Concept:
True strain (ϵ_{T}) = ln (1 + ϵ)
Calculation:
Given:
Engineering strain \((\epsilon) = 0.100\% = \frac{{0.1}}{{100}} = 0.001\)
True strain (ϵ_{T}) = ln (1 + ϵ)
ϵ_{T} = ln (1 + 0.001)
ϵ_{T} = 0.099%Concept:
Whenever the bar is subjected to the axial tensile load, there will be an increase in the length of the bar along the direction of the loading.
The longitudinal strain is defined as the ratio of increase in the length of the bar in the direction of applied load to that of the original length.
\(Strain~=~\frac{Change~in~length}{Original~length}~=~\frac{\Delta L}{L}\)
Calculation:
Given:
Length of cylinder = L meters, deformation = l cm
deformation = ΔL
\(Strain= \frac{{{\rm{Δ }}L}}{L} \)
\(Strain= \frac{l}{L}\frac{{cm}}{m}\)
\(Strain= \frac{l}{{L \times 100}}\frac{{m}}{{\;m}}\)
\(Strain= \frac{0.01\ l}{L}\)
Explanation:
Poisson's ratio is the ratio of transverse strain to longitudinal strain.
\(\mu =  \frac{{{\epsilon_{(lateral)}}}}{{{\epsilon_{(longitudinal)}}}}\)
For perfectly isotropic elastic material, Poisson’s ratio is 0.25 but for most of the materials, the value of Poisson's ratio lies in the range of 0 to 0.5.
Poisson's ratio for various materials are:
An elastic isotropic body is in a hydrostatic state of stress as shown in the figure. For no change in the volume to occur, what should be its Poisson's ratio?
Given,
The elastic body is in hydrostatic state of stress
∴ According to Pascal's law the body has equal stresses on the body
i.e σ_{x }= σ_{y} = σ_{z}
We know that the Volumetric strain is given by
\(ϵ{_v} = \left( {\frac{{{σ _x} + {σ _y} + {σ _z}}}{E}} \right)\left( {1  2\mu } \right)\)
ϵ_{v} = volumetric strain = δV/V
\(\therefore \frac{{δ V}}{V} = \left( {\frac{{{σ _x} + {σ _y} + {σ _z}}}{E}} \right)\left( {1  2\mu } \right)\)
As no change in the volume is given
δV = 0
Either \({σ _x} + {σ _y} + {σ _z} = 0\ or\;1  2\mu = 0\)
As stresses is not zero
1 – 2μ = 0
μ = 0.5The threedimension state of stress at a point is given by
\(\left[ \sigma \right] = \left[ {\begin{array}{*{20}{c}} {30}&{10}&{  10}\\ {10}&0&{20}\\ {  10}&{20}&0 \end{array}} \right]MPa\)
The shear stress on the xface in ydirection at the same point is then equal toConcept:
The threedimension state of stress is given by:
\(\left[ \sigma \right] = \left[ {\begin{array}{*{20}{c}} {{\sigma _1}}&{{\tau _{xy}}}&{{\tau _{xz}}}\\ {{\tau _{yx}}}&{{\sigma _2}}&{{\tau _{yz}}}\\ {{\tau _{zx}}}&{{\tau _{zy}}}&{{\sigma _3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {30}&{10}&{  10}\\ {10}&0&{20}\\ {  10}&{20}&0 \end{array}} \right]\)
where, first row shows the x face
second row shows the y face
third row shows the z face And
first column shows the x direction,
second column shows the y direction,
third column shows the z direction,
Therefore, the shear stress on the xface in ydirection at the same point is, τ_{xy} = 10 MN/m^{2}
In a tension test on a mild steel rod of 12 mm diameter, following observations are made.
a) Load at yield point = 28.5 kN
b) breaking load = 42 kN
Find out Yield stress.
Concept:
Yield stress = Load at yield point / Area
Area, \(A = \frac{\pi }{4}{d^2}\)
Calculation:
Given,
d = 12 mm
Load at yield point = 28.5 kN
Breaking load = 42 kN
\(A = \frac{\pi }{4}{d^2}\)
\(A = \frac{\pi }{4} \times {12^2} = 113.09m{m^2}\)
Yield stress = Load at yield point / Area
\({\sigma _y} = \frac{{28.5 \times {{10}^3}}}{{113.09}} = 252.01N/m{m^2}\)
(1 N/mm^{2} = 1 MN/m^{2})
Yield stress = 252.01 MN/m^{2}
So, the correct answer is 252.5 MN/m2.
Concept:
Relationship between stress and strain is given by:
σ = Eϵ
where σ = stress, ϵ = strain and E = Young's modulus of elasticity
Calculation:
Given:
L = 100 mm, ΔL = 0.05 mm, E = 200 GPa = 200 × 10^{9} Pa
Strain produced is:
\(\varepsilon = \frac{{Δ L}}{L} = \frac{0.05}{100}=5 × \mathop {10}\nolimits^{  4} \)
Stress Produced is:
σ = Eϵ
σ = 200 × 10^{9} × 5 × 10^{4}
σ = 100 MPa
Explanation:
Brittle materials:
the materials which have less than 1% elongation at fracture on a gauge length of 100 are classified as brittle materials e.g. cast iron, stone, and glass, etc.
In such types of materials, the fracture occurs without any noticeable change in the rate of elongation which implies that there is no difference between the ultimate strength and the rupture strength.
Fig shows the rupture of brittle material. It can be noted that there is no necking of the specimen as in the case of a ductile fracture shown below 
In the case of brittle fracture, the rupture occurs along a surface perpendicular to the load.Explanation:
The stress is given as, \(Stress = \frac{{Load\;}}{{Area}}\)
where, stress is the ratio of ultimate stress to the factor of safety, i.e. \({\rm{Working\;stress\;}} = \frac{{{\rm{Ultimate\;stress\;}}}}{{{\rm{Factor\;of\;safety}}}}\)
Calculations:
Given:
Axial load P = 200000 kg, Ultimate stress σ_{u }= 4800 kg/cm^{2}, Factor of safety n = 4
Now,
\({\rm{Working\;stress\;}} = \frac{{{\rm{Ultimate\;stress\;}}}}{{{\rm{Factor\;of\;safety}}}}\)
\(Working\;stress = \frac{{4800}}{4}\)
Working stress = 1200 kg/cm^{2}
^{\(Stress = \frac{{Load\;}}{{Area}}\)}
\(Area = \frac{{load\;}}{{working\;stress}}\)
\(A = \frac{{200000}}{{1200}} = 166.66\)
A = 166.66 cm^{2}
Explanation:
The chart shows the relation between stressstrain in different materials.
StressStrain Curve 
Type of Material or Body 
Examples 
Ideally plastic material. 
Viscoelastic (elastoplastic) 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 
Nondilatant material, (water) ideal fluid, etc. 

Nonlinear elastic material 
Natural rubbers, elastomers, and biological gels, etc 
Explanation:
Tensile Stress and Strain
Compressive Stress and Strain
A 8 mm thick Copper sheet is cut with a 9 cm diameter round punch. If the punch exerts a force of 16 kN, Find the shear stress in the sheet
Concept:
Shear stress: Shear stress is the ratio of resisting force to the crosssectional area along the direction of the force.
Area of the Copper sheet after cutting, A = circumference of copper sheet × thickness of the sheet
A = 2πrt
Shear stress = Exerted Force/Area of copper sheet
Calculation:
Given,
Exerted force= 16 KN = 16000 N
Copper sheet diameter cut by round punch = 9 cm = 90 mm
Radius of sheet = 90/2 = 45 mm
the thickness of the copper sheet = 8 mm
A = 2 × π × 45 × 8 = 2261.95 mm^{2}
Hence,
Shear stress = exerted force/area of copper sheet
Shear stress = 16000/2261.95 = 7.0735 N/mm2
Shear stress of copper sheet = 7.0735 N/mm^{2} = 7.0735 mpa
A steel rod of crosssectional area 10 mm^{2} is subjected to loads at points P, Q, R and S as shown in the figure below:
If E_{steel} = 200 GPa, the total change in length of the rod due to loading is
Concept:
Change in length due to axial loading is,
\(\Delta l = \frac{{PL}}{{AE}}\)
Where, P = axial load, L =initial length, A = crosssectional area of bar, E = young’s modulus of elasticity
Calculation:
Total change in length ∆l = ∆l_{PQ} + ∆l_{QR} + ∆l_{RS}
For the section PQ: l = 500 mm, P = 200 N
For the section QR: l = 1000 mm, P = 200 N
For the section RS: l = 500 mm, P = 100 N
\(\Delta l = \;\frac{{200 \times 500}}{{10 \times 2 \times {{10}^5}}} + \frac{{  200 \times 1000}}{{10 \times 2 \times {{10}^5}}} + \frac{{100 \times 500}}{{10 \times 2 \times {{10}^5}}} = \;  25 \times {10^{  3}}mm\)
In a simple stressstrain test, the volumetric strain is equal to ________ strain.
Explanation:
Volumetric strain:
It is defined as the ratio of change in volume to the initial volume of the body. It is equal to the sum of strain in all directions.
Equal normal forces are acting on all faces of the cube, so strain will be equal in all directions (ϵx = ϵy = ϵz).
ϵv = ϵx + ϵy + ϵz
ϵv = 3ϵx
Hence, the Volumetric strain is 3 times the linear strain.
The true stress (σ) – true strain (ε) diagram of a strain hardening material is shown in figure. First, there is loading up to point A, i.e., up to stress of 500 MPa and strain of 0.5. Then from point A, there is unloading up to point B, i.e., to stress of 100 MPa. Given that the Young’s modulus E = 200 GPa, the natural strain at point B (ε_{B}) is _________ (correct to three decimal places).
Concept:
In the given situation, when unloading is done then the material recovers the elastic strain whose slope is equal to the elastic constant, Young’s Modulus.
Calculation:
Given E = 200 GPa
Here along the line (AB) elastic recovery will take place. So slope of line (AB) is equal to the young’s Modulus.
Consider ΔACB
Slope of AB \(= \frac{{AC}}{{BC}} = \frac{{500  100}}{{BC}}\)
\(200 \times {10^3} = \frac{{400\;}}{{BC}} \Rightarrow BC = 0.002\)
From figure; BC = (0.5  ϵ_{B})
⇒ 0.002 = 0.5  ϵ_{B}
ϵ_{B} = 0.5 – 0.002 = 0.498
Points to remember: Don’t get confused because the diagram is given for (true stress – true strain). No need for the conversion of (true – strain → engineering strain) Or (true stress → engineering stress)
As in the question (true strain) is asked.
Explanation:
where,
Point 2 = proportional limit
It is clearly shown
Fracture stress (at point 5) > Yield Stress (at point 3)
Important Points
The largest internal force in the bar shown in the figure is –
Explanation
By making free body diagram of each element explicitly –
∴ The largest internal force in the bar is 13 kN (CD Portion)
A prismatic bar PQRST is subjected to axial loads as shown in the figure. The segments having maximum and minimum axial stresses, respectively, are
Concept:
Stress: Force (F) applied per unit area (A) is called stress.
\(σ = {F \over A}\)
Unit of stress: N/m2.
Since the given beam is prismatic i.e. constant crosssectional area, the stress produced is directly proportional to force acted on the section.
Calculation:
Given:
Draw the FBD of the given beam to analyze the force acting on each separate section.
Therefore we can see that the force acting on the section PQ, QR, RS, and ST are tensile whose magnitude is 10 kN, 20 kN, 5 kN, and 25 kN respectively.
∴ the maximum force is 25 kN on ST and the minimum force is 5 kN on RS respectively.
∵ σ ∝ F
∴ σ_{max} is on ST and σ_{min} is on RS respectively.
Explanation:
Hooke’s law:
It States that under direct loading, within proportionality limit, stress is directly proportional to strain.
i.e stress ∝ strain
σ = E∈
where, E is a constant of proportionality and known as Young’s Modulus of elasticity.
Additional Information
1. Hooke’s law is valid up to the limit of proportionality. However, for mild steel proportionally limit and elastic limit are almost equal. But for other metals, the elastic limit may be higher than the proportionality limit. E.g Rubber.
2. The slope of the stressstrain curve is called the modulus of elasticity (E). The modulus of elasticity (E) is the constant of proportionality which is defined as the intensity of stress that causes unit strain. Thus, the modulus of elasticity (E) has the unit the same as that of stress.
A thin plate of uniform thickness is subject to pressure as shown in the figure below
Under the assumption of plane stress, which one of the following is correct?
Explanation:
Plane stress
\(\sigma = \left[ {\begin{array}{*{20}{c}} {{\sigma _{xx}}}&{{\tau _{xy}}}&0\\ {{\tau _{yx}}}&{{\sigma _{yy}}}&0\\ 0&0&0 \end{array}} \right]\)