Question

Problem example

Answer 1

Surry99 I'd like to discuss this example problem we did in class together when you have time. I have not got around to doing a similar problem in the homework assignment since I am trying to figure out another problem that I might have questions later if I am unable to figure out today. Thanks

Answer 2

ok

Answer 3

So here is the problem I want to discuss a little. I have a problem similar I will be working on a in a few. Once I get home I will scan and attach the problem. Are we allowed to put x' and y' wherever we want to on the diagram?

Answer 4

So basically I have a better understand of what is going on now. Now that I feel a little comfortable with the formulas and have a better idea of what is going on I will use your method instead. Can you please check to see why my Ix isn't 1.874×10^6mm4

Answer 5

Yes, I've been waiting!

Answer 6

ok , where do you want to start?

Answer 7

So did I approach the problem appropriately? Look at my latest attachments please

Answer 8

img 174 and img 173?

Answer 9

yeah. And my x bar and y bar are fine. I checked the answers.

Answer 10

But now I am trying to find Ix, the inertia

Answer 11

ok let me take a look

Answer 12

So I compared my answer to the solutions manual. The solutions manual doesn't cut the isosceles triangle in half like I did. I don't think that would make a difference of me cutting it in half.

Answer 13

Hey, if I don't reply soon it's because I am writing a report for another class

Answer 14

ok

Answer 15

your mistake is on page 2...
you have Ix' = Ix1 + Ix2 - Ix3...
Moments of inertia can never be negative values...
it should have been...
Ix' = Ix3-Ix1-Ix2...try that and you should get the correct answer.

Answer 16

why Ix3 first then Ix1 sec? Does order matter?

Answer 17

Yes it absolutely does...if you look at the section based on the basic shapes of rectangesl and squares, the way the shape is created is by:
1) creating the large rectangle first (I3)
2) then you cut away the two smaller triangles (I1 and I2)
As a result, your equation needs to reflect this...I3 -I1-I2

This is just like the first problem we did yesterday.

Answer 18

I will be back in an hour or so....post any further questions

Answer 19

oh i see because the triangles are subtracting from the total area of the rectangle?

Answer 20

yes...that is absolutely critical to see to get these problems correct.

Answer 21

okay thanks. i will try to get started on the next problem and if not able to figure out post questions here. thanks

Answer 22

your welcome

Answer 23

problem. I got 1.91x10^6 answer is 1.874x10^6

Answer 24

i got using your numbers 1.874E06...
I3 -I1-I2 = 1944000- 2* 34992 =

Answer 25

can you just add the two triangles togther then subtact from the rectangle?

Answer 26

yeah i noticed you multiplied the triangle by 2 since they are identical. But I am trying to understand the math portion of this rather than taking short cuts. If you ad the two negative triangles together you obtain a negative answer correct? Then you subtract from the rectangle

Answer 27

i don't get 34992 i get 31104... hmmm

Answer 28

2(1/3*72*18^3+1/2*72*18*6)

Answer 29

look carefully at the first and second term where you where you were trying to calculate distance squared..
(1/3(18))^2 is what you have for Ix1...but for Ix2 you have (1/3(18)^2)..
now the first way you expressed it is correct...the second is not...my guess is the error is in there
1/36(72)(18)^3 + 1/2(72)(18) ((1/3(18))^2 = 34992

Answer 30

I see!

Answer 31

for y what formula do i use? Can I use Iy=1/3*b*h^3

Answer 32

never mind I got it figured out

Answer 33

@surry99

Answer 34

I am here...

Answer 35

I am trying to find the problem I need help on. I thought the question was on here but it isnt

Answer 36

hold on a sec

Answer 37

so the dimension for the diagram are given.

Answer 38

let me give you the properties for the C and W

Answer 39

So I am looking for Ix so I have Ix bar=Ixw+2(Ixc)

Why is Ixc bar 1.31in^4

So I would have 2(Ix+Ad^2) Why would the Ix=1.31 and not 32.5?

Answer 40

let me have a look

Answer 41

and then for Iy bar=Iyw+2(Iyc+Ad^2) Iyw=37.1in^4

Answer 42

and based on the given diagram are the axes x' and y' or are those the original axes? I am kind of confused...

Answer 43

so your first question is:
for the C 8 x 11.5 section why is Iy bar = 1.31 in^4?

Answer 44

no why is Ixc=1.31in^4

Answer 45

here look at the solution

Answer 46

look at Ixchan they use 1.31 instead of the 32.5, not sure why

Answer 47

I am missing the column headings at the top of the second page...can you rescan please

Answer 48

thanks...you must look very carefully at the coordinate system on the sections vs the coordinate system on the composite sections

Answer 49

the composite section is on the mini diagrams I just sent you?

Answer 50

Ix=1.31in^4 is the distance between the x axis and x' axis?

Answer 51

i mean Iy

Answer 52

Got the composite image...
For the C8 x 11.5 relative to its own coordinate system...Ix bar = 32.5 , Iy bar = 1.31
Agreed?

Answer 53

okay

Answer 54

Great...but when you place in in the composite section, it is being rotated...agreed?

Answer 55

when you say composite section your talking about the image given in the problem?

Answer 56

yes, that is called a composite (made of of several sections)

Answer 57

so yes I agree it rotates

Answer 58

and that is the reason why we use the Iy because the coordinates flip

Answer 59

Yes...

Answer 60

so basically x becomes y and y becomes x

Answer 61

yes

Answer 62

okay i thought that was the case but I needed more explanation as to why i had that thought, thanks

Answer 63

welcome

Answer 64

My attempt. I have three equations with Tcf and Tde can I use two of those equations and solve for either tension? I tried using two of those equations and didn't obtain correct answer. Does my lengths have to be in terms of m instead of mm? On the bottom of the page I was going to take moments about D, but if I do that then I will obtain two more unknowns, therefore I would have to find two more equations?

Answer 65

Hint: For a body in 3D equilibrium, you have a total of 6 equations available.
Forces in x, Forces in y, Forces in z, Moments about x, Moments about y and Moments about z
The cable tensions are each one unknown but you have four unknowns at A
Ax, Ay, Mxa, Mxz.
So go back to your FBD and make sure you have the four unknowns at A.
Then apply your 6 equations of equilibrium.

Answer 66

I dont have a reaction in the y direction, only z and x have a reacton.

Answer 67

four unknowns at A? there is only 2 unkowns at A.

Answer 68

what do you mean by Mxa and Mxz? Moments about x and z?

Answer 69

I have four unknowns. Two reactions at A and two unknown tensions

Answer 70

Sorry for the typo...the four unknown reactions at A:
1) Ax, Az, Mxa, Mza....(think about why these two moments must be there)
2) Then you have the unknowns in each cable, Tde and Tcf
3) Those are the six unknowns you must solve for by applying the six equations of equilibrium.

Answer 71

These moments are caused by the tensions?