Last Updated on September 15, 2026 by Maged kamel
Solved problem for the design of a steel section for Continuous Beam, Part 4/4.
The negative moments are due to dead Uniform loads and concentrated loads.
We will substitute the span and Load values; the general form of the Matrix, as we have done, lets us obtain bending Moment values for different span lengths and loading values.
The values of span lengths are for the first span length L1 = 30 ‘, for the second span length L2 = 40 ‘, and for the third span length L3 = 30 ‘.
We will substitute in the first Matrix: 2*L1 = 60, L2 = 40′, L1 = 30′, 2*(L1 + L2) = 2*(30 + 40) = 2*70 = 140′, 2*(L1 + L2) = 2*(40 + 30) = 2*70 = 140′. We will put the zeros where they are located in the Matrix.
For the elastic reactions Matrix for the uniform-load vector, we have -L1^3/4 = -(30^3/4)& -L2^3/4 = -(40^3/4), and-L3^3/4 = -(30^3/4).
For the elastic reaction Matrix for the concentrated-load vector Matrix, we have only one parameter: -(3/8)*L1^2 = -(3/8)*(30)^2, which we will substitute on the next slide. These values are shown for the Matrix; the vector Matrix for moments (MA, MB, MC, MD).
I used an Excel sheet to solve the problem, and these are the elastic reaction parameter values for the vector Matrix under Uniform loads. These are the original parameters as shown, -L1^3/4=-(30)^3/4=-6750& -L2^3/4=-(40)^3/4=-16000& -L3^3/4=-(30)^3/4=-6750.
We need to estimate the inverse of the Matrix to solve for the negative moments. We can compute the inverse of a Matrix using Excel or other programs; the symbol -1 denotes the inverse.
When we multiply on the left by the inverse Matrix, we get the identity Matrix, with 1s on the diagonal and zeros elsewhere.
We will multiply the inverse Matrix by the elastic reaction parameters, again by the concentrated Load column vector. Where P1=15 kips, P2=P3=P4=0, the vector column Matrix is( 15 0 0 0). We have the vector column Matrix on the left-hand side (MA, MB, MC, MD) and two matrices added together.
The next slide shows the bending Moment values for dead loads.
The next slide shows the bending Moment values for Live loads.
The Bending Moment for the Total service loads.
We add the negative bending moments due to dead and live loads to obtain the negative values at each support. The Moment at support A, MA = -394.14 ft-kips. The moment at support B is MB = -505.0 ft-kips (positive bending Moment).
The Moment at support C: Mc = -505.0 ft-kips for the positive bending moments. There is no negative Moment at support D, so Md = 0.
The next slide shows the positive values for spans AB, BC, and CD.
On the next slide, we will verify that the Shear at the end span CD is zero at the point of maximum positive Moment. The maximum Moment Value is 233.0 Ft.kips.
The elastic redistribution for a Moment, or the 0.90 rule.
For the maximum positive Value of Moment, we have span AB with M+ve = +263.0 Ft-Kips, span BC with M+ve = +295.0 Ft-Kips, and span CD with M+ve = +233.00 Ft-Kips. The largest Value is +295.0 Ft-Kips, so we add *average of (505.0).
The Moment diagram shows that the maximum negative Moment equals 505 ft-kips, while the maximum positive Moment is 295.0 ft-kips. These values are before using the 0.90 rule.
For the 10% reduction of the negative Moment at span BC, we multiply 0.1*(505), and we get 50.50; the final Value of the negative moments at supports B and C will be -+-455 ft. kips. We add the drop to the positive Value of the Moment at span BC; the final Value will be equal to (50.50 + 295) = 345 Ft-Kips.
For the added values of the edge supports, we add 10% of the average negative Moment, as shown in the next slide.
Design of a steel section for a continuous Beam based on the maximum Moment.
We have two maximum values: a maximum positive Moment of 345 ft-kips and a maximum negative Moment of -455 ft-kips. We check the larger Value of 455 Ft.kips.
We estimate the Nominal Moment by dividing 455 by omega(1.67). The nominal Moment equals 760 ft-kips. We use Table 3.2 to select the appropriate lightest W section. For the continuous Beam, the required Zx is 182.36 in³.
The selected W section is W24x76, with Zx = 200.0 in³. The Zx Value exceeds the required Zx.
This is a Link to the previous Post: Post 39, part 3-4.
You can review or download the PDF for this Post from the following document.
The next Post: Practice problem 5-6-1-find the total service Load for W12x65
Provide more information about the structural analysis – III.
For more details on the upper and lower bounds, see Post 33.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14th ed.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15th ed.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16th ed.