Rectangular RC Nominal Moment

ϕ M_{n} = ϕ A_{s} f_{y} (d - \frac{a}{2}) ϕ = 0.90 (tension-controlled)

Lever arm between T and Cc is (d − a/2). For tension-controlled sections (εt ≥ 0.005) φ = 0.90. Check ρ ≤ ρmax so the section stays tension-controlled.

$M_{n}$ (N·m): Nominal moment
$d$ (mm): Effective depth
$a$ (mm): Whitney depth
$ϕ$: Strength reduction

Step-by-step solved example

As fy = 180 k, d = 16 in, a = 4.41 in. φMn?

φMn = φ As fy (d − a/2). Lever arm jd = d − a/2.

1. Arm
d − a/2 = 16 − 2.205 = 13.80 in.
2. Mn
Mn = 180 × 13.80 = 2484 k·in = 207 k·ft; φMn = 0.90×207 = 186 k·ft.
$ϕ M_{n} = 186 k\cdotft$

Answer: 186 k·ft

10 practice questions

0/10 correct

1.Tension-controlled φ for flexure is

2.Compression-controlled φ (flexure/axial) is

3.jd ≈ d − a/2 is the

4.εt = 0.005 is the

5.d is measured to

6.If a decreases, Mn

7.Mu must be

8.ρ = As/(bd). Minimum ρ exists to

9.Cover and bar diameter affect

10.Grade 60 bars have fy =