Question

(20%) Question 6. The elastic modulus, yield strength and ultimate strength of a certain material afe 110 GPa 2400 MPa and 265 MPal respectively. If a cylindrical bar of this material 380 mm long is subjected to a uniaxial tension test and its length is increased by an amount áf O.50 mm under a tensile load ót 6660 N.At the same time its diameter decreased by0,28 mm Calculate; a) The initial diameter of the bar, (8%) d b) Under an axial tensile load of 13270 N, is there any plastic strain in the bar? (3% ) c) Calculate the diameter of the material to carry a tensile force of 13270 N without any plastic deformation, (4 % ) a) The true strain at yielding, (3%) d) The engineering and true stresses at 6660 N (6 % )

Answer 1

Ans) Given,

Modulus of elasticity (E) = 110 GPa or 110,000 N/mm²

Yield strength ($\dpi{100} \sigma$_y)= 240 MPa

Ultimate strength ($\dpi{100} \sigma$_u) = 3265 MPa

Length = 380 mm

a) We know,

$\dpi{100} \delta$ = P L /A E

where, P = Load

$\dpi{100} \delta$ = deflection

A = cross sectional area

Let initial diameter be D, then Area = ($\dpi{100} \pi$/4)(D - 0.28)²

Putting values,

0.50 = 6660 x 380 / [($\dpi{100} \pi$/4)(D - 0.28)² x 110000]

=> ($\dpi{100} \pi$/4)(D - 0.28)² = 46.015

=> D = 7.94 mm

b) Stress = Load / area

= 13270 / ($\dpi{100} \pi$/4)(7.94)²

= 268.14 N/mm²

Strain = Stress / E

= 268.14 / 110000

= 0.00243

Yield stress = 240 MPa or 240 N/mm²

  => elastic strain = 240/11000

= 0.00218 < 0.0243

Since, elastic strain is less then strain at 13270 N , the strain is plastic

c) To resist the load of 13270 N without any plastic deformation, put strain = 0.00218

We know, stress = strain x modulus of elasticity

=> Stress = 0.0218 x 110000 = 240 N/mm²

also, stress = Load/Area

=> Area = Load /stress = 13270/240

=> ($\dpi{100} \pi$/4)(D)² = 55.29

=> D= 8.39 mm

d) True strain at yielding = stress / E

  We know,   $\dpi{100} \delta$ = P L /A E

=> $\dpi{100} \delta$ = P L /A E

Putting values,

  $\dpi{100} \delta$ = 240 x 380 / 110000 (P/A = 240 N/mm²)

= 0.829 mm

=> Actual length = 380 + 0.829 = 380.829

=> True strain = $\dpi{100} \delta$ /L = 0.829 / 280.829     

= 0.00295

Ans e) Engineering stress = Load / Initial area

= 6660 / ($\dpi{100} \pi$/4)(7.94)²

= 134.57 N/mm²

True stress = Load / Instantaneous area

= 6660 / ($\dpi{100} \pi$/4)(7.94 - 0.28)²

= 144.59 N/mm²