Doubt Regarding Shearing Stresses In a Beam

1. Dec 27, 2013

Abhishekdas

Lets say we have a beam which is simply supported at the two extreme ends(support conditions dont matter in my question). A concentrated transverse load is applied at the halfway point. Now lets say we take a section at x= L/3 where is L the length of the beam. Now we know that transverse shear stress at that section is max at the neutral axis and minimum(0) at the top. Now I define the axes
x- along the axis of the beam
y- along towards top
z- coming out of the plane of paper
let y vary from t/2 to -t/2. Now take a very small element at the top of the section at y = t/2. Now at this element tau(xY) (shear stress on this face(perpendicular to x axis) in the Y direction) is zero according to " transverse shear stress is minimum(0) at the top". But how is this possible because apparently even at the top the material will tend to be sheared due to the transverse load. How am I wrong? I would really appreciate is someone could help me out.

Last edited: Dec 27, 2013
2. Dec 27, 2013

pukb

the surface at top and bottom are free and hence inorder to have forces balanced, on taking a small element at top of the beam ; since there are no shear stresses towards atmosphere there cannot be any shear stress inside.

3. Dec 28, 2013

Abhishekdas

SO basically its right when i say that the shear stresses in the xy(towards the downward portion) is actually zero. Which means maybe the material out there does not tend to get sheared?
Maybe true but hard to digest. I am having problems as to how to visualize it physically that the material wont be/tend to be sheared.

4. Dec 28, 2013

AlephZero

The shear is only 0 on at the top and bottom surface. For a rectangular cross section, the shear along a line through the beam from top to bottom is a quadratic function.

This is fairly easy to show if you understand stress in 3 dimensions. The axial direct stress varies with the depth of the beam. $\sigma = My/I$ where y is distance from the neutral axis. To maintain equilibrium, the shear has to be the integral of that stress component. The "arbitrary constant" in the integral is fixed by the boundary conditions that pukb described, i.e. the shear is zero at the top and bottom surface of the beam.

5. Jan 7, 2014

Abhishekdas

Thank you AlephZero for the reply, but could you please elaborate which integral you are talking about. The derivation for shear stress the way I know it is that you take forces due to moments (axial) for a small element in either direction. If the element length is dx the change in moment is dM and this difference in forces in either direction is accounted for by the shear force at that element. So you end up with tau=dM/dx/Ib*(∫ydA) integrated from y1 to C and here b is the width of the element at y=y1. Is it the same thing you are referring to?

6. Jan 7, 2014