The problem with steel beam design is at first glance it is really really simple. Then you get into lateral torsional buckling, web buckling, etc and it's like you're trying to read a book while it's on fire. The 9th edition AISC steel manual did a decent job at breaking down how to analyze for the various failure criteria, but with the 13th edition it became very difficult to get things done right (not to mention the fact that at times the steel manual will reference other books if you'd like to proceed). But that is neither here nor there, I will do my best to explain how best to solve for a beam by hand.
Steel beam design has many different checks and criteria, the common ones will be explained further here:
Steel design is broken up into a variety of steps. You have to check to make sure the beam is braced often enough, you have to check to make sure the web does not buckle, you have to check to make sure there isn't too much shear going through the beam. All of these checks are normally done automatically with whatever structural design program you use, but if analyzing a beam by hand (which with today's Steel Codes have become quite the tedious process), it can take quite some time to find the actual capacity of a steel beam. That being said, the
Design of a simple beam (using ASD?) can be done one of two ways:
1) Find the allowable moment acting on the beam:
For Compact Sections with adequate lateral and vertical support, the nominal flexural strength of a steel beam is:
where:
The Nominal Moment will be modified by the Safety Factor (Ω_{b}) to become:
where:
Given an A36 Wmember that is compact and properly braced, with a Plastic Section Modulus of 41 in^{3} find the allowable Moment.
Note that A36 steel has a yield strength of 36 ksi and then:
1) Take M_{n} = F_{y} * Z_{x} = 36 ksi * 41 in^{3} = 1476 kin
2) Using the Safety Factor Ω_{b}: M\le {M_n\over \Omega_b} = 1476 kin\over {1.67} = 884 kin
The Answer is 884 kin
2) Compare the stress acting on a beam to the Allowable Stress: The more common way to solve for the allowable bending in a steel beam is to check the beam against its allowable stress. Reworking the equation in Section 1 above you can come up with:
Most common simplified bending equation: f_b = {M\over Z_x} \le F_b \approx {0.6*F_y}

where:
Given an A50 Wmember that is compact and properly braced, with a Plastic Section Modulus of 18.5 in^{3} and a moment of 141 kin. Does the member work? What is it's design ratio?
Note that A50 steel has a yield strength of 50 ksi and then:
1) Take f_{b} = M / Z_{x} = 141 kin / 18.5 in^{3} = 7.62 ksi
2) Using the Safety Factor Ω_{b}: F_b= {F_y\over \Omega_b} = 50 ksi \over {1.67} = 30 ksi
3) Since f_{b} < F_{b} then the member's demand is less than capacity and therefore it is within the allowable bending stress of the beam.
4) To find the Design ratio take f_{b} / F_{b} = 7.62 ksi / 30 ksi = 0.254
The member is at 25.4% capacity