86/3848–12FE Reference Handbook 10.6 · Statics · p. 99 right-triangle rows: Ixc = bh³/36, Ix = bh³/12 about the base
1. Moment & Equilibrium2. Centroid of a Composite Area3. Coulomb Friction4. Force Resultant5. Rectangle Centroidal Moment of Inertia6. Parallel-Axis Theorem7. Distributed Load Resultant8. Method of Joints9. Two-Force Member10. Ideal Pulley11. Moment of a Force12. Simply Supported Beam Reactions13. Load–Shear–Moment Relations14. Method of Sections (Truss)15. Sagging Cable (Small Sag)16. Overhanging-Beam Reactions17. Radius of Gyration18. Triangle Area Moment of Inertia19. Triangular Distributed Load Resultant20. Three-Force Member21. Couple Moment22. Frames: Multi-Force Members23. Zero-Force Truss Members

Handbook formula

Triangle Area Moment of Inertia

Both right-triangle rows on p. 99 give Ixc = bh³/36 about the centroidal axis parallel to the base and Ix = bh³/12 about that base. The height centroid is h/3 from the base on both rows. The x-centroid is 2b/3 or b/3 depending on which row matches the right angle — do not copy xc from the wrong figure. Rectangle centroidal inertia is also bh³/12; that match with the triangle’s base-axis formula is a coincidence.

Base
Height

Step-by-step solved example

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

b = 6 inh = 9 inh/3Ī = b h³ / 36 = 6×729 / 36Ī = 121.5 in⁴Replay
Triangle: centroid at h/3 from the base; Ī = bh³/36.

  1. 1. Formula

    Ī=bh³/36=6×729/36=121.5 in⁴.

  2. 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