - 1 - Question 14

A beam cross-section is used in two different orientations as shown in figure given below:

Detailed Solution for Strength of Materials - 1 - Question 14

From the simple bending theory equation

If σb is the maximum bending stresses due to bending moment M on shaft:

For same bending moment:

 

 

 

 

Strength of Materials - 1 - Question 15

The curve ABC is the Euler’s curve for stability of column. The horizontal line DEF is the strength limit. With reference to this figure Match List-I and select the correct answer using the codes below the lines

Detailed Solution for Strength of Materials - 1 - Question 15

Slenderness ratio is the ratio of the length of a column and the least radius of gyration of its cross section. It is used extensively for finding out the design load as well as in classifying various columns in short/intermediate/long.

Long column: buckling occurs elastically before the yield stress is reached.

Short column: material failure occurs inelastically beyond the yield stress.

 

Strength of Materials - 1 - Question 16

With one fixed end and other free end, a column of length L buckles at load, P1. Another column of same length and same cross-section fixed at both ends buckles at load P2. Value of P2/P1 is

Detailed Solution for Strength of Materials - 1 - Question 16

The maximum load at which the column tends to have lateral displacement or tends to buckle is known as buckling or crippling load. Load columns can be analysed with the Euler’s column formulas can be given as

  • For both end hinged, n = 1
  • For one end fixed and other free, n = ½
  • For both end fixed, n = 2
  • For one end fixed and other hinged, n = √2
Strength of Materials - 1 - Question 17

A body having weight of 1000 N is dropped form a height of 10 cm over a close-coiled helical spring of stiffness 200 N/cm. The resulting deflection of spring is nearly

Detailed Solution for Strength of Materials - 1 - Question 17

100 +10x - x2 = 0

x2 -10x - 100 = 0

x = 16 cm

Strength of Materials - 1 - Question 18

The diameter of shaft A is twice the diameter of shaft B and both are made of the same material. Assuming both the shafts to rotate at the same speed, the maximum power transmitted by B is

Detailed Solution for Strength of Materials - 1 - Question 18

Strength of Materials - 1 - Question 19

A bar having length L and uniform cross-section with area A is subjected to both tensile force P and torque T. If G is the shear modulus and E is the Young’s modulus, the internal strain energy stored in the bar is

Detailed Solution for Strength of Materials - 1 - Question 19

The elastic strain energy stored in a member of length s (it may be curved or straight) due to axial force, bending moment, shear force and torsion is summarized below:

Strain energy due to combined effect of load P and torque T:

Strength of Materials - 1 - Question 20

Resilience for mild steel under uni-axial tensile loading is given by the shaded portion of the stress-strain diagram as shown in

Detailed Solution for Strength of Materials - 1 - Question 20

The ability of a material to absorb energy in the elastic region. This is given by the strain energy per unit volume which is the area of the elastic region.

 

 Ability to absorb energy in the plastic range. This is given by the total area under the stress-strain curve.

 

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