# Crane, I-beam, max. deflection. HELP!

## Homework Statement

Ok. So, i have been giving the assignment of designing and testing an engine room crane for use on ship. I have designed a crane which has 2 i-beams which a trolley travels along. My questions are:-

-How do i find the second moment of area of the I-beams? The equation i am using is giving me negative answers EVERYtime.
-How do i convert the weight of the crane trolley (tonnes) into a force (Newtons).
-How do i calculate the max. deflection with a modulus of elasticity of 210G/nm squared.

I think i have the right equations for the last 2 but the answer i get for the first question is messing everything else up.

## Homework Equations

The equations i am using are:-

Ix = (bh)cubed - 2((b-tw)/2) x (h1)cubed / 12

WL = (W/No. of Beams) x (Lx - x squared)

Max Deflection = 1.25 WL ^4 / 38 x 4EI

## The Attempt at a Solution

Relevant equations can be found/confirmed using Google. I'll give you a headstart:
For second moment of area:
http://en.wikipedia.org/wiki/Second_moment_of_area
For units conversion:
http://www.onlineconversion.com/force.htm
The maximum deflection for a uniformly distributed load is 5wL^4/384EI. You are looking for the maximum deflection of a point load, which has a different formula than yours. I suggest you do a little research to find the right formula.

Also, once you have obtained the correct formulas, you could post your attempts at a solution and tell us what bothers you, if anything.

Finally, a little advice: if you intend to pursue your career in structural or mechanical analyses, you may be interested to know that the trade requires A LOT of care in numbers. Formulas that you obtain from any source must be verified and validated before using. Results must be validated using an independent method. You will understand that the failure or collapse of a structure such as a crane will have serious consequences such as loss of human life. Such calculations must be taken very seriously.

Its the formula on wikipedia thats giving me the problems.

The measurements are as follows h=300mm b=230mm h1= 220mm and tw=30mm.

Unfortunately when putting these numbers into the second moment of area equation i come up with -1.5009x10^-4. This has a knock on effect with the other equations.

As you can probably tell this is not my area of expertise at all as i am a marine engineer. This is part of a college project where the actual crane design is only 8% of the project while the other 92% is the planning and problem solving aspect.

> second moment of area equation i come up with -1.5009x10^-4.

Could you reproduce the formula you found from the Wiki article, which is a little long and has numerous formulae. After that, I'd suggest you provide your detailed calculations so that we can try to spot whether the formula is not correct, or its interpretation or calculations.

You have shown your data in mm, but it seems that you are working in metres. Please specify any units conversion you have applied, if any.

We're here to help, but you have to show us how you got your numbers.

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The second moment of area equation on the wiki article is:-

(b*h)^3 - 2(b-tw)/2 * h1^3 and then the whole equation divided by 12.

So when i placed the numbers in the equation was

(0.23*0.3)^3 -2 (0.23-0.03)/2 * 0.22^3 then the answer divided by 12

I have tried inputting the numbers in both mm and m with no success. There are also limitations on the changes i can make to the i beam as it is meant to be compatible with a piece of machinery.

Thanks for any help

> (0.23*0.3)^3 -2 (0.23-0.03)/2 * 0.22^3 then the answer divided by 12

In the article, the formula reads
(0.23*0.3^3 -2 (0.23-0.03)/2 * 0.22^3)/12

So that should solve your first part of the problem. It's OK to calculate in metres, as long as the rest of your calculations use the same units.

Post us the results of the second moment of area, and also the conversion of tons to newtons.
Note that only you can tell us what tonnes they are, namely short tons (2000 lbs), long tons (2240 lbs) or metric tons (1000 kg).

But that is my problem. The second moment of area equation i gave you in the last post gives me the answer of -1.5009x10^-4. Is this ok to be minus??

The load on the beams is 15 tonnes (metric) so that is 147112.5

nvn
Homework Helper
enginecadet88: That's not OK. The formula is I = [b*h^3 - (b-tw)*h1^3]/12. Try it again, and post your calculation. You can use mm, if you wish. Your beam load value is close enough. What value are you using for g?

Did you notice that I put in bold the correction required to your interpretation of the formula. In the original formula, there was no parentheses around b and h, which means that only h is raised to the third power. You have inserted parentheses around b and h, which makes the first term much smaller, hence the negative value.
No, second moment of area is never negative for a real object.

If you have read the whole article on the second moment of area, you will notice that for the second moment of area about the x-axis is simply the difference of two rectangles of size b and h, and (b-tw) and h1. Since the second moment of area of a rectangle is bh^3/12 (also shown in the article), therefore the second moment of area of an I-beam is simply bh^3/12 - (b-tw)h1^3/12.

Please post your new value of the second moment of area, and 15 tonnes (metric) in newtons.

before i try again i think i may have solved my problem.

In the first part of the equation i have been using (bh)^3. Is this supposed to be h^3 then multiplied by b?

Right on!

Ah. Now i might be onto the right track.

I have recalculated and the result for the second moment of area is now = 340033333.3

Does this not seem a little high? also, do you know the units?

The 15 tonnes (metric) in newtons is 147100.5 n.

nvn
Homework Helper
Excellent work, enginecadet88. Both answers are correct. The units are mm^4, and capital N, not lowercase n. The unit name is newton, with lowercase n, but the unit symbol is capital N. By the way, there should always be a space between the numeric value and its following unit symbol. E.g., 300 mm, not 300mm. See international standard for writing units; i.e., ISO 31-0. Your modulus of elasticity would be written 210 GPa, not 210G/nm squared.

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Thanks mathmate and nvn. Been struggling with this for a while.

I worked out all equations with a final deflection of 13.48mm. Due to regulations on cranes the max deflection is 1/750th of the total length so my max allowed was 21.313mm.

This is great and will surely help me on my project that has been in the making for 8 months now.

So once again thanks!

Great!
Now you're onto the last stretch.
From your description of the problem, it is not clear to me whether the beams are simply supported at the ends, or are the ends fixed to a wall or some other structure.
If it is the former case, the formula you are looking for is the maximum deflection of a simply support beam under a point load (the trolley weight W is distributed 50% to each beam). The length is L, and the distance of the load from one of the ends is x.
Where do you think the trolley should be to give the maximum deflection?

http://physics.uwstout.edu/statStr/statics/Beams/bdsnp412b.htm
http://www3.hi.is/~thorstur/teaching/cont/Continuum_CommonBeamFormulas.pdf

Post any time if you have difficulties finding the right formula.

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By backtracking your data, it seems that the span L is 16 m (=750*0.021313), force due to half trolley is 147100.5 N, and the maximum deflection occurs at mid-span where the load is also situated.
For a simply supported beam, the maximum deflection is
PL^3/48EI
=147100.5*16^3/(48*210*10^9*0.00034)
=0.176 m
=176 mm >>21.3 mm

You may want to double check data and/or formulae used.

The proposal layout stated that the main engine room dimensions hadnt been decided so it is not known where it will go.

So i've used the formula 5 WL^4/384EI.

Now i need to find the power required to move the trolley along the beam so i can choose a motor.

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The formula 5WL^4/384EI is for a uniformly distributed load (along the beam) such as the weight of the beam. In general a trolly of a crane is considered a point load that moves along the beam and acts at (almost) a single point. The maximum bending moment and deflection occurs in the middle when the crane is at mid-span, where the deflection is given by PL^3/48EI. (note that W is un N/m, and P is in N, and the powers of L are respectively 4 and 3).

nvn
Homework Helper
enginecadet88: It sounds like you have a bridge crane. It appears your I-beam cannot be longer than 5573 mm. What is the length (L) of your I-beam?

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My crane can't be classed as a uniformly distributed or a point load as the trolley is 4 m long itself. Therefore the weight is spread over 4 m of the total length at any time.

> What is the length (L) of your I-beam?

Formulae are available for a partial uniformly distributed load. However, since the trolley is likely on wheels, it is a closer approximation to model the load by a number of point loads, say 5 over 4 m. length, spaced at 1 m intervals.

<< solution deleted by berkeman >>

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berkeman
Mentor
> What is the length (L) of your I-beam?

Formulae are available for a partial uniformly distributed load. However, since the trolley is likely on wheels, it is a closer approximation to model the load by a number of point loads, say 5 over 4 m. length, spaced at 1 m intervals.

<< solution deleted by berkeman >>

Please do not do the OP's homework for them. It is against the PF Rules. Thank you.

Dear Berkeman,
Message well received. I am sorry that my offer of help went too far. As you can see by the history of the posts, my intention was to help him help himself. I should have noticed since the 3rd post that the intention of the question was to get the project over and done with.

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berkeman
Mentor
Dear Berkeman,
Message well received. I am sorry that my offer of help went too far. As you can see by the history of the posts, my intention was to help him help himself. I should have noticed since the 3rd post that the intention of the question was to get the project over and done with.

No worries. You're doing great. You are one of the many folks that help the PF be such a great resource. Thanks.