Structural Resource

Reinforced Concrete Beam Design

Step-by-step reinforced concrete beam design to AS 3600 — flexure, shear, deflection, and detailing, with a worked example.

Table of contents

Reinforced concrete beam design combines flexural strength, shear strength, deflection control, and reinforcement detailing into a single member design, governed in Australia by AS 3600.

Design Steps

Determine design actions ($M^*$, $V^*$) from AS 1170 load combinations
Select trial section (width $b$, depth $D$, effective depth $d$)
Design flexural reinforcement ($A_{st}$)
Check ductility ($k_u \leq 0.36$)
Design shear reinforcement (stirrups)
Check deflection (serviceability)
Detail reinforcement (cover, spacing, laps, curtailment)

Flexural Design

For a singly reinforced rectangular beam:

$$ M_u = A_{st} f_{sy} \left(d - \frac{\gamma k_u d}{2}\right) $$ $$ k_u = \frac{A_{st} f_{sy}}{\gamma \alpha_2 f'_c b d} $$

$f'_c$ (MPa)	$\alpha_2$	$\gamma$
20–50	$1.0 - 0.003f'_c$ (min 0.67)	$1.05 - 0.007f'_c$ (min 0.67)

Design requirement: $\phi M_u \geq M^*$, with $\phi = 0.85$ for well-detailed ductile sections.

Minimum and Maximum Reinforcement

$$ A_{st,min} = \frac{0.20(D/d)^2 f'_{ct.f}}{f_{sy}} bd $$

Maximum reinforcement is governed indirectly by the ductility limit $k_u \leq 0.36$ rather than a fixed percentage.

Shear Design

$$ V_u = V_{uc} + V_{us} \leq V_{u,max} $$

Concrete contribution:

$$ V_{uc} = \beta_1 \beta_2 \beta_3 b_v d_o \left(\frac{A_{st} f'_c}{b_v d_o}\right)^{1/3} $$

Stirrup contribution:

$$ V_{us} = \frac{A_{sv} f_{sy.f} d_o}{s} $$

Minimum stirrups are required unless $V^* \leq 0.5\phi V_{uc}$.

Deflection Control

Two approaches are permitted:

Method	When Used
Span-to-depth ratio (deemed to comply)	Simple, uniformly loaded members within standard spans
Direct calculation (effective $I$, creep, shrinkage)	Long spans, heavy loads, or sensitive finishes

Worked Example

Problem: Design a simply supported rectangular beam, span 6 m, $b = 300$ mm, $D = 500$ mm, $d = 450$ mm, $f'_c = 32$ MPa, $f_{sy} = 500$ MPa, $M^* = 280$ kNm, $V^* = 190$ kN.

Step 1: Trial reinforcement:

Assume $\gamma = 0.85$, $\alpha_2 = 0.85$ for $f'_c = 32$ MPa.

Solving $M_u = A_{st} f_{sy}(d - \gamma k_u d/2) = M^*/\phi$ iteratively for $A_{st}$:

$M_u = 280/0.85 = 329$ kNm

Try $A_{st} = 1800$ mm² → $k_u = \frac{1800 \times 500}{0.85 \times 0.85 \times 32 \times 300 \times 450} = 0.34$ (< 0.36, OK)

$M_u = 1800 \times 500 \times (450 - 0.85 \times 0.34 \times 450 / 2) = 335$ kNm > 329 kNm ✓

Use 4-N24 bars ($A_{st} = 1800$ mm²).

Step 2: Shear check:

$$ V_{uc} = 1.0 \times 1.0 \times 1.0 \times 300 \times 450 \times \left(\frac{1800 \times 32}{300 \times 450}\right)^{1/3} \approx 120\ \text{kN} $$ $\phi V_{uc} = 0.7 \times 120 = 84$ kN < $V^* = 190$ kN → stirrups required. $$ V_{us} = \frac{190/0.7 - 120}{1} \Rightarrow \text{solve for } A_{sv}/s $$

Provide N12 stirrups at 150 mm centres (2-leg): check against calculated $V_{us}$ demand.

Detailing Checklist

Minimum cover per exposure classification (see AS 3600 Concrete Structures Design)
Bar spacing: minimum of 25 mm or bar diameter between bars
Development length and lap lengths per AS 3600 Section 13
Anchorage of bottom bars at simple supports

Practical Notes

Iterating $A_{st}$ against $k_u \leq 0.36$ converges quickly: start with an estimate from $A_{st} \approx M^*/(\phi f_{sy} \times 0.9d)$ and refine.
Shear often governs stirrup spacing near supports even when flexure is comfortably satisfied at midspan.
Deflection, not strength, frequently controls beam depth on long-span, lightly loaded floors.