If we think of a beam as a collection of small elements, then each of those can store a small amount of energy (which is the product of internal force and deformation). The actual deflection of the beam does not appear in the expression for internal energy. That deflection exist only for the entire system, not for the individual infinitesimal elements. When we add up the energy for the individual elements in the beam, we get the total energy. Suppose we expressed this internal energy in terms of the applied load, since the internal force in each member is a function of that applied load. But, since that energy is not directly defined in terms of the system displacement, how could that displacement result when we differential the energy expression with respect to the applied load?

Dr. Structure for any structure solvable by work energy principle is giving the same result when I am differentiating the internal energy w.r.t the applied load..and to calculate the displacement in any other location we are assuming an imaginary load at that point and calculating the displacement and in the final expression the imaginary load is substituted by 0. Now it is very similar to unit load method and seems to be an application of virtual work principle..but I am not getting the actual connection between these two.

I am not quite sure what you are doing, how you are solving the problem. Feel free to write down your solution in a step-wise manner and email it to me (dr.structure@educativetechnologies.net). Once I better understand your solution, we can have a discussion around the topic.

Dr. Structure I am sending you the photo of solution of one problem sir

What you are using is called Castigliano’s Second Theorem. The theorem is derived from the observation that change in external work equals to change in internal energy, say with respect to load P. Let U = internal energy. P be an applied load, and D the displacement under P. Since there may be other loads applied to the structure, the expression for U could involved other loads and displacements. Say, there are two applied loads, P and Q with the corresponding displacements D and E. Then we can write: U = (1/2) P D + (1/2) Q E. Now, suppose we make an infinitesimal change to P, call it dP. This causes a change in internal energy which we refer to as dU. This additional energy is the result of dP traveling through D. That is: dU = dP D. If you are asking wouldn’t dP cause additional displacement in direction D? If we denote change in that displacement as dD, then it can be said that there is also a (1/2)(dP)(dD) change in internal energy due to dP. But this term is very small and can be neglected. So, dU = dP D. But since dU is a function in terms of other loads and displacements, we can write is in terms of partial derivate this way: Let r denote the partial derivative operator, then (rU/rP)(dP) = dP D Note that the left hand side of the equation is dU. Simplifying the above equation, we get: rU/rP = D. That is, the target displacement equals to the partial derivative of U with respect to P. Intuitively, if the expression of external work and internal energy is: (1/2) P D = (1/2) Integral of (M m dx/EI) Then taking the (partial) derivative of both sides with respect with gives us: D = d(W)/dP where W is the integral expression.

Just proportional to number of engineers

Is this also Castigliano's Second Theorem?

In principle, yes.

Dr. Structure I see. Do you have sample problem about Castigliano's Second Theorem?

Sorry, don't have any ready examples for that.

Dr. Structure Oh, it's alright. Thank you!!

+Structures Question Since the load increment is very small, (f times delta) is a good approximation for the work done by the force. According to the table, total work equals: f(delta) + 2f(delta)+3f(delta)+...+n f(delta). This can be written as: SUM (i=1 to n) (i)(f)(delta). Since SUM(i=1 to n) i = n(n+1)/2, then the above sum becomes: n(n+1)(f)(delta)/2 Since n(f) = F and n(delta) = D, then we can write: total work = (1/2)(F)(D + delta). But since, delta is quite small, it can be neglected. So, work = (1/2)(F)(D).

A load of 2kN is applied at mid span. That makes the left and right vertical reactions 1 kN each. A reaction force of 1 kN at the left end of the beam results in a bending moment of (1 kN) (3 m) = 3 kN.m three meters away from the reaction force.

Thank you so much. Got it!

At t2, the added f causes an additional displacement delta. So, we get f*delta. But, the other f, the one that was there from step t1 also goes through the new delta doing work. The work done by that f in step t2 is also f*delta. So, the total work increment in step t2 is the sum of the two f*deltas, or, 2f*delta.

@@DrStructure okay got it thank you so much

Think of it this way: a very small force applied to the beam causes a very small deflection. As the load increases so does the deflection. This relationship between load and deflection, if we graph it, takes the shape of an inclined line, To graph the line we can use two points. Assuming the x-axis represents deflection and the y-axis represents load, one of the points could be the origin (meaning that when the load is zero, the deflection is zero). And, when the load is P, say the deflection is D. Note that such an inclined line defines a triangle. That is, the area under the line is that of a triangle. Since work done by P can be defined as the area under the P-D diagram, then we can write work as (P)(D)/2. Without 1/2, we end up calculating the area of a rectangle. That would mean that the relationship between P and D is constant. That is, no matter how small or large the load, the beam deflects the same amount. This is like saying the beam deflects 10mm regardless of the magnitude of the load. But that kind of behavior is not consistent with how structural materials behave.

@@DrStructure Thank you sir! I got it.

In classical beam analysis, we generally ignore internal energy due to shear deformation since it is often small and negligible. Shear deformation however becomes an issue in deep beams, thus should not be ignored in such cases.

External work comes to develop under applied loads only, no matter how the structure deforms. If the beam was subjected to one moment only, then only that moment contributes to external work. If the moment at the left side did not exist (if it was zero), then the expression for work would have become: Work = (1/2) ( 0 theta_1 + M theta_2) Or simply, Work = (1/2) (M theta_2)

I am assuming your are referring to the un-deformed vs deformed length. When analyzing the beam, we always use the un-deformed length of the beam.

The area of the triangle represents the sum of the work done by the load increments. That is, the sum of the entries in the last column of the table at 5:21 Linear material are ones in which the ratio of force to displacement, (or stress to strain) is linear. So, when we graph force vs displacement, we get a straight line. Elastic material is one in which when the load is removed from the member, its displacement goes back to zero. So, displacements are not permanent, they disappear when the load disappears. Pull a spring, and then release it, and you would see elastic behavior in action. Pull the spring too much, the elongation becomes permanent, the material (steel) no longer behaves elastically when a large force is being used. So, steel could be treated as elastic material up to a certain point before it becomes inelastic. If we apply a force to an object that moves unopposed, like pushing a cart on the friction-less surface, the work done by force P equals P * Delta where Delta is the amount of displacement the car goes through. Beams, however, don't behave like that. Beams want to resist displacement, that is why when the load is removed, the displacement vanishes. In the case of the cart, that is not the case. That internal resistance to displacement in structural members is the cause for (1/2) factor in the expression of work.

thank you so much, Dr

You are welcome!

The idea of integrating (P)(delta) over the length of the beam does not work here. Why? Because P is a concentrated load, not distributed over the length of the beam, work involves only P and the displacement under it. The beam displacements away from P do not contribute to work done. If the load was distributed so that we could write it as w(x) and then displacement in the beam was expressed as d(x), then the integration idea would work, but not in our case. Although, we can think about work due to a concentrated load as an integration problem too, but integrating over time. Say, we define P as a function of time. p(t) where at each point along time, there is a load increment, say, p(1), p(2),…p(n) are load increments at time, 1, 2, …, n. And at each such increment, displacement under the load is d(1), d(2), …, d(n). If we graphed this as p vs d, then we get that triangular area. Why the area is triangle and not rectangle, why is that (1/2) factor present when calculating the area under the graph? This has to do with how much work is being done per time increment. If we take p as a small load increment, then in increment 1 work done is (p)(d), in increment 2 it is (p)(2d), then (p)(3d), and so on until we reach the last increment at which the work is (p)(nd). The total work then is the sum (or the integral) over all the time increments. That sum is: (1/2)(P)(D) where P = np and D = nd.

No. The 1/2 factor is a function of the shape of the P-Delta diagram (triangular). It does not matter how many loads there are on the beam, if they are applied in increments, we get the same triangular shape. This is under the assumption that we have linear elastic behavior. To lessen the confusion, consider applying the method of superposition, when thinking about this problem. According to this principle, external work for P1 is (1/2)(P1)(D1) and external work due to P2 is (1/2)(P2)(D2). So, total work is the sum of the two: (1/2)(P1)(D1) + (1/2)(P2)(D2). Also, keep in mind that (1/2) does not show up in the incremental steps, it shows up at the end when we are adding up all the increments.

I just read a proof on maxwell betti reciprocal theorem that they first apply p1 and then p2 and while p2 was increasing p1 work was without the 0.5 factor plus the half when it was first appleyed.but thank a lot.

That animation is done using Adobe Animate. VideoScribe cannot create that type of animation.

+Dr. Structure thank you very much

+Dr. Structure thank you very much

Imagine the beam consists of N infinitesimal elements each being deformed by some small angle (d theta_i) due to an internal bending moment M_i. Since the internal energy associate with each element is (M_i)(d theta_i)/2, the total internal energy can be written as the sum of element internal energies, or: SUM (M_i)(d theta_i)/2 for i = 1 to N. Since M is a continuous function, the above discrete expression, mathematically speaking, can be replaced by the continuous integral expression: INTEGRAL M (d theta) /2

@@DrStructure why the N is 10 ? 0 to 10?

@@febrianimariyaningrum7013 Are you referring to the limit of the integral? That 10 is the length of the beam. That is, we are integrating the expression over the length of the beam.

The derivation for the work done by a single force was presented earlier in the video. A single force (P) going through displacement (D) results in (P)(D)/2 work. If we have multiple forces acting on a body each going through a specific displacement, then the total work done by these forces is the sum of the work done by the individual forces. So, if two forces (P1 and P2) go through displacements D1 and D2 respectively, the total work would equal to: (P1)(D1)/2 + (P2)(D2)/2.

@@DrStructure @Dr. Structure what makes me confused is: Firstly, I assume, P1 is working @A, causing displacement D1, and after P1 worked, P2 involved (let me know if Im wrong here, if both are working at the same time unlike I assumed, then how will the calculation be?). If my first assumption is correct, when P1 is working @ point A, it causes some displacement not only @ point A, but at the other point B too!! Let that be D2. So far so good, Now when another force P2 is working @B and generated a certain displacement (let D3), MY CONFUSION IS the total displacement @B (letD3) after P2 involved includes not only the displacement by force P2 itself, but also from previous force P1. In my sense displacement by P2@B will be found by subtracting the displacement @B by caused by P1 itself...so the formula be like 1/2(P1*D1+ P2(D3-D2))does it work that way?? I don't know...plz correct me and plz give me some concept too. I hope I could make you understand my confusion😇. It will be very helpful for me if you plz show me some derivation for "1/2(P1*D1+P2*D2)" like you did for "1/2P*delta"...

Yes, you are correct, D1 and D2 are dependent on both P1 and P2. So, it is not the case that P1 only causes D1 and P2 only causes D2. Calculating D1 and D2 due to P1 and P2 is a different conversation than what is the total work done by P1 and P2. Here, we are not using the expression for work to calculate D1 and D2. We are simply stating that if the structure has displaced by D1 (say, at point 1) and there is a displacement of D2 at point 2, and these displacement develop under load system P1 and P2. Then, the total work equals (P1)(D1)/2 + (P2)(D2)/2.

@@DrStructure owww....now I got it😊. Is there any video or book that you can refer me to know more about how these formulas are derived from the scratch? Actually your awesome explanation and creative videos made me so stubborn to know about this topic from very deep😊. This cerdit is yours! Thank you very much again. May I know from which country is this brilliant channel's guy from?😊

You want to look into energy methods. Most textbooks on introductory structural analysis devote at least a chapter or two to energy methods. The KZfaq channel is being maintained (technical questions answered) by a few people distributed across the world. The channel itself originates from the United States.

Energy is stored work. In beams where beam segments deform (by some angle) due to the internal force (moment), internal work = stored energy = the integral of (angle/rotation) times moment. This is analogous to external work which is the product of displacement and force.

@@DrStructure why is that moment*rotation is analogous to displacement*Force?

is there any geometrical basis or something ?

I understand that moment (you refer as force) causes rotation(deformation), but the fact that moment is not a force, and rotation is not a change in length confuses me.

Rotation *is* a displacement. Moment *is* a force. It is true that we can distinguish between linear vs rotational displacements/forces (the same way that we can distinguish between linear vs angular velocity), but that distinction does not imply that only one type of displacement/force produces work/energy. Work/energy is the product of displacement and force, be they linear or angular/rotational.

Hopefully this video is just a part of your learning experience, and you have other resources at your disposal to help you develop a good understanding of structural analysis concepts and method.