Strength of Materials MCQ Quiz - Objective Question with Answer for Strength of Materials - Download Free PDF

Explanation:

Given, P = 120 N at C

Free body diagram

Let R_A and R_B are the reaction at the fixed end A and B. 

RA + R_B = P

Sign convetion: +ve(Tension), -ve(Compression)

As the Beam AB is fixed, So

Δ_AB = 0

Δ_AC + Δ_CB = 0

\(\frac{(P\ - \ R_B)L}{AE} \ +\ \frac{- R_B\times 2L}{AE} = 0\)

After solving,

R_B = P/3 = 120/3 = 40 N

R_A = P - R_B = 120 - 40

R_A = 80 N

Explanation:

Maximum shear stress theory (Guest & Tresca’s Theory):

• According to this theory, failure of the specimen subjected to any combination of a load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation:

• \({{\rm{\tau }}_{{\rm{max}}}} \le \frac{{{{\rm{\sigma }}_{\rm{y}}}}}{2}\) For no failure
• \({{\rm{\sigma }}_1} - {{\rm{\sigma }}_2} \le \left( {\frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{\rm{FOS}}}}} \right)\) For design
• σ1 and σ2 are maximum and minimum principal stress respectively.
• Here, τmax = Maximum shear stress
• σy = permissible stress
• This theory is justified but a conservative theory for ductile materials. It is an uneconomical theory. 

Additional Information

Maximum shear strain energy / Distortion energy theory / Mises – Henky theory: 

• It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.
• \(\frac{1}{2}\left[ {{{\left( {{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}} \right)}^2} + {{\left( {{{\rm{\sigma }}_2} - {{\rm{\sigma }}_3}} \right)}^2} + {{\left( {{{\rm{\sigma }}_3} - {{\rm{\sigma }}_1}} \right)}^2}} \right] \le {\rm{\sigma }}_{\rm{y}}^2\) for no failure
• \(\frac{1}{2}\left[ {{{\left( {{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}} \right)}^2} + {{\left( {{{\rm{\sigma }}_2} - {{\rm{\sigma }}_3}} \right)}^2} + {{\left( {{{\rm{\sigma }}_3} - {{\rm{\sigma }}_1}} \right)}^2}} \right] \le {\left( {\frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{\rm{FOS}}}}} \right)^2}\) For design

• It is the best suitable theory for ductile material.
• It cannot be applied to the material under hydrostatic pressure.​

Maximum principal stress theory (Rankine’s theory):

• According to this theory, the permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.
• For the design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.
• \({{\rm{\sigma }}_{1,2}} \le {{\rm{\sigma }}_{\rm{y}}}\) for no failure
• \({{\rm{\sigma }}_{1,2}} \le \frac{{\rm{\sigma }}}{{{\rm{FOS}}}}\) for design
• Note: For no shear failure τ ≤ 0.57 σy

Graphical representation:

• For brittle material, which does not fail by yielding but fail by brittle fracture, this theory gives a satisfactory result.
• The graph is always square even for different values of σ1 and σ2.

Maximum principal strain theory (ST. Venant’s theory):

• According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.
• \({\epsilon_{1,2}} \le \frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{{\rm{E}}_1}}}\) For no failure in uniaxial loading.
• \(\frac{{{{\rm{\sigma }}_1}}}{{\rm{E}}} - {\rm{\mu }}\frac{{{{\rm{\sigma }}_2}}}{{\rm{E}}} - {\rm{\mu }}\frac{{{{\rm{\sigma }}_3}}}{{\rm{E}}} \le \frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{\rm{E}}}\) For no failure in triaxial loading.
• \({{\rm{\sigma }}_1} - {\rm{\mu }}{{\rm{\sigma }}_2} - {\rm{\mu }}{{\rm{\sigma }}_3} \le \left( {\frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{\rm{FOS}}}}} \right)\) For design, Here, ϵ = Principal strain
• σ1, σ2, and σ3 = Principal stresses   

Graphical Representation:

This theory overestimates the elastic strength of ductile material.

Maximum strain energy theory (Haigh’s theory):

• According to this theory, a body complex stress fails when the total strain energy at the elastic limit in simple tension.
• Graphical Representation:
• \(\left\{ {{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2 + {\rm{\sigma }}_3^2 - 2{\rm{\mu }}\left( {{{\rm{\sigma }}_1}{{\rm{\sigma }}_2} + {{\rm{\sigma }}_2}{{\rm{\sigma }}_3} + {{\rm{\sigma }}_3}{{\rm{\sigma }}_1}} \right)} \right\} \le {\rm{\sigma }}_{\rm{y}}^2\)  for no failure
• \(\left\{ {{\rm{\sigma }}_1^2 + {\rm{\sigma }}_2^2 + {\rm{\sigma }}_3^2 - 2{\rm{\mu }}\left( {{{\rm{\sigma }}_1}{{\rm{\sigma }}_2} + {{\rm{\sigma }}_2}{{\rm{\sigma }}_3} + {{\rm{\sigma }}_3}{{\rm{\sigma }}_1}} \right)} \right\} \le {\left( {\frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{\rm{FOS}}}}} \right)^2}\) for design
• This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.

Important Points

• For Brittle material:- Maximum  Principal Stress Theory (Rankine criteria) is used.
• Maximum Shear Stress Theory (Tresca theory), Total strain energy theory, Maximum Distortion Energy Theory (von Mises) useful for a ductile material.
• Tresca's theory fails in the hydrostatic state of stresses.
• All theories will give the same results if loading is uniaxial.

Explanation:

Longitudinal Stress in a Thin Closed Cylinder:

• In the context of thin-walled pressure vessels, such as a thin closed cylinder containing a hydrostatic fluid, longitudinal stress refers to the stress experienced along the length of the cylinder due to the internal pressure exerted by the fluid. Understanding the nature of this stress is crucial for designing safe and efficient pressure vessels.
• When a thin-walled cylindrical vessel is subjected to internal hydrostatic fluid pressure, it experiences stress in both the longitudinal (axial) and circumferential (hoop) directions. The longitudinal stress is the stress along the length of the cylinder, and it is caused by the internal pressure pushing the ends of the cylinder apart.

In a thin closed cylinder with internal fluid pressure, longitudinal stress arises due to pressure acting on the end caps, trying to pull them apart.

Hence, the nature of longitudinal stress is tensile.

Formulas:

• Hoop stress: \( \sigma_h = \frac{p d}{2 t} \)
• Longitudinal stress: \( \sigma_l = \frac{p d}{4 t} \)

Explanation:

Thermal Expansion of a Steel Bar

• Thermal expansion refers to the increase in the dimensions of a material when it is subjected to a temperature rise. For a linear material like a steel bar, this expansion can be calculated using the coefficient of linear expansion.

Given Data:

• Coefficient of linear expansion (α) = 12 × 10⁻⁶/°C
• Original length of the steel bar (L₀) = 15 cm = 150 mm
• Expansion of the bar (ΔL) = 0.3 mm

Formula:

The formula for linear expansion is given by:

ΔL = α × L₀ × ΔT

Where:

• ΔL = Change in length (0.3 mm)
• α = Coefficient of linear expansion (12 × 10⁻⁶/°C)
• L₀ = Original length (150 mm)
• ΔT = Change in temperature (°C)

Calculation:

Rearranging the formula to solve for ΔT:

ΔT = ΔL / (α × L₀)

Substituting the given values:

ΔT = 0.3 / (12 × 10⁻⁶ × 150)

ΔT = 0.3 / (1.8 × 10⁻³)

ΔT = 166.67 °C

Explanation:

Charpy Impact Test

• The Charpy impact test, also known as the Charpy V-notch test, is a standardized high strain-rate test that determines the amount of energy absorbed by a material during fracture. This absorbed energy is a measure of the material's toughness and acts as an indicator of its ability to resist brittle fracture.

Working Principle: In the Charpy impact test, a notched specimen is struck by a swinging pendulum hammer at a fixed speed. The specimen is typically supported at both ends, and the hammer impacts the specimen at the notch. The energy absorbed by the specimen during fracture is measured by the height to which the pendulum rises after breaking the specimen. This energy is directly related to the toughness of the material.

Procedure:

• A rectangular bar specimen, usually with a V-notch or U-notch in the middle, is prepared according to standardized dimensions.
• The specimen is placed horizontally between two supports in the testing machine.
• The pendulum hammer is released from a known height to strike the specimen at the notch.
• The pendulum swings through the specimen, breaking it and rising to a height determined by the energy absorbed during fracture.
• The difference in the height of the pendulum before and after the impact is used to calculate the absorbed energy.

Explanation:

Ductile material fails along the principal shear plane as they are weak in shear and brittle material fails along with principal normal stress.

In Brittle materials under tension test undergoes brittle fracture i.e their failure plane is 90° to the axis of load and there is no elongation in the rod that’s why the diameter remains same before and after the load. Example: Cast Iron, concrete etc.

But in case of ductile materials, material first elongates and then fail, their failure plane is 45° to the axis of the load. After failure cup-cone failure is seen. Example Mild steel, high tensile steel etc.

Explanation:

Perfectly Plastic Material:

For this type of material, there will be only initial stress required and then the material will flow under constant stress.

The chart shows the relation between stress-strain in different materials. 

Stress-Strain Curve	Type of Material or Body	Examples
	Rigidly Perfectly Plastic Material	No material is perfectly plastic
	Ideally plastic material.	Visco-elastic (elasto-plastic) material.
	Perfectly Rigid body	No material or body is perfectly rigid.
	Nearly Rigid body	Diamond, glass, ball bearing made of hardened steel, etc
	Incompressible material	Non-dilatant material, (water) ideal fluid, etc.
	Non-linear elastic material	Natural rubbers, elastomers, and biological gels, etc

Explanation:

• Change in the temperature causes the body to expand or contract.
• Thermal stress is created when a change in size or volume is constrained due to a change in temperature.
• So an increase in temperature creates compressive stress and a decrease in temperature creates tensile stress.

Explanation:

ϵv = ϵx + ϵy + ϵz

\( {ϵ_x} = \frac{1}{E}\left[ {{σ _x} - ν \left( {{σ _y} + {σ _z}} \right)} \right] \)

\({ϵ_{\rm{y}}} = \frac{1}{{\rm{E}}}\left[ {{σ _y} - ν \left( {{σ _x} + {σ _z}} \right)} \right] \)

\({ϵ_{\rm{z}}} = \frac{1}{{\rm{E}}}\left[ {{σ _z} - ν \left( {{σ _x} + {σ _y}} \right)} \right]\)

Total strain or volumetric strain is given by 

\( {ϵ_v} = \frac{1}{E} [ {σ_x} + {σ_y} + {σ_z} ](1-2ν) \)

There will be no change in volume if volumetric strain is zero.

ϵ_v = 0 ⇒  ν = 0.5

Explanation:

Since all the faces are free to expand the stresses due to temperature rise is equal to 0.

If the cube is constrained on all six faces, the stress produced in all three directions will be the same.

∴ thermal strain in x-direction = -α(ΔT) = \(\frac{{{\sigma _x}}}{E} - \nu \frac{{{\sigma _y}}}{E} - \nu \frac{{{\sigma _z}}}{E}\)

σ_x = σ_y = σ_z = σ

\(\sigma = - \frac{{\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2\nu } \right)}}\)

Concept:

Let RA and RB be the reaction at support A and B respectively.

Free body diagram of the system is:

\(R_A=\frac{Pb}{L}\;\&\;R_B=\frac{Pa}{L}\)

Calculation:

Given:

As per figure P = 10 kN, a = 1 m and b = 2 m.

\(R_A=\frac{Pb}{L}\)

\(R_A=\frac{10\times2}{3}=\frac{20}{3}\;kN\)

\(R_B=\frac{10\times1}{3}=\frac{10}{3}\;kN\)

Explanation:-

Resilience

• The total strain energy stored in a body is commonly known as resilience. Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence resilience is also defined as the capacity of a strained body for doing work on the removal of the straining force.
• It is the property of materials to absorb energy and to resist shock and impact loads.
• It is measured by the amount of energy absorbed per unit volume within elastic limit this property is essential for spring materials.
• The resilience of material should be considered when it is subjected to shock loading.

Proof resilience

• The maximum strain energy, stored in a body, is known as proof of resilience. The strain energy stored in the body will be maximum when the body is stressed upto the elastic limit. Hence the proof resilience is the quantity of strain energy stored in a body when strained up to the elastic limit.
• It is defined as the maximum strain energy stored in a body.
• So, it is the quantity of strain energy stored in a body when strained up to the elastic limit (ability to store or absorb energy without permanent deformation).

Modulus of resilience

• It is defined as proof resilience per unit volume.
• It is the area under the stress-strain curve up to the elastic limit.

Toughness:

• Toughness is defined as the ability of the material to absorb energy before fracture takes place.
• This property is essential for machine components that are required to withstand impact loads.
• Tough materials have the ability to bend, twist or stretch before failure takes place.
• Toughness is measured by a quantity called modulus of toughness. Modulus of toughness is the total area under the stress-strain curve in a tension test.
• Toughness is measured by Izod and Charpy impact testing machines.
• When a material is heated it becomes ductile or simply soft and thus less stress is required to deform the material and the stress-strain curve will shift down and the area under the curve decreases thus toughness decreases.
• Toughness decreases as temperature increases.

Hardness:

• Hardness is a measure of the resistance to localized plastic deformation induced by either mechanical indentation or abrasion. 
• Hardness Testing measures a material’s strength by determining resistance to penetration.
• There are various hardness test methods, including Rockwell, Brinell, Vickers, Knoop and Shore Durometer testing.

When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue.

The slow and continuous elongation of a material with time at constant stress and high temperature below the elastic limit is called creep.

Explanation:

Elastic recovery/strain: The strain recovered after the removal of the load is known as elastic strain.

Plastic strain: The permanent changes in dimension after the removal of load is known as plastic strain.

The load is removed when the stress was 200 MPa and the corresponding strain was 0.03

After the removal of load, the body recovered and the final strain found was 0.01.

∴ Elastic strain = 0.03 - 0.01 ⇒ 0.02 and Plastic strain = 0.01 respectively.

Concept:

Stress at any section of the bar is given by,

\(stress, \sigma =\frac{{Load ~(P)}}{{Cross-sectional ~area~(A)}}\)

Calculation:

Given:

Load in section BC, P = 30 kN (compressive), 

cross-sectional area, A = 15 m² = 15 × 10⁶ mm²

\(stress~ in ~section ~BC, \sigma =\frac{{30~\times~10^3}}{{15~\times ~10^6}}=0.002~N/mm^2\)

Explanation:

Strain energy:

When a body is subjected to gradual, sudden, or impact load, the body deforms and work is done upon it. If the elastic limit is not exceeded, this work is stored in the body. This work-done or energy stored in the body is called strain energy.

Strain energy = Work done.

Case-I:

When a rigid body is slowly dropped on another body, it is a case of gradual loading:

Work-done on the bar = Area of load-deformation diagram \(⇒\frac{1}{2}\;×\;P\;×\;δ l\)

Work stored in the bar = Area of resistance deformation diagram

\(⇒\frac{1}{2}\;×\;R\;×\;δ l\)

\(⇒\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\;\;\;[\because R=\sigma A]\)

We can write;

\(⇒\frac{1}{2}\;×\;P\;×\;δ l=\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\)

\(\sigma_{gradual}=\frac{P}{A}\)

Case-II:

Work-done on the bar = Area of load-deformation diagram ⇒ P × δl

Work stored on the bar = Area of resistance deformation diagram

\(⇒\frac{1}{2}\;×\;R\;×\;δ l\)

\(⇒\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\;\;\;[\because R=\sigma A]\)

We can write;

\(P\times\delta l=\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\)

\(\sigma_{sudden}=\frac{2P}{A}\)

\(\therefore \frac{\sigma_{sudden}}{\sigma_{gradual}}=2\)

∴ maximum stress intensity due to suddenly applied load is twice the stress intensity produced by the load of the same magnitude applied gradually.

Impact Loading:

When a load is dropped from a height before it commences to load the body, such loading is known as Impact loading.

The ratio of the stress or deflection produced due to impact loading to the stress or deflection produced due to static or gradual loading is known as the Impact factor.

\(IF=\frac{\sigma_{impact}}{\sigma_{gradual}}=\frac{\Delta_{impact}}{\Delta_{gradual}}\)

\(IF=\frac{\sigma_{sudden}}{\sigma_{gradual}}=\frac{\Delta_{sudden}}{\Delta_{gradual}}=2\)

∴ deflection due to sudden loading is twice that of gradual loading.