CELE Structural Theory & Analysis — Loads and Load Combinations (NSCP)Exam Answer Templates

Marks

1

Topic

Load Types — Dead Load

Difficulty

easy

Template Id

T1

Examiner Tip

One clear, complete sentence that includes 'permanent' and lists concrete examples is sufficient. Do not over-explain; VSA answers must be concise.

Model Answer

Dead load (D) is the permanent gravity load due to the self-weight of the structural members, architectural finishes, fixed mechanical/electrical equipment, and all other permanently attached components of a building, as defined under NSCP 2015 Section 204.

Question Type

very_short_answer

Answer Structure

• Single sentence: define dead load by naming its sources (self-weight, finishes, fixed equipment) and its permanent nature [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining dead load as only 'weight of the structure' without mentioning finishes or fixed equipment — too narrow for full credit
• Calling dead load 'variable' or 'changing' — factually wrong, no mark awarded
• Omitting the concept of permanence

Key Phrases To Include

• permanent
• self-weight
• fixed equipment
• architectural finishes
• NSCP 2015

Marks

1

Topic

NSCP LRFD Load Combinations

Difficulty

easy

Template Id

T2

Examiner Tip

Memorise all seven LRFD combinations verbatim. For a 1-mark VSA the equation alone (with correct factors) is the full answer.

Model Answer

The governing LRFD combination for gravity (D and L) is: U = 1.2D + 1.6L (NSCP 2015 Section 203.3, Combination 2, with no roof live load or wind/seismic components present.)

Question Type

very_short_answer

Answer Structure

• State the combination equation exactly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 1.4D + 1.7L — old ACI/UBC factors, not current NSCP 2015 LRFD
• Writing 1.2D + 1.6L + 0.5Lr without being asked about roof load — adds unnecessary terms
• Confusing this with the ASD combination D + L

Key Phrases To Include

• 1.2D
• 1.6L
• LRFD
• NSCP 2015
• governs

Marks

1

Topic

NSCP LRFD Load Combinations — Uplift/Overturning

Difficulty

medium

Template Id

T3

Examiner Tip

The key insight examiners look for is the word 'uplift' or 'overturning' — it shows you understand the physical scenario this combination addresses.

Model Answer

The 0.9D factor reduces the stabilising effect of dead load to its minimum probable value, thereby producing the worst-case net uplift or overturning effect when wind load W acts upward or laterally against the structure. It ensures that dead load is not over-credited as a restoring force.

Question Type

very_short_answer

Answer Structure

• State the purpose: worst-case uplift/overturning check by reducing stabilising dead load [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying '0.9D is a safety factor on dead load' without explaining why it is less than 1.0 — incomplete reasoning
• Confusing this with a load reduction factor for live load

Key Phrases To Include

• uplift
• overturning
• stabilising effect
• minimum dead load
• worst-case

Marks

2

Topic

Design Philosophies — LRFD vs ASD

Difficulty

easy

Template Id

T4

Examiner Tip

Use a side-by-side structure: LRFD paragraph, then ASD paragraph. End with the key distinction sentence to show deeper understanding and secure both marks.

Model Answer

LRFD (Load and Resistance Factor Design): Loads are multiplied by load factors greater than 1.0 to obtain factored (ultimate) loads (e.g., U = 1.2D + 1.6L). The structure's nominal resistance is then multiplied by a strength-reduction factor φ (< 1.0). Design is acceptable when factored demand ≤ φ × nominal strength. ASD (Allowable Stress Design): Loads are used at service (unfactored) levels (e.g., D + L). The allowable capacity is the nominal strength divided by a factor of safety (FS). Design is acceptable when service demand ≤ nominal strength / FS. Key distinction: LRFD amplifies loads and reduces resistance separately for each load type, providing more uniform reliability; ASD uses a single FS and is simpler but less precise in reliability.

Question Type

short_answer

Answer Structure

• Point 1: Define LRFD — factored loads, φ-factor on resistance [1 mark]
• Point 2: Define ASD — service loads, FS on nominal strength [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating LRFD uses a 'factor of safety' — wrong terminology; LRFD uses φ-factors and load factors
• Failing to mention that ASD uses service (unfactored) loads — this is the defining distinction
• Giving only one definition without comparing the two approaches

Key Phrases To Include

• load factors
• factored loads
• strength-reduction factor φ
• service loads
• factor of safety
• nominal strength
• ultimate

Marks

2

Topic

Tributary Area — Beam Line Loads

Difficulty

easy

Template Id

T5

Examiner Tip

Always write the formula w = q·s explicitly before substituting. Show wD and wL as separate lines before combining — this earns partial credit even with arithmetic errors.

Model Answer

Given: Beam spacing, s = 3.5 m (interior beam → tributary width = 3.5 m) Dead load intensity, qD = 4.8 kPa Live load intensity, qL = 2.4 kPa Tributary line load formula: w = q × s Dead load line load: wD = qD × s = 4.8 × 3.5 = 16.8 kN/m Live load line load: wL = qL × s = 2.4 × 3.5 = 8.4 kN/m ASD service load combination (NSCP 2015 ASD Combo 2: D + L): w_service = wD + wL = 16.8 + 8.4 = 25.2 kN/m ∴ Service (ASD) line load on interior beam = 25.2 kN/m

Question Type

numerical

Answer Structure

• Step 1: List given data and identify tributary width = beam spacing for interior beam [0.5 mark]
• Step 2: Apply w = q × s to compute wD and wL separately, with units [1 mark]
• Step 3: State ASD combination D + L and compute total service load [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using half the spacing for an interior beam — interior beams carry full spacing as tributary width
• Forgetting to multiply by the spacing and just stating qD = 4.8 kN/m (area load ≠ line load)
• Omitting units (kN/m) on the final answer

Key Phrases To Include

• tributary width
• w = q × s
• ASD combination D + L
• kN/m
• interior beam

Marks

2

Topic

LRFD Load Combinations — Beam Design

Difficulty

easy

Template Id

T6

Examiner Tip

Check: does Combo 1 (1.4D = 1.4×12 = 16.8 kN/m) govern over Combo 2 (24.0 kN/m)? No — always show this comparison to demonstrate rigor.

Model Answer

Given: Span, L = 6 m; wD = 12 kN/m; wL = 6 kN/m Governing LRFD gravity combination (NSCP 2015 Section 203.3, Combo 2): wu = 1.2wD + 1.6wL = 1.2(12) + 1.6(6) = 14.4 + 9.6 = 24.0 kN/m Factored midspan moment (simply supported, UDL): Mu = wuL²/8 = 24.0 × (6)² / 8 = 24.0 × 36 / 8 = 108 kN·m ∴ wu = 24.0 kN/m and Mu = 108 kN·m

Question Type

numerical

Answer Structure

• Step 1: Write LRFD Combo 2 formula and substitute values to get wu [1 mark]
• Step 2: Apply Mu = wuL²/8 and compute Mu with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 1.4D instead of 1.2D for gravity with live load present — Combo 1 (1.4D) is only valid when there is no live load
• Forgetting to square L in the moment formula — computing wuL/8 instead of wuL²/8
• Reporting Mu in kN instead of kN·m

Key Phrases To Include

• 1.2D + 1.6L
• LRFD Combo 2
• wu = 1.2(12) + 1.6(6)
• Mu = wuL²/8
• kN·m

Marks

3

Topic

Tributary Area — Column Axial Load

Difficulty

medium

Template Id

T7

Examiner Tip

Always compute PD and PL separately before applying load factors. Applying a single factor to the combined service load is a common and serious error that loses 2 marks.

Model Answer

Given: Tributary area, At = 49 m² qD = 5 kPa; qL = 3 kPa Step 1 — Service axial loads: PD = qD × At = 5 × 49 = 245 kN PL = qL × At = 3 × 49 = 147 kN Step 2 — ASD service combination (NSCP 2015 ASD Combo 2: D + L): P = PD + PL = 245 + 147 = 392 kN Step 3 — LRFD factored combination (NSCP 2015 LRFD Combo 2: 1.2D + 1.6L): Pu = 1.2PD + 1.6PL = 1.2(245) + 1.6(147) = 294.0 + 235.2 = 529.2 kN Also check LRFD Combo 1: 1.4PD = 1.4(245) = 343 kN < 529.2 kN → Combo 2 governs. ∴ ASD service load P = 392 kN; LRFD factored load Pu = 529.2 kN

Question Type

numerical

Answer Structure

• Step 1: Compute service loads PD and PL using P = q × At with units [1 mark]
• Step 2: ASD combination P = PD + PL = 392 kN [1 mark]
• Step 3: LRFD combination Pu = 1.2PD + 1.6PL = 529.2 kN, verify Combo 1 does not govern [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using At = 7 × 7/4 (quarter area) — this is wrong; interior column takes full tributary area
• Skipping the Combo 1 check (1.4D) — examiners look for this comparison
• Computing Pu = 1.2(392) — incorrectly applying load factor to the combined service load instead of to D and L separately

Key Phrases To Include

• P = q × At
• tributary area
• ASD Combo 2: D + L
• LRFD Combo 2: 1.2D + 1.6L
• governs
• kN

Marks

3

Topic

NSCP LRFD — Wind Load Combinations

Difficulty

hard

Template Id

T8

Examiner Tip

For any problem with lateral loads (W or E), ALWAYS write out Combo 6 (LRFD) or Combo 7 (ASD) even if it seems obvious dead load won't be overcome. The examiner's rubric specifically checks for this.

Model Answer

Given: D = 200 kN, L = 120 kN, W = ±90 kN (axial; + = compression, − = tension/uplift) Evaluate LRFD Combinations (NSCP 2015 Section 203.3): Combo 1: U = 1.4D = 1.4(200) = 280.0 kN (compression) Combo 2: U = 1.2D + 1.6L = 1.2(200) + 1.6(120) = 240 + 192 = 432.0 kN (compression) Combo 4: U = 1.2D + 1.0W + 1.0L = 1.2(200) + 1.0(+90) + 1.0(120) = 240 + 90 + 120 = 450.0 kN (compression, W additive) = 1.2(200) + 1.0(−90) + 1.0(120) = 240 − 90 + 120 = 270.0 kN (compression, W subtractive) Combo 6: U = 0.9D + 1.0W = 0.9(200) + 1.0(+90) = 180 + 90 = 270.0 kN (compression) = 0.9(200) + 1.0(−90) = 180 − 90 = 90.0 kN (compression; check for net tension if W > 0.9D → not the case here) Comparison: Maximum compression: Combo 4 (+W) → Pu = 450.0 kN ← GOVERNS for compression design Minimum (possible uplift check): Combo 6 (−W) → Pu = 90.0 kN (still compression, no net uplift) ∴ Governing Pu = 450.0 kN (compression), from LRFD Combo 4 with additive wind.

Question Type

numerical

Answer Structure

• Step 1: Evaluate gravity-only combos (1 and 2) [1 mark]
• Step 2: Evaluate Combo 4 (1.2D + 1.0W + 1.0L) for both ±W [1 mark]
• Step 3: Evaluate Combo 6 (0.9D + 1.0W) for uplift check, identify governing value [1 mark]

Scoring Breakdown

Common Mark Deductions

• Evaluating only one sign of wind load — both +W and −W must be checked
• Omitting Combo 6 (0.9D + 1.0W) — the most commonly missed combination on board exams
• Selecting Combo 2 as governing without checking wind combinations

Key Phrases To Include

• 0.9D + 1.0W
• uplift check
• ±W
• LRFD Combo 4
• governs
• 450.0 kN

Marks

3

Topic

Earthquake Load — Philippine Context

Difficulty

medium

Template Id

T9

Examiner Tip

This 3-mark conceptual question follows a classic Why-What-How structure: Why is E large in PH? What does the code say? How does it affect design? One paragraph per mark point is ideal.

Model Answer

The Philippines lies along the Pacific Ring of Fire and is one of the most seismically active countries in the world, situated at the convergence of the Philippine Sea Plate, the Eurasian Plate, and the Sunda Plate. Consequently, NSCP 2015 assigns high seismic zone factors and requires buildings to be designed for significant lateral seismic forces. From a load combination standpoint, the NSCP 2015 LRFD combination involving earthquake is: U = 1.2D + 1.0E + 1.0L (Combo 5) U = 0.9D + 1.0E (Combo 7 — for uplift/overturning due to seismic) Unlike wind, which varies with height and exposure, seismic force E is derived from the building's mass and the ground acceleration (per NSCP 2015 Section 208), meaning heavier structures attract larger seismic forces. In most Philippine buildings, the combination with E produces the largest lateral force demand on columns, shear walls, and connections, causing these combinations to govern over gravity-only or wind combinations in lateral design. This is why Philippine structural engineers must prioritise seismic design in proportioning lateral force-resisting systems.

Question Type

short_answer

Answer Structure

• Point 1: Geographic/tectonic context — Philippines on the Ring of Fire, high seismicity [1 mark]
• Point 2: Cite the NSCP LRFD combinations involving E (Combos 5 and 7) [1 mark]
• Point 3: Explain why E governs — mass-proportional force, high ground accelerations, lateral demand on structural members [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'earthquake load is large' without explaining why (tectonic setting) — lacks technical justification
• Citing the correct combination but with wrong factors (e.g., 1.6E) — loses the combination mark
• Discussing seismic design in general without linking back to load combinations

Key Phrases To Include

• Ring of Fire
• seismically active
• 1.2D + 1.0E + 1.0L
• 0.9D + 1.0E
• lateral force
• NSCP 2015 Section 208
• governs

Marks

5

Topic

Tributary Area + LRFD Combinations — Beam Analysis

Difficulty

medium

Template Id

T10

Examiner Tip

Structure your answer with clear part labels (a), (b), (c). Examiners mark part by part, so a mistake in (b) does not cost you marks in (c) if you carry forward your wu value correctly and show the moment formula.

Model Answer

Given: Span, L = 7 m Tributary width, b = 4 m Area dead load, qD = 3.5 kPa Area live load, qL = 2.0 kPa (a) Service Line Loads: wD = qD × b = 3.5 × 4 = 14.0 kN/m wL = qL × b = 2.0 × 4 = 8.0 kN/m Total service load (ASD, D + L): w = 14.0 + 8.0 = 22.0 kN/m (b) LRFD Factored Line Load: Check Combo 1: 1.4wD = 1.4(14.0) = 19.6 kN/m Check Combo 2: wu = 1.2wD + 1.6wL = 1.2(14.0) + 1.6(8.0) = 16.8 + 12.8 = 29.6 kN/m ← GOVERNS (Combo 2 > Combo 1) ∴ wu = 29.6 kN/m (c) Factored Midspan Moment: For a simply supported beam with UDL: Mu = wuL²/8 = 29.6 × (7)²/8 = 29.6 × 49/8 = 1,450.4/8 = 181.3 kN·m Summary: (a) wD = 14.0 kN/m, wL = 8.0 kN/m, w_service = 22.0 kN/m (b) wu = 29.6 kN/m (LRFD Combo 2 governs) (c) Mu = 181.3 kN·m

Question Type

numerical

Answer Structure

• Part (a): Apply w = q × b to get wD and wL, state service total (ASD) [1.5 marks]
• Part (b): Evaluate LRFD Combo 1 and Combo 2, identify governing wu [2 marks]
• Part (c): Apply Mu = wuL²/8 with correct substitution and units [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Skipping Combo 1 check — the examiner expects both combinations to be evaluated
• Using L = 7 m without squaring in the moment formula (computing 29.6 × 7 / 8 instead of 29.6 × 49 / 8)
• Reporting the moment in kN/m instead of kN·m — dimensional error deducts half a mark

Key Phrases To Include

• w = q × b
• tributary width
• Combo 1: 1.4D
• Combo 2: 1.2D + 1.6L
• governs
• Mu = wuL²/8
• kN/m
• kN·m

Marks

5

Topic

LRFD Gravity Combinations with Roof Live Load

Difficulty

hard

Template Id

T11

Examiner Tip

The Lr factor swap between Combo 2 (0.5Lr companion) and Combo 3 (1.6Lr primary) is a classic board-exam trap. Write out both combinations fully to demonstrate you understand the distinction.

Model Answer

Given: D = 300 kN, L = 180 kN, Lr = 50 kN (roof live), W = 0, E = 0 NSCP 2015 LRFD Gravity Combinations (Section 203.3): Combo 1: U = 1.4D = 1.4(300) = 420.0 kN Combo 2: U = 1.2D + 1.6L + 0.5Lr [since only Lr is present, use 0.5Lr] = 1.2(300) + 1.6(180) + 0.5(50) = 360 + 288 + 25 = 673.0 kN Combo 3: U = 1.2D + 1.6Lr + 1.0L [f1 = 1.0 as conservative default] = 1.2(300) + 1.6(50) + 1.0(180) = 360 + 80 + 180 = 620.0 kN (Alternative using 0.5L for non-assembly occupancy: 1.2(300) + 1.6(50) + 0.5(180) = 360 + 80 + 90 = 530 kN) Comparison: Combo 1: 420.0 kN Combo 2: 673.0 kN ← MAXIMUM Combo 3: 620.0 kN (conservative, f1 = 1.0) ∴ Governing Pu = 673.0 kN from LRFD Combo 2 (1.2D + 1.6L + 0.5Lr).

Question Type

numerical

Answer Structure

• Identify applicable combinations: Combos 1, 2, and 3 (gravity only, no wind/seismic) [1 mark]
• Correctly evaluate Combo 1: U = 1.4D = 420.0 kN [1 mark]
• Correctly evaluate Combo 2: U = 1.2D + 1.6L + 0.5Lr = 673.0 kN [1.5 marks]
• Correctly evaluate Combo 3: U = 1.2D + 1.6Lr + 1.0L = 620.0 kN [1 mark]
• Compare all three results and state governing Pu = 673.0 kN [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using the wrong factor on Lr in Combo 2 (using 1.6Lr instead of 0.5Lr in Combo 2) — the 0.5 companion factor applies when Lr accompanies 1.6L
• Evaluating only Combo 2 and skipping Combo 3 — both must be compared
• Confusing Lr (roof live load) with L (floor live load) — these take different positions in different combinations

Key Phrases To Include

• 1.4D
• 1.2D + 1.6L + 0.5Lr
• 1.2D + 1.6Lr + 1.0L
• f1 factor
• governs
• 673.0 kN
• NSCP 2015 Combo 2

Marks

3

Topic

Tributary Area — Column Load Distribution

Difficulty

medium

Template Id

T12

Examiner Tip

A small sketch showing a 3×3 column grid with tributary areas shaded for interior, edge, and corner columns communicates the concept instantly and can substitute for a lengthy written explanation — it also earns the diagram bonus mark if the rubric allows.

Model Answer

Tributary area is the floor area that is assumed to channel its load to a particular structural member — it is bounded by lines drawn midway between adjacent parallel members. For a regular bay with column spacing Sx in the x-direction and Sy in the y-direction: • Interior column: receives load from all four surrounding bays → tributary area = Sx × Sy (full bay area). • Edge column (on one exterior face): receives load from one full span in one direction and a half-span in the other → tributary area = (Sx/2) × Sy or Sx × (Sy/2) depending on which edge. • Corner column: receives only a quarter-bay contribution from each adjacent bay → tributary area = (Sx/2) × (Sy/2). Design axial load: P = q × At, where q is the area load intensity (kPa) and At is the tributary area. Example: Sx = Sy = 6 m, q = 8 kPa Interior: At = 36 m² → P = 288 kN Edge: At = 18 m² → P = 144 kN Corner: At = 9 m² → P = 72 kN This shows that interior columns carry the largest axial load and must be proportioned accordingly.

Question Type

short_answer

Answer Structure

• Define tributary area clearly with the midpoint-boundary concept [1 mark]
• Compare interior vs. edge vs. corner column tributary areas with correct fractions [1 mark]
• Apply P = q × At with a numerical example showing the difference [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating tributary area without explaining the midpoint-boundary rule — too vague
• Giving only the interior column case without comparing to edge/corner columns
• Not applying the formula P = q × At with an example — explanation alone is insufficient for full marks

Key Phrases To Include

• tributary area
• midway between adjacent members
• interior column: Sx × Sy
• edge column: Sx/2 × Sy
• corner column: Sx/2 × Sy/2
• P = q × At

Marks

2

Topic

Live Load Values by Occupancy

Difficulty

easy

Template Id

T13

Examiner Tip

Board questions often ask for code table values — memorise at least residential (1.9 kPa), office (2.4 kPa), corridors (4.8 kPa), and storage (5.0–12 kPa) live loads from NSCP Table 205-1.

Model Answer

(a) Residential occupancy minimum live load: 1.9 kPa (NSCP 2015 Table 205-1) (b) School classroom minimum live load: 1.9–2.9 kPa depending on the specific classroom type (e.g., 1.9 kPa for typical classrooms; up to 4.8 kPa for areas with fixed seats and possible crowd loading — NSCP 2015 Table 205-1) Live loads are specified by occupancy because the actual distribution and magnitude of people, furniture, and movable equipment are highly variable and cannot be determined exactly at the design stage. Code-specified minimum values represent statistically derived envelopes of the maximum loads expected over the life of the building for a given use, ensuring consistent and conservative design without requiring building-by-building occupancy surveys.

Question Type

short_answer

Answer Structure

• Part (a): State residential live load = 1.9 kPa with code table reference [0.5 mark]
• Part (b): State classroom live load range 1.9–2.9 kPa with code table reference [0.5 mark]
• Explain why occupancy-based specification: statistical nature, variability, design-stage uncertainty [1 mark]

Scoring Breakdown

Common Mark Deductions

• Giving an incorrect value for residential live load (e.g., 2.5 kPa) — no mark for that part
• Explaining only what live loads are instead of why they are specified by occupancy

Key Phrases To Include

• 1.9 kPa
• NSCP 2015 Table 205-1
• variable
• statistical
• occupancy
• design stage uncertainty

Marks

5

Topic

NSCP ASD Load Combinations

Difficulty

hard

Template Id

T14

Examiner Tip

ASD Combo 4 (D + 0.75L + 0.75Lr) is often overlooked because reviewees memorise only D + L as the governing gravity case. Always run all four ASD gravity combos when Lr is given.

Model Answer

Given: wD = 10 kN/m, wL = 8 kN/m, wLr = 3 kN/m, L = 7 m, simply supported NSCP 2015 ASD Load Combinations (Section 203.4): Combo 1: w = D = 10.0 kN/m Combo 2: w = D + L = 10 + 8 = 18.0 kN/m Combo 3: w = D + Lr = 10 + 3 = 13.0 kN/m Combo 4: w = D + 0.75L + 0.75Lr = 10 + 0.75(8) + 0.75(3) = 10 + 6.0 + 2.25 = 18.25 kN/m (Note: Combos 5, 6, 7 involve W or E which are zero here → not applicable) Comparison: Combo 1: 10.0 kN/m Combo 2: 18.0 kN/m Combo 3: 13.0 kN/m Combo 4: 18.25 kN/m ← GOVERNS ∴ Governing ASD design UDL = 18.25 kN/m Design moment (simply supported, UDL): M = wL²/8 = 18.25 × (7)²/8 = 18.25 × 49/8 = 894.25/8 = 111.8 kN·m ∴ ASD design moment M = 111.8 kN·m

Question Type

numerical

Answer Structure

• Identify applicable ASD combinations (1–4, omit 5–7 due to W = E = 0) [1 mark]
• Evaluate Combo 1 (D) and Combo 2 (D + L) correctly [1 mark]
• Evaluate Combo 3 (D + Lr) and Combo 4 (D + 0.75L + 0.75Lr) correctly [1.5 marks]
• Identify governing combination as Combo 4 at 18.25 kN/m [0.5 mark]
• Compute design moment M = wL²/8 = 111.8 kN·m with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using LRFD factors (1.2D, 1.6L) in an ASD problem — fundamental conceptual error, no marks for those steps
• Not evaluating Combo 4 and assuming Combo 2 always governs in ASD — Combo 4 (0.75 companions) can govern when Lr is present
• Computing M = wL/8 without squaring L

Key Phrases To Include

• ASD Combo 4
• D + 0.75L + 0.75Lr
• 18.25 kN/m
• governs
• M = wL²/8
• 111.8 kN·m
• NSCP 2015

Marks

2

Topic

LRFD Companion Live-Load Factor f1

Difficulty

medium

Template Id

T15

Examiner Tip

The f1 factor is a small but frequently tested NSCP detail. Memorise the trigger conditions: assembly, garages, or L > 4.8 kPa → f1 = 1.0. Everything else → f1 = 0.5.

Model Answer

In NSCP 2015 LRFD Combinations 3, 4, and 5 (where L appears alongside a primary variable load such as Lr, W, or E), the live load term is written as f1L, where f1 is the companion live-load factor: • f1 = 1.0 for: (a) places of public assembly (e.g., auditoriums, stadia, gymnasiums), (b) garages and parking structures, and (c) any occupancy where the design live load L > 4.8 kPa. • f1 = 0.5 for: all other occupancies (e.g., typical offices, residences, classrooms). Rationale: For high-occupancy or storage areas where full live load is likely to coexist with the primary variable load event, f1 = 1.0 applies. For typical occupancies, statistical evidence shows that full L and full Lr/W/E are unlikely to occur simultaneously, so the reduced factor f1 = 0.5 is used. In practice, many review centres default to f1 = 1.0 as the conservative value.

Question Type

short_answer

Answer Structure

• Point 1: State that f1 = 1.0 applies to assembly areas, garages, and L > 4.8 kPa [1 mark]
• Point 2: State that f1 = 0.5 applies to all other occupancies, with a brief rationale [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating f1 = 0.5 always — ignoring the 1.0 condition for assembly/garages
• Confusing f1 with the load factor on L in Combo 2 (which is always 1.6, not f1)

Key Phrases To Include

• f1 = 1.0
• f1 = 0.5
• assembly areas
• garages
• L > 4.8 kPa
• companion factor
• simultaneous occurrence