34- Easy approach to the Plastic Moment for a Continuous Beam

Last Updated on September 15, 2026 by Maged kamel

The Plastic Moment For a Continuous Beam

The plastic Moment for continuous beams by the statical method theorem.

How do we estimate the bending Moment for continuous beams?

  We can estimate the bending Moment using the statical method, the Kinematic method, or the virtual work method. For the end span, we have one roller support with one reaction and, at the other end, a fixed support with three reactions; adding them together (3+1=4), we have 3 equations of equilibrium.

The sum of X=0, the sum of Y=0, and the sum of moments=0. Subtract these equation numbers from 4 to get 4-3=1; then the system is statically determinate. We will deduct 1 to convert the system to an unstable system.

The number of plastic hinges=3+1-3+1=2. The proposed location for the first hinge is at the fixed support; the second plastic hinge is a distance x from the roller support.

We have a positive Moment in the first span that is a parabola. We assume that the hinge is a distance x apart from the left support. As a reminder of this method, we have a negative Moment Mp and a positive Mp, and we lay the two diagrams together.

We will deduct. I will draw another sketch to show it more clearly. This MP is a negative Moment; we set the Value to zero, then draw the parabola as W*L^2/8.

We do not know the exact location of the plastic hinge in the first span, or whether it is located in the mid-span; in that case, the Mp positive Value will be (W*L^2/8-0.50*MP).
The remaining ordinate is the Mp positive Value.

Two spans can represent the end span: the first span is loaded under w/lineaw/lineare, and the second span is subjected to Moment M acting on the right support in the clockwise direction. For the reactions, they can be considered as the superposition of the two spans; for a roller support, RA=w*L/2-M/L.

For the other support, b, RB = w*L/2 + M/L. For M= Mp at the end support, the Moment at distance x  from the left support can be estimated as the Moment from reactions minus the Moment from the uniform loading Mx=RA*x-w*x^2/2-Mp/L*x.

This Moment at the plastic hinge should = Mp if this point is the exact location of the plastic hinge.
Note that at any other location, Mx will not equal Mp; we will differentiate with respect to x.

Consider w = wp, since we are estimating at the collapse stage. dM/dx=0, 0=wp*L/2 -2*wp*x-Mp/L, then,  Mp/L=wp*(L/2-x), then Mp=wp*L(L/2-x) as equation number 2. Let Mx = tp in equation Number 1; we will substitute the Value of Mp/L.

In the next slide, go back to equation no. 1: Mx = Wp*L*x/2 – wp*x^2/2 – (Mp/L)*x, where Mp = wp*L (L/2- x).

Let Mx= Mp,( Mp+Mp*x/L)=w*P/2*(Lx-x^2). The term wp will be eliminated by wp; L goes with L, multiply by 2, and readjust the terms.

Finally, x^2+2x*L-L^2=0 is a second-degree formula for which the x Value is the distance to the plastic hinge from the left support. x = L ± sqrt(2) L; x = 0.414L, not at mid-span.

To get the MP Value, we will substitute it with the Value of x = 0.414*L: Mp = wp*L (L/2-0.414L)=wp*L(0.086). This is the Value of Mp, the Plastic Moment at the end span, in terms of wp, the plastic Uniform Load.

The plastic Moment for continuous beams at the Intermediate span.

The intermediate span, with a uniformly distributed Load wp, can be represented as a span with two end moments, each Mp, under a uniform Load of wp.

We still use the lower-bound theorem: 2Mp = Wp*L^2/8; Mp = wp*L^2/16; then the Load at collapse, wp = 16*Mp/L^2.

This Value is bigger than the Mp Value estimated at the end span.

The third span is symmetrical to the first span, so the Plastic Moment Value is the same.

The plastic Moment for continuous beams by the Upper bound theorem.

What is the external work for a uniformly distributed Load on a simply supported Beam by the upper bound theorem?

First, let us derive a relation between the UUniformLoad w and the external work due to that Load for a supported Beam of length L.

If we divide the Beam into two halves, the total external work is twice the work done on one half, treating the uniform Load as a concentrated Load of wL/2. For work multiply w(l/2)Δ/2=W(L/4)Δ. So the total work is due to a uniform Load on a Beam: W_e = 2W (L/4) Δ = 0.50WL. Δ is half the span, Ul*delta.

What is the plastic Moment for continuous beams under a uniformly distributed Load at the end span?

To obtain the plastic Moment for continuous beams at the end span using the upper-bound theorem.

We will create a mechanism. We assume the plastic hinge is located at a distance x from the left support; the remaining distance to the right support is L-x.

For the deflected shape,  at the plastic hinge, the Deflection Value is Δ  at a distance x from the left support; the Moment at the right support =Mp, and there is another Moment Mp+ve at the plastic hinge.

We equate the external work with the internal work: Wp*(L/2)*Δ=Mp*β+Mp(α+ β); α=tan α=Δ/L, β=Δ/L-x,α+β=(Δ/x)+(Δ/L-x).

Checking the equation, we have Mp*(α + 2β).

We combine all together (α+2*β)=(Δ/x)+2*Δ/(L-x)=Δ(L-x+2x)/(x)*(L-x)=Δ(L+x)/(x)*(L-x). We substitute in the equation for external work, wp*L/2*Δ=Mp*Δ(L+x)/(x)*(L-x),Δ goes with Δ, then Mp=wp*L/2*(L-x)/(L+x).

We could make a table to show the relation between x and Mp as a diagram, instead of plotting and giving different values of x; for each increment, say x0=0*L, x1=0.20*L, etc., we get the corresponding Mp Value.
The graph will have a minimum point where the slope equals 0.

We will differentiate dMp/dx = 0, using the laws of differentiation for a fraction. We continue until we get x^2 + 22x*L L^2 = 0.

The Value of x is 0.4142*L; this is the same result we obtained using the statistical method or the lower bound theorem.

We substitute x into the Mp relation, with x = 0.4142*L. The Mp Value is equal to 0.08578 *wp*L^2, which is the Value of the Plastic Moment for Continuous Beams at the end span. The wp is equal to (Mp)*(1/0.08578L^2)=11.657 Mp/L^2.

We have satisfied the three conditions, which are equilibrium, mechanism, and uniqueness, as there is no higher Value other than Mp, by using differentiation.

What is the plastic Moment for continuous beams under a uniformly distributed Load on the Intermediate span?

The last requirement for the interior span is to find the Mp values using the upper-bound theorem.

We will get the same result for Mp = wp*L^2/16, the Plastic Moment for Continuous Beams at the intermediate span, which is similar to the lower-bound theorem result, so we get a unique solution.

We have completed the estimation of the plastic Moment for continuous beams with three equal spans under a uniformly distributed plastic Load.

You can view or download the PDF file for this Post from the following Document.

For a useful Link on plastic analysis, refer to Structural Analysis – Link to III.

This is the next Post, Solved Problem 8-22 for Plastic Moment.

Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14^th ed.

Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15^th ed.

Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16^th ed.