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2 years ago

What is true about the normal stresses when the in-plane shear stress is maximum? select all that apply.

Engineering

1 answer:

Darina [25.2K]2 years ago

3 0

When the in-plane shear stress is at its highest, it is true that the normal stresses are equal.

Their combined value is σn+σt.

<h3>What is in-plane shear stress?</h3>

6.3 ksi is the maximum in-plane shear stress. 10.2 ksi is the maximum out-of-plane shear stress.
By examining Mohr's circle, we may understand why out of plane shear stress is the highest.
Maximum in-plane shear stress is indicated by the inner blue circle radius (6.3 ksi).
Maximum shear stress planes are 45 degrees from the major planes.
The principle stress is the highest stress that may be created in a plane, and the principal stress refers to the principal plane when shear stress is zero.

The complete question is:

What is true about the normal stresses when the in-plane shear stress is maximum? Select all that apply.

a) They are equal.

b) They are zero.

c) Their sum is equal to σn+σt.

d) They are maximum.

e) They are equal to shear stress.

To learn more about in-plane shear stress, refer to:

brainly.com/question/15517350

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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

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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 $0.435$ .

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