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CIV-224 · Lesson 2 of 12 · 26 min

Design loads and load combinations

Before Owen can analyse a single truss for the operations centre, he needs to know what loads it must carry. Loads come from the weight of the structure itself, from people and contents, from snow, rain, wind and earthquakes, and from equipment such as the garage's bridge crane. In Canada, the minimum loads and the rules for combining them are set by the National Building Code of Canada (NBC), and by the provincial codes based on it. Ontario's Building Code follows the national structural provisions closely. This lesson uses the NBC 2020 format, which most provinces had adopted by 2026; the NBC 2025 (the sixteenth edition) keeps the same limit states framework but revises several climatic provisions, so always confirm which edition applies to your project.

Kinds of load

Loads are classified by how they arise and how variable they are, because the code treats them differently.

  • Dead load (D) is the weight of the structure and everything permanently attached to it: members, deck, slabs, roofing, ceilings, ducts and partitions where their location is fixed. It is known fairly accurately and changes little over time.
  • Live load (L) comes from the use of the building: people, furniture, stored goods, movable partitions and light vehicles. It is specified by occupancy as a uniform area load and, for some occupancies, a concentrated load that must be checked separately.
  • Snow and rain loads (S, and R in US practice) depend on climate, roof shape, exposure and the possibility of drifting or ponding.
  • Wind load (W) is a pressure or suction on the building surfaces that depends on the site's reference wind speed, terrain, height and building shape.
  • Earthquake load (E) arises from ground shaking and depends on the seismic hazard, soil, building mass and the ductility of the lateral system.
  • Other loads include soil and water pressure, thermal effects, settlement, impact, and the dynamic effects of moving equipment such as cranes.

Dead loads from unit weights

Dead loads are computed from member sizes and material unit weights. You will refine them as the design develops, because the member sizes you assumed at the start rarely survive unchanged.

Typical unit weights for estimating dead loads. Manufacturers' data govern for proprietary products.
MaterialUnit weight
Reinforced concreteabout 24 kN/m³
Structural steelabout 77 kN/m³
Softwood lumberabout 4–6 kN/m³
Steel roof deck, 38 mmabout 0.1 kPa
Built-up or membrane roofing with insulationabout 0.3–0.6 kPa
Suspended ceiling and servicesabout 0.3–0.5 kPa

Worked problem

Dead load on a garage roof truss

The garage roof consists of membrane roofing and insulation (0.45 kPa), 38 mm steel deck (0.10 kPa), open-web steel joists at 1.8 m (0.15 kPa) and an allowance of 0.25 kPa for sprinklers, lights and ducts hung below. The joists bear on trusses spaced 7.5 m apart. Estimate the dead load per metre of truss, including an allowance of 0.6 kN/m for the truss's own weight.

Show solution
  1. Sum the area loads: \(0.45 + 0.10 + 0.15 + 0.25 = 0.95\ \text{kPa}\).
  2. \[w_D = 0.95 \times 7.5 + 0.6 = 7.13 + 0.6 = 7.7\ \text{kN/m}\]

Answer: About 7.7 kN/m of dead load. Because the joists deliver it as point loads at the truss panel points, Owen will convert it to joint loads in lesson 4.

Live loads and live load reduction

The NBC specifies live loads by use and occupancy. Examples include 1.9 kPa for residential areas, 2.4 kPa for offices above the first storey, 4.8 kPa for assembly areas without fixed seats, corridors and the first-storey floors of offices, and higher values for storage and vehicle repair areas. Roofs carry a minimum specified load of 1.0 kPa, but in most of Canada snow governs. Concentrated loads, such as 9 kN over a small area for office floors, must also be checked because they can govern for short members and slabs.

Large areas are unlikely to be fully loaded at once, so the NBC allows the live load on a member supporting a large tributary area to be reduced. For most occupancies other than assembly and storage, where the tributary area B exceeds 20 m², the specified live load may be multiplied by \(0.3 + \sqrt{9.8/B}\). For a column supporting 80 m² this factor is \(0.3 + 0.35 = 0.65\). Different rules apply for assembly and storage occupancies, where full load is more likely.

Snow loads

Snow is often the largest variable load on a Canadian roof. The NBC gives the specified snow load as \(S = I_S\,[S_s (C_b C_w C_s C_a) + S_r]\) where \(S_s\) is the ground snow load and \(S_r\) the associated rain load from the climatic table for the location, \(I_S\) is the importance factor (1.0 for normal buildings at ultimate limit states, higher for post-disaster buildings), \(C_b\) is the basic roof factor (usually 0.8), \(C_w\) a wind exposure factor, \(C_s\) a slope factor and \(C_a\) an accumulation factor that rises well above 1 where snow drifts against a higher roof or parapet. A municipal operations centre that houses emergency vehicles may be classed as a post-disaster building, which raises its importance factors, and Priya has asked the city to confirm the classification.

Worked problem

Roof snow load on the garage

Take \(S_s = 2.9\ \text{kPa}\) and \(S_r = 0.3\ \text{kPa}\) for this example (values of this order apply in northwestern Ontario; a real design uses the code's climatic table), with \(I_S = 1.0\), \(C_b = 0.8\), \(C_w = 1.0\), \(C_s = 1.0\) for the low-slope roof and \(C_a = 1.0\) away from drifts. Find the snow load per metre of truss for trusses at 7.5 m.

Show solution
  1. \[S = 1.0\,[2.9(0.8)(1.0)(1.0)(1.0) + 0.3] = 2.62\ \text{kPa}\]
  2. \[w_S = 2.62 \times 7.5 = 19.7\ \text{kN/m}\]

Answer: The snow load, about 19.7 kN/m, is more than twice the dead load. Near the office wing, where the garage roof sits below the higher office roof, drift accumulation would raise \(C_a\) and the local load much further.

Wind and earthquake in brief

Wind pressure in the NBC takes the form \(p = I_W\,q\,C_e C_t C_g C_p\), where \(q\) is the reference velocity pressure for the site, \(C_e\) an exposure factor that grows with height and open terrain, \(C_t\) a topographic factor, \(C_g\) a gust factor and \(C_p\) a pressure coefficient that depends on the surface and wind direction. Earthquake loads come from the site's spectral accelerations, the building's period and the ductility of its lateral system. Both are covered in later design courses; here it is enough to know that each produces lateral loads applied at roof and floor levels.

Limit states design and load factors

Canadian structural design uses limit states design. At the ultimate limit states (ULS), the factored resistance of each member must equal or exceed the effect of factored loads: \(\phi R \ge \sum \alpha_i S_i\). Load factors above 1 cover the chance that loads exceed their specified values and the uncertainties of analysis; resistance factors below 1 cover variability in material strength and workmanship. At serviceability limit states (SLS), such as deflection and vibration, specified (unfactored) loads are used.

Because the maximum values of different variable loads are unlikely to occur at the same moment, the codes combine a principal load at its full factored value with companion loads at reduced factors.

NBC 2020 load combinations for ultimate limit states (simplified; storage and some other occupancies have additional rules). The 0.9D factor applies where dead load resists the other loads, as in uplift or overturning.
CasePrincipal loadsCompanion loads
11.4D—
2(1.25D or 0.9D) + 1.5L1.0S or 0.4W
3(1.25D or 0.9D) + 1.5S1.0L or 0.4W
4(1.25D or 0.9D) + 1.4W0.5L or 0.5S
51.0D + 1.0E0.5L + 0.25S

You must check every combination and use the one that produces the largest effect in each member. Different members, and even different points in the same member, can be governed by different combinations. A roof truss is usually governed by dead plus snow; a column base or anchor rods by dead plus wind with the low dead load factor, because wind uplift tries to lift the roof off.

Worked problem

The governing combination for the truss

Using the dead load of 7.7 kN/m and the snow load of 19.7 kN/m, plus a roof live load of 1.0 kPa × 7.5 m = 7.5 kN/m (which cannot act with full snow), find the governing factored gravity load on the truss. Wind on this low roof produces net uplift of about 4 kN/m.

Show solution
  1. Case 1: \(1.4(7.7) = 10.8\ \text{kN/m}\).
  2. Case 2 with roof live load as principal: \(1.25(7.7) + 1.5(7.5) = 9.6 + 11.3 = 20.9\ \text{kN/m}\).
  3. Case 3 with snow principal: \(1.25(7.7) + 1.5(19.7) = 9.6 + 29.6 = 39.2\ \text{kN/m}\).
  4. Uplift check, case 4 with low dead load factor: \(0.9(7.7) - 1.4(4.0) = 6.9 - 5.6 = +1.3\ \text{kN/m}\) downward, so no net uplift on the truss, though the deck fasteners must still be checked for local suction.

Answer: Dead plus snow governs, at about 39 kN/m factored. The truss will be analysed for this load, and for an unbalanced snow case with full snow on one side and reduced snow on the other.

Sofia automates the combination check so that it can be repeated whenever the loads change. Writing the combinations as data, rather than as separate formulas, makes them easy to review against the code table.

Checking NBC load combinations for the garage truss
# NBC 2020 ULS combinations (simplified) for one member; loads in kN/m, downward +
D, L, S, W = 7.7, 7.5, 19.7, -4.0      # roof live load treated as L; wind uplift negative

combos = {
    "1: 1.4D":            1.4 * D,
    "2: 1.25D+1.5L":      1.25 * D + 1.5 * L,
    "2: 1.25D+1.5L+1.0S": 1.25 * D + 1.5 * L + 1.0 * S,   # a roof cannot carry both; shown for comparison
    "3: 1.25D+1.5S":      1.25 * D + 1.5 * S,
    "3: 1.25D+1.5S+0.4W": 1.25 * D + 1.5 * S + 0.4 * max(W, 0),
    "4: 0.9D+1.4W":       0.9 * D + 1.4 * W,
    "4: 1.25D+1.4W+0.5S": 1.25 * D + 1.4 * max(W, 0) + 0.5 * S,
}
for name, w in combos.items():
    print(f"{name:22s} {w:7.2f} kN/m")
realistic = {k: v for k, v in combos.items() if "1.5L+1.0S" not in k}
gov = max(realistic, key=realistic.get)
low = min(realistic, key=realistic.get)
print("Governing downward:", gov, f"{realistic[gov]:.1f} kN/m")
print("Least downward (uplift check):", low, f"{realistic[low]:.1f} kN/m")
Output
1: 1.4D                  10.78 kN/m
2: 1.25D+1.5L            20.88 kN/m
2: 1.25D+1.5L+1.0S       40.58 kN/m
3: 1.25D+1.5S            39.17 kN/m
3: 1.25D+1.5S+0.4W       39.17 kN/m
4: 0.9D+1.4W              1.33 kN/m
4: 1.25D+1.4W+0.5S       19.48 kN/m
Governing downward: 3: 1.25D+1.5S 39.2 kN/m
Least downward (uplift check): 4: 0.9D+1.4W 1.3 kN/m

Crane and impact loads

The garage's 5-tonne bridge crane runs on runway beams carried by brackets on the columns. Moving loads like this are increased for dynamic effects: the wheel loads are multiplied by an impact allowance, and lateral and longitudinal forces from acceleration and braking are added. CSA S16, the steel design standard, and the crane manufacturer's data give the values. Lesson 9 shows how to find the position of moving wheel loads that produces the largest moment in a runway beam.

Serviceability load cases

Strength is not the only requirement. A roof that sags enough to pond water, a floor that bounces when people walk, or a frame that sways enough to crack the cladding has failed its users even if no member is close to breaking. Serviceability checks use specified loads without load factors, and usually only the variable load that causes the movement being checked. Typical deflection limits for beams are span/360 under live or snow load where ceilings or brittle finishes are attached, and span/240 or span/180 where they are not, with stricter limits for members supporting masonry or crane rails. Lessons 10 and 11 develop the methods for computing these deflections.

Keep the two sets of numbers separate in your calculations. A common source of error in hand calculations is carrying factored loads into a deflection check, which overestimates the deflection by roughly 40 to 60 per cent and can lead to members that are heavier than they need to be. Labelling every load as specified or factored, and noting which combination it comes from, avoids the confusion.

Common mistakes

  • MistakeApplying the full factor to every variable load at once.

    InsteadUse one principal load at its full factor and the others at their companion factors, and check each combination.

  • MistakeUsing only the largest factored gravity load and forgetting uplift.

    InsteadCheck the combination with the low dead load factor (0.9D) with wind, which governs anchors, connections and members that reverse.

  • MistakeTreating code load values as exact.

    InsteadThey are specified minimums with uncertainty. Use engineering judgement for unusual loads and check drifts, ponding and concentrated loads.

  • MistakeForgetting to update the dead load after resizing members.

    InsteadRecompute self-weight when sizes change, especially for concrete members.

Summary and key terms

  • Loads are classified as dead, live, snow and rain, wind, earthquake and others; each has its own code provisions.
  • Dead loads come from unit weights; live loads from the occupancy, with reductions allowed for large tributary areas.
  • NBC snow load is \(S = I_S[S_s(C_b C_w C_s C_a) + S_r]\); drifting can raise the load far above the basic value.
  • Strength design compares factored resistance with the effect of factored load combinations.
  • One principal variable load at full factor is combined with companion loads at reduced factors.
  • Check every combination, including low dead load with wind for uplift and overturning.
Dead load
The permanent weight of the structure and fixed attachments.
Live load
Load from the use and occupancy of a building.
Tributary area reduction
A code reduction in live load for members supporting large areas.
Load factor
A multiplier above or below 1 applied to a specified load in a load combination.
Load combination
A code-prescribed sum of factored loads to be checked in design.
Principal load
The variable load taken at its full factor in a combination.
Companion load
A variable load taken at a reduced factor in a combination.
Drift
Accumulation of snow at steps, parapets and obstructions on a roof.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. An office floor has D = 4.0 kPa and L = 2.4 kPa. Find the factored load for NBC case 2.

    Show answer

    \(1.25(4.0) + 1.5(2.4) = 5.0 + 3.6 = 8.6\ \text{kPa}\).

  2. Find the NBC live load reduction factor for a column with a tributary area of 45 m².

    Show answer

    \(0.3 + \sqrt{9.8/45} = 0.3 + 0.467 = 0.77\).

  3. Why is the dead load factor below 1 in one set of combinations?

    Show answer

    Where dead load resists other effects (uplift, overturning, sliding), overestimating it would be unconservative.

  4. A roof truss is governed by dead plus snow. Which combination is likely to govern its anchor bolts?

    Show answer

    Low dead load plus wind uplift.

  5. Why do companion loads have smaller factors than principal loads?

    Show answer

    Their maximum values are unlikely to occur at the same time as the maximum of the principal load.

  6. Name two reasons a roof's local snow load can exceed the basic value.

    Show answer

    Drifting against a higher roof or parapet, sliding from a higher sloped roof, and unbalanced loading on gable roofs.

Lesson quiz · 5 questions

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