What checks does ACI 318-19 require for a concrete beam?

Four primary checks: (1) flexural strength using the Whitney rectangular stress block (Ch. 22.2, Table 21.2.2), (2) shear strength from concrete and stirrup contributions (Table 22.5.5.1, 22.5.10.5.3), (3) crack control bar spacing (24.3.2), and (4) deflection under service loads using the Bischoff effective moment of inertia (Table 24.2.3.5) with long-term multiplier (24.2.4.1.1).

How do you calculate the effective moment of inertia for deflection?

ACI 318-19 Table 24.2.3.5 uses the Bischoff equation: Ie = Icr / [1 - ((2/3) Mcr / Ma)^2 (1 - Icr / Ig)], where Mcr is the cracking moment, Ma is the service-level moment, Ig is the gross moment of inertia, and Icr is the cracked moment of inertia. This applies when Ma exceeds (2/3) Mcr; otherwise Ie equals Ig.

What is the long-term deflection multiplier in ACI 318-19?

The multiplier lambda_delta = xi / (1 + 50 rho'), where xi depends on sustained-load duration (Table 24.2.4.1.3): 1.0 at 3 months, 1.2 at 6 months, 1.4 at 12 months, and 2.0 at 5 years or more. rho' is the compression reinforcement ratio. The total incremental deflection is lambda_delta times the immediate dead-load deflection plus the immediate live-load deflection.

How to design a concrete beam: ACI 318-19 worked example

This page walks through a complete rectangular beam design per ACI 318-19, covering all four strength and serviceability checks. Every intermediate value is pinned by the same regression tests that guard the live beam calculator. For the theory behind each individual check, see the beam methodology hub.

Problem statement

Design a simply-supported rectangular concrete beam for the following loads and geometry.

Section	12 in. wide × 24 in. deep, rectangular
Span	L = 20 ft, simply-supported
Concrete	f′c = 4,000 psi, normal-weight (λ = 1.0)
Steel	Grade 60 (f_y = f_yt = 60 ksi)
Longitudinal bars	3-#8 (A_b = 0.79 in²), 1 layer, uncoated
Stirrups	#3 @ 8 in. (2 legs, A_v = 0.22 in²)
Clear cover	1.5 in. (Table 20.6.1.3.1, not exposed to weather)
Factored loads	M_u = 150 k-ft, V_u = 30 kip
Service loads	w_D = 1.0 klf, w_L = 0.6 klf
Sustained load	5-year duration, no compression steel (ρ′ = 0)

Step 1: Effective depth

The effective depth d is measured from the extreme compression fiber to the centroid of the tension reinforcement. Cover is measured to the outside face of the stirrup (Table 20.6.1.3.1):

d = h - \text{cover} - d_{\text{stirrup}} - \tfrac{d_b}{2} = 24 - 1.5 - 0.375 - 0.500 = 21.625\ \text{in}

Step 2: Flexural strength (ACI Ch. 22.2, Table 21.2.2)

Compute the steel area and Whitney stress-block depth (see the flexure methodology):

A_s = 3 \times 0.79 = 2.37\ \text{in}^2

a = \frac{A_s \cdot f_y}{0.85\,f'_c\,b} = \frac{2.37 \times 60}{0.85 \times 4 \times 12} = 3.485\ \text{in}

Neutral axis depth and net tensile strain:

\beta_1 = 0.85 \quad \text{(Table 22.2.2.4.3, } f'_c = 4{,}000 \text{ psi)}

c = \frac{a}{\beta_1} = \frac{3.485}{0.85} = 4.100\ \text{in}

\varepsilon_t = 0.003 \times \frac{d - c}{c} = 0.003 \times \frac{21.625 - 4.100}{4.100} = 0.01282

Since ε_t = 0.01282 far exceeds 0.005, the section is tension-controlled and φ = 0.9 (Table 21.2.2).

\phi M_n = \phi\,A_s\,f_y\!\left(d - \tfrac{a}{2}\right) = 0.9 \times 2.37 \times 60 \times (21.625 - 1.743) = 2{,}544.5\ \text{k-in} = 212.0\ \text{k-ft}

\text{DCR} = \frac{M_u}{\phi M_n} = \frac{150}{212.0} = 0.707

Flexure: PASS (DCR = 0.707).

Step 3: Minimum reinforcement (ACI 9.6.1.2)

Minimum steel ensures the section does not fail in a brittle manner before the steel yields:

A_{s,\min} = \max\!\left(\frac{3\sqrt{f'_c}\,b\,d}{f_y},\; \frac{200\,b\,d}{f_y}\right) = \max(0.821,\; 0.865) = 0.865\ \text{in}^2

A_s = 2.37 in² > 0.865 in²: OK.

Step 4: Shear strength (ACI Table 22.5.5.1, 22.5.10.5.3)

Concrete contribution V_c from Table 22.5.5.1(a) (see the shear methodology):

V_c = 2\,\lambda\,\sqrt{f'_c}\,b\,d = \frac{2 \times 1.0 \times \sqrt{4{,}000} \times 12 \times 21.625}{1{,}000} = 32.82\ \text{kip}

Stirrup contribution V_s (§22.5.10.5.3):

V_s = \frac{A_v\,f_{yt}\,d}{s} = \frac{0.22 \times 60 \times 21.625}{8} = 35.68\ \text{kip}

\phi V_n = \phi\,(V_c + V_s) = 0.75 \times (32.82 + 35.68) = 51.38\ \text{kip}

\text{DCR} = \frac{V_u}{\phi V_n} = \frac{30}{51.38} = 0.584

Spacing checks:

s_max = min(d/2, 24) = min(10.81, 24) = 10.81 in.; s = 8 in. < 10.81: OK
V_s = 35.68 < 4√f′c·b·d/1000 = 65.6 kip, so no halving of s_max (Cl 9.7.6.2.2)
A_v/s = 0.22/8 = 0.0275 in²/in > A_v,min/s = 0.01 in²/in (Table 9.6.3.3): OK

Shear: PASS (DCR = 0.584).

Step 5: Crack control (ACI 24.3.2)

The maximum center-to-center bar spacing is limited to control flexural crack widths (see the crack control methodology). The steel stress is taken as (2/3)f_y per §24.3.2.1:

f_s = \tfrac{2}{3}\,f_y = \tfrac{2}{3} \times 60 = 40\ \text{ksi}

Clear cover c_c is measured from the tension face to the nearest surface of the bar closest to that face:

c_c = \text{cover} + d_{\text{stirrup}} = 1.5 + 0.375 = 1.875\ \text{in}

s_{\max} = \min\!\left(15\,\frac{40{,}000}{f_s} - 2.5\,c_c,\; 12\,\frac{40{,}000}{f_s}\right) = \min(10.31,\; 12) = 10.31\ \text{in}

Actual center-to-center spacing:

s_{\text{actual}} = \frac{b - 2\,c_c - d_b}{n_{\text{bars}} - 1} = \frac{12 - 2(1.875) - 1.0}{2} = 3.625\ \text{in}

s_actual = 3.625 in. < s_max = 10.31 in.: PASS.

Step 6: Deflection (ACI Table 9.3.1.1, Table 24.2.3.5, 24.2.4.1)

First check minimum depth (Table 9.3.1.1 for a simply-supported beam: L/16). Then compute deflections using the Bischoff effective moment of inertia (see the deflection methodology):

h_{\min} = \frac{L}{16} = \frac{20 \times 12}{16} = 15.0\ \text{in}

h = 24 in. > 15 in.: OK. Proceed to compute deflections for elements supporting non-structural elements likely to be damaged.

6a. Material and section properties

E_c = 57{,}000\,\sqrt{f'_c} = 57{,}000\,\sqrt{4{,}000} = 3{,}605\ \text{ksi}

f_r = 7.5\,\sqrt{f'_c} = \frac{7.5 \times \sqrt{4{,}000}}{1{,}000} = 0.4743\ \text{ksi}

I_g = \frac{b\,h^3}{12} = \frac{12 \times 24^3}{12} = 13{,}824\ \text{in}^4

M_{cr} = \frac{f_r\,I_g}{y_t} = \frac{0.4743 \times 13{,}824}{12} = 546.4\ \text{k-in} = 45.5\ \text{k-ft}

6b. Cracked moment of inertia

The modular ratio n and cracked neutral axis kd:

n = \frac{E_s}{E_c} = \frac{29{,}000}{3{,}605} = 8.044

\frac{b\,(kd)^2}{2} = n\,A_s\,(d - kd) \quad\Rightarrow\quad 6\,(kd)^2 + 19.064\,(kd) - 412.26 = 0

kd = 6.852\ \text{in}

I_{cr} = \frac{b\,(kd)^3}{3} + n\,A_s\,(d - kd)^2 = 1{,}287 + 4{,}161 = 5{,}448\ \text{in}^4

6c. Service moments

For a uniformly loaded simply-supported beam, the midspan moment is wL²/8:

M_{a,D} = \frac{w_D\,L^2}{8} = \frac{1.0 \times 20^2}{8} = 50.0\ \text{k-ft} = 600\ \text{k-in}

M_{a,D+L} = \frac{(w_D + w_L)\,L^2}{8} = \frac{1.6 \times 20^2}{8} = 80.0\ \text{k-ft} = 960\ \text{k-in}

6d. Effective moment of inertia (Bischoff, Table 24.2.3.5)

The Bischoff equation applies when M_a > (2/3)M_cr; otherwise I_e = I_g:

\tfrac{2}{3}\,M_{cr} = \tfrac{2}{3} \times 546.4 = 364.3\ \text{k-in}

M_a,D = 600 > 364.3 and M_a,D+L = 960 > 364.3, so Bischoff applies to both:

I_e = \frac{I_{cr}}{1 - \left(\dfrac{\frac{2}{3}\,M_{cr}}{M_a}\right)^{\!2}\!\left(1 - \dfrac{I_{cr}}{I_g}\right)}

I_{e,D} = \frac{5{,}448}{1 - (364.3/600)^2\,(1 - 5{,}448/13{,}824)} = 7{,}014\ \text{in}^4

I_{e,D+L} = \frac{5{,}448}{1 - (364.3/960)^2\,(1 - 5{,}448/13{,}824)} = 5{,}968\ \text{in}^4

6e. Immediate and long-term deflections

Immediate deflections for a simply-supported beam with uniform load (δ = 5wL⁴/384E_cI_e):

\delta_D = \frac{5\,w_D\,L^4}{384\,E_c\,I_{e,D}} = 0.142\ \text{in}

\delta_L = \frac{5\,w_L\,L^4}{384\,E_c\,I_{e,D+L}} = 0.100\ \text{in}

Long-term multiplier (ACI 24.2.4.1.1). With ξ = 2.0 for 5-year sustained load (Table 24.2.4.1.3) and no compression steel (ρ′ = 0):

\lambda_\Delta = \frac{\xi}{1 + 50\,\rho'} = \frac{2.0}{1 + 0} = 2.0

\delta_{LT} = \lambda_\Delta \cdot \delta_D + \delta_L = 2.0 \times 0.142 + 0.100 = 0.385\ \text{in}

6f. Deflection limits

Limit	Threshold (in.)	Actual (in.)	Status
L/180 (immediate live)	1.333	0.100	PASS
L/360 (immediate live, floors)	0.667	0.100	PASS
L/480 (long-term, damage-sensitive)	0.500	0.385	PASS
L/240 (long-term, general)	1.000	0.385	PASS

Summary

All checks pass. The 12 × 24 in. beam with 3-#8 bars and #3 stirrups @ 8 in. is adequate for M_u = 150 k-ft, V_u = 30 kip on a 20 ft simply-supported span.

Check	ACI reference	DCR / Value	Status
Flexure	Ch. 22.2, Table 21.2.2	0.707	PASS
Min. reinforcement	9.6.1.2	2.37 > 0.865	OK
Shear	Table 22.5.5.1	0.584	PASS
Stirrup spacing	Cl 9.7.6.2.2	8 < 10.81	OK
Crack control	24.3.2	3.625 < 10.31	PASS
Deflection (L/480)	Table 24.2.3.5	0.385 < 0.500	PASS