# Plastic analysis of a beam,virtual work?

1. Apr 4, 2009

### em07189

hI!

I'm trying to understand here a little bit of plastic analyses,but i can understand in the page of the book bellow,why the work at the plastic hinge is equal to MPx2theta?why MP and not MA?
can someone explain me this with some angle relation or trigonometrics?

the second is why the displacement of the working load is equal to W(L/2xtheta), why this result?

Is the work at the plastic hinge, the external work?
Is the displacement of the working load the internal work?

Thanks.

http://img441.imageshack.us/img441/754/plasticanalysis.png [Broken]

Last edited by a moderator: May 4, 2017
2. Apr 5, 2009

### nvn

em07189: The elastic moment is Ma, if the beam is not yielding. If you continue increasing the applied load, the moment will increase to the plastic moment Mp, when the beam collapses. The work done at the plastic hinge is Mp*2*theta, because a midspan moment Mp rotates through an angle 2*theta, where theta is just an arbitrary rotation angle, shown in the diagram. For a simply-supported beam having an applied midspan point load W, the support reaction force is 0.5*W, located at a distance 0.5*L from the beam midspan. Therefore, the external work done by each support reaction force on the beam midspan is 2(0.5*L)(0.5*W)(theta) = 0.5*W*L*theta. The work of Mp at the plastic hinge is the internal work. The work due to displacement of the applied load W is the external work.

3. Apr 5, 2009

### em07189

HI nvn!

And thanks for you help.

But i'm still not understanding why the work at the plastic hing gives MP*2theta?

why the workdone at the tree plastic hinges is =MP(theta+2*theta+theta)?

and why work done by the displacement of the load=W/L*L/2*L/2*theta? why this reaults?

thanks again.

4. Apr 5, 2009

### nvn

em07189: Keep in mind, posts 1 and 2, and equations 2.24 and 2.25, refer to a preceding diagram, which you did not post. Only equations 2.26 and 2.27 refer to figure 2.20. Work is defined as a force translating through a distance, or a torque rotating through an angle. For the unshown beam, the beam rotates through two angles theta; therefore, the internal work is Mp*2*theta.

Your last two questions in post 3 refer to figure 2.20. The beam rotates through four angles theta. Therefore, the internal work is Mp*4*theta. I currently don't have an explanation for your last question, because the author does not show his derivation. And the way he has written W is unclear. If the uniform load were w, having units of force per unit length, then equation 2.27 would be Mp = w*(L^2)/16, which is the correct answer. If W = w*L, then this becomes Mp = W*L/16. But I currently don't have an explanation for equation 2.26.b.