# Maxima minima problem

Discussion in 'Math' started by amilton542, Jun 21, 2012.

1. ### amilton542 Thread Starter Active Member

Nov 13, 2010
496
64
I'm stuck on the following question. I feel like I have computed the correct results but because my units have cancelled I'm a bit lost.

Anyway, here's the question:

If a beam of length L is fixed at the ends and loaded in the centre of the beam by a point load of F Newtons the deflection at a distance x from one end is given by:

$y = \frac{F}{48EI} (3L^{2}x - 4x^{3})$

Where E = Young's Modulus and I = Second Moment of Area of the beam. Find the position and the value of the maximum deflection of the beam if:

L = 10m

EI = $30_{10^{6}}Nm^{2}$

F = 100kN

$y = \left(\frac{100kN}{(48)(30_{10^{6}}Nm^{2})}\right) ((3)(10m)^{2}x - 4x^{3})$

$y = \left(\frac{1}{14400}\right) (300x - 4x^{3})$

$y = \frac{1}{48} (x) - \frac{1}{3600} (x^{3})$

$\frac{dy}{dx} = \frac{1}{48} - \frac{1}{1200} (x^{2}) = 0\ for\ a\ maximum\ or\ minimum\ value$

$x = \sqrt{\frac{1200}{48}} = \pm5$

$\frac{d^{2}y}{dx^{2}} = -600x$

$\frac{d^{2}y}{dx^{2}} = - (600)(5) = -3000\ for\ a\ maximum\ value$

Substituting x= (+5) into the initial equation for a maximum deflection:

$y = \left(\frac{1}{14400}\right) ((300)(5) - (4)(5^{3})) = \frac{5}{72}$

If I envisage this geometrically, I'd say the position of the beam is (+5) for a maximum deflection, could some one please confirm this.

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
The answer of 5 certainly makes sense intuitively. The ends are fixed and the beam is loaded in the center. This means that the ends can not move, and there is symmetry about the center. Loading in the center would seem to make the center deflect the most, and hence gives only one answer of 5 m as reasonable. The value of -5 m is not allowed since the beam does not exist at that position.

By the way, your answer does have units of meters, since both x and y have units of meters given the context of the question.

EDIT: I just noticed something strange about the question. It says that the ends are fixed. This means that y=0 at x=0 and x=L. However, the equation for y says that the deflection at x=0 is y=0 as expected, while for x=L, one gets a non zero value for y. Hence, I'm either misinterpreting the question, or there is an error in the question.

Last edited: Jun 21, 2012
amilton542 likes this.
3. ### WBahn Moderator

Mar 31, 2012
23,381
7,094
Note that this formula is only valid for 0<x<L/2. For x > L/2, use the fact that it is a symmetrically loaded beam and thus, y(L-x) = y(x).

Due to symmetry, the greatest deflection has to be at x = L/2.

Do NOT blindly differentiate and set the derivative to zero because the behavior of the equation beyond x = L/2 is NOT guaranteed to be consistent or even well behaved. Thus, remember that you must evaluate not only every place the derivative goes to zero, but also allow for the possibility that the max occurs at the one end of the range or the other. In this case, symmetry demands that the slope go to zero at x=L/2, so blindly setting the derivative to zero will work in this case, but only by coincidence.

Are you aware that you are typesetting the 10^6 as a subscript of the 30?

You went about this the long way. Note in the original equation, only the factor in parentheses is dependent on x, so you only need to find where the derivative of that vanishes. Also, don't go plugging in specific values unless you need to. Instead, work symbolically and get a more general answer that is easier to sanity check.

$
\frac{dy}{dx} \ = \ 0 \ = \ \frac{d}{dx}\left(3L^2x \ - \ 4x^3 \right)
3L^2\ - \ 12x^2 \ = \ 0
12x^2 \ = \ 3L^2
x^2 = \frac{L^2}{4}
x = \frac{L}{2} \ \text{negative solution not within range of interest}
$

Do yourself a really, really big favor and start tracking your units from beginning to end of every problem you ever do for the rest of your life. Most errors you make will screw up the units, but if you don't diligently track the units, you throw away what is perhaps the most valuable and potent error detection tool you have.

Substituting x = 5m into the equation for y:

$y = \frac{F}{48EI} (3L^{2}x - 4x^{3})
\
y = \frac{100kN}{48(30\times10^6 Nm^2)} \left( 3(10m)^{2}(5m) - 4(5m)^{3} \right)
\
y = \frac{1}{14400m^2} (1500m^3 - 500m^3)
\
y = \frac{1000m^3}{14400m^2}
\
y = \frac{10}{144}m
\
\
y = \ 0.0694m \ = \ 69.4mm
$

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The attachment might shed some light.

File size:
169.8 KB
Views:
20
5. ### amilton542 Thread Starter Active Member

Nov 13, 2010
496
64
How have you extracted the inequalities? I can't see them.

That's exactly what I felt like I was doing. How can you see this by looking at the formula?

Yes I know. It's just a dirty habit I've picked up with standard form for clarity. Maybe this is not so for others.

Ok, will do.

I must of made an error in the beginning from doing it the long way. My units cancelled at the start.

Looks good to me.

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
Actually, this still doesn't resolve my confusion. The problem given has two fixed points, while the attachment is for a one-end attachment of the beam.

The problem says the beam is "fixed at both ends" which to me means that the very ends (x=0 and x=L) are attached to a fixed object. However, the equation given implies that x=0 is fixed and x=L sqrt(3)/2 is fixed. This results in the end x=L deflecting upward while the loading point x=L/2 deflects downward.

However, either way, the point L/2 has the greatest deflection, so it's not critical which interpretation is used.

Last edited: Jun 22, 2012
7. ### steveb Senior Member

Jul 3, 2008
2,433
469
Oops, I recommend the OP double check this. I just calculated the deflection at x=L and it deflects upward the same amount as x=L/2 deflects downward.

So, the interpretation is important, although the fact that the deflections are equal allows the answer of the deflection amount to be correct, but the point where it occurs is unclear (strictly both points are valid mathematically).

8. ### WBahn Moderator

Mar 31, 2012
23,381
7,094
I've already pointed out that the equation given only applies for 0<x<L/2. It does not apply at x=L. If you want defections beyond the midpoint, you use symmetry.

9. ### WBahn Moderator

Mar 31, 2012
23,381
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Because when I took Strengths of Materials, we had to derive most of the standard equations and the ones for symmetric systems commonly were of this form. Consequently, you get in the habit of assuming any symmetric system is in this form because, even if it is valid over the entire range, applying symmetry still obviously holds.

The quick check to see if the answer is only valid for the first half is to see if the necessary constraints are met at the other end. In this case they aren't, so it is almost certainly a formula that is valid over the first half.

10. ### steveb Senior Member

Jul 3, 2008
2,433
469
But, how do you know that? The problem doesn't say that. Seems ambiguous to me. What bothers me is that it says "the distance from one end", not "from either end".

11. ### WBahn Moderator

Mar 31, 2012
23,381
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Whichever end you choose to measure your distance from is x=0. If you choose the right end, then x increases moving to the left.

I agree that the question is missing some pretty important information. I don't know the context in which the formula was provided to the OP, so I don't know if they failed to include them or if they weren't given all of the information they should have been from the beginning.

This might help, at least as far as establishing that I'm not making this up:

Look at the top case on the second page.

12. ### WBahn Moderator

Mar 31, 2012
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Strictly speaking, asking for a "maximum" would require the positive answer since, strictly speaking, the negative answer is the "minimum". But the only reasonable interpretation of the question is that it is asking for the point at which the magnitude of the deflection is a maximum, and were the deflection equation valid over the entire length of the beam, then there are two answers.

Even so, the two deflections aren't equal. One is positive and the other is negative. They are not the same. In most engineering fields, thinking of positive and negative quantities that have the same magnitude as being "the same" is a very dangerous thing. A large compressive load and a large tensile load applied to a concrete member has very different consequences. We would never let someone get away with saying that a large positive voltage and a large negative voltage applied to an electrolytic cap are in any way the same. So just as a question asking for the amount of current flowing in a diode is not correct unless the direction is correct, so too is the answer to a question asking for the amount of a deflection not correct unless the direction of the deflection is correct.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The formula of interest is on the second page of the pdf attachment.

This one ....

• ###### Beam deflection formula.jpg
File size:
86 KB
Views:
9
Last edited: Jun 22, 2012
14. ### steveb Senior Member

Jul 3, 2008
2,433
469
I thank you both. I missed the fact that there is a second page to that document. I agree that most likely the constraint 0<x<L/2 is intended and left out, or just implied from the context of the book or class. Anyway, it's up to the OP to figure this out for sure.

15. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
Hello Steve,

Actually WBahn is half right. The equation as given is invalid anywhere beyond (to the right) of x.

The way to write a differential equation valid over the whole range is to use Macaulay Brackets. Then you can perform the (double) integration to solve it for y at any x.

go well

16. ### WBahn Moderator

Mar 31, 2012
23,381
7,094
I don't follow. The value of x is a variable that the user selects. So, for this 10m long beam, they could choose x=1m to find the deflection 1m from the end of the beam. The equation is still valid for another 4m beyond x. If they choose x=7m, then it isn't valid at x and hasn't been valid for the prior 2m.

The equation, as written, is valid for all values of x such that x <= 0 <= L/2.

17. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
The differential equation governing a beam, length L, restrained at both ends and loaded with point load W at the midpoint is

$EI\frac{{{d^2}y}}{{d{x^2}}} = {M_A} + {V_A}x - W\left\langle {x - \frac{L}{2}} \right\rangle$

The third term is a Macaulay bracket which means that the term is ignored for any value of x that makes the bracket negative.

http://en.wikipedia.org/wiki/Macaulay's_method

This can then be integrated twice with the constants of integration being determined by the conditions y = dy/dx = 0 at x = 0 Steve.

Then the original plus two integrated equations provide three simultaneous equations that can be solved for M, V and y.

Last edited: Jun 24, 2012
18. ### WBahn Moderator

Mar 31, 2012
23,381
7,094
But the question had nothing to do with the differential equation that led up to the equation given by the OP and that equation, which was the only equation under discussion, is valid for the constraints I provided. I'm still unclear why my information was only half right and also unclear what is meant by the equation not being valid beyond x as apposed to not being valid beyond L/2.

19. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
You guys had been busy solving the original equation whilst I was away sailing on the Zuider Zee, so I did not want to add to that.

Steve expressed remaining worries and I was addressing these since there is no reason for an electrical engineer to have detailed knowledge of beam theory.

The standard proceedure is for x to represent a moving section travelling from left to right and dividing the beam into two free bodies.

Only the left hand free body is considered. As a result the central load (sorry I called it W, not F) should not appear until the section passes its point of application - that is x > L/2. The Macaulay brackets apply a step function to implement this.

I would not use the formula method as given in the OP imply because it only works with W applied in the middle. You would need another formula for every possible point of application it is just simpler to solve the equation and anyway loadings are generally much more complicated.

As an aside I'm not sure what you mean by 'minimum deflection is a negative.'
The minimum deflection is zero and occurs at each support.

It is not a good idea to use differential calculus directly to find min and max deflections in beams, simply because the min or max may or may not occur at a point of inflection.

Last edited: Jun 24, 2012
20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
My inferiority complex is coming back to haunt me.

I recollect being told as an electrical engineering undergraduate to make my way to the mechanical engineering school and broaden my narrow education by taking a course in mechanical studies - no choice in the matter.

I also recollect sitting in mech engineering class with a very friendly teacher looking at us with some bemusement and saying "I thought you electrical engineers were meant to be good at maths ...". That's when it began - my feeling of inferiority in relation to mechanical engineers. Doubt had crept through the door of my troubled mind. Later I recollect watching a documentary on the construction of the Sydney Harbour (that's how we spell it here) Bridge. A kindly older gentleman was talking about how as a young engineer on the project, he became so skilled in the pen and paper mechanical engineering calculations he was promoted to an important supervisory role. Again my feelings of inferiority welled up.

I also had a friend studying physics at university at the same time. He was a brain and made me also feel inferior.

Then WBahn tells me most electrical engineers are lazy and sloppy - more inferiority.

Anyway ...

I note the OP's problem is a special case of a more general problem as one might observe in the attachment [see one of my earlier posts] - case 6 is a special case of case 7. This addresses studiot's comment in part, about needing another more general formula to cover single point loading at some arbitrary location along the beam.

It seems to me one could therefore use this more general relationship to determine the location and value of maximum deflection. It would also include the special case of the loading being exactly at the beam center.