37a – Solved Problem 8-34 for a nominal Uniform Load.

Last Updated on September 15, 2026 by Maged kamel

Solved Problem 8-34 for a nominal Uniform Load.

A solved Problem 8- 34: we have a continuous Beam with three spans; the first span length=24′ and the second span length=36′, while the last span length = 24 ‘. The Beam has a uniformly distributed Load; the section is W27x84, Fy = 50 ksi. The plastic Modulus, Zx, = 244 in^3. Estimate the nominal Load for the beams that can develop plastic hinges and collapse.

We need to find the indeterminacy number for the beams; we do not consider the horizontal loads. We have six unknowns: two moments and four unknown vertical loads. Based on the summation of Y and Moment equals zero. We will deduct 2. The indeterminacy number is equal to 6-2=4.

We have 7 hinges to be placed at spans AC, CD, and DE. The number of mechanisms is 7 – 4 = 3.

From the given W27x84, we can find the Zx Value and estimate the plastic Moment, equal to 50* 244 = 12200 inches. Kips. The final Value of Mp is t-kips.

We can get the plastic Moment Value by multiplying 50*244, and the result is 12200 in-kips; to convert it to foot-kips, divide by 12. The Mp Value will be 1016.66 ft-kips. This Value applies to the third span due to symmetry.

The three possible mechanisms for the three beams are shown in the next slide. Each mechanism is created from three plastic hinges.

Nominal Uniform Load for the first span by the lower bound method.

We can estimate the plastic Moment for the first span by equating the moments from the reactions and the Uniform Load to the plastic Moment at the middle of the first span. The Wn Value obtained using the lower-bound method is 28.24 kips/ft. Please refer to the image on the next slide for more details.

Nominal Uniform Load for the second span by the lower-bound method.

For the second span, we can determine the Value of Wn that creates three plastic hinges: two at the supports and one at the mid-span. Again, we equate the Moment from the reaction at the left support with the Moment from the Uniform Load to Mp.

We estimate Wnn at 12.551 ft/kip.

We will use the Upper-bound theory to find Wn for both the first and second spans.

Solved problem 8-34: Plastic nominal Uniform Load Wn for the first span by the upper-bound theorem.

To get the plastic Moment from the upper-bound theorem, we equate the external work done by the Uniform Wn to the internal work due to the plastic moments. We will estimate end slopes in terms of delta.
The slope of each side due to Deflection Δ is θ = Δ/12. External work = internal work; external work = Wn*(0.50*δ*24) = 12*Wn*Δ. We have three MP hinges. The internal work=Mp*θ+Mp*θ+Mp*(θ+θ).
The internal work=Mp*θ*412*Wn*Δ=Mp*θ*4=Mp*4*Δ/12. Δ goes with Δ.Wn=1*Mp/36.

The plastic Moment Value is known to be 1016.66 ft-kips; we can get the Wn Value as Wn = 1 Mp/36. = 1016.66/36 = 28.24 kips/ft. The same result can be obtained by considering that 2Mp equals Wn*L^2/8.

Using the upper-bound theorem for the second span, solve problem 8-34.

Due to symmetry, the Deflection is Δ at the mid-span.The slope at each angle=Δ/18. The external work = internal work. W_n*36*(0.50*Δ)=18*Wn*Δ.

The internal work equals 4Mp*θ, but θ = tanθ = Δ/18/18, and the internal work = 2*Mp*Δ/9p*Δ/9. We can estimate Wnd; its Value is 112 ft-ki.ps. The same Value of Wn can be verified by the equation Wn l^2/8 to two mp. Please refer to the image on the next slide.

We will examine all possible nominal Load values from the different mechanisms and select the lowest, Wn = 12.55 kips/ft.

Nominal Uniform Load for the first span using MASTAN 2.

We can use MASTAN 2, the first inelastic analysis, to determine Wn, the nominal Uniform Load. The Modulus of Elasticity for steel, E, is 29000 ksi, and Fy = 50 ksi. These two parameters are necessary to find the plastic nominal Uniform Load. Convert the Beam lengths to inches.

The working Load is in kips/inch, and I selected 1.00 kips/inch. There are 15 increments, each 0.10.

According to the next slide, three plastic hinges have been developed. The first hinge is located between the two supports, and one is at the Midpoint of the second span. Wn is 1.046 kip/inch, which matches our Wn in kips/ft after the conversion.

Two plastic hinges were created at a lower Value of Wn equal to 10.932 Kips /ft or 0.911 kips/inch.

With MASTAN 2, we can determine the Moment values for the three beams. Please refer to the image on the next slide for more information.

The Shear values for the beams and the nominal Load can also be estimated based on the program. The sum of vertical forces and reactions is equal to zero. Please refer to the calculations attached herewith. Thank You all.

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Provide more information about the structural analysis – Link to III.
For the next Post, 38-solved problem 10-1-Design of steel section for continuous Beam part-1/3.

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.