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I have a solids problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track, (perhaps there was an issue setting up my deflection equations?). Any assistance would be highly appreciated.

Question:

enter image description here

My work is as follows:

Castigliano's theorem: deflection = partial derivative of strain energy U wrt a force F

I split the beam into two parts, A - B and C - B (So I don't have to deal with the reaction forces at C)

Edit:

X is the distance from the free end for Mab and Uab

And it is the distance from F for Mbc and Ubc

My first moment:

Mab = Fx

2nd moment:

Mbc = F(x-b)

U = integral from o to x of (M^2/(2EI))

Uab = integral from 0 to b of ((f^2*x^2)/2EI) = (f^2/2EI)*[x^3/3] (0-b)

Uab = (f^2/2EI)*(b^3/3) = ((F^2*b^3)/6EI)

Ubc = integral from 0 to b of ((F^2*(x-b)^2)/2EI) = (F^2/2EI)*[(x - b)^3/3] (L-b)

Ubc = ((F^2*(L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)

Therefore

Utot = ((F^2*b^3)/6EI) + ((F^2*(L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)

Utot = ((F^2*(b^3+L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)

Utot = ((F^2*(L^3 - (3b(L^2))+(3L(b^2))))/6EI)

Deflection = dUtot/dF = ((2*F*(L^3 - (3b(L^2))+(3L(b^2))))/6EI)

Deflection = ((F*(L^3 - (3b(L^2))+(3L(b^2))))/3EI)

Clearly not the answer, I am not sure but maybe I went wrong somewhere in the first couple of steps, perhaps either setting up my moment equations or setting them up in the integral. But I can't figure it out. Can someone please help ID my error and let me know what I did wrong?

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  • $\begingroup$ you need to actually label the figure. Whats x? $\endgroup$ – agentp Mar 31 '18 at 13:16
  • $\begingroup$ Hi I have edited the question, and have described my labels. I can physically go label the picture if necessary. $\endgroup$ – gunter Mar 31 '18 at 13:23
  • $\begingroup$ there is no moment in the free end $\endgroup$ – agentp Mar 31 '18 at 13:27
  • $\begingroup$ Hi, thanks for your reply, can you please clarify where I should be taking moments about? $\endgroup$ – gunter Mar 31 '18 at 13:40
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Castigliano's theorem says that we can apply a dummy load $P$ at any point of interest; the deflection in that location is $$y=\int_L\frac{M}{EI}\frac{\partial M}{\partial P}\,dx$$ letting $P\to 0$.

Apply the dummy load at the free end of the beam (call this $x=0$). I'm only going to consider the region of the beam to the right of the applied force $F$, since the other part stores no real strain energy. The bending moment in this region is $M=Px+F(x-b)$; thus, $\frac{\partial M}{\partial P}=x$. So we have $$y=\int_b^L\frac{[Px+F(x-b)]x}{EI}\,dx$$ which can be verified for $P\to 0$ to give $y=\frac{F}{6EI}(2L^3-3L^2b+b^3)$, which matches the answer given in the problem.

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  • $\begingroup$ thank you so much for your insightful response, its much appreciated. You really helped me understand this. $\endgroup$ – gunter Apr 3 '18 at 10:09

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