# What is the work done by a force acting on a deformable body?

In our mechanical engineering course, we have encountered the subject Mechanics of deformable bodies. I am familiar with the work energy theorem for point masses and rigid bodies. But my concepts are not developed on distributed elastic masses.

The textbook here expects us to take it as said.

My doubt is, the displacement at any point is because of the influence of all the forces. And each force influences displacement at every point of the body. So how can we say that work done is equal to force at the point×displacement at the point, for any point? Help me out with an interpretation, or point me to reading sources please. Thanks

• Plastic deformation takes energy to accomplish (if it didn't, the material would not be a solid). – Jon Custer Oct 10 '18 at 13:54
• @JonCuster: In this case the text is discussing elastic deformation, which also absorbs energy. – James Oct 10 '18 at 15:12
• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – user191954 Oct 11 '18 at 13:20
• @Chair that would be a sizeable bit of work – Rohit Oct 11 '18 at 15:43
• @Rohit Well the site policies I linked indicate an extremely strong consensus that this is necessary. It looks like you've used a pdf; you could copy-paste and attribute. Otherwise, you need to put the work into the question if you want people to answer it. Remember the reasons I mentioned earlier: it's a serious accessibility and view-ability issue: screen readers don't catch that, and you loose readers who don't want to zoom about until it's legible. The fact that you have an extended quotation does NOT exempt your post from the guidelines. – user191954 Oct 11 '18 at 15:45

If the forces were constant as the deformations increase from zero to the fully deformed state, then the work done at each force would be

$$W=F\delta$$

However, the text is making the assumption that each force is increasing from zero to 100% while at the same time the deformation is increasing from zero to 100%. Therefore, the work done at each force is the average force times the deformation.

$$W=F\delta{}/2$$

The above is true for elastic deformation, where the ratio of force and displacement is constant.

For plastic deformation, the average force will not be half of the total force. In this case, a material stress vs. strain curve would be needed to calculate the average force.

The text also mentions that we assume the force application is gradual. This simplifies the situation by removing inertia considerations.

A force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force. Notice that the definition of work doesn't say that the displacement was caused by the force.

The reason the works at each point can be summed up to get the total work done is because each force is ramped from zero to its maximum value during the same time period. This is what is meant in the text about a fraction of each load being applied at the same time. This load application process ensures that each point moves only from zero deflection to its maximum deflection.

Assume instead that $$F_1$$ is fully applied first and then $$F_2$$ is applied. In this case, the length of the path traveled by point 2 would likely be greater, and the work done at point 2 would be correspondingly greater.

In other words, the work done on a body is path dependent. In a similar way, the work done lifting an object directly up one flight of stairs is less than the total work done lifting it up two flights and then back down one flight.

• I have edited the question to narrow down my doubts.... – Rohit Oct 11 '18 at 0:46