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kicyunya [14]
1 year ago
8

Is annealed polycrystalline copper a material that work hardens significantly or a material that exhibits a low work-hardening r

ate
Engineering
1 answer:
Olegator [25]1 year ago
6 0

Those that harden under strain, such as the aluminum-magnesium alloys used in beverage cans and the copper-zinc alloy, brass, used for cartridges, which show more strain hardening than pure copper or aluminum, respectively.

When a material is deformed under a substantial amount of strain, strain hardening is seen as a strengthening process. Lamellar crystals and chain molecule orientation on a vast scale are the culprits. When plastic materials are stretched past their yield point, this phenomena is frequently seen. When a metal is stretched past its yield point, strain hardening occurs. The metal appears to get stronger and harder to deform as more stress is needed to cause additional plastic deformation. Strain hardening is directly related to fatigue.

Learn more about strain hardening here-

brainly.com/question/15058191

#SPJ4

You might be interested in
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
How did Atlantis benefit from lessons learned in construction of earlier orbiters?
Alenkasestr [34]

Answer:

Atlantis benefited from lessons learned in the construction and testing of Enterprise, Columbia and Challenger. ... The Experience gained during the Orbiter assembly process also enabled Atlantis to be completed with a 49.5 percent reduction in man hours (compared to Columbia).

Explanation:

8 0
3 years ago
A 2-bit positive-edge triggered register has data inputs d1, d0, clock input clk, and outputs q1, q0. Data inputs d1d0 are 01 an
ale4655 [162]

Answer:

  q1q1 ⇒ 01

Explanation:

The outputs of a positive edge triggered register will match the inputs after a rising clock edge.

  q1q1 ⇒ 01 . . . . matching d1d0 = 01

7 0
3 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferre
docker41 [41]

Answer:

Initial temperature = 170. 414 °C

Total mass = 94.478 Kg

Final volumen = 33.1181 m^3

Diagram  = see picture.

Explanation:

We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.  

The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).  

For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.  

to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.

For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:

Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg

Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg

if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.

Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg

Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg  

Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg

If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.  

If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.  

Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg

We already know that this a close system so the mass in it keeps constant during the process.

 

If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.  

Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3

You can the P-v diagram for this system in the picture.  

For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.  

Quality of steam = mass of steam / total mass of the system

Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.  

Since this a low value we can say that we are very close the saturated liquid point in the diagram.  

6 0
3 years ago
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