Stress and Strain Caused by Axial Loads
Stress
Learning Objectives:

Understand basic concepts of stress, strain and their relations based on linear elasticity.

Study material behaviors due to different types of loading.

Understand and know how to calculate stresses and deformation of a bar due to an axial loading under uniform and nonuniform conditions.

Understand basic stressstrain response of engineering materials.

Calculate stresses and deformation of a bar due to an axial loading.

When a structural member is under load, predicting its ability to withstand that load is not possible merely from the reaction force in the member.

It depends upon the internal force, cross sectional area of the element and its material properties.

Thus, a quantity that gives the ratio of the internal force to the cross sectional area will define the ability of the material in with standing the loads in a better way.

That quantity, i.e., the intensity of force distributed over the given area or simply the force per unit area is called the stress.
\[\mathrm{stress} = \dfrac{F}{A}\] In SI units, force is expressed in newtons (N) and area in square meters. Consequently, the stress has units of newtons per square meter (N/m\(^2\)) or Pascals (Pa). The stress defined in the above equation is obtained by dividing the force by the cross sectional area and hence it represents the average value of the stress over the entire cross section.
Solved Example: 41101
______ is defined as load per unit area.(MP Sub Engg Sep 2018  Shift II)
A. Strain
B. Rigidity
C. Stress
D. Pressure
Correct Answer: C
Strain
Learning Objectives:

Summarize normal strain under axial loading.

Calculate deformations of members under axial loading.

The structural member and machine components undergo deformation as they are brought under loads.

To ensure that the deformation is within the permissible limits and do not affect the performance of the members, a detailed study on the deformation assumes significance.

A quantity called strain defines the deformation of the members and structures in a better way than the deformation itself and is an indication on the state of the material.
Consider a rod of uniform cross section with initial length as \(L_0\). Application of a tensile load P at one end of the rod results in elongation of the rod by \(\delta\).
After elongation, the length of the rod is L. As the cross section of the rod is uniform, it is appropriate to assume that the elongation is uniform throughout the volume of the rod. If the tensile load is replaced by a compressive load, then the deformation of the rod will be a contraction.
The deformation per unit length of the rod along its axis is defined as the normal strain. It is denoted by \(\epsilon\).
\[\mathrm{strain} = \frac{\Delta L}{L_0} = \frac{L  L_0}{L_0}\]
Though the strain is a dimensionless quantity, units are often given in \(mm/mm\), \(\mu m/m\).
Solved Example: 41201
A steel rod of original length 200 mm and final length of 200.2 mm after application of an axial tensile load of 20 kN, what will be the strain developed in the rod? (SSC JE ME Jan 2018 Morning)
A. 0.01
B. 0.1
C. 0.001
D. 0.0001
Correct Answer: C
Poissons Ratio
Learning Objectives:

Describe and define Poisson’s ratio.
Consider a rod under an axial tensile load P such that the material is within the elastic limit. The normal stress on x plane is \(\sigma_{xx} = \dfrac{P}{A}\) and the associated longitudinal strain in the x direction can be found out from \(\epsilon_x = \dfrac{\sigma_{xx}}{E}\). As the material elongates in the x direction due to the load P, it also contracts in the other two mutually perpendicular directions, i.e., y and z directions. Hence, despite the absence of normal stresses in y and z directions, strains do exist in those directions and they are called lateral strains.
The ratio between the lateral strain and the axial/longitudinal strain for a given material is always a constant within the elastic limit and this constant is referred to as Poisson’s ratio.
It is denoted by \(nu\). Since the axial and lateral strains are opposite in sign, a negative sign is introduced in the definition to make \(\nu\) positive. \[\nu =  \dfrac{\mathrm{lateral\ strain}}{\mathrm{longitudinal\ strain}}\]
Poisson’s ratio can be as low as 0.1 for concrete and as high as 0.5 for rubber. In general, it varies from 0.25 to 0.35 and for steel it is about 0.3.
Young’s modulus = E = \(\dfrac{\mathrm{tensile\ stress}}{\mathrm{tensile\ strain}}\)
Young’s modulus = E = \(\dfrac{\mathrm{compressive\ stress}}{\mathrm{compressive\ strain}}\)
Shear modulus = G = \(\dfrac{\mathrm{shear\ stress}}{\mathrm{shear\ strain}}\)
Bulk modulus = K = \(\dfrac{\mathrm{Volumetric\ stress}}{\mathrm{volumetric\ strain}}\)
K= \(\dfrac{\mathrm{pressure}}{\mathrm{volumetric\ strain}}\) = \(\dfrac{p}{(\delta V/ V)}\)
Solved Example: 41301
The value of Poisson’s ratio for steel is between:
A. 0.01 to 0.1
B. 0.18 to 0.20
C. 0.25 to 0.33
D. 0.4 to 0.6
Correct Answer: C
Solved Example: 41302
Poisson’s ratio is defined as the ratio of:
A. Longitudinal stress and longitudinal strain
B. Longitudinal stress and lateral stress
C. Lateral stress and longitudinal stress
D. Lateral stress and lateral strain
Correct Answer: C
Solved Example: 41303
A bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension on gauge length of 200 mm is 0.1 mm and change in diameter is 0.004 mm. Calculate Poisson’s ratio.
A. 0.267
B. 0.733
C. 3.75
D. 2.75
Correct Answer: A
Solved Example: 41304
Poisson's ratio of a material is 0.5. Percentage change in its length is 0.04%. What is the change in percentage of diameter?
A. 0.04%
B. 0.03%
C. 0.02%
D. 0.01%
Correct Answer: C
Solved Example: 41305
Which of the following statements is NOT true?
A. Coefficient of thermal expansion is independent of original beam length.
B. Since wood is not a metal, its Young's Modulus is dependent on the plastic zone elongation
C. While designing a beam, subjected to pure bending, shear stresses are assumed to be negligible.
D. Concrete has different material characteristics in tension and compression.
Correct Answer: B
Solved Example: 41306
Which of the following describes the concept of Poisson's ratio most accurately?
A. Ratio of change in volume to original volume
B. Ratio of lateral stress to lateral strain
C. Ratio of lateral strain to longitudinal strain
D. Ratio of Bulk Modulus to Modulus of Elasticity
Correct Answer: C
Solved Example: 41307
What is the unit of the modulus of elasticity?
A. Nm
B. Unitless
C. Pa
D. Nm/s
Correct Answer: C
Solved Example: 41308
Within elastic limit, the volumetric strain is proportional to the hydrostatic stress. What is the constant that relates these two quantities called?
A. Modulus of rigidity
B. Modulus of elasticity
C. Young's modulus
D. Bulk modulus
Correct Answer: D
Solved Example: 41309
What is another term for modulus of rigidity?
A. Shear modulus
B. Young’s modulus
C. Bulk modulus
D. Modulus of elasticity
Correct Answer: A
Solved Example: 41310
The ratio of lateral strain to the linear strain within elastic limit is known as:
A. Young’s modulus
B. Modulus of rigidity
C. Modulus of elasticity
D. Poisson’s ratio
Correct Answer: D
Solved Example: 41311
Young’s modulus is defined as the ratio of:
A. Volumetric stress and volumetric strain
B. Lateral stress and lateral strain
C. Longitudinal stress and longitudinal strain
D. Shear stress to shear strain
Correct Answer: C
Solved Example: 41312
The materials having same elastic properties in all directions are called:
A. Ideal materials
B. Uniform materials
C. Isotropic materials
D. Practical materials
Correct Answer: C
Solved Example: 41313
The value of modulus of elasticity for mild steel is of the order of:
A. 2.1 $\times$ l0$^5$ kg/cm$^2$
B. 2.1 $\times$ 10$^6$ kg/cm$^2$
C. 2.1 $\times$ 10$^7$ kg/cm$^2$
D. 0.1 $\times$ 10$^6$ kg/cm$^2$
Correct Answer: B
Solved Example: 41314
A metallic rod of 500 mm length and 50 mm diameter, when subjected to a tensile force of 100 kN at the ends, experiences, an increase in its length by 0.5 mm and a reduction in its diameter by 0.015 mm. The Poisson's ratio?
A. 3.33
B. 0.3
C. 0.2
D. 0.25
Lateral strain $=\dfrac {0.015}{50}=0.003$
Longitudinal strain $=\dfrac {0.5}{500}=0.001$
Poisson's ratio $\nu =\dfrac {0.0003}{0.001}=0.3$
Correct Answer: B
Elastic Constants
Learning Objectives:

Summarize Hooke’s Law.

Discuss engineering/elastic constants and their relationship to each other.
E, G, K and \(\mu\) are called elastic constants because these are applicable only within elastic limit and their values are constant for a particular material within elastic limits.
Modulus of Elasticity or Young’s Modulus:
A tensile test is generally conducted on a standard specimen to obtain the relationship between the stress and the strain which is an important characteristic of the material. In the test, the uniaxial load is applied to the specimen and increased gradually. The corresponding deformations are recorded throughout the loading. Stressstrain diagrams of materials vary widely depending upon whether the material is ductile or brittle in nature. If the material undergoes a large deformation before failure, it is referred to as ductile material or else brittle material.
Initial part of the loading indicates a linear relationship between stress and strain, and the deformation is completely recoverable in this region for both ductile and brittle materials.
This linear relationship, i.e., stress is directly proportional to strain, is popularly known as Hooke’s law. \[E = \frac{\mathrm{stress}}{\mathrm{strain}}\] The coefficient E is called the modulus of elasticity or Young’s modulus.
Most of the engineering structures are designed to function within their linear elastic region only.
Relation between Elastic Constants:
\[E = 2G (1 + \nu)\] \[E = 3K (1  2\nu)\] \[E = \dfrac{9KG}{(3K+G)}\]
Factor of Safety
Learning Objectives:

Understand the importance of selection of Factor of Safety in Design.
A good design of a structural element or machine component should ensure that the developed product will function safely and economically during its estimated life time. The stress developed in the material should always be less than the maximum stress it can withstand which is known as ultimate strength. During normal operating conditions, the stress experienced by the material is referred to as working stress or allowable stress or design stress.
The ratio of ultimate strength to allowable stress is defined as factor of safety. For brittle materials, \[\mathrm{Factor\ of\ safety} = \frac{\mathrm{Ultimate\ tensile \ strength}}{\mathrm{Maximum\ working\ stress}}\] For ductile materials, \[\mathrm{Factor\ of \ safety} = \frac{\mathrm{Yield \ stress}}{\mathrm{maximum\ working\ stress}}\] Factor of safety take care of the uncertainties in predicting the exact loadings, variation in material properties, environmental effects and the accuracy of methods of analysis. If the factor of safety is less, then the risk of failure is more and on the other hand, when the factor of safety is very high the structure becomes unacceptable or uncompetitive.
Hence, depending upon the applications the factor of safety varies. It is common to see that the factor of safety is taken between 2 and 3.
Solved Example: 41501
What can understand by the factor of safety equal to one?
A. It means that the structure will fail under load
B. It means that the structure will only support the actual load
C. It means that the structure will support more than the actual load
D. There is no relation between factor safety and load application
When the factor of safety is one it means that the ultimate stress is equal to the working stress and therefore the body can only support load up to actual load and no more before failing
Correct Answer: B
Solved Example: 41502
Which of the following is the numerator of factor of safety formula?
A. Working stress
B. Shear stress
C. Tensile stress
D. Ultimate stress
Factor of safety is defined as ratio of ultimate stress and working stress. It is also called as factor of ignorance. The factor of safety is dependent on the type of load.
Correct Answer: D
Solved Example: 41503
Factor of safety in a ductile material is:
A. Design stress/working stress
B. Ultimate stress / allowable stress
C. Yield stress/allowable stress
D. None
Correct Answer: C
Solved Example: 41504
In a brittle material, factor of safety is:
A. Design stress/working stress
B. Ultimate stress / allowable stress
C. Yield stress/allowable stress
D. None
Correct Answer: B
Solved Example: 41505
A 30m long Aluminum bar is subjected to a tensile stress of 175 MPa. Determine the elongation if E = 69116 MPa.
A. 78 mm
B. 76 mm
C. 74 mm
D. 72 mm
Correct Answer: B
Solved Example: 41506
High strength steel band saw, 20 mm wide and 0.8 mm thick runs over the pulley 600 mm in diameter of pulleys can be used without exceeding the flexural stress of 400 MPa? Note: E = 200 GPa.
A. 250 cm
B. 325 mm
C. 400 mm
D. 150 in
Correct Answer: C
Solved Example: 41507
A hollow shaft has an inner diameter of 0.035 m and an outer diameter of 0.06 m. Compute the torque if the shear stress is not exceed 120 MPa.
A. 4,500 Nm
B. 4,300 Nm
C. 5,500 Nm
D. 3,450 Nm
Correct Answer: A
Solved Example: 41508
A thin mild steel wire is loaded by adding loads in equal increments till it breaks. The extensions noted with increasing loads will behave as under:
A. Uniform throughout
B. Increase uniformly
C. First increase and then decrease
D. Increase uniformly first and then increase rapidly
Correct Answer: D
Solved Example: 41509
Modulus of rigidity is defined as the ratio of:
A. Longitudinal stress and longitudinal strain
B. Volumetric stress and volumetric strain
C. Lateral stress and lateral strain
D. Shear stress and shear strain
Correct Answer: D
Solved Example: 41510
If the radius of wire stretched by a load is doubled, then its Young’s modulus will be:
A. Doubled
B. Halved
C. Become four times
D. Remain unaffected
Correct Answer: D
Solved Example: 41511
The ultimate tensile stress of mild steel compared to ultimate compressive stress is:
A. Same
B. More
C. Less
D. More or less depending on other factors
Correct Answer: B
Solved Example: 41512
Tensile strength of a material is obtained by dividing the maximum load during the test by the:
A. Area at the time of fracture
B. Original crosssectional area
C. Average of (A) and (B)
D. Minimum area after fracture
Correct Answer: B
Solved Example: 41513
The impact strength of a material is an index of its:
A. Toughness
B. Tensile strength
C. Capability of being cold worked
D. Hardness
Correct Answer: A
Solved Example: 41514
The Young’s modulus of a wire is defined as the stress which will increase the length of wire compared to its original length:
A. Half
B. Same amount
C. Double
D. Onefourth
Correct Answer: B
Solved Example: 41515
Percentage reduction of area in performing tensile test on cast iron may be of the order of:
A. 50%
B. 25%
C. 0%
D. 15%
Correct Answer: C
Solved Example: 41516
The intensity of stress which causes unit strain is called:
A. Unit stress
B. Bulk modulus
C. Modulus of rigidity
D. Modulus of elasticity
Correct Answer: D
Solved Example: 41517
The change in the unit volume of a material under tension with increase in its Poisson’s ratio will,
A. Increase
B. Decrease
C. Remain same
D. Increase initially and then decrease
Correct Answer: B
Solved Example: 41518
If a material expands freely due to heating it will develop:
A. Thermal stresses
B. Tensile stress
C. Compressive stress
D. No stress
Correct Answer: D
Solved Example: 41519
In a tensile test, near the elastic limit zone, the
A. Tensile strain increases more quickly
B. Tensile strain decreases more quickly
C. Tensile strain increases in proportion to the stress
D. Tensile strain decreases in proportion to the stress
Correct Answer: A
Solved Example: 41520
The stress necessary to initiate yielding is
A. Considerably greater than that necessary to continue it
B. Considerably lesser than that necessary to continue it
C. Greater than that necessary to stop it
D. Lesser than that necessary to stop it
Correct Answer: A
Solved Example: 41521
In the tensile test, the phenomenon of slow extension of the material, i. e. stress increasing with the time at a constant load is called:
A. Creeping
B. Yielding
C. Breaking
D. Plasticity
Correct Answer: A
Solved Example: 41522
The ratio of direct stress to volumetric strain in case of a body subjected to three mutually perpendicular stresses of equal intensity, is equal to:
A. Young's modulus
B. Bulk modulus
C. Modulus of rigidity
D. Modulus of elasticity
Correct Answer: B
Solved Example: 41523
The stress at which extension of the material takes place more quickly as compared to the increase in load is called:
A. Elastic point of the material
B. Plastic point of the material
C. Breaking point of the material
D. Yielding point of the material
Correct Answer: D
Solved Example: 41524
The energy absorbed in a body, when it is strained within the elastic limits, is known as:
A. Strain energy
B. Resilience
C. Proof resilience
D. Modulus of resilience
Correct Answer: A
Solved Example: 41525
Resilience of a material is considered when it is subjected to:
A. Frequent heat treatment
B. Fatigue
C. Creep
D. Shock loading
Correct Answer: D
Solved Example: 41526
The maximum strain energy that can be stored in a body is known as:
A. Impact energy
B. Resilience
C. Proof resilience
D. Modulus of resilience
Correct Answer: C
Solved Example: 41527
The total strain energy stored in a body is termed as:
A. Resilience
B. Proof resilience
C. Modulus of resilience
D. Toughness
Correct Answer: A
Solved Example: 41528
Proof resilience per unit volume of material is known as:
A. Resilience
B. Proof resilience
C. Modulus of resilience
D. Toughness
Correct Answer: C
Solved Example: 41529
The strain energy stored in a body due to suddenly applied load compared to when it is applied gradually is:
A. Same
B. Twice
C. Four times
D. Eight times
Correct Answer: C
Solved Example: 41530
A material capable of absorbing large amount of energy before fracture is known as:
A. Ductility
B. Toughness
C. Resilience
D. Shock proof
Correct Answer: B
Solved Example: 41531
A beam is loaded as cantilever. If the load at the end is increased, the failure will occur:
A. In the middle
B. At the tip below the load
C. At the support
D. Anywhere
Correct Answer: D
Solved Example: 41532
The value of shear stress which is induced in the shaft due to the applied couple varies:
A. From maximum at the center to zero at the circumference
B. From zero at the center to maximum at the circumference
C. From maximum at the center to minimum at the circumference
D. From minimum at the center to maximum at the circumference
Correct Answer: B
Solved Example: 41533
A key is subjected to side pressure as well at shearing forces. These pressures are called:
A. Bearing stresses
B. Fatigue stresses
C. Crushing stresses
D. Resultant stresses
Correct Answer: A