## Steel Beam Design Breakdown

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.

### By Subject:

Steel beam design has many different checks and criteria, the common ones will be explained further here:

## The Basics to Steel Beam Design: Fig. 1: Moment Capacity vs. Unbraced Length

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:

M_n = F_yZ_x

where:

Mn = The Nominal Moment (note this is not the design moment)
Fy = The Yield Strength of the Steel (e.g. 36 ksi, 46 ksi, 50 ksi)
Zx = The Plastic Section Modulus in the x or strong axis. Zx is similar to the Section Modulus of a member (it is usually a minimum of 10% greater than the Section Modulus) (in3)

The Nominal Moment will be modified by the Safety Factor (Ωb) to become:

M \le {M_n\over \Omega_b} = {M_n\over 1.67}

where:

Ωb = The Safety Factor for Elements in Bending = 1.67

#### Example #1:

Given an A36 W-member that is compact and properly braced, with a Plastic Section Modulus of 41 in3 find the allowable Moment.

Note that A36 steel has a yield strength of 36 ksi and then:
1) Take Mn = Fy * Zx = 36 ksi * 41 in3 = 1476 k-in

2) Using the Safety Factor Ωb: M\le {M_n\over \Omega_b} = 1476 k-in\over {1.67} = 884 k-in

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:

f_b = {M\over {Z_x}}
F_b = {F_y\over {\Omega_b}} = {F_y\over 1.67} \approx 0.6*F_y
f_b \le F_b
 Most common simplified bending equation: f_b = {M\over Z_x} \le F_b \approx {0.6*F_y}

where:

fb = The computed stress in the beam in bending
M = The maximum moment acting on the beam
Zx = The Plastic Section Modulus in the x or strong axis. Zx is similar to the Section Modulus of a member (it is usually a minimum of 10% greater than the Section Modulus) (in3)
Fb = The allowable stress of the beam in bending
Fy = The Yield Strength of the Steel (e.g. 36 ksi, 46 ksi, 50 ksi)
Ωb = The Safety Factor for Elements in Bending = 1.67

#### Example #2:

Given an A50 W-member that is compact and properly braced, with a Plastic Section Modulus of 18.5 in3 and a moment of 141 k-in. 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 fb = M / Zx = 141 k-in / 18.5 in3 = 7.62 ksi

2) Using the Safety Factor Ωb: F_b= {F_y\over \Omega_b} = 50 ksi \over {1.67} = 30 ksi

3) Since fb < Fb 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 fb / Fb = 7.62 ksi / 30 ksi = 0.254

The member is at 25.4% capacity

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