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36-Solved problem 8-32 for plastic nominal Uniform load.

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Solved problem 8-32 for plastic nominal Uniform load.

Solved problem 8-32 for plastic Nominal Uniform load by the lower bound theorem.

We have a Statically determinate system, a beam that is hinged at one support and has rolled in the other support, and loaded under a partially uniform load. The uniform load occupies a length of 16′ out of the total span, which is  32′.

The section of the beam is W18x40 with Fy=50 Ksi. The beam is a statically determinate system.

We have a given section of W18x40, use a table to get Zx, the plastic section modulus for that section, Zx value=78.40 inch3.
The Nominal Mn= Mp=FyZx=5078.40 inch. kips, then divide the value of Mp by 12 to convert it into Ft.kips.Mn=Mp=50*78.40/12)=326.66 Ft.kips.

Solved problem 8-32 for  Nominal Uniform load.

To convert the system to an unstable system, we need to add a hinge at the midspan. We need to estimate the positive moment of the beam at the plastic hinge at C. To get the value of reactions to estimate the plastic moment, The load weight=Wn*16.
The Reaction at A takes half of this value, RA=0.50*16W=8Wn, and the same reaction value RB for support at B.
The bending moment is a line from zero to the edge.

To estimate the moment to the edge of the uniform load, we will multiply RA*8.

The moment value at the edge=8*8*Wn=64*Wn, this value is the same value at the other edge of the uniform load. A parabola of w*(16)^2/8 will be added to get the value of the moment at point C, which is =64*Wn+32*Wn=96*Wn, which is the maximum positive moment value.

The Mp, the plastic moment at the plastic hinge  =96*Wn, and also equals 326.66 Ft.kips. The plastic nominal uniform load Wn=Mp/96=326.66/96=3.40 Kips/ft, is the value Wn as estimated by the lower bound theorem or the statical method.

Solving the example by the upper bound theorem.

Another method of estimating the plastic nominal uniform load is the upper bound theorem or the mechanism method.
Since we have a partially uniform load,  the external work of the uniform load is the value of this load multiplied by the average deflection underneath the uniform load.

This rule applies if the uniform load is occupying the full length of the span, then the external work will be Wn*0.50*Δ, but in this case, a different treatment will be used, at support A, there is a slope due to deflection at point C, due to collapse, the angle at A=θ.

The same angle will be present about the roller support and its value is =θ. The positive Plastic moment at the mid-span will have an angle=2*θ, which is internal work. How can we estimate the external work in the case of a partially loaded beam?

From Prof. BHVAKati’s book, Structural Analysis – Vol. 2, in the case of partially Uniform load, consider the span as composed of the fully-loaded uniformly loaded span acting downwards minus partially loaded two parts of span in the upward direction for the places of zero loadings as shown in the sketch.

The external work can be estimated as the sum of case 1 for the fully loaded span, Wi=0.50*Wn*32*Δ, minus the external work for the two parts for the uniform loading acting upwards.

The deflection at the edge=Δ/2, since the distance to the edge/mid span=0.50. The external work for that case =2*Wn*8*(0.50)*Δ/2.

The total external work for the system=0.50*Wn*32*Δ-2*Wn*8*(0.50)*Δ/2=16*Wn*Δ-4*Wn*Δ=12*Wn*Δ.
This is the external work for the system, while for the internal work=Mp*2*θ, where θ=Δ/16.(12*Wn*Δ)=Mp*2*(Δ/16)= Wn, Δ goes with Δ, Wn=2*Mp/(16*12)= Mp/96. Wn=Mp/96, since Mp=Mn=326.66 Ft. kips, then wn=326.66/96=3.40 kips/ft.

Solving the example by the upper bound theor

The value of plastic nominal uniform load =3.40 Kips/ft, this is the same value obtained by using the lower bound method, or the statical method.

Solving problem 8-32 by MASTAN 2.

The beam was modeled using MASTAN 2, by the first inelastic analysis, using a uniform load of 1 kip/inch, the node distances are all in inches. As we can see from the next slide image

There is a plastic hinge located at the mid-span due to a Wn load of 0.284 kip/inch. The Value of Wn as given by the program MASTAN 2 matches with our estimated Wn value of 3.402 Kips/ ft.

The Wn load By MAStA

The next slide image shows the moment distribution done By MASTAN 2 for the partially loaded beam. The plastic moment obtained is 3920 inch. kips which is the same as estimated earlier the Wn value Wn value is 3.403 K/ft.

This is the pdf file used in the illustration of this post, the pdf does not include the MASTAN 23 additional data..

Have more information about the structural analysis –III.
The next post is Plastic Nominal Uniform load for partially loaded beam.

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