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.
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.
D rahul said:
9 years ago
@Sandeep
The correct method for max. shear stress is (max. principal stress - min. principal stress)/2.
i.e answer must be 5.
The correct method for max. shear stress is (max. principal stress - min. principal stress)/2.
i.e answer must be 5.
Sandeep m said:
1 decade ago
Max. Shear stress = Tensile stress - Compressive stress\2.
Compressive stress taken as negative.
Answer is 15.
Compressive stress taken as negative.
Answer is 15.
Gouse said:
1 decade ago
I didn't get that can you elaborate please.
Sagar m said:
1 decade ago
Answer is right. Because,
Compressive stress value will be taken as negative. So,
20-(-10)/2 = (20+10)/2=15KN/m2.
Compressive stress value will be taken as negative. So,
20-(-10)/2 = (20+10)/2=15KN/m2.
Pavan said:
1 decade ago
Answer is wrong maximum shear stress is 20-10/2 = 5KN/m2.
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