Last Updated on January 27, 2026 by Maged kamel
List of Steel Beam Posts-part-4a.
Solved Problem 10-1-Design Of Steel Section For Continuous Beam Part-1/4.
This is the 38th post in the Steel beam Posts-part-4a series, which includes solved problem 10-1 for a given continuous beam. It is required to select the lightest W-section available using the plastic analysis, assuming full lateral support is provided for both flanges.
Two methods are used: the first is the statistical method, or the lower-bound method, and the second is the Mechanism, or the upper-bound method. Both methods will lead to the same results.
The following slide shows how we can obtain the Ultimate loads for the given first-end beam.
In part 1 of 4, a detailed estimate of the plastic moment for the first end span AB is provided.
This is the link for post-38: Solved problem 10-1 for Design Of Steel Section For Continuous Beam – Part-1/4.
Solved problem 10-1-design of steel continuous beam-2/4.
This is post number 38a, one of the Steel beam Posts-part-4a posts, which explains how to derive a general expression for the moment values of a continuous beam using the three-moment equations. This is the second part of the solved problem 10-1. In part 2 of 4, a detailed estimate of the plastic moment for the middle and end spans is provided.
The following slide image shows the Plastic moment value for the middle beam BC.
After estimating the plastic moment, we select the maximum value (466 ft-kips) as the nominal moment and determine the plastic section modulus Zx.The value of Zx=159.84 inch3. The section suitable for the beam is W21x68, with a Zx value of 160 in³, which is larger than the required Zx.
We can design the continuous beam using LRFD. The following slide shows the detailed design of the beam.
This is the link for post-38a: Solved problem 10-1-design of steel continuous beam –Part 2/4
39-Solved problem 10-1-use three-moment equations 3/4.
This is the 39th post of the Steel beam Posts-part-4a, which includes how to derive a general expression for the moment values for a continuous beam using the three-moment equations. This is the third part of the solved problem 10-1—a step-by-step guide to using the three-moment equation.
The following slide shows how to write the three-moment equation at point A and how we use matrices to relate moment values to loads.
This is the link for post-39: Solved problem 10-1-use three-moment equations 3/4.
40-Solved problem 10-1-design of steel sections 4/4.
This is the 40th post of the Steel beam Posts-part-4, which includes the design of the beam, using elastic analysis with service loads and the 0.90 rule, and assuming full lateral support is provided for both flanges.
This is the fourth part of the solved problem 10-1. The next slide image shows the final bending moment diagram after using the 0.90 rule, in which the negative moments at the supports are reduced. and the redistribution is done by adding moments to the existing dead loads. The maximum positive value is 345 Ft.kips, and the maximum negative value is 455 Ft.kips. A W24x76 is selected, and the section has a Zx value of 200 inch3. The detailed design of the ASD is illustrated in this post.
The following slide image shows the detailed design of the continuous beam based on the ASD design. We use Table 3-2, where w sections are arranged by Zx value. The selected beam section is W24x76, which carries a moment of 499 ft-kips, which is larger than the estimated maximum of 455 ft-kips.
This is the link for post-40: Easy Design of steel section for continuous beam-4-4.
A very useful external resource A Beginner’s Guide to Structural Engineering