The Structural Toolbox - ACI Development Length

Bar
d_b	0.375	in	bar diameter.
λ
f'_c		psi	Specified 28-day compressive strength of concrete in which the bar is to be developed.
f_y		Yield strength of bar to be developed.

↑ ACI 318-19 Straight Compression Development Length, L_dc:

Ψ_r

Other

$L_{d c} = max [\begin{matrix} (a) (\frac{f_{y} ψ_{r}}{50 λ \sqrt{f_{c}^{'}}}) d_{b} (25.4.9.3a) \\ (b) 0.0003 f_{y} ψ_{r} d_{b} (25.4.9.3b) \\ 8 in \end{matrix}] [25.4.9.1]$
$(\frac{f_{y} ψ_{r}}{50 λ \sqrt{f_{c}^{'}}}) d_{b} = (\frac{(40000.00) (1.00)}{50 (1.00) (54.77)}) 0.375 = 5.477 in$
$0.0003 f_{y} ψ_{r} d_{b} = (0.003) (40000.00) (1.00) (0.375) = 4.500 in$
$L_{d c} = max [\begin{matrix} (a) 5.477 \\ (b) 4.500 \\ (c) 8 in \end{matrix}] = 8.000 in [25.4.9.1]$

↑ ACI 318-19 Straight Tension Development Length, L_d:

Ψ_t		More than 12 in. of fresh concrete placed below horizontal reinforcement.
Ψ_e		Epoxy-coated or zinc and epoxy dual-coated reinforcement with clear cover less than 3d_b or clear spacing less than 6d_b.
Ψ_s		No. 6 and smaller bars and deformed wires
Ψ_g		Grade 40 or Grade 60
c_c		in	distance from center of bar to nearest concrete surface.
s_b		in	center-to-center spacing of bars being developed.
c_b	1.5	in
A_tr		in²	Total cross-sectional area of all transverse reinforcement within spacing s that crosses the potential plane of splitting through the reinforcement being developed. If unsure of value use 0.
s		in	maximum center-to-center spacing of transverse reinforcement within L_d.
n			number of bars being developed.
K_tr	0.000

$L_{d} = max [\begin{matrix} (a) (\frac{3}{40} \frac{f_{y}}{λ \sqrt{f_{c}^{'}}} \frac{ψ_{t} ψ_{e} ψ_{s} ψ_{g}}{(\frac{c_{b} + K_{t r}}{d_{b}})}) d_{b} (25.4.2.4a) \\ (b) 12 in \end{matrix}] [25.4.2.1]$
$K_{t r} = \frac{40 A_{t r}}{s n} = \frac{40 (0.00)}{(10.00) (0.00)} = 0.000 (25.4.2.4b)$
$\frac{c_{b} + K_{t r}}{d_{b}} = min [\begin{matrix} \frac{1.50 + 0.00}{0.375} \\ 2.5 \end{matrix}] = min [\begin{matrix} 4.00 \\ 2.5 \end{matrix}] = 2.500 [25.4.2.4]$
$(\frac{3}{40} \frac{f_{y}}{λ \sqrt{f_{c}^{'}}} \frac{ψ_{t} ψ_{e} ψ_{s} ψ_{g}}{(\frac{c_{b} + K_{t r}}{d_{b}})}) d_{b} = (\frac{3}{40} \frac{40000.00}{(1.00) (54.77)} \frac{(1.7) (0.80) (1.00)}{(2.50)}) 0.375 = 11.174 in (25.4.2.4a), where ψ_{t} ψ_{e} need not exceed 1.7$
$L_{d} = max [\begin{matrix} 11.174 in \\ 12 in \end{matrix}] = 12.000 in$

↑ ACI 318-19 Hooked Development Length, L_dh:

Ψ_e		Epoxy-coated or zinc and epoxy dual-coated reinforcement
Ψ_r		Other
Ψ_o		Other
Ψ_c		f^'_c<6,000 psi = f^'_c/15,000 + 0.6

$L_{d h} = max [\begin{matrix} (a) (\frac{f_{y} ψ_{e} ψ_{r} ψ_{o} ψ_{c}}{55 λ \sqrt{f_{c}^{'}}}) d_{b}^{1.5} \\ (b) 8 d_{b} \\ (c) 6 in \end{matrix}] [25.4.3.1]$
$(\frac{f_{y} ψ_{e} ψ_{r} ψ_{o} ψ_{c}}{55 λ \sqrt{f_{c}^{'}}}) d_{b}^{1.5} = (\frac{(40000.00) (1.20) (1.60) (1.25) (0.8000)}{55 (1.00) (54.77)}) {0.375}^{1.5} = 5.854 in$
$L_{d h} = max [\begin{matrix} (a) 5.854 \\ (b) 3.000 \\ (c) 6 in \end{matrix}] = 6.000 in [25.4.3.1]$
⌀_inner = 2.250 in r_inner = 1.125 in ⌀_outer = 3.000 in r_outer = 1.500 in L_ext = 4.500 in
⌀_inner = 2.250 in r_inner = 1.125 in ⌀_outer = 3.000 in r_outer = 1.500 in L_ext = 2.500 in