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 ³ V_{s} / [0.95 x (f'_{c})^{1}^{/2}]
(English)
Shear reinforcement, A_{v} / s ³
V_{s} / (0.4 x f_{s} 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 ³ V_{u} / [f x 2 x (f'_{c})^{1}^{/2}] (English)
Shear reinforcement, A_{v} / s ³
V_{u} / (f x f_{s} x
d)
b.
AASHTO LRFD (use the simplified method with d_{v}
= 0.9d)
Note: The
simplified method is technically not permitted for members with no shear
reinforcement and thicknesses > 16"
Effective
depth, d ³ V_{u} / [f x 0.9 x 2 x (f'_{c})^{1}^{/2}] (English)
Shear reinforcement, A_{v} / s ³
V_{u} / (f x 0.9 x f_{s}
x d)
c.
CAN/CSA-S6 (use the simplified method with d_{v}
= 0.9d)
Effective
depth, d ³ 1000 x V_{u} / [0.9 x (600 x f_{c} x f_{r}
- V_{u})] (Metric)
Shear reinforcement, A_{v} / s ³
V_{u} / (f_{s} x 0.9 x f_{s}
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 M_{cr}
but not to exceed 4 x M_{u} / 3. Cracking moment is calculated based on the
modulus of rupture, f_{r}.
CRITERIA |
f_{r} |
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)
C_{1} = 0.5 x f x (f_{r}
x f_{s})^{2}
/ (a_{1} x f_{c} x f'_{c})
C_{2} = -f x f_{r}
x f_{s} x d
C_{3} = M_{design}
A_{s} = (-C_{2} - ((C_{2})^{2}
- 4 * C_{1} * C_{3})^{1/2}) / (2 * C_{1})
b.
Determine maximum reinforcement limits
i.
AASHTO and AREMA (r_{max} = 0.75 r_{bal})
e_{reinf} = f_{r} x f_{s}
/ E_{s}
y_{bal} = e_{max}
x d / (e_{max} + e_{reinf})
a_{bal} = y_{bal}
x b_{1}
A_{smax} = 0.75 x a_{1}
x f_{c} x f'_{c} x a_{bal} / (f_{r} x f_{s})
ii.
AASHTO LRFD and CAN/CSA-S6 (limit c / d ratio)
A_{block} = A_{s}
x f_{r} x f_{s} / (a_{1}
x f_{c} x f'_{c})
c / d = A_{block}
/ (b_{1} 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 = A_{s} / d
n = E_{s} / E_{c}
k = (2rn + (rn)^{2})^{1/2} - rn
j = 1 - k / 3
f_{c} = 2M / (d^{2}kj)
f_{s} = M / (A_{s}jd)
b.
Calculate required flexural reinforcement area as required for AASHTO and
AREMA service load design (unit width b = 1)
Allowable reinforcing stress = f_{sa}
= 0.4 x f_{s} x Fsls
(increase factor)
Allowable concrete stress = f_{ca} = 0.4 x f'_{c}
x Fsls
k = f_{ca}
/ (f_{sa} / n + f_{ca})
j = 1 - k / 3
Required reinforcement area = A_{s} > M /
(f_{sa} 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 = f_{s} = M
/ (A_{s} 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:
f_{cr} = z / (d_{c} x A)^{1/3}
£ f_{max}
A = 2 x d_{c} x s (spacing)
f_{max}_{ }= 0.6 x f_{s}
(AASHTO and AASHTO LRFD)
f_{max} = 0.5 x f_{s}
(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:
A_{s} (required)
> A_{s} x Stress / f_{cr}
Solve for spacing with same reinforcement per unit
width (not applicable if Stress > f_{max}):
s £
(z)^{3} / (2 x (f_{cr})^{3} x
(d_{c})^{2})
iii.
Calculate required area and/or spacing for CAN/CSA-S6
Calculate crack control factor:
b_{2} = (0.9 x s_{c} +
100) x f_{s} x [1 - (M_{w} / M_{s})^{2}]
b_{2} £
50,000 N/mm (moderate)
b_{2} £
30,000 N/mm (severe)
f_{s} £
f_{max} = 0.6 x f_{s}
s_{c} = clear spacing between
reinforcement
M_{w} = 0.4 x f_{r}
x (Thick)^{2} / 6
f_{r} = modulus of rupture = 0.4
x (f'_{c})^{1/2}
Solve for area of reinforcement (not applicable if
Stress > f_{max}):
f_{s} = b_{2}
/ {(0.9 x s_{c} + 100) x [1 - (M_{w} / M_{s})^{2}]}
A_{s} (required)
> A_{s} x Stress / f_{cr}
Solve for spacing with same reinforcement per unit
width:
s_{c} £ b_{2} / {0.9 x f_{s} x [1 -
(M_{w} / M_{s})^{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 |
A_{s} |
AASHTO |
0.125
in^{2}/ft per face |
AREMA |
0.25
in^{2}/ft per face |
CAN/CSA-S6 |
500
mm^{2}/m per face |
AASHTO
LRFD |
For
T £ 48" A_{s} = 55 x A_{g} / f_{y} < 0.0015 x A_{g} / 2 per face For
T > 48" A_{s} = 3 x s /
100 ≥ 0.44 in^{2} s = spacing (in) (3 is from 2d_{c} + d_{b} £
3") |