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SCORPION-xisa [38]
3 years ago
10

What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?

Engineering
1 answer:
Fofino [41]3 years ago
4 0

Answer:

DRAWING LOAD IS  3.67 A_{O}\sigma

Explanation:

wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire

drawing load for wire drawing is given as P = A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

A_{f} = A_{O}- 0.6A_{O}

    = 0.4 A_{O}

Due to change in area ,drawing load p is

p = 0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})

p = 3.67 A_{O}\sigma

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Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)
Novosadov [1.4K]

Answer:

a)  ∝  and β

   The phase compositions are :

    C_{\alpha } = 5wt% Sn - 95 wt% Pb

    C_{\beta } =  98 wt% Sn - 2wt% Pb

b)

The phase is; ∝  

The phase compositions is;   82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

The phase compositions are :

C_{\alpha } = 5wt% Sn - 95 wt% Pb

C_{\beta } =  98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝  

The phase compositions is;  82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

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3 years ago
What is an example of a product made of textile?
Otrada [13]

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4 0
2 years ago
Consider a simply supported rectangular beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Norm
yulyashka [42]

Answer:

beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Normal weight concrete is used for the beam. This beam carries a total factored load of 9.4 kips. The beam is reinforced with tensile steel, which continues uninterrupted into the support. The concrete has a strength of 4000 psi, and the yield strength of the steel is 60,000 psi. Using No. 3 bars and 60,000 psi steel for stirrups, do the followings:

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3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

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Answer:

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Explanation:

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