A Solved Problem 2-22 For Shear Stress-part-A.
The video I used for illustration.
In this video, we will discuss how to estimate the shear stresses for given sections, the Rectangle section, and the I section, step-by-step guide for the calculations.
You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.
The shear stress distribution for a rectangular section is shown in the next slide, The shear stress formula can be written as τ max =V*Q/Ixx*b. where V is the acting force while Q is the statical first moment of area about the neutral axis, for the portion enclosed by the upper edge and the neutral axis. Ix is the inertia of the section at the x-axis, and b is the breadth of the section.
Derive the expression for the shear stress for a rectangular section.
We know Ix= b*h^3/12, where h is the height of the section, The statical first moment of area Q=1*0.50(b*h)*(h/4)=b*h^2/8. The maximum shear stress at the neutral axis of the rectangular section can be estimated as, τ max =V*(b*h^2/8)/(b*h^3/12)*b)=(3/2)*v/A. A is the area of the rectangle.
How to get the value for the First moment of area for the I section?
If we use the expression Q as an indication for the first moment, for the portion included between the upper edge and the neutral axis, this portion is in a form of the T- section, or half of the complete I section.
Then we have two items. The first moment of area for the I section consists of Q for Flange+ and we will add Q for web, at the N.A Qf=area of flange*y bar=bf*tf*(d/2-tf/2). The term d is the overall height of the I section, while the bf is the flange breadth.
The first moment of area for the Q web at the N>A will be (1/2*hw*tw)*hw/4=1/8*(b*h^2w).
Adding these values together we can get the final first moment of area for the I beam. While Ix for the same section is shown in the next slide.
Part A of the solved problem 2.22 for shear stress.
We have solved problem 2.22 for which it is required to estimate the maximum shearing stress for the following sections when the external shear force V=75.0 kips. For part A when the shear force is given as V=75 kips for a rectangular section the expression for the maximum shear stress can be written as τ max =3/2*(V/A).
We will substitute as τ max =3/2*(75/(12*4))=2344 psi or divide by 1000 to get the value as τ max=2.344 ksi. In the next post, we will continue the solution to the same problem for the other shapes.