Area moment of inertia of a triangle

[andmcg]

Homework Statement

What is the moment of inertia WRT to the x-axis (I_x) of a right triangle with base b and height h where the legs of the triangle coincide with the x and y axis?
It would be helpful to know both the single integral and double integral method.

Homework Equations

I^x=int(y^2 dA)
where y is the distance from the y-axis to the centroid of area element dA.

The Attempt at a Solution

I let dA=b dy and y=y so the integral looks like this:

int(y^2*b dy) integrating from 0 to h

I get (1/3)*h^3*b but the answer is (1/12)*h^3*b

I'm pretty sure I'm doing something wrong with dA but I'm not sure.

 

[rock.freak667]

Well consider a small rectangular element of width 'dx' and height 'y'. The moment of inertia of this element about its own center is dI_x=(1/12)y³dx.

So the total inertia of the triangle is

I_x= ∫ (1/12)y³ dx.


So now if the triangle starts at the origin and has base length 'b' and height 'h', can you find an expression for 'y'?

 

[Liquidxlax]

Well consider a small rectangular element of width 'dx' and height 'y'. The moment of inertia of this element about its own center is dI_x=(1/12)y³dx.

So the total inertia of the triangle is

I_x= ∫ (1/12)y³ dx.So now if the triangle starts at the origin and has base length 'b' and height 'h', can you find an expression for 'y'?

not sure if this would be the same but

I_x= ∫y²dM = ∫∫y²ρdydx

where ρ is density, so you would integrate over volume... well area in this case

y goes 0 to y=h-x

x goes 0 to b

 

[rock.freak667]

not sure if this would be the same but

I_x= ∫y²dM = ∫∫y²ρdydx

where ρ is density, so you would integrate over volume... well area in this case

y goes 0 to y=h-x

x goes 0 to b

Yes that would work as well but since your shape is in 2D and you are finding the area moment it would be like this

I_x = ∫∫_A y² dA = ∫∫ y² dydx.

You're correct in the x is from x=0 to x=b.

But for the limits for y - if it starts at y=0, it will end at y=(h/b)x.

I should amend by previous post as you wanted the area inertia about the x-axis, for the single integral, for a rectangular element of width 'dx' and height 'h' the moment of inertia about the x-axis is

dI_x= (1/3)y³ dx in which the limits are x=0 to x=b, giving the same answer as the double integral way.

 

[Liquidxlax]

Yes that would work as well but since your shape is in 2D and you are finding the area moment it would be like this

I_x = ∫∫_A y² dA = ∫∫ y² dydx.

You're correct in the x is from x=0 to x=b.

But for the limits for y - if it starts at y=0, it will end at y=(h/b)x.

I should amend by previous post as you wanted the area inertia about the x-axis, for the single integral, for a rectangular element of width 'dx' and height 'h' the moment of inertia about the x-axis is

dI_x= (1/3)y³ dx in which the limits are x=0 to x=b, giving the same answer as the double integral way.

my mistake, I was thinking that b=h for some stupid reason, it would be a centroid right? so the density would cancel out?

 

[rock.freak667]

my mistake, I was thinking that b=h for some stupid reason, it would be a centroid right? so the density would cancel out?

No you don't need the density as you already have the 'dA' in it.