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olchik [2.2K]
3 years ago
13

IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The follow

ing properties of annealed copper are given: strength coefficient is 90,000 psi; true strain at the onset of non-uniform deformation is 0.45; and, engineering strain at yield is 0.11. The coefficient of friction between strip and roll is 0.2. The roll radius is 14inches and the rolls rotate at 150 rpm. Calculate the roll-strip contact length. Calculate the absolute value of thetrue strain that the strip undergoes in this operation. Determine the average true stress of the strip in theroll gap. Calculate the roll force. Calculate the horsepower required.
Engineering
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, $h_0=2.2$ inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

$\Delta h = H_0-h_f=\mu^2R$

     $=0.2^2 \times 14 = 0.56$ inches

$h_f = 2.2 - 0.56$

     = 1.64 inches

Roll strip contact length (L) = $\sqrt{R(h_0-h_f)}$

                                             $=\sqrt{14 \times 0.56}$

                                             = 2.8 inches

Absolute value of true strain, $\epsilon_T$

$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$

Average true stress, $\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$ Psi

Roll force, $L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$

                                 = 788,900 lb

For SI units,

Power = $\frac{2 \pi FLN}{60}$  

           $=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$

           = 10399.81168 W

Horse power = 13.9357

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