Reinforced Concrete Design

 

Reinforced concrete design calculations are performed in accordance with the following procedure.

 

1.                   Calculate effective depth of shear reinforcement required to satisfy serviceability limit state for shear. (unit width b = 1)

a.                   AASHTO and AREMA service load design

Effective depth, d Vs / [0.95 x (f'c)1/2] (English)

 

Shear reinforcement, Av / s Vs / (0.4 x fs x d)

 

2.                   Calculate effective depth of shear reinforcement required to satisfy ultimate limit state for shear. (unit width b = 1)

a.                   AASHTO and AREMA ultimate strength design

 

Effective depth, d Vu / [fx 2 x (f'c)1/2] (English)

 

Shear reinforcement, Av / s Vu / (f x fs x d)

 

b.                  AASHTO LRFD (use the simplified method with dv = 0.9d)

 

Note: The simplified method is technically not permitted for members with no shear reinforcement and thicknesses > 16"

 

Effective depth, d Vu / [fx 0.9 x 2 x (f'c)1/2] (English)

 

Shear reinforcement, Av / s Vu / (f x 0.9 x fs x d)

 

c.                   CAN/CSA-S6 (use the simplified method with dv = 0.9d)

 

Effective depth, d 1000 x Vu / [0.9 x (600 x fcx fr - Vu)] (Metric)

 

Shear reinforcement, Av / s Vu / (fs x 0.9 x fs x d)

 

3.                   Calculate flexural reinforcement required to satisfy ultimate limit state (not required for AASHTO and AREMA service load design).

a.                   Calculate minimum design moment as 1.2 x Mcr but not to exceed 4 x Mu / 3. Cracking moment is calculated based on the modulus of rupture, fr.

 

CRITERIA

fr

AASHTO

& AREMA

7.5(f'c)1/2 (English)

7.5(f'c)1/2 / 12 (Metric)

AASHTO LRFD

240(f'c / 1000)1/2 (English)

0.63(f'c)1/2 (Metric)

CAN/CSA-S6

4.8(f'c)1/2 (English)

0.4(f'c)1/2 (Metric)

 

a.                   Calculate required flexural reinforcement area (unit width b = 1)

 

C1 = 0.5 x f x (fr x fs)2 / (a1 x fc x f'c)

C2 = -f x fr x fs x d

C3 = Mdesign

As = (-C2 - ((C2)2 - 4 * C1 * C3)1/2) / (2 * C1)

 

b.                  Determine maximum reinforcement limits

i.                     AASHTO and AREMA (rmax = 0.75 rbal)

 

ereinf = fr x fs / Es

ybal = emax x d / (emax + ereinf)

abal = ybal x b1

Asmax = 0.75 x a1 x fc x f'c x abal / (fr x fs)

 

ii.                   AASHTO LRFD and CAN/CSA-S6 (limit c / d ratio)

 

Ablock = As x fr x fs / (a1 x fc x f'c)

c / d = Ablock / (b1 x d)


AASHTO LRFD: c / d limit = 0.42

CAN/CSA-S6: c/d limit = 0.5

 

4.                   Calculate flexural reinforcement spacing required to satisfy serviceability limit state

a.                   Calculate stress (unit width b = 1)

 

p = As / d

n = Es / Ec

k = (2rn + (rn)2)1/2 - rn

j = 1 - k / 3

fc = 2M / (d2kj)

fs = M / (Asjd)

b.                  Calculate required flexural reinforcement area as required for AASHTO and AREMA service load design (unit width b = 1)

 

Allowable reinforcing stress = fsa = 0.4 x fs x Fsls (increase factor)

Allowable concrete stress = fca = 0.4 x f'c x Fsls

k = fca / (fsa / n + fca)

j = 1 - k / 3

Required reinforcement area = As > M / (fsa x j x d)

 

c.                   Calculate required flexural reinforcement area and/or spacing for crack control (not required for AASHTO and AREMA service load design).

 

i.                     Calculate stress in flexural reinforcement

 

k = (2 x r x n + (r x n)2)1/2 - r x n

j = 1 - k / 3

Stress = fs = M / (As x j x d)

 

ii.                   Calculate required area and/or spacing for AREMA, AASHTO and AASHTO LRFD

 

Note: A 2" or 50 mm maximum cover is assumed for calculating the cracking stress even though the AASHTO LRFD code is the only one to mention this.

 

Calculate cracking stress:

 

fcr = z / (dc x A)1/3 fmax

 

A = 2 x dc x s (spacing)

fmax = 0.6 x fs (AASHTO and AASHTO LRFD)

fmax = 0.5 x fs (AREMA)

z = 170,000 lb/in (30,000 N/mm) moderate

z = 130,000 lb/in (23,000 N/mm) severe

z = 100,000 lb/in (17,500 N/mm) buried

 

Estimate area of reinforcement at same spacing:

 

As (required) > As x Stress / fcr

 

Solve for spacing with same reinforcement per unit width (not applicable if Stress > fmax):

 

s (z)3 / (2 x (fcr)3 x (dc)2)

 

iii.                  Calculate required area and/or spacing for CAN/CSA-S6

 

Calculate crack control factor:

 

b2 = (0.9 x sc + 100) x fs x [1 - (Mw / Ms)2]

 

b2 50,000 N/mm (moderate)

b2 30,000 N/mm (severe)

fs fmax = 0.6 x fs

sc = clear spacing between reinforcement

Mw = 0.4 x fr x (Thick)2 / 6

fr = modulus of rupture = 0.4 x (f'c)1/2

 

Solve for area of reinforcement (not applicable if Stress > fmax):

 

fs = b2 / {(0.9 x sc + 100) x [1 - (Mw / Ms)2]}

As (required) > As x Stress / fcr

 

Solve for spacing with same reinforcement per unit width:

 

sc b2 / {0.9 x fs x [1 - (Mw / Ms)2]} - 100

 

Notes: Crack control calculations for CAN/CSA-S6 are included even though Section 8.5.2.1 states that "The requirements of Clause 8.12 (crack control) shall be met except for tensile surfaces of components that are permanently covered with 500 mm or more of earth."

 

Crack control calculations for CAN/CSA-S6 assume Mw is based on 0.4fcr to be consistent with the original formulas in the OHBDC even though the CAN/CSA-S6 definitions state that Mw is based on fcr.

 

5.                   Calculate required temperature and shrinkage reinforcement

CRITERIA

As

AASHTO

0.125 in2/ft per face

AREMA

0.25 in2/ft per face

CAN/CSA-S6

500 mm2/m per face

AASHTO LRFD

For T 48"

 

As = 55 x Ag / fy

< 0.0015 x Ag / 2 per face

 

For T > 48"

 

As = 3 x s / 100 ≥ 0.44 in2

s = spacing (in)

(3 is from 2dc + db 3")