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

Design actions

These two actions never occur at the same cross-section. On a simply supported beam under a UDL the moment peaks at midspan, where the shear is zero; the shear peaks at the support, where the moment is at or near zero. ACS checks every action in a combination simultaneously, so entering both into one combination asks it to design a section carrying 236.3 kN.m and 157.5 kN — a load case the beam never sees. That is a modelling error, not a conservative assumption: it manufactures a longitudinal-chord demand (AS 3600 Cl 8.2.7) the member does not have.

Enter a separate ULS combination per critical section instead, each carrying the actions that genuinely coexist there. That is what this example does — and the next section is about choosing where “there” is, because on the shear side the standard gives you a choice and both actions depend on it.

Choosing the critical section for shear

AS 3600:2018 Cl 8.2.3.2 offers two places to take the maximum transverse shear near a support: (a) the face of the support, or (b) a distance $d_v$ from the face, provided

(i) the member is directly supported and diagonal cracking cannot take place at the support or extend into it; and (ii) the transverse shear reinforcement required at $d_v$ from the support is continued unchanged to the face of the support.

Both conditions hold here. The beam bears directly on its supporting columns (that is why the support condition is in the given data — on a beam framing into the side of a girder the load hangs from the web, the support is indirect, and route (b) is not available). And the N10-200 fitment is uniform for the whole span, so whatever is required at $d_v$ runs unchanged to the face, satisfying (ii) by the detailing already chosen rather than by a promise.

The same clause screens for deep-component behaviour: if the distance from the point of zero shear to the face of the support were less than $2 d_v$, the member would be designed to Section 12 instead. Here the point of zero shear is midspan, 3000 mm from the face, against $2 d_v = 972$ mm — clear by a factor of three.

So this example designs at route (b). With $d_v = 486$ mm (Cl 8.2.1.9, derived in Step 6):

Relocating the critical section moves both actions, not just the shear. At the face the moment is zero; 486 mm in it is 70.3 kN.m. The smaller $V^*_y$ is the obvious half of the change and the smaller half: it is the moment that unloads the compression chord in the Cl 8.2.8.3 check of Step 7, taking it from a 1.05 failure to 0.10. Reading Cl 8.2.3.2(b) as a shear reduction alone leaves that check looking like a failure the member does not have, and invites a reinforcement remedy it does not need.

ACS applies no critical-section relocation of its own: it checks the shear you enter, at the section you entered it for. Choosing the section is the engineer’s job — and so is entering the moment that belongs to it.

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 and member type

• Design code: AS 3600
• Concrete grade: N40 ($f'_c = 40$ MPa)
• Rebar grade: D500N ($f_{sy} = 500$ MPa)
• Cover: 40 mm all sides (manual mode)
• Stress model: Rectangular
• Member type: Beam

Member type is not cosmetic — it selects which checks run and which AS 3600 Section 5 axis distance applies. ACS refuses to analyse a section that has none saved rather than assuming “beam” on your behalf.

Step 3: Place reinforcement

The effective depth to the bottom reinforcement is:

(cover + fitment 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 along the top edge to carry the fitment cage and the negative-moment chord force. Their depth from the compression face is

Step 4: Configure the fitments

This section is a convex rectangle, so the whole outline is a single cell and the Cell Tie tool opens its dialog with no cell to pick first — the banner reads “Configure the perimeter tie — it applies to the whole section.” (Bar Wrap also works: select all six bars and ACS infers a closed tie from three or more, but it takes more clicks.)

In the dialog:

• Stirrup Diameter: 10 mm (N10)
• Grade: D500N
• Spacing (mm): 200
• Hook ends: 135° both ends

There is no “legs” input. $A_{sv}$ is derived from the tie geometry you drew — a closed perimeter tie on a rectangle crosses any horizontal cut twice, so

Step 5: Enter the design combinations

In the Load Combinations panel, add the ULS combinations below. Note the shear column is $V^*_y$ — vertical shear, the axis that pairs with $M^*_x$. $V^*_x$ is horizontal shear across the web width and is checked separately; entering this beam’s shear there designs the 300 mm dimension as the shear depth and returns capacities for a member you have not built.

ULS-1 and ULS-2 are the design. ULS-2a is the Cl 8.2.3.2(a) route — the same support checked at the face instead of at $d_v$ — and it is here only so the rest of the page can show what that choice costs. It is not what this beam is designed to. Enter the first two if you only want the design; enter all three to reproduce the comparison in Step 7 — but expect the Design Summary to report an overall Fail if you do, because ULS-2a is a live combination and ACS has no notion of one you entered for illustration.

Step 6: Review results

Flexure — ULS-1 (midspan)

The section is adequate for flexure with about 13% reserve.

$k_u$ is the neutral-axis parameter $c/d_o$, where $d_o$ is the depth to the outermost layer of tensile reinforcement (AS 3600:2018 Cl 1.7). It is small here for two reasons that pull the same way: the section is lightly reinforced for its depth, and the two N16 compression bars carry part of the compressive force, so less concrete is needed to balance the tension steel and the neutral axis rises. Compression steel always reduces $k_u$.

Shear — ULS-2 (support, at $d_v$ from the face)

Web crushing does not govern: $V_{u,max} = 1373$ kN, more than three times $V_u$. Cl 8.2.3.2 requires that bound to be satisfied at the face of the support whichever critical section you design to — “notwithstanding the above” — and at 157.5 kN against 1373 kN it is, by nearly nine times.

ULS-2 now carries a moment as well, so ACS also runs the flexural check on it: at 70.3 kN.m against the same $\phi M_u = 270.9$ kN.m the utilisation is 0.260, nowhere near midspan’s 0.872. The moment at this section does not size anything. It matters for one reason only, and Step 7 is that reason.

Step 7: The check the shear table does not cover

Shear does not only load the web. The truss diagonals it is carried on anchor into the longitudinal bars, so AS 3600:2018 Cl 8.2.7 adds a force $\Delta F_{td}$ to both chords, and Cl 8.2.8.2 / 8.2.8.3 then check each chord against the steel that is actually there.

ACS runs that check on every combination with shear, and reports it in the Design Summary as Longitudinal (Shear). For ULS-2 both sides pass:

The $\pm M^*_x / z$ term is the whole story of this check. $\Delta F_{td}$ lands on both chords with the same sign and the same size; it is the flexural chord force that separates them, and it is signed: the tension chord gets $M^*_x/z$ added (Cl 8.2.8.2), the compression chord gets the same quantity subtracted (Cl 8.2.8.3). At $d_v$ from the face, $M^*_x/z = 70.3 / 0.486 = 144.7$ kN, which cancels almost all of the 162.2 kN the shear delivers — leaving the compression chord at 17.6 kN, about a tenth of what its 2-N16 can carry.

The tension chord takes the other half of that trade, rising to 306.9 kN, and it is the side that now governs the check at 0.575. That is the correct place for the demand to land: it is the bottom steel, continuous through the support and fully anchored past it, that the truss diagonals hang from. Detailing those bars to continue through and be fully anchored past the support — normal practice, and required by Cl 8.2.8 in any case — is what makes that number real.

What the face of the support would have said

Had this example designed at Cl 8.2.3.2(a) instead — combination ULS-2a — the chord check would not have passed:

At the face $M^*_x = 0$, so neither chord gets any relief and both carry the full $\Delta F_{td}$ — which is 11% larger as well, because $\Delta F_{td}$ scales with $V^*$. The 2-N16 compression chord is then 5% short.

That failure is an artefact of the section chosen, not a property of the beam. Route (a) is the conservative option and it is always available; on a member that does not meet conditions (i) and (ii) it is the only option. But choosing it here and then reinforcing for its result would buy top steel the member has no use for. Note that at $M^*_x = 0$ the split into a “tension” and a “compression” chord is a naming convention rather than a physical compression zone — but it does not change the verdict. Both chords carry the same $\Delta F_{td}$, ACS checks each against the steel actually in its half, and reports the worse of the two, so the governing number is $\Delta F_{td}$ divided by the smaller capacity whichever half is called which. Here that is the 2-N16 at 1.05; the 4-N20 side sits at 0.34 and never governs. The 1.05 is a real result, not an artefact of labelling — what is conventional at zero moment is only which chord gets named in a per-chord breakdown.

Two relaxations ACS does not take

Cl 8.2.8.2 carries two allowances that ACS deliberately leaves on the table. Both omissions are conservative, and neither changes this beam’s answer:

• The member-level cap. $T_{td}$ “need not be more than that required at the section with the maximum tension force demand for flexure, axial force and torsion”. Here that section is midspan, where $M^*_x / z = 236.3 / 0.486 = 486.2$ kN with no $\Delta F_{td}$ — well above the 306.9 kN at $d_v$, so the cap does not bind. It would on a member whose shear peak sits closer to its moment peak.
• The deemed-to-comply detailing route. For reinforced members with no axial tension or torsion and no sudden change in the calculated tension force, Cl 8.2.8.2 may instead be satisfied by extending the flexural tensile reinforcement a distance $d_v \cot \theta_v = 486 \times 1.376 = 669$ mm beyond the point at which it is no longer required for flexure. That is a detailing rule, not a section check: a section analyser cannot see where a bar stops, so ACS evaluates Eq 8.2.8.2(1) and reports its number instead. If you detail to the 669 mm extension, Cl 8.2.8.2 is satisfied by that route regardless of what the arithmetic one reports.

This check is currently reported in the Design Summary only; the ULS panel does not yet break it down (tracked as #5907). Read the PDF report’s shear section for the full chord breakdown in the meantime.

Results summary

The section is adequate as detailed: 4-N20 bottom, 2-N16 top, N10-200 fitments. Flexure at midspan governs the design at 0.872.

The conservative alternative, for contrast only — not the design:

Discussion

Flexure at midspan governs, at 0.872 — about 13% reserve. Web shear at $d_v$ sits at 0.439, four times that reserve, and the longitudinal chord check — which never appears in a $V_u$ table — at 0.575 on the tension side.

The example carries that chord check through to the end anyway, because it is the check the choice of critical section decides. Move the section 486 mm and shear falls 16% while the compression chord falls by a factor of ten — the shear term moves a little, the $-M^*_x/z$ term appears from nothing. A conclusion drawn at the face and a conclusion drawn at $d_v$ differ here not by a margin but by a verdict, and by two top bars nobody needs.

The ductility margin is wide ($k_{uo} = 0.133$ against a limit of 0.36), confirming a tension-controlled failure: the tension steel yields long before the concrete crushes, so the member gives visible warning.

If flexural utilisation were too high, in order of effectiveness:

• Increase the beam depth — moment capacity scales roughly with $d^2$
• Add tension reinforcement — effective up to the ductility limit
• Increase the concrete grade — moderate effect for flexure

Note that the third option moves shear down as well as flexure up: $\alpha_2$ and $\gamma$ both decrease with $f'_c$ under Cl 8.1.3, so a grade increase buys less flexural capacity than the change in $f'_c$ suggests.

The 200 mm fitment spacing could be relaxed away from the supports on shear grounds, but check Cl 8.2.1.7 minimum shear reinforcement and Cl 8.3.2.2 crack-control spacing before you do — the $k_v = 0.15$ and $\phi = 0.75$ this example enjoys both depend on $A_{sv}/s \geq A_{sv.min}/s$.

Hand calculation verification

Flexure

Rectangular stress block per AS 3600:2018 Cl 8.1.3, for $f'_c = 40$ MPa:

These are the 2018 bending factors. The superseded 2009 edition gave $\gamma = 1.05 - 0.007 f'_c$, and $1.0 - 0.003 f'_c$ is $\alpha_1$ — the uniform-compression factor for the squash load $N_{u0}$ in Cl 10.6.2.2, not a bending parameter at all. Using either in place of the values above overstates $\alpha_2$ by 11% and understates $\gamma$ by 11%.

Neutral axis depth from force equilibrium. Ignoring the compression bars for a first estimate:

The 2-N16 top bars sit inside that depth, so they take compression too and the neutral axis rises. Including them — at strain $\varepsilon_{sc} = 0.003(c - d')/c$, elastic at this depth, and deducting the concrete they displace:

Moment capacity, taking moments about the tension steel. With $\gamma c = 62.7$ mm the concrete block runs from the top face down to $y = 237.3$ mm; deducting the bar holes at $y = 242$ mm lifts its resultant slightly, to $\bar{y} = 269.3$ mm above the section centroid:

Design capacity. $k_{uo} = 0.133 \leq 0.36$, so Table 2.2.2(b) gives $\phi = 0.85$:

Every figure matches the ACS result above. The one residual is $c$: ACS reports 71.8 mm against 72.03 mm by hand, because the hand calculation deducts each N16’s whole area from the concrete block while ACS deducts only the part of it that actually lies inside the 62.7 mm block depth — the bars straddle its soffit, so about 16% of each sits below it. That is a 0.3% difference in $c$ and under 0.01% in $M_u$; it is a stated difference in how the two treat bar-hole geometry, not a disagreement about the method.

Shear

Simplified method per AS 3600:2018 Cl 8.2.4.3 (Amd 2:2021). First confirm the fitments reach the Cl 8.2.1.7 minimum, because both $k_v$ and $\phi$ depend on it:

Satisfied, so $k_v = 0.15$ and $\theta_v = 36^\circ$ (Cl 8.2.4.3(2)).

Effective shear depth, Cl 8.2.1.9:

Concrete contribution, Cl 8.2.4.1 (with $\sqrt{f'_c} \leq 8$ MPa):

Fitment contribution, Cl 8.2.5.2(1), vertical fitments:

Web-crushing bound, Cl 8.2.3.3(1) as amended (the 0.9 efficiency factor is Amd 2:2021):

Design capacity. $A_{sv}/s \geq A_{sv.min}/s$ and web crushing does not govern, so Table 2.2.2(e)(i) gives $\phi = 0.75$:

$V_u$, $\phi V_u$ and $d_v$ are section properties — they do not depend on which critical section you checked. Only the demand moved.

Longitudinal chord force

Cl 8.2.7, Eq 8.2.7(2), with $P_v = 0$ and no torsion, at the Cl 8.2.3.2(b) section:

The internal lever arm, Cl 8.2.1.9 form (the same expression as $d_v$, and the same value here):

Cl 8.2.8.2 and Cl 8.2.8.3 then give the two chords, with $N^* = 0$:

Ratios 0.575 and 0.103 — matching ACS, the tension side governing.

At the face of the support instead, $V^*_y = 157.5$ kN and $M^*_x = 0$:

Both chords carry the same force because there is no moment to split them, and the compression chord’s 2-N16 is 5% short of it. The two calculations differ by 486 mm of beam.

Related pages

• Section analysis — overview of all analysis types
• Moment-curvature theory — M-$\kappa$ analysis background
• RC column example — column design with biaxial bending