# Calculating shear stress in a system having friction

I am trying to create a shear stress for an introduction to mechanics course. All the questions I have encountered with shear stress assume a free body which is transmitting force. Typically friction is ignored but that may be to idealize the problem.

My question, what effect does friction have on calculating shear stress?

Is the shear stress Applied Force / Area

or is it Net Force / Area

or is something else at play?

Question with numbers: A garbage can (0.75 m in diameter) is filled with 25kg of paper. The garbage can sits on a floor u(s) = 0.3 and a 125N lateral force is applied.

Area = pi*(d^2)/4 = 9/64*pi = 0.442 m^2

Force Normal = (25 kg)(9.81 m/s^2) = 245 N

Maximum Force of Friction = (0.3)(245 N) = 73.5 N

Net force = applied force - friction force = 125 N - 73.5N = 51.5N

Assumptions: Shear stress is generated when the lateral force is applied. Since Net force is not equal to 0.

Shear stress = Force parallel to the surface/Area

So... Is the shear stress

Applied Force / Area

(125 N) /(0.442 m^2)

or the

Net Force / Area

(51.5 N) / (0.442 m^2)

or is something else at play.

There is something else at play. The problem you gave is a little complicated, so let's consider a simpler problem in two dimensions. Image an infinite 2D beam sitting on the ground. The beam is infinite in the horizontal direction, which we will call the $x$ direction. The beam has a height $h$ in the vertical direction, which we will call the $y$ direction.

Now let's suppose the beam has a mass density $\rho$, so that the force per unit length in the $y$ direction is $-\rho g h \hat{y}$. If the beam is at rest on the ground, this means the stress in the beam is $$\newcommand{\s}{\boldsymbol{\sigma}}\s_\textrm{rest}=\rho g (h-y) \hat{y}\otimes \hat{y}.$$ This is consistent with the body force distribution $-\rho g \hat{y}$, since $\nabla \cdot \s = -\rho g \hat{y}$.

Now let's consider what happens if we apply a force on the upper surface at height $h$, in the $\hat{x}$ direction. Let's make the magnitude $f_x$ of this force small enough that it does not overcome static friction. Then the stress in the beam becomes $$\s_\textrm{static} = \rho g (h-y) \hat{y}\otimes \hat{y} + f_x\left(\hat{x}\otimes \hat{y} + \hat{y}\otimes \hat{x}\right).$$ This should still all be old hat to you because the tangential force per unit surface was $f_x$, and this is exactly the shear stress. In fact the form given above still holds even when the block is moving at a constant velocity so that the force on the top is exactly balanced by the kinetic friction force on the bottom.

So now lets come to the case where the force per unit length on the top $f_{xt}$ overcomes the kinetic friction force on the bottom $f_{xb}$. In this case we can see that the shear stress is not uniform throughout the beam. We know that the shear stress is $\sigma_{xy}$, and therefore the shear stress is $\hat{x} \cdot \s \cdot \hat{y}$, and since the surface force is $\s \cdot \hat{y}$, we have that the shear stress at a surface is the $x$ component on the surface force. But at the top surface, this $x$ component is $f_{xt}$, and at the bottom surface this $x$ component is $f_{xb}$. Since these two are not equal, the shear stress must not be uniform.

The non-uniformity of the stress is directly related to the fact that the beam is accelerating in the $x$ direction. The acceleration $a$ of the beam must be given by $f_{xt}-f_{xb} = \rho h a$. This acceleration can only be caused by an unbalanced stress distribution in the beam. Specifically, we must have that $$- \nabla \cdot \s = \rho a \hat{x} + \rho g \hat{y}.$$ Above the $\rho g \hat{y}$ term is the unbalanced stress that serves to cancel gravity. Since the system is uniform in the $x$ direction, we must have that $\partial_x$ acts to give zero so that $\left( \nabla \cdot \s\right)_j = \partial_y \sigma_{yj}$, so that in particular, setting $j=x$, we get $$\partial_y \sigma_{yx} = -\rho a.$$ This tells us that the acceleration in the $x$ direction must be caused by a shear stress gradient. Since the acceleration is uniform, this shear stress gradient is constant, so the shear stress must be a linear function of $y$. In fact, we were able to calculate $a$ using newton's second law, and we can plug this $a$ in to get $$\partial_y \sigma_{yx} = \frac{f_{xt}-f_{xb}}{h}.$$ We can now do a cross-check to see if this shear stress gradient is consistent with our boundary conditions that the shear stress at the top is $f_{xt}$ and the shear stress at the bottom is $f_{xb}$. Let's check if the integral of the stress gradient we calculated from Newton's law is indeed the difference in shear stress: $$f_{xt}-f_{xb} \stackrel{?}{=} \int_0^h \partial_y \sigma_{yx} dy = \int_0^h \frac{f_{xt}-f_{xb}}{h} dy =f_{xt}-f_{xb}.$$ Since the far left and right hand sides are equal, we see that the stress gradient caused by the acceleration does exactly account for the differences between the force applied at the top and at the bottom. So the answer in this case is that the shear stress is equal to the applied force at the top, but the net force at the bottom.

Now your problem with the trash can is more complicated. In my toy problem, the translation symmetry in the $x$ direction simplified things greatly. In the trash can problem, I don't think you will have a simple constant shear stress gradient. In practice, you would have to solve a partial differential equation to get the stress distribution. But hopefully I have given you a qualitative conceptual idea of why the shear stress is not constant when there is acceleration.

• Thank-you. Main take away was that the force is non linear. at the top of the body, shear stress = applied force (as the object has to keep itself together). At the bottom, – user150972 Apr 5 '17 at 17:17
• Thank-you. Main take away was that the force is non linear. In the simpler case you presented. At the top of the body, shear stress = applied force for that portion of the object (as the object has to keep itself together and you are assuming a distributed force). At the bottom, shear stress = net force (because the floor exerts a force as well) – user150972 Apr 5 '17 at 17:26