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 :)


closed as off-topic by Jack D'Aurizio, Daniel W. Farlow, Namaste, kingW3, Smylic Jun 14 '17 at 22:38

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  • $\begingroup$ 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 $a$. $\endgroup$ – Jack D'Aurizio Jun 11 '17 at 16:58
  • $\begingroup$ You should at least pass your units to the metric system, as most of the world has. $\endgroup$ – DonAntonio Jun 11 '17 at 16:58
  • $\begingroup$ @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... $\endgroup$ – DonAntonio Jun 11 '17 at 16:59
  • $\begingroup$ @JackD'Aurizio That's the problem with the three only countries in the world who hasn't addopted the metric system... $\endgroup$ – DonAntonio Jun 11 '17 at 17:00
  • $\begingroup$ @DonAntonio: how many inches give a feet? I do not know that. $\endgroup$ – Jack D'Aurizio Jun 11 '17 at 17:01

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.

  • $\begingroup$ That would mean that a = 1/147456, according to your information. If that's not correct, please tell me why. Thanks for the response. $\endgroup$ – Mezex Jun 11 '17 at 17:16
  • 1
    $\begingroup$ $147456/12=12288$. $\endgroup$ – hypergeometric Jun 11 '17 at 17:25
  • $\begingroup$ @Mezex That seems to be accurate...but that's only calculations. $\endgroup$ – DonAntonio Jun 11 '17 at 17:31

$y$ 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 $y=\frac {-1}{36}$

  • $\begingroup$ 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. $\endgroup$ – Mezex Jun 14 '17 at 15:15
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Jun 14 '17 at 15:28
  • $\begingroup$ 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. $\endgroup$ – Mezex Jun 14 '17 at 16:33
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Jun 14 '17 at 17:26
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Jun 14 '17 at 18:26

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