14.2 Doubly Reinforced Beams and T-Beams

14.2 Doubly Reinforced Beams and T-Beams

When structural beam dimensions are constrained by architectural headroom or clearance limits, tension reinforcement alone may exceed the maximum permissible steel ratio ($\rho_{\max}$), resulting in a non-ductile compression failure. To overcome this limitation, structural engineers add compression reinforcement ($A'_s$), creating a doubly reinforced beam. Furthermore, in cast-in-place reinforced concrete floor systems, floor slabs are poured monolithically with the supporting beams, forming flanged sections known as T-beams (interior bays) and L-beams (spandrel/edge beams). Understanding the flexural mechanics of both systems is a staple of the PRC Civil Engineering Licensure Examination.

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1. Role and Behavior of Compression Steel ($A'_s$)

Although compression steel increases the ultimate flexural capacity of a beam, that is rarely its primary design objective. The key engineering reasons for providing compression steel include:

1. Long-Term Deflection Control: Concrete undergoes creep and drying shrinkage under sustained dead and live loads. Compression reinforcement carries a progressively increasing portion of the compressive force, relieving the concrete and dramatically suppressing long-term deflections. Under NSCP 2015 Section 424.2.4.1.1, the long-term deflection multiplier is: $\lambda_\Delta = \frac{\xi}{1 + 50 \rho'}$ where $\rho' = A'_s / (b d)$ and $\xi$ is a time-dependent factor ($2.0$ for $\ge 5$ years, $1.4$ for 12 months, $1.2$ for 6 months, $1.0$ for 3 months).
2. Ductility Enhancement: By carrying compressive forces, compression steel allows the concrete compression block depth ($c$) to decrease. This shifts the neutral axis upward, significantly increasing the net tensile strain ($\epsilon_t$) in the tension steel and guaranteeing tension-controlled ductile failure ($\phi = 0.90$).
3. Seismic Moment Reversal: During severe earthquake shaking, lateral sway induces positive moments at beam ends that normally experience negative moments, requiring bottom reinforcement to act in tension and top reinforcement in compression.
4. Fabrication Rigidity: Compression bars provide secure top anchoring points for shear stirrups and prevent distortion of the reinforcement cage during concrete casting.

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2. Strain Compatibility and Yield Verification for Compression Steel

Unlike tension reinforcement—which is designed to yield—compression reinforcement may or may not yield before the concrete reaches its crushing strain of $\epsilon_u = 0.003$.

From the linear strain diagram across the section:

$\frac{\epsilon'_s}{c - d'} = \frac{0.003}{c} \implies \epsilon'_s = 0.003 \left(\frac{c - d'}{c}\right)$

Where $d'$ is the distance from the extreme compression fiber to the centroid of the compression steel.

Condition for Compression Steel Yielding

The compression reinforcement yields if its strain exceeds the yield strain: $\epsilon'_s \ge \epsilon_y = f_y / E_s$.

Setting $\epsilon'_s = \frac{f_y}{200,000}$:

$0.003 \left(\frac{c - d'}{c}\right) \ge \frac{f_y}{200,000} \implies c \ge c_{\text{yield}} = \frac{600 d'}{600 - f_y}$

In terms of reinforcement ratios, compression steel yields at nominal strength if and only if:

$(\rho - \rho') \ge \bar{\rho}_{cy} = \frac{0.85 f'_c \beta_1 d'}{f_y d} \left(\frac{600}{600 - f_y}\right)$

Where $\rho = A_s / (b d)$ and $\rho' = A'_s / (b d)$.

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3. Equilibrium and Flexural Capacity of Doubly Reinforced Beams

Case 1: Compression Steel Yields ($f'_s = f_y$)

When $c \ge c_{\text{yield}}$:

• Concrete compression: $C_c = 0.85 f'_c a b$
• Steel compression: $C_s = A'_s (f_y - 0.85 f'_c)$ (or approximately $C_s = A'_s f_y$ if concrete displaced by steel is neglected)
• Steel tension: $T = A_s f_y$

Enforcing horizontal equilibrium ($\sum F_x = 0$):

$C_c + C_s = T \implies 0.85 f'_c a b + A'_s f_y = A_s f_y$

$a = \frac{(A_s - A'_s) f_y}{0.85 f'_c b}$

The nominal moment capacity ($M_n$) is the sum of the concrete couple ($M_{n1}$) and the compression steel couple ($M_{n2}$):

$M_n = M_{n1} + M_{n2} = 0.85 f'_c a b \left(d - \frac{a}{2}\right) + A'_s f_y (d - d')$

Case 2: Compression Steel Does Not Yield ($f'_s < f_y$)

When $c < c_{\text{yield}}$, the stress in the compression steel is governed by Hooke's Law:

$f'_s = E_s \epsilon'_s = 200,000 \times 0.003 \left(\frac{c - d'}{c}\right) = 600 \left(\frac{c - d'}{c}\right)$

Substituting into horizontal equilibrium ($0.85 f'_c a b + A'_s f'_s = A_s f_y$) with $a = \beta_1 c$:

$0.85 f'_c (\beta_1 c) b + A'_s \left[600 \left(\frac{c - d'}{c}\right)\right] = A_s f_y$

Multiplying through by $c$ produces a quadratic equation in $c$:

$(0.85 f'_c \beta_1 b) c^2 + (600 A'_s - A_s f_y) c - 600 A'_s d' = 0$

Solving this quadratic equation yields the exact neutral axis depth $c$. The depth of the stress block is $a = \beta_1 c$, and the compression steel stress is $f'_s = 600 (c - d') / c$. Nominal moment capacity is:

$M_n = 0.85 f'_c a b \left(d - \frac{a}{2}\right) + A'_s f'_s (d - d')$

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4. T-Beams and Flanged Sections: Geometry and Effective Flange Width

In cast-in-place floor systems, beams and slabs are cast as a single monolithic unit. Under positive bending, the slab acts as a wide compression flange. However, due to shear deformation in the slab (known as shear lag), longitudinal compressive stresses decrease with distance from the web ($b_w$). To account for this, codes define an effective flange width ($b_e$) over which compressive stresses are assumed to be uniform.

NSCP 2015 Criteria for Effective Flange Width (Section 406.3.2)

NSCP 2015 follows ACI 318-14 Table 6.3.2.1, which limits the overhang on each side of the web. Here $s_w$ is the clear distance to the adjacent web and $\ell_n$ is the clear span of the beam.

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5. Flexural Analysis of T-Beams

Analysis of a flanged section begins by determining whether the compression zone penetrates into the web.

   <-------------- b_e -------------->
   ===================================  ^ 
   |             Flange              |  | h_f
   ===================================  v
           |      Web      |
           |     (b_w)     |

Compute the maximum compressive capacity that the entire flange can provide:

$C_{f,\max} = 0.85 f'_c b_e h_f$

Calculate the total tensile force at yield: $T = A_s f_y$.

Condition A: Rectangular Beam Behavior ($T \le C_{f,\max} \implies a \le h_f$)

If $T \le C_{f,\max}$, the equivalent rectangular stress block depth $a$ is less than or equal to slab thickness $h_f$. The neutral axis lies within the flange. The tension concrete in the web is neglected anyway, so the beam behaves identically to a rectangular beam of width $b = b_e$:

$a = \frac{A_s f_y}{0.85 f'_c b_e} \le h_f$

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

Condition B: True T-Beam Behavior ($T > C_{f,\max} \implies a > h_f$)

If $T > C_{f,\max}$, the compression block penetrates into the web ($a > h_f$). The compression zone consists of two distinct parts:

1. Overhanging Flanges ($b_e - b_w$): Carries compressive force $C_f$: $C_f = 0.85 f'_c (b_e - b_w) h_f$ Balanced by tension steel area $A_{sf} = \frac{C_f}{f_y}$. Moment arm: $(d - h_f / 2)$.
2. Web Compression ($b_w \times a$): Carries compressive force $C_w$: $C_w = 0.85 f'_c a b_w$ Balanced by remaining tension steel: $A_{sw} = A_s - A_{sf}$. Depth of web compression block: $a = \frac{A_{sw} f_y}{0.85 f'_c b_w} = \frac{(A_s - A_{sf}) f_y}{0.85 f'_c b_w}$ Moment arm: $(d - a / 2)$.

Total nominal flexural strength is:

$M_n = M_{nf} + M_{nw} = C_f \left(d - \frac{h_f}{2}\right) + C_w \left(d - \frac{a}{2}\right)$

$M_n = 0.85 f'_c (b_e - b_w) h_f \left(d - \frac{h_f}{2}\right) + 0.85 f'_c a b_w \left(d - \frac{a}{2}\right)$

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6. Comprehensive Worked Examples

Worked Example 1: Analysis of a Doubly Reinforced Beam

Problem: A rectangular beam has $b = 350\text{ mm}$, effective depth $d = 600\text{ mm}$, and compression steel depth $d' = 65\text{ mm}$. The beam is reinforced with $6 - \phi 32\text{ mm}$ tension bars ($A_s = 4825.5\text{ mm}^2$) and $2 - \phi 25\text{ mm}$ compression bars ($A'_s = 981.7\text{ mm}^2$). Material strengths are $f'_c = 28\text{ MPa}$ and $f_y = 420\text{ MPa}$. Determine whether the compression steel yields and calculate the nominal moment capacity $M_n$.

Solution:

• Step 1: Check Yield Condition of Compression Steel: For $f'_c = 28\text{ MPa}$, $\beta_1 = 0.85$. The yield limit for neutral axis depth is: $c_{\text{yield}} = \frac{600 d'}{600 - f_y} = \frac{600 \times 65}{600 - 420} = \frac{39,000}{180} = 216.67\text{ mm}$

  Assume compression steel yields ($f'_s = f_y = 420\text{ MPa}$): $a = \frac{(A_s - A'_s) f_y}{0.85 f'_c b} = \frac{(4825.5 - 981.7) \times 420}{0.85 \times 28 \times 350} = \frac{3843.8 \times 420}{8330} = 193.80\text{ mm}$ $c = \frac{a}{\beta_1} = \frac{193.80}{0.85} = 228.00\text{ mm}$

  Check assumption: Since $c = 228.00\text{ mm} > c_{\text{yield}} = 216.67\text{ mm}$, the compression steel has yielded ($f'_s = f_y = 420\text{ MPa}$). Assumption verified!

• Step 2: Check Tension Steel Ductility: $\epsilon_t = 0.003 \left(\frac{d - c}{c}\right) = 0.003 \left(\frac{600 - 228.0}{228.0}\right) = 0.00489$ Since $0.002 < \epsilon_t = 0.00489 < 0.005$, the beam is in the transition zone: $\phi = 0.65 + (0.00489 - 0.002) \times \left(\frac{250}{3}\right) = 0.65 + 0.2408 = 0.891$

• Step 3: Calculate Nominal Moment Strength ($M_n$): $C_c = 0.85 f'_c a b = 0.85 \times 28 \times 193.80 \times 350 \times 10^{-3} = 1614.35\text{ kN}$ $C_s = A'_s f_y = 981.7 \times 420 \times 10^{-3} = 412.31\text{ kN}$ $M_n = C_c \left(d - \frac{a}{2}\right) + C_s (d - d')$ $M_n = 1614.35 \times \left(600 - \frac{193.80}{2}\right) \times 10^{-3} + 412.31 \times (600 - 65) \times 10^{-3}$ $M_n = 1614.35 \times 0.5031 + 412.31 \times 0.535 = 812.18 + 220.59 = 1032.77\text{ kN}\cdot\text{m}$ $\phi M_n = 0.891 \times 1032.77 = 920.20\text{ kN}\cdot\text{m}$

Worked Example 2: Flexural Capacity of a True T-Beam

Problem: An interior T-beam has an effective flange width $b_e = 900\text{ mm}$, flange thickness $h_f = 110\text{ mm}$, web width $b_w = 300\text{ mm}$, and effective depth $d = 550\text{ mm}$. Reinforcement consists of $A_s = 5200\text{ mm}^2$. Material strengths are $f'_c = 21\text{ MPa}$ and $f_y = 420\text{ MPa}$. Determine the nominal moment capacity $M_n$.

Solution:

• Step 1: Check Flange Capacity vs. Tension Force: $T = A_s f_y = 5200 \times 420 = 2,184,000\text{ N} = 2184.0\text{ kN}$ $C_{f,\max} = 0.85 f'_c b_e h_f = 0.85 \times 21 \times 900 \times 110 = 1,766,970\text{ N} = 1767.0\text{ kN}$ Since $T = 2184.0\text{ kN} > C_{f,\max} = 1767.0\text{ kN}$, the stress block penetrates the web ($a > h_f$). This is a true T-beam.

• Step 2: Flange Component ($M_{nf}$): $C_f = 0.85 f'_c (b_e - b_w) h_f = 0.85 \times 21 \times (900 - 300) \times 110 = 1,178,100\text{ N} = 1178.1\text{ kN}$ $A_{sf} = \frac{C_f}{f_y} = \frac{1,178,100}{420} = 2805.0\text{ mm}^2$ $M_{nf} = C_f \left(d - \frac{h_f}{2}\right) = 1178.1 \times \left(550 - \frac{110}{2}\right) \times 10^{-3} = 1178.1 \times 0.495 = 583.16\text{ kN}\cdot\text{m}$

• Step 3: Web Component ($M_{nw}$): $A_{sw} = A_s - A_{sf} = 5200 - 2805.0 = 2395.0\text{ mm}^2$ $a = \frac{A_{sw} f_y}{0.85 f'_c b_w} = \frac{2395.0 \times 420}{0.85 \times 21 \times 300} = \frac{1,005,900}{5355} = 187.84\text{ mm}$ (Note: $a = 187.84\text{ mm} > h_f = 110\text{ mm}$, confirming web penetration.) $M_{nw} = 0.85 f'_c a b_w \left(d - \frac{a}{2}\right) = 1005.9\text{ kN} \times \left(550 - \frac{187.84}{2}\right) \times 10^{-3}$ $M_{nw} = 1005.9 \times 0.45608 = 458.77\text{ kN}\cdot\text{m}$

• Step 4: Total Nominal Moment Capacity: $M_n = M_{nf} + M_{nw} = 583.16 + 458.77 = 1041.93\text{ kN}\cdot\text{m}$

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7. Licensure Exam Pitfalls & Review Notes