Second moment of area
The second moment of area measures how a section's material is spread about a bending axis. The further the material sits from the axis, the stiffer the beam. For a solid rectangle about its centroid:
I =b·h³12
- I
- second moment of area
- m⁴
- b
- width, parallel to the axis
- m
- h
- depth, across the axis
- m
- A
- area
- m²
- Z
- elastic section modulus
- m³
- y
- distance from the axis
- m
- d
- diameter, or inner diameter
- m
- D
- outer diameter
- m
- s
- distance between parallel axes
- m
| A = bh | 7,500 mm² |
| Z = bh²/6 | 1.88 × 10⁵ mm³ |
| Iy = hb³/12 | 1.56 × 10⁶ mm⁴ |
| Ix = bh³/12 | 1.41 × 10⁷ mm⁴ |
Worked example
Uses the values in Fig. 2.5-1. Change them and the working updates.
Given
- bwidth50 mm
- hdepth150 mm
Find
The second moment of area about the horizontal centroidal axis, Ix.
Solution
Ix = bh³/12
= 50 mm × (150 mm)³ / 12
= 1.406 × 10⁷ mm⁴
= 1.406 × 10⁻⁵ m⁴
Answer
Ix = 1.41 × 10⁷ mm⁴
About the vertical axis, hb³/12 = 1.56 × 10⁶ mm⁴.
Other sections
About a centroidal axis unless stated.
-
Solid circle
I = πd⁴64
-
Tube
I = π(D⁴ − d⁴)64
-
Hollow rectangle
I = BH³ − bh³12
-
Parallel axis, distance s away
I = Ic + As²
Assumptions
- I is taken about an axis through the centroid, for a section of one material.
- For a rectangle the centroidal axes are also the principal axes, so bending about x–x causes no sideways bending.
- Dimensions are those of the real section. Rounded corners and holes reduce I.
Derivation
By definition I = ∫ y² dA, where y is the distance of each element of area dA from the axis.
- For a rectangle, take strips of width b and thickness dy:
dA = b dy. I = ∫ by² dy, taken fromy = −h/2toy = h/2.I = b[y³/3]between those limits, which isb(h³/24 + h³/24) = bh³/12.
Depth enters cubed and width only once. That is why joists stand on edge: doubling h makes a beam eight times stiffer, while doubling b only doubles it.