Show that the shearing stress developed at the neutral axis of a beam with circular cross section is τmax = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.
Solution :
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
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)
Q = 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.
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.
The width b will be dependent over the value of y
(b/2)2 = R2- y2
b = 2x SQRT(R2 -y2)
The area dA of the strip of thickness dy is
dA = b.dy = 2.SQRT(R2 -y2).dy
By substitute the values of (A. ȳ ) and width b. in the shear stress equation, we will have following expression for shear stress 
When y1 = R , the fiber is in outer surface,
the shear stress is τ = 0
If y1 = 0 ,the fiber is at neutral axis,and I= ПR4/4
The shear stress maximum τmax = 
Average shear stress, τav= Shear force/ Area
Average shear stress, τav= F/ПR2
The max. shear stress τmax = 4/3 τave
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. The shear stress distribution in the circular section as shown in below
Thank you for the article :)
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