Homework Help: Thin walled Beams Question

1. Aug 4, 2012

tigerstyle

1. The problem statement, all variables and given/known data
To find the centroid and second moment of area of hollow thin walled trapezoidal.

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Aug 5, 2012

pongo38

You haven't shown your attempt at a solution. In this sort of question I find it helpful to draw up a table showing ( for the four members) area, distance of centroid from a reference axis. moment of area, etc. Can you try something like that?

3. Aug 6, 2012

tigerstyle

in the question it says find the centroid and second moment of area about the horizontal axis at the centroid. i am confused in the distance of centroid from reference axis for each rectangle, when doing the table. can you please guide me how to do this question. i am using the centroid formula : c=first moment of area/ total area of the section

4. Aug 6, 2012

rock.freak667

The centroid of a rectangle would be at its geometric centre. So if the length is a, the width is b, the centroid would be at a length of a/2 and a width of b/2.

Draw the centroids of each member in separate diagrams. If your reference axis is side BD, then where you drew your centroids, calculate the distance of that centroid to the axis.

Take y as vertical and x as horizontal.

For the rectangle BD, the 'y' distance from the reference axis along BD is 4 mm (8/2)

Do the same for each member.

5. Aug 7, 2012

tigerstyle

but what about the side DE and AB because they are at an angle 120 degree. or do i use sin rule at the centroid of DE to find the height?

Thanks

6. Aug 7, 2012

tigerstyle

i have tried doing a solution and the centroid in y axis is coming as 0.444m. i dont know if that is right or wrong. can you please check it.
thanks

7. Aug 7, 2012

tigerstyle

and for the second moment of area i have got 15x10^-3 m^4. can anyone check if thats right please

thanks

8. Aug 7, 2012

rock.freak667

What distance did you use for side AB and DE?

9. Aug 7, 2012

tigerstyle

I used sin rule to calculate it and it came as 0.866m.

10. Aug 7, 2012

rock.freak667

0.866 m would be the height of triangle formed, you didn't account for the centroid being halfway down the rectangle.

11. Aug 7, 2012

tigerstyle

what i did for rectangle DE and AB: the midpoint will be 0.5m. and then drew a triangle and used sin rule to find the height of centroid vertically. is that right?

12. Aug 7, 2012

tigerstyle

i mean to find the y distance from the base to the midpoint of the rectangle AB and DE

13. Aug 7, 2012

rock.freak667

Yes that is right. But when you consider where the centroid is, the hypotenuse is 0.5 mm, so the vertical distance is 0.5sin(60) mm.

14. Aug 8, 2012

tigerstyle

I have got the centroid and the second moment for this problem. and have found the shear flow in the walls. now the question is asking me to evaluate the shear force in the walls. i dont know what that means. can you help please

thanks

15. Aug 8, 2012

tigerstyle

and sketch the shear flow in the cross section. i am really struggling on this as i dont know how to do it. your help will be much appreciated.

thanks

16. Aug 8, 2012

rock.freak667

How did you calculate the shear flow? Shear flow is related to the shear force, first moment of area and second moment of area.

17. Aug 8, 2012

tigerstyle

i used da formula q=q0 - (Vy/Ixx) Integral t*y*ds and calculated it for qOA, qAB and qBC. qBC = 0 as it lies on the axis of symmetry.

18. Aug 8, 2012

tigerstyle

i used da formula q=q0 - (Vy/Ixx) Integral t*y*ds and calculated it for qOA, qAB and qBC. qBC = 0 as it lies on the axis of symmetry.

19. Aug 8, 2012

tigerstyle

where Vy=Vertical shear force given in the diagram and Ixx =second moment of area calculated as 101.25 x 10^8 mm4, t=thickness and y is the distance to the centroid

20. Nov 14, 2013

rushit_31

how did u calculate the second moment of area??