CELE Reinforced & Prestressed Concrete — Reinforced Concrete Fundamentals: WSD and USDStudy Notes
Full study notes for Reinforced Concrete Fundamentals: WSD and USD — built specifically for the CELE 2026. These notes cover every concept, definition, formula, and worked example you need for the Reinforced & Prestressed Concrete subtest of the CELE, structured in the order Professional Regulation Commission (PRC) — Board of Civil Engineering typically tests them.
Exam context
On the CELE 2026, the Reinforced & Prestressed Concrete subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Reinforced Concrete Fundamentals: WSD and USD lands at position 1st out of 7 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 Reinforced & Prestressed Concrete on a typical CELE paper.
Reinforced Concrete Fundamentals: WSD and USD - Study Notes
Reinforced concrete represents one of the most important composite materials in structural engineering, combining the compressive strength of concrete with the tensile strength of steel reinforcement. The Philippine civil engineer must master two complementary design philosophies: Working Stress Design (WSD), an elastic, allowable-stress method still appearing in legacy structures and board exams, and Ultimate Strength Design (USD), formally called Limit State Design or LRFD, which is the mandatory approach in NSCP 2015. This chapter establishes the theoretical foundation for all subsequent reinforced concrete design: the material properties, the equivalent stress block, strength-reduction factors, and the load-factoring conventions that dominate modern Philippine practice. Understanding both methods—and knowing when to apply each—is essential for licensure success.
Summary
This chapter establishes the theoretical foundation for reinforced concrete design in the NSCP 2015 framework. The two design philosophies—Working Stress Design (WSD) and Ultimate Strength Design (USD)—represent different approaches to safety and structural behavior. WSD is elastic and allowable-stress based, treating the structure as linear-elastic under service loads; it remains historically important but is now secondary. USD is the NSCP 2015 mandate: it factors up loads (e.g., 1.2D + 1.6L), reduces nominal strength by a member-type-specific factor φ, and checks that φM_n ≥ M_u. The equivalent rectangular (Whitney) stress block, with intensity 0.85f'_c and depth a = β₁c, is the cornerstone of USD moment capacity calculations. The factor β₁ depends critically on concrete strength: β₁ = 0.85 for f'_c ≤ 28 MPa, then decreases linearly to 0.65 at f'_c ≥ 55 MPa. Material properties (E_c = 4700√f'_c, E_s = 200,000 MPa, modular ratio n) are essential inputs. Strength-reduction factors φ vary by member type and failure mode: φ = 0.90 for tension-controlled flexure (most beams), φ = 0.75 for shear and spiral columns, φ = 0.65 for tied columns. The transition between compression-controlled (φ = 0.65/0.75) and tension-controlled (φ = 0.90) is marked by a net tensile strain ε_t = 0.005 (five times the yield strain). Understanding both methods, applying the correct φ, correctly computing β₁, and distinguishing between factored and unfactored loads are essential for success on the PRC Civil Engineer Licensure Examination. Common pitfalls include confusing WSD and USD, misapplying β₁, using the wrong φ for columns, and failing to check all load combinations. Mastery of these fundamentals enables confident design of beams, columns, and other reinforced concrete elements in subsequent chapters.
Sections
Reinforced concrete (RC) is a composite material where concrete resists compression and steel bars resist tension. Concrete alone is brittle and weak in tension; steel alone is expensive and prone to corrosion. Together, they create a material that is durable (when properly detailed), economical, and highly versatile—the backbone of Philippine infrastructure. Two competing design frameworks exist: **Working Stress Design (WSD, also called Alternate Design or Elastic Method):** - Assumes linear-elastic behavior under **service loads** (unfactored dead load D and live load L). - Keeps computed stresses below allowable fractions of the material strength (e.g., concrete stress ≤ 0.45f'_c, steel stress ≤ 0.5f_y or 0.4f_y depending on loading). - Uses the **modular ratio** n = E_s/E_c to convert steel area into an equivalent concrete section. - Conservative and conceptually simple, but does not directly address safety factors against ultimate failure. - Still encountered in older Philippine designs and legacy code references (e.g., earlier ACI editions). **Ultimate Strength Design (USD, also LRFD—Limit State Design or Load and Resistance Factor Design):** - Factors **up** the loads: $M_u = 1.2M_D + 1.6M_L$ (and other combinations per NSCP 2015 Section 203). - Reduces the nominal (ultimate) strength by a **strength-reduction factor** φ: $$\phi M_n \ge M_u, \quad \phi V_n \ge V_u$$ - Replaces the real curved concrete stress distribution with an **equivalent rectangular stress block** of intensity 0.85f'_c and depth a = β₁c. - Directly designed for failure; the φ factors account for variability in material and construction, and differences between assumed and actual stress distributions. - **NSCP 2015 mandates USD** for all new design; WSD is an alternative but secondary method, retained for educational and legacy-code comparisons. In practice, USD is more realistic because it acknowledges that materials do not behave perfectly elastically at high stress levels and that failure is a discrete, observable event. The PRC Civil Engineer Licensure Examination expects fluency in both approaches, with heavy emphasis on USD.
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1. Introduction to Reinforced Concrete and the Two Design Philosophies
Examples
Label
Example 1.1 — Understanding Load Factoring
Problem
A building beam supports a dead load moment M_D = 100 kN·m (including self-weight) and live load moment M_L = 80 kN·m. Compute the factored design moment M_u under gravity loading per NSCP 2015.
Solution
Under NSCP 2015 Section 203 (Limit State of Strength—Vertical Loads), the governing combination for gravity loads is: $$M_u = 1.2M_D + 1.6M_L = 1.2(100) + 1.6(80) = 120 + 128 = 248 \text{ kN·m}$$ The beam section must be proportioned so that φM_n ≥ 248 kN·m. For tension-controlled flexure, φ = 0.90, so M_n ≥ 248/0.90 = 275.6 kN·m is required.
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Example 1.2 — WSD vs USD Mindset
Problem
Explain the conceptual difference between WSD and USD for a concrete beam under the same loading.
Solution
WSD Analysis: Take the service loads (D + L) as given. Compute internal stresses: concrete stress f_c and steel stress f_s. Check that f_c ≤ 0.45f'_c and f_s ≤ allowable (typically 0.4f_y or 0.5f_y). Design by increasing section or reinforcement until these are satisfied. Safety is built in indirectly by the conservative allowable fractions. USD Analysis: Factor the loads up (1.2D + 1.6L). Compute the section's nominal (ultimate) moment capacity M_n using a plastic stress distribution (0.85f'_c rectangular block). Check that φM_n ≥ M_u. If not, add steel or increase section. Safety is explicit: the φ factor and load factors jointly ensure that failure probability is acceptably low (typically 1 in 1000–2000 for normal buildings). USD is more rational because it directly addresses the failure condition and accounts for the nonlinear behavior of concrete at high strain.
Key Points
- Reinforced concrete: concrete resists compression, steel resists tension.
- WSD: service-load elastic design with allowable stresses (historical, but still tested).
- USD: factored-load design with strength reduction factor φ; NSCP 2015 default.
- USD formula: φM_n ≥ M_u, where M_u = 1.2M_D + 1.6M_L (typical gravity loads).
- Equivalent stress block: 0.85f'_c at depth a = β₁c; the key to USD moment capacity.
**Concrete Compressive Strength f'_c (MPa):** The 28-day cylinder compressive strength is the standard design parameter in NSCP 2015. Common grades in Philippines: f'_c = 20.7, 24.5, 27.6, 34.5 MPa (and metric equivalents). In design, always use the specified strength; do not assume "typical" or "average" unless explicitly stated in a problem. **Modulus of Elasticity E_c:** For normal-weight concrete (density ρ ≈ 2400 kg/m³), NSCP 2015 Section 421 (based on ACI 318) gives: $$E_c = 4700\sqrt{f'_c} \text{ (MPa, where } f'_c \text{ is in MPa)}$$ This formula is empirical and applies to normal-weight concrete. For lightweight concrete, a reduction factor is used. In the range f'_c = 20–55 MPa, E_c typically ranges from ~21,000 to ~35,000 MPa. **Steel Reinforcement Yield Strength f_y (MPa):** Two main grades are used in Philippines (per ASTM A615 / PH equivalent): - Grade 275 MPa (older, less common now): f_y = 275 MPa, E_s = 200,000 MPa. - Grade 415 MPa (standard): f_y = 415 MPa, E_s = 200,000 MPa. In USD design, the stress-strain curve is idealized as elastic-perfectly-plastic: linear up to f_y, then constant at f_y. The yield strain is: $$\varepsilon_{ty} = \frac{f_y}{E_s} \quad \text{(typical: } \varepsilon_{ty} = \frac{415}{200,000} = 0.002075 \approx 0.0021 \text{)}$$ A reinforcing bar is said to be "at yield" when the strain in the bar reaches ε_ty. In a beam under positive moment, the bottom steel yields first. **Modular Ratio n (WSD only):** WSD uses the modular ratio to transform a section into an equivalent concrete section: $$n = \frac{E_s}{E_c}$$ For example, one bar with area A_s transforms to an equivalent concrete area nA_s, because concrete and steel have different stiffnesses. This allows treating the composite section as if it were all concrete, with the "transformed" steel area. The modular ratio depends on f'_c and must be calculated for each strength grade (unlike f_y, which is fixed by the reinforcement manufacture).
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2. Material Properties: Concrete and Steel
Examples
Label
Example 2.1 — Computing E_c and Modular Ratio
Problem
A concrete beam is designed with f'_c = 28 MPa and Grade 415 reinforcing steel. Compute (a) E_c, and (b) the modular ratio n.
Solution
(a) Using NSCP formula: $$E_c = 4700\sqrt{f'_c} = 4700\sqrt{28} = 4700 \times 5.292 = 24,872 \text{ MPa} \approx 24,900 \text{ MPa}$$ (b) Modular ratio: $$n = \frac{E_s}{E_c} = \frac{200,000}{24,872} = 8.04 \approx 8$$ In WSD analysis, 1 cm² of steel is equivalent to 8 cm² of concrete (in terms of stiffness). This is used in the transformed-section method.
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Example 2.2 — Yield Strain and the Transition to Plastic Behavior
Problem
A Grade 415 steel bar reaches a strain of ε = 0.003. Has it yielded? If so, by how much?
Solution
Yield strain: ε_ty = f_y / E_s = 415 / 200,000 = 0.002075. Applied strain: ε = 0.003. Since 0.003 > 0.002075, yes, the bar has yielded. Excess strain (plastic strain): Δε = 0.003 − 0.002075 = 0.000925 = 0.0925%. Once yielded, the steel stress remains constant at 415 MPa, even if strain increases further. This is the basis for USD plastic analysis.
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Example 2.3 — Material Properties Across Strength Grades
Problem
Compute E_c and n for three common Philippine concrete strengths: f'_c = 21, 35, and 55 MPa.
Solution
For f'_c = 21 MPa: $$E_c = 4700\sqrt{21} = 4700 \times 4.583 = 21,540 \text{ MPa}$$ $$n = 200,000 / 21,540 = 9.29 \approx 9.3$$ For f'_c = 35 MPa: $$E_c = 4700\sqrt{35} = 4700 \times 5.916 = 27,805 \text{ MPa}$$ $$n = 200,000 / 27,805 = 7.19 \approx 7.2$$ For f'_c = 55 MPa: $$E_c = 4700\sqrt{55} = 4700 \times 7.416 = 34,855 \text{ MPa}$$ $$n = 200,000 / 34,855 = 5.74 \approx 5.7$$ Note: As concrete strength increases, E_c increases (stiffer), so n decreases. Higher-strength concrete deforms less, so steel represents a smaller equivalent area relative to concrete.
Key Points
- f'_c: 28-day cylinder compressive strength (MPa), the primary concrete design parameter.
- E_c = 4700√f'_c (MPa): formula for normal-weight concrete modulus.
- E_c ranges ~21,000–35,000 MPa for typical design grades (f'_c = 20–55 MPa).
- f_y: 275 or 415 MPa; E_s = 200,000 MPa (always, for mild steel).
- Yield strain ε_ty = f_y/E_s; for Grade 415: ε_ty ≈ 0.00208 or 0.208%.
- Modular ratio n = E_s/E_c; WSD uses this to transform steel into equivalent concrete.
- n ranges ~6–10 depending on f'_c; smaller n for higher-strength concrete (stiffer).
One of the most important simplifications in USD is the **equivalent rectangular stress block** (also called the Whitney stress block, after one of its developers). Instead of using the real, curved concrete stress–strain distribution, USD assumes: - **Constant stress intensity:** 0.85f'_c throughout the compression zone (not the full f'_c, but 0.85 times it). - **Depth of the block:** a = β₁c, where c is the neutral-axis depth and β₁ is a factor that accounts for the shape and nonlinearity of the real stress distribution. - **Location of the resultant force:** at the centroid of the block, which is a/2 from the top fiber (or c from the neutral axis, since a = β₁c). This simplification is valid because: 1. Concrete does not behave as a linear-elastic material at near-failure stresses; the stress–strain curve is nonlinear and curved. 2. The factor 0.85 accounts for the difference between the peak stress in the real distribution and the rectangular approximation. 3. The factor β₁ adjusts for the varying shape as concrete strength changes. **The β₁ Factor:** The value of β₁ depends on f'_c and decreases as concrete strength increases. NSCP 2015 Section 421.1.3.3.1 specifies: $$\beta_1 = 0.85 \quad \text{for } f'_c \le 28 \text{ MPa}$$ $$\beta_1 = 0.85 - 0.05 \frac{f'_c - 28}{7} \quad \text{for } 28 < f'_c \le 55 \text{ MPa}$$ $$\beta_1 = 0.65 \quad \text{for } f'_c \ge 55 \text{ MPa}$$ **Why does β₁ decrease?** Higher-strength concrete has a stiffer, narrower stress block. The neutral axis moves closer to the compression face, so the stress block depth a becomes a smaller fraction of c. Using a constant β₁ = 0.85 for all strengths would overestimate the block depth and thus overestimate moment capacity at high f'_c. **Practical Insight:** For f'_c ≤ 28 MPa, β₁ = 0.85 is constant. Above 28 MPa, β₁ decreases linearly at a rate of 0.05 per 7 MPa of increase in f'_c, reaching 0.65 at f'_c = 55 MPa. Above 55 MPa, β₁ remains 0.65. This formula is critical in computing neutral-axis depth c and moment capacity M_n.
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3. The Equivalent Rectangular Stress Block and β₁
Examples
Label
Example 3.1 — Computing β₁ for Common Strengths
Problem
Determine β₁ for (a) f'_c = 21 MPa, (b) f'_c = 35 MPa, (c) f'_c = 42 MPa, and (d) f'_c = 55 MPa.
Solution
(a) f'_c = 21 MPa ≤ 28 MPa: $$\beta_1 = 0.85$$ (b) f'_c = 35 MPa, which is between 28 and 55 MPa: $$\beta_1 = 0.85 - 0.05 \frac{35 - 28}{7} = 0.85 - 0.05(1) = 0.80$$ (c) f'_c = 42 MPa, between 28 and 55 MPa: $$\beta_1 = 0.85 - 0.05 \frac{42 - 28}{7} = 0.85 - 0.05(2) = 0.75$$ (d) f'_c = 55 MPa ≥ 55 MPa: $$\beta_1 = 0.65$$ Note the linear decrease from 0.85 to 0.65 as f'_c rises from 28 to 55 MPa. This is one of the most frequently tested formulas on the PRC exam.
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Example 3.2 — Relationship Between c and a
Problem
A beam section has neutral-axis depth c = 200 mm and concrete strength f'_c = 35 MPa. Determine the equivalent stress-block depth a.
Solution
From Example 3.1(b), β₁ = 0.80 for f'_c = 35 MPa. $$a = \beta_1 c = 0.80 \times 200 = 160 \text{ mm}$$ The concrete stress block extends 160 mm from the top fiber. The resultant compression force acts at a/2 = 80 mm from the top, with intensity 0.85f'_c.
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Example 3.3 — Stress-Block Intensity
Problem
What is the concrete compressive stress at the neutral axis (or anywhere within the block) for f'_c = 28 MPa and f'_c = 42 MPa?
Solution
In the USD equivalent stress block, the stress is constant at 0.85f'_c throughout the depth a, regardless of the location within the block (unlike the curved real distribution). For f'_c = 28 MPa: $$f_{block} = 0.85 \times 28 = 23.8 \text{ MPa}$$ For f'_c = 42 MPa: $$f_{block} = 0.85 \times 42 = 35.7 \text{ MPa}$$ The block intensity depends only on f'_c, not on the section depth or geometry. This simplification is what makes USD analysis tractable and why it is so widely adopted.
Key Points
- Equivalent stress block: constant intensity 0.85f'_c, depth a = β₁c.
- β₁ accounts for nonlinearity and shape of real stress distribution.
- β₁ = 0.85 for f'_c ≤ 28 MPa (constant).
- β₁ = 0.85 − 0.05(f'_c − 28)/7 for 28 < f'_c ≤ 55 MPa (decreasing).
- β₁ = 0.65 for f'_c ≥ 55 MPa (constant, minimum).
- The block resultant acts at depth a/2 from the top, or equivalently, distance (d − a/2) from the tension steel.
- The factor 0.85 (not 1.0) applied to f'_c reflects test data on real stress distributions under near-failure conditions.
The strength-reduction factor φ ("phi") is a multiplier applied to the nominal (ultimate) strength to account for: 1. Uncertainty in material properties (concrete strength varies, steel yield varies). 2. Variability in construction (reinforcement placement, concrete compaction). 3. Differences between assumed and actual stress distributions. 4. Type of failure mode (sudden brittle vs. ductile behavior). Different failure modes and member types have different φ values because they have different levels of warning before failure. Ductile failures (e.g., beam flexure where steel yields first) are more forgiving; brittle failures (e.g., shear, column buckling) are sudden. Thus: $$\phi M_n \ge M_u \quad (\text{flexure}), \quad \phi V_n \ge V_u \quad (\text{shear}), \quad \phi P_n \ge P_u \quad (\text{columns})$$ **NSCP 2015 Strength-Reduction Factors:** For **flexure (bending):** - Tension-controlled sections: φ = 0.90. This includes most beam designs where the bottom steel is in tension and yields prior to concrete crushing. The beam bends visibly before failure, providing warning. - Compression-controlled sections: φ = 0.65 (tied columns) or 0.75 (spiral columns). These fail when concrete is crushed; less ductile, more sudden. - Transition zone (between tension- and compression-controlled): φ is interpolated linearly between 0.65/0.75 and 0.90 as the net tensile strain ε_t increases. For **shear and torsion:** - φ = 0.75 (all cases). Shear and torsional failures are typically sudden and brittle, offering little warning. For **compression (columns):** - Spiral reinforcement: φ = 0.75. Spiral confinement provides lateral restraint, increasing ductility. - Tied reinforcement: φ = 0.65. Ties are less effective at confinement; failure is more brittle. For **bearing on concrete:** - φ = 0.65. Bearing is a localized compression failure, often sudden. For **plain concrete:** - φ = 0.60. Plain (unreinforced) concrete has no ductility reserve. **Transition Between Compression- and Tension-Controlled (ACI/NSCP):** A section is: - **Compression-controlled** if the net tensile strain ε_t ≤ ε_ty (the yield strain of the reinforcement). In this case, φ = 0.65 (tied) or 0.75 (spiral). - **Tension-controlled** if ε_t ≥ 0.005 (5 times the yield strain). In this case, φ = 0.90. - **In transition** if ε_ty < ε_t < 0.005. In this case, φ is linearly interpolated. For Grade 415 steel: ε_ty = 415/200,000 = 0.002075. Thus a section is tension-controlled if the tensile strain in the extreme steel layer exceeds 0.005 = 5%. For **tied columns**, the φ interpolation formula is: $$\phi = 0.65 + 0.25 \frac{\varepsilon_t - \varepsilon_{ty}}{0.005 - \varepsilon_{ty}}$$ This is tested frequently on the licensure exam, especially in column design problems.
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4. Strength-Reduction Factors φ (NSCP 2015)
Examples
Label
Example 4.1 — Identifying φ for Different Member Types
Problem
State the appropriate strength-reduction factor φ for each of the following: (a) A reinforced concrete beam in positive flexure, where the tension steel yields before concrete crushes. (b) A tied reinforced concrete column under axial compression. (c) A beam designed for shear (stirrups). (d) A spiral column. (e) A bearing pad under a column.
Solution
(a) Tension-controlled flexure (most common beam design): φ = 0.90. (b) Tied column: φ = 0.65. (c) Shear: φ = 0.75. (d) Spiral column: φ = 0.75. (e) Bearing: φ = 0.65. Note: The higher φ for beams (0.90) reflects that yielding of steel provides ductility and warning. The lower φ for columns (0.65–0.75) and shear (0.75) reflects more sudden failure modes.
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Example 4.2 — Tension-Controlled vs. Compression-Controlled
Problem
A section has a net tensile strain ε_t = 0.006 in the extreme tension layer. Is it tension-controlled or compression-controlled? What is φ?
Solution
Compare ε_t to the threshold values for Grade 415 steel: - ε_ty = f_y / E_s = 415 / 200,000 = 0.002075 ≈ 0.0021. - Tension-controlled threshold: 0.005. Since ε_t = 0.006 > 0.005, the section is **tension-controlled**. Therefore, φ = 0.90 (for flexure) or φ = 0.90 (for compression members in high eccentricity, if applicable). The steel has yielded and stretched well beyond yield; the concrete is not the limiting factor. This is desirable because steel provides warning by visibly bending.
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Example 4.3 — Interpolation in the Transition Zone
Problem
A tied reinforced concrete column has a net tensile strain ε_t = 0.003. The steel is Grade 415 (ε_ty = 0.002075). Is it in the transition zone? If so, compute φ.
Solution
Checking the bounds: - ε_ty = 0.002075 < ε_t = 0.003 < 0.005. Yes, the section is in the **transition zone**. Using the NSCP/ACI interpolation formula for tied sections: $$\phi = 0.65 + 0.25 \frac{\varepsilon_t - \varepsilon_{ty}}{0.005 - \varepsilon_{ty}} = 0.65 + 0.25 \frac{0.003 - 0.002075}{0.005 - 0.002075}$$ $$= 0.65 + 0.25 \frac{0.000925}{0.002925} = 0.65 + 0.25(0.3163) = 0.65 + 0.0791 = 0.729 \approx 0.73$$ So φ ≈ 0.73 for this section, lying between 0.65 and 0.90.
Key Points
- φ: strength-reduction factor, accounts for uncertainty, variability, and failure mode.
- Flexure (tension-controlled, most common beams): φ = 0.90.
- Flexure (compression-controlled, rare): φ = 0.65 (tied) or 0.75 (spiral).
- Shear and torsion: φ = 0.75 (always).
- Columns: φ = 0.65 (tied) or 0.75 (spiral).
- Bearing: φ = 0.65.
- Plain concrete: φ = 0.60.
- Tension-controlled threshold: ε_t ≥ 0.005 (for all bar sizes).
- Compression-controlled threshold: ε_t ≤ ε_ty = f_y / E_s ≈ 0.002 for Grade 415.
- Transition zone: φ interpolates linearly between the two limiting values.
- Higher φ (e.g., 0.90) rewards ductile behavior; lower φ (e.g., 0.65) penalizes brittle behavior.
USD design requires that the section be proportioned so that the nominal (ultimate) strength, reduced by φ, exceeds the factored (design) loads. The factored loads are obtained by multiplying service loads (dead D, live L, earthquake E, wind W, etc.) by prescribed load factors. NSCP 2015 Section 203 specifies the combinations. **Primary Load Factors (Limit State of Strength):** For **gravity loads only** (most common in building design): $$U = 1.2D + 1.6L \quad (\text{Combination 1, all positive})$$ Alternative gravity combinations: $$U = 1.2D + 1.0L + 1.0E \quad (\text{with seismic})$$ $$U = 1.2D + 1.0W + 1.0L \quad (\text{with wind, reduced live})$$ $$U = 0.9D + 1.6W \quad (\text{wind-only, minimal dead load})$$ $$U = 0.9D + 1.0E \quad (\text{seismic-only, minimal dead load})$$ Where: - **D** = dead load (self-weight, permanent fixtures). - **L** = live load (occupancy, temporary loads). - **E** = earthquake load (seismic forces). - **W** = wind load. For **design**, the largest U (or M_u, V_u, P_u) from all applicable combinations governs. In most residential and office buildings, the 1.2D + 1.6L combination controls for positive moment, while 0.9D + 1.6W or 0.9D + 1.0E might control for negative moment or stability. **Physical Meaning:** The factors 1.2 and 1.6 are chosen such that: - Dead load is loaded at 1.2 (fairly high confidence in self-weight estimate, hence lower factor). - Live load is loaded at 1.6 (greater uncertainty in live load, hence higher factor). - The combination represents a mean-plus-variability approach: a rare event where D and L occur simultaneously at their mean plus some multiple of standard deviation. **Application in Reinforced Concrete:** For a beam or slab: $$M_u = 1.2M_D + 1.6M_L \quad (\text{moment due to factored loads})$$ $$V_u = 1.2V_D + 1.6V_L \quad (\text{shear due to factored loads})$$ The section is then designed so that: $$\phi M_n \ge M_u, \quad \phi V_n \ge V_u$$ where M_n and V_n are computed based on the geometric and material properties of the section.
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5. Load Factors and Factored Load Combinations (NSCP 2015)
Examples
Label
Example 5.1 — Factored Moment from Service Loads
Problem
A floor beam carries a dead load moment M_D = 90 kN·m (including self-weight) and a live load moment M_L = 70 kN·m. Compute the factored design moment M_u.
Solution
Using the primary gravity combination per NSCP 2015: $$M_u = 1.2M_D + 1.6M_L = 1.2(90) + 1.6(70) = 108 + 112 = 220 \text{ kN·m}$$ The beam must be proportioned so that φM_n ≥ 220 kN·m. For tension-controlled flexure, φ = 0.90, so M_n ≥ 220/0.90 = 244.4 kN·m is required.
Label
Example 5.2 — Multiple Load Combinations
Problem
A beam must be checked against three combinations: (a) 1.2D + 1.6L, (b) 1.2D + 1.6W, and (c) 0.9D + 1.6W. Service loads: D = 30 kN/m, L = 20 kN/m, W = 15 kN/m. Span L_beam = 6 m (pin supports, uniform load). Compute factored moment at midspan for each combination and identify which governs.
Solution
For a simply supported beam with uniform load, midspan moment M = wL²/8. (a) Combination 1.2D + 1.6L: $$M_u = \frac{1}{8}[1.2(30) + 1.6(20)](6)^2 = \frac{1}{8}(36 + 32)(36) = \frac{68 \times 36}{8} = 306 \text{ kN·m}$$ (b) Combination 1.2D + 1.6W: $$M_u = \frac{1}{8}[1.2(30) + 1.6(15)](6)^2 = \frac{1}{8}(36 + 24)(36) = \frac{60 \times 36}{8} = 270 \text{ kN·m}$$ (c) Combination 0.9D + 1.6W: $$M_u = \frac{1}{8}[0.9(30) + 1.6(15)](6)^2 = \frac{1}{8}(27 + 24)(36) = \frac{51 \times 36}{8} = 229.5 \text{ kN·m}$$ **Governing moment: M_u = 306 kN·m** (from combination 1.2D + 1.6L, where L dominates). This is the design moment; the section must satisfy φM_n ≥ 306 kN·m.
Label
Example 5.3 — Understanding Load Factors
Problem
Why is the load factor for live load (1.6) higher than for dead load (1.2)?
Solution
Dead load consists of structural self-weight, which is known with high precision (concrete density ~2400 kg/m³, steel density 7850 kg/m³, and dimensions are fixed). Engineers can estimate D accurately, so a lower factor (1.2) is appropriate. Live load is much less certain. A floor rated for 5 kPa live load might experience temporary clustering of people, equipment, or storage. Live load can vary significantly from the assumed design value. This greater uncertainty justifies a higher factor (1.6). Another way to think about it: the 1.2D and 1.6L factors, combined with variability in material strengths and resistance, define a target reliability level (probability of failure) of roughly 1 in 1,000–2,000 for normal buildings. The higher L factor accounts for the fact that occupancy loads are less predictable than dead load.
Key Points
- Factored loads: multiply service loads by load factors per NSCP 2015 Section 203.
- Primary gravity combination: U = 1.2D + 1.6L.
- D = dead load (self-weight); L = live load (occupancy, temporary).
- Load factor for D: 1.2 (confidence in weight estimate).
- Load factor for L: 1.6 (greater uncertainty, heavier loading).
- Alternative combinations with wind (W = 1.6) and seismic (E = 1.0) reduce L to 1.0 or 0.5.
- Minimal gravity (0.9D) used when live or wind/seismic loads might reduce total load.
- Design capacity must satisfy φM_n ≥ M_u for all governing combinations.
- The governing (largest) factored load from all combinations drives the design.
- NSCP 2015 provides specific combinations for different loading scenarios.
Though **NSCP 2015 mandates USD for new designs**, WSD remains important because: 1. Older Philippine buildings (pre-2000s) were designed to WSD codes. 2. The PRC Civil Engineer Licensure Examination tests both methods for comparison and historical knowledge. 3. WSD provides intuitive understanding of how concrete and steel work together under service loads. 4. Some designers still use WSD for preliminary sizing due to its simplicity. **WSD Philosophy:** WSD assumes the structure remains in the **linear-elastic range** under service loads (D + L, unfactored). Concrete and steel are treated as elastic materials with their respective moduli E_c and E_s. The key steps are: 1. Assume a cracked section (most practical for beams in tension). 2. Use the modular ratio n = E_s/E_c to convert steel area into equivalent concrete area. 3. Compute the neutral axis depth c from the condition that the first moment of area = 0. 4. Calculate concrete stress f_c and steel stress f_s from the moment using the elastic formula. 5. Check that f_c ≤ allowable f_c and f_s ≤ allowable f_s. **Typical Allowable Stresses (WSD):** Allowable concrete compressive stress: $f_c \le 0.45 f'_c$ (maximum at extreme fiber). Allowable steel stress: $f_s \le 0.4 f_y$ or $0.5 f_y$ (varies by loading and rebar type). **Transformed Section (WSD):** The composite section is transformed into an "equivalent concrete section" by replacing steel area A_s with nA_s. The neutral axis is then found from the centroid of the transformed area. Moment capacity is computed using $f = M \cdot y / I$, where y is the distance from the neutral axis and I is the moment of inertia of the transformed section. **Limitations of WSD:** - Does not directly address failure; relies on indirect "factors of safety" embedded in the allowable stresses. - Cannot easily account for variations in material strength or construction variability. - Linear-elastic assumption breaks down near failure; high-strength concrete and ductility are poorly represented. - Allowable stresses vary by code and year, creating confusion. Despite these limitations, understanding WSD is essential for: - Analyzing heritage structures. - Comparing with USD to see why the shift was made. - Solving exam problems that explicitly ask for WSD design or analysis. **Key Difference from USD:** WSD: service loads → check stresses ≤ allowables. USD: factored loads → check φ × nominal strength ≥ factored demand.
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6. Working Stress Design (WSD) — Elastic, Allowable-Stress Method
Examples
Label
Example 6.1 — WSD Transformed Section
Problem
A rectangular concrete beam has b = 300 mm, d = 500 mm (effective depth). Concrete strength f'_c = 28 MPa, steel Grade 415. The beam is reinforced with 4 bars of 20 mm diameter (A_s = 4 × 314.2 = 1256.8 mm²) near the bottom. Find the neutral-axis depth c in the transformed (cracked) section.
Solution
Step 1: Compute modular ratio. E_c = 4700√28 = 24,872 MPa (from Example 2.1). n = 200,000 / 24,872 = 8.04 ≈ 8. Step 2: Transformed section. The equivalent concrete area of steel is: A_s(equiv) = nA_s = 8 × 1256.8 = 10,054.4 mm². Step 3: Set first moment of area = 0 about the neutral axis. Concrete area above NA: b × c = 300c. Steel (below NA) at distance (d − c) from NA: $$300c \cdot \frac{c}{2} = 10,054.4(500 - c)$$ $$150c^2 = 10,054.4 \times 500 - 10,054.4c$$ $$150c^2 + 10,054.4c - 5,027,200 = 0$$ $$c^2 + 67.03c - 33,514.7 = 0$$ Using the quadratic formula: $$c = \frac{-67.03 + \sqrt{67.03^2 + 4(33,514.7)}}{2} = \frac{-67.03 + \sqrt{4493.0 + 134,058.8}}{2}$$ $$= \frac{-67.03 + \sqrt{138,551.8}}{2} = \frac{-67.03 + 372.23}{2} = \frac{305.2}{2} = 152.6 \text{ mm}$$ Neutral-axis depth: **c ≈ 153 mm**. This is the depth below the top fiber where the bending stress changes from compression (in concrete) to tension (in steel).
Label
Example 6.2 — WSD Stress Check
Problem
For the beam in Example 6.1, the service moment is M = 150 kN·m. Compute the bending stresses and verify they are within WSD allowables.
Solution
Step 1: Compute the moment of inertia of the transformed section about the neutral axis. Concrete contribution (rectangular): $$I_c = \frac{bc^3}{3} + A_c \left(\frac{c}{2}\right)^2$$ where $A_c = 300 \times 153 = 45,900$ mm², $\bar{y}_c = c/2 = 76.5$ mm from top. $$I_c = \frac{300 \times 153^3}{3} + 45,900(76.5)^2 = 117,351,900 + 268,609,350 = 385,961,250 \text{ mm}^4$$ Steel contribution (concentrated at d − c = 500 − 153 = 347 mm from NA): $$I_s = A_s(equiv) \times (d - c)^2 = 10,054.4 \times 347^2 = 10,054.4 \times 120,409 = 1,211,093,596 \text{ mm}^4$$ Total: $I = 385,961,250 + 1,211,093,596 = 1,597,054,846$ mm⁴. Step 2: Compute concrete stress at extreme compression fiber (top, y = c = 153 mm): $$f_c = \frac{M \cdot c}{I} = \frac{150 \times 10^6 \text{ N·mm} \times 153}{1,597,054,846} = \frac{22,950 \times 10^6}{1,597,054,846} = 14.37 \text{ MPa}$$ Allowable: $f_c \le 0.45 f'_c = 0.45 \times 28 = 12.6$ MPa. ✗ **Stress exceeds allowable!** The concrete stress (14.37 MPa) exceeds 12.6 MPa, so this section is **inadequate under WSD**. The section would need to be enlarged (larger b or d) or concrete strength increased. Step 3: Steel stress at depth (d − c) = 347 mm from NA: $$f_s = n \cdot f_c \cdot \frac{d - c}{c} = 8 \times 14.37 \times \frac{347}{153} = 8 \times 14.37 \times 2.268 = 259.9 \text{ MPa}$$ Allowable (Grade 415): $f_s \le 0.4 f_y = 0.4 \times 415 = 166$ MPa (lower bound) or 0.5 × 415 = 207.5 MPa (higher bound, rare). ✗ **Steel stress also exceeds allowable** under the conservative 0.4f_y limit (though it is under 0.5f_y). Conclusion: The section is under-designed for M = 150 kN·m under WSD with f'_c = 28 MPa and 4-20 bars. More reinforcement or a larger section is needed.
Key Points
- WSD: elastic, allowable-stress method; assumes linear behavior under service loads.
- Service loads: D + L unfactored (not multiplied by load factors).
- Cracked-section assumption: concrete below neutral axis is ineffective in tension.
- Modular ratio n = E_s/E_c transforms steel area to equivalent concrete area.
- Transformed section: replace A_s with nA_s; find neutral axis and moment of inertia.
- Check stresses: f_c ≤ 0.45f'_c (concrete), f_s ≤ 0.4f_y or 0.5f_y (steel).
- WSD does not explicitly address failure; safety is embedded in allowable fractions.
- Older Philippine designs (pre-2015 NSCP) may use WSD; still tested on licensure exams.
- WSD is simpler conceptually than USD but less realistic at near-failure conditions.
- USD has superseded WSD in modern codes because it is more rational and explicit.
This section consolidates the essential formulas and concepts from the chapter. Mastery of these relationships is critical for both theory questions and numerical problems on the PRC Civil Engineer Licensure Examination. **Material Properties:** $$E_c = 4700\sqrt{f'_c} \quad (\text{MPa; normal-weight concrete})$$ $$n = \frac{E_s}{E_c} = \frac{200,000}{E_c} \quad (\text{modular ratio, WSD})$$ $$\varepsilon_{ty} = \frac{f_y}{E_s} \quad (\text{yield strain, typically } 0.002\text{–}0.003)$$ **Stress-Block Factor β₁:** $$\beta_1 = \begin{cases} 0.85 & \text{if } f'_c \le 28 \text{ MPa} \\ 0.85 - 0.05 \frac{f'_c - 28}{7} & \text{if } 28 < f'_c \le 55 \text{ MPa} \\ 0.65 & \text{if } f'_c \ge 55 \text{ MPa} \end{cases}$$ Equivalent rectangular stress block depth: $$a = \beta_1 c \quad (\text{depth of 0.85} f'_c \text{ intensity block})$$ **Strength-Reduction Factors φ (NSCP 2015):** | Component | Condition | φ | |---|---|---| | Flexure | Tension-controlled (ε_t ≥ 0.005) | 0.90 | | Flexure | Compression-controlled (ε_t ≤ ε_ty) | 0.65 (tied) / 0.75 (spiral) | | Flexure | Transition (ε_ty < ε_t < 0.005) | Linear interpolation | | Shear | All | 0.75 | | Columns | Tied | 0.65 | | Columns | Spiral | 0.75 | | Bearing | All | 0.65 | **Load Factoring (NSCP 2015, Gravity Loads):** $$M_u = 1.2 M_D + 1.6 M_L \quad (\text{primary combination})$$ $$V_u = 1.2 V_D + 1.6 V_L$$ Other combinations with wind, seismic, or reduced live load apply; use governing (largest) factored demand. **USD Design Inequality (All Member Types):** $$\phi M_n \ge M_u, \quad \phi V_n \ge V_u, \quad \phi P_n \ge P_u$$ where M_n, V_n, P_n are the nominal (ultimate) capacities based on the section geometry and reinforcement. **WSD Allowable Stresses:** $$f_c \le 0.45 f'_c \quad (\text{concrete compression})$$ $$f_s \le 0.4 f_y \text{ or } 0.5 f_y \quad (\text{steel tension, varies by code})$$ Under service loads M = M_D + M_L (unfactored).
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7. Key Relationships and Summary of Formulas
Examples
Label
Example 7.1 — Complete Material Properties for a Design
Problem
An engineer is designing a reinforced concrete beam for a building in Metro Manila. The concrete strength is f'_c = 35 MPa, and Grade 415 steel is used. Compute all key material properties needed for subsequent design calculations.
Solution
Concrete properties: $$E_c = 4700\sqrt{35} = 4700 \times 5.916 = 27,805 \text{ MPa}$$ $$\beta_1 = 0.85 - 0.05 \frac{35 - 28}{7} = 0.85 - 0.05(1) = 0.80$$ Modular ratio (for WSD): $$n = \frac{200,000}{27,805} = 7.19 \approx 7.2$$ Steel properties: $$f_y = 415 \text{ MPa}$$ $$E_s = 200,000 \text{ MPa}$$ $$\varepsilon_{ty} = \frac{415}{200,000} = 0.002075$$ Tension-controlled threshold: $\varepsilon_t \ge 0.005$ Compression-controlled threshold: $\varepsilon_t \le 0.002075$ Design factors (for USD flexure): - Tension-controlled (most beams): φ = 0.90. - Compression-controlled (rare): φ = 0.65 (tied) or 0.75 (spiral). These values are used in all subsequent beam and column design calculations.
Label
Example 7.2 — Choosing the Governing Load Combination
Problem
A building floor beam must be checked under three load combinations: (1) 1.2D + 1.6L, (2) 1.2D + 0.5L + 1.6W, (3) 0.9D + 1.6W. Moments due to service loads: M_D = 80 kN·m, M_L = 60 kN·m, M_W = 40 kN·m. Determine M_u and φ.
Solution
Compute factored moment for each combination: (1) 1.2D + 1.6L: $$M_u = 1.2(80) + 1.6(60) = 96 + 96 = 192 \text{ kN·m}$$ (2) 1.2D + 0.5L + 1.6W: $$M_u = 1.2(80) + 0.5(60) + 1.6(40) = 96 + 30 + 64 = 190 \text{ kN·m}$$ (3) 0.9D + 1.6W: $$M_u = 0.9(80) + 1.6(40) = 72 + 64 = 136 \text{ kN·m}$$ **Governing (largest) moment: M_u = 192 kN·m** (from combination 1). Assuming the section is tension-controlled (typical for flexure), φ = 0.90. Required nominal moment: $M_n = M_u / \phi = 192 / 0.90 = 213.3$ kN·m. The beam section and reinforcement must be designed to provide at least 213.3 kN·m of nominal moment capacity.
Key Points
- E_c = 4700√f'_c is the fundamental relation for concrete modulus.
- β₁ = 0.85 for f'_c ≤ 28 MPa; decreases linearly above 28 MPa; minimum 0.65 at 55 MPa.
- φ = 0.90 for most beam flexure (tension-controlled); 0.75 for shear; 0.65–0.75 for columns.
- Load factors: 1.2 for D, 1.6 for L (gravity); adjust with wind/seismic combinations.
- USD inequality φM_n ≥ M_u is the core design check; select section size and reinforcement to satisfy.
- WSD: check stresses against allowables under unfactored service loads; less realistic but simpler.
- Transition zone (tied columns): φ interpolates linearly between 0.65 and 0.90 as ε_t increases from ε_ty to 0.005.
- Yield strain ε_ty = f_y/E_s ≈ 0.002 for Grade 415; 0.00138 for Grade 275.
- All formulas must be memorized and applied correctly; common exam pitfalls include wrong β₁, wrong φ, and mixing WSD/USD logic.
The following errors are frequently committed by exam takers. Awareness and avoidance are essential for success. **Pitfall 1: Confusing WSD and USD Logic** Error: Computing service-load stresses (like in WSD) but then comparing to φ-reduced strengths (like in USD). Correct Approach: **WSD uses unfactored service loads and allowable stresses**. **USD uses factored loads and φ-reduced nominal strengths**. Choose one method and apply it consistently. **Pitfall 2: Wrong β₁ for a Given f'_c** Error: Using β₁ = 0.85 for all concrete strengths, or incorrectly computing β₁ for 28 < f'_c ≤ 55 MPa. Correct Approach: Always check f'_c first. If f'_c ≤ 28, use β₁ = 0.85. If f'_c > 28, use the linear formula β₁ = 0.85 − 0.05(f'_c − 28)/7. Many students forget to check the threshold and apply the formula when f'_c = 21 or 24 MPa, yielding an incorrect (too-low) β₁. **Pitfall 3: Using 0.85 Instead of 0.85f'_c in the Stress Block** Error: Computing the compression force as C = 0.85 × b × a, forgetting to multiply by f'_c. Correct Approach: The block intensity is **0.85f'_c** (in MPa), so the compression force is $C = 0.85 f'_c \cdot b \cdot a$ (in newtons, when b and a are in mm and f'_c is in MPa, divide by 1000). **Pitfall 4: Incorrect Strength-Reduction Factor φ for Columns** Error: Using φ = 0.90 for a column (beams use 0.90, so students assume columns do too). Correct Approach: Columns are NOT tension-controlled in flexure. Use φ = 0.65 for tied columns and φ = 0.75 for spiral columns. Only flexure (beams, slabs, deep beams) use φ = 0.90 when tension-controlled. **Pitfall 5: Not Checking All Load Combinations** Error: Assuming the 1.2D + 1.6L combination always governs. Correct Approach: NSCP 2015 requires checking multiple combinations. Depending on the relative magnitudes of D, L, W, and E, the governing factored load might come from 0.9D + 1.6W or other combinations. Always compute all applicable combinations and use the largest demand. **Pitfall 6: Misidentifying Tension-Controlled vs. Compression-Controlled** Error: Confusing the strain threshold (0.005) or misinterpreting what "tension-controlled" means. Correct Approach: A section is **tension-controlled** if the net tensile strain ε_t ≥ 0.005 (5 times the yield strain, for any bar size). At this high strain, the concrete is not the limiting factor; steel yields and strains extensively. **Compression-controlled** means ε_t ≤ ε_ty (the yield strain), so the concrete crushes before steel yields; failure is abrupt. **Pitfall 7: Wrong Modular Ratio or Confused Transformed Section** Error: Using n = E_s / E_c incorrectly in the transformed section, or forgetting that only the steel is transformed (not the concrete). Correct Approach: The modular ratio n = E_s / E_c (always > 1 because steel is stiffer than concrete). In the transformed section, concrete area remains b × h, and steel area A_s is replaced by n × A_s. The neutral axis depth is found by setting the first moment of area = 0. **Pitfall 8: Load Factor Order or Sign** Error: Applying 1.2 to live load and 1.6 to dead load (backwards), or using negative factors. Correct Approach: Load factors are always positive. The typical gravity combination is **1.2D + 1.6L** (not the reverse). Mnemonic: "Live load gets the bigger factor" because it is more uncertain than dead load. **Pitfall 9: Forgetting to Reduce Capacity by φ in the Design Check** Error: Comparing nominal capacity M_n directly to factored demand M_u, without multiplying M_n by φ. Correct Approach: The USD inequality is $\phi M_n \ge M_u$, not $M_n \ge M_u$. Always reduce the nominal capacity by φ before comparing to the factored demand. φ is always less than 1, so φM_n < M_n; this is the "penalty" for uncertainty.
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8. Common Exam Pitfalls and Misconceptions
Examples
Label
Example 8.1 — β₁ Errors
Problem
A student computes β₁ for f'_c = 24 MPa by incorrectly applying the linear formula: β₁ = 0.85 − 0.05(24−28)/7 = 0.85 + 0.0286 = 0.879. Is this correct?
Solution
**No, this is wrong.** Since f'_c = 24 MPa ≤ 28 MPa, the correct answer is simply: $$\beta_1 = 0.85$$ The linear formula **only applies for 28 < f'_c ≤ 55 MPa**. Below 28 MPa, β₁ is constant at 0.85. The student's error: (a) applied the formula when it should not be used; (b) also got the sign wrong inside the formula (should be minus, not plus, but that is irrelevant since the formula does not apply). Mnemonic: "If f'_c ≤ 28, just use 0.85. No formula needed."
Label
Example 8.2 — Mixing WSD and USD
Problem
A beam has M_D = 100 kN·m and M_L = 80 kN·m. A student computes M_u = 1.2(100) + 1.6(80) = 248 kN·m (correct USD factoring), then checks the concrete stress using the WSD formula f_c = M × y / I (unfactored moment in the stress formula). Is this approach valid?
Solution
**No, this mixes the two methods and is invalid.** Here is what went wrong: The student correctly factored the loads (USD step), but then used WSD stress-checking logic (E_c-based linear analysis of unfactored moment). These cannot be combined. Correct approach for USD: Factor the loads to get M_u = 248 kN·m. Compute the section's nominal moment capacity M_n using the equivalent rectangular stress block (0.85f'_c). Then check φM_n ≥ M_u. Correct approach for WSD (if required): Use unfactored M = M_D + M_L = 100 + 80 = 180 kN·m. Compute stresses in the transformed section. Check stresses against allowables (0.45f'_c for concrete, 0.4f_y for steel). The two must not be intermixed.
Key Points
- WSD and USD are incompatible; choose one and apply it consistently.
- β₁ = 0.85 for f'_c ≤ 28 MPa; use the linear formula for 28 < f'_c ≤ 55 MPa; minimum 0.65 for f'_c ≥ 55 MPa.
- Stress-block intensity is 0.85f'_c (not just 0.85); include f'_c in all calculations.
- Column φ: 0.65 (tied) or 0.75 (spiral); not 0.90.
- Beam φ: 0.90 (tension-controlled, typical); 0.65–0.75 (compression-controlled, rare).
- Check all governing load combinations; 1.2D + 1.6L does not always control.
- Tension-controlled threshold: ε_t ≥ 0.005; compression-controlled: ε_t ≤ ε_ty ≈ 0.002.
- Modular ratio n = E_s / E_c; used only in WSD transformed-section analysis.
- Load factors: 1.2 for D, 1.6 for L (gravity). Never swap them.
- USD check: φM_n ≥ M_u (reduce nominal capacity by φ before comparing).
- WSD check: f_c ≤ 0.45f'_c and f_s ≤ allowable (use unfactored service loads).
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