When does a T-beam behave as a rectangular beam?

When the Whitney stress-block depth a is less than the flange thickness hf, the neutral axis stays in the flange and the section is analyzed as rectangular with width bf instead of bw.

How does T-beam flexure differ when the neutral axis falls in the web?

The compression zone splits into two parts: the flange overhangs (Asf, moment arm d - hf/2) and the web (Asw, moment arm d - a_web/2). The total moment capacity is the sum of both contributions.

Does the T-beam flange affect shear capacity?

No. ACI Table 22.5.5.1 computes Vc using the web width bw and effective depth d, regardless of flange geometry. The flange contributes to flexure but not shear.

T-beam vs rectangular beam: ACI 318-19 comparison

A beam cast monolithically with a slab uses the slab as a compression flange. ACI 318-19 Table 6.3.2.1 limits the effective flange width. When the compression zone (stress block depth a) stays within the flange, the section behaves as a wider rectangular beam. When a exceeds the flange thickness, a two-part calculation accounts for the overhang and web separately. This page compares both cases side by side using the same engine that powers the live T-beam calculator. For effective flange width rules, see the flange width spoke.

Effective flange width (brief)

ACI Table 6.3.2.1 limits how much slab acts as flange:

Interior T-beam: b_f = min(L/4, b_w + 16 h_f, b_w + clear spans / 2)
Edge L-beam: half of the interior overhang on the free side

Common inputs

Both cases share these properties:

Web width	b_w = 12 in.
Total depth	h = 24 in.
Clear cover	1.5 in. (Table 20.6.1.3.1)
Concrete	f′c = 4,000 psi, normal-weight (λ = 1.0)
Steel	Grade 60 (f_y = 60 ksi)
Stirrups	#3 @ 8 in., 2 legs (A_v = 0.22 in²)
Factored loads	M_u = 250 k-ft, V_u = 40 kip

Case A: neutral axis in the flange (a ≤ h_f)

4-#8 bars (1 layer), b_f = 48 in., h_f = 6 in.

d = 24 - 1.5 - 0.375 - 0.500 = 21.625\ \text{in}

A_s = 4 \times 0.79 = 3.16\ \text{in}^2

Whitney stress block using the full flange width:

a = \frac{A_s\,f_y}{0.85\,f'_c\,b_f} = \frac{3.16 \times 60}{0.85 \times 4 \times 48} = 1.162\ \text{in}

Since a = 1.162 in. < h_f = 6.0 in., the neutral axis stays in the flange. The section behaves as rectangular with width b_f = 48 in.

c = \frac{a}{\beta_1} = \frac{1.162}{0.85} = 1.367\ \text{in}

\varepsilon_t = 0.003 \times \frac{d - c}{c} = 0.003 \times \frac{21.625 - 1.367}{1.367} = 0.04447

ε_t = 0.04447 far exceeds 0.005, so the section is tension-controlled and φ = 0.9 (Table 21.2.2).

\phi M_n = 0.9 \times 3.16 \times 60 \times (21.625 - 0.581) = 3{,}591.0\ \text{k-in} = 299.2\ \text{k-ft}

\text{DCR} = \frac{M_u}{\phi M_n} = \frac{250}{299.2} = 0.835

When a < h_f, ignore the web entirely for flexure. The wide flange provides so much compression area that the stress block is shallow. This is the common case for typical slab-beam construction.

Case B: neutral axis in the web (a > h_f)

6-#8 bars (2 layers), b_f = 24 in., h_f = 3 in.

d = 20.625\ \text{in (2 layers: centroid shifts down)}

A_s = 6 \times 0.79 = 4.74\ \text{in}^2

First try the rectangular assumption with the full flange width:

a = \frac{A_s\,f_y}{0.85\,f'_c\,b_f} = \frac{4.74 \times 60}{0.85 \times 4 \times 24} = 3.485\ \text{in}

a = 3.485 in. > h_f = 3.0 in., so the neutral axis drops into the web. Split the compression zone into flange overhangs and web:

Two-part decomposition

A_{sf} = \frac{0.85\,f'_c\,(b_f - b_w)\,h_f}{f_y} = \frac{0.85 \times 4 \times (24 - 12) \times 3}{60} = 2.04\ \text{in}^2

A_{sw} = A_s - A_{sf} = 4.74 - 2.04 = 2.70\ \text{in}^2

a_{\text{web}} = \frac{A_{sw}\,f_y}{0.85\,f'_c\,b_w} = \frac{2.70 \times 60}{0.85 \times 4 \times 12} = 3.971\ \text{in}

Moment contributions

M_{n,f} = A_{sf}\,f_y\!\left(d - \frac{h_f}{2}\right) = 2.04 \times 60 \times (20.625 - 1.5) = 2{,}340.9\ \text{k-in}

M_{n,w} = A_{sw}\,f_y\!\left(d - \frac{a_{\text{web}}}{2}\right) = 2.70 \times 60 \times (20.625 - 1.986) = 3{,}019.6\ \text{k-in}

M_n = M_{n,f} + M_{n,w} = 5{,}360.5\ \text{k-in}

Strain check:

c = \frac{a_{\text{web}}}{\beta_1} = \frac{3.971}{0.85} = 4.671\ \text{in}

\varepsilon_t = 0.003 \times \frac{20.625 - 4.671}{4.671} = 0.01025 > 0.005

Tension-controlled, φ = 0.9.

\phi M_n = 0.9 \times 5{,}360.5 = 4{,}824.5\ \text{k-in} = 402.0\ \text{k-ft}

\text{DCR} = \frac{250}{402.0} = 0.622

When a > h_f, the overhang compression and web compression have different moment arms. A_sf acts at (d − h_f/2) and A_sw acts at (d − a_web/2). The engine iterates because the initial a (assuming full b_f width) must be rechecked against h_f.

Shear: same formula, but ρ_w matters

For shear, ACI Table 22.5.5.1 uses b_w (web width) regardless of section type. The T-beam flange does not contribute to V_c. However, Table 22.5.5.1 provides two equations, and the one that governs depends on the longitudinal steel ratio ρ_w = A_s / (b_w d):

Eq (a): V_c = 2 λ √f′c b_w d (conservative lower bound, ρ_w-independent)
Eq (b): V_c = 8 λ (ρ_w)^1/3 √f′c b_w d (rewards higher ρ_w)

When minimum stirrups are present, the engine uses max(V_c,a, V_c,b).

Case A (4-#8, d = 21.625)

\rho_w = \frac{3.16}{12 \times 21.625} = 0.01218

V_c,a = 32.82 kip, V_c,b = 30.21 kip. Eq (a) governs: V_c = 32.82 kip.

Case B (6-#8, d = 20.625)

\rho_w = \frac{4.74}{12 \times 20.625} = 0.01915

V_c,a = 31.31 kip, V_c,b = 33.50 kip. Eq (b) governs: V_c = 33.50 kip.

With more longitudinal steel (higher ρ_w), eq (b) can exceed eq (a). T-beams that need more tension steel for the two-part flexure check may also get a higher V_c.

Crack control and deflection

Crack control (ACI 24.3.2): same formula for both section types. Bar spacing is computed from b_w, not b_f. T-beams with more bars in the web can have tighter spacing.
Deflection: T-beam I_g uses the full T-section (web + flanges). Higher I_g leads to lower deflections, giving T-beams an advantage for serviceability.

Summary comparison

Property	Case A (NA in flange)	Case B (NA in web)
Flange width b_f	48 in.	24 in.
Flange thickness h_f	6 in.	3 in.
Bars	4-#8, 1 layer (A_s = 3.16)	6-#8, 2 layers (A_s = 4.74)
Effective depth d	21.625 in.	20.625 in.
Stress block a	1.162 in. (< h_f)	3.971 in. (> h_f)
Flexure method	Rectangular with b_f	Two-part: A_sf + A_sw
φM_n	299.2 k-ft	402.0 k-ft
Flexure DCR	0.835	0.622
V_c controlling eq	Table 22.5.5.1(a)	Table 22.5.5.1(b)
V_c	32.82 kip	33.50 kip
φV_n	51.38 kip	50.65 kip
Shear DCR	0.779	0.790