So in one of my classes, we very non-rigorously derived the 1D stress field for a bar with a longitudinal surface loading(adsbygoogle = window.adsbygoogle || []).push({}); T(x)like show below:

By taking out a differential element (assuming constant cross-sectional area) and applying good old Newton's Second, we end up with:

[tex]A \,d\sigma=-T(x)dx[/tex]

which is easily separable to find [itex]\sigma(x)[/itex].

My problem now is to do the same assuming that a concentrated loading is applied to one end like shown below:

Now I feel like this should be simple in that we are only adding on a constant force. But I am not sure where to add it? That is, when I take out an arbitrary differential element, do I apply the F_{A}to the element? That is,do I assume that Flike in elementary statics? Like I have shown below:_{A}acts along the entire length of the bar

I ask because if I do this, I end up with the following differential EQ:

[tex]A \,\frac{d\sigma}{dx} = -T(x)-\frac{F_A}{dx}[/tex]

which I am not sure what to do with (though I admit, I have not triedtoohard).

Any insight would be appreciated.

~Casey

EDIThmmm.....maybe I am a little slow today I think that by multiplying through bydxmight help out.

EDIT 2maybe not.....if I multiply through by 'dx' I end up with

[tex]d\sigma = -\frac{T(x)dx}{A} - \frac{F}{A}[/tex]

which I cannot figure out how to integrate. The last term [tex]\frac{F}{A}[/tex] does not have a 'differential' to which I would integrate with respect too.

As a matter of fact, it should act to remain as a constant stress; and it seems like it 'wants to' but I cannot get the mathematics to show it.

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# Deriving 1D Stress Field in a Bar

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