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Lecture 1 stresses and strains
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.
  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
Syllabus
  Unit – 4
    Thin Cylinders and Spheres
    Thick cylinders

  Unit – 5
    Curved beams
    Unsymmetrical Bending
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
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
                     σ=
                        A
P = External force or load

A = Cross-sectional area

  €
Strain
  Strain
     Change in dimension of an object under application of
      external force is strain
                      dL
                   ε=
                       L
dL = Change in length      L = Length


      €
Types of Stresses and
       Strains
     Stress         Strain
   Tensile stress Tensile strain

   Compressive Compressive
     stress       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
                                                            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


                                              €                  €
€
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
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
€

                  €
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
                           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

                              €
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
                                                         €

                              €
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 ⎦



                    €
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.
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.


                P = P1 + P2
              P = σ1 A1 + σ2 A2                1           2


                  ε1 = ε 2
    €               σ1 σ 2
                      =
                    E1 E 2                             P
€
        €
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
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    ⎠
                         €

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Lecture 1 stresses and strains

  • 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 ⎠ €