Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 2 (Q.No. 39)
39.
A body is subjected to two normal stresses 20 kN/m2 (tensile) and 10 kN/m2 (compressive) acting perpendicular to each other. The maximum shear stress is
5 kN/m2
10 kN/m2
15 kN/m2
20 kN/m2
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 2 of 3.

Akshay tc said:   8 years ago
5 or 15?

Please explain the answer.

Jyoshna said:   8 years ago
Tensile(+) ; comp(-ve) formula max shear stress is (sigma1 - sigma2)÷2.
Sigma1=20, sigma2=10.
Now substitute [+20 -(-10)]÷22 =15.

Mdsufiyan said:   8 years ago
How to calculate it?

Ch v gopinath said:   8 years ago
[20-(-10)]/2.

The y-axis is compressive.

Abhishek Aryan said:   8 years ago
@Hardik.

IN THIS CASE, sigma(max) is 20 and sigma(min) is -10, and the formula to get max shear stress is [sigma(max)- sigma(min)]/2 so it would give us 15 or [20-(-10)]/2 = 15.

Hardik said:   8 years ago
Is that necessary to keep the numbers in a single equation? Couldn't it be possible that its like (20-10)-2, and if it's wrong what the reason behind this. Is there any formula?

Adhikary saroj said:   9 years ago
Answer is c. I agree.

Gaurav said:   9 years ago
Max. shear stress = radius of mohr's circle = [{(sigma1 - sigma2)/2}^2 + tow^2]^0.5.

Tahir said:   9 years ago
They are acting perpendicular to each other. What does this mean?

Ajay said:   9 years ago
Sign conventions: Tensile +ve and Compressive is -ve.

So, both Sagar and Sandeep answers are correct.


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