Basic Tension Development Length

ℓ_{d} = \frac{f _{y} ψ _{t} ψ _{e}}{20 λ f _{c}^{'}} d_{b} (simplified FE form)

Bars must be embedded ℓd so the bond can develop fy. ψt (casting position), ψe (epoxy), λ (lightweight), and db increase ℓd. Hooks can reduce the required length. Always use the handbook form given on the exam.

$ℓ_{d}$ (in): Development length
$d_{b}$ (in): Bar diameter
$ψ_{t}, ψ_{e}$: Casting and coating factors
$λ$: Lightweight factor

Step-by-step solved example

fy = 60 ksi, f'c = 4 ksi, ψt=ψe=λ=1, db = 1.0 in (#8). Approximate ℓd = fy db / (20 √f'c).

Tension bar must be embedded ℓd (or hooked ℓdh) to develop fy.

1. Root
√f'c = √4000 = 63.2 psi.
2. ld
ℓd = 60000/(20×63.2) × 1.0 = 47.5 in.
$ℓ_{d} = 47.5 in$

Answer: about 48 in

10 practice questions

0/10 correct

1.ℓd is proportional to

2.Epoxy-coated bars typically

3.Top bars (ψt) have larger ℓd because of

4.A standard 90° hook can

5.Lightweight concrete (λ < 1) makes ℓd

6.Splices in tension are typically

7.√f'c in the US customary form uses f'c in

8.Available embedment must be

9.#6 bar has db =

10.Compression development is generally