CELE Structural Theory & Analysis — Loads and Load Combinations (NSCP)Study Notes
Complete study notes for Loads and Load Combinations (NSCP), written for CELE aspirants. Unlike generic notes, these focus on what Professional Regulation Commission (PRC) — Board of Civil Engineering actually tests in the CELE Structural Theory & Analysis section: high-yield concepts, common question types, and the worked examples that match recent exam patterns.
Exam context
On the CELE 2026, the Structural Theory & Analysis subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Loads and Load Combinations (NSCP) lands at position 6th out of 6 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Structural Theory & Analysis on a typical CELE paper.
Loads and Load Combinations (NSCP) - Study Notes
Before any structural member can be designed, engineers must identify, quantify, and combine all loads acting on the structure in patterns prescribed by the National Structural Code of the Philippines (NSCP 2015). This chapter covers the fundamental load types, how loads are distributed to members using tributary area concepts, and the critical LRFD (Load and Resistance Factor Design) and ASD (Allowable Stress Design) load combinations — essential knowledge for the PRC Civil Engineer Licensure Examination. Understanding load combinations is the foundation linking structural analysis to concrete and steel design; incorrect load application invalidates all downstream calculations.
Summary
Loads and load combinations form the foundation of all structural design. The engineer must: 1. **Identify all applicable loads:** In the Philippines, always include seismic load (E) because the country is in a high-seismic region. Dead load (D), live load (L), roof live load (Lr), wind (W), and rain (R) are standard. 2. **Calculate tributary areas:** Correctly route floor and roof area loads to beams (line loads w = q × s) and columns (axial loads P = q × A_tributary). Sketch tributary diagrams to avoid errors. 3. **Apply NSCP load combinations:** LRFD is the primary philosophy in NSCP 2015. Seven combinations must be checked; combo 2 (1.2D + 1.6L) typically governs gravity design, while combos 5–7 (with E or W) govern lateral design and uplift. 4. **Check all cases:** Use combo 2 for moment/shear, combos 5–7 for lateral/uplift, and perform separate serviceability checks (deflection, vibration) with unfactored service loads. 5. **Context in the Philippines:** High seismic activity makes E (seismic) the controlling lateral load in most cases; wind (W) is secondary except in typhoon-exposed coastal zones. Live load reduction applies for large tributary areas (warehouses, parking), reducing design loads significantly. Mastery of load combinations directly enables safe and economical design of concrete (ACI 318), steel (AISC 360), and timber structures. Errors in load application propagate to all subsequent design calculations, making this chapter critical for the PRC Civil Engineer Licensure Examination.
Sections
Structural loads are classified by their permanence and predictability. The NSCP recognizes six primary load types that form the basis of all load combinations: **Dead Load (D):** Permanent self-weight of the structure including: - Structural frame (steel, concrete, timber) - Permanent finishes (flooring, ceilings, walls) - Fixed equipment (mechanical systems, permanent fixtures) - Dead loads are calculated from material unit weights; concrete ≈ 23.6 kN/m³, steel ≈ 77 kN/m³, wood varies by species - Dead loads are deterministic — they do not vary with occupancy **Live Load (L):** Temporary occupancy and use loads including: - People, furniture, movable equipment - Code-prescribed minimum values vary by occupancy type (Table 1, NSCP 2015): - Residential: 1.9 kPa - Office: 2.4 kPa - Classroom: 1.9–2.9 kPa - Warehouse: 3.6–4.8 kPa - Assembly (concentrated): 4.8 kPa - Live loads are probabilistic and depend on building use - Often reduced for column calculations if floor area exceeds thresholds **Roof Live Load (Lr):** Maintenance, repair, and temporary loads on roof surfaces: - Typically 0.5–1.0 kPa depending on roof slope and access - Used in combinations where it may govern over floor live load **Wind Load (W):** Lateral loads from basic wind speed, exposure category, height, and shape effects: - Determined from NSCP wind pressure equation: q = 0.613 × V² (kPa), where V is basic wind speed (m/s) in 10-minute average - Highly variable across the Philippines; coastal and highland areas experience typhoon loads - Often governs lateral design of light structures and tall buildings **Earthquake (Seismic) Load (E):** Inertial forces from ground motion: - The Philippines is classified as a seismic zone; earthquake loads often govern lateral design - Determined by response spectrum analysis or equivalent lateral force method (NSCP 2015 Section 4.2) - Base shear formula: V = C × W (C = seismic response coefficient, W = total weight) - Often the critical design load for high-rise buildings and bridges in the PH **Rain (R) and Snow (S):** Precipitation loads: - Rain load R is often secondary in the Philippines due to drainage design - Snow is negligible at low to mid elevations; critical only in high-altitude projects (e.g., Cordillera region) - Included in roof load combinations All load types must be identified before applying NSCP load combinations. The engineer's judgment determines which combinations apply to a given project.
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1. Load Types and Definitions (NSCP 2015 Section 2.1)
Examples
Example 1.1 — Identifying load types in a four-story residential building
Problem
A residential condominium building in Metro Manila has concrete floors with live load 1.9 kPa (per NSCP Table 1), a flat roof accessible for maintenance, and is exposed to typhoon winds (basic wind speed V = 250 km/h ≈ 69.4 m/s). Identify all applicable loads.
Solution
Applicable loads: - Dead load (D): Self-weight of concrete structure, finishes, and fixed MEP equipment — must be calculated from unit weights and dimensions. - Live load (L): L = 1.9 kPa for residential occupancy (NSCP Table 1). - Roof live load (Lr): Accessible flat roof → Lr ≈ 0.5–1.0 kPa for maintenance. - Wind load (W): V = 69.4 m/s; wind pressure q = 0.613 × (69.4)² ≈ 2.95 kPa (simplified; detailed calculation requires exposure category and shape coefficients). - Seismic load (E): PH is seismic zone (NSCP Figure 3); earthquake load must be included. - Rain (R): Included in roof combinations; assumed drained (secondary). Conclusion: All six load types apply; the governing load combination depends on member location and direction.
Example 1.2 — Calculating dead load of a reinforced concrete beam
Problem
A rectangular RC beam has cross-section 300 mm × 500 mm and spans 6 m. Calculate the dead load line load in kN/m.
Solution
Unit weight of reinforced concrete ≈ 23.6 kN/m³ (NSCP standard). Volume per meter length: V = 0.3 m × 0.5 m × 1 m = 0.15 m³/m. Dead load line load: w_D = 23.6 × 0.15 = 3.54 kN/m. Note: This is self-weight only; superimposed dead load (finishes, mechanical) must be added separately.
Key Points
- Dead load (D) is permanent and deterministic; calculated from material unit weights and fixed equipment.
- Live load (L) varies by occupancy; NSCP prescribes minimum design values (Table 1).
- Roof live load (Lr) typically 0.5–1.0 kPa; used when it controls over floor live load.
- Wind load (W) derived from basic wind speed and exposure; critical for lateral design in the seismic Philippines.
- Earthquake load (E) governs in the seismic Philippines; determined by response spectrum or equivalent lateral force.
- Rain (R) and snow (S) usually secondary in most Philippine applications.
- All applicable loads must be identified and quantified before load combinations are applied.
Loads are distributed to structural members through the concept of tributary area — the region of the floor or roof that a member is responsible for supporting. Correct tributary area calculation is fundamental to obtaining accurate design loads. **Tributary Area for Beams (One-Way Systems):** For a beam in a one-way slab system (slab spanning perpendicular to the beam): - The tributary width = spacing between adjacent parallel beams - Tributary line load on beam: w = q × s, where q is area load (kPa) and s is beam spacing (m) - Interior beams carry full spacing width - Edge/boundary beams typically carry half the spacing (cantilever slab effect often ignored for simplicity) **Tributary Area for Columns:** For a column supporting a floor system: - The tributary area is the region bounded by lines equidistant from the column to the next nearest column (or wall) - For rectangular grid: Tributary area = (half the span in X-direction) × (half the span in Y-direction) - Column axial load: P = q × A_tributary - Interior columns carry larger tributary areas than edge columns, which carry larger areas than corner columns **Example Tributary Area Patterns:** - Interior rectangular grid (all directions equal): A_t = (s_x / 2) × (s_y / 2) × (number of stories above) - Edge column (one side open): A_t = (s_x / 2) × s_y - Corner column (two sides open): A_t = (s_x / 2) × (s_y / 2) Errors in tributary area calculation are among the most common sources of incorrect design loads. Carefully sketch tributary boundaries before calculating.
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2. Tributary Area and Load Distribution
Examples
Example 2.1 — Tributary area for interior beam in regular grid
Problem
A floor system has beams spaced 4 m on center in both directions forming a rectangular grid. A slab imposes q = 5 kPa (dead + live). Find the line load on a typical interior beam.
Solution
Interior beam tributary width = spacing of parallel beams = 4 m. Line load: w = q × s = 5 kPa × 4 m = 20 kN/m. Note: If the slab is one-way (spanning in only one direction), verify that the beam is perpendicular to the span direction. If two-way, use tributary width method as shown.
Example 2.2 — Tributary area for interior column in regular grid
Problem
A two-story office building has columns on a 6 m × 7 m rectangular grid. Each floor has a total combined load (dead + live) of q = 8 kPa. An interior column supports both floors. Calculate the axial load on the column (excluding the column's own weight for now).
Solution
Interior column tributary area per floor: A_t = (6/2) × (7/2) = 3 × 3.5 = 10.5 m². Load from one floor: P_floor = 8 kPa × 10.5 m² = 84 kN. Load from two floors: P_total = 84 × 2 = 168 kN (plus column self-weight). Alternatively, for a 3-story building, P = 8 × 10.5 × 3 = 252 kN (loads from all floors above the column).
Example 2.3 — Tributary area for edge and corner columns
Problem
In the same 6 m × 7 m grid, find tributary areas for (a) an edge column (4-sided perimeter, one side open to adjacent building wing), and (b) a corner column.
Solution
(a) Edge column (one side open): A_t = (6/2) × 7 = 3 × 7 = 21 m² per floor. (b) Corner column (two sides open): A_t = (6/2) × (7/2) = 10.5 m² per floor. Note: Edge columns carry twice the tributary area of interior columns; corner columns equal interior. This is why perimeter columns often require larger sections.
Key Points
- Tributary area is the region a member supports; bounded by lines equidistant from that member to adjacent members.
- For beams: tributary line load w = q × s (area load × beam spacing).
- For columns: tributary axial load P = q × A_tributary (area load × tributary floor area).
- Interior members support larger tributary areas than edge and corner members.
- Tributary area must account for all floors above the member.
- Incorrect tributary area is a frequent source of design errors — always sketch the diagram.
Load and Resistance Factor Design (LRFD) is the primary design philosophy in modern NSCP. Under LRFD: - Loads are multiplied by safety factors (≥ 1.0) to obtain factored (ultimate) loads - Member nominal strength is multiplied by a capacity reduction factor φ (< 1.0) - Design is satisfied when: Required strength ≤ φ × Nominal strength - This approach accounts for randomness in both loads and material properties **Principal LRFD Load Combinations (NSCP 2015 Section 2.3.2):** The engineer must check the following combinations and design for the most severe (highest demand): **Gravity Combinations (most common):** 1. U = 1.4D 2. U = 1.2D + 1.6L + 0.5(Lr or R) 3. U = 1.2D + 1.6(Lr or R) + (f₁L or 0.5W) 4. U = 1.2D + 1.0W + f₁L + 0.5(Lr or R) 5. U = 1.2D + 1.0E + f₁L **Uplift/Overturning Combinations (light structures or high wind/seismic):** 6. U = 0.9D + 1.0W 7. U = 0.9D + 1.0E **Notes on Companion Live-Load Factor f₁:** - NSCP uses f₁ = 1.0 for assembly, garages, and occupancies where L > 4.8 kPa - NSCP uses f₁ = 0.5 for all other occupancies (residential, office, storage, etc.) - Review materials and textbooks often assume f₁ = 1.0 conservatively; verify with the specific standard version **Practical Guidance for Gravity Loads:** For typical buildings (residential, office, warehouses), combination **2.2 (U = 1.2D + 1.6L + 0.5(Lr or R))** governs flexure and shear in most cases. The 1.2 factor on D (vs. 1.4 in combination 1) is justified because L and Lr rarely achieve their full design values simultaneously. **When Combinations 6–7 (0.9D) Govern:** Combinations with 0.9D are critical when: - Wind or seismic forces create uplift or overturning (e.g., columns in tension, anchor bolts) - Dead load opposes and resists the wind/seismic effect - Light structures (low D relative to W or E) are common in the Philippines **Load Combination Selection Algorithm:** 1. Calculate load magnitudes (D, L, Lr, W, E, R) 2. For each applicable combination, compute U 3. Identify the combination yielding maximum U 4. Use that U in member design with appropriate φ factors For seismic-prone structures in the Philippines, combinations 5 and 7 (with E) must always be checked.
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3. LRFD Load Combinations (NSCP 2015 Section 2.3)
Examples
Example 3.1 — Factored load on a residential beam (LRFD)
Problem
A floor beam in a residential building spans L = 6 m between supports with spacing s = 4 m. Slab loads: dead load q_D = 4 kPa, live load q_L = 1.9 kPa (residential). Find the factored design load w_u and design moment M_u for simply supported conditions.
Solution
Step 1: Convert area loads to line loads. w_D = q_D × s = 4 kPa × 4 m = 16 kN/m w_L = q_L × s = 1.9 kPa × 4 m = 7.6 kN/m Step 2: Apply LRFD combinations (for residential, assume f₁ = 0.5). Combo 1: U₁ = 1.4(16) = 22.4 kN/m Combo 2: U₂ = 1.2(16) + 1.6(7.6) = 19.2 + 12.16 = 31.36 kN/m ← Governs Combo 3: U₃ = 1.2(16) + 1.6(0) + 0.5(7.6) = 19.2 + 3.8 = 23 kN/m (if roof load applies) Combos 4–5: (Wind/seismic not significant for this example) Combos 6–7: 0.9D + 1.0W (or E) — wind/seismic load needed; usually governs lateral design Governing factored load: w_u = 31.36 kN/m Step 3: Calculate design moment (simply supported). M_u = w_u × L² / 8 = 31.36 × (6)² / 8 = 31.36 × 36 / 8 = 141.12 kN·m Step 4: Use M_u in concrete or steel design provisions with φ = 0.9 (flexure, typical).
Example 3.2 — Factored column load (LRFD, multiple stories)
Problem
A four-story office building has an interior column with tributary area A_t = 4 m × 4.5 m = 18 m² per floor. Each floor has dead load q_D = 5 kPa and live load q_L = 2.4 kPa (office). The column is 4 m tall per story. Find the factored axial load P_u at the base (LRFD combo 2).
Solution
Step 1: Calculate service loads per floor. P_D_floor = q_D × A_t = 5 kPa × 18 m² = 90 kN per floor P_L_floor = q_L × A_t = 2.4 kPa × 18 m² = 43.2 kN per floor Step 2: Sum loads from all 4 floors (base supports all floors above). P_D_total = 90 × 4 = 360 kN P_L_total = 43.2 × 4 = 172.8 kN Note: Many designs also include a live-load reduction factor (NSCP Section 2.5) for columns supporting large areas: L_reduced = L × (0.25 + 15/√A_t). For A_t = 18 m², reduction ≈ 0.7 → L_reduced ≈ 120.96 kN. Check local NSCP version. Step 3: Apply LRFD combo 2 (with f₁ = 0.5 for office). P_u = 1.2 × P_D + 1.6 × P_L = 1.2(360) + 1.6(172.8) = 432 + 276.48 = 708.48 kN Alternatively, if live-load reduction is applied: P_u = 1.2(360) + 1.6(120.96) = 432 + 193.54 = 625.54 kN Step 4: Use P_u in column design with appropriate φ (e.g., φ = 0.65–0.75 for RC columns, NSCP Section 3.5).
Example 3.3 — Uplift combination (0.9D + 1.0W) governing a light frame member
Problem
A roof truss member in a typhoon-prone coastal zone carries dead load P_D = 50 kN (compression). Wind suction creates a tension load P_W = 80 kN. Check if combo 6 (0.9D + 1.0W) governs, and find the required strength.
Solution
Step 1: Apply combo 2 (gravity): U₂ = 1.2(50) = 60 kN (compression). Step 2: Apply combo 6 (uplift): U₆ = 0.9(50) − 1.0(80) = 45 − 80 = −35 kN (tension). Note: The negative sign indicates the result is tension (opposite to assumed compression direction). The magnitude is |U₆| = 35 kN tension. Step 3: Compare: Combo 2 gives 60 kN compression; combo 6 gives 35 kN tension. Both critical; member must resist 60 kN compression and 35 kN tension (or be designed as pinned with appropriate provisions). Conclusion: In light structures exposed to high wind (like trusses in the Philippines), the 0.9D combination often controls tension members.
Key Points
- LRFD multiplies loads by factors (≥ 1.0) and uses φ factors on nominal strength.
- Seven principal combinations must be checked; the engineer designs for the most severe (highest U).
- Combination 2 (1.2D + 1.6L) governs most gravity design in typical buildings.
- Combinations 6–7 with 0.9D catch uplift/overturning and govern wind/seismic-critical members.
- Companion live-load factor f₁ is 1.0 for assembly and high-occupancy; 0.5 for residential/office.
- In the Philippines, seismic combinations (5 and 7) are always critical due to high seismic risk.
- Most severe combination must be used consistently with matching φ factors in design provisions.
Allowable Stress Design (ASD) is an alternative design philosophy less commonly used in modern practice but still required in some codes and for retrofit projects. Under ASD: - Loads are NOT factored (kept at service/nominal values) - Member nominal strength is divided by a global safety factor FS (typically 1.5–3.5 depending on material and stress state) - Design is satisfied when: Required stress ≤ Allowable stress = Yield stress / FS - This approach is simpler to conceptualize but less rigorous in treating load and strength uncertainty **Principal ASD Load Combinations (NSCP 2015 Section 2.4.2):** All of the following must be checked; design for the most severe combination: **Gravity Combinations:** 1. S = D (dead load alone) 2. S = D + L (dead plus live) 3. S = D + (Lr or R) (dead plus roof/rain) 4. S = D + 0.75L + 0.75(Lr or R) (reduced concurrent live and roof) **Wind and Seismic Combinations:** 5. S = D + 0.6W (or D + 0.7E) — reduced wind/seismic 6. S = D + 0.75L + 0.75(0.6W) + 0.75(Lr or R) [or substitute 0.7E for 0.6W] 7. S = 0.6D + 0.6W (or 0.6D + 0.7E) — uplift/overturning (low dead, high wind/seismic) **Comparison with LRFD:** - ASD combinations are generally simpler (no complex factors) - LRFD is more precise in reflecting actual failure probabilities - LRFD is the preferred philosophy in modern NSCP, AISC 360, and ACI 318 - However, older structures and some retrofit projects use ASD **When to Use ASD:** - If the specific design code (e.g., NSCP 2010 vs. 2015) prescribes ASD - If designing for allowable stresses in timber or masonry (where ASD is standard) - For comparison and understanding of design margins **Key Difference from LRFD:** In LRFD, the 1.2 and 1.6 factors on D and L amplify the worst-case load scenarios. In ASD, these reductions (0.75, 0.6, 0.7) acknowledge that simultaneous occurrence of full D, L, W, and E is rare. However, the engineer must verify that stress levels are truly acceptable across all use cases.
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4. ASD Load Combinations (NSCP 2015 Section 2.4)
Examples
Example 4.1 — ASD design load for a beam (comparison with LRFD)
Problem
Using the same residential beam from Example 3.1 (w_D = 16 kN/m, w_L = 7.6 kN/m, L = 6 m), find the ASD service load w and moment M, then compare to LRFD.
Solution
ASD Combo 2 (gravity): w = D + L = 16 + 7.6 = 23.6 kN/m ASD moment: M = w × L² / 8 = 23.6 × 36 / 8 = 106.2 kN·m LRFD moment (from Example 3.1): M_u = 141.12 kN·m Ratio: M_u / M = 141.12 / 106.2 ≈ 1.33 This means LRFD design moment is about 33% higher than ASD. The difference reflects: - LRFD factors (1.2 on D, 1.6 on L) amplify loads - φ factor (0.9 for flexure) reduces nominal strength - Net ratio ≈ (1.2 + 1.6)/1.33 × 0.9 ≈ 1.33 (simplified) Conclusion: LRFD members are generally larger or use higher-strength materials than ASD-designed members for the same loads.
Example 4.2 — ASD column load (uplift combination 0.6D + 0.6W)
Problem
A light roof column in a warehouse has dead load P_D = 100 kN (compression) and wind lateral load P_W = 150 kN. Check ASD combo 5 (uplift) and gravity combo 2.
Solution
Combo 2 (gravity): S = D = 100 kN (compression, no live load on roof structure) Combo 5 (uplift): S = 0.6D + 0.6W = 0.6(100) + 0.6(150) = 60 + 90 = 150 kN Note: Combo 5 here is interpreted as combined compression and lateral load. If wind creates upward (tension) suction: S = 0.6D − 0.6W = 0.6(100) − 0.6(150) = 60 − 90 = −30 kN (tension, magnitude 30 kN) Conclusion: Gravity dominates (100 kN compression) unless wind suction is very large. Both gravity and uplift combos must be checked.
Key Points
- ASD uses unfactored (service) loads and divides nominal strength by a safety factor.
- Seven principal ASD combinations must be checked; the most severe governs design.
- ASD combination S = D + L is typical for gravity design in most occupancies.
- Reduced factors (0.75, 0.6, 0.7) in ASD acknowledge low probability of concurrent maximum loads.
- ASD is less commonly used in modern NSCP; LRFD (with φ factors) is the standard.
- Uplift combo (0.6D + 0.6W or 0.7E) addresses wind/seismic tension in light structures.
- ASD is still applicable for timber, masonry, and retrofit design in some jurisdictions.
**Live Load Reduction (NSCP Section 2.5):** For columns, beams supporting large tributary areas (typically > 20–30 m²), the full live load rarely acts simultaneously. NSCP allows a reduction: L_reduced = L × [0.25 + 15/√A_t] where A_t is tributary area in m². Example: A_t = 100 m² → L_reduced = L × (0.25 + 15/10) = L × 1.75. This exceeds L, so no reduction; the formula implies a minimum tributary area beyond which reduction applies. For A_t = 36 m² → L_reduced = L × (0.25 + 15/6) = L × 2.75. Again, exceeds 1.0, so no reduction. For A_t = 400 m² → L_reduced = L × (0.25 + 15/20) = L × 1.0. Reduced to 1.0 (no benefit). **Roof Live Load vs. Floor Live Load:** Which controls? - If roof is sloped (drainage), Lr (typically 0.5–1.0 kPa) is less than floor L - If roof is flat but occasionally accessed, consider both - LRFD combo 3 (1.2D + 1.6Lr + other factors) typically shows when Lr governs **Wind and Seismic in Philippine Context:** The Philippines experiences: - Typhoons (tropical cyclones) with sustained winds up to 200+ km/h; basic wind speeds vary by region (see NSCP Figure 4 and Table 2) - High seismic activity (NSCP Figure 3 shows seismic zones); earthquake load E often exceeds wind load - Combinations 5 and 7 (with E) are almost always critical for lateral design **Uplift and Anchor Bolts:** When 0.9D or 0.6D combinations yield tension (negative values), the structure experiences uplift: - Shallow foundations may lift; deep foundations and anchor bolts must resist tension - In the Philippines, typhoon-induced uplift and sliding are common failure modes - Check combo 6 (LRFD: 0.9D + 1.0W) and combo 7 (0.9D + 1.0E) for all exposed structures **Load Combination Matrix for Designers:** A practical approach is to construct a table: | Combo | D | L | Lr | W | E | U or S | Controls? | |-------|---|---|----|----|---|--------|----------| | LRFD 1 | 1.4 | 0 | 0 | 0 | 0 | 1.4D | Rarely | | LRFD 2 | 1.2 | 1.6 | 0.5 | 0 | 0 | Best gravity | Usually | | LRFD 6 | 0.9 | 0 | 0 | 1.0 | 0 | Uplift | Wind-critical | | LRFD 7 | 0.9 | 0 | 0 | 0 | 1.0 | Uplift | Seismic-critical | For residential/office buildings in the Philippines: Use LRFD combo 2 for gravity, combos 6–7 for lateral (wind/seismic). **Impact Load and Dynamic Effects:** NSCP Section 2.2 notes that impact loads (e.g., equipment, vehicle parking) may require dynamic amplification (impact factor). For typical floors, code prescribes minimum 25–50% increase above static load. This is often embedded in the prescribed live load table values. **Serviceability Checks (Deflection, Vibration):** While load combinations drive ultimate/strength design, deflection and vibration are checked using **service loads** (without LRFD factors): - Deflection: Use w = w_D + w_L or (D + 0.5L) depending on code guidance - Vibration: Use w = w_D + w_L for floor systems - These checks use nominal (unfactored) loads and are independent of the strength design combination **Summary Table: When Each Load Type Matters** | Load | Typical Structures | Controls | Method | |------|-------------------|----------|--------| | D | All | Always | Self-weight calc | | L | Buildings (floors) | Gravity design | Table 1, NSCP | | Lr | Roofs (accessible) | Roof design | Usually secondary | | W | Tall, exposed, light | Lateral design, uplift | Wind pressure eq. | | E | Seismic zones (PH!) | Lateral design | Response spectrum | | R | Watersheds, drains | Secondary | Often ignored if drained |
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5. Special Considerations and Practical Applications
Examples
Example 5.1 — Live load reduction for a large tributary area (column)
Problem
A warehouse column (4-story) has tributary area A_t = 225 m² per floor. Prescribed live load L = 3.6 kPa. Does live load reduction apply? If so, calculate the reduced L_red and the effect on column design load.
Solution
Check live load reduction formula (NSCP Section 2.5): L_reduced = L × [0.25 + 15/√A_t] = 3.6 × [0.25 + 15/√225] = 3.6 × [0.25 + 15/15] = 3.6 × [0.25 + 1.0] = 3.6 × 1.25 = 4.5 kPa Result: L_reduced = 4.5 kPa > original L = 3.6 kPa. No reduction (formula indicates load increases or remains at 1.0L). Alternatively, if A_t = 900 m²: L_reduced = 3.6 × [0.25 + 15/30] = 3.6 × [0.25 + 0.5] = 3.6 × 0.75 = 2.7 kPa (25% reduction) Service load per floor (without reduction): P_L = 3.6 × 225 = 810 kN Service load with reduction (if applicable): P_L = 2.7 × 225 = 607.5 kN For 4 floors: P_L_total = 810 × 4 = 3240 kN (no reduction) or 2430 kN (if reduction applied) Conclusion: Large warehouses often qualify for live load reduction, reducing column and foundation sizes significantly.
Example 5.2 — Seismic load combination governing in the Philippines
Problem
A 6-story office building in Quezon City (seismic zone, E load critical) has base shear V = 1.2 MN (from response spectrum). An interior column at the base carries: - Dead load (all 6 floors): P_D = 2500 kN - Live load (all 6 floors): P_L = 1200 kN - Seismic overturning moment: M_overturn = 150 MN·m (base moment) The column base area is 0.5 m × 0.6 m. Check if combo 7 (0.9D + 1.0E) or combo 5 (1.2D + 1.0E + 1.0L) governs, and assess uplift risk on the foundation anchor bolts.
Solution
Step 1: Calculate axial forces in seismic combos. Combo 5 (LRFD): P_u = 1.2(2500) + 1.0(1200) + 1.0(E contribution to axial) = 3000 + 1200 = 4200 kN (plus any vertical component of seismic inertia, usually small). Combo 7 (LRFD, uplift): P_u = 0.9(2500) + 1.0(E) = 2250 kN (axial) Step 2: Seismic moment creates stress distribution. Using base moment and section modulus: Stress = P/A ± M/Z Base area A = 0.5 × 0.6 = 0.30 m² Section modulus (about major axis): Z ≈ b × h² / 6 = 0.5 × (0.6)² / 6 ≈ 0.03 m³ Combo 7 creates tension at far edge: Stress_tension = −2250/300 + 150,000/(0.03 × 1000) = −7.5 + 5000 = +4992.5 kPa (actually compression at this edge due to moment dominating) Note: The calculation shows moment creates significant tension/compression gradient. Detailed stability analysis and foundation design are required. Step 3: Assessment. Combos 5 and 7 both critical for seismic-prone structures. Combo 7 (0.9D + 1.0E) often governs anchor bolt tension in shallow foundations. Deep piles and anchor bolts must resist uplift. Conclusion: In the Philippines, seismic load combinations always govern lateral design and often control foundation anchorage.
Example 5.3 — Serviceability deflection check (service loads, not LRFD)
Problem
For the residential beam in Example 3.1 (w_D = 16 kN/m, w_L = 7.6 kN/m, L = 6 m), check deflection under service load. Assume RC beam with EI = 1200 MN·m² (in SI units, 1200 × 10⁶ N·m²). Code limit: Δ_max = L/240 = 6000/240 = 25 mm.
Solution
Step 1: Determine service load for deflection check. (NSCP typically uses D + live load at service level for deflection; often taken as D + L without factors.) w_service = w_D + w_L = 16 + 7.6 = 23.6 kN/m = 23,600 N/m Step 2: Calculate deflection for simply supported beam. Δ = (5 × w × L⁴) / (384 × EI) Δ = (5 × 23,600 × (6)⁴) / (384 × 1200 × 10⁶) Δ = (5 × 23,600 × 1296) / (384 × 1200 × 10⁶) Δ = (152,928,000) / (460,800 × 10⁶) Δ = 152,928,000 / 460,800,000,000 Δ ≈ 0.00033 m = 0.33 mm (extremely small; likely underestimated EI or overestimated stiffness) Alternative calculation (check): Δ = 5 × 23.6 × 1296 / (384 × 1200) ≈ 5 × 23.6 × 1296 / 460,800 ≈ 152,928 / 460,800 ≈ 0.33 mm Assume EI = 12 MN·m² (more realistic): Δ = 152,928 / 4,608 ≈ 33 mm > 25 mm limit → Beam is too flexible; increase section or strength. Conclusion: Serviceability (deflection) is checked separately using service loads, independent of LRFD factors. This often drives member sizing in flexible structures.
Key Points
- Live load reduction applies for large tributary areas (check NSCP Section 2.5 formula for applicability).
- Roof live load (Lr) typically less than floor live load; which controls depends on the structure and combinations.
- In the Philippines, seismic load (E) and wind load (W) are often critical; combinations 5–7 must always be checked.
- Uplift combos (0.9D + 1.0W or E) control tension in shallow foundations, anchor bolts, and light exposed structures.
- Serviceability (deflection, vibration) uses service loads, not LRFD factored loads.
- Load combination tables and matrices help engineers systematically check all cases.
- Dynamic/impact effects are usually embedded in prescribed live load values; check code for special cases.
This section provides complete, detailed solutions to complex problems typical of the PRC Civil Engineer Licensure Examination and professional review courses. Each solution follows a structured format: Given → Find → Solution → Answer → Discussion. **Problem 6.1: Multi-Story Building Load Combination (Comprehensive)** Given: - A 5-story office building in Manila (seismic zone, basic wind speed V = 220 km/h ≈ 61.1 m/s) - Floor system: One-way slab on beams; floor spacing s = 5 m, beam span L = 6 m - Loads per floor: q_D = 5 kPa (including structure + finishes), q_L = 2.4 kPa (office) - Roof: Accessible, q_Lr = 0.75 kPa - Wind pressure (simplified): p_w = 0.613 × (61.1)² × C_p × C_e ≈ 1.5 kPa net (assume combined effects) - Seismic base shear (from spectral analysis): V = 0.15 × W_total (W_total = total dead weight) Find: (a) Line load on an interior floor beam (service and LRFD) (b) Design moment M_u for the beam under LRFD combo 2 (c) Factored axial load P_u on an interior column (4 typical floors above base) under LRFD combo 5 (seismic) (d) Assess whether combo 6 (0.9D + 1.0W) or combo 7 (0.9D + 1.0E) governs uplift Solution: (a) **Service and LRFD line loads on floor beam:** Service (ASD): w = w_D + w_L = (5 × 5) + (2.4 × 5) = 25 + 12 = 37 kN/m LRFD (combo 2): w_u = 1.2 w_D + 1.6 w_L = 1.2(25) + 1.6(12) = 30 + 19.2 = 49.2 kN/m Alternatively, with roof live load concern (combo 3 or other), check but typically combo 2 governs floors. (b) **Design moment M_u (simply supported, L = 6 m, LRFD combo 2):** M_u = w_u × L² / 8 = 49.2 × (6)² / 8 = 49.2 × 36 / 8 = 1771.2 / 8 = 221.4 kN·m (c) **Factored axial load P_u on interior column (4 floors above, combo 5 with seismic):** Tributary area per floor: A_t = (5 m / 2) × (6 m / 2) = 7.5 m² (assuming regular 5 m × 6 m grid; alternatively assume 5 m × 5 m → 6.25 m²) Wait: If floor beams are spaced 5 m on center and span 6 m, the column grid is likely 5 m × 6 m or similar. For an interior column: A_t = (distance to mid-span in X) × (distance to mid-span in Y) Assuming regular grid 5 m × 6 m: A_t = 2.5 × 3 = 7.5 m² per floor Alternatively, if grid is 5 m × 5 m: A_t = 2.5 × 2.5 = 6.25 m² Use A_t = 7.5 m² (conservative for 5 m × 6 m grid). Service loads per floor: P_D = 5 kPa × 7.5 m² = 37.5 kN P_L = 2.4 kPa × 7.5 m² = 18 kN For 4 floors (column at 5th story, supporting stories 2–5): P_D_total = 37.5 × 4 = 150 kN P_L_total = 18 × 4 = 72 kN Note: Alternatively, a 5-story building has 5 floors; a column at the base supports all 5 floors: P_D_total = 37.5 × 5 = 187.5 kN P_L_total = 18 × 5 = 90 kN Assume the problem means 4 typical floors above a column position (e.g., at 2nd story). Combo 5 (LRFD, seismic): P_u = 1.2 P_D + 1.0 P_L + 1.0 E The seismic load E on axial capacity is typically small (E is lateral); vertical component of E ≈ 0 unless specified. However, NSCP may include E in the combination directly. Assuming E contributes negligibly to vertical axial force: P_u ≈ 1.2(150) + 1.0(72) = 180 + 72 = 252 kN (Note: If E includes vertical inertia, add; if E is lateral only, the above is correct.) (d) **Uplift combos (0.9D + 1.0W and 0.9D + 1.0E):** Combo 6 (wind uplift): U_6 = 0.9 P_D + 1.0 P_W Wind load effect on a single column: This requires wind force distribution (typically base shear divided among columns by their tributary area). For a rough estimate: Total wind force at this floor level: F_w ≈ (0.15 × W_total) / number of floors × height factor ≈ (1.5 kPa slab equivalent) × tributary area in plan. For simplicity, assume wind creates a tributary lateral force ≈ 1.5 kPa × 7.5 m² = 11.25 kN (lateral, not vertical). Vertical wind component ≈ 0 (wind is horizontal). However, the combination 0.9D + 1.0W with W including uplift (wind suction on roof/horizontal surfaces): If roof experiences upward suction force F_suction ≈ 1.5 kPa × roof area, this creates tension at anchor bolts. For this column (interior, not anchored at roof): combo 6 and 7 are critical for shallow foundations or edge columns that experience uplift. Combo 7 (seismic, potential uplift if E has vertical component): U_7 = 0.9 P_D + 1.0 E_vertical If seismic inertia is primarily lateral: U_7 ≈ 0.9(150) = 135 kN (compression) Conclusion: For an interior floor column, combo 7 (0.9D) gives 135 kN compression (not uplift). Uplift would occur at foundation or edge columns if lateral overturning moment creates tension at anchors. **Answer:** (a) Service w = 37 kN/m; LRFD w_u = 49.2 kN/m (b) M_u = 221.4 kN·m (c) P_u ≈ 252 kN (under combo 5; may be adjusted for E vertical component) (d) Combo 7 (0.9D + 1.0E) gives 135 kN compression (interior column). Uplift governs shallow foundations and edge anchors, requiring detailed overturning moment analysis. **Discussion:** This problem integrates tributary area, LRFD combinations, and seismic considerations critical in the seismic Philippines. Key points: - Combo 2 (1.2D + 1.6L) governs floor beam design for moment and shear - Combo 5 (seismic) includes E in the combination; vertical E is often neglected for interior columns - Combo 7 (0.9D + 1.0E) applies to uplift; foundation anchorage design is critical - Wind (combo 6) and seismic (combos 5, 7) must always be checked in the Philippines
Heading
6. Worked Board-Exam Problems (Full Solutions)
Examples
Example 6.1 — Complete LRFD design load determination for a floor beam
Problem
A residential apartment building in Makati has a floor system with beams spaced 4 m on center and spanning 7 m. Each floor has q_D = 4.5 kPa (slab + finishes + MEP), q_L = 1.9 kPa (residential, f₁ = 0.5), and roof has q_Lr = 0.75 kPa. Wind is q_W ≈ 1.2 kPa (simplified). Apply all applicable LRFD combinations for a typical interior floor beam and identify the governing combination.
Solution
Step 1: Convert area loads to line loads (tributary width s = 4 m for interior beam). w_D = 4.5 × 4 = 18 kN/m w_L = 1.9 × 4 = 7.6 kN/m w_Lr = 0.75 × 4 = 3 kN/m w_W = 1.2 × 4 = 4.8 kN/m (lateral; treated separately but included for completeness) Step 2: Apply LRFD combinations (residential, so f₁ = 0.5 for companion L; check NSCP version). Combo 1: w_u = 1.4 × 18 = 25.2 kN/m Combo 2: w_u = 1.2(18) + 1.6(7.6) + 0.5(3) = 21.6 + 12.16 + 1.5 = 35.26 kN/m ← Likely governs gravity Combo 3: w_u = 1.2(18) + 1.6(3) + 0.5(7.6) = 21.6 + 4.8 + 3.8 = 30.2 kN/m (roof controls if it exists) Combo 4: w_u = 1.2(18) + 1.0(4.8) + 0.5(7.6) + 0.5(3) = 21.6 + 4.8 + 3.8 + 1.5 = 31.7 kN/m (wind + reduced live) Combo 5: Seismic (E not given for this example; if present, would include E + reduced L) Combo 6: w_u = 0.9(18) + 1.0(4.8) = 16.2 + 4.8 = 21 kN/m (uplift/wind; less than combo 2) Combo 7: Similar to combo 6 with E; not applicable without seismic data Governing combination: **Combo 2, w_u = 35.26 kN/m** (or round to 35.3 kN/m) Step 3: Calculate factored moment (simply supported, L = 7 m). M_u = w_u × L² / 8 = 35.26 × (7)² / 8 = 35.26 × 49 / 8 = 1,727.74 / 8 = 215.97 kN·m Round: **M_u ≈ 216 kN·m** Step 4: Additional checks (narrative). - Shear design: V_u = w_u × L / 2 = 35.26 × 7 / 2 = 123.41 kN. Design web/stirrups for this shear with φ = 0.75 (ACI 318). - Deflection (serviceability): Use service load w = w_D + w_L = 18 + 7.6 = 25.6 kN/m. Check Δ ≤ L/240 = 7000/240 ≈ 29 mm (adjust based on code and member type). - Torsion, diagonal cracking, other local effects: Check if edge beams or corner conditions apply. Conclusion: The 7 m residential beam requires M_u = 216 kN·m, V_u ≈ 123 kN under LRFD combo 2. Select RC or steel section accordingly with φ = 0.9 (flexure), φ = 0.75 (shear).
Problem Continuation
Then calculate the resulting M_u for a simply supported 7 m span. Finally, note any additional checks (deflection, shear) needed.
Example 6.2 — Seismic load combination dominance (PH context)
Problem
A 10-story office tower in Quezon City (seismic zone, moderate-to-high seismic demand) has: - Total dead weight W_D ≈ 5000 kN (lumped estimate for illustration) - Live load capacity: W_L ≈ 1500 kN (all floors combined at time of design) - Seismic design acceleration: S_a (spectral acceleration) ≈ 0.4g → Response modification factor R = 8 → Seismic response coefficient C_s = S_a / R ≈ 0.05 → Seismic base shear V_E ≈ 0.05 × W ≈ 0.05 × (5000 + some fraction of L) ≈ 250–300 kN (simplified) - Wind base shear (basic wind speed 220 km/h, exposure C): V_W ≈ 150 kN (rough estimate for 10-story office) Apply LRFD combinations 2, 5, and 7 to the base shear demand. Which governs lateral design?
Solution
Step 1: Clarify base shear for lateral combinations. Combo 2 (gravity): No lateral load; M_u and V_u from gravity only (from previous examples). Combo 5 (1.2D + 1.0E + f₁L): Seismic base shear V_E ≈ 250 kN (lateral demand) Combo 6 (0.9D + 1.0W): Wind base shear V_W ≈ 150 kN (lateral demand) Combo 7 (0.9D + 1.0E): Seismic base shear V_E ≈ 250 kN (lateral demand) Step 2: Compare lateral demands. V_W = 150 kN < V_E = 250 kN → **Seismic governs** (typical for the Philippines) Step 3: Story shear distribution and column design. For a simplified 10-story tower, seismic force is distributed with higher concentration at upper floors (per response spectrum). Example: Top 5 stories might experience 60% of base shear → 250 × 0.6 = 150 kN distributed among perimeter and interior columns. Lower 5 stories: 250 × 0.4 = 100 kN. Each column carries proportional shear based on tributary area. For an interior column: Story shear at level 5 (mid-height) ≈ 150 kN (top 5 stories) / 12 columns ≈ 12.5 kN per column (rough) Combined demand (axial + lateral shear in member): P_u (from gravity) + V_u (from seismic) → Design for combined axial and shear (interaction diagram for RC columns or interaction formulas for steel) Conclusion: In this seismic-prone structure, **combo 5 or 7 (seismic E) governs lateral design** with V_E ≈ 250 kN >> wind V_W = 150 kN. Columns must be designed for combined P_u (gravity + seismic axial inertia) and V_u (shear transfer). This is typical for the Philippines; wind design is often secondary.
Key Points
- Complex problems require systematic tributary area calculation, then application of all applicable load combinations.
- Interior members: Combos 2 (gravity), 5–7 (lateral) typically govern.
- Always identify which combination produces the maximum demand in each direction (vertical, horizontal, moment).
- Seismic combos (5 and 7) are nearly always critical in the Philippines.
- Uplift combos (0.9D + 1.0W/E) govern shallow foundations and anchor bolts in light or tall structures.
- Service loads (unfactored) are used separately for deflection and vibration checks.
- Solving multi-story buildings requires clarity on which floor/column level is being analyzed.
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