Loads and Load Combinations (NSCP) - Misconception Buster
In the PRC Civil Engineer Licensure Examination, the topic of Loads and Load Combinations is a perennial source of lost marks — not because the formulas are complex, but because reviewees carry subtle but fatal misconceptions from classroom shortcuts, misread notes, and conflated LRFD/ASD rules. A single wrong load combination can cascade into an incorrect design moment, an under-designed section, or a completely invalid answer choice. This guide targets the specific wrong beliefs that trap even well-prepared examinees, explains exactly why those beliefs feel correct, and provides trap questions that mirror actual board-exam item formats. Master these corrections and you eliminate an entire class of preventable errors.
Common misconceptions
The governing LRFD gravity combination is always 1.2D + 1.6L — no need to check the other combinations.
Exam impact
Board exam problems often provide W or E values precisely to test whether the examinee applies Combo 4/5/6/7. An examinee who only uses 1.2D + 1.6L will select a larger-than-correct Pu for gravity (missing uplift) or a smaller-than-correct Pu when wind dominates.
The reality
NSCP requires checking ALL applicable load combinations and designing for the MOST SEVERE. When roof live load Lr or wind W is significant, Combo 3 (1.2D + 1.6Lr + 1.0L) or Combo 4 (1.2D + 1.0W + 1.0L + 0.5Lr) can govern. For uplift-sensitive structures, Combo 6 (0.9D + 1.0W) or Combo 7 (0.9D + 1.0E) will govern. Blindly using only Combo 2 will MISS uplift conditions entirely.
Trap question
Question
A column carries dead load D = 200 kN, live load L = 120 kN, and wind load W = ±90 kN. Using NSCP LRFD, what is the governing factored axial COMPRESSION load Pu?
Combo 4 = 1.2(200) + 1.0(90) + 1.0(120) + 0.5(0) = 240 + 90 + 120 = 450 kN > 432 kN from Combo 2. The wind term increases the factored load beyond Combo 2. Always evaluate all combinations when lateral loads are present.
Wrong answer
432 kN (from 1.2D + 1.6L only)
Correct answer
450 kN (from Combo 4: 1.2D + 1.0W + 1.0L = 1.2×200 + 1.0×90 + 1.0×120)
Correct vs incorrect approach
Correct approach
Evaluate all applicable combos: C2: 1.2(200)+1.6(120)=432 kN; C4: 1.2(200)+1.0(90)+1.0(120)=450 kN; C6: 0.9(200)+1.0(−90)=180−90=90 kN (net tension/uplift check). Governing compression = 450 kN (Combo 4); governing uplift = 90 kN net (Combo 6). Both must be satisfied.
Incorrect approach
Given D=200 kN, L=120 kN, W=±90 kN. Student computes only: Pu = 1.2(200)+1.6(120) = 240+192 = 432 kN and stops there.
Why students believe it
In most textbook gravity examples, Combo 2 (1.2D + 1.6L) produces the largest factored load, so reviewees memorize it as 'the' LRFD formula and skip evaluating the other six combinations. Review centers that rush through the topic reinforce this by only solving examples where Combo 2 indeed governs.
LRFD and ASD load combinations can be mixed — e.g., use factored loads (1.2D+1.6L) but then divide by a safety factor to get allowable stress.
Exam impact
An examinee who mixes methods will compute a design force that is neither the correct factored load nor the correct service load, leading to wrong member selection and wrong answers on both concrete (ACI 318/NSCP 400) and steel (NSCP 500/AISC 360) design problems.
The reality
LRFD and ASD are two SEPARATE and INCOMPATIBLE design philosophies. LRFD: factored (amplified) loads are compared against factored (reduced) nominal strength φRn. ASD: service (unfactored) loads are compared against nominal strength divided by a safety factor Rn/Ω. Mixing them — applying factored loads to an allowable stress check — produces a meaninglessly conservative or dangerously unconservative result. NSCP Section 203 is explicit: choose one method and apply it consistently throughout a design.
Trap question
Question
An engineer computes Pu = 1.2D + 1.6L = 480 kN and then checks if this is less than the nominal capacity Pn = 600 kN. Is this a valid design check?
In LRFD, factored demand Pu must be ≤ factored resistance φPn. Comparing Pu to the bare Pn ignores the reliability reduction factor φ and leads to an unconservative, unsafe conclusion.
Wrong answer
Yes, because 480 kN < 600 kN, the member is safe.
Correct answer
No. The correct LRFD check is Pu ≤ φPn. Using Pn without the φ-factor is not valid LRFD. With φ = 0.65 (compression): φPn = 0.65×600 = 390 kN < 480 kN — the member FAILS.
Correct vs incorrect approach
Correct approach
LRFD: Use Pu = 340 kN and compare with φPn (e.g., φ = 0.90 for tension, 0.65 for columns). ASD: Use Pa = D+L = 150+100 = 250 kN and compare with Pn/Ω. Never combine factored loads with ASD resistance, or service loads with φRn.
Incorrect approach
Student computes Pu = 1.2(150)+1.6(100) = 340 kN (LRFD factored), then divides by safety factor 1.67 to get 203.6 kN for 'allowable' design — mixing LRFD loads with ASD resistance.
Why students believe it
Students see both sets of combinations in the same NSCP table and assume they are interchangeable or can be blended. The distinction between 'ultimate strength' and 'service' is not always emphasized clearly in review classes.
The 0.9D term in LRFD combinations 6 and 7 is a mistake or a minor adjustment — dead load should always be amplified, not reduced.
Exam impact
Board exam problems on retaining walls, cantilevered structures, or anchor design specifically test Combo 6 or 7. An examinee who ignores these combinations will compute an incorrect (too high) net resisting force and conclude a structure is safe when it may overturn.
The reality
The 0.9D factor is deliberate and critically important. In combinations 6 (0.9D + 1.0W) and 7 (0.9D + 1.0E), the dead load RESISTS overturning or uplift caused by wind or seismic loads. Using the full dead load would be unconservative because the actual dead load could be 10% less than estimated. The 0.9 factor accounts for that variability. These combos are the governing check for: anchor bolts in tension, foundation uplift, retaining wall overturning, and cantilevered structures.
Trap question
Question
A light steel column supports D = 90 kN axial compression. Wind creates an uplift (tension) of W = 95 kN. Using NSCP LRFD, is the column in net tension or compression under the critical uplift combination?
Combo 6 (0.9D + 1.0W) specifically uses the reduced dead load factor 0.9 to find the worst-case uplift. Using 1.2D unconservatively inflates the resisting dead load and masks a genuine uplift condition.
Wrong answer
Compression: 1.2(90) − 1.0(95) = 108 − 95 = +13 kN compression. No uplift problem.
Correct answer
NET TENSION of 14 kN: Apply Combo 6: 0.9(90) − 1.0(95) = 81 − 95 = −14 kN. The column is in tension and the connection must be designed for 14 kN uplift.
Correct vs incorrect approach
Correct approach
Apply Combo 6: 0.9D + 1.0W = 0.9(80) + 1.0(−100) = 72 − 100 = −28 kN (net TENSION/uplift). The column anchor must resist 28 kN tension. This is the critical uplift check that Combo 2 completely misses.
Incorrect approach
A roof column carries D = 80 kN (compression) and wind uplift W = −100 kN. Student uses 1.2D + 1.6L + 1.0W = 1.2(80) + 0 + 1.0(−100) = −4 kN. Since it's small, student ignores uplift concern.
Why students believe it
Students are conditioned to think load factors > 1.0 always make design more conservative. Using 0.9 on dead load seems to 'reduce safety' and feels counterintuitive, so many reviewees ignore or forget these combinations entirely.
Tributary width for an edge (exterior) beam is the same as for an interior beam — equal to the full beam spacing.
Exam impact
This error directly doubles the edge beam's line load and bending moment, leading to selection of a much heavier section or a computed moment far from the correct answer in a multi-part problem.
The reality
Tributary area is defined as the area CLOSER to a member than to any adjacent member. For an INTERIOR beam at spacing s: tributary width = s (half-spacing on each side = s/2 + s/2). For an EDGE beam: tributary width = s/2 (only one side, the half-spacing toward the interior). If beams are at 3 m spacing, interior beam TW = 3 m; edge beam TW = 1.5 m. Using full spacing for an edge beam DOUBLES the load — a 100% error.
Trap question
Question
A one-way slab system has beams spaced 4 m on center and a slab area load (D+L) of 8 kPa. What is the service line load on an EXTERIOR edge beam?
An exterior edge beam receives load only from one side — the half-bay toward the interior. Its tributary width is s/2 = 4/2 = 2 m, giving w = 8 × 2 = 16 kN/m, exactly half that of an interior beam.
Wrong answer
32 kN/m (using full 4 m tributary width: 8 × 4 = 32)
Correct answer
16 kN/m (using half-spacing: 8 × 4/2 = 8 × 2 = 16 kN/m)
Correct vs incorrect approach
Correct approach
Edge beam: TW = 3/2 = 1.5 m (only one side contributes). Line load = 6 kPa × 1.5 m = 9 kN/m. Interior beam: TW = 3 m. Line load = 6 kPa × 3 m = 18 kN/m.
Incorrect approach
Beams at 3 m spacing. Edge beam: TW = 3 m (full spacing). Line load = 6 kPa × 3 m = 18 kN/m. WRONG.
Why students believe it
Students apply the rule 'tributary width = beam spacing' to all beams. The distinction between interior and exterior is taught but quickly forgotten under exam pressure. Some review notes show only interior beam examples.
ASD Combination 2 (D + L) always governs for gravity loading under allowable stress design.
Exam impact
On roof beam or combined gravity+lateral problems in ASD format, the examinee who only checks D+L will miss the governing combination and select an incorrect — potentially under-designed — value.
The reality
NSCP ASD combinations include D + 0.75L + 0.75Lr (Combo 4), which can exceed D + L when the roof live load is significant relative to L. For roof structures and light occupancy floors combined with large Lr, Combo 4 may govern. Moreover, for lateral loads, Combo 5 (D + 0.6W or D + 0.7E) and Combo 6 must be evaluated. Always check all seven ASD combinations.
Trap question
Question
Using NSCP ASD, a roof beam carries: D = 10 kN/m, L = 4 kN/m (occupancy), Lr = 6 kN/m (roof live). What is the governing ASD service load intensity?
ASD Combo 4 = D + 0.75L + 0.75Lr = 17.5 kN/m governs over D+L = 14 kN/m and D+Lr = 16 kN/m. Roof structures must always be checked with this combination when both L and Lr are present.
Wrong answer
14 kN/m (D + L = 10 + 4)
Correct answer
17.5 kN/m (Combo 4: D + 0.75L + 0.75Lr = 10 + 0.75×4 + 0.75×6 = 10 + 3 + 4.5)
Correct vs incorrect approach
Correct approach
C2: D+L = 10+4 = 14 kN/m; C3: D+Lr = 10+6 = 16 kN/m; C4: D+0.75L+0.75Lr = 10+3+4.5 = 17.5 kN/m. Governing = 17.5 kN/m from Combo 4. D+L was NOT the worst case.
Incorrect approach
D = 10 kN/m, L = 4 kN/m, Lr = 6 kN/m. Student uses only D+L = 14 kN/m as the governing ASD service load.
Why students believe it
D + L is the most intuitive and simplest ASD combination. Students assume that since D and L are the primary gravity loads, their simple sum must produce the worst case. They overlook that the fractional-factor combos like D + 0.75L + 0.75Lr can sometimes govern when Lr is large.
The live load factor in LRFD combination 3, 4, and 5 is always 1.0L.
Exam impact
On precision problems where the 0.5 vs 1.0 factor changes which combination governs, the examinee using 1.0L universally will compute a higher combined load from combos 3–5 and may incorrectly conclude one of those combos governs over Combo 2.
The reality
NSCP explicitly uses f1·L in combos 3, 4, and 5, where f1 = 1.0 for: public assembly areas, parking garages, and occupancies where L > 4.8 kPa; and f1 = 0.5 for ALL OTHER occupancies (residential, office, classroom, etc.). For an office floor (L = 2.4 kPa), the companion live load in Combo 4 is 0.5L, not 1.0L. Using 1.0L for a residential building is conservative but technically incorrect per the code provision.
Trap question
Question
A residential floor beam (L = 1.9 kPa) carries D = 12 kN/m, L = 6 kN/m, and W = 10 kN/m. Applying NSCP LRFD Combo 4 with the correct companion factor, the factored load is:
For residential occupancy (L = 1.9 kPa < 4.8 kPa, not assembly or parking), f1 = 0.5. Combo 4 = 1.2D + 1.0W + 0.5L = 14.4 + 10 + 3 = 27.4 kN/m. Using f1 = 1.0 overestimates the factored load by 3 kN/m.
Wrong answer
1.2(12) + 1.0(10) + 1.0(6) = 14.4 + 10 + 6 = 30.4 kN/m
Correct answer
1.2(12) + 1.0(10) + 0.5(6) = 14.4 + 10 + 3 = 27.4 kN/m
Correct vs incorrect approach
Correct approach
Office floor (L = 8 kN/m < 4.8 kPa may apply per area; verify): f1 = 0.5. Combo 4: 1.2(15)+1.0(12)+0.5(8) = 18+12+4 = 34 kN/m. Combo 2: 1.2(15)+1.6(8) = 18+12.8 = 30.8 kN/m. Combo 4 (34) still governs but the value is different from using 1.0L (38).
Incorrect approach
Office building: D = 15, L = 8, W = 12 kN/m. Combo 4: 1.2(15)+1.0(12)+1.0(8)+0.5(0) = 18+12+8 = 38 kN/m. Student concludes Combo 4 governs.
Why students believe it
Most review center notes print 1.0L in combos 3–5 as a conservative default, and students memorize this without knowing the NSCP provision for the companion action factor f1.
For a two-way slab column strip, tributary area is computed the same way as for a one-way slab beam (using only spacing in one direction).
Exam impact
Using only one-direction spacing for a column in a two-way system can halve the tributary area and thus halve the computed column axial load — a catastrophically unconservative error on column design problems.
The reality
For a column supporting a two-way slab, the tributary area is a RECTANGLE (or irregular polygon) extending halfway to each adjacent column in BOTH directions. A column at spacing L1 × L2 has tributary area At = (L1/2 + L1/2) × (L2/2 + L2/2) = L1 × L2 for an interior column. Edge columns take L1 × (L2/2) and corner columns take (L1/2) × (L2/2). The column axial load is P = q × At, not just q × spacing-in-one-direction.
Trap question
Question
An interior column in a 6 m × 8 m bay two-way slab system supports a total area load of 9 kPa. What is the axial service load on the column?
An interior column collects load from the full tributary rectangle: half the bay in each direction on all four sides. For a 6 m × 8 m grid, At = 6 × 8 = 48 m². P = 9 × 48 = 432 kN. Using only one dimension (6 m) produces a 8-fold error in area and hence in load.
Wrong answer
54 kN (using P = 9 × 6 = 54, only one bay dimension)
Correct answer
432 kN (At = 6 × 8 = 48 m²; P = 9 × 48 = 432 kN)
Correct vs incorrect approach
Correct approach
Interior column: At = 5 × 6 = 30 m². P = 8 × 30 = 240 kN. Edge column (along 5 m direction): At = 5 × (6/2) = 5 × 3 = 15 m². P = 8 × 15 = 120 kN. Corner column: At = (5/2) × (6/2) = 2.5 × 3 = 7.5 m². P = 8 × 7.5 = 60 kN.
Incorrect approach
Column grid: 5 m × 6 m bays, area load q = 8 kPa. Student computes P = 8 × 5 = 40 kN/m (treating it like a beam) — dimensionally incorrect.
Why students believe it
The formula w = q × s is so drilled for one-way systems that students apply it even to two-way systems and columns, forgetting that a column or two-way panel receives loads from two directions.
Dead load (D) includes all loads that don't move — including live loads that are 'practically permanent' like heavy storage.
Exam impact
If heavy storage is classified as D instead of L, the load factors differ (1.6 vs 1.2) and the governing combination may change. This affects both the factored demand and the ASD service check.
The reality
NSCP defines dead load strictly as the WEIGHT OF ALL PERMANENT CONSTRUCTION: structural members, floors, roofs, fixed partitions, fixed service equipment, and finish materials. Live loads are defined by OCCUPANCY AND USE — they represent loads that vary in magnitude and position during the life of the structure, including storage, occupants, and movable equipment. Even a warehouse with steel racks permanently bolted down carries the rack weight as dead load and the STORED MATERIALS as live load (NSCP Table 205-1: storage, heavy = 12 kPa). Misclassifying live load as dead load changes the governing combination and the load factor applied.
Trap question
Question
A storage warehouse floor has a concrete slab self-weight of 3.6 kPa and stores goods at 12 kPa (heavy storage per NSCP). What is the correct LRFD factored floor load wu?
Stored goods are LIVE LOAD (NSCP Table 205-1), not dead load. The 1.6 factor on L versus 1.2 on D produces a higher factored load. Combo 1 (1.4D = 1.4×3.6 = 5.04 kPa) does not govern. Correct wu = 23.52 kPa from Combo 2.
Wrong answer
1.4(3.6 + 12) = 1.4 × 15.6 = 21.84 kPa (treating all as dead load)
Correct answer
23.52 kPa: D = 3.6 kPa, L = 12 kPa; wu = 1.2(3.6) + 1.6(12) = 4.32 + 19.2 = 23.52 kPa
Correct vs incorrect approach
Correct approach
D = slab self-weight = 3.5 kPa (permanent construction only). L = stored goods = 12 kPa (live load per NSCP Table 205-1). LRFD: wu = 1.2(3.5) + 1.6(12) = 4.2 + 19.2 = 23.4 kPa. Correct governing load is higher because L carries factor 1.6.
Incorrect approach
A warehouse stores heavy goods (12 kPa). Student classifies: D = slab self-weight (3.5 kPa) + stored goods (12 kPa) = 15.5 kPa. Factored: 1.4D = 1.4 × 15.5 = 21.7 kPa. WRONG classification.
Why students believe it
The word 'dead' intuitively means 'not moving,' leading students to classify heavy but movable storage, water tanks (when full), and even crowded occupancy as dead load because they seem fixed in practice.
LRFD Combination 1 (1.4D) only applies when no live load is present — otherwise skip it.
Exam impact
On problems with high D to L ratios, omitting Combo 1 leads to an underestimated factored load and an unconservative design. Board exams sometimes set up D >> L scenarios specifically to test this.
The reality
Combo 1 (1.4D) must always be checked. While it rarely governs when significant live load is present, it CAN govern in structures with very heavy dead load and minimal live load — such as heavy concrete transfer beams, massive retaining walls, or heavily finished floors with low occupancy live loads. For example: D = 20 kN/m, L = 1 kN/m → Combo 2 = 1.2(20)+1.6(1) = 25.6 kN/m; Combo 1 = 1.4(20) = 28 kN/m → Combo 1 GOVERNS.
Trap question
Question
A prestressed concrete beam carries D = 20 kN/m and L = 1.5 kN/m. Using NSCP LRFD, the governing factored load wu is:
When D is very large relative to L, Combo 1 (1.4D) can exceed Combo 2. Here, 1.4×20 = 28 kN/m > 26.4 kN/m. Always evaluate Combo 1 — it governs heavy dead load, low live load scenarios.
Wrong answer
26.8 kN/m (Combo 2: 1.2×20 + 1.6×1.5 = 24 + 2.4 = 26.4 kN/m — note: some may compute slightly differently)
Correct answer
28.0 kN/m (Combo 1: 1.4×20 = 28 kN/m > Combo 2: 1.2×20 + 1.6×1.5 = 24 + 2.4 = 26.4 kN/m)
Correct vs incorrect approach
Correct approach
Combo 1: 1.4(18) = 25.2 kN/m. Combo 2: 1.2(18)+1.6(2) = 24.8 kN/m. Combo 1 governs: wu = 25.2 kN/m. The difference of 0.4 kN/m matters in a precise member selection.
Incorrect approach
D = 18 kN/m, L = 2 kN/m. Student skips Combo 1, uses Combo 2: wu = 1.2(18)+1.6(2) = 21.6+3.2 = 24.8 kN/m.
Why students believe it
Students observe that Combo 2 (1.2D + 1.6L) almost always exceeds 1.4D when live load is non-zero, so they conclude Combo 1 is irrelevant when L exists and habitually skip it.
Wind load W and earthquake load E can be simply added together in the same load combination.
Exam impact
An examinee who adds W and E gets a fictitiously large lateral load, leading to an over-designed (wrong) section. On a multiple-choice exam, adding W+E will produce a value that is not among the answer choices.
The reality
NSCP load combinations NEVER combine W and E simultaneously in the same combination. Wind and seismic loads are treated as INDEPENDENT HAZARDS with low probability of simultaneous maximum occurrence. The NSCP LRFD combinations include EITHER W (Combos 4 and 6) OR E (Combos 5 and 7), never both. ASCE 7 (which NSCP is patterned after) takes the same approach. The designer evaluates both sets separately and uses the worse case, but never W+E in a single combination.
Trap question
Question
A structure is subject to D = 300 kN, L = 150 kN, W = 100 kN, E = 120 kN. A student proposes Pu = 1.2D + 1.0W + 1.0E + 1.0L. What is the correct NSCP LRFD governing Pu?
NSCP combinations never include both W and E simultaneously. Evaluate Combo 4 (W) and Combo 5 (E) separately, then use the larger. Here Combo 5 (seismic) governs at 630 kN, not the invalid W+E sum of 730 kN.
Wrong answer
1.2(300) + 1.0(100) + 1.0(120) + 1.0(150) = 360 + 100 + 120 + 150 = 730 kN
Correct answer
570 kN: Combo 5 (with E): 1.2(300)+1.0(120)+1.0(150) = 360+120+150 = 630 kN; Combo 4 (with W): 1.2(300)+1.0(100)+1.0(150) = 360+100+150 = 610 kN; Governing = 630 kN (Combo 5).
Correct vs incorrect approach
Correct approach
Wind governs lateral: Combo 4 with W: 1.2(200)+1.0(80)+1.0(100) = 240+80+100 = 420 kN. Seismic governs: Combo 5 with E: 1.2(200)+1.0(90)+1.0(100) = 240+90+100 = 430 kN. Governing Pu = 430 kN (seismic combo). W and E are never summed.
Incorrect approach
D = 200 kN, L = 100 kN, W = 80 kN, E = 90 kN. Student uses: 1.2D + 1.0W + 1.0E + 1.0L = 1.2(200)+80+90+100 = 240+270 = 510 kN. WRONG.
Why students believe it
Students see both W and E listed as lateral loads in textbooks and assume they can appear simultaneously in the same combination the way D and L do. The logic seems to be: 'more loads = more conservative.'
In ASD, the safety factor (Ω) is applied to the LOAD, not to the RESISTANCE — so the allowable load = Pu / Ω.
Exam impact
This misconception affects steel connection and member checks where Ω values are explicitly given. Computing Pa/Ω and comparing to Rn produces the wrong check and wrong conclusion about adequacy.
The reality
In ASD, the factor of safety Ω is applied to the NOMINAL RESISTANCE (strength): Allowable resistance Ra = Rn / Ω. The check is: applied SERVICE load ≤ Rn/Ω. The load is NOT divided by Ω; the strength IS divided by Ω. For example, if a bolt has Rn = 50 kN and Ω = 2.0, the allowable load capacity is Ra = 50/2 = 25 kN, and the applied service load must be ≤ 25 kN. Dividing the applied load by Ω and comparing to Rn would produce a non-code-compliant, potentially unconservative result.
Trap question
Question
An ASD check for a steel bolt gives: applied service shear = 35 kN, nominal shear strength Rn = 60 kN, Ω = 2.0. Is the bolt adequate?
ASD divides RESISTANCE by Ω, not the load. Allowable Ra = 60/2 = 30 kN. Since the service load (35 kN) exceeds the allowable resistance (30 kN), the bolt is inadequate. The incorrect approach gives a false safety indication.
Wrong answer
Yes: 35/2.0 = 17.5 kN < 60 kN. Passes.
Correct answer
No. Allowable = Rn/Ω = 60/2.0 = 30 kN. Applied = 35 kN > 30 kN. The bolt FAILS the ASD check.
Correct vs incorrect approach
Correct approach
Allowable resistance = Rn/Ω = 50/2 = 25 kN. Check: Pa = 30 kN > Ra = 25 kN → FAILS. The correct ASD check reveals the connection is inadequate; the wrong approach falsely passes it.
Incorrect approach
Applied service load Pa = 30 kN, Rn = 50 kN, Ω = 2.0. Student checks: Pa/Ω = 30/2 = 15 kN < Rn = 50 kN. 'Passes!' — but this check is meaningless.
Why students believe it
The phrase 'factor of safety' makes students think you divide the applied load by FS to get an 'allowable applied load.' This confuses ASD (where Rn/Ω ≥ applied service load) with a simplistic over-stress check concept.
Quick self-check
LRFD Combination 1 (1.4D) needs to be checked even when live load L is present.
Answer
True
Combo 1 can govern when dead load is very large relative to live load (e.g., D = 20 kN/m, L = 1 kN/m: 1.4×20 = 28 > 1.2×20+1.6×1 = 25.6). Always evaluate all combinations.
An edge beam at the perimeter of a floor system has the same tributary width as an interior beam at the same spacing.
Answer
False
An edge beam has tributary width = s/2 (only one side contributes). An interior beam has tributary width = s (half-spacing on each side). The edge beam carries half the line load of the interior beam.
NSCP LRFD Combinations 6 and 7 (with 0.9D) are used to check uplift and overturning conditions.
Answer
True
The 0.9D factor conservatively reduces the dead load that resists uplift or overturning from wind (Combo 6) or seismic (Combo 7). These combinations are critical for anchor design, foundation uplift, and light structures.
In a load combination problem, it is acceptable to include both wind (W) and earthquake (E) loads in the same NSCP combination.
Answer
False
NSCP load combinations treat W and E as separate hazards. Combos 4 and 6 use W; Combos 5 and 7 use E. They are never combined simultaneously in the same load combination.
In LRFD design, the factored load Pu is compared against φRn (phi times nominal resistance), not against Rn alone.
Answer
True
LRFD requires: Pu ≤ φRn. The φ factor (< 1.0) reduces the nominal resistance to account for variability. Comparing Pu to bare Rn ignores this reduction and is unconservative.
A stored warehouse load of 12 kPa should be classified as dead load D because the goods are permanently stored.
Answer
False
NSCP defines dead load as permanent construction elements (slab, beams, finishes). Stored goods are live load L per NSCP Table 205-1 (heavy storage = 12 kPa), even if they remain in place for extended periods. They carry load factor 1.6, not 1.2.
An area load of 8 kPa can be directly assigned as 8 kN/m to a beam without any further conversion.
Answer
False
kPa = kN/m². To get the line load (kN/m) on a beam, multiply the area load by the tributary width: w = q × s. For s = 3 m: w = 8 × 3 = 24 kN/m, not 8 kN/m. Omitting this step is a dimensional error.
In ASD, the factor of safety Ω is divided into the NOMINAL RESISTANCE to get the allowable resistance (Ra = Rn/Ω), and the applied service load must not exceed Ra.
Answer
True
ASD check: applied service load ≤ Rn/Ω. The Ω factor is applied to resistance, not to the load. This gives an allowable (permissible) resistance value that the service-level demand must stay within.
Diagrams
NSCP LRFD Load Combination Selection — Step-by-Step Decision Process
A decision flowchart guiding the correct evaluation sequence for all NSCP LRFD load combinations, emphasizing that ALL combinations must be checked, W and E are never combined, and the 0.9D uplift combos are mandatory for lateral load scenarios.
Tributary Area — Correct Load Routing for Interior, Edge, and Corner Members
A flowchart illustrating how to correctly determine tributary width or area for interior, edge, and corner members in both one-way and two-way systems, and how to convert area loads (kPa) to line loads (kN/m) or point loads (kN).
NSCP Load Types and Combination Overview — Concept Map
A comprehensive mind map organizing all NSCP load types, LRFD combinations, ASD combinations, and key principles into a visual hierarchy, helping students see the full picture and relationships between concepts.
LRFD vs ASD — Correct Method Application and Common Mixing Error
A side-by-side flowchart contrasting the WRONG mixed-method approach with the correct LRFD (factored loads vs φRn) and correct ASD (service loads vs Rn/Ω) design checks, clearly showing that the two methods must never be combined.
In summary
The most exam-critical mistakes in Loads and Load Combinations fall into five categories: (1) Incomplete combination checking — always evaluate ALL NSCP combinations, including Combos 1, 6, and 7, which are frequently neglected but specifically targeted in board exams. (2) Confusing LRFD and ASD — these are entirely separate philosophies; factored loads go with φRn, service loads go with Rn/Ω, and mixing them produces invalid results. (3) Tributary area errors — edge beams carry s/2, not s; columns in two-way systems carry L1×L2; and area loads (kPa) MUST be multiplied by tributary width or area before use. (4) Treating W and E as additive — NSCP never combines wind and seismic in the same load combination; evaluate them separately and use the worse result. (5) Misclassifying loads — stored goods, heavy equipment, and occupancy loads are LIVE loads (factor 1.6), not dead loads (factor 1.2). In seismic-prone Philippines, always evaluate the seismic combinations (Combos 5 and 7) — E frequently governs lateral design and uplift. Master these five correction areas and you eliminate the most common and most costly errors in the structural analysis and design sections of the PRC board examination.