1. UNIT-01. SIMPLE STRESSES and STRAINS Lecture Number - 02 Prof. M. N. CHOUGULE MECHANICAL DEPARTMENT SIT LONAVALA Strength of Materials

2. AGENDA Modulus of Elasticity(E) Modulus of Rigidity(G) Bulk Modulus(K) Relation between E, G,K. Factor of Safety Numerical Strength of Materials

3. Modulus of Elasticity(E) Within elastic limit of material, in which Hook's law obeyed Stress is directly proportional to strain. Where, E=Constant of proportionality (Modulus of Elasticity) Strength of Materials

4. Modulus of Rigidity(G) Within elastic limit, Shear stress α Shear strain Where, G =constant of Proportionality (Modulus of Rigidity) Strength of Materials

5. Bulk Modulus(K) Ratio of Direct stress or Hydrostatic pressure and volumetric strain produced is called bulk modulus. Strength of Materials A P VV V A (surface area)

6. Relation between Modulus of Elasticity(E) Modulus of Rigidity(G) and Bulk Modulus(K) Relation between Modulus of Elasticity(E) Modulus of Rigidity(G) is, Relation between Modulus of Elasticity(E) and Bulk Modulus(K) is, Relation between Modulus of Elasticity(E) Modulus of Rigidity(G) and Bulk Modulus(K) is, Strength of Materials

7. Factor of Safety The load which any member of a machine carries is called working load, and stress produced by this load is the working stress. This working stress is also called the permissible stress or the allowable stress or the design stress. Strength of Materials

8. Numericals Strength of Materials Q.1.For a certain material E=210 GPa,μ=0.3.Calculate values of other two elastic constants. Ans. 210 = 2G(1+0.3) 210 =3K(1-2x0.3) G=80.77 GPa. K=175 GPa.

9. Strength of Materials Q.2.A bar of cross section 8 mm x 8 mm is subjected to axial pull of 7 kN. The lateral dimensions of bar are found to have reduced by 1.5 x 10 -3 mm. Find Poisson's ratio and Modulus of Elasticity, Assuming G=80 GPa. Ans. Find -normal stress=109.375 MPa; Longitudinal strain =stress/E=109.375/E. Lateral strain = Change in dimension/ original dim. = 0.0015/8 =1.875x 10 -4 Poison's ratio (μ) = lateral strain / longitudinal strain we get, E=583.33 x 10 3 μ………..(i) E=160 x 103 (1+μ)…..(ii) Equating i,ii we get, μ=0.377,E=220.47 GPa.

10. Q.3.A material has modulus of rigidity equal to 0.4x10 5 N/mm 2 and Bulk modulus equal to 0.95 x10 5 N/mm 2 then find value of Young’s modulus in GPa. Ans. Use, E=105 GPa Strength of Materials