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.

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?

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.

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

The y-axis is compressive.

Mdsufiyan said:   8 years ago
How to calculate it?

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.

Akshay tc said:   8 years ago
5 or 15?

Please explain the answer.

Qasim said:   7 years ago
{20-(-10)}/2=15 as 10kN is compressive force (-ve).

Prateek kumar said:   7 years ago
The maximum shear stress is the radius of Mohr's circle.

ie, √(σx- σy/2)^2 + τxy^2.

Thus σy is compressive so negative.
σx-(- σy)/2 root and square get cancelled.
So, (20+10)/2 = 15 KN/m^2 ans.

MANOJ said:   7 years ago
5 is the answer.

Tmax=""(σx - σy)^2/2.
""(20-10)^2/2=5kN/m2.

Bijo rajan said:   7 years ago
Actually maximum means to add them, and the compression is in a negative sign it gets reduced and becomes minus.


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