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.
The diagram at the link is suspiciously simplified. It is in fact an isosceles trapezoid.
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.