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3 years ago

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Problem 2. The length of a side of the square block is 4 in. Under the application of the load V, the top edge of the block disp

laces 1 16 in. (dashed lines show displacement). Determine the shear strain at corner A and the shear strain at angle COD.

1 answer:

White raven [17]3 years ago

8 0

Answer and Explanation:

The answer is attached below

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Steam undergoes an isentropic compression in an insulated piston–cylinder assembly from an initial state where T1 5 1208C, p1 5

balu736 [363]

Answer:

temperature T2 = 826.9°C

$\frac{W}{m}$ = -1142.7 kJ/kg

Explanation:

given data

initial state temperature = 120°C

final state pressure p1 = 1 bar

pressure p2 = 100 bar

solution

we use here super heated water table A6 that is

specific internal energy u1 = 2537.3 kJ/kg

specific entropy s1 = 7.4668 kJ/kg.K

and

here specific entropy stage 1 = stage 2

so for specific entropy and pressure 100 bar

specific internal energy u2 = 3680 kJ/kg

and temperature T2 = 826.9°C

so here now we get specific work of steam is

ΔU = -W ...........1

m ( u1 - u2) = W

$\frac{W}{m}$ = u1 - u2

$\frac{W}{m}$ = 2537.3 - 3680

$\frac{W}{m}$ = -1142.7 kJ/kg

3 0

3 years ago

7 to 1 inch above the stock

Marysya12 [62]

B. Pass a written test on safety and operating procedures of the table saw with a ... 1. Blade guard. 8. Fence lock. 2. Table insert. 9. Fence carriage. 3. Table. 10. ... Never raise the saw blade more than ¼ inch above the material being cut. ... should extend no more than ______ inches above the stock being cut. a. ¼ b. 1/8 c.

6 0

3 years ago

The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a

Arturiano [62]

Answer:

Explanation:

Given that:

Torque T = 2300 lb - ft

Bending moment M = 1500 lb - ft

axial thrust P = 2500 lb

yield points for tension σY= 100 ksi

yield points for shear τY = 50 ksi

Using maximum-shear-stress theory

$\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}$

where;

$A = \pi c^2$

$I = \dfrac{\pi}{4}c^4$

$\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}$

$\tau_A = \dfrac{T_c}{\tau}$

where;

$\tau = \dfrac{\pi c^4}{2}$

$\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{55200 }{\pi c^3}}$

$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)$

Let say :

$|\sigma_1 - \sigma_2| = \sigma_y$

Then :

$2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)$

$(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6$

$6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0$

According to trial and error;

c = 0.75057 in

Replacing c into equation (1)

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma _1 = 22193 \ Psi$

$\sigma_2 = -77807 \ Psi$

The required diameter d = 2c

d = 1.50 in or 0.125 ft

6 0

3 years ago

For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True

Verdich [7]

Answer:

B.

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table for a NAND gate with 2 inputs is as follows:

0 0 1

0 1 1

1 0 1

1 1 0

The truth table for an OR gate, is as follows:

0 0 0

0 1 1

1 0 1

1 1 1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

1 1 0 0 0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

3 0

3 years ago

If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head

FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.

8 0

3 years ago