Strength of materials tells which

materials are strong in


bending /torsion/buckling. Hence

depending upon the application, one can

select the material for that application.

Application includes chair, table, rod,

shaft or any item one can think of. The

selection is possible after performing

tests under various kinds of

forces/moments/torques etc. Tests help

to know which material is stronger under

which type of load.

Examples strength of materials

  1. wires/ropes/belts/chains should be of materials strong in tension

  2. Short columns/chair legs/table legs/walls should be of materials strong in compression

  3.  Beams should be of materials strong in tension as well as in compression

  4.  Shafts should be of materials strong in tension, compression, shear and should have high endurance strength

  5.  Medium and long columns should be of materials strong in tension and compression

  Sources of force in strength of materials

Symbols used for loads are W or P or F.

  1. Self weight of the component/body under consideration
  2. Weight being supported by the component /body being considered
  3. Wind Load
  4. Seismic load –due to earth quake
  5. Thermal load
  6. Pressure load
  7. Inertia force
  8. Centrifugal force
  9. Frictional force


It is the internal resistance of a material against the external force applied. Since it is an internal resistance, it is invisible.

Mathematically stress = force/area =N/mm 2.


(i) Tensile stress

(ii) Compression stress

(iii) Shear stress

Symbols of simple stresses in strength of materials

For tensile stress            = σ (sigma) = tensile force/Area perpendicular to the force = + in nature

compression stress symbol  = σ (sigma)= compression force/Area perpendicular to the force = – in nature

Symbol for shear stress               = τ (tou) = shear force/area parallel to force = (+/- ) in nature

Shear stress is always in a pair. One is applied and the other is induced. Induced is perpendicular to the applied. Thus, it is accompanied by a complementary shear stress. Complementary shear stress is always at right angle to the applied shear shear stress. This complementary shear stress is also called the induced shear stress.

Normal stress:   Stress due to a weight is a normal stress.

Tensile stress is a normal stress. Compression stress is a normal stress due to weight.

Bending stresses are normal stresses. Buckling stress is a normal stress.


Each and every stress unit is N/mm2.

1 MN/m2 = 1 N/mm2 = 1 M Pa

Unit Stress or Engineering stress

Engineering stress=  Force/original area =nominal stress = Unit stress

In this, stress is calculated with the original area.



Proportional limit stress, σp,,

Elastic limit stress, σe,

Upper yield point stress, σuyp,

Lower yield point stress, σlyp,

Ultimate stress, σult,

Breaking stress OR failure stress, σf

Yield Point Stress

A Yield point stress is that stress below which strain increases with decrease of stress. There are upper and lower yield stresses. It is found only in a ductile material. It is much lower than the ultimate stress for a ductile material. There is no yield point stress for a brittle material.

Ultimate stress

An Ultimate stress is the maximum stress which can be induced in a material. Ultimate and elastic limit stress are equal for a brittle material. Ultimate stress is much greater than the elastic limit stress and yield  for a ductile material.

Stress due to suddenly applied load

It is two times the stress if the same load was applied gradually. Peq.g = 2 Psud,

σsudden = 2σgal

σgal is the stress due to gradually applied load

Stress due to impact load

It is many times the stress if the same load was applied gradually. It can be 50, 100 or any times the nominal stress.

Peq. g = Pi[1+ (1+2hAE/PiL)0.5], σi = Pi[1+ (1+2hAE/PiL)0.5] / A

Stresses in a composite section

σ1= Ptotal E1/ (E1A1 + E2A2)

σ2= Ptotal E2/ (E1A1 + E2A2)


Thermal stress (σth) is there only when contraction due to fall of temperature is restricted——It is a tensile stress.=EαΔt

Thermal stress (σth) is there only when expansion due to rise of temperature is restricted——- It is a compressive stress.= EαΔt

Thermal strain = εth= σth / E=  αΔt

Where  is the coefficient of thermal expansion

Thermal stress in a single material   σth= Eα Δt,  εth = α Δt


COMMON THERMAL LOAD =  Pth= [(α1 –α2 ) Δt] /(1/A1E1 +1/A2E2)

σth1  = Pth /A1 =  [(α1 –α2 ) Δt] /A1(1/A1E1 +1/A2E2)

σth2  = Pth /A2 =  [(α1 –α2 ) Δt] /A2(1/A1E1 +1/A2E2)

Definition of strain

Strain is change/original

Strain is change in length/original length


  1. Tensile strain,

  2. Compressive strain

  3. Shear strain

  4. Linear strain

  5. Lateral strain

  6. Volumetric strain

1-dimensional Strain–Strain is change in width/original width, Strain is change in thickness/original thickness.

2-dimensional Strain      Strain is change in area/original area

3-dimensional Strain- Strain is change in volume/original volume.


tensile strain             =Є (epsilon)     = δL/L = + in nature

compressive strain = Є (epsilon)     = δL/L

shear strain               =γ (gamma), it is an angular strain.

Volumetric strain = change in volume/ original volume= δV/ V. It is used in pressure vessels.

Engineering strain OR Unit Strain

Engineering strain= Change in dimension/original dimension=Nominal strain=Unit strain

It is called an unit strain If strain is calculated with the original length


There is no units for each and every strain since strain is a ratio of length vs length or volume vs volume = δL/L  or  δV/ V


Young’s Modulus

Ratio of tensile stress to tensile strain OR Ratio of compressive stress to compressive strain. Its  symbol is E.

Young’s modulus = E =  tensile stress/ tensile strain

Young’s modulus = E =  compressive stress/ compressive strain

 Modulus of rigidity OR Shear modulus 

Shear modulus  is a ratio of shear stress to shear strain. Its symbol is G.  G = τ / γ  Its units are GN/m2. Its value for steel is 80 GN/m2. But when used in solving problems it is used as 80 x 1000 N/mm2

Shear modulus    = G =  shear stress/ shear strain

 Bulk modulus

Bulk modulus is the ratio of volumetric stress to volumetric strain. Its symbol is ‘K’. Its units are GN/m2. It is due to pressure exerted by a fluid on a solid causing change in volume of a solid. Fluid pressure causes change in volume and thus volumetric strain in a solid. It is given as

Bulk modulus = K =Volumetric stress/ volumetric strain= pressure/volumetric strain=

K =  p / δV/V

Units of each modulus is GN/m2. and Modulus in N/mm2 = (GN/m2 ) x 1000


It is ratio of lateral strain to linear strain.

Lateral strain is in a direction to perpendicular direction of force. Thus it will be less than linear strain.

Linear strain is in the direction of the force. Thus it will be more than lateral strain.

Lateral strain is of opposite type than the linear strain.

Lateral strain is compressive when linear strain is tensile.

Therefore, there is a negative sign with the Poisson’s Ratio.

Its symbol is  μ (mu). Its value is less than 1.

Elastic Constants

E,K,G and μ are called elastic constants since these are applicable only within elastic limit.

Relation Between Elastic Constants

Firstly E = 2G (1 + μ)

Secondly E = 3K (1 – 2μ)

Thirdly E = 9KG/(3K+G)

St. Venant’s principle

 Strain is not uniform near the area where forces are applied. Yet, strain becomes uniform at some perpendicular distance from the area on which force is applied. Say 5 % of the length from either end of a beam in simple tension. Say the length of the beam under simple tension is 2 m. Then strain is not uniform for 100 mm on either side of the rod in simple tension. Similarly it applies to simple compression and simple strain cases also.

Factor of Safety

 Factor of safety accounts for uncertainties in the actual situation and the assumed situation. Because of many uncertainties, factor of safety is FACTOR OF IGNORANCE.

(i) First, uncertainty about the homogeneous nature of the material

(ii) Second, Uncertainty about the isotropic nature of the material

(iii)Third,  Uncertainty about the continuous nature of the material

(iv) Fourth, uncertainty about the method of manufacture of the material component

(v) Fifth, uncertainty about the exact type of loading

(vi)Sixth,  Uncertainty about the load application-truly axial or eccentric

The value of Factor of safety will be

For Ductile materials            FOS = σyp all

For Brittle Materials               FOS = σult all

σyp = yield point stress for ductile materials

 Its value is very near to proportional limit stress

σall =  allowable stress = 1/3 to 2/3 of elastic limit stress

σult  = ultimate stress

Normal factor of safety

(i) Gradually applied load               2 to 4

(ii) Suddenly applied load               4 to 8

(iii) Impact load                                8 to 16

Stress concentration factor k

k is strain energy= Stress at a discontinuity / Nominal stress

It is always >1


Ductility=Either  % elongation OR  % reduction in area of cross section at fracture   

 Toughness = Energy required to fracture a specimen under an impact load= Area of load extension curve up to the point of fracture.

Hardness is the resistance to indentation.

Resilience is strain energy per unit volume. It is represented by ‘u’.

 u = σ2 /2E

u=σt2 /2E   in tension

u= σc2 /2E  in compression

u= τ 2 /2G    in shear

Proof resilience=ue = σe2 /2E

σe is the elastic limit stress.

https://www.mesubjects.net/wp-admin/post.php?post=2310&action=edit        Q. Ans. Simple stresses

https://www.mesubjects.net/wp-admin/post.php?post=7658&action=edit         MCQ Simple stresses

https://www.mesubjects.net/wp-admin/post.php?post=3905&action=edit         Testing of materials

Similar Posts