We were discussing section modulus of the beam and derivation for beam bending equation in our previous session. We have also discussed assumptions made in the theory of simple bending and expression for bending stress in pure bending during our last session.

Now we are going ahead to start new topic i.e. Shear stress distribution in circular section in the strength of material with the help of this post.

Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box which is provided below this post. So let us come to the main subject i.e. Shear stress distribution in circular section.

In our previous session, we were discussing the bending stress produced in a beam which is subjected to a pure bending. We have assumed there that beam will be subjected with a pure bending moment and shear force will be zero and hence shear stress will also be zero.

In actual practice, beam will be subjected with shear force also and therefore shear stress too. Shear force and hence shear stress will vary section to section.

We will see here the shear stress distribution across the various sections such as shear stress distribution in rectangular section, circular section, I section and T section. In this post, we will see shear stress distribution in circular section.

Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam.
We have following information from above figure
R = Radius of the circular section of the beam
F = Shear force acting on the circular section of the beam
N.A: Neutral axis of the beam section
EF:  Layer of the beam section at a distance y1 from the neutral axis of the circular section of the beam
A= Area of the section, where shear stress is to be determined
ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section

#### Shear stress at a section will be given by following formula as mentioned here

Where,
F = Shear force (N)
τ = Shear stress (N/mm2)
A = Area of section, where shear stress is to be determined (mm2)
ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m)
A. ȳ = Moment of the whole shaded area about the neutral axis
I = Moment of inertia of the given section about the neutral axis (mm4)
For circular cross-section, Moment of inertia, I = ПR4/4
b= Width of the given section where shear stress is to be determined (m)

Let us consider one strip of thickness dy and area dA at a distance y from the neutral axis of the section of the beam.

We can see here that width b will be dependent over the value of y and let us first determine the value of b and after that we will determine the value of area of small strip of thickness dy i.e. dA.

(b/2)2 = R2- y2
Let us secure the value of the area dA of the strip of thickness dy and we can write it as mentioned here
Let us find the value of moment of this small strip area dA about the neutral axis and after that we will integrate the equation of moment of this area dA about the neutral axis between limits y1 to R and that result will be the moment of whole shaded area about the neutral axis i.e. A. ȳ.

Let us see the following figure, where we have determined the moment of area dA about the neutral axis and also we have integrated the secured equation between limits y1 to R in order to secure the value of moment of whole shaded area about the neutral axis i.e. A. ȳ.
As we have all values such as value for A. ȳ and width b. Hence we will use the formula for shear stress at a section, as displayed above in figure, and we will have following expression for shear stress for a beam with circular cross-section.
We can easily say from above equation that maximum shear stress will occur at y1 = 0 or maximum shear stress will occur at neutral axis.

Value of shear stress will be zero for the area at the extreme ends because at extreme ends y1 = R and therefore shear stress will be zero at extreme ends.

We will also find the value of maximum shear stress and it could be easily calculated by using the value of y1 = 0 and therefore we will have following formula for maximum shear stress as displayed here in following figure.
As we know that average shear stress or mean shear stress will be simply calculated by dividing shear force with area and therefore we can say that

Average shear stress, τav= Shear force/ Area
Average shear stress, τav= F/ПR2
Therefore we can say that for a circular section, value of maximum shear stress will be equal to the 4/3 times of mean shear stress.

We can say, from equation of shear stress for a circular section, that shear stress distribution diagram will follow parabolic curve and we have drawn the shear stress distribution diagram for a circular section as displayed in following figure.
We will discuss another topic i.e. in the category of strength of material in our next post.

### Reference:

Strength of material, By R. K. Bansal
1. 