CIV-330 · Lesson 1 of 12 · 22 min
Concrete structures and limit states design
Reinforced concrete is the most widely used structural material in the world. It is made from local aggregates, cement, water and steel bars, it can be cast into almost any shape, it resists fire and weather well, and it needs little maintenance when it is detailed properly. This course teaches how to design the common members of a concrete building: beams, one-way and two-way slabs, columns and footings. The design rules come from CSA A23.3, the Canadian standard for the design of concrete structures, used together with the loads of the National Building Code of Canada (NBC).
This first lesson sets the stage. It explains why concrete needs steel, introduces the project that runs through the whole course, describes the documents that govern design, and develops the limit states philosophy that every later calculation follows. It ends with the load combinations that produce the design forces for two members of the project.
Why concrete needs reinforcement
Concrete is strong in compression and weak in tension. A typical structural mix reaches a specified compressive strength \(f'_c\) of 25 to 40 MPa at 28 days, but its tensile strength is only about a tenth of that, and it is unreliable because shrinkage, temperature changes and small accidents of construction can crack the concrete before any load arrives. A plain concrete beam fails suddenly as soon as the bottom fibres crack. Steel bars placed near the tension face carry the tension after cracking, while the concrete above the neutral axis carries the compression. Deformed bars, with ribs rolled onto their surface, lock into the hardened concrete so that the two materials strain together.
The partnership works for three further reasons. Steel and concrete have nearly the same coefficient of thermal expansion, so temperature changes do not break the bond between them. The alkaline cement paste protects the steel from corrosion as long as the cover is adequate and the concrete is dense. And concrete insulates the steel in a fire, giving reinforced concrete members long fire-resistance ratings without added protection.
The consequence for design is that a reinforced concrete member is expected to crack in service. Cracks of a few tenths of a millimetre are normal and harmless; the designer controls their width, keeps deflections acceptable, and makes sure that at the ultimate limit the steel yields before the concrete crushes so that failure is gradual and visible.
The project: the municipal operations centre
The course continues the project from CIV-224 and CIV-321. Kestrel Bay Engineering is the structural consultant for a new municipal operations centre in Thunder Bay, Ontario. The site holds a two-storey office and administration building, a fleet garage with a covered upper parking level for light vehicles, a maintenance bay with a hoist, and a salt and sand storage dome. In CIV-321 the team analysed the frames; now they must size the concrete. Priya Raman, P.Eng., leads the structural design. Owen Lachance, EIT, prepares the calculations and drawings, and Sofia Martel, a co-op student, writes the short Python programs that check each step.
| Lesson | Member designed | Main check |
|---|---|---|
| 3 | Office floor beam | Flexural reinforcement |
| 4 | Garage T-beam and a depth-limited office beam | Flanged and doubly reinforced sections |
| 5 | Office floor beam | Deflection and crack control |
| 6 | Office beam and maintenance bay girder | Stirrup design |
| 7 | All beams and slabs | Bar development, hooks and splices |
| 8 | Garage parking slab | One-way slab design |
| 9 | Administration wing flat plate | Two-way moments and punching shear |
| 10 | Ground-floor interior column | Interaction diagram |
| 11 | Tall garage column | Slenderness |
| 12 | Interior column footing | Footing design and project summary |
The documents that govern design
Structural design in Canada is regulated by the provinces. Each province adopts a building code, usually based closely on the NBC, and that code references the standards that set out design methods. For concrete the key standards are:
- CSA A23.3, Design of concrete structures: the analysis and design rules for members and connections. The 2024 edition has been published; adopted provincial codes may still reference the 2019 edition, and designers use the edition that the applicable building code calls up.
- CSA A23.1 and A23.2: concrete materials and methods of construction, and the test methods, including exposure classes, cover and curing.
- CSA G30.18: carbon-steel bars for concrete reinforcement, including the common 400R and 400W grades.
- NBC Part 4: the loads (dead, live, snow, wind, earthquake) and the load combinations used with A23.3. Ontario's building code follows the same approach for a project in Thunder Bay.
A design standard is written in mandatory language and organized by member type. It states requirements and limits, but it does not teach. This course explains the mechanics behind each requirement so that the rules become logical, and it paraphrases the clauses rather than quoting them. When you design real structures you work from the standard itself.
Limit states design
A structure must do two things. It must be safe: it must not collapse, overturn or lose its load path under the largest loads it is likely to see. And it must be serviceable: under everyday loads it must not deflect, crack or vibrate enough to upset the people who use it or damage the finishes. Each way the structure can fail to do its job is called a limit state.
| Category | Examples | Loads used |
|---|---|---|
| Ultimate limit states (strength) | Flexural or shear failure, crushing of a column, loss of anchorage, instability, overturning | Factored loads |
| Serviceability limit states | Excessive deflection, wide cracks, vibration, durability loss | Specified (service) loads |
| Fatigue and special states | Repeated loads on bridges and crane girders, fire | As defined for the case |
For an ultimate limit state, the NBC and A23.3 require that the factored resistance be at least equal to the effect of the factored loads. In symbols, for a beam in bending,
The load factors \(\alpha_i\) are greater than 1 for the principal load, to account for the chance that loads exceed their specified values and for errors in analysis. The resistance side carries its own factors. A23.3 applies them to the materials rather than to the whole member: the strength of concrete is multiplied by \(\phi_c = 0.65\) and the strength of reinforcing bars by \(\phi_s = 0.85\) wherever they appear in the equations. The resulting \(M_r\) is called the factored resistance.
The figure shows the reasoning. Both the load effect and the resistance are random. The load factors push the design value of the load effect up into its upper tail, and the resistance factors pull the design value of the resistance down into its lower tail. As long as the factored resistance exceeds the factored load effect, the probability that the real load effect exceeds the real resistance is very small. Calibration studies chose the factors so that the reliability index, a measure of the distance between the two distributions, is about 3 to 3.5 for typical members over a 50-year life.
The material factors of A23.3 have a clear physical logic. Concrete is batched, placed and cured on site, and its in-place strength varies more than that of steel, so \(\phi_c\) is lower. Reinforcing bars are rolled in a mill under quality control, so \(\phi_s\) is higher. Because the factors are applied inside the equations, a member that fails by steel yielding automatically has a higher effective resistance factor than one that fails by concrete crushing. A23.3 also uses \(\phi_p = 0.90\) for prestressing tendons and other values for special materials.
| Material or action | CSA A23.3 resistance factor |
|---|---|
| Concrete | \(\phi_c = 0.65\) |
| Reinforcing bars | \(\phi_s = 0.85\) |
| Prestressing tendons | \(\phi_p = 0.90\) |
| Structural steel in composite members | \(\phi_a = 0.90\) |
| Members in a precast plant with certified quality | \(\phi_c = 0.70\) |
Loads and load combinations
The NBC defines specified loads: dead load D (self-weight and permanent fixtures), live load L from use and occupancy, snow load S, wind load W and earthquake load E. Specified live loads come from tables; for example, 2.4 kPa for office floors above the first storey and 2.4 kPa for parking areas for cars. The strength load combinations of NBC Part 4 take one principal load with its full factor plus companion loads with smaller factors, because the maximum values of different loads are unlikely to occur together.
| Case | Principal loads | Companion loads |
|---|---|---|
| 1 | 1.4D | — |
| 2 | (1.25D or 0.9D) + 1.5L | 1.0S or 0.4W |
| 3 | (1.25D or 0.9D) + 1.5S | 1.0L or 0.4W |
| 4 | (1.25D or 0.9D) + 1.4W | 0.5L or 0.5S |
| 5 | 1.0D + 1.0E | 0.5L + 0.25S |
Importance factors in the NBC scale the snow, wind and earthquake loads according to the consequences of failure. A municipal operations centre that houses snow-clearing equipment and emergency coordination is likely to be classed as a post-disaster building, which increases these loads; the team confirms the importance category with the municipality before calculating the environmental loads.
Design forces for two members
Owen starts with two members. The office floor beam spans 7.5 m between columns and carries a 5 m wide strip of floor. Its specified dead load, including the slab, the beam stem, ceilings, services and a partition allowance, is 20 kN/m; the live load is 2.4 kPa × 5 m = 12 kN/m. The garage roof beam spans 9.0 m over the parking level and carries a dead load of 18 kN/m and a snow load of 13.5 kN/m. For a first estimate he treats both as simply supported; the continuous analysis of CIV-321 refines the moments later.
Worked problem
Factored load on the office floor beam
Find the governing factored uniform load, moment and shear for the office floor beam (D = 20 kN/m, L = 12 kN/m, span 7.5 m, simply supported).
Show solution
- Case 1 gives
- \[w_f = 1.4(20) = 28.0\ \text{kN/m}\]
- Case 2 with L as the principal load gives
- \[w_f = 1.25(20) + 1.5(12) = 25.0 + 18.0 = 43.0\ \text{kN/m}\]
- There is no snow on an interior floor, so the companion load adds nothing. Case 2 governs. The factored moment and shear are
- \[M_f = \frac{w_f L^2}{8} = \frac{43.0(7.5)^2}{8} = 302\ \text{kN·m} \qquad V_f = \frac{w_f L}{2} = 161\ \text{kN}\]
Answer: \(w_f = 43.0\) kN/m, \(M_f = 302\) kN·m and \(V_f = 161\) kN. These values are used to design the beam in lessons 3, 5 and 6.
Sofia's program repeats the calculation for both members so that the governing combination can be read at a glance. For the garage roof beam the snow case governs, as expected for a roof in a snowy climate.
# NBC strength load combinations for two members of the operations centre
# Specified loads per metre of beam (kN/m)
members = {
"office floor beam": {"D": 20.0, "L": 12.0, "S": 0.0, "span": 7.5},
"garage roof beam": {"D": 18.0, "L": 0.0, "S": 13.5, "span": 9.0},
}
# (name, factor on D, factor on L, factor on S)
combos = [
("1 1.4D", 1.4, 0.0, 0.0),
("2 1.25D + 1.5L + 1.0S", 1.25, 1.5, 1.0),
("3 1.25D + 1.5S + 1.0L", 1.25, 1.0, 1.5),
]
for name, m in members.items():
print(name, "- span", m["span"], "m")
best = None
for label, fd, fl, fs in combos:
w = fd * m["D"] + fl * m["L"] + fs * m["S"]
print(f" {label:24s} wf = {w:6.1f} kN/m")
if best is None or w > best[1]:
best = (label, w)
w = best[1]; L = m["span"]
Mf = w * L**2 / 8; Vf = w * L / 2
print(f" governs: {best[0].strip()} Mf = {Mf:.0f} kN.m Vf = {Vf:.0f} kN")office floor beam - span 7.5 m 1 1.4D wf = 28.0 kN/m 2 1.25D + 1.5L + 1.0S wf = 43.0 kN/m 3 1.25D + 1.5S + 1.0L wf = 37.0 kN/m governs: 2 1.25D + 1.5L + 1.0S Mf = 302 kN.m Vf = 161 kN garage roof beam - span 9.0 m 1 1.4D wf = 25.2 kN/m 2 1.25D + 1.5L + 1.0S wf = 36.0 kN/m 3 1.25D + 1.5S + 1.0L wf = 42.8 kN/m governs: 3 1.25D + 1.5S + 1.0L Mf = 433 kN.m Vf = 192 kN
Two features of the output are worth noting. First, the factored loads are 30 to 40 per cent above the specified loads, which is the margin the load factors provide. Second, the governing combination changes from member to member, which is why design software checks every combination at every section. A23.3 does not add any further load factor; all of its safety beyond the load factors lies in \(\phi_c\) and \(\phi_s\).
Serviceability and durability from the start
Strength is only half of the design. Under specified loads the beam must not deflect enough to crack partitions, and its cracks must stay narrow enough not to look alarming or admit water and chlorides. A23.3 controls deflection either by minimum member thicknesses or by calculated deflections compared with span-based limits, and it controls cracking by the distribution of reinforcement. Lesson 5 covers both.
Durability decisions are made early because they set the concrete strength and cover. CSA A23.1 assigns exposure classes. The office interior is a mild exposure, but the garage parking level is exposed to de-icing salts tracked in on vehicles, which puts it in a chloride exposure class (C-1 for a structurally reinforced parking slab). That class requires a minimum strength of 35 MPa, a low water-to-cementing materials ratio, air entrainment and larger cover, and the parking structure standard CSA S413 adds requirements for membranes and drainage. The team uses 30 MPa for the office and 35 MPa for the garage.
Common mistakes
MistakeApplying \(\phi_c = 0.65\) to the whole moment resistance of a beam.
Instead\(\phi_c\) multiplies the concrete strength and \(\phi_s\) the steel strength inside the equations; there is no single member factor.
MistakeDesigning for service loads at the strength limit state.
InsteadStrength checks always use factored loads; service loads are for deflection and cracking.
MistakeAssuming the largest load factor always gives the governing combination.
InsteadCheck every combination; the governing one depends on the ratio of the loads, as the roof beam shows.
MistakeLeaving durability until the details are drawn.
InsteadExposure sets strength, cover and air content, which change member sizes, so decide it first.
Summary and key terms
- Reinforced concrete combines concrete for compression with deformed steel bars for tension; cracking in service is expected and controlled.
- CSA A23.3 with NBC loads governs design in Canada; the edition used is the one referenced by the adopted building code.
- Limit states design checks ultimate (strength) limit states with factored loads and serviceability limit states with service loads.
- A23.3 applies material resistance factors \(\phi_c = 0.65\) and \(\phi_s = 0.85\) inside the resistance equations.
- For the office beam, \(w_f = 1.25D + 1.5L = 43.0\) kN/m and \(M_f = 302\) kN·m.
- Exposure classes decide concrete strength, air content and cover, so they are set at the start of design.
- Limit state
- A condition beyond which a structure no longer fulfils a design requirement.
- Ultimate limit state
- A limit state involving collapse or failure, checked with factored loads.
- Serviceability limit state
- A limit state affecting function or appearance in normal use, checked with service loads.
- Factored resistance
- The resistance computed with material resistance factors applied to concrete and steel strengths.
- Load combination
- A set of loads, each multiplied by a factor, that may act together.
- Companion load
- A load acting with the principal load in a combination, given a reduced factor.
- Exposure class
- A category describing the environment of a concrete element, used to set durability requirements.
- Deformed bar
- A reinforcing bar with surface ribs that interlock with concrete.
Practice problems
Work each problem on paper first, then reveal the answer to check.
A floor beam carries D = 15 kN/m and L = 10 kN/m. Find \(w_f\).
Show answer
Case 2: \(1.25(15) + 1.5(10) = 33.8\) kN/m; case 1 gives 21.0 kN/m, so 33.8 kN/m governs.
Why is \(\phi_c\) smaller than \(\phi_s\)?
Show answer
Concrete strength in place varies more than mill-produced steel, and concrete failure is brittle.
Name one ultimate and one serviceability limit state for a parking slab.
Show answer
Ultimate: flexural or punching shear failure. Serviceability: wide cracks that let chlorides reach the steel, or excessive deflection.
For a roof beam with D = 10 kN/m and S = 8 kN/m, which case governs?
Show answer
Case 3: \(1.25(10) + 1.5(8) = 24.5\) kN/m versus case 1, 14.0 kN/m. Case 3 governs.
Why are companion loads given smaller factors than principal loads?
Show answer
The maximum values of different variable loads are unlikely to occur at the same time.
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