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State of plain stresses

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- 1. 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
- 2. 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)
- 3. Strength of Material Why we need to study this course.
- 4. Strength of Material Why we need to study this course.
- 5. Grading Policy 10 marks class attendance. 10 marks for teacher assessment. 30 marks for internal sessional tests. 100 marks external university exam.
- 6. 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
- 7. Syllabus Unit – 4 Thin Cylinders and Spheres Thick cylinders Unit – 5 Curved beams Unsymmetrical Bending
- 8. Unit 1- Stress and StrainTopics Covered Lecture -1 - Introduction, state of plane stress Lecture -2 - Principle Stresses and Strains Lecture -3 - Mohrs Stress Circle and Theory of Failure Lecture -4- 3-D stress and strain, Equilibrium equations and impact loading Lecture -5 - Generalized Hooks law and Castiglionos
- 9. 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)
- 10. Stresses Stress Force of resistance per unit area offered by a body against deformation P σ= AP = External force or loadA = Cross-sectional area €
- 11. Strain Strain Change in dimension of an object under application of external force is strain dL ε= LdL = Change in length L = Length €
- 12. Types of Stresses and Strains Stress Strain Tensile stress Tensile strain Compressive Compressive stress strain Shear stress Shear strain
- 13. 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 € €€
- 14. 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
- 15. 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€ €
- 16. PROBLEMPROBLEM – 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) Stress2) Strain3) Elongation of the rod
- 17. 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 €
- 18. 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 € €
- 19. 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 ⎦ €
- 20. Analysis of bars of varying sections Section 3 Section 2 Section 1 P=35000 D3=2cm D3=3cm D3=5cm P=35000N 20cm 25cm 22cmPROBLEM – An axial pull of 35000N is acting on a bar consisting of three lengthsas shown in fig above. if Young’s modulus =2.1x105 N/mm2 determine:1) Stresses in each section2) Total extension in bar.
- 21. Principal of superposition When number of loads are acting on a body the resulting strain will be sum of strains caused by individual loads.
- 22. 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€ €
- 23. 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 bar15 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
- 24. 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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