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 1 of 3.
Aleem said:
2 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)
ANIL said:
2 years ago
maximum shear stress = (stress1- stress2)/2.
= (20-(-10))/2.
= 30/2,
= 15.
= (20-(-10))/2.
= 30/2,
= 15.
(3)
Tapas said:
4 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.
Basar sk said:
5 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)
Suresh said:
6 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).
Sumit said:
6 years ago
Here, it comes (20+10/2).
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.
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.
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.
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.
Qasim said:
7 years ago
{20-(-10)}/2=15 as 10kN is compressive force (-ve).
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