Question

ive already gotten 2 wrong answers this is my last attempt please be correct!!!!!!!!!!!!!!

Incorrect Consider the brass alloy the stress-strain behavior of which is shown in the Animated Fioure 2.12. A cylindrical specimen of this alloy 15 mm in diameter and 214 mm long is to be pulled in tension Calculate the force necessary to cause 0.00788 mm reduction in diameter. Assume a value of 0.35 for Poisson's ratio. 26507.1 the tolerance is +/-2% Click if you would like to Show Work for this question Oen Show.Work References Tensile strength 450 MPa (65,000 psi) Strain = 0 Stress = 0 MPA Stress -0 psi 500 70 Strain 0 Stress 0 MPA |Stress = 0 psi 60 - 400 103 psi -50 MPa 40 F Yield strength 300 250 MPa (36,000 psi) 40 200 30 30 200 20 F 100 20 10 100 10 0.30 0.40 0.20 0.10 Strain

Answer 1

Solution:

For the cylindrical specimen

Diameter, d = 15 mm

Length, L = 214 mm

Change in diameter, $\Delta$d = 0.00788 mm

Poission's ratio, $\mu$= 0.35

Lateral strain,

e_la = ($\Delta$d/d) = (0.00788 mm) /(15 mm) = 0.0005253333

longitudinal strain, e_lo = e_la/$\mu$

e_lo = 0.0005253333/0.35

e_lo = 0.001501

At elongation strain e_lo =0.001501

Stress = 128 MPa

Therefore, required force, F

= (128)$\pi$d²/4

= (128)$\pi$(15)²/4

= 22619.5 N