Solved problem 8-32 for plastic Nominal Uniform load.
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Solved problem 8-32 for the plastic Nominal Uniform load. It is required to evaluate the Plastic nominal uniform load for a simple beam that has hinged support and roller support the span is 32 feet. The beam is a steel beam with a given section W18x40 with Fy=50 Ksi, use the table to get the necessary data of Zx.
The beam is acted upon by a partial uniform load that occupies a length of 16′.
Two methods are used to get the Plastic nominal load, the first method is the statical method or the lower bound theory in which the plastic hinges are assumed and the two-moment diagrams are laid together, the place of a maximum positive moment to be estimated at that point one plastic hinge to be placed, the other two hinges are located at the places of maximum negative moments, at the fixed supports.
.The second method is the Upper bound method, in which the external work is equivalent to the internal moment, step by step procedure is explained. A relation between the plastic nominal load and the plastic moment Mp, which is considered as M nominal, hence the value of Plastic Nominal uniform load can be estimated.
The example was further modified to include another situation where the uniform load is shifted to the left, to occupy half of the full length of the beam. The value of the plastic Nominal uniform load is again estimated by using the two methods.
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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.
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.
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=Fy*Zx=50*78.40 inch.kips, then divide the value of Mp by 12 to convert into Ft.kips.Mn=Mp=50*78.40/12)=326.66 Ft.kips.
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 of 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 uniformly 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=326.66KFt.kips, then wn=326.66/96=3.40 kips/ft.
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.
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