What's the difference between W = 0.5 P Δ and W = P Δ for external work?

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Homework Statement


both of these notes are in the same chapter . So , they are relevant to each other ... In the first picture, what's the difference between the 2 circled part ? The first one is W = 0.5 P Δ , the second one is W = P Δ..

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The Attempt at a Solution



For the second picture , I think we should use W = 0.5 P Δ instead of W = P Δ ..
(For the external work, the author use 1Δ , so 1 = P so , it's clear that the author use W = P Δ and not W = 0.5 P Δ)

I think W = 0.5 P Δ is more appropriate because W = force x displacement , which is also area under the graph ... Since the area under the graph is triangle , so 0.5 is necessary [/B]
 

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The author seems very sloppy with his maths. The first two lines should be:
P ∝ ∆
P = k∆

and he should integrate P.d∆ not P1.d∆1

This is analogous to doing work against a spring, so the factor ½ is required and accounts for the area under the line being triangular.

I can't figure out how the equation in your red rectangle is relevant here. (It would apply in the case where P was fixed regardless of ∆.)

Your second image is beyond me.
 
In the first image, I do not understand what it means to say "P1 is located for displacement Δ1".
If it means it is constant over that displacement then it makes sense.

For the second image, there seems to be missing context. It refers to "the" unit load being applied, as though this is continuing from some prior description. Can you supply that background?
 
Can I say that for the trusses member, constant force is applied so that the trusses deform ( displacement occur ) ?? So, p (Delta ) is used ??
 
fonseh said:
Can I say that for the trusses member, constant force is applied so that the trusses deform ( displacement occur ) ?? So, p (Delta ) is used ??
If you apply a constant force to, say, a spring that starts off relaxed then you will get a substantial initial acceleration. The KE developed would mean that you would overshoot the equilibrium position.
However, if the situation is that there is already a significant load and we are just applying a small extra load then the force can be largely constant over that deformation.