Triangle Area Moment of Inertia

I_{base} = \frac{b h ^{3}}{12}, I ˉ_{centroid} = \frac{b h ^{3}}{36}

Handbook: about the base, bh³/12; about a centroidal axis parallel to the base, bh³/36. Centroid sits at h/3 from the base (h/3 from the apex is wrong — it is 2h/3 from the apex).

$b$: Base
$h$: Height

Step-by-step solved example

Triangle b=6 in, h=9 in. Centroidal I parallel to the base?

Triangle: centroid at h/3 from the base; Ī = bh³/36.

1. Formula
Ī=bh³/36=6×729/36=121.5 in⁴.
$I ˉ = 121.5 in^{4}$
2. Check via PAT
I_base=bh³/12=364.5, d=h/3=3, Ad²=(27)(9)=243, 364.5−243=121.5.

Answer: 121.5 in⁴

10 practice questions

0/10 correct

1.Centroid of a triangle is

2.I_base / Ī_centroid =

3.Area of the triangle is

4.For b=h=6, Ī =

5.Parallel-axis from centroid to base uses d=

6.A triangular distributed load on a beam is NOT the same as

7.Right triangle about its vertical leg (height h) uses the same formulas with

8.Composite T: subtract a triangle by using

9.bh³/12 is also I of a rectangle about

10.If h doubles, Ī_centroid scales by