Last Updated on August 22, 2026 by Maged kamel
Plastic nominal uniform load for a partially loaded beam.
We will adjust the solution in the previous post by placing the uniform load at the midpoint of the span on the left side.
For the steel beam with a W18x40 section and a yield strength Fy = 50 ksi, the Plastic Nominal Uniform load for the partially loaded beam, Wn, must be estimated.
What is the distance x for a point of maximum moment?
In the case of working load w, we can find the position of the point of maximum moment by using the summation of shear forces to be equal to zero.
First, we estimate the reactions at A and B. The RA value equals 12w, where w is the working uniform load. Similarly, Rb will be equal to 6w
If x is, the distance from the left support to the point of the maximum moment, its value will be equal to RA/w =12 feet. The value of the maximum moment will be equal to 72 w. Please refer to the next slide image for more information.
Solving the modified example by the lower bound theorem.
When w is equal to the nominal load Wn, the maximum moment will be equal to Mn, and we can equate Mn to 72*wn. We have a given W18x40 section; use a table to find Zx, the plastic section modulus for that section. Zx value = 78.40 in^3. The yield stress is 50 ksi.
The nominal Mn = Mp = Fy*Zx = 50*78.40 in. kips. We have Mn = 326.66 kips/ft, so the nominal load wn is 4.537 Kips/ft. Please refer to the next slide image for more details.
Solving problem 8-32 using the upper bound theorem.
For the same modified problem, to estimate the Plastic Nominal Uniform load for a partially loaded beam, we will create a mechanism and follow the same procedure.
Partial loading can be considered the sum of two cases: the first, where the span is under full downward load, and the second, where partial loads act upward.
The plastic hinge is located at the place of the maximum moment at x=12′ from the left support.
The deflection at that point is Δ, the angle at A due to that deflection=θ, and the angle at support B is θ1.
The internal work=Mp*(θ+θ1). θ=tan θ=Δ/12, while θ1=tan θ1=Δ/20.
For the external work=We1=0.50*Wn*32*Δ, then We=32*Wn/2-We2; we will estimate the deflection value at the edge, which has a distance of 16′ from support B. The value of Δc can be estimated from Δc/Δ=16/20, Δc=0.80*Δ.
We estimate the average deflection at point C, Δc, to be 0.40*Δ. The value of We=Wn*Δ*16-Wn*16*0.40*Δ, We=16*Wn(1-0.40)=9.6*Wn*Δ.
For the internal work Wi=Mp*(θ+θ1), θ+θ1=(Δ/12)+(Δ/20)=32*Δ/240.
We=9.60*Wn*Δ)=Mp*32*Δ/240. Wn=32Mp/(2409.60), Wn=32326.66/(2409.60)=4.54 kips/ft. We can find that the value of the Plastic Nominal Uniform load is 4.54 kips/ft.
We get the same value of 4.54 kps/ft for the Plastic Nominal Uniform load for the partially loaded beam from both the lower- and upper-bound methods, which are identical.
Solving the modified example by MASTAN 2.
The partially loaded beam under a uniform load has been solved in Mastan-2 using 1.0 kips/inch and increments, but at a load of 0.378 kips/inch, a plastic hinge forms at 144 inches from the left support.
The value of 0.378 kip/inch matches our solution of 4.54 kips/ft. The deflected shape of the beam is shown on the next slide, along with the location of the plastic hinge.
On the next slide, we see the Bending moment diagram for the partially loaded beam, and we note that the Plastic moment is 3920 in. kips, which is 326.66 ft. kips, and the nominal load is 4.537 kips /ft.
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Have more information about the structural analysis – Link to III.
The next post is solved problems 8-33&34 for a nominal uniform load.
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.