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 y

_{1}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 y

_{1}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*

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

Where,

F = Shear force (N)

τ = Shear stress (N/mm

^{2})
A = Area of section, where shear stress is to be
determined (mm

^{2})
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 (mm

^{4})
For circular cross-section, Moment of inertia, I =
ПR

^{4}/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}= R^{2}- y^{2}
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 y

_{1}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 y

_{1}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 y

_{1}= 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 y

_{1}= 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 y

_{1}= 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, τ

_{av}= Shear force/ Area
Average shear stress, τ

_{av}= F/ПR^{2}
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. Shear stress distribution in I section
in the category of strength of
material in our next post.

###
**Reference:**

Strength
of material, By R. K. Bansal

Image
Courtesy: Google

Timely

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