RC beam design

Worked example: design a simply supported reinforced concrete beam for ULS flexure and shear, with hand calculation verification.

Problem statement

A simply supported reinforced concrete beam spans 6.0 m and carries a uniformly distributed load. The beam is an interior beam in an office building, supporting a 200 mm thick slab. The exposure classification is B1 (interior, dry environment).

Design the beam for ultimate flexure and shear to AS 3600:2018.

Given data

Parameter	Value	Units	Source
Span	6000	mm	Given
Beam width ( $b$ )	300	mm	Assumed
Beam depth ( $d_{total}$ )	600	mm	Assumed
Concrete grade	N40	—	AS 3600
$f'_c$	40	MPa	N40 grade
Rebar grade	D500N	—	AS/NZS 4671
$f_y$	500	MPa	D500N
Cover	40	mm	AS 3600 Table 4.10.3.2, exposure B1
Dead load ( $G$ )	25	kN/m	Includes self-weight + slab
Live load ( $Q$ )	15	kN/m	Office loading
ULS combination	$1.2G + 1.5Q$	—	AS/NZS 1170.0
Stirrup size	N10	—	Assumed
Stirrup legs	2	—	Standard
Stirrup spacing	200	mm	Assumed

Design actions

w^* = 1.2 \times 25 + 1.5 \times 15 = 52.5 \text{ kN/m}

M^* = \frac{w^* L^2}{8} = \frac{52.5 \times 6.0^2}{8} = 236.3 \text{ kN.m}

V^* = \frac{w^* L}{2} = \frac{52.5 \times 6.0}{2} = 157.5 \text{ kN}

Step-by-step solution

Step 1: Define section geometry

In ACS, create a new section and apply the Rectangular template with $b = 300$ mm, $d = 600$ mm.

Step 2: Set materials

Design code: AS 3600
Concrete grade: N40 ( $f'_c = 40$ MPa)
Rebar grade: D500N ( $f_y = 500$ MPa)
Cover: 40 mm all sides (manual mode)
Stress model: Rectangular (default)

Step 3: Place reinforcement

The effective depth for bottom reinforcement is:

d = 600 - 40 - 10 - 20/2 = 540 \text{ mm}

(cover + stirrup diameter + half bar diameter)

Use the Edge Pattern tool on the bottom edge:

Bar diameter: 20 mm (N20)
Number of bars: 4
Offset: automatically computed from cover settings

This gives $A_{st} = 4 \times 314 = 1257$ mm $^2$ .

Add 2 $\times$ N16 compression bars along the top edge for crack control and to support the stirrup cage.

Step 4: Configure stirrups

Diameter: 10 mm (N10, $A_{bar} = 78.5$ mm $^2$ )
Legs: 2
Spacing: 200 mm

$A_{sv} = 2 \times 78.5 = 157$ mm $^2$

Step 5: Enter design loads

In the Applied Loads panel:

Member type: Beam
ULS 1: $M^*_x = 236.3$ kN.m, $V^* = 157.5$ kN

Step 6: Review results

Flexure

ACS computes:

Result	Value	Units
$M_u$	~310	kN.m
$\phi M_u$	~264	kN.m
Utilisation	~0.90	—
$k_u$	~0.19	—
Ductility limit ( $k_u \leq 0.36$ )	Pass	—

The beam is adequate for flexure with approximately 10% reserve capacity.

Shear

Result	Value	Units
$V_{uc}$	~105	kN
$V_{us}$	~212	kN
$V_u$	~317	kN
$\phi V_u$	~238	kN
Utilisation	~0.66	—

Shear is adequate with significant reserve.

Results summary

Check	Demand	Capacity	Utilisation	Status
Flexure ( $M^*_x$ )	236.3 kN.m	~264 kN.m	~0.90	Pass
Shear ( $V^*$ )	157.5 kN	~238 kN	~0.66	Pass
Ductility ( $k_u$ )	0.19	$\leq 0.36$	—	Pass

Discussion

The design is governed by flexure with a utilisation ratio of approximately 0.90. The section has adequate ductility ( $k_u = 0.19$ , well below the limit of 0.36), confirming tension-controlled failure.

Shear capacity has more reserve (utilisation ~0.66). The stirrup spacing of 200 mm could potentially be increased away from the supports, but 200 mm is a practical minimum that simplifies construction.

If the flexural utilisation were too high, you could:

Increase the beam depth (most effective — moment capacity scales approximately with $d^2$ )
Add more tension reinforcement (effective up to the ductility limit)
Increase the concrete grade (moderate effect for flexure)

Hand calculation verification

Verify the flexural capacity using the rectangular stress block method per AS 3600 Cl. 8.1:

Stress block parameters for $f'_c = 40$ MPa:

\alpha_2 = 1.0 - 0.003 \times 40 = 0.88

\gamma = 1.05 - 0.007 \times 40 = 0.77

Neutral axis depth from force equilibrium ( $C_c = T_s$ ):

\alpha_2 f'_c \gamma c \cdot b = A_{st} f_y

0.88 \times 40 \times 0.77 \times c \times 300 = 1257 \times 500

c = \frac{628{,}500}{8{,}131.2} = 77.3 \text{ mm}

Check ductility:

k_u = \frac{c}{d} = \frac{77.3}{540} = 0.143

This is well below 0.36 — pass. (Note: ACS reports $k_u \approx 0.19$ , which accounts for the compression steel contribution to the equilibrium.)

Moment capacity:

M_u = A_{st} f_y \left(d - \frac{\gamma c}{2}\right) = 1257 \times 500 \times \left(540 - \frac{0.77 \times 77.3}{2}\right)

M_u = 628{,}500 \times (540 - 29.8) = 628{,}500 \times 510.2 = 320.7 \text{ kN.m}

Design capacity ( $\phi = 0.85$ for tension-controlled):

\phi M_u = 0.85 \times 320.7 = 272.6 \text{ kN.m}

Utilisation:

\frac{M^*}{\phi M_u} = \frac{236.3}{272.6} = 0.87

The hand calculation gives a utilisation of 0.87, which is close to the ACS result of ~0.90. The small difference is because ACS accounts for the compression reinforcement and uses a more precise iterative solution for the neutral axis.

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