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

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)

Strength of Material   Why we need to study this course.

Strength of Material   Why we need to study this course.

Grading Policy   10 marks class attendance.

  10 marks for teacher assessment.

  30 marks for internal sessional tests.

  100 marks external university exam.

Syllabus   Unit -1

  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

Syllabus  Unit – 4

 Thin Cylinders and Spheres

 Thick cylinders

 Unit – 5

 Curved beams

 Unsymmetrical Bending

Unit 1- Stress and Strain

  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

Topics Covered

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)

Stresses   Stress

  Force of resistance per unit area offered by a body against deformation

P = External force or load

A = Cross-sectional area

€

σ =PA

Strain   Strain

  Change in dimension of an object under application of external force is strain

dL = Change in length L = Length

€

ε =dLL

Types of Stresses and Strains

Stress Strain

Tensile stress Tensile strain

Compressive stress

Compressive strain

Shear stress Shear strain

Shear Stress   Shear Stress

  Stress induced when body is subjected to equal and opposite forces that are acting parallel to resisting surface.

Strain

dl = Transversal displacement

Stress

€

φ =DD1AD

=dlh

€

φ

D

A

D1 C

B l

h

€

φ

P dl

€

τ =PL

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.

€

E =StressStrain

E = Young’s Modulus or Modulus of Elasticity

Elasticity   Shear Modulus/Modulus of rigidity – ratio of shear

stress to shear strain.

  Young’s modulus/Modulus of elasticity- ratio of tensile or compressive stress to tensile or compressive strain.

  Factor of safety =

€

C =Shear _ stressShear_ strain

=τφ

€

E =Tensile_ stressTensile_ strain

=Compressive_ stressCompressive_ strain

=σe

€

Max _ stressesWorking_ stresses

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

Poisson ratio   Ratio of lateral strain to longitudinal strain

€

υ =Lateral_ strain

Longitudinal_ strain

€

σ1€

σ2

€

σ3

Stress will produce strain in x-direction =

€

σ1

€

σ1E

Stress y and z direction due to =

€

σ1

€

−υσ1E

Negative sign is because the strain in y and z direction will be compressive

Stress will produce strain in y-direction =

€

σ2

€

σ2E

Stress x and z direction due to =

€

−υσ 2

E

€

σ2Stress will produce strain in z-direction =

€

σ3E

Stress x and y direction due to =

€

−υσ 3

E

€

σ3

€

σ3

3-Dimensional Stress System

Poisson ratio   Ratio of lateral strain to longitudinal strain

€

σ1€

σ2

€

σ3

Total Strain in x-direction due to

€

σ1,σ2,σ3

€

=σ1E−υ

σ2E−υ

σ3E

Total Strain in z-direction due to

Total Strain in y-direction due to

€

σ1,σ2,σ3

€

σ1,σ2,σ3

€

=σ2E−υ

σ1E−υ

σ3E

€

=σ3E−υ

σ1E−υ

σ2E

3-Dimensional Stress System

Analysis of bars of varying sections

A1 A2 A3

Section 1 Section 2

Section 3

P P

L1 L2 L3

€

dL = P L1E1A1

+L2E2A2

+L3E3A3

⎡

⎣ ⎢

⎤

⎦ ⎥ Total change in length of bar

Analysis of bars of varying sections

D3=5cm

Section 1 Section 2

Section 3

P=35000 P=35000N

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.

D3=3cm D3=2cm

20cm

Principal of superposition

  When number of loads are acting on a body the resulting strain will be sum of strains caused by individual loads.

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.

1 2

P

€

P = P1 + P2

€

P =σ1A1 +σ2A2

€

ε1 = ε2

€

σ1E1

=σ 2

E2

Analysis of bars of composite sections

1 2

P=45000N

PROBLEM – A steel rod of 3cm diameter is enclosed centrally in a hollow copper tube of external diameter 5cm and internal diameter of 4cm. The composite bar 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 2)  Load carried by each bar

Take E for steel =2.1x105 N/mm2 and E for copper = 1.1x105 N/mm2

5 cm

4 cm

3 cm 15 cm

Thermal Stresses

€

dL = αTL

€

stress = strain * E

L dL

A B B’ Stresses are induced when temperature of the body changes.

When rod is free to expand the extension in the rod

€

α =Coefficient of linear expansion

T = Rise in temperature

€

stress = α × T × E

Stress and strain when supports yield = expansion due to rise in temp - yielding

€

= αTL −δ

Strain

€

=αTL −δ

LStress

€

=αTL −δ

L⎛

⎝ ⎜

⎞

⎠ ⎟ × E