# Area moment of inertia of trapezium about y-axis

Good day!

As stated in the title, I need to obtain the area moment of inertia, also sometimes called the second moment of area of a trapezium as shown in the figure below, about the y-axis:

\begin{align} I_y&=\int x^2 dA \\ &=\int_a^b x^2y \ dx + \int_b^c x^2y \ dx +\int_c^d x^2y \ dx \\ &=\int_a^b x^2 \frac{(x-a)}{(b-a)}h \ dx + \int_b^c x^2 h \ dx +\int_c^d x^2\frac{(x-d)}{(c-d)}h \ dx \\ &=\frac{h(b-a)(3b^2+2ab+a^2)}{12} + \frac{h(c^3-b^3)}{3} + \frac{h(d-c)(d^2+2cd+3c^2)}{12} \\ &=\frac{h}{12}[(d^3+c^3)-(a^3+b^3)+ cd(c+d) -ab(a+b)] \tag{1} \\ \end{align}

I also referred to this wikipedia page which gives that the second moment of area about the y-axis for any simple polygon assumed to have $$n$$ vertices, numbered in counter-clockwise fashion can be calculated as: $$I_y = \frac{1}{12}\sum\limits_{i = 1}^n {(x_i \ y_{i+1} - x_{i+1} \ y_{i})({x_i}^2 + {x_{i+1}}^2} + x_i \ x_{i+1})$$ where $$\ x_{i}, \ y_{i} \$$ are the coordinates of the $$i-th$$ polygon vertex, for $$1\leq i\leq n$$. Also, $$\ x_{n+1}, \ y_{n+1} \$$ are assumed to be equal to the coordinates of the first vertex, i.e., $$\ x_{n+1}=x_{1} \$$ and $$\ y_{n+1}=y_{1} \$$. On calculating $$I_y$$ using above formula for trapezium $$ABCD$$ as in the figure, I got back the same answer as (1).

However a different result shows up here as:

$$I_y = h \frac {(a+b)(a^2 + 7b^2))} {48}$$

I am stuck as to what the actual output should be, kindly help. Thanks in advance.

• Something seems off about the last result you linked: For one, it only depends on the 'h', 'a' and 'b' that they have defined- while MoI should indeed depend on at least one other parameter. Your answer is correct afais. Commented Oct 1, 2022 at 7:28

You have calculated the moment for a general trapezoid with four parameters $$a,b,c,d$$ besides the height – and in two different ways, which should have convinced you enough that your working is correct. To reproduce the isosceles model in the link with only two parameters $$a',b'$$ and the height the following substitutions must be made: $$a=0,b=b'/2-a'/2,c=b'/2+a'/2,d=b'$$ Once this is done, simplifying gives the moment of this specialised trapezoid as $$\frac{h(a'+b')(7b'^2+a'^2)}{48}$$ matching the link.