Say, I have to design a simply-supported steel beam and am given the following constraints:

I am also given the following information:

Length, L = 50 ft Steel density, rsteel = 490 lbs/ft3 Area Live Load = 40 lb/ft2

First, I need to estimate the dead load. There are a couple of ways I can go about doing this. The first way is would be to guess how deep the beam is going to have to be to satisfy my requirements. Then, design the beam based on this guess. So, I will guess that the beam has to be 6 inches deep. Therefore,

The live load must be transformed into a linear force. Therefore,

Now, I can figure that the

So, my total uniform load q will be = 2,055 lb/ft or 2.055 kips/ft. After I calculate my support reactions and draw my shear and moment diagrams, I find that my maximum moment will be

Now, I can calculate my maximum tension in the bottom of the beam. Therefore,

Well, I found that this is no good, because I am only allowed to have 20 ksi. Therefore, I must start over and redesign the beam. I may do it again and try a beam depth of 6.5 inches and see if that works. Or, I can try another method and get the exact value.

Instead of guessing a beam depth, let me calculate the uniform load as a function of h. This first requires calculating the dead load as a function of h. I do this as follows:

Therefore,

Now, I can calculate the maximum moment as a function of h:

Let’s convert this value to an expression in terms of kips and inches. Therefore,

M_max(h) = (1071.9 kips)h + 1275.6 in·kips

Now, I can calculate my maximum tension in the bottom of the beam as a function of h.

If I set this equal to my constraint value, I will be able to find the exact h where the necessary conditions are met. Thus,

(107.2 kips/in)h + 127.6 kips] = (20 ksi)h²

or

h² - (5.36 in)h -6.38 in² = 0

I can solve for h by using the quadratic equation:

	= 2.68 in ± 3.68 in
	= 6.36 in or -1.00 in

Well, it’s obvious that the only one of these answers that makes sense is 6.36 in. Therefore, I know that my beam depth must be greater than this value. But, is this depth enough? I will have to check the deflection constraint to make sure.

The equation for the maximum deflection of a simply-supported beam under a uniform load is

I am going to plug in all the values that I know and leave h as the independent variable in order to get vmax as a function of h. Therefore,

vmax(h)	=
	= [(267.8 in3)h + 318.4 in4] / h3

Once again, if I set this equal to my constraint value, I will be able to find the exact h where the necessary conditions are met. Thus,

[(267.8 in³)h + 318.4 in⁴] / h³ = 2 in

or

h³ - (133.9 in²)h - 159.2 in³ = 0

Now, I use some method to solve this cubic equation and find that

h = -10.92, -1.20, and 12.13 in.

The only positive answer is 12.13 in. So, I find that the deflection constraint governs my design and I must design h to be greater than 12.13 in.

Say, I choose a value of 12.2 inches for h. Now, I go back and check my results.

Note that s_max = 9.64 ksi which is < 20 ksi.

and

v_max = 1.97 in which is < 2 in. Therefore, the design is OK.