SA56: Moment Influence Line for Indeterminate Beams: Part 1

This lecture deals with a quantitative approach for constructing moment influence lines for indeterminate beams.

[@weskentocy]

thanks for the video, and it will be great if more example for the reaction, shear and moment influence line for statically indeterminante structure, either solve by conjugate beam method or virtual work method

[@highlightsontheway7012]

Please provide lecture on deflection of inderminate structure basic. Love from India

[@DARVESHKAUSHIKHR09]

Please make video lectures on mechics of solid

[@SajjadAhmed-nq3kz]

on the right side due to the roller support the influence line will be straight or curved?? i think it will be a straight for indeterminate beam with a fixed and roller support!

[@DrStructure]

The influence line is going to be nonlinear (curve) all over.

[@thyathnarongny7967]

Although the given moving load is not distributed to which this imply it need to be curved but there are one thing to remember Fixed support keep the beam from rotating around making deformed shape of the beam to curve like to satify its support condition

[@saicharangarrepalli9590]

For the propped cantilever shouldn't the influence line be straight to the right side of C i.e. towards the roller support?

[@DrStructure]

Although the shear diagram for the beam is linear, the shear/reaction/moment influence line for indeterminate beams is nonlinear. For the propped beam, the reactions at the fixed end in terms of the position of the unit load are: M = x - (3 x^2/2L) + (x^3/ 2L^2) V = 1 + x^2 (x - 3L)/2 L^3 Note that both shear/reaction and moment at the wall vary non-linearly with respect to x (unit load position). The above equations indicate that shear at any point to the right of the fixed support (regardless of x value) is nonlinear as well.

[@saicharangarrepalli9590]

@@DrStructure But isn't the right part going to rotate freely upon introduction of an internal hinge?

[@DrStructure]

That is not a real hinge in the beam. The real behavior of the beam is not influenced by the fictitious hinge we've placed there. The actual shear value remains nonlinear in terms of x (per the above equations), unless we have a real hinge in which case the beam becomes statically determinate (and the above equations become linear in terms of x). The principle that we used to draw shear influence line for determinate beams (a diagram consisting of straight lines) does not hold true for indeterminate beams.

[@dylandewilzen1864]

I would just like to find out from you if im on the correct path. The question is for an indeterminate beam: if you have a 3 span beam, (end supports are fixed and the 2 supports separating span 1 and 2 as well as separating span 2 and 3 are roller) span 1 =EI, span 2 = 2EI =2 and span 3 -EI. How would you approach it to determine a point of the beam using the Muller Breslau principle. Would you combine the Muller Breslau principle with the Double integration method (Macaulay method) to take the varying EI into consideration? Hope that made some sense

[@DrStructure]

Alright, we have a continuous beam with varying moment of inertia. But what is it that we want to determine? Moment influence line? Deflection? reactions? ...

[@chungken8496]

Excuse me, @0825, why the do moment at C be negative when m1-[(1-4m+3m^2)^0.5/(1-m)]? How do we know the moment is positive or negative?

[@DrStructure]

At 3:20, we can see that the influence line drops below the x-axis, it becomes negative. Knowing that moment could become negative, now we want to figure out: (1) for what m values that happens and (2) where exactly the diagram crosses the x axis. The derivation tells us than the negative region develops only when m < 1/3. It also gives us the equation (the one you specified) for determining the zero-point of the influence line. So, we can determine the positive and negative regions of the diagram qualitative, then use the derivation to determine their exact locations.

[@sainathreddyvarikuti4835]

Can u do examples on Deflection of beams by moment area method

[@chungken8496]

Excuse me, @1119, I have 3 questions. 1. What is the physical meaning of the area under the curved scaled? 2. What is the physical meaning of the actual area? 3. Why do the max. negative moment at C equal to the UDL times the actual area?

[@DrStructure]

This goes back to the definition of influence line. Say, we have constructed the moment influence line for point C. Now, we divide the beam into finite number of points. We can refer t a typical point as x_i. So, the points on the beam are x_0, x_1, x_2,...x_k where the k is the total number of points. At each point the influence line has a height, let's refer to the height at x_i as h_i. What is h_i? It is the magnitude of the moment at C when a unit load is placed at x_i. Now, suppose the beam is actually subjected to a uniformly distributed load (say, with the magnitude w), meaning there is a concentrated load applied at each point on the beam. If we assume a unit distributed load, then at each x_i we find a unit load, meaning there would be a non-zero h_i (i.e., there is a moment at C due to that load at x_i). But since there is more than one unit load on the beam, then the total moment at C would be Sum (h_i) for i = 0 to k. This sum, we compute by integrating the influence line function which is the same as calculating the area under the curve. We then multiply the area by w (since the influence line is drawn for a unit load) to get the total moment at C due to the actual distributed load placed on the beam. When only a part of the beam is loaded, say, only the negative area is loaded in order to determine the maximum negative moment at C, then we integrate the influence over the negative area only, meaning we are not placing the distributed load over the entire length of the beam, rather we are pacing it over the negative region for maximum effect. As to your first question, you can view our derivation as if we first scaled the beam to [0,1] interval, the beam has a length of 1. Or, the location of the unit load varies between 0 and 1 only. We do our calculation based on the scaled model, then at the end we multiply the result by the actual length of the beam to get the actual moment value.

[@samiulislamniloy6785]

Dear sir, m

[@DrStructure]

Correct, the mathematical expressions are derived for this particular beam; we get different expressions for different beams ...

[@thearaksmey6237]

Where is the part 2 of this ?

[@DrStructure]

At this point, we have two lectures available on influence lines for indeterminate beams, this one (SA56) and SA57 (kzbin.info/www/bejne/kKfWnqWJebBjq5o ).

[@thearaksmey6237]

I have already watched the SA 57. Thanks for your sharing. I think u help me a lot.

[@geostructureacademy]

Why are you making this simple problem so complex??

[@DrStructure]

Please feel free to present alternative solutions for this statically indeterminate influence line problem.