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
Discussion:
26 comments Page 3 of 3.
Sumit said:
7 years ago
Here, it comes (20+10/2).
Suresh said:
7 years ago
Max shear stress = √((sigma max +sigma min)/2)^2 + (shear stress)^2.
Here, σ max 20 and σ min -10(compressive). Shear stress is zero (if not given).
Here, σ max 20 and σ min -10(compressive). Shear stress is zero (if not given).
Basar sk said:
6 years ago
Unlike in nature i.e. sigma 1=20 and sigma 2 = -10.
So max shear stress = (20 - (-10) )/2 = 15.
So max shear stress = (20 - (-10) )/2 = 15.
(1)
Tapas said:
5 years ago
Since the tensile and compressive stresses are in opposite direction, therefore the sign convention of these stresses should also be in opposite direction.
ANIL said:
3 years ago
maximum shear stress = (stress1- stress2)/2.
= (20-(-10))/2.
= 30/2,
= 15.
= (20-(-10))/2.
= 30/2,
= 15.
(3)
Aleem said:
3 years ago
I agree with you @Sagar M.
Answer is 15KN/m2 because compressive stress will be taken as negative.
Answer is 15KN/m2 because compressive stress will be taken as negative.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers