Whitney Equivalent Stress Block

a = \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b} β_{1} = 0.85 for f_{c}^{'} \leq 4000 psi

The rectangular compression block has depth a = β1 c and intensity 0.85 f'c. Horizontal equilibrium: 0.85 f'c a b = As fy for tension-controlled singly reinforced beams (ignore compression steel).

$a$ (in): Block depth
$A_{s}$ (in²): Tension steel area
$f_{y}$ (psi): Steel yield
$f_{c}^{'}$ (psi): Concrete strength
$b$ (in): Width

Step-by-step solved example

b = 12 in, As = 3.0 in², fy = 60 ksi, f'c = 4 ksi. Find a.

ACI equivalent rectangle: depth a = As fy / (0.85 f'c b).

1. T
T = As fy = 3.0×60 = 180 k.
2. a
a = 180 / (0.85×4×12) = 180/40.8 = 4.41 in.
$a = 4.41 in$

Answer: a = 4.41 in

10 practice questions

0/10 correct

1.The 0.85 in 0.85 f'c is

2.β1 for f'c = 3 ksi is

3.c related to a by

4.As fy must equal Cc for

5.Doubling As (still tension-controlled) approximately

6.f'c = 5 ksi → β1 is

7.Minimum β1 is

8.Cc = 0.85 f'c a b acts at

9.If a computed > h the steel is

10.fy in the a formula must match units of