Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 5 (Q.No. 1)
1.
The maximum shear stress in a thin cylindrical shell subjected to internal pressure p is
Discussion:
30 comments Page 3 of 3.
Sanjay said:
8 years ago
I think C is right.
Sajith said:
9 years ago
C is the correct.
You are right. @sivakumar.
You are right. @sivakumar.
Baljinder singh said:
9 years ago
C is the correct answer.
Siva Kumar G said:
9 years ago
Radial stress(= 0) must be considered as the third principal stress.
And the maximum shear stress = (σ1 - σ3)/2
where
σ1 = Pd/2t ..........hoop stress
σ2 = Pd/4t ..........longitudinal stress
σ3 = 0 ................radial stress
So, the answer would be C.
And the maximum shear stress = (σ1 - σ3)/2
where
σ1 = Pd/2t ..........hoop stress
σ2 = Pd/4t ..........longitudinal stress
σ3 = 0 ................radial stress
So, the answer would be C.
Swapnil said:
9 years ago
Yes right @Renjith.
Renjith said:
9 years ago
Max.shear stress = Max. Principal shear stress due to hoop stress (pd/2t) in circumferential direction and maximum axial stress (pd/4t) in axial direction. i.e., (pd/2t - pd/4t) / 2 = pd/8t. Answer (D) is correct.
Manojhna said:
9 years ago
Sorry for the previous one.
?1 = pd/2t.
?2 = pd/4t.
Max shear = max of { (pd/2t-pd/4t) /2, (pd/2t) /2, (pd/4t) /2}.
?1 = pd/2t.
?2 = pd/4t.
Max shear = max of { (pd/2t-pd/4t) /2, (pd/2t) /2, (pd/4t) /2}.
Debasish said:
9 years ago
It should be pd/4t, 1/2 of hoop stress. So option C is correct.
Arpit Ghatiya said:
1 decade ago
The maximum shear stress is in the plane containing '$1' and 'p' and is equal to 1/2*($1) = pd/4t. So the answer is [D]. Here, $1 is the circumferential stress (sigma 1).
Gaurav said:
1 decade ago
How it should be B?
i.e. Hoop's stress = pd/2t.
i.e. Hoop's stress = pd/2t.
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