14.1 Ultimate Strength Design (USD) Principles and Singly Reinforced Beams

14.1 Ultimate Strength Design (USD) Principles and Singly Reinforced Beams

Reinforced concrete design in the Philippines is governed by the National Structural Code of the Philippines (NSCP 2015, Volume 1, 7th Edition), which largely mirrors the provisions of ACI 318M-14. In modern structural practice and the PRC Civil Engineering Licensure Examination (CELE), the Ultimate Strength Design (USD) method—also known as the Strength Design Method—is the standard for proportioning structural concrete members. USD replaces the historical Working Stress Design (WSD) by applying explicit load factors to service loads and strength reduction factors ($\phi$) to nominal member resistances, providing a consistent margin of safety against structural collapse.

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1. Ultimate Strength Design (USD) Philosophy

The fundamental safety criterion of Ultimate Strength Design dictates that the design strength provided by a structural component must equal or exceed the required strength generated by factored service loads:

$\phi R_n \ge U \quad \implies \quad \phi M_n \ge M_u$

Where:

• $M_u$ = factored design bending moment calculated from structural analysis under governing load combinations.
• $M_n$ = nominal flexural moment strength computed using theoretical equilibrium and strain compatibility.
• $\phi$ = strength reduction factor accounting for structural importance, dimensional tolerances, and the relative ductility of the failure mode.

NSCP 2015 Basic Factored Load Combinations

Under NSCP 2015 Section 203.3 / Section 405.3, the primary factored gravity and lateral load combinations include:

1. $U = 1.4 D$
2. $U = 1.2 D + 1.6 L + 0.5 (L_r \text{ or } R)$
3. $U = 1.2 D + 1.0 L + 1.0 W$
4. $U = 1.2 D + 1.0 L + 1.0 E$
5. $U = 0.9 D + 1.0 W$
6. $U = 0.9 D + 1.0 E$

Where $D$ is dead load, $L$ is live load, $L_r$ is roof live load, $R$ is rain load, $W$ is wind load, and $E$ is earthquake load.

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2. Fundamental Assumptions for Flexure

Flexural analysis of reinforced concrete relies on five foundational physical assumptions specified in NSCP 2015 Section 422:

1. Strain Linearity: Plane sections perpendicular to the longitudinal axis before bending remain plane after bending (Navier-Bernoulli hypothesis). Consequently, strain in both concrete and steel varies directly with distance from the neutral axis ($c$).
2. Maximum Usable Concrete Strain: The maximum usable compressive strain at the extreme concrete compression fiber is taken as $\epsilon_u = 0.003$ at ultimate limit state.
3. Tensile Strength Neglected: The tensile strength of concrete is completely ignored in flexural strength calculations because concrete cracks at low tensile strains (tensile rupture modulus $f_r = 0.62 \lambda \sqrt{f'_c}\text{ MPa}$).
4. Elastoplastic Steel Behavior: Reinforcing steel behaves as a linear-elastic, perfectly plastic material. For strains less than yield strain $\epsilon_y = f_y / E_s$, steel stress is $f_s = E_s \epsilon_s$. For $\epsilon_s \ge \epsilon_y$, steel stress remains constant at $f_s = f_y$. Modulus of elasticity of steel is taken as $E_s = 200,000\text{ MPa}$.
5. Perfect Bond: Strain compatibility exists between reinforcing bars and surrounding concrete; no slip occurs prior to ultimate capacity.

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3. Whitney's Equivalent Rectangular Stress Block

At ultimate flexural capacity, the true stress distribution in the concrete compression zone is non-linear and parabolic. Charles S. Whitney proposed an equivalent rectangular stress block that yields identical resultant compressive force ($C_c$) and line of action to the actual stress envelope.

Under NSCP 2015 Section 422.2.2.4:

• An equivalent uniform compressive stress of $0.85 f'_c$ is assumed to act over an equivalent depth $a$ bounded by edges of the cross-section and a line parallel to the neutral axis at distance $a = \beta_1 c$ from the extreme compression fiber.
• The resultant compressive force of the concrete is:

$C_c = 0.85 f'_c a b$

• The resultant acts at a distance of $a/2$ from the extreme compression fiber.

Stress Block Factor $\beta_1$

The factor $\beta_1$ relates equivalent stress block depth $a$ to true neutral axis depth $c$ ($a = \beta_1 c$):

$\beta_1 = 0.85 \quad \text{for } 17\text{ MPa} \le f'_c \le 28\text{ MPa}$

$\beta_1 = 0.85 - \frac{0.05 (f'_c - 28)}{7} \quad \text{for } f'_c > 28\text{ MPa}$

$\beta_1 \ge 0.65 \quad (\text{absolute lower limit})$

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4. Strain Compatibility and Strength Reduction Factor ($\phi$)

The behavior and ductility of a beam cross-section are governed by the net tensile strain ($\epsilon_t$) in the extreme layer of longitudinal tension steel at nominal strength.

From similar triangles in the linear strain diagram:

$\frac{\epsilon_u}{c} = \frac{\epsilon_t}{d_t - c} \implies \epsilon_t = 0.003 \left(\frac{d_t - c}{c}\right)$

where $d_t$ is the distance from extreme compression fiber to the centroid of the extreme tension steel layer (equal to effective depth $d$ for a single layer of steel).

   COMPRESSION FIBER
   -----------------  <-- Strain = ε_u = 0.003
          |      /
          |     / 
          |    /  
        c |   /   (Neutral Axis)
   - - - -|- / - - - - - - - - - - - - 
          | /     
          |/      
   (d - c)|       
          |\      
          | \     
   -------|--\----    <-- Strain = ε_t (Tension Steel)

Section Classification and $\phi$ Values (NSCP 2015 Table 421.2.1)

1. Tension-Controlled Section ($\epsilon_t \ge 0.005$):
   • Steel yields substantially before concrete reaches $\epsilon_u = 0.003$.
   • Warning of failure through large deflections and wide cracking.
   • $\phi = 0.90$.
2. Transition Zone ($0.002 \le \epsilon_t < 0.005$ for Grade 420 steel):
   • Steel yields, but ductility is reduced.
   • Strength reduction factor is linearly interpolated: $\phi = 0.65 + (\epsilon_t - 0.002) \left(\frac{250}{3}\right) = 0.65 + 0.25 \left(\frac{\epsilon_t - 0.002}{0.003}\right)$
3. Compression-Controlled Section ($\epsilon_t \le \epsilon_y \approx 0.002$):
   • Brittle, explosive crushing of concrete before steel yields.
   • $\phi = 0.65$ for members with transverse ties.
   • $\phi = 0.75$ for members with continuous circular spirals.

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5. Reinforcement Ratios: Balanced, Maximum, and Minimum

The steel reinforcement ratio is defined as:

$\rho = \frac{A_s}{b d}$

Internal Force Equilibrium

For a singly reinforced rectangular beam with tension steel yielding ($f_s = f_y$):

$C_c = T \implies 0.85 f'_c a b = A_s f_y$

$a = \frac{A_s f_y}{0.85 f'_c b} = \frac{\rho f_y d}{0.85 f'_c}$

$c = \frac{a}{\beta_1} = \frac{\rho f_y d}{0.85 \beta_1 f'_c}$

Balanced Reinforcement Ratio ($\rho_b$)

A balanced condition occurs when the extreme concrete compression fiber reaches strain $\epsilon_u = 0.003$ simultaneously as the tension reinforcement reaches its initial yield strain $\epsilon_y = f_y / E_s$:

$\frac{c_b}{d} = \frac{\epsilon_u}{\epsilon_u + \epsilon_y} = \frac{0.003}{0.003 + f_y / 200000} = \frac{600}{600 + f_y}$

Equating $C_c = T$ at balanced conditions yields:

$\rho_b = \frac{0.85 f'_c \beta_1}{f_y} \left(\frac{600}{600 + f_y}\right)$

Maximum Reinforcement Ratio ($\rho_{\max}$)

To ensure ductile behavior and prevent sudden brittle compression failures:

• NSCP 2015 / ACI 318 Tension-Controlled Limit: The design requires $\epsilon_t \ge 0.005$ to use $\phi = 0.90$. Setting $\epsilon_t = 0.005$ at $d_t = d$: $\frac{c}{d} = \frac{0.003}{0.003 + 0.005} = \frac{3}{8} = 0.375$ $\rho_{\max} = \frac{0.85 f'_c \beta_1}{f_y} (0.375) = \frac{0.85 f'_c \beta_1}{f_y} \left(\frac{3}{8}\right)$
• NSCP 2001 (Legacy Code Limit): Often still referenced in CELE problems: $\rho_{\max} = 0.75 \rho_b$

Minimum Reinforcement Ratio ($\rho_{\min}$)

If a beam has too little reinforcement, the flexural capacity of the uncracked concrete section ($M_{cr}$) may exceed the nominal capacity of the reinforced cracked section ($M_n$), leading to immediate catastrophic rupture upon initial cracking. NSCP 2015 Section 409.6.1.2 mandates:

$A_{s,\min} = \frac{\sqrt{f'_c}}{4 f_y} b_w d \ge \frac{1.4}{f_y} b_w d$

$\rho_{\min} = \max\left(\frac{\sqrt{f'_c}}{4 f_y}, \quad \frac{1.4}{f_y}\right)$

(Note: $\frac{\sqrt{f'_c}}{4 f_y}$ governs when $f'_c > 31.36\text{ MPa}$; otherwise $\frac{1.4}{f_y}$ governs.)

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6. Nominal and Design Flexural Capacity

Once the depth of the compression block $a$ is obtained, the internal moment arm is $(d - a/2)$. The nominal flexural strength is:

$M_n = T \left(d - \frac{a}{2}\right) = A_s f_y \left(d - \frac{a}{2}\right)$

Expressed in terms of the reinforcement ratio $\rho$:

$M_n = \rho f_y b d^2 \left(1 - 0.59 \frac{\rho f_y}{f'_c}\right) = R_n b d^2$

Where the coefficient of resistance $R_n$ is:

$R_n = \rho f_y \left(1 - 0.59 \frac{\rho f_y}{f'_c}\right)$

The design moment capacity is:

$\phi M_n = \phi R_n b d^2 \ge M_u$

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7. Step-by-Step Analysis vs. Design Procedures

Analysis of a Given Cross-Section ($b, d, A_s, f'_c, f_y$ known):

1. Calculate $\rho = A_s / (b d)$. Verify that $\rho \ge \rho_{\min}$.
2. Compute $a = \frac{A_s f_y}{0.85 f'_c b}$ and determine $\beta_1$.
3. Calculate neutral axis depth $c = a / \beta_1$.
4. Calculate extreme tension steel strain $\epsilon_t = 0.003 \frac{d_t - c}{c}$.
   • If $\epsilon_t \ge \epsilon_y$, steel has yielded ($f_s = f_y$).
   • If $\epsilon_t < \epsilon_y$, recalculate $c$ using quadratic equilibrium: $0.85 f'_c \beta_1 b c^2 + 600 A_s c - 600 A_s d = 0$.
5. Determine $\phi$ based on $\epsilon_t$ ($\phi = 0.90$ if $\epsilon_t \ge 0.005$).
6. Compute $M_n = A_s f_y (d - a/2)$ and design strength $\phi M_n$.

Design of Beam Reinforcement ($M_u, b, d, f'_c, f_y$ given):

1. Assume tension-controlled behavior ($\phi = 0.90$).
2. Compute required resistance factor: $R_n = \frac{M_u}{\phi b d^2}$.
3. Solve for required reinforcement ratio $\rho$: $\rho = \frac{0.85 f'_c}{f_y} \left(1 - \sqrt{1 - \frac{2 R_n}{0.85 f'_c}}\right)$
4. Verify that $\rho_{\min} \le \rho \le \rho_{\max}$. If $\rho > \rho_{\max}$, enlarge the beam cross-section or design as a doubly reinforced beam.
5. Compute required steel area $A_s = \rho b d$ and select suitable bar diameter and quantity.

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8. Comprehensive Worked Examples

Worked Example 1: Analysis of a Singly Reinforced Rectangular Beam

Problem: A reinforced concrete beam has a width of $b = 300\text{ mm}$, an overall depth of $h = 550\text{ mm}$, and an effective depth of $d = 485\text{ mm}$. It is reinforced with $4 - \phi 25\text{ mm}$ bars in a single layer. Specified material strengths are $f'_c = 28\text{ MPa}$ and $f_y = 420\text{ MPa}$. Determine: (a) depth of the equivalent stress block, (b) extreme tension steel strain $\epsilon_t$, (c) nominal flexural strength $M_n$, and (d) design moment capacity $\phi M_n$.

Solution:

• Step 1: Section Properties and Steel Area: $A_s = 4 \times \left(\frac{\pi}{4} \times 25^2\right) = 4 \times 490.87 = 1963.5\text{ mm}^2$ $\rho = \frac{A_s}{b d} = \frac{1963.5}{300 \times 485} = 0.013495 \quad (1.35\%$

• Step 2: Check Code Limits: $\rho_{\min} = \max\left(\frac{\sqrt{28}}{4 \times 420}, \frac{1.4}{420}\right) = \max(0.00315, 0.00333) = 0.00333$ Since $\rho = 0.0135 > 0.00333$, minimum steel requirement is satisfied.

  For $f'_c = 28\text{ MPa} \implies \beta_1 = 0.85$. $\rho_b = \frac{0.85(28)(0.85)}{420} \left(\frac{600}{600 + 420}\right) = 0.048167 \times 0.588235 = 0.02833$ $\rho_{\max} (\epsilon_t = 0.005) = \frac{0.85(28)(0.85)}{420} \times 0.375 = 0.01806$ Since $\rho = 0.0135 < 0.01806$, the section is guaranteed to be tension-controlled.

• Step 3: Depth of Equivalent Stress Block ($a$) and Neutral Axis ($c$): $a = \frac{A_s f_y}{0.85 f'_c b} = \frac{1963.5 \times 420}{0.85 \times 28 \times 300} = \frac{824,670}{7,140} = 115.50\text{ mm}$ $c = \frac{a}{\beta_1} = \frac{115.50}{0.85} = 135.88\text{ mm}$

• Step 4: Check Tensile Strain and Determine $\phi$: With a single layer of steel, $d_t = d = 485\text{ mm}$: $\epsilon_t = 0.003 \left(\frac{d_t - c}{c}\right) = 0.003 \left(\frac{485 - 135.88}{135.88}\right) = 0.003 \left(\frac{349.12}{135.88}\right) = 0.00771$ Since $\epsilon_t = 0.00771 > 0.005$, the beam is fully tension-controlled $\implies \phi = 0.90$.

• Step 5: Nominal and Design Flexural Strengths: $M_n = A_s f_y \left(d - \frac{a}{2}\right) = 1963.5 \times 420 \times \left(485 - \frac{115.50}{2}\right) \times 10^{-6}$ $M_n = 824.670 \times 427.25 \times 10^{-3} = 352.34\text{ kN}\cdot\text{m}$ $\phi M_n = 0.90 \times 352.34 = 317.11\text{ kN}\cdot\text{m}$

Worked Example 2: Flexural Design of a Singly Reinforced Beam

Problem: A simply supported rectangular beam with $b = 300\text{ mm}$ and effective depth $d = 500\text{ mm}$ must resist a factored moment of $M_u = 260\text{ kN}\cdot\text{m}$. Material strengths are $f'_c = 21\text{ MPa}$ and $f_y = 420\text{ MPa}$. Design the required tension reinforcement area $A_s$.

Solution:

• Step 1: Compute Required Resistance Factor ($R_n$): Assuming a tension-controlled section ($\phi = 0.90$): $R_n = \frac{M_u}{\phi b d^2} = \frac{260 \times 10^6}{0.90 \times 300 \times 500^2} = \frac{260 \times 10^6}{67.5 \times 10^6} = 3.8519\text{ MPa}$

• Step 2: Calculate Required Reinforcement Ratio ($\rho$): $\rho = \frac{0.85(21)}{420} \left(1 - \sqrt{1 - \frac{2(3.8519)}{0.85(21)}}\right) = 0.0425 \times \left(1 - \sqrt{1 - 0.43158}\right)$ $\rho = 0.0425 \times (1 - \sqrt{0.56842}) = 0.0425 \times (1 - 0.75394) = 0.010458 \quad (1.05\%$

• Step 3: Check Limits: $\rho_{\min} = \frac{1.4}{420} = 0.00333$ $\rho_{\max} (\epsilon_t = 0.005) = \frac{0.85(21)(0.85)}{420} \times 0.375 = 0.01355$ Since $0.00333 \le \rho = 0.01046 \le 0.01355$, the tension-controlled assumption $\phi = 0.90$ is fully validated.

• Step 4: Compute Required Area of Steel: $A_s = \rho b d = 0.010458 \times 300 \times 500 = 1568.7\text{ mm}^2$ Selection: Using $28\text{ mm}$ bars ($A_b = 615.75\text{ mm}^2$): $N = \frac{1568.7}{615.75} = 2.55 \implies \text{Provide } 3 - \phi 28\text{ mm} \quad (A_s = 1847\text{ mm}^2)$

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9. Licensure Exam Pitfalls & Review Notes