2. Deepak in a nutshell
Academic
MBA, Digital Business (IE Business School, Spain)
MS, Mechanical Engineering (Purdue University, USA)
B.E, Mechanical Engineering (Delhi College of Engg)
Professional
Founder, perfectbazaar.com
Application Engineer ( Robert Bosch, USA)
Controls Engineer (Cummins Engine Company, USA)
HAPPY TO CHAT ANYTIME
3. What is Strength of
Materials?
Study of internal effects (stresses and strains) caused by external
loads (forces and moments) acting on a deformable body/
structure.
Also known as: Strength of Materials or Mechanics of Solids
Determines:
1. Strength (determine by stress at failure)
2. Deformation (determined by strain)
3. Stiffness (ability to resist deformation; load needed to cause
a specific deformation; determined by the stress- strain
relationship)
4. Stability (ability to avoid rapidly growing deformations
caused by an initial disturbance; e.g., buckling)
6. Grading Policy
10 marks class attendance.
10 marks for teacher assessment.
30 marks for internal sessional tests.
100 marks external university exam.
7. Unit -1
Syllabus
Compound Stress and Strains
3-D Stress, Theories of failure
Unit -2
Stresses in Beam
Deflection of Beams
Unit – 3
Helical and Leaf Spring
Column and Struts
8. Syllabus
Unit – 4
Thin Cylinders and Spheres
Thick cylinders
Unit – 5
Curved beams
Unsymmetrical Bending
9. Unit 1- Stress and Strain
Topics Covered
Lecture -1 - Introduction, state of plane stress
Lecture -2 - Principle Stresses and Strains
Lecture -3 - Mohr's Stress Circle and Theory of
Failure
Lecture -4- 3-D stress and strain, Equilibrium
equations and impact loading
Lecture -5 - Generalized Hook's law and Castigliono's
10. What is Strength of
Materials?
Study of internal effects (stresses and strains) caused by external
loads (forces and moments) acting on a deformable body/
structure.
Also known as: Strength of Materials or Mechanics of Solids
Determines:
1. Strength (determine by stress at failure)
2. Deformation (determined by strain)
3. Stiffness (ability to resist deformation; load needed to cause
a specific deformation; determined by the stress- strain
relationship)
4. Stability (ability to avoid rapidly growing deformations
caused by an initial disturbance; e.g., buckling)
11. Stresses
Stress
Force of resistance per unit area offered by a body against
deformation
P
σ=
A
P = External force or load
A = Cross-sectional area
€
12. Strain
Strain
Change in dimension of an object under application of
external force is strain
dL
ε=
L
dL = Change in length L = Length
€
13. Types of Stresses and
Strains
Stress Strain
Tensile stress Tensile strain
Compressive Compressive
stress strain
Shear stress Shear strain
14. Shear Stress
Shear Stress
Stress induced when body is subjected to equal and
opposite forces that are acting parallel to resisting
dl P
surface.
D D1 C
DD1 dl
Strain φ = =
AD h
dl = Transversal displacement
h
P
Stress τ= φ φ
€ L
A l B
€ €
€
15. Hooke’s Law
Hooke’s law – Stress is proportional to strain within
elastic limit of the material.
The material will recover its shape if stretched
to point 2.
There will be permanent deformation in the
Material if the object is stretched to point 4.
Upto point 2 stress is proportional to strain.
Stress
E=
Strain
E = Young’s Modulus or Modulus of
Elasticity
16. Elasticity
Shear Modulus/Modulus of rigidity – ratio of shear
stress to shear strain.
Shear _ stress τ
C= =
Shear _ strain φ
Young’s modulus/Modulus of elasticity- ratio of
tensile or compressive stress to tensile or compressive
strain.
€
Tensile _ stress Compressive _ stress σ
E= = =
Tensile _ strain Compressive _ strain e
Factor of safety =
Max _ stresses
Working _ stresses
€
€
17. PROBLEM
PROBLEM – A rod 150cm long and diameter 2.0cm is
subjected to an axial pull of 20kN. If modulus of
elasticity of material of rod is 2x105 n/mm2
determine:
1) Stress
2) Strain
3) Elongation of the rod
18. Poisson ratio
Ratio of lateral strain to longitudinal strain
Lateral _ strain
υ=
Longitudinal _ strain
3-Dimensional Stress System
σ1
σ2 Stress σ1 will produce strain in x-direction =
σ1 E
€ Stress y and z direction due to σ1 = −υ
E
Negative sign is because the strain in y and z
€ € direction will be compressive
σ1 €
σ3 € strain in y-direction = σ2
Stress σ2 will produce €
σ2 E
Stress x and z direction due to σ2= −υ
€ E
σ
Stress σ3 will produce strain in z-direction = 3
€ €
Stress x and y direction due to σ3 = €−υ σ 3 E
€ € E
€
19. Poisson ratio
Ratio of lateral strain to longitudinal strain
3-Dimensional Stress System
σ2
Total Strain in x-direction due to σ1,σ2 ,σ3
σ1 σ σ
= −υ 2 −υ 3
€ E E E
σ1
Total Strain in y-direction due to σ1,σ2 ,σ3
€
σ2 σ1 σ3
σ3 = −υ −υ
€ E E E
€
€
Total Strain in z-direction due to σ1,σ2 ,σ3
σ3 σ σ
€ = −υ 1 −υ 2
€ E E E
€
€
20. Analysis of bars of
varying sections
Section 3
Section 2
Section 1
P A1 A2 A3 P
L1 L2 L3
⎡ L1 L2 L3 ⎤
Total change in length of bar dL = P ⎢ + + ⎥
⎣ E1 A1 E 2 A2 E 3 A3 ⎦
€
21. Analysis of bars of
varying sections
Section 3
Section 2
Section 1
P=35000
D3=2cm D3=3cm D3=5cm P=35000N
20cm 25cm 22cm
PROBLEM – An axial pull of 35000N is acting on a bar consisting of three lengths
as shown in fig above. if Young’s modulus =2.1x105 N/mm2 determine:
1) Stresses in each section
2) Total extension in bar.
22. Principal of
superposition
When number of loads are acting on a body the
resulting strain will be sum of strains caused by
individual loads.
23. Analysis of bars of
composite sections
Bar made up of 2 or more bars of equal length but of
different materials rigidly fixed with each other.
P = P1 + P2
P = σ1 A1 + σ2 A2 1 2
ε1 = ε 2
€ σ1 σ 2
=
E1 E 2 P
€
€
24. Analysis of bars of
composite sections
PROBLEM – A steel rod of 3cm diameter is enclosed
centrally in a hollow copper tube of external diameter
3 cm 5cm and internal diameter of 4cm. The composite bar
15 cm 1 2 is then subjected to an axial pull of 45000N. If the
length of each bar is equal to 15cm. Determine
1) Stresses in the rod and the tube and
4 cm 2) Load carried by each bar
5 cm Take E for steel =2.1x105 N/mm2 and
E for copper = 1.1x105 N/mm2
P=45000N
25. Thermal Stresses
Stresses are induced when temperature of A B B’
the body changes.
When rod is free to expand the extension
in the rod L dL
dL = αTL
α = Coefficient of linear expansion
stress = strain * E
T = Rise in temperature
€
stress = α × T × E
€Stress and strain when supports yield = expansion due to rise in temp - yielding
€
= αTL − δ
€ αTL − δ ⎛ αTL − δ ⎞
Strain = Stress = ⎜ ⎟ × E
L ⎝ L ⎠
€