
#1
Dec2312, 07:39 AM

P: 24

how do calculate send moment of area.
here is the exampel. I do understand the way one calculate Ixx and Iyy and Ixy. the smaller part has thikness t and the biggest part 3t. I get : if you place global coordinate system on the top you will get the position of center of gravity to CG=(3a/8,a/4). Now I do: define variable s which is the road Ixx=∫y˛dA=∫t(s sin30)˛ds=t[(Sł/3)(1/4)]=tał/12 and we do the same with the right part. and we will get 3tał/12 Adding these two gives tał/3. but in the solution they have (tał/3)*(1/2)˛. I dont get where (1/2)˛ is comming from. This is not homework! This is an example from a exam! 



#2
Dec2312, 05:57 PM

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well, sin(30)=1/2 ... which may give us a clue.
From your results: the first part gives: $$\frac{ta^3}{12}=\frac{ta^3}{3}\frac{1}{2^2}$$  the second part gives: $$\frac{3ta^3}{12}=\frac{3ta^3}{3}\frac{1}{2^2}$$... adding them together gives: $$\frac{ta^3}{3}\frac{1}{2^2}+\frac{3ta^3}{3}\frac{1}{2^2}=\frac{4ta^3} {3}\frac{1}{2^2}$$... so the question is not so much where the (1/2)^{2} comes from  but where the factor of 4 went. I see you have two beams length a, one of thickness t and the other of thickness 3t, which meet at an angle which is not specified on the diagram. There is a "30" on the diagram which I take to mean that one of the beams makes and angle of 30 degrees to something but the "something" is not specified. From your analysis I'm guessing you are trying to find the polar moment J_{xx} by integrating over each beam and adding them  thus: you have oriented your coordinate axis so the yaxis is parallel to the bending force? if ##dA=t.ds## and ##y=s\sin(30)## then I guess the beams meet at 90 degrees to each other and the light (thickness 1t) beam forms an angle of 30 degrees to the horizontal. It this correct? I'd rather not guess! I think we'd need this information. 



#3
Dec2412, 03:38 AM

P: 24

Hi simon! thank you for answer. Yes the right answer according to solution is 4tał/12*(1/2˛). sorry if I wrote it wrong at the first place. my bad. I still don't get where (1/2) ˛ is comming from. I write it again how I do it: yes the angles is 30 on the both sides and is not 90 at the top No! yaxis is parallell to ##∫y˛.dA=∫t.(s \sin30)˛ds=[t.(sł/3 )sin30˛]=[tał/12]##. so (1/2) just went. ##∫y˛.dA=∫3t.(s \sin30)˛ds=[3(t.sł/3 )sin30˛]=[3tał/12]##. so (1/2) just went. Adding these two: tał/3 which is not correct . the answer is : ##(tał/3).(1/2˛)## 



#4
Dec2412, 08:15 PM

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Second moment of area
sin^{2}(30) = (1/2)^{2} like I said  look carefully at the derivation I showed you at the start of post #2 ... you have exactly the same result they have only you have multiplied out the known values... that's where the (1/2) "just went". i.e. $$\int y^2dA = \int (s\sin(30))^2tds = \int \frac{ts^2}{2^2}ds$$... keep the ##(1/2)^2## like that instead of evaluating it and see what happens.




#5
Dec2612, 04:44 AM

P: 24

$$ \frac{ts^3}{3}$$ while their is : $$ \frac{ts^3}{3}\frac{1}{4}$$ 



#6
Dec2612, 05:21 AM

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What you need to do is identify and report the correct solution and also go back over your integration showing the intermediate steps. You should also provide the information requested in post #2. 



#7
Dec2612, 01:20 PM

P: 24

they are all the same.
what i get from my integration is : ta^3 /3 and the slution in the exam says that the result should be: (4ta^3 / 12)(1/4) which is (ta^3/3)(1/4) which is (ta^3/3)(1/2^2) so all of the same the problemnis that just get ta^3/3which is 4 times bigger than that in the solution. 



#8
Dec2712, 05:59 PM

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If you do not provide consistent information I cannot help you.
If you do not answer questions I cannot help you. The three quoted values in post #6 are not all the same  please reread carefully: the last one has an s in it which is not present in the other two ... I included the first two because the form of the solution appears to be contributing to the confusion. In post #7 you are telling me something different for your answer as well. How did you get ##\frac{ta^3}{3}## ? In post #6 I have asked to see this working  do you not want to supply it? In post #2 I asked about the angle between the two beams  the working shown in post #1 suggests that both beams may be 30deg from the horizontal  is that correct? As drawn in post #1 there is actually not enough information to solve the problem. Please be more careful with your typing  typos in the math just add to the confusion. 



#9
Dec2812, 04:04 AM

P: 24

For simplyfing and save time I paste the question and answer.
http://i47.tinypic.com/19pe1k.jpg Solution: http://i50.tinypic.com/2ce6d61.jpg would you please tell and show me how they calculate the moment of inertia? thanks a lot 



#10
Dec2812, 06:00 PM

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OK  now I see.
The model answer is just using a rule  knowing the second moment for a beam and how to combine them. That way they could do it in one step. Simple addition, that you used, does not always hold. I think you should look at the examples in the wikipedia article on "second moment of inertia" to see what I mean then go through the calculation again without leaving out any steps. Meantime, I'll see if I can't produce a useful model answer for you. 



#11
Dec2912, 12:49 PM

P: 24

I will be very thankfull if you could explain it to me. Whatever I do I cant get where the (1/2)˛ is coming from. Will you do me some calculations? please? I have an exam soon and it is a real pain in the ***.




#12
Jan113, 09:55 PM

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Sorry for the delay  I've been somewhat busy myself.
Preparing for exams over xmas/newyear is a bummer! Anyhow  I had a go looking for the 1step approach used by the model answer and got nowhere. Someone with more recent experience should do better ... I can only conclude that the examiner was expecting people to use a remembered result from class. You should look through your notes although it is possible that the particular method was not used this year. It is also possible that the model answer or the question is somehow in error :) Anyway  the way to check would be to use the transformations. Divide the area into two rectangular areas. Let area 1 be the thin rectangle and area 2 be the fat one. Use the following observations: (iirc) For a rectangular area with extent a along the x axis and t along the y axis, the CG moments are: $$J_{xx}=\frac{at^3}{12}\; ; \; J_{yy}=\frac{a^3t}{12}\; ; \; J_{xy}=0$$ After a rotation angle ##\phi## in the xy plane: $$[J_{xx}]_{rot}=\frac{1}{2}(J_{xx}+J_{yy})+\frac{1}{2}(J_{xx}J_{yy})\cos(2\phi)J_{xy}\sin(2\phi)$$ ...etc. ... but all that is for an axis through the centroid. If the axis is a distance d from the centroid, then you use the parallel axis theorum: $$J_{xx}=[J_{xx}]_{CG}+Ad^2$$ ... for you, for ##J_{xx}##, ##A=4ta## and ##d=a/4## (if the xaxis passes through the apex of the triangle formed by the beams.) ... you'll have to modify for ##J_{yy}## but you already know the center of mass coordinates. There is going to be some approximation here  you'll have to use your judgement. I was working with the origin through the center of one end of each  which means there is an overlap of area ##3t^2/4## which I hope is small. You may want to use a different geometry. For the small area I get: $$J_{xx}=\frac{3}{4}\frac{at^3}{12}+\frac{ta^3}{12}$$ ... which seems suggestive. It's been a while since I had to do these though so check yourself. 



#13
Jan213, 01:30 PM

P: 24

Amazing. thank you very much Simon!!



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