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3 years ago

Describe in your own words the three strengthening mechanisms

Engineering

1 answer:

TEA [102]3 years ago

3 0

Answer:

a) Decreasing grain size decreases the amount of possible pile up at the boundary, increasing the amount of applied stress necessary to move a dislocation across a grain boundary. ... Grain sizes can range from about 100 μm (0.0039 in) (large grains) to 1 μm (3.9×10−5 in) (small grains).

b) Solid solution strengthening is a type of alloying that can be used to improve the strength of a pure metal. The technique works by adding atoms of one element (the alloying element) to the crystalline lattice of another element (the base metal), forming a solid solution.

c) Work hardening, also known as strain hardening, is the strengthening of a metal or polymer by plastic deformation. ... This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material.

Explanation:

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X cotx expansion using maclaurins theorem.

Lemur [1.5K]

It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.

<h3>What is Maclaurin Expansion?</h3>

The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

<h3>What is the explanation for the above?</h3>

as indicated above, the Maclaurin infinite series expansion is given as:

F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

If F(0) = Cot 0

F(0) = ∝ = 1/0

This is not definitive,

Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.

Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
#SPJ1

4 0

1 year ago

A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

Mazyrski [523]

Answer:

P > 142.5 N (→)

the motion sliding

Explanation:

Given

W = 959 N

μs = 0.3

If we apply

∑ Fy = 0 (+↑)

Ay + By = W

If Ay = By

2*By = W

By = W / 2

By = 950 N / 2

By = 475 N (↑)

Then we can get F (the force of friction) as follows

F = μs*N = μs*By

F = 0.3*475 N

F = 142.5 N (←)

we can apply

P - F > 0

P > 142.5 N (→)

the motion sliding

6 0

3 years ago

If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam

Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

$\zeta =\dfrac{C}{C_c}$

Where C is the damping coefficient and Cc is the critical damping coefficient.

$C_c=2\sqrt{mK}$

So from above we can say that

$\zeta =\dfrac{C}{2\sqrt{mK}}$

$\zeta \alpha \dfrac{1}{\sqrt K}$

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0

3 years ago

• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma

elena-s [515]

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

$T(n) = 10\times 2n$

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

$\frac{T(k)}{T(n)} = 64$

$\frac{20k}{20n} = 64$

k = 64n

Thus the new machine will process 64 inputs in the time duration T

8 0

3 years ago

1. What affects the weight of an airplane?<br> a. Mass<br> b. Matter<br> c. Materials<br> d. Speed

BARSIC [14]

Mass affects the weight of an airplane because everything has mass

5 0

2 years ago

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