Mechanical Engineering - Strength of Materials - Discussion

I think C is the correct one, when not mentioned properly, we take absolute shear stress and absolute shear stress will be the maximum of all three shear stresses which we will get after subtracting hoop stress(pd/2t), longitudinal stress(pd/4t) and the radial stress (ZERO in this case) with each other one by one and then dividing it by 2. Here we are getting maximum absolute shear stress when we will subtract hoop stress with radial stress and then divide it by 2 i.e. (pd/2t - 0)/2 = pd/4t. So option C is correct.

Max shear stress = 1/2 (max. Principal-min. Principal stress).

Now max. Principal stress = hoops (pd/2t).

Also, min principal stress= radial (-P= approx 0 as compared to other stresses, so it is taken as 0).

So now, solve for this and we get : max shear stress as = pd/4t which is also known as absolute maximum shear stress.

And pd/8t is max. Wall shear stress.

Maximum shear stress in the plane is pd/8t but here it is not mentioned that it is in the plane. There are two shear stresses one is in the plane and the other is out of the plane. So the maximum shear stress is pd/4t that is out of plane stress. That's why the correct answer is pd/4t that is option C.

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.

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.

C is the right answer. If it was asked that maximum shear stress in the plan then pd/8t is the maximum shear stress in the plane. Otherwise, pd/4t is absolute maximum shear stress.

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).

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}.

If you say plane max shear stress than C is correct, otherwise of only max shear stress D is correct.

Maximum shear stress = 1/2(max. Principal stress- Min. Principal stress).
So, the right answer is D.