Strength of Materials

UNIT-01. SIMPLE STRESSES and STRAINS

Lecture Number - 02Prof. M. N. CHOUGULE

MECHANICAL DEPARTMENT SIT LONAVALA

Strength of Materials

AGENDA

• Modulus of Elasticity(E)• Modulus of Rigidity(G)• Bulk Modulus(K)• Relation between E, G,K.• Factor of Safety• Numerical

Strength of Materials

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)

StrainStress

EConstStrain

Stress 2mm

N

Strength of Materials

Modulus of Rigidity(G)

• Within elastic limit,Shear stress α Shear strain

Where, G =constant of Proportionality (Modulus of Rigidity)

GConstnShearStrai

sShearStres

2mm

N

Strength of Materials

Bulk Modulus(K)

• Ratio of Direct stress or Hydrostatic pressure and volumetric strain produced is called bulk modulus.

F

AF

F

P

DV

V

A (surface area)

ve

Strain Volumetric

StressDirect =K

2mm

N

Strength of Materials

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,

)1(2E G

)21(3E K

)3(

9E

GK

KG

Strength of Materials

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.

w

ut

Stress or WorkingDesign

Stress Yieldor Ultimate=Safty ofFactor

Strength of Materials

Numericals

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)

)1(2E G

G=80.77 GPa.

)21(3E K

K=175 GPa.

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 103 μ………..(i)

E=160 x 103 (1+μ)…..(ii)Equating i ,ii we get, μ=0.377,E=220.47 GPa.

)1(2E G

Strength of Materials

Q.3.A material has modulus of rigidity equal to 0.4x105 N/mm2 and Bulk modulus equal to 0.95 x105 N/mm2 then find value of Young’s modulus in GPa.

Ans. Use,

E=105 GPa)3(

9E

GK

KG