Consider this triangle with its centroid at $C$:

Text-book triangle

Then this is how I believe we can calculate the Second Moment of Area along the $x_{C}$ and $y_{C}$ axis:

$$\begin{align} I_{xc} &= \frac{bh^3}{36} \\ I_{yc} &= \frac{hb(b^2-st)}{36} \end{align}$$

So far so good.

But then we have this triangle:

Slanted triangle


  • Is there any similar way of calculating the second moment of area for a slanted triangle, like the one above?
  • Will even the same equations work?
  • How is $s$ and $t$ derived in that case? Can they be negative?

Please assume we can NOT rebase the triangle — we must use $LR$ as the base.

Source for equations


3 Answers 3


Oscar, I'm not sure what your math background is, but there are a lot of different ways to get there. The most direct is to just reason from the properties of affine transforms, specifically shear mapping.

During a horizontal shear, as in from triangle 1 to triangle 2, the distances moved by different points are proportional to their y coordinate. This tells us, in effect, that if your Iyc equation is a general solution for acute triangles, it must also be a general solution for obtuse triangles, since they are generated from the same transform. The only difference is the magnitude of the transform. So you just have to be consistent with how you measure s and t. s starts at L and goes to T, and t starts at T and goes to R, measuring horizontally. This is easy to verify by comparing two triangles that are almost right angles (tiny t value), one of each type.

  • $\begingroup$ I'd say my math experience is optimistic hobbyist, so I can't say I understand why the equation must work—but I'm buying this explanation and reasoning : ) Thanks! $\endgroup$
    – 0scar
    Commented Jan 4, 2018 at 16:00

The short answer is that the same equations are applicable in both cases for $I_{xc}$. For $I_{yc}$, the equation you have is applicable only with a sign change if you assume that $t$ is positive in both cases. However it is possible to have an equation in $b$ and $s$ that is valid in both cases.

A long math-based answer. I am going to call the vertex of right angled vertex 0. In both cases, the final triangle TLR is obtained from triangles TOL and TOR.

The area, centroid, and second moment of areas about the centroid for each triangle:

\begin{array}{cc} & area & centroid & moment \\ TOL & \frac{h s}{2} &\left\{\frac{2 s}{3},\frac{h}{3}\right\} & \left\{\frac{h^3 s}{36},\frac{h s^3}{36}\right\} \\ TOR & \frac{h t}{2} & \left\{\frac{2 s+b}{3},\frac{h}{3}\right\} & \left\{\frac{h^3 t}{36},\frac{h t^3}{36}\right\} \\ \end{array}

I will now use the parallel axis theorem to compute the moment about the centroid of TLR which is at $\left\{\frac{b+s}{3},\frac{h}{3}\right\}$. (The formula is $I=I_c+Ad^2$)

The new moments of TOL are

$$\left\{\frac{h^3 s}{36},\frac{h s^3}{36}\right\}+\frac{1}{2} h s \left\{\left(\frac{h}{3}-\frac{h}{3}\right)^2,\left(\frac{b+s}{3}-\frac{2 s}{3}\right)^2\right\}$$

which simplifies to $$ \left\{I_{XTOL}, I_{YTOL}\right\} = \left\{\frac{h^3 s}{36},\frac{1}{36} h s \left(2 b^2-4 b s+3 s^2\right)\right\}$$

Similarly the new moments of TOR are

$$ \left\{\frac{h^3 t}{36},\frac{h t^3}{36}\right\}+\frac{1}{2} h t \left\{\left(\frac{h}{3}-\frac{h}{3}\right)^2,\left(\frac{1}{3} (b+2 s)-\frac{b+s}{3}\right)^2\right\} $$

and that simplifies to

$$ \left\{I_{XTOR}, I_{YTOR}\right\} = \left\{\frac{h^3 t}{36},\frac{1}{36} h t \left(2 s^2+t^2\right)\right\}$$

$I_{xc}$ for TLR

Case 1 ($t+s=b$): $$I_{xc} = I_{XTOL}+I_{XTOR}= \frac{h^3 s}{36}+\frac{h^3 t}{36}= \frac{h^3 b}{36}$$

Case 2 ($s-t=b$): $$I_{xc} = I_{XTOL}-I_{XTOR}= \frac{h^3 s}{36}-\frac{h^3 t}{36}= \frac{h^3 b}{36}$$

$I_{yc}$ for TLR

Case 1 ($t=b-s$): $$I_{yc} = I_{YTOL}+I_{YTOR}= \frac{1}{36} b h \left(b^2-b s+s^2\right)$$

Case 2 ($t=s-b$): $$I_{yc} = I_{YTOL}-I_{YTOR}= \frac{1}{36} b h \left(b^2-b s+s^2\right)$$


For $I_{xc}$ you can use $\frac{h^3 b}{36}$ for both cases.

For $I_{yc}$ you can use $ \frac{1}{36} b h \left(b^2-b s+s^2\right)$ for both cases.

If you want to use the expression you have for case 1, then for case 2 you need to use $\frac{1}{36} b h \left(b^2+st\right)$ if you assume $t$ is positive, or you can use the same expression for both cases but must have the proper sign for $t$.


I used Mathematica for the rather tedious computations.

enter image description here


Slanted triangle can be obtained by cutting off piece of right triangle, and you can calculate moment ot inertia or second moment of area by subtracting smaller right triangle from the larger one using Steiner's rule. You can express all dimensions considering your base and edges using trigonometric functions.


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