 # 19- Review of shear stresses for steel beams.

## Review Of Shear Stresses for steel beams.

### The video I used for illustration.

Introduction to shear stress for beams, and proof that both vertical and horizontal shear stresses are equal.

You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.

The topics included are a review of the shear stress derivation and how to estimate the shear value for a beam.

### How to derive the expression for shear stress?

If we take two sections at a beam apart by a distance=dx, at elevation =y, due to the moment difference, there will be two compressions forces due to M and M+dM.

The stress developed from the known formula f=M*y/I will be multiplied by dA and create two compressive forces, these forces are not equal due to the difference at the moment which is=dM. to create a balance shear force will be created acting horizontally.

As sum of Fx =0, shear force is developed to create a balance, this force=shear stress* area=τ*dx*b=σ’*dA-σ*dA, but (σ’-σ)*dA=(dM)*y/Ix, then τ*dx*t=dM*y/I.

Making integration, we get an expression for the stress for shear τ=∫dM*y*dA/(dx*t)=V*Q/IT, where V is the shear. Q is the first moment of the area at the point of interest, I moment of inertia for the section finally,t is the breadth of the section.

### Horizontal shear stress for beams.

The horizontal stress of shear should be accompanied by vertical stress due to shear for the balance of the element of a beam, as shown in the next slide.

It is proven that the value of τh=τv, where τv is the vertical stress due to shear, while τh is the horizontal stress due to shear by multiplying by area and taking moment at the edge

### The expression for shear stress for a rectangular section.

We are going to substitute in the general equation for stress as τ=VQ/(Ix*t),  then at the neutral axis where the first moment of area V=b*(h/2)*h/4)=b*h^2/8, while Ix=bh^3/12.

At the neutral axis we have  τ= Q*(b*h^2/8)/b(b*h^3/12) =Q/A*(4/3), then τ=1.5* average stress at the neutral axis, the stress distribution is shown in the next slide.

The next post will contain a solved problem for stress estimation for different shapes.

This is the pdf file used in the illustration of this post.
For more detailed shear in bending, please read link-Shear in Bending
This is a link for the next post, A solved problem for shear stress.

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