Last Updated on January 27, 2026 by Maged kamel
Solved problem 10-1 for the Design of Steel Section for Continuous Beam-Part-4/4.
The negative moments are due to dead uniform loads and concentrated loads.
We will substitute the values of spans and loads; the general form of the matrix, as we have done, enables us to use it to obtain the 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.
As for the elastic reactions matrix for the uniform loads vector matrix, we have -L1^3/4=-(30^3/4)& -L2^3/4=-(40)^3/4 &-L3^3/4=-*(30)^3/4.
As for the elastic reactions matrix for the concentrated load’s vector matrix, we have only one parameter, which is -(3/8)*L1^2=-(3/8)*(30)^2, which we will substitute in the next slide. These values are shown for the matrix 4×4, 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 inverse is denoted by the symbol -1.
When multiplying on the left by the inverse matrix, we get the unity matrix, with a 1 on the diagonal and zeros elsewhere.
We will multiply the inverse of the matrix by the elastic reaction parameters, again by the concentrated load’s column-vector matrix. 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, MB=-505.0 Ft.kips. For the positive bending moments.
The moment at support c: Mc = -505.0 ft-kips. For the positive bending moments. There is no negative moment at the end support D, so Md = 0.
The next slide image 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 the 0.10*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: one for the maximum positive moment, 345 Ft. kips, and the other for the maximum negative moment, -455 Ft. kips. We check the bigger 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 steel section for the continuous beam, Zx required 182.36 inch3.
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.
The PDF file for this post can be reviewed or downloaded 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 for upper bound and lower bound, this is a link to post 33 ; the upper bound and lower bound