Calculate the maximum value of shear flow, , in the web at a section 1m from the free end of the beam.

Answer:

See explanation

Explanation:

Since no figure was given, I'll explain how to do this problem theoretically. The formula for shear flow is [tex]q=\frac{VQ}{I}[/tex] where V is the shear force, Q is the moment of area (more on this later), and I is the moment of inertia.

The first step to solve this problem is to find the resultant internal forces of the beam. This can be done in several ways, but the easiest is to solve the beam statically and draw a shear diagram to determine the maximum shear force V.

The second step to solving this problem is to determine the location of the neutral axis of the cross section if it is not given. The formula for the neutral axis is  [tex]NA = \frac{\sum y*A}{\sum A}[/tex]. The y in this equation represents the middle of the small shapes that the web is divided into. An I-beam can be thought of as 3 rectangles, while a T-beam can be thought of as 2. The A in this formula represents the area of each of the rectangles (an I-beam will have 3 of these and a T-beam will have 2).

The third step for this problem is to find the moment of inertia. There are several formulas for moment of inertia depending on the shape of the cross section. I-beam's and T-beams both can be thought of as multiple rectangles, so they have the same base formula of [tex]I=\frac{1}{12}bh^3[/tex] where b is the base of the rectangles and h is the height. For I-beams, the easiest way to calculate moment of inertia is to think of the entire cross section as a big rectangle that had two smaller rectangles cut out of it. The formula for this moment of inertia becomes [tex]I=\frac{1}{12} b_{big}h^{3} _{big}-\frac{1}{6}b_{small}h^{3}_{small}[/tex]. Note that this form of moment of inertia already takes into account subtracting 2 small rectangles. For T-beams, this approach will not work, so the parallel axis theorem must be used. The moment of inertia for the T-beam becomes [tex]I=\frac{1}{12}b_{1} h^{3}_{1} +b_{1}h_{1}dy_{1}^{2} +\frac{1}{12}b_{2} h^{3}_{2} +b_{2}h_{2}dy_{2}^{2}[/tex] where the terms with the subscript 1 represent the first rectangle and the terms with the subscript 2 represent the second rectangle. The dy terms represent the distance from the center of that specific rectangle to the neutral axis.

The fourth step for this problem is to find Q. The formula to find Q is [tex]Q=\sum y'A'[/tex] where y' represents the distance from the neutral axis to the center of the "wanted" point and A' is the area of the rectangle that has the wanted point at its center. (This would be the area above or below the thickness (t) if you were solving for maximum shear [tex]\tau=\frac{VQ}{It}[/tex]).

The last step for this problem is to substitute the found values into the formula for shear flow [tex]q=\frac{VQ}{I}[/tex]. V came from step 1, Q came from step 4, and I came from step 3.

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