Thin walled Beams Question

[tigerstyle]

Homework Statement

To find the centroid and second moment of area of hollow thin walled trapezoidal.

Homework Equations

The Attempt at a Solution

 

[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?

 

[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

 

[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.

 

[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

 

[tigerstyle]

i have tried doing a solution and the centroid in y-axis is coming as 0.444m. i don't know if that is right or wrong. can you please check it.
thanks

 

[tigerstyle]

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

thanks

 

[rock.freak667]

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

What distance did you use for side AB and DE?

 

[tigerstyle]

What distance did you use for side AB and DE?

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

 

[rock.freak667]

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

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

 

[tigerstyle]

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

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?

 

[tigerstyle]

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

 

[rock.freak667]

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?

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.

 

[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 don't know what that means. can you help please

thanks

 

[tigerstyle]

and sketch the shear flow in the cross section. i am really struggling on this as i don't know how to do it. your help will be much appreciated.

thanks

 

[rock.freak667]

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 don't know what that means. can you help please

thanks

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

 

[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.

 

[tigerstyle]

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

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.

 

[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

 

[rushit_31]

how did u calculate the second moment of area??
Can you please help me?

 

[SteamKing]

how did u calculate the second moment of area??
Can you please help me?

What have you tried? Do you understand the calculation?

 

[rushit_31]

I am not understanding how to calculate the second moment of area for the inclined surfaces??
is it calculated using bh3^3/12 or by using integration...

 

[SteamKing]

You can calculate the moment of inertia of the inclined surface as if it were not inclined, and then transform the values using a process similar to the parallel axis theorem.

The following article describes the process (pages A-16 and A-17):

http://mbarkey.eng.ua.edu/courses/AEM250/CRAIG_007-021_APPC.pdf

 

[rushit_31]

how did u get the second moment of inertia for the DE and AB??

 

[SteamKing]

Did you read the attachment in Post #23? It contains the formulas you need to do the calculation. You take the length and thickness of the side plates and calculate Ix and Iy as if the plates were standing vertical. Then, using the angles from the cross-section diagram, you transform Ix and Iy for the sides into Ix' and Iy' to be used in the cross-section inertia calculation.