Question 9.SP.1: Determine the moment of inertia of a triangle with respect t...
Determine the moment of inertia of a triangle with respect to its base.
STRATEGY: To find the moment of inertia with respect to the base, it is expedient to use a differential strip of area parallel to the base. Use the geometry of the situation to carry out the integration.
MODELING: Draw a triangle with a base b and height h, choosing the x axis to coincide with the base (Fig. 1). Choose a differential strip parallel to the x axis to be dA. Since all portions of the strip are at the same dis-tance from the x axis, you have
ANALYSIS: Using similar triangles, you have
\frac{l}{b}=\frac{h-y}{h} \quad l=b \frac{h-y}{h} \quad d A=b \frac{h-y}{h} d yIntegrating dI_{x} from y = 0 to y = h, you obtain
\begin{aligned}&I_{x}=\int y^{2} d A=\int_{0}^{h} y^{2} b \frac{h-y}{h} d y=\frac{b}{h} \int_{0}^{h}\left(h y^{2}-y^{3}\right) d y\\&=\frac{b}{h}\left[h \frac{y^{3}}{3}-\frac{y^{4}}{4}\right]_{0}^{h}\\ &I_{x}=\frac{b h^{3}}{12}\end{aligned}REFLECT and THINK: This problem also could have been solved using a differential strip perpendicular to the base by applying Eq. (9.2) to express the moment of inertia of this strip. However, because of the geometry of this triangle, you would need two integrals to complete the solution.
\begin{gathered}d A=b d y \quad d I_{x}=y^{2} b d y \\I_{x}=\int_{0}^{h} b y^{2} d y=\frac{1}{3} b h^{3}\end{gathered} (9.2)
