37-Solved problem 8-33 for Plastic nominal Uniform Load.

Last Updated on September 15, 2026 by Maged kamel

Solved problem 8-33 for a plastic nominal Uniform Load.

Solved problem 8-33; estimate the nominal Load Wn.

In solved problem 8-33. The given Beam section is W16x26. To find W_n, the plastic nominal Uniform Load, from Table 1-1 we get the plastic section Modulus, Zx = 44.20 in^3.

The specified yield stress for steel is 50 ksi; see the second slide.

The nominal Moment, Mn, equals Fy*ZxFy*Zx44.50*44.2015.166 ft-kips. Because we will use MASTAN 2 to compare with our estimate, we set Mn to 2210 kips.

A List of different types of steel and their Yield stresses.

Solved problem 8-33-Number of possible mechanisms.

We need to estimate the number of indeterminacies in the given Beam system. There are two possible ways. The first way is to disregard the horizontal loads and find the number of vertical and Moment unknowns.

The number is 4. We will deduct the number of equilibrium equations, which are ∑ Y=0 and ∑ M=0. The difference will be equal to 4-2=2.

The second way is to disregard the horizontal loads and use the equation 2m+r-2n, where m is the number of members. This the renumber of actions, ,( )whand is the number of nodes, h(

The ID number will equal 6+4-8=2, the same Value as estimated from the first method.

The next question is: what is the number of possible plastic hinges? The possible hinges are 5. Deduct 2 from the 5; we get three, which is the number of mechanisms.

The next slide shows the details of the three mechanisms that we will investigate. The first and last Mechanisms are similar.

What is the xx distance Value for Shear for the first span, and the Wn Value?

What is the distance x from the left support to the point of maximum Mp? It will be the point of zero Shear.

We have a Beam AB with a plastic Moment Mp at the right end. The reaction RA equals 2wn*(L/2)-Mp/L. To have zero Shear, we get the relation of wn*L-Mp/L=2*Wn*x. We adjust the equation to obtain a Value for Mp.

We can derive a relationship between Mp and Wn by examining the plastic hinge. And equate it to Mp. We substitute the obtained Value of x to derive x. The x Value will be equal to 0.414*L

Substitute x, and we get Mp = 184.167 ft·kips. Finally, we get Wn =4.193 Kips/ft.

Check Wn using the upper-bound method.

If we have a span distance L = 1616 ‘, then the x distance to the plastic hinge = 0.414414*( 1 6)=6.624’. Due to the failure mechanism, a Deflection of Δ will occur.

For the first span, the distance to the maximum Deflection is 6.624′ from the left support.
The remaining distance to the right support is (16-6.624)= 9.376′. We have two angles θ1 and θ2. θ1=Δ/0.414L, for θ2= Δ/0.586 L.

We equate the external work with the internal work. The next slide image shows the full data for the Wn estimate, which equals 4.193 Kips/ft.

Solved problem 8-33: estimate the plastic nominal Load Wn for the 2nd span.

The second span can be treated as a continuous Beam on both sides.

Due to symmetry, we have 2*Mp at the two edges and one Mp at the mid-span. We will check the internal work. The second span = 16 ‘. We have three MPs acting as shown in the sketch. Due to Deflection, we will have an angle θ at both sides. For θ =Δ/8.The Value of 2*Δ=Δ/4.

The external work is = Loa d * span * averagdeflection = Wn* 16* Δ* 0.50Δ* 0.50. The internal work= Mp(θ) +Mp(θ)+Mp(2*θ).

The total Value of the internal work =4*Mp*θ. The external work=internal work, W_n=Mp*(1/16). The nominal Wn is estimated from the second span: Wn = Mp/16 = 0.0625*Mp. Mp is the M-nominal of the given section W16*26. The Wn Value is 11.51 kips/ft.

We have two values for Wn: 4.193 kips/ft and 11.51 kips/ft. We will select the lower Value, 4.1933 kips/ft. This is the nominal Uniform Load for the three spans. But the first and third spans will carry a nominal Uniform Load of 2*41934 = 8.3868 kips/ft.

Investigate the middle span; we have no plastic hinge there.

We have a plastic hinge in the middle of the second span, Bc, because the Moment Value is (+184.176-134.176=49.991 ft-kips, which is less than Mp.

I have included the Shear force diagram for the three spans due to the nominal Wn loads.

Use MASTAN 2 to estimate the Value of Wn.

We can use MASTAN 22 to find the nominal Uniform Load. E = 29000 ksi, and Fy = 50 ksi. To find Wn, we must convert the Beam spans to inches and the Uniform loads to kips/inch.

I consider a working Load of 0.20 kips/ft, 0.40 kips/inch for the first and third spans, and 0.20 kips/inch for the mid-span.

After making the first order inelastic, the lambda Value needed to create a mechanism is 1.747. The nominal Load Wn equals 2*1.747 = 0.3494 kips/inch, or 4.1928 kips/ft, which matches our calculation.

The first and second hinges form under a Load of 0.3322 kips/inch. Please refer to the next slide for Deflection and the locations of plastic hinges.

We can confirm first- and second-order plastic hinges when the Bending Moment equals Mp at Wn = 7.9728 kips/ft.

The next slide shows the Moment values for the second span.

In the next slide, we can check the Shear Value of the three beams using MASTAN 2. Thanks a lot, and I hope this Post is useful.

For the PDF file for this Post, you can view or download it from the next document.

Provide more information about the structural analysis – Link to III.
This is the next Post, 37a-solved problem 8-34– How to get Wn Value for a three-span Beam?

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.