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Tamiku [17]
3 years ago
11

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

, given that the included die angle is 300 with a coefficient of friction (μ) of 0.08 and the yield stress for copper is 350 MPa.
Engineering
1 answer:
dybincka [34]3 years ago
8 0

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

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denis23 [38]

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature T_\infity = 15 degree celcius

T_S =  220 + 273 = 473 K

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

Q = hA(T_S - T_{\infty}) \eta

= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8

Q = 125.538 W

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3 years ago
Discuss in detail the following methods used to redistribute income and wealth in cash grants?​
aev [14]

Answer:

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Redistribution tax policy should not be confused with predistribution policies. "Predistribution" is the idea that the state should try to prevent inequalities from occurring in the first place rather than through the tax and benefits system once they have occurred. For example, a government predistribution policy might require employers to pay all employees a living wage, not just a minimum wage, as a "bottom-up" response to widespread income inequalities or high poverty rates.

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7 0
3 years ago
The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

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s_max = 0.8394m

8 0
3 years ago
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26
scZoUnD [109]

Answer:

The strength coefficient is 625 and the strain-hardening exponent is 0.435

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  (K) and the (n).

\sigma =K\epsilon^n

We will plug the values in the formula.

250=K\times (0.12)^n\\350=K\times (0.26)^n

We will solve these equation.

K=\frac{250}{(0.12)^n} plug this value in 350=K\times (0.26)^n

350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\  \\1.4=(2.17)^n

Taking a natural log both sides we get.

ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435

Now, we will find value of K

K=\frac{250}{(0.12)^n}

K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625

So, the strength coefficient is 625 and the strain-hardening exponent is 0.435.

5 0
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Hatshy [7]

Answer:

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