Ask question

Ask question

All categories

2 years ago

12

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

GPa (15.0 ×106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in2 ) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?

1 answer:

Alona [7]2 years ago

3 0

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

You might be interested in

A steel plate shine but wooden vessel desnt

Tcecarenko [31]

<u>Answer</u> :

A new steel plate shines but an wooden vessel does not shine because new steel plate has lustre. ... Because steel absorbs very little amount of light and reflects back the light but in case of wooden vessel it has dull appearence and reflects very little amount of light so it does not shine.

Please mark as brainliest. . . ☺

4 0

2 years ago

Read 3 more answers

A tank contains 100 gal of water and 50 oz of salt.water containing a salt concentration of 1 4 (1 1 2 sin t) oz/gal flows into

Alchen [17]

Answer:

Explanation:

Heres the possible full question and solution:

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

a. Find the amount of salt in the tank at any time.

b. Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.

c. The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

solution

a)

Consider the tank contains 100gal of water and 50 oz of salt

Assume that the amount of salt in the tank at time t is Q(t).

Then, the rate of change of salt in the tank is given by $\frac{dQ}{dt}$ .

Here, $\frac{dQ}{dt}$ =rate of liquid flowing in the tank - rate of liquid flowing out.

Therefore,

$Rate_{in} =2gal/min \times \frac{1}{4} (1+ \frac{1}{2}sin t)oz/gal\\\\\\ \frac{1}{2} (1+ \frac{1}{2}sin t)oz/min\\\\\\Rate_{out}=2gal/min \times\frac{Q}{100}oz/gal\\\\\frac{Q}{50}oz/min$

Therefore,

$\frac{dQ}{dt}$ can be evaluated as shown below:

$\frac{dQ}{dt}=\frac{1}{2}(1+\frac{1}{2}\sin t)-\frac{Q}{50}\\\\\\\frac{dQ}{dt}+\frac{1}{50}Q=\frac{1}{2}+\frac{1}{4}\sin t$

The above differential equation is in standard form:

$\frac{dy}{dt}+Py=G$

Here, $P=\frac{1}{50},G=\frac{1}{2}+\frac{1}{4}\sin t$

The integrating factor is as follows:

$\mu(t)=e^{\int {P}dt}\\\mu(t)=e^{\int {\frac{1}{50}}dt}\\\mu(t)=e^{\frac{t}{50}}$

Thus, the integrating factor is $\mu(t)=e^{\frac{t}{50}}$

Therefore, the general solution is as follows:

$y\mu(t)=\int {\mu (t)G}dt\\\\Qe^{\frac{t}{50}}=\int {e^{\frac{t}{50}}(\frac{1}{2}+\frac{1}{4}\sin t) dt}\\\\Qe^{\frac{t}{50}}=\frac{1}{2}\int {e^{\frac{t}{50}}dt + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}\\\\\Qe^{\frac{t}{50}}=25 {e^{\frac{t}{50}} + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}+C...(1)$

Here, C is arbitrary constant of integration.

Solve $\int {\sin te^{\frac{t}{50}}} dt}$

Here $u = e^{\frac{t}{50}}$ and $v =\sin t$ .

Substitute u , v in the below formula:

$\int{u,v}dt=u\int{v}dt-\int\frac{du}{dt}\int{v}dt\dot dt\\\\\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{1}{50}\int{e^{\frac{t}{50}}\cos t}dt...(2)$

Now, take $u = e^{\frac{t}{50}}$ , $v =\sin t$

Therefore, $\int{e^{\frac{t}{50}}\cos t} dt=\int {e^{\frac{t}{50}}\sin t}dt - \frac{1}{50}\int{e^{\frac{t}{50}}\sin t}dt...(3)$

Use (3) in equation(2)

$\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{e^{\frac{t}{50}}}{50}\sin t - \frac{1}{2500}\int{e^{\frac{t}{50}}\sin t}dt\\\\\frac{2501}{2500}\int{e^{\frac{t}{50}}\sin t}dt={e^{\frac{t}{50}}\cos t}+\frac{e^{\frac{t}{50}}}{50}\sin t\\\\\int{e^{\frac{t}{50}}\sin t}dt=\frac{2500}{2501}{e^{\frac{t}{50}}\cos t}+\frac{50}{2501}e^{\frac{t}{50}}\sin t...(4)$

Use (4) in equation(l) .

$Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+C$

Apply the initial conditions $t =0, Q = 50$ .

$50=25-\frac{625}{2501}+c\\\\c=\frac{63150}{2501}$

So, $Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}$

Therefore, the amount of salt in the tank at any time is as follows:

$Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}e^{\frac{-t}{50}}$

b)

sketch the solution curve as shown in attachment as graph 1:

CHECK COMMENT FOR C

3 0

2 years ago

PLEASE HELPPP 100 POINTS HURRY !!!!Which diagram best illustrates the magnetic field of a bar magnet? A bar magnet with a north

Serggg [28]

I think this is right I hope this is right for you

7 0

2 years ago

Read 2 more answers

A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at

Stells [14]

The correct answer is <span>3) $K_f = \frac{1}{2}mv_0^2 + mgh$ .
</span>
In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
</span> $E=U_i+K_i=mgh + \frac{1}{2}mv_0^2$
<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
</span> $E=K_f$ <span>
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
</span> $K_f = mgh + \frac{1}{2}mv_0^2$ <span>
</span>

7 0

2 years ago

The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0

2 years ago