# Caulculating maximum shear stress at certain points

I have a structures problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track. Any assistance would be highly appreciated.

My work is as follows:

At point B:

$T = Fd$, where $d = 300\text{ mm}$ and $F = 100\text{ N}$, therefore $T = 30,000\text{ Nmm}$

From torsion, shear stress $\tau = Tr/J$

$$J = \dfrac{\pi}{2}\cdot(r_1^4 - r_2^4)$$

Therefore $\tau = \dfrac{30,000 \cdot 30 \cdot 2}{\pi(30^4 - 25^4)} = 1.37\text{ MPa}$

However then for point A:

$T = Fd$, I believe $d$ now changes to 330 mm as you have to add the outer radius of the bar so $T = 33,000\text{ Nmm}$

Following the same process used for B:

$$t = \dfrac{33,000 \cdot 30 \cdot 2}{\pi(30^4 - 25^4)} = 1.50\text{ MPa}$$

This is clearly not the answer my lecturer gave. I am a confused as to where I went wrong. I am aware that I did not use the value 250mm but I am unsure where it applies. What makes points A and B different, aside from their distance from the force? I would really appreciate any assistance in the matter, thank you so much for reading.

• You need to add the transversal shear too, which is zero at B and if I remember correctly 4/3 tau (average), at A. After all the pipe is acting as a cantilever beam too. Also the torque lever arm is same 300 mm in both cases. – kamran Mar 9 '18 at 17:17

The reason is, that the pipe is also subject to transverse shear.

This is due to the fact, that this lever doesn't apply a "pure" torque moment, but an eccentric force, which results in torque and shear. Now you get two different shear stresses when you consider A and B. As this pipe is subject to the shear force $V_z=100N$, the transverse shear "flows" from B over A towards the opposite side of B.

(note: There would be a bending moment $M_y=100N\cdot L_{pipe}$ the base of the pipe as well, but it doesn't result in transverse shear.)

You can determine the transverse shear stress, using the following formula $$\tau=\frac{V_z\cdot Q_y}{I_y\cdot t}$$ Take a look at this question for further explanation of transverse shear.

further explanation [edited]

Take a look at this graphic, especially at the transformed coordinated systems. (By convention, a dot in a circle depicts a force/direction facing away from you, and a cross in a circle is a force/direction facing towards you)

In sketch $II$ you can see, how the $100N$ force does create a bending moment, the orange line is a sketch of the deformation curve, that said force would cause.

On the other hand, in sketch $III$ you can see two opposing eccentric forces, resulting in the same torque $T=2\cdot(50N\cdot 0.3m)=30Nm$, which would not result in a bending moment, as they cancel each other out. $M_y=50N\cdot L_{pipe}-50N\cdot L_{pipe}=0$. Furthermore, there would be no transverse shear stress in the pipe as well.

• I am sorry, but I see neither the transverse stress or the bending moment. The question seems to indicate that the 100N force is being used for torsion. So I don't see where the transverse stress is (as I do not see where the shear load is) nor how the pipe bends. I also don't understand why the torsion at points A and B is the same when they are different distances away from the force. Thank you so much for your assistance. – gunter Mar 11 '18 at 7:30
• I edited the answer and added further detail at the bottom. I hope it is more comprehensible now. – Andrew Mar 11 '18 at 13:04
• Thank you so much for your in depth explanation and assistance it is much appreciated. – gunter Mar 14 '18 at 13:34