CHAPTER 22
DESIGN AND ANALYSIS FLOWCHARTS
FLOWCHARTS
To help the student and design engineer to prepare their own programs, flowcharts are
given for most chapters.
1. Flowcharts 22.1, 22.2, and 22.3 explain the analysis of single, double, and T-sections
(Chapter 3).
2. Flowcharts 22.4, 22.5, and 22.6 explain the design of single, double, and T-sections
(Chapter 4).
3. Flowchart 22.7 explains the calculation of development length (Chapter 7).
4. Flowchart 22.8 explains shear design (Chapter 8).
5. Flowchart 22.9 explains the analysis of rectangular columns at balanced condition
(Chapter 11).
6. Flowchart 22.10 explains the analysis of rectangular columns (Chapter 11).
7. Flowchart 22.11 explains the design of rectangular or square footings (Chapter 13).
8. Flowchart 22.12 explains the design for combined shear and torsion (Chapter 15).
Given: b, d, As, f’c, fy
Required: φMn
Let ρ = As / bd
No
Yes
ρ > ρmin
(1)
Yes
Increase ρ
ρ ≤ ρmax
(2)
a = Asfy/ (0.85f’cb)
c = a / β1
εt = 0.003(dt – c)/c
εt ≥ 0.005
φ = 0.9
φMn = φAsfy(d-a/2)
End
Flowchart 22.1 Analysis of single reinforced rectangular section.
(1) ρ min = 3
f 'c / fy ≥ 200 / fy
 87
(2) ρb = (0.85β1)(f’c / fy) 
 87 + f y

0.003
+
f
/
E

y
s 
ρ max= 
 ρb
0.008





No
Reduce ρ
Given: b, d, d’, As, A’s, f’c, fy
Required: φMn
Let ρ = As / bd, ρ’ = A’s / bd
No
ρ ≥ ρmin
Increase ρ
No
Yes
Yes
ρ - ρ' ≥ K
(1)
f’s < fy
f’s = fy
Solve for c:
A1c2 + A2c + A3 = 0
A1 = 0.85β1f’cb
A2 = A’s(87 – 0.85f’c) - Asfy
A3 = -87A’sd’
Yes
No
ρmin ≤ ρ-ρ’≤ρmax
a=(As–A’s)fy/(0.85f’cb)
Change ρ
Calculate f’s < fy;
ε’s = 0.003(c-d’)/c
εt = 0.003(dt -c)/c
f’s = Esε’s = 87 (c − d ')
c
Yes
No
εt ≥ 0.005
No
εt ≤ 0.002
Yes
φ = 0.9
φ = 0.65+(εt – 0.002)(250/3)
(other members)
a=
φ = 0.65 (other
members)
( As f y − A 's f 's )
0.85 f 'c b
φMn = φ[Asfy – A’sf’s)(d-a/2) + A’sf’s(d-d’)]
Flowchart 22.2 Analysis of double reinforced rectangular section.
 87 
(1) K = (0.85β1)(f’c / fy)(d’ / d) 
 87 − f y 


(2) Refer to flowchart 22.1 to calculate ρmin and ρmax.
End
Given: b, bw, d, t, As, f’c, fy
Required: φMn
No
Yes
As ≤ As max
(1)
Yes
Reduce As
No
As ≥ As min
(2)
For T-section, be is the smallest value of: Span / 4 or Center to
center of adjacent slabs or (bw + 16t); t = slab thickness.
For L-section, be is the smallest value of: Span / 12 or half clear
distance to next web + web width or (bw + 6t); t = slab thickness
Increase As
Let a’ = Asfy/(0.85f’cbe)
No
εt =0.003(dt-c)/c
εt ≥ 0.005
Enlarge
section
Yes
Yes
a’≤ t
No
T-section analysis
Rectangular analysis
a = a’
φMn = φAsfy (d - a/2)
End
Asf = 0.85f’ct(b-bw) / fy
a = (As – Asf)fy/(0.85f’cbw)
φMn = φ[(As – Asf)fy(d-a/2) + Asffy(d-t/2)]
End
Flowchart 22.3 Analysis of T- and L-sections.
(1) As max = 0.6375(f’c/fy)[t(b-bw) + 0.375bwβ1d]
(2)
3 fc' 
 200
As min = 
bwd ≥  bwd
 fy 
 f y 
Given: Mu, f’c, fy
Required: As
Yes
b and d are given
Calculate
Ru = Mu/(bd2)
ρ=
No
Assume b (12-20) in.
Or d / b ≈ 2
0.85 f 'c 
1 − 1 − (2 Ru / 0.85 f 'c φ ) 
fy 
ρmin ≤ ρ ≤ ρmax
(1)
No
Assume ρ.
ρmin ≤ ρ ≤ ρmax
or let ρ ≈ ρb / 2
Yes
Calculate Ru:
Ru = φρfy (1-ρfy/1.7f’c)
As = ρbd
Change section
d = M u / Ru b
As = ρbd
Choose bars
h = d+2.5 in. (one row of bars)
h = d+3.5 in. (two rows of bars)
Round h to the next higher inch.
End
Flowchart 22.4 Design of single reinforced rectangular section.
(1) Refer to flowchart 22.1 to calculate ρ min and ρ max
Given: Mu, b, d, d’, f’c, fy
Required: As, A’s
Calculate ρmin and ρmax
(1)
Calculate Ru max
Ru max = φρmaxfy(1-ρmaxfy/1.7f’c)
Calculate φMn = Mu1 as singly reinforced
Mu1 = Ru maxbd2
No
Yes
Mu ≥ Mu1
Comp. steel is not
required.
As1 = ρmaxbd
Mu2 = Mu - Mu1
Go to flowchart 22.4
As2 = Mu2 / φfy(d-d’)
Total As = As1 + As2
As ≤ As max
No
εt =0.003(dt-c)/c
εt ≥ 0.005
Yes
Change
section
Yes
f’s ≥ fy
Check if comp. steel yields.
Let a = As1fy / 0.85f’cb
c = a / β1
Calculate f’s = 87(c - d’) / c
No
f's = fy
A’s= As2
f's < fy
A’s= As2fy/ f’s
End
End
Flowchart 22.5 Design of rectangular sections with compression steel.
(1) Refer to flowchart 22.1 to calculate ρ min and ρ max
Given: Mu, b, bw, d, t, f’c, fy
Required: As
No
Yes
d is given
Calculate Muft (total flange).
a=t
Cft = 0.85ϕf’cbt
Muft = φ Cft(d – t/2)
Assume a = t
d = Mu / ϕ(0.85f’cbt) + t/2
Yes
As = Mu / [φ fy (d - t/2)]
If steel is high,assume
a < t or a ≈ t / 2
d = Mu / (0.85φf’cba) + a/2
Mu ≤ Muft
Rectangular section
design, a ≤ t
a = a’
T-section design,
a>t
Ru = Mu / bd2
Cf = 0.85f’ct(b – bw)
Asf = Cf / fy
ρ = (0.85f’c/fy)[1- 1 − 2 Ru / 0.85φ f 'c ]
As = Mu / [φfy(d – a/2)]
No
Muf = φCf (d – t/2)
Muw (web) = Mu - Muf
As = ρbd
Ruw = Muw / bwd2
ρw =(0.85f’c/fy)[1- 1 − 2 Ruw / 0.85φ f 'c ]
As = Asf + Asw
No
Asmin≤As≤Asmax
Or
εt = (dt - c)/c ≥ 0.005
Change section
Flowchart 22.6 Design of T- or L-sections.
Asw = ρwbwd
Yes
As is OK
Given: bw, d, f’c, fy, bars
Required: Development length
f 'c ≤ 100 psi
Compression bars
Tension bars
(for no 7 and large bars)
ldc = (0.02dbfy) / λ f 'c
ldc ≥ 0.0003dbfy
ld1 / db = ψ tψ e f y / 20λ
f c'
(for no-6 and smaller bars)
ld1 / db = ψ tψ e f y / 25λ
f c'
ld = ldc*(Rs or Rs1, if applicable)
ld ≥ 8 in.
No
Rs = As (required) / As (provided)
Rs1 = 0.75 for spiral columns
ld = 1.5 ld1 ≥ 12 in.
Flowchart 22.7 Calculation of development length
(1) c = clear cover
s = clear spacing
Is
c ≥ db
s ≥ db
or s > 2db
(1)
Yes
ld = ld1 ≥ 12 in.
Given: bw, d, f’c, fy, Vu
Required: shear reinforcement
Vc = 2λ
f 'c bwd
φ = 0.75
No
Yes
Vu ≥ φVc/2
Yes
No shear
reinforcement
is required
Vu > φVc
Vs = (Vu - φVc) / φ
No
Vs > 4Vc
Av = VsS / fyd
Or
S = Avfyd/Vs
Yes
Vs > 2Vc
S ≤ d/4 ≤ 12 in.
End
Flowchart 22.8 Shear Design.
Yes
Increase
section
No
S ≤ d/2 ≤ 24 in.
S ≤ Avfy / 50bw
End
No
Choose minimum
shear reinforcement.
Av ≥ 50bwS / fy
Av =0.75 f 'c (bwS/fy)
S ≤ d/2 ≤ 24 in.
Minimum stirrups
no.3 at Smax
Given: b, d, d’, As, A’s, f’c, fy
Required: Pb, Mb, eb
cb = 87d / (87 + fy)
ab = β1cb
f’s = [87(cb – d’) / cb]
f’s ≤ fy (Ksi)
Cs = A’s (f’s – 0.85f’c)
Cc = 0.85f’cab
T = Asfy
Pb = Cc + Cs - T
Mb = Cc(d – ab/2 – d’’) + Cs(d – d’ – d’’) + Td’’
eb = Mb / Pb
End
Flowchart 22.9 Balanced load, moment, and eccentricity for rectangular
column sections
(Use Ksi for f’c and fy; d’’ = distance from the plastic centroid to As).
Given: b, d, d’, As, e
Required: Pn, Mn
No
Yes
e > eb
Compression failure
fs < fy
a > ab
φ = 0.65
Tension failure
fs = fy
a < ab
0.65 ≤ φ ≤ 0.9
Assume f’s = fy
Solve for a:
Aa3 + Ba2 + Ca + D = 0
A = 0.85f’cb/2
B = 0.85f’cb(e’ – d)
C = A’s(fy– 0.85f’c)(e’– d + d’) + 87Ase’
D = -87Ase’β1d
e’ = e + d’’, c = a / β1
Solve for a:
Aa2 + Ba + C = 0
A = 0.425f’cb
B = 2A(e’ – d)
C = A’s(f’s– 0.85f’c)(e’– d + d’) - Asfye’
e’ = e + d’’, c = a / β1
Solve by trial or calculator
No
f’s = [87(c – d’)/c] ≤ fy
fs = [87(d – c)/c] ≤ fy
ε’s ≥ εy
(1)
f’s = fy
f’s = [87(c – d’)/c] ≤ fy
T = Asfs
T = Asfy
Cc = 0.85f’cab
Cs = A’s(f’s– 0.85f’c) ≥ 0
Pn = Cc + Cs - T
Mn = Pn e
End
Flowchart 22.10 Analysis of rectangular columns
(Use Ksi for f’c and fy)
(1) ε’s = 0.003(c-d’/c); εy = fy / Es
Yes
Given: PD, PL, H, qa, f’c, fy
Column size c, and bars
Assume footing depth h (in.)
Let Wc = 150 pcf, Ws = 100 pcf
qe = qa – h(150)/12 – (H - h)(100)/12 ; psf
Footing area A = (PD + PL)/qe
Side = A for square footing
Or L x B for rectangular footing
Choose L / B < 2
Pu = 1.2 PD + 1.6 PL
qu = Pu / A
Let average d = h – 4.5 in.
Check two-way shear: d2 is the largest value of:
d2 = Vu2/(4λ f 'c b0)
d2 = Vu2/λ[(αsd/b0) + 2]
d2 = Vu2/λ[(4/β + 2]
Check one-way shear:
d1 = Vu1/(2λ f 'c b)
f 'c b0
f 'c b0
αs = 40, 30, 20 for interior, edge and corner columns
b0 = 4(c + d) for square columns
b0 = 2(c1+d) + 2(c2 + d) for rectangular columns
φ = 0.75
Long direction MuL = 0.5qu (L/2 – c/2)2
Short direction MuS = 0.5qu (B/2 – c/2)2
Calculate AsL and AsS
Choose bars and check ld
Yes
d = the larger of d1 and d2
h’ = d + 4.5 ≥ h
If h’ < h, increase h and repeat.
N1 = φ(0.85f’c)A1 (column)
N2 = N1( A2 / A1 ) ≤ 2N1
φ = 0.65
Pn > N1
No
Dowel bars
Min. Asd = 0.005 A1
Use 4 bars
Pex = Pu – N1
Asd = Pex/fy ≥ 0.005A1
Choose dowel bars and check ld in compression.
End
Flowchart 22.11 Design of rectangular or square footings .
Given: b, h, Vu, Tu
Required: Closed stirrups and A1
Let x0 = b, y0 = h, φ = 0.75
Let x1 = (b – 3.5 in.), y1 = (h – 3.5 in.)
Acp = x0y0, Pcp = 2(x0 + y0)
A0 = 0.85 x1y1 = 0.85A0h
Ph = 2(x1 + y1)
Let Q = (φλ f 'c )A2cp/ Pcp
Yes
No
Yes
Tu > 4Q
Tu is neglected.
Check for Vu
(flowchart 22.7)
Tu = 4Q
for compability Tu
Use Tu
Yes
No
Tu > Q
No
(Vu / bw d ) 2 + (Tu Ph /1.7 A2 0 h ) 2 ≤ φ[(Vc / bw d ) + (8 f 'c )]
Increase
section
At/S = Tn/(2A0fytcotθ)
A1 = (At/S) Ph (fyt/fy) cot2θ
A1 min = (5 f 'c Acp)/fy – (At/S) Ph (fyt/fy)
Al min ≥ 25bw / f yt
Min. bar diameter =0.042S ≥ no. 3
Total area of closed stirrups
Avt = 2At + Av
Avt ≥ (50bwS/fyt)
(Av from shear)
Spacing of stirrups = area of bar/Atot
Max. S = Ph/8 ≤ 12 in.
End
Flowchart 22.12 Design for combined shear and torsion.