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 2 of 3.
Ajay said:
9 years ago
Sign conventions: Tensile +ve and Compressive is -ve.
So, both Sagar and Sandeep answers are correct.
So, both Sagar and Sandeep answers are correct.
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)
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)
Gaurav said:
9 years ago
Max. shear stress = radius of mohr's circle = [{(sigma1 - sigma2)/2}^2 + tow^2]^0.5.
ANIL said:
3 years ago
maximum shear stress = (stress1- stress2)/2.
= (20-(-10))/2.
= 30/2,
= 15.
= (20-(-10))/2.
= 30/2,
= 15.
(3)
MANOJ said:
7 years ago
5 is the answer.
Tmax=""(σx - σy)^2/2.
""(20-10)^2/2=5kN/m2.
Tmax=""(σx - σy)^2/2.
""(20-10)^2/2=5kN/m2.
Tahir said:
9 years ago
They are acting perpendicular to each other. What does this mean?
Pavan said:
1 decade ago
Answer is wrong maximum shear stress is 20-10/2 = 5KN/m2.
Qasim said:
7 years ago
{20-(-10)}/2=15 as 10kN is compressive force (-ve).
Gouse said:
1 decade ago
I didn't get that can you elaborate please.
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