Respuesta :
Answer:
Explanation:
A) we know that volume is given as V
[tex]V =\frac{\pi}{4} D^2 h[/tex]
where D = 1.5 in , h = 2.0 in
so [tex]V = \frac{\pi}{4} 1.5^2\times 2 = 3.53in^3[/tex]
[tex]Area =\frac{\pi}{4} D^2 = \frac{\pi}{4} \times 1.5^2 = 1.76 in^4[/tex]
yield strenth is given as[tex] \sigma_y = \frac{force}{area} = \frac{140,000}{1.76}[/tex]
[tex]\sigma_y = 79.224 ksi[/tex]
b)
elastic strain[tex] \epsilon = \frac{\sima_y}{E} = \frac{79.224}{30\times 10^3} = 0.00264[/tex]
strain offsets = 0.00264 + 0.002 = 0.00464 [where 0.002 is offset given]
[tex]\frac{\delta}{h} = 0.00464[/tex]
[tex]\frac{h_i -h_o}{h_o} = 0.00464[/tex]
[tex]h_i = 2\times(1-0.00464) = 1.99 inch[/tex]
area [tex]A = \frac{volume}{height} = \frac{3.534}{1.9907} = 1.775 in^2[/tex]
True strain[tex] \sigma = \frac{force}{area} = \frac{140,000}{1.775 in^2} = 78,862 psi[/tex]
At P= 260,000 lb ,[tex] A = \frac{3.534}{1.6} = 2.209 inc^2[/tex]
true stress [tex]\sigma = \frac{260,000}{2.209} = 117,714 psi[/tex]
true strain [tex]\epsilon = ln\frac{2}{1.6} = 0.223[/tex]
flow curve is given as \sigma = k\epsilon^n
[tex]\sigma_1 = 78,862 psi[/tex]
[tex]\epsilon_1 = 0.00464[/tex]
[tex]\sigma_2 = 117,714 psi[/tex]
[tex]\epsilon_2 = 0.223[/tex]
so flow curve is
[tex]78,868 = K 0.00464^n[/tex] .........1
[tex]117,714 = K 0.223^n[/tex] .........2
Solving 1 and 2
we get
n = 0.103
and K =137,389 psi
Strength coffecient = K = 137.389ksi
strain hardening exponent = n = 0.103
