CELE Structural Theory & Analysis — Loads and Load Combinations (NSCP)Detailed Explanation
Detailed explanations for CELE Structural Theory & Analysis — Loads and Load Combinations (NSCP). This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Loads and Load Combinations (NSCP) questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Structural Theory & Analysis section sits under a "Core" weighting, and Loads and Load Combinations (NSCP) is the 6th chapter in the 6-chapter CELE Structural Theory & Analysis rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Structural Theory & Analysis.
Loads and Load Combinations (NSCP) - Detailed Explanation
Every structural design begins with a single critical question: what forces must the structure resist? The National Structural Code of the Philippines (NSCP 2015, 7th Edition) provides the definitive answer by specifying the types of loads a structure must carry, their minimum magnitudes, and — crucially — how they must be combined for design. This chapter covers the complete load taxonomy recognised by NSCP, the concept of tributary area for load routing, and the LRFD and ASD load combination sets that you will apply in every concrete, steel, and timber design problem in the licensure examination. Whether you are sizing a floor beam, checking a column for axial load, or verifying overturning stability against wind, the correct answer starts here.
Concepts
Load Types and Code-Specified Minimum Values
NSCP 2015 Section 204 to 208 classifies structural loads into six principal categories, each reflecting a distinct physical source and statistical character. **Dead Load (D):** The permanent weight of the structural system itself — concrete slabs, beams, columns, walls — plus all superimposed permanent items such as floor finishes, ceiling systems, mechanical/electrical conduit, and fixed partitions. Dead load is the most predictable load type; its coefficient of variation is low (~0.10). Typical values: normal-weight concrete 24 kN/m³; ceramic tile finish ≈ 0.5 kPa; 100 mm hollow-core block wall ≈ 1.9 kPa. **Live Load (L):** Loads due to building occupancy — people, furniture, stored goods, and movable equipment. NSCP Table 205-1 prescribes minimum uniformly distributed live loads by occupancy: residential (dwelling) 1.9 kPa; classrooms 1.9–2.9 kPa; offices 2.4 kPa; retail stores 4.8 kPa; assembly areas with fixed seats 2.9 kPa; heavy storage 7.2 kPa and up. Live load has higher variability (~0.25) than dead load, justifying its larger load factor in LRFD (1.6 vs. 1.2). **Roof Live Load (Lr):** Temporary loads on roofs from maintenance workers and equipment. NSCP 206 — minimum 1.0 kPa on accessible flat roofs; reduced values for sloped roofs. **Wind Load (W):** Computed from basic wind speed (V, m/s), exposure category, height, and pressure/shape coefficients per NSCP Chapter 207. The Philippines has one of the world's highest recorded wind speeds (Typhoon Haiyan ≈ 87 m/s gust), so wind load is a critical lateral-force consideration. **Earthquake (Seismic) Load (E):** The Philippines sits on the Pacific Ring of Fire with numerous active fault systems (e.g., Philippine Fault Zone, Marikina Valley Fault). Seismic load is computed using the equivalent static force method or dynamic analysis per NSCP Chapter 208. E often governs lateral design in most Philippine localities and must always be checked. **Rain Load (R):** Ponding of water on flat/slightly sloped roofs where drainage may be blocked. Typically 1.0 kPa minimum or computed from roof geometry. **Snow (S):** Essentially zero in the Philippines; appears only in the NSCP combination equations for completeness.
Examples
Always separate D and L since the code applies different factors and different load combinations to each. The 24 kN/m³ unit weight for normal-weight concrete is standard per NSCP Section 204.2.
Scenario
A 150 mm normal-weight reinforced concrete slab with a 50 mm ceramic tile finish supports a residential occupancy. Compute the total design area load (service level).
Solution
Dead load from slab: wD,slab = 0.150 m × 24 kN/m³ = 3.60 kPa Dead load from tile finish: wD,tile = 0.50 kPa (from NSCP Table 204 or typical value) Total dead load: wD = 3.60 + 0.50 = 4.10 kPa Live load (residential, NSCP Table 205-1): wL = 1.90 kPa Total service load: w_service = D + L = 4.10 + 1.90 = 6.00 kPa
Applications
- Selecting minimum live load values from NSCP Table 205-1 for a given occupancy type
- Computing self-weight of structural members using NSCP unit weights
- Establishing whether wind or seismic governs the lateral design of a building
- Identifying rain load as a potential design control for flat-roof structures
Misconceptions
- Confusing dead load with live load for fixed partitions — NSCP allows fixed partitions to be treated as dead load (minimum 1.0 kPa added to D for movable partitions per NSCP 205.3).
- Using the same live load value for all occupancies — always check the occupancy type against NSCP Table 205-1.
- Treating seismic load E as only horizontal — NSCP 208 includes a vertical seismic component (0.2 SDS D) in the definition of E.
- Omitting roof live load Lr when a roof is accessible — accessible roofs must include Lr = 1.0 kPa minimum.
Related Concepts
- Tributary Area and Load Distribution
- LRFD Load Combinations
- ASD Load Combinations
- Moment and Shear in Beams due to UDL
Common Exam Questions
Example
A 200 mm RC slab carries offices. What is the total area dead load if a 75 mm concrete topping and 0.30 kPa ceiling are also present? (D = 0.200×24 + 0.075×24 + 0.30 = 4.80 + 1.80 + 0.30 = 6.90 kPa; L = 2.4 kPa for offices)
Approach
Read the problem statement for occupancy type and structural system. Assign D from geometry × unit weight; assign L from NSCP Table 205-1. State units clearly (kPa for area loads, kN/m for line loads, kN for point loads).
Question Type
Load identification and quantification
Key Points To Remember
- NSCP 2015 is the governing code for load specification in the Philippines — not ASCE 7 directly, though NSCP is heavily based on it.
- Dead load D uses actual computed weights or NSCP Table 204 material unit weights.
- Live load minimum values are from NSCP Table 205-1 by occupancy type — memorise the most common values (1.9, 2.4, 4.8, 7.2 kPa).
- Wind and earthquake are lateral (horizontal) loads; they also cause net uplift and overturning moments.
- The seismic load E already includes both horizontal and vertical seismic components per NSCP 208.
- Snow load S ≈ 0 in the Philippines but appears in combination equations — do not omit it in formula recall, just set S = 0 for local problems.
- In practice, roof live load Lr and rain load R are mutually exclusive in a given combination — use the critical one.
Tributary Area and Load Distribution to Members
Once area loads (in kPa) are established, structural members must be assigned their share of those loads. The governing principle is **tributary area**: each member supports the load from the region that is geometrically closer to it than to any neighbouring member. **For a beam in a one-way or two-way slab system:** If parallel floor beams are spaced s (m) apart and carry an area load q (kPa), the tributary width of an interior beam is s (the full spacing), yielding a uniformly distributed line load: w = q × s (kN/m) An edge beam carries half the spacing: tributary width = s/2. **For a column:** An interior column at grid spacing Lx × Ly carries the area load over its tributary area: At = Lx × Ly (m²) P = q × At (kN) An edge column: At = (Lx/2) × Ly; a corner column: At = (Lx/2) × (Ly/2). **Multi-storey columns:** Axial load accumulates storey by storey. For n identical storeys: P_total = n × (q × At) **Two-way slab load distribution (yield-line / 45° method):** For beams on a two-way slab panel, the load distribution is trapezoidal on the long-span beams and triangular on the short-span beams. However, for board exam purposes at this level, uniform distribution with tributary widths is the standard approach unless the problem explicitly asks for the two-way distribution. **Self-weight of beam:** Do not forget to add the beam's own self-weight as a line dead load: w_beam = b × h × γ_c (kN/m) where b and h are the beam cross-section dimensions and γ_c = 24 kN/m³.
Examples
Beam self-weight must be added as a dead load line load — a very common board exam pitfall. The slab area load is already the net area load excluding the beam web below slab.
Scenario
Floor beams span 8 m and are spaced 3.5 m on centre. The slab carries a dead load of 5 kPa (including slab self-weight and finishes) and a live load of 3 kPa (offices). The beam cross-section is 300 mm × 600 mm (b × h). Find the total service dead and live line loads on an interior beam.
Solution
Tributary width: s = 3.5 m (interior beam) Slab dead load: wD,slab = 5 × 3.5 = 17.5 kN/m Beam self-weight: wD,beam = 0.30 × 0.60 × 24 = 4.32 kN/m Total dead: wD = 17.5 + 4.32 = 21.82 kN/m Live load: wL = 3 × 3.5 = 10.5 kN/m
In reality, NSCP 205.4 allows live load reduction for large tributary areas and multiple storeys, but board exam problems typically use full unreduced live loads unless explicitly stated otherwise.
Scenario
A typical interior column on a 6 m × 7.5 m bay grid supports 4 storeys. Each floor has a superimposed dead load of 4 kPa and live load of 2.4 kPa (offices). Ignore column self-weight. Find the total service axial load at the base.
Solution
Tributary area (interior column): At = 6.0 × 7.5 = 45 m² Per floor: PD = 4 × 45 = 180 kN; PL = 2.4 × 45 = 108 kN Total for 4 storeys: PD,total = 4 × 180 = 720 kN PL,total = 4 × 108 = 432 kN Service: P = 720 + 432 = 1152 kN
Applications
- Converting slab area loads to beam line loads for bending moment and shear calculations
- Computing column design axial loads for multi-storey buildings
- Determining reactions at supports for frame analysis
- Sizing members in preliminary design based on tributary load
Misconceptions
- Using full spacing for edge beams — edge beams carry only half the spacing as tributary width.
- Forgetting beam self-weight — always compute b×h×24 kN/m³ and add to dead load.
- Using tributary area for column = bay area (ignoring that edge/corner columns have reduced tributary areas).
- Treating live load reduction (NSCP 205.4) as mandatory — apply only when explicitly permitted or required in the problem.
Related Concepts
- Load Types and Code-Specified Minimum Values
- LRFD Load Combinations
- Reactions and Free Body Diagrams
- Shear and Moment Diagrams
Common Exam Questions
Example
Beams at 2.5 m spacing, q = 8 kPa, beam 250×500 mm. Interior beam line load: w = 8×2.5 + 0.25×0.50×24 = 20 + 3 = 23 kN/m.
Approach
Identify whether the beam is interior or edge. Multiply area load (kPa) by tributary width (m) to get line load (kN/m). Add beam self-weight if cross-section is given.
Question Type
Tributary width / line load conversion
Example
3-storey corner column, bay 5×5 m, D=3 kPa, L=2 kPa per floor. At = 2.5×2.5 = 6.25 m². PD = 3×6.25×3 = 56.25 kN; PL = 2×6.25×3 = 37.5 kN.
Approach
Compute At for interior/edge/corner as appropriate. Multiply by storey load per storey and by number of storeys.
Question Type
Multi-storey column accumulation
Key Points To Remember
- Interior beam: tributary width = full beam spacing s.
- Edge beam: tributary width = s/2 (half the spacing to the adjacent beam).
- Interior column: tributary area = Lx × Ly (product of bay dimensions in both directions).
- Edge column: one tributary dimension is halved; corner column: both are halved.
- Always add beam self-weight as an additional dead load line load.
- Tributary area concept applies to ANY type of load (D, L, W pressure, etc.).
- For multi-storey problems, accumulate column loads floor by floor.
LRFD Load Combinations (NSCP 2015)
Load and Resistance Factor Design (LRFD) amplifies the characteristic loads by **load factors** (γ > 1.0 for loads that add demand; as low as 0.9 for loads that resist demand) and reduces the nominal member strength by a **resistance factor** φ (< 1.0). The design condition is: φ Rn ≥ U (factored demand) The factored load U (also called the required strength) is the most severe of the seven combinations prescribed by NSCP 2015 Section 203.3 (identical in form to ACI 318-14 Section 5.3 and aligned with ASCE 7): 1. U = 1.4D 2. U = 1.2D + 1.6L + 0.5(Lr or S or R) 3. U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.5W) 4. U = 1.2D + 1.0W + 1.0L + 0.5(Lr or S or R) 5. U = 1.2D + 1.0E + 1.0L 6. U = 0.9D + 1.0W 7. U = 0.9D + 1.0E **Why seven combinations?** - Combo 1: Dead load only (e.g., during construction before live load is applied). - Combo 2: Governs most typical gravity floor designs (full D and L, reduced roof/rain). - Combo 3: Roof load governs (e.g., maintenance loading on roof with minimal floor live). - Combo 4: Wind accompanies gravity (wind amplifies demand; 1.0L already factored). - Combo 5: Seismic accompanies gravity. - Combo 6: Wind can cause net uplift — dead load *resists* uplift, so it is reduced to 0.9D. - Combo 7: Seismic uplift check — same logic as Combo 6. **The companion live load factor note:** In combinations 3, 4, and 5, the factor on L is technically f₁L where f₁ = 1.0 for assembly occupancies, garages, and cases where L > 4.8 kPa, and f₁ = 0.5 for all others. For board exam problems, use 1.0L unless told otherwise (conservative). **Gravity design (most common board exam scenario):** For a typical floor with only D and L, Combo 2 almost always governs: U = 1.2D + 1.6L Compare with Combo 1: U = 1.4D — governs only if 1.4D > 1.2D + 1.6L, i.e., when L < 0.125D (rarely true for occupied floors).
Examples
Always evaluate all applicable combos numerically — do not assume Combo 2 governs without checking Combo 1, especially when D is large and L is small.
Scenario
A simply supported beam of span 6 m carries service dead load wD = 12 kN/m and service live load wL = 8 kN/m (office). No wind or seismic. Find the governing factored line load wu and the factored midspan moment Mu.
Solution
Combo 1: wu = 1.4 × 12 = 16.8 kN/m Combo 2: wu = 1.2(12) + 1.6(8) = 14.4 + 12.8 = 27.2 kN/m ← GOVERNS (No Lr, R, W, or E present, so other combos do not increase demand.) Mu = wu L² / 8 = 27.2 × (6)² / 8 = 27.2 × 36 / 8 = 122.4 kN·m
Roof problems often require Combo 3 to be checked since Lr appears with factor 1.6. In this case, Combo 3 governs over Combo 2 because roof live load is significant relative to floor live load.
Scenario
A roof beam with wD = 10 kN/m, wL = 4 kN/m (floor live), and wLr = 5 kN/m (roof live load) over span 5 m. Find wu.
Solution
Combo 1: 1.4(10) = 14.0 kN/m Combo 2: 1.2(10) + 1.6(4) + 0.5(5) = 12 + 6.4 + 2.5 = 20.9 kN/m Combo 3: 1.2(10) + 1.6(5) + 1.0(4) = 12 + 8 + 4 = 24.0 kN/m ← GOVERNS Combo 4 (no wind): N/A Governing wu = 24.0 kN/m Mu = 24.0(5)²/8 = 75.0 kN·m
When wind produces tension (negative), Combo 6 (0.9D + 1.0W) checks net tension. The 0.9 factor on D is the code's way of being conservative about how much dead load is actually present to counteract uplift.
Scenario
A column carries PD = 300 kN, PL = 200 kN, and wind compression/tension PW = ±150 kN. Find the governing Pu.
Solution
Combo 1: 1.4(300) = 420 kN Combo 2: 1.2(300)+1.6(200) = 360+320 = 680 kN ← CHECK Combo 4 (W compression): 1.2(300)+1.0(150)+1.0(200) = 360+150+200 = 710 kN ← GOVERNS compression Combo 6 (W uplift): 0.9(300)+1.0(-150) = 270-150 = 120 kN (net compression, not uplift) If PW = -150 kN (tension/uplift): Combo 6: 0.9(300)+1.0(-150) = 270-150 = +120 kN compression (OK, column stays in compression) Governing Pu = 710 kN (compression from Combo 4)
Applications
- Computing Mu and Vu for RC beam design per ACI 318 / NSCP Section 400
- Computing Pu for RC column design (interaction diagram)
- Computing factored forces for steel member design per AISC 360 / NSCP Section 502
- Checking overturning stability of retaining walls and footings under wind/seismic
Misconceptions
- Assuming Combo 2 always governs — check Combo 1 for cases with heavy dead load and negligible live load (construction stages).
- Using 1.6L in Combo 4 — the factor on L in Combo 4 is 1.0L, not 1.6L.
- Adding wind or seismic to Combo 2 — Combo 2 is for gravity loads only; wind/seismic appear in Combos 3–7.
- Using LRFD factored loads (e.g., Pu = 1.2D+1.6L) with an ASD allowable capacity — this is a fundamental design error.
- Forgetting the 0.9D uplift combos (6 and 7) for light or anchor bolt design problems.
Related Concepts
- ASD Load Combinations
- Resistance (Strength Reduction) Factors φ in ACI 318
- Nominal vs. Design Strength
- Load Types and Code-Specified Minimum Values
Common Exam Questions
Example
Given D, L only: compare 1.4D vs 1.2D+1.6L. If D=20, L=5 kN/m: Combo1=28, Combo2=32 → Combo2 governs. If D=20, L=0: Combo1=28, Combo2=24 → Combo1 governs.
Approach
Evaluate all applicable combos numerically. The combination yielding the maximum demand (maximum positive moment, maximum compression, or maximum tension for uplift check) governs.
Question Type
Governing LRFD combination identification
Example
wu=24 kN/m, L=6m: Mu = 24×36/8 = 108 kN·m
Approach
Compute wu = governing combo, then apply Mu = wuL²/8 (UDL, simply supported) or other moment formula as applicable.
Question Type
Factored moment computation
Key Points To Remember
- Seven LRFD combinations per NSCP 2015 Section 203.3 — memorise all seven.
- Combo 2 (1.2D + 1.6L) governs most gravity floor design problems.
- Combo 1 (1.4D) governs only when live load is very small (< 12.5% of dead load).
- Combos 6 and 7 use 0.9D — the 0.9 factor reduces the stabilising dead load to check uplift/overturning.
- LRFD factors must be used with φ-strength (strength reduction factor); never mix LRFD loads with ASD allowable stresses.
- For seismic E in combo 5/7, E per NSCP 208 already includes both horizontal and vertical components.
- Roof terms (Lr, S, R): only the largest applies in one combination — pick the governing one.
ASD Load Combinations (NSCP 2015)
Allowable Stress Design (ASD) — also called Working Stress Design (WSD) — keeps loads at **service (unfactored) level** and instead reduces the material's allowable capacity to a fraction of its strength using a factor of safety (FS). The design condition is: f_actual ≤ F_allowable = F_nominal / FS NSCP 2015 Section 203.4 specifies the following ASD load combinations: 1. D 2. D + L 3. D + (Lr or S or R) 4. D + 0.75L + 0.75(Lr or S or R) 5a. D + (0.6W) — wind case D + (0.7E) — seismic case 6a. D + 0.75L + 0.75(0.6W) + 0.75(Lr or S or R) D + 0.75L + 0.75(0.7E) + 0.75(Lr or S or R) 7a. 0.6D + 0.6W — wind uplift 0.6D + 0.7E — seismic uplift **Note on the wind/seismic ASD factors:** - The 0.6W factor (not 1.0W) in ASD reflects that wind loads in the ASD format are already at a nominal level. NSCP aligns with ASCE 7 where wind load W at the ASD level equals 0.6 × LRFD wind load W. - Similarly, the 0.7E factor approximates 0.7 × LRFD seismic E. - In older Philippine practice (pre-NSCP 7th Edition), a one-third stress increase was used for load combinations including wind or seismic — this is no longer applicable in the current NSCP. **Comparison: ASD vs. LRFD** For gravity-only (D + L), the two approaches should yield comparable designs when properly calibrated: - ASD: Design for D + L (service) - LRFD: Design for 1.2D + 1.6L (factored), with φ typically 0.90 for flexure For a beam with wD = 10 kN/m and wL = 10 kN/m: - ASD service: w = 20 kN/m → M_service = wL²/8 - LRFD factored: wu = 1.2(10)+1.6(10) = 28 kN/m → Mu = 28L²/8 - Ratio: 28/20 = 1.4 — the LRFD system effectively applies a combined load factor of 1.4 for equal D and L.
Examples
For gravity-only ASD problems with D and L, Combo 2 (D+L) almost always governs — it is the simple sum of dead and live loads.
Scenario
A beam has service dead load wD = 14 kN/m and service live load wL = 10 kN/m, span = 7 m (simply supported). Find the governing ASD service moment.
Solution
Combo 1: w = 14 kN/m → M = 14(7)²/8 = 85.75 kN·m Combo 2: w = 14+10 = 24 kN/m → M = 24(7)²/8 = 147.0 kN·m ← GOVERNS Combo 3 (no Lr): = Combo 1 or less Governing M_ASD = 147.0 kN·m
Combo 7 with 0.6D is the critical uplift check. The 0.6 factor on D accounts for uncertainty in the dead load being fully present to resist uplift — the code conservatively assumes only 60% of D provides resistance.
Scenario
A footing carries PD = 400 kN, PL = 200 kN, and lateral wind load creating an overturning moment equivalent to P_wind = ±100 kN vertical component. Check ASD Combos 5 and 7.
Solution
Combo 5 (D + 0.6W with compression): P = 400 + 200 + 0.6(100) = 660 kN (max compression) Combo 5 (D + 0.6W with tension): P = 400 + 200 + 0.6(-100) = 540 kN (still compression) Combo 7 (0.6D + 0.6W uplift): P = 0.6(400) + 0.6(-100) = 240 - 60 = 180 kN (compression — no uplift) If wind tension were -450 kN: Combo 7: 0.6(400) + 0.6(-450) = 240 - 270 = -30 kN ← NET UPLIFT → design for tension/anchor bolts
Applications
- Computing allowable stress design moments and shears for timber or steel members
- Checking foundation bearing capacity at service load level
- Structural analysis for steel connections using ASD per AISC 360 ASD provisions
- Comparing ASD and LRFD design results for calibration
Misconceptions
- Using ASD Combo 5 as D + W (unfactored wind) — the ASD wind factor is 0.6W, not 1.0W.
- Applying ASD combinations with LRFD (φ-factor) capacities — the two design philosophies are mutually exclusive.
- Using the old one-third increase in allowable stress for wind/seismic — this provision is NOT in NSCP 2015.
- Confusion between 0.6D (ASD uplift) and 0.9D (LRFD uplift) — different codes, different values.
Related Concepts
- LRFD Load Combinations
- Factor of Safety in ASD
- Allowable Bending Stress in Steel (AISC 360 ASD)
- Modular Ratio Method for RC (WSD)
Common Exam Questions
Example
wD=8, wL=6 kN/m: ASD w=14 kN/m; LRFD wu=1.2(8)+1.6(6)=9.6+9.6=19.2 kN/m. LRFD/ASD = 1.37.
Approach
Compute both service (ASD D+L) and factored (LRFD 1.2D+1.6L) demands. Note that LRFD demand is always larger but is checked against φRn, while ASD demand is smaller and checked against Rn/FS.
Question Type
ASD vs LRFD comparison
Example
D=500 kN, W=-700 kN (tension): Combo 7 = 0.6(500)+0.6(-700) = 300-420 = -120 kN → net uplift of 120 kN governs.
Approach
Use Combo 7: 0.6D ± 0.6W (or 0.7E). If the result is negative (tension), uplift governs and anchor bolts must be designed.
Question Type
Wind/seismic ASD uplift check
Key Points To Remember
- ASD uses service (unfactored) loads; the safety margin is in the allowable capacity.
- Seven ASD combinations per NSCP 2015 Section 203.4.
- Combo 2 (D + L) governs most gravity ASD floor designs.
- The ASD wind factor is 0.6W (not 1.0W) and seismic is 0.7E.
- Combo 7 uses 0.6D for uplift — same logic as LRFD Combo 6 (0.9D).
- Do NOT apply the old one-third stress increase with NSCP 2015 combinations — it is obsolete.
- ASD and LRFD must not be mixed — use one philosophy with its matching capacity framework.
Applying Load Combinations — Step-by-Step Procedure
The board exam will present a structural member with given load values and ask for the governing design force effect (moment, shear, or axial). The systematic procedure ensures no combination is overlooked and the correct design value is identified. **Step 1 — Identify applicable load types.** From the problem statement, determine which of D, L, Lr, W, E, R are present and non-zero. Mark the rest as zero. **Step 2 — Compute member loads for each load type.** Using tributary area (for distributed loads) or direct assignment (for point loads), convert area loads to line loads or point loads. **Step 3 — Evaluate all applicable LRFD or ASD combinations.** Write out each combination numerically. For LRFD, this means computing wu (or Pu, Vu) for each combination. For ASD, compute w (or P, V) for each. **Step 4 — Identify the governing combination.** The combination producing the maximum demand (or most critical tension/uplift) governs. **Step 5 — Compute the force effect.** Using the governing factored (LRFD) or service (ASD) load, apply statics: - Simply supported UDL: Mmax = wL²/8, Vmax = wL/2 - Cantilever UDL: Mmax = wL²/2, Vmax = wL - Simply supported point load at midspan: Mmax = PL/4 **Step 6 — Match with design capacity.** For LRFD: Mu ≤ φMn; Vu ≤ φVn; Pu ≤ φPn For ASD: M ≤ Mall = Mn/FS; V ≤ Vall; P ≤ Pall This six-step procedure covers the complete load-to-design chain tested in the board exam.
Examples
Step 2 is where most mistakes occur — always add beam self-weight and all dead components before factoring. Combo 2 governs as expected for a typical floor with significant live load.
Scenario
COMPLETE BOARD-STYLE PROBLEM: A simply supported floor beam spans 5 m with beam spacing of 2.5 m. Slab dead load = 4.5 kPa, superimposed dead = 1.2 kPa, live load = 3.6 kPa (retail). Beam self-weight = 0.8 kN/m. No wind or seismic. Using LRFD, find Mu.
Solution
Step 1: Loads present: D, L only (no W, E, Lr, R) Step 2: Line loads on beam: wD,slab = (4.5+1.2) × 2.5 = 5.7 × 2.5 = 14.25 kN/m wD,self = 0.80 kN/m wD = 14.25 + 0.80 = 15.05 kN/m wL = 3.6 × 2.5 = 9.0 kN/m Step 3: LRFD Combos: Combo 1: wu = 1.4(15.05) = 21.07 kN/m Combo 2: wu = 1.2(15.05)+1.6(9.0) = 18.06+14.40 = 32.46 kN/m ← GOVERNS (Combos 3–7: no Lr, W, or E → not governing) Step 4: Governing wu = 32.46 kN/m Step 5: Mu = wu L²/8 = 32.46 × (5)²/8 = 32.46 × 25/8 = 101.4 kN·m Answer: Mu = 101.4 kN·m
Applications
- Full structural design workflow: from occupancy loads to Mu/Vu for beam sizing
- Column design: from tributary area to Pu for interaction diagram check
- Foundation design: from column loads to bearing pressure check
- Frame analysis: input factored loads for lateral drift and P-delta checks
Misconceptions
- Computing Mu from service loads instead of factored loads in LRFD problems.
- Using Mu = wL²/8 for cantilever beams — cantilever Mmax = wL²/2.
- Skipping beam self-weight in multi-step problems.
- Evaluating only Combo 2 without checking Combo 1 (could govern for construction loads).
Related Concepts
- LRFD Load Combinations
- ASD Load Combinations
- Tributary Area and Load Distribution
- Shear and Moment Diagrams for Beams
Common Exam Questions
Example
Standard board exam format: 'A beam with wD=__, wL=__, span=__ m. Find Mu (LRFD).' Apply Combo 2 = 1.2D+1.6L, then Mu = wuL²/8.
Approach
Follow the 6-step procedure: identify loads, compute line loads (including self-weight), evaluate all combos, find governing wu, compute Mu = wuL²/8 or appropriate formula.
Question Type
Complete beam design problem (multi-step)
Key Points To Remember
- Always evaluate ALL applicable combinations numerically — never guess which one governs.
- For gravity-only (D+L, no wind/seismic): LRFD Combo 2 typically governs; ASD Combo 2 governs.
- For wind/seismic: also evaluate uplift combos (LRFD 6/7; ASD 7).
- Convert area loads to line loads BEFORE evaluating combinations (w = q × s).
- Use the correct moment formula for the beam support conditions given in the problem.
- Maintain consistent units throughout: kN and m → kN·m for moments.
Practice Problems
Retail occupancy has one of the higher live loads (4.8 kPa per NSCP Table 205-1). Note that L = 4.8 kPa is exactly at the threshold where f₁ = 1.0 (no companion factor reduction), confirming the 1.6 factor in Combo 2. Beam self-weight contributes ~22% of total dead load — never neglect it.
Problem
Problem 1 (Tributary Load — Beam): A one-way slab system has floor beams spaced 3.0 m on centre spanning 7.5 m. The slab carries: (a) slab self-weight = 3.84 kPa, (b) floor finish = 0.75 kPa, (c) live load = 4.8 kPa (retail). The beam cross-section is 300 mm wide × 550 mm deep (overall). For an interior beam using LRFD, find: (i) the total dead and live line loads, (ii) the governing factored line load wu, and (iii) the maximum factored moment Mu.
Solution
(i) Line loads: wD,slab = (3.84 + 0.75) × 3.0 = 4.59 × 3.0 = 13.77 kN/m wD,beam = 0.300 × 0.550 × 24 = 3.96 kN/m wD (total) = 13.77 + 3.96 = 17.73 kN/m wL = 4.8 × 3.0 = 14.40 kN/m (ii) LRFD Combinations (no Lr, W, E): Combo 1: wu = 1.4(17.73) = 24.82 kN/m Combo 2: wu = 1.2(17.73) + 1.6(14.40) = 21.28 + 23.04 = 44.32 kN/m ← GOVERNS (iii) Mu = wu L² / 8 = 44.32 × (7.5)² / 8 = 44.32 × 56.25 / 8 = 311.3 kN·m Final Answer: wD = 17.73 kN/m; wL = 14.40 kN/m; wu = 44.32 kN/m; Mu = 311.3 kN·m
LRFD Pu / ASD P = 1524.6 / 1138.5 = 1.339. This ratio reflects the combined effect of the 1.2 and 1.6 load factors weighted by the D:L ratio. The ratio approaches 1.4 when D dominates and approaches 1.6 when L dominates.
Problem
Problem 2 (Column Load — LRFD vs ASD): A 5-storey building has typical floor bays of 5.5 m × 6.0 m. An interior column carries: dead load 4.5 kPa per floor, live load 2.4 kPa per floor (office). Ignore column self-weight. Find: (i) PD and PL at the base column, (ii) the governing LRFD Pu, and (iii) the ASD service P.
Solution
(i) Tributary area (interior column): At = 5.5 × 6.0 = 33.0 m² Per floor: PD,floor = 4.5 × 33.0 = 148.5 kN; PL,floor = 2.4 × 33.0 = 79.2 kN For 5 storeys: PD = 5 × 148.5 = 742.5 kN PL = 5 × 79.2 = 396.0 kN (ii) LRFD Combinations: Combo 1: Pu = 1.4(742.5) = 1039.5 kN Combo 2: Pu = 1.2(742.5) + 1.6(396.0) = 891.0 + 633.6 = 1524.6 kN ← GOVERNS (iii) ASD: P_service = PD + PL = 742.5 + 396.0 = 1138.5 kN Final Answers: PD = 742.5 kN; PL = 396.0 kN; Pu(LRFD) = 1524.6 kN; P(ASD) = 1138.5 kN
The systematic evaluation of all seven combinations is essential. Combo 2 governs compression. Combo 6 is the uplift check — in this problem dead load is sufficient to prevent uplift, but the example demonstrates the methodology. Board exam problems sometimes add a follow-up: 'What minimum dead load prevents uplift?' → set Combo 6 ≥ 0: 0.9PD ≥ |PW| → PD ≥ |PW|/0.9.
Problem
Problem 3 (Wind + Gravity — LRFD All Combos): A column carries PD = 280 kN, PL = 160 kN, PLr = 30 kN (roof live), and wind-induced PW = ±90 kN. Evaluate ALL seven LRFD combinations and identify: (i) the governing compression Pu, and (ii) whether net tension (uplift) occurs.
Solution
Combo 1: Pu = 1.4(280) = 392.0 kN Combo 2: Pu = 1.2(280)+1.6(160)+0.5(30) = 336+256+15 = 607.0 kN Combo 3 (+): Pu = 1.2(280)+1.6(30)+1.0(160) = 336+48+160 = 544.0 kN Combo 4 (W comp): Pu = 1.2(280)+1.0(90)+1.0(160)+0.5(30) = 336+90+160+15 = 601.0 kN Combo 4 (W tension): Pu = 1.2(280)+1.0(-90)+1.0(160)+0.5(30) = 336-90+160+15 = 421.0 kN Combo 5 (no E given): N/A Combo 6 (W comp): Pu = 0.9(280)+1.0(90) = 252+90 = 342.0 kN Combo 6 (W tension): Pu = 0.9(280)+1.0(-90) = 252-90 = +162.0 kN (still compression) Combo 7: N/A (no E) (i) Governing compression: Combo 2 → Pu = 607.0 kN (ii) No net tension — minimum Pu = +162.0 kN (compression). No uplift occurs. Note: If PW were -350 kN: Combo 6 = 0.9(280)+(-350) = 252-350 = -98 kN → 98 kN NET TENSION → uplift governs.
For roof beams where Lr is the primary live load and there is no floor live load, ASD Combo 3 (D + Lr) governs. Note that ASD Combo 4 includes the 0.75 reduction factor on Lr, making it less critical than Combo 3 for this load ratio.
Problem
Problem 4 (ASD — Roof System): A roof beam on a 6 m span (simply supported) has wD = 9 kN/m and wLr = 5 kN/m (no floor live load). Using ASD load combinations, find the governing service moment M.
Solution
Applicable ASD Combos (no L, W, E present): Combo 1: w = 9 kN/m → M = 9(6)²/8 = 40.5 kN·m Combo 2: w = 9+0 = 9 kN/m (L=0) → M = 40.5 kN·m Combo 3: w = 9+5 = 14 kN/m → M = 14(6)²/8 = 63.0 kN·m ← GOVERNS Combo 4: w = 9+0.75(0)+0.75(5) = 9+3.75 = 12.75 kN/m → M = 57.4 kN·m Governing M (ASD) = 63.0 kN·m (Combo 3: D + Lr) For comparison, LRFD Combo 3: wu = 1.2(9)+1.6(5) = 10.8+8.0 = 18.8 kN/m → Mu = 18.8(36)/8 = 84.6 kN·m
Seismic problems require evaluating both seismic compression (additive to gravity) and seismic uplift (subtractive from gravity). Combo 5 governs compression; Combo 7 governs potential uplift. The 0.9D factor in Combo 7 is the NSCP's conservative acknowledgment that not all dead load may be present when seismic uplift occurs.
Problem
Problem 5 (Comprehensive — Mixed Loads with Seismic): A shear wall carries at its base: D = 500 kN (compression), L = 200 kN (compression), and seismic load E = ±300 kN (vertical). Evaluate LRFD Combos 5 and 7 and determine the governing axial force for design.
Solution
Seismic load E = ±300 kN (positive = additional compression; negative = uplift/tension) Combo 5 (seismic compression): Pu = 1.2D+1.0E+1.0L = 1.2(500)+1.0(300)+1.0(200) = 600+300+200 = 1100 kN (compression) Combo 5 (seismic tension): Pu = 1.2(500)+1.0(-300)+1.0(200) = 600-300+200 = 500 kN (compression) Combo 7 (seismic uplift): Pu = 0.9D+1.0E = 0.9(500)+1.0(-300) = 450-300 = +150 kN (compression — no uplift) Combo 2 (gravity): Pu = 1.2(500)+1.6(200) = 600+320 = 920 kN Governing: - Maximum compression: Combo 5 (E comp) → Pu = 1100 kN (governs for compression design) - Minimum force: Combo 7 (E uplift) → Pu = +150 kN (compression; no uplift here) If E were -500 kN: Combo 7 = 0.9(500)+(-500) = 450-500 = -50 kN → 50 kN NET TENSION → design wall/foundation for tension.
Exam Preparation Tips
- Memorise all seven LRFD and seven ASD load combinations from NSCP 2015 Section 203 — write them from memory on your scratch paper at the start of the exam.
- For gravity-only problems (D + L, no wind/seismic), LRFD Combo 2 (1.2D + 1.6L) almost always governs — verify by checking Combo 1 (1.4D) numerically.
- Never forget beam self-weight: w_self = b × h × 24 kN/m³. It is a dead load that must be added BEFORE factoring.
- Memorise tributary area rules: interior beam = full spacing s; edge beam = s/2; interior column = Lx × Ly; edge column = (Lx/2) × Ly; corner column = (Lx/2) × (Ly/2).
- For the uplift/overturning check, use LRFD Combo 6 (0.9D + 1.0W) or Combo 7 (0.9D + 1.0E); for ASD, use Combo 7 (0.6D + 0.6W or 0.6D + 0.7E).
- Key NSCP live load values to memorise: residential 1.9 kPa, offices 2.4 kPa, classrooms 1.9–2.9 kPa, retail 4.8 kPa, heavy storage 7.2 kPa.
- LRFD vs ASD: LRFD uses φRn ≥ U (factored loads); ASD uses Rn/FS ≥ (service loads). Never cross-apply — factored LRFD loads with ASD allowable capacity is a fatal error.
- The LRFD wind/seismic companion factor on L in Combos 3–5 is 1.0L for assembly, garages, and L > 4.8 kPa; 0.5L otherwise — but most exam problems use 1.0L as the conservative value.
- For multi-storey columns, simply multiply per-floor loads by the number of floors (assuming identical floors) — live load reduction per NSCP 205.4 is applied only if explicitly stated.
- Board exam problems often include unit conversion traps: always convert kPa × m = kN/m and kPa × m² = kN consistently.
- When evaluating LRFD Combo 2 for a beam with both floor live (L) and roof live (Lr): verify whether Combo 3 (1.2D + 1.6Lr + 1.0L) exceeds Combo 2 — roof problems frequently test this.
- Practice the complete 6-step load-to-design procedure until it is automatic: (1) identify load types, (2) compute member loads, (3) evaluate all combos, (4) identify governing combo, (5) compute force effect, (6) check against capacity.
In summary
The identification, quantification, and combination of structural loads is the foundation upon which all structural design rests. As a Philippine civil engineer, you must be thoroughly familiar with NSCP 2015's load taxonomy (D, L, Lr, W, E, R), the tributary area principle for converting area loads into beam line loads and column axial loads, and the seven LRFD and seven ASD load combinations that determine the governing design force effects. The most frequently tested scenarios in the PRC licensure examination are: (1) computing factored beam moments using LRFD Combo 2 (1.2D + 1.6L), (2) accumulating column loads over multiple storeys using tributary area, and (3) checking the uplift/overturning combinations (LRFD Combo 6: 0.9D + 1.0W; ASD Combo 7: 0.6D + 0.6W). Mastery of these three scenarios, combined with the ability to add beam self-weight correctly and select tributary widths for interior vs. edge members, will allow you to solve the vast majority of load combination problems in the examination. As a future licensed civil engineer practising under RA 544, you carry the professional responsibility to apply these code provisions correctly — not just for examination success, but to ensure that the structures you design can safely serve the Filipino public throughout their design life, especially given the Philippines' extreme exposure to typhoons and earthquakes. The NSCP load combinations are the first line of defence in that responsibility.
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