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A simply supported beam is 64 feet long and has a load at the center (see figure). The deflection (bending) of the beam at its center is 1 inch. The shape of the deflected beam is parabolic. https://www.webassign.net/larprecalcaga5/10-1-090.gifenter image description here

(a) Find an equation of the parabola. (Assume that the origin is at the center of the beam. Express x and y in feet.)

(b) How far from the center of the beam is the deflection equal to 1/3 inch? (Round your answer to one decimal place.)

I know that the answer to a is y=(1/12288)x^2, but I have absolutely no idea why.

I have no idea how to go about solving b.

Any information is much appreciated, thanks :)

DonAntonio
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marc.soda
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  • You do not have to do much more than following instructions. If $y=ax^2$ and when $x=32\text{ feet}$ you have $y=1\text{ inch}$, there are not many chances for the value of $a$. – Jack D'Aurizio Jun 11 '17 at 16:58
  • You should at least pass your units to the metric system, as most of the world has. – DonAntonio Jun 11 '17 at 16:58
  • @JackD'Aurizio That is what I thought at first, but it is trickier: it is one inch and 34 feet...and I think a feet is $;12;$ inches, or some similar nonsense... – DonAntonio Jun 11 '17 at 16:59
  • @JackD'Aurizio That's the problem with the three only countries in the world who hasn't addopted the metric system... – DonAntonio Jun 11 '17 at 17:00
  • @DonAntonio: how many inches give a feet? I do not know that. – Jack D'Aurizio Jun 11 '17 at 17:01
  • @JackD'Aurizio I think it is $;12;$ since an inch is approx. $;2.54;$ cms., whereas a feet is something like $;30.4;$ cms. – DonAntonio Jun 11 '17 at 17:02
  • Bad problem: It's not really a (quadratic) parabola, but a piecewise cubic, according to Bernoulli's theory of beams. One could place two equal loads at equal distances from the center; then the piece between them would be a parabola. – ccorn Jun 12 '17 at 10:46
  • Note that Mezex posted the same question, twice. – amWhy Jun 14 '17 at 21:35
  • Do. Not. Repost. Ever. There is no upside (the interested answerers will easily find the first edition). There are lot of downsides. The material is scattered, the volunteer answerers may waste their time reproducing old stuff etc. For such reason reposting is considered rude, and we have a rule against doing that. The rule is strictly upheld, and the chances of a repost falling thru the cracks is very low. – Jyrki Lahtonen Jun 15 '17 at 10:37

2 Answers2

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$y$ should be smallest in the center. You should specify your units, as the question talks both about feet and inches. I assume the units are feet. I will take $y=0$ to be the position of the beam before the deflection. Then the beam is at $y=0$ at the ends, which are $x=\pm 32$ and $y=-\frac 1{12}$ at the center. The equation is then $y=-\frac 1{12}+\frac {x^2}{12288}$. The deflection at any point is just $|y|$, so solve $y=\frac {-1}{36}$

Ross Millikan
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  • The vertex is at the origin. The proper equation is y=1/12288x^2. My instructor says so. So, given that, when is the deflection equal to 1/3 inches. Thank you for the quick reply though. – marc.soda Jun 14 '17 at 15:15
  • That is fine, I defined $y$ as the starting height before the deflection. With your coordinate system, where did the beam start out in $y$?. The deflection is downward from that. – Ross Millikan Jun 14 '17 at 15:28
  • I'm not sure what the answer is based on the information you have given me. Sorry, I'm just a bit confused. I'm kinda new at all this. – marc.soda Jun 14 '17 at 16:33
  • Deflection is measured from the unloaded condition. Without the load, the beam would be straight. What $y$ value would it be in your coordinates? Then you want the loaded $y$, which is your equation, to be $\frac 13$ inch lower than that. That gives you the loaded $y$ and you need to find an $x$ that corresponds. – Ross Millikan Jun 14 '17 at 17:26
  • We are both getting confused between feet and inches. The undeflected beam is at $y=1/12$ because we are using feet. You are looking for the point $1/36$ below this. – Ross Millikan Jun 14 '17 at 18:26
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Since the parabola's vertex is the origin and we're talking of an upwards parabola, it looks like $\;y=ax^2\;,\;\;a>0\;$ . Now observe the parabola passes through the point $\;(32\cdot12,1)\;$ (in inches and assuming we indeed have $\;12\;$ inches in one feet)..

Find now $\;a\;$ and yoour parabola's formula.

DonAntonio
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