STRENGTH OF MATERIALS INTRODUCTION CLASS NOTES FOR MECHANICAL ENGINEERING
STRENGTH OF MATERIALS
INTRODUCTION CLASS NOTES FOR
MECHANICAL ENGINEERING
Strength of materials tells which
materials are strong in
tension/compression/
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

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

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

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

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

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.
 Self weight of the component/body under consideration
 Weight being supported by the component /body being considered
 Wind Load
 Seismic load –due to earth quake
 Thermal load
 Pressure load
 Inertia force
 Centrifugal force
 Frictional force
STRESS IN STRENGTH OF MATERIALS
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}.
TYPES OF STRESSES IN STRENGTH OF MATERIALS: Three
(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.
UNITS OF STRESS IN STRENGTH OF MATERIALS
Each and every stress unit is N/mm^{2}.
1 MN/m^{2} = 1 N/mm^{2} = 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.
STRESSES IN THE STRESSSTRAIN CURVE
Fig. STRESS STRAINCURVE FOR MILD STEEL
Proportional limit stress, σ_{p,},
Elastic limit stress, σ_{e},
Upper yield point stress, σ_{uyp},
Lower yield point stress, σ_{lyp},
Ultimate stress, σ_{ult},
Breaking stress OR failure stress, σ_{f}
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. P_{eq.g} = 2 P_{sud},
σ_{sudden} = 2σ_{gal}
σ_{gal }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.
P_{eq. g} = Pi[1+ (1+2hAE/P_{i}L)^{0.5}], σ_{i} = Pi[1+ (1+2hAE/P_{i}L)^{0.5}] / A
Stresses in a composite section
σ_{1}= P_{total} E_{1}/ (E_{1}A_{1} + E_{2}A_{2})
σ_{2}= P_{total} E_{2}/ (E_{1}A_{1} + E_{2}A_{2})
THERMAL STRESS
Thermal stress (σ_{th}) is there only when contraction due to fall of temperature is restricted——It is a tensile stress.=EαΔt
Thermal stress (σ_{th}) is there only when expansion due to rise of temperature is restricted—— It is a compressive stress.= EαΔt
Thermal strain = ε_{th}= σ_{th }/ E= αΔt
Where is the coefficient of thermal expansion
Thermal stress in a single material σ_{th}= Eα Δt, ε_{th} = α Δt
THERMAL STRESSES IN A COMPOSITE MATERIAL
COMMON THERMAL LOAD = P_{th}= [(α1 –α2 ) Δt] /(1/A_{1}E_{1} +1/A_{2}E_{2})
σ_{th1 } = P_{th }/A_{1 }= [(α1 –α2 ) Δt] /A_{1}(1/A_{1}E_{1} +1/A_{2}E_{2})
σ_{th2 } = P_{th }/A_{2 }= [(α1 –α2 ) Δt] /A_{2}(1/A_{1}E_{1} +1/A_{2}E_{2})
Definition of strain
Strain is change/original
Strain is change in length/original length
TYPES OF STRAINS

Tensile strain,

Compressive strain

Shear strain

Linear strain

Lateral strain

Volumetric strain
1dimensional Strain–Strain is change in width/original width, Strain is change in thickness/original thickness.
2dimensional Strain Strain is change in area/original area
3dimensional Strain Strain is change in volume/original volume.
SYMBOLS
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
UNITS
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
Modulus
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/m^{2}_{.} Its value for steel is 80 GN/m^{2}. But when used in solving problems it is used as 80 x 1000 N/mm^{2}_{. }
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/m^{2}. 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/m^{2}. and Modulus in N/mm^{2} = (GN/m^{2} ) x 1000
POISSON’S RATIO
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 applicationtruly axial or eccentric
The value of Factor of safety will be
For Ductile materials FOS = σ_{yp }/σ_{all}
For Brittle Materials FOS = σ_{ult }/σ_{all}
σ_{yp }= yield point stress for ductile materials
Its value is very near to proportional limit stress
σ_{all} = allowable stress = 1/3 to 2/3 of elastic limit stress
σ_{ult } = 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
MECHANICAL PROPERTIES
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 = σ^{2} /2E
u=σ_{t}^{2} /2E in tension
u= σ_{c}^{2} /2E in compression
u= τ^{ 2} /2G in shear
Proof resilience=u_{e} = σ_{e}^{2} /2E
_{σe is the elastic limit stress.}
https://www.mesubjects.net/wpadmin/post.php?post=2310&action=edit Q. Ans. Simple stresses
https://www.mesubjects.net/wpadmin/post.php?post=7658&action=edit MCQ Simple stresses
https://www.mesubjects.net/wpadmin/post.php?post=3905&action=edit Testing of materials