State of plain stresses

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)

4. Strength of Material
 Why we need to study this course.

5. Strength of Material
 Why we need to study this course.

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 ⎠
€