Design parameters for longitudinal bars (ACI 318)
For each of these parameters, any user defined limits (as specified on the appropriate Reinforcement Settings page within Design Settings) are considered in addition to any ACI code recommendations.
Minimum and maximum diameter of reinforcement
IF torsional reinforcement is required, there shall be at least one longitudinal bar in every corner of the stirrups. Longitudinal bars shall have a diameter at least 0.042 times the stirrup spacing, but not less than 9 mm (3/8 in).
The maximum diameters of reinforcement to be used in the various locations is set by the user.
Standard hooks for stirrups and ties are limited to No.8 bars, d_{b}=25mm (1.0in.) and smaller.
And the 90degree hook with 6d_{b} extension is further limited to No. 5, d_{b}=16mm (0.625in.) bars and smaller.
For primary reinforcement there is no limit on bar size.
Minimum distance between bars
The minimum clear horizontal distance between individual parallel bars, s _{cl,min}, is given by;
s_{cl,min}  ≥  MAX [d_{b}, 25 mm]  metric units 
s_{cl,min}  ≥  MAX [d_{b}, 1in]  US units 


IF the above check fails then a Warning is displayed.
Where parallel reinforcement is placed in two or more layers, bars in the upper layers shall be placed directly above the bars in the bottom layer with clear distance between layers not less than 25mm (1in.).
Maximum spacing of tension bars
The spacing of reinforcement closest to the tension face, s is given by;^{1}
s  ≤  MIN[380mm*280MPa/fs2.5*c_{c}, 300mm*(280MPa/f_{s})]  metric units 
s  ≤  MIN[15in*40000psi/f_{s}2.5*c_{c}, 12in*(40000psi/f_{s})]  US units 


f_{s}  =  calculated stress in reinforcement at service load; it shall be permitted to take 
=  (2/3)*f_{y}*(A_{s,reqd}/A_{s,prov})  
c_{c}  =  the least distance from surface of reinforcement to the tension face 

IF the above check fails then a Warning is displayed
IF torsional reinforcement is required:
the longitudinal reinforcement required for torsion shall be distributed around the perimeter of the closed stirrups with maximum spacing of 300mm (12 in.)^{2}
Minimum area of beam reinforcement
The minimum area of longitudinal tension reinforcement, A _{s,min}, is given by;^{3}
A_{s,min}  ≥  MAX[ (f'_{c}^{0.5}/(4*f_{y}))*b_{w}*d, 1.4MPa*b_{w}*d/ f_{y} ]  metric units 
A_{s,min}  ≥  MAX[ (3* f'_{c}^{0.5}/(f_{y})*b_{w}*d, 200Psi*b_{w}*d/ f_{y} ]  US units 


f'_{c}  =  specified compressive strength of concrete 
f_{y}  =  specified yield strength of reinforcement 
b _{w}  = 
web width; for statically determinate members with a flange in tension 
= 
MIN(2*b_{w}, b_{eff}) ^{4} 

d  =  distance from extreme compression fiber to centroid of longitudinal compression reinforcement 

The above equation is used wherever reinforcement is needed, except where such reinforcement is at least onethird greater than that required by analysis, in which case A_{s,min} not required.
Minimum area of slab reinforcement
For ACI 31808 and ACI 31811^{5}
For structural slabs of uniform thickness the minimum area of tensile reinforcement in the direction of the span is:
For USunits:
IF Grade 40 to 50 deformed bars are used
A_{s,min,reqd} ≥ b*h*0.0020
IF Grade 50 to 60 deformed bars or welted wire reinforcement are used
A_{s,min,reqd} ≥ b*h*0.0018
For metric units:
IF Grade 280 to 350 deformed bars are used
A_{s,min,reqd} ≥ b*h*0.0020
IF Grade 350 to 420 deformed bars or welted wire reinforcement are used
A_{s,min,reqd} ≥ b*h*0.0018
IF yield stress exceeding 420 MPa
A_{s,min,reqd} ≥ b*h* [MAX(0.0014 , 0.0018*420/f_{y})]
Maximum area of reinforcement
Net tensile strain in extreme layer of longitudinal tension steel, ε_{t} should not be less than 0.004;
ε_{t}  ≥  0.004 
A_{s,max}  ≤  0.85*(f'_{c} / f_{y})*β_{1}*b_{w}*d*[0.003/(0.003+0.004)] ^{6} 
≤  0.85*(f'_{c} / f_{y})*β_{1}*b_{w}*d*(3/7)  

A_{g}  =  the gross area of the concrete section  
β_{1}  =  stress block depth factor ^{7}  
=  0.85  for f`_{c} ≤ 28MPa 
metric units 

=  0.85  0.05 *[(f'_{c}  28MPa)/7MPa]  for 28MPa < f`_{c} < 55MPa  
=  0.65  for f`_{c} ≥ 55MPa  



=  0.85  for f`_{c} ≤ 4000 psi 
US units 

=  0.85  0.05 *[(f'_{c}  4ksi)/1ksi]  for 4000 psi < f`_{c} < 8000psi  
=  0.65  for f`_{c} ≥ 8000 psi 
Footnotes