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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).

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No

YesNo

Given: b, d, As, fc, fyRequired: Mn

Let = As / bd

> min(1)

Increase max(2)

Yes

Reduce

a = Asfy/ (0.85fcb)

c = a /1

= 0.9

Mn = Asfy(d-a/2)

End

Flowchart 22.1 Analysis of single reinforced rectangular section.

(1) min = 3 'cf / fy 200 / fy

(2) b = (0.851)(fc / fy)87

87y

f

+

max=0.003 /

0.008

y s

b

f E

+

t = 0.003(dt c)/c

t 0.005

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932/1000

No

No

No

No

No

Given: b, d, d, As, As, fc, fyRequired: Mn

Let = As / bd, = As / bd

- ' K(1)

fs < fy

min -maxYes

Yes

Change

a = ( ' ' )0.85 '

s y s s

c

A f A ff b

Flowchart 22.2 Analysis of double reinforced rectangular section.

(1) K = (0.851)(fc / fy)(d / d)87

87yf

(2) Refer to flowchart 22.1 to calculate min and max.

fs = fy

Solve for c:A1c

2 + A2c + A3 = 0

A1 = 0.851fcbA2 = As(87 0.85fc) - AsfyA3 = -87Asd

Calculate fs < fy;

s = 0.003(c-d)/c

fs = Ess =( ')

87c d

c

Mn = [Asfy Asfs)(d-a/2) + Asfs(d-d)] End

min Yes

Increase

t = 0.003(dt -c)/c

t 0.005Yes

t 0.002

= 0.9

a=(AsAs)fy/(0.85fcb)

Yes

= 0.65 (othermembers)

= 0.65+(t 0.002)(250/3)(other members)

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933/1000

No

No

No

No

Given: b, bw, d, t, As, fc, fyRequired: Mn

As A

s max(1)

Reduce As As As min(2)

Yes

Yes

Increase As

Let a = Asfy/(0.85fcbe)

Mn = [(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(fc/fy)[t(b-bw) + 0.375bw1d]

(2) dbf

dbf

fA

w

y

w

y

c

s

=

2003'

min

Rectangular analysis

a = a

Mn = Asfy (d - a/2)

End

a t

T-section analysis

Yes

Asf= 0.85fct(b-bw) / fy

a = (As Asf)fy/(0.85fcbw)

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

t =0.003(dt-c)/c

t 0.005

Yes

Enlarge

section

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YesNo

NoYes

Given: Mu, fc, fyRequired: As

b and d are given

Calculate

Ru = Mu/(bd2)

minmax(1)

Assume .minmaxor let b / 2

Assume b (12-20) in.

Or d / b 2

As = bd

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.

Flowchart 22.4 Design of single reinforced rectangular section.

(1) Refer to flowchart 22.1 to calculate min and max

= 0.85 ' 1 1 (2 / 0.85 ' )c u cy

fR f

f

Change section

Calculate Ru:

Ru = fy (1-fy/1.7fc)

d = /u uM R b

As = bdChoose bars

End

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No

No

No

Given: Mu, b, d, d, fc, fyRequired: As, As

Calculate min and max

(1)

Mu M

u1

Comp. steel is not

required.

fs fyYes

Yes

Mu2 = Mu - Mu1

As2 = Mu2/fy(d-d)

f's < fy

As= As2fy/ fs

End

As1 = maxbd

Check if comp. steel yields.

Let a = As1fy / 0.85fcb

c = a /1Calculate fs = 87(c - d) / c

Go to flowchart 22.4

End

Calculate Ru maxRu max = maxfy(1-maxfy/1.7fc)

Calculate Mn = Mu1 as singly reinforcedMu1 = Ru maxbd

2

Total As = As1 + As2As As max

f's = fy

As= As2

Flowchart 22.5 Design of rectangular sections with compression steel.

(1) Refer to flowchart 22.1 to calculate min and max

t =0.003(dt-c)/c

t 0.005

Changesection

Yes

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No

No

No

Given: Mu, b, bw, d, t, fc, fyRequired: As

d is given

Assume a = t

Mu MuftYes

Yes

Change section

Flowchart 22.6 Design of T- or L-sections.

Rectangular sectiondesign, a t

As = bd

AsminAsAsmax

Ort = (dt - c)/c 0.005

T-section design,a > t

Yes

d = Mu/(0.85fcbt) + t/2

As = Mu/ [ fy (d - t/2)]

If steel is high,assume

a < t or a t / 2

Calculate Muft (total flange).

a = tCft = 0.85fcbt

Muft = Cft(d t/2)

Cf= 0.85fct(b bw)

Asf= Cf/ fy

Muf= Cf(d t/2)Muw (web) = Mu - Muf

Ru = Mu / bd

As = Mu / [fy(d a/2)]

d = Mu / (0.85fcba) + a/2

Ruw = Muw / bwd

= (0.85fc/fy)[1- 1 2 / 0.85 'u cR f ]

w =(0.85fc/fy)[1- 1 2 / 0.85 'uw cR f ]

Asw = wbwdAs = Asf+ Asw

As is OK

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937/1000

No

Given: bw, d, fc, fy, bars

Required: Development length

Compression bars

Isc db

s dbor s > 2db(1)

Yes

ld = ld1 12 in.

(for no 7 and large bars)

ld1 / db ='20/

cyetff

(for no-6 and smaller bars)

ld1 / db ='25/

cetff

Flowchart 22.7 Calculation of development length

(1)c = clear covers = clear spacing

Tension bars

ldc = (0.02dbfy) / 'cf

ldc 0.0003dbfy

ld = ldc*(Rs or Rs1, if applicable)

ld 8 in.

Rs = As (required) / As (provided)

Rs1 = 0.75 for spiral columns

'cf 100 psi

ld = 1.5 ld1 12 in.

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No

No

No

Yes No

Given: bw, d, fc, fy, VuRequired: shear reinforcement

Vc = 2 'cf bwd = 0.75

VuVc/2

No shearreinforcement

is required

Vu > VcYes

Yes

Choose minimumshear reinforcement.

Av 50bwS / fy

Av =0.75 'cf (bwS/fy)

S d/2 24 in.Minimum stirrups

no.3 at Smax

Vs = (Vu - Vc) /

Av = VsS / fyd

OrS = Avfyd/Vs

End

Flowchart 22.8 Shear Design.

Vs > 4Vc Yes

Increase

section

Vs > 2Vc

S d/4 12 in. S d/2 24 in.

S Avfy/ 50bw

End

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cb = 87d / (87 + fy)ab = 1cb

fs = [87(cb d) / cb]

fs fy (Ksi)

Mb = Cc(d ab/2 d) + Cs(d d d) + Td

End

Flowchart 22.9 Balanced load, moment, and eccentricity for rectangularcolumn sections

(Use Ksi for fc and fy; d = distance from the plastic centroid to As).

Cs = As (fs 0.85fc)Cc = 0.85fcab

T = Asfy

eb = Mb / Pb

Pb = Cc + Cs - T

Given: b, d, d, As, As, fc, fyRequired: Pb, Mb, eb

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940/1000

No

Given: b, d, d, As, eRequired: Pn, Mn

e > eb YesNo

Tension failure

fs = fya < ab

0.65 0.9

Cc = 0.85fcab

Cs = As(fs 0.85fc) 0

Flowchart 22.10 Analysis of rectangular columns(Use Ksi for fc and fy)

(1) s = 0.003(c-d/c); y = fy / Es

Solve for a:

Aa2 +

Ba + C = 0A = 0.425fcb

B = 2A(e d)C = As(fs 0.85fc)(e d + d) - Asfye

e = e + d, c = a / 1

fs = [87(c d)/c] fy

T = Asfy

End

Compression failure

fs < fya > ab = 0.65

Solve for a:

Aa3 +

Ba2

+ Ca + D = 0A = 0.85fcb/2

B = 0.85fcb(e d)C = As(fy 0.85fc)(e d + d) + 87Ase

D = -87Ase1d

e = e + d, c = a /1Solve b trial or calculator

fs = [87(c d)/c] fyfs = [87(d c)/c] fy

T = Asfs

Pn = Cc +Cs - T

Mn = Pn e

Assume fs = fy

sy(1)

fs = fy

Yes

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941/1000

No

Given: PD, PL, H, qa, fc, fyColumn size c, and bars

Assume footing depth h (in.)

Let Wc = 150 pcf, Ws = 100 pcf

Pu = 1.2 PD + 1.6 PLqu = Pu / A

Let average d = h 4.5 in.

Pn > N1Yes

Check two-way shear: d2

is the largest value of:

d2 = Vu2/(4 'cf b0)

d2 = Vu2/[(sd/b0) + 2] 'cf b0

d2 = Vu2/[(4/ + 2] 'cf b0

s = 40, 30, 20 for interior, edge and corner columnsb0 = 4(c + d) for square columnsb0 = 2(c1+d) + 2(c2 + d) for rectangular columns

= 0.75

Check one-way shear:

d1 = Vu1/(2 'cf b)

Dowel bars

Min. Asd = 0.005 A1Use 4 bars

Pex = Pu N1Asd = Pex/fy 0.005A1

d = the larger of d1 and d2

h = d + 4.5 h

If h < h, increase h and repeat.

End

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 footingChoose L / B < 2

Long direction MuL = 0.5qu (L/2 c/2)

Short direction MuS = 0.5qu (B/2 c/2)2

Calculate AsL and AsSChoose bars and check ld

Choose dowel bars and check ld in compression.

lowchart 22.11 Desi n of rectan ular or s uare footin s .

N1 = (0.85fc)A1 (column)

N2 = N1( 2 1/A A ) 2N1

= 0.65

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942/1000

No

No

NoYes

Given: b, h, Vu, TuRequired: Closed stirrups and A1

Let x0 = b, y0 = h, = 0.75

Let x1 = (b 3.5 in.), y1 = (h 3.5 in.)

Tu > QYes

Yes

Spacing of stirrups = area of bar/Atot

Max. S = Ph/8 12 in.

Flowchart 22.12 Design for combined shear and torsion.

Let Q = ( 'cf )A2

cp/ Pcp

Acp = x0y0, Pcp = 2(x0 + y0)

A0 = 0.85 x1y1 = 0.85A0h

Ph = 2(x1 + y1)

Tu is neglected.

Check for Vu

(flowchart 22.7)Use Tu Tu = 4Q

for compability Tu

Increase

section

Total area of closed stirrups

Avt = 2At + AvAvt (50bwS/fyt)

(Av from shear)

At/S = Tn/(2A0fytcot)

End

Tu > 4Q

2 2 2

0( / ) ( /1.7 ) [( / ) (8 ' )]

u w u h h c w cV b d T P A V b d f + +

A1 = (At/S) Ph (fyt/fy) cot2

A1 min = (5 'cf Acp)/fy (At/S) Ph (fyt/fy)

ytwlfbA /25

min

Min. bar diameter =0.042S no. 3

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943/1000

1

CONTENTS

Solutions Manual Chapter

Chapter

2 Properties of Reinforced Concrete3 Flexural Analysis of Reinforced Concrete Beams4 Flexural Design of Reinforced Concrete Beams6 Deflection and Control of Cracking7 Development Length of Reinforcing Bars8 Shear and Diagonal Tension9 One Way Slabs10 Axially Loaded Columns11 Members in Compression and Bending12 Slender Columns13 Footings14 Retaining Walls15 Design for Torsion16 Continuous Beams and Frames17 Design of Two-Way Slabs18 Stairs19 Introduction to Prestressed Concrete20 Seismic Design of Reinforced Concrete Structures21 Beams curved in Plan

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2

CHAPTER 2

PROPERTIES OF REINFORCED CONCRETE

2.1- 2.8 refer to the relative section in text

2.9 Calculate the modulus of elasticity; Ec (see the table below)

2.10 Calculate the modular ratio; n and the modulus of rupture; fr:

Density fc Ec n fr

160 pcf 5000 psi. 4,723,000 psi. 6.14 530.3 psi.

145 pcf 4000 psi. 3,644,000 psi. 7.96 474.3 psi.

125 pcf 2500 psi. 2,306,000 psi. 12.58 375.0 psi.

2400 kg/m3

35 MPa 29,910 Mpa 6.69 3.668 MPa

2300 kg/m3

30 MPa 25,980 Mpa 7.70 3.396 MPa

2100 kg/m3

25 MPa 20,690 Mpa 9.67 3.10 MPa

2.11a.) See figure below.

b.) Secant modulus (at fc/2 = 1910 psi.)

Ec = 1910 / 6.1010-4

= 3130 ksi.

Approximate Initial Modulus = 2.6(ksi.) / 5.4510-4

= 4771 ksi.

(Possible range 4600 5200)c.) Ec (ACI formula) = 57000 ' 57000 3820 3523 ksi.fc = =

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3

CHAPTER 3

FLEXURAL ANALYSIS OF REINFORCED CONCRETE BEAMS

Problem 3.1

DIM. No.a No.b No.c No.d No.e No.f No.g No.h

b (in.) 14 18 12 12 16 14 10 20

d (in.) 22.5 28.5 23.5 18.5 24.5 26.5 17.5 31.5

As (in.2)

4#10

5.08

6#10

7.62

4#9

4.00

4#8

3.16

5#10

6.35

5#9

5.00

3#9

3.00

4#9

4.00

a(in.) 6.4 7.47 5.88 4.65 7.00 6.3 5.29 3.53

Mn(k.ft) 441.2 849.1 370.1 230.0 600.0 525.3 200.5 535.2

0.0161 0.0149 0.0142 0.0142 0.0162 0.0135 0.0171 0.0063

max

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4

Problem 3.3

DIM. No.a No.b No.c No.d .No.e No.f No.g No.h

b ( in.) 54 48 72 32 44 50 40 42

bw ( in.) 14 14 16 16 12 14 16 12

t ( in. ) 3 4 4 3 4 3 3 3

d ( in. ) 17.5 16.5 18.5 15.5 20.5 16.5 16.5 17.5

As (in.2)

4#10

5.08

4#9

4.00

8#10

10.16

6#9

6.00

8#9

8.00

7#9

7.00

5#10

6.35

6#9

6.00

a ( in. ) 2.213 1.961 3.32 4.41 4.27 3.29 3.74 3.36

Rect. or T R R R T T T T T

Mn (k.ft) 374.7 279.4 769.9

Asmin< As < Asmax Yes Yes Yes

Asf (in.2) 2.04 5.44 4.59 3.06 3.83

a ( in. ) 4.94 5.02 4.05 4.84 4.25

Mn (k.ft) 293.4 660.1 466.8 415 425.9

Asmin< As < Asmax No Yes Yes Yes Yes

Problem 3.4

DIM. No.a No.b No.c No.d No.e No.f No.g No.h

fc 3 ksi 4 ksi 4 ksi 5 ksi 30 Mpa 20 Mpa 30 Mpa 25 Mpa

fy 40 ksi 60 ksi 75 ksi 60 ksi 400 Mpa 300 Mpa 500 Mpa 300 Mpa

b 12 in. 12 in. 12 in. 12 in. 300 mm 300 mm 300 mm 300 mm

d 20 in. 20 in. 20 in. 20 in. 500 mm 500 mm 500 mm 500 mm

As

(in.2)

(mm

2

)

4#8

3.14

4#7

2.41

4#9

4.00

4#9

4.00

3*30 mm

2121mm

2

3*25 mm

1473

4*25 mm

1963

4*20 mm

12571 0.85 0.85 0.85 0.8 0.85 0.85 0.85 0.85

0.037 0.0285 0.0207 0.0335 0.0325 0.0321 0.0236 0.04

max 0.020 0.0181 0.0145 0.0212 0.0203 0.01805 0.0162 0.0225

Rumax0.614

ksi

0.821

ksi

0.822

ksi

0.973

ksi

6.14

Mpa

4.11

Mpa

6.13

Mpa

5.11

Mpa

0.013 0.0100 0.0167 0.0167 0.01414 0.00982 0.01309 0.0084

Ru0.422

7 ksi

0.4942

ksi

0.9182

ksi

0.7941

ksi

4.52

Mpa

2.42

Mpa

5.13

Mpa

2.13

Mpa

a/d ratio 0.205 0.1772 0.3676 0.2353 0.2218 0.1732 0.2567 0.118

(a/d) max 0.318 0.3194 0.3199 0.2993 0.3184 0.3194 0.3176 0.3176

min

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947/1000

5

Problem 3.5:

3.5 a:

As = 1256.6mm2

; = 0.00914 ; Ru = 3.6 Mpa ; Mn = 272.3 KN.m

b = 0.02024 ; max = 0.6375b = 0.0129 Ok

3.5 b:

As = 1472 mm2

; = 0.0107 ; Ru = 3.51 Mpa ; Mn = 265.4 KN.mb = 0.02024 ; max = 0.6375b = 0.0129

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6

Problem 3.10

3.10 a:

= 0.011157 ; = 0.00556 ; - = 0.005597 < max = 0.0203

- < k = 0.0139, Compression steel doesnt yield.

c = 3.07 in. ; a = 1c = 6.61 in. ; fs= 16.15 ksi fy

C1 = 79.85 k ; C2 = 16.32 k ; As1 = 1.996 in.2

; 1 = 0.009242 k = 0.0139, Compression steel yields.

- 60 ; Compression steel yields, fs = 60 ksiC1 = 451.66 k ; C2 = 79.2 k ; T = 530.86 k ; As1b = 7.53 in.

2

As max = 0.63375As1b + As= 6.09 in.2

(2) = 0.02307 ; = 0.005 ; - = 0.01807 > k, Compression steel yields.

- = max ; Ok ; a = 7.01 in. ; Mn = 512.8 k.ft ; Wl = 8.516 k/ft

Problem 3.12:

= 0.01805 ; = 0.0031 ; - = 0.015 < max = 0.01806 - < k = 0.0194 , Compression steel doesnt yield.

c = 6.44 in. ; a = 5.47 in. ; fs= 53.23 ksi < fy

C1 = 186 k ; C2 = 30.4 k ; As1 = 3.1 in.2; 1 = 0.0155 t ; T section ; Asmax = 4.99 in.2

> As Ok

(2) Asf= 2.55 in.2

; a = 5.08 in. ; Mn = 339.6 k.ft

Problem 3.15:a = 3.69 in. > t, T-section, Asf= 2.55 in.

2, Asw = 2.16 in.2, As(max) = 5.80 in.

2,

a = 5.08 in., Mu = 466.78 k-ft.

Problem 3.16:

Same analysis and answer as 3.14

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7

CHAPTER 4

DESIGN OF REINFORCED CONCRETE BEAMS

Problem 4.1:

DIM. No.a No.b No.c No.d No.e No.f No.g No.h

Mu (k.ft) 272.7 969.2 816 657 559.4 254.5 451.4 832

b (in) 12 18 16 16 14 10 14 18

d (in) 21.5 32 29.52 26.5 24.5 21.5 21.75 28

Ru psi 589.9 631 780 700 799 660.7 818 708

(%

) 1.168 1.206 1.7 1.5 1.75 1.115 1.8 1.27

AS (in2) 3.013 6.947 8.029 6.36 6.0 2.39 5.48 6.42

Bars

rows

5#7

One

7#9

two

7#10

two

5#10

one

6#9

two

6#6

one

6#9

two

8#9

two

h (in) 24 36 32 29 28 24 25 32

DIM. No.i No.j No.k No.l

Mu(k.ft) 345 510 720 605

b (in) 15 12 12 16

d (in) 18.5 24.9 29.67 23.6

(%) 1.77 1.81 1.8 1.8

AS (in)2

4.91 5.4 6.4 6.8

Barsrows

5#9one

6#9two

5#10two

7#9two

h (in.) 21 29 33 28

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8

Problem 4.2:

Assume c/dt = 0.375 for fy = 60Ksi,(all problem)dt = (d + 1)in. for 2 rows of bars; d = 2.5 in.

DIM.Prob.

No.a

Prob.

No.b

Prob.

No.c

Prob.

No.d

Prob.

No.e

Prob.

No.f

Prob.

No.g

Prob.

No.h

Prob.

No.i

Prob.

No.j

Prob.

No.k

Prob.

No.l

Mu(k.ft) 554 790 448 520 765 855 555 300 400 280 290 400

b (in) 14 16 12 12 16 18 16 12 16 12 14 14

d (in) 20.5 24.5 18.5 20.5 20.5 22 18.5 16.5 16.5 16.5 14.5 17.5

c (in) 8.39 9.156 6.91 7.66 7.66 8.22 3.258 6.166 6.166 6.187 5.41 6.54

a (in) 7.13 7.78 5.87 6.511 6.512 6.988 2.769 5.24 5.241 5.26 4.606 5.56

Mu1(k.ft) 402 656.3 280.6 344.6 459.5 595.3 374.2 223.2 297.7 223.2 201.1 293.0

Mu2(k.ft) 152 133.7 167.4 175.4 305.5 259.7 180.8 76.76 102.3 56.76 88.86 107.02

As1(in2) 5.44 7.37 3.996 4.428 5.904 7.128 5.328 3.564 4.75 3.576 3.654 4.41

As2(in2) 1.68 1.13 2.324 2.165 3.77 2.959 2.511 1.22 1.62 1.35 1.645 1.585

As (in2) 7.12 8.50 6.38 6.66 9.674 10.09 7.839 4.784 6.376 4.925 5.299 5.995

fs (ksi) 60 60 55.52 58.6 58.6 60 20.24 51.73 51.73 51.84 46.8 53.74

As (in2) 1.68 1.35 2.51 2.22 3.862 2.96 7.44 1.413 1.88 1.0425 2.12 1.77

Tension

Barsrows

6#10

Two

2#8

Two

2#10

Two

2#10

Two

9#6

Two

10#5

Two

6#10

Two

5#5

Two

10#4

Two

2#7

Two

7#5

Two

4#6

Two

Comp. bars 1#9 2#7 2#10 2#9 4#9 3#9 3#8 2#9 2#8 2#6 2#9 2#8

h (in) 24 28 22 24 24 26 22 20 20 20 18 21

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Problems 4.3:

DIM. No.a No.b No.c No.d No.e No.f No.g No.h

Mu (ft) 394 800 250 327 577 559 388 380

b (in) 48 60 44 50 54 48 44 46

bw (in) 14 16 15 14 16 14 12 14

t (in) 3 4 3 3 4 4 3 3

d (in) 18.5 19.5 15 13 18.5 17.5 16 15

Mn(k.ft) 468.2 803.3 340.8 329.9 681.6 569.2 366 356.3

R or_T R R R R R R T T

Ru (psi) 288 421 303 464 375 456

(%) 0.0063 0.769 0.646 0.558 0.76 0.804AS (in

2) 5.577 9.006 4.264 3.631 7.59 6.754

Bars

rows

10#7

Two

4#14

one

4#9

one

12#5

two

6#10

two

3#14

one

6#9

two

5#10

two

c/ dt 0.375 Yes Yes Yes Yes Yes Yes Yes Yes

Cf(k) 244.8 244.8

ASf(in2) 4.08 4.08

Muf(ft) 266.2 247.9

Muw (ft) 121.8 132.1

Ru (web) (psi) 476 503

w(%) 1.0 1.07

ASw (in2) 1.92 2.25

AS (in2) 6.0 6.33

AS (max) (in2) 7.85 11.7 6.75 7.06 10.50 9.10 6.68 6.93

h (in.) 21 23 18 18 22 21 20 19

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Problems 4.3: (continued)

DIM. No.i No .j No.k No.l No.m No.n No.o No.p

Mu (ft) 537 515 361 405 378 440 567 507

b (in) 60 54 44 50 44 36 48 46

bw (in) 16 16 15 14 16 16 12 14

t (in) 3 3 3 3 3 4 3 3

d (in) 16.5 17.5 15 15.5 13.5 18 22.5 18

Mn(k.ft) 516.4 495.7 340.8 401.6

R or_T T T T T R R R T

Ru (psi) 379 452 280

Barsrows

8#9two

6#10two

6#9two

7#9two

6#9two

5#10one

6#9two

7#9two

c/ dt 0.375 Yes Yes Yes Yes Yes Yes Yes Yes

Cf(k) 336.6 290.7 221.9 275.4

ASf(in2) 5.61 4.85 3.7 4.59

Muf(ft) 378.7 348.8 224.6 289.2

Muw (ft) 158.3 166.2 136.44 115.8

Ru (web) (psi) 436.12 407 485 413

w(%) 0.9 0.84 1.02 0.85

ASw (in2) 2.38 2.35 2.3 1.84

AS (in2) 7.99 7.2 6.0 6.43

AS (max) (in2) 9.19 8.64 6.75 7.53 6.49 7.30 8.25 3.58

h (in.) 20 21 19 19 20 21 26 22

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Problem 4.4:

(a) = max , d = 18.4 in. ; AS = 3.31 in.2

; Use 4#9 bars ( two rows. AS = 4.0 in.2

);h = 22 in. ; a = 5.84 in.; c = 6.87 in.; dt = 19.5 in. => c / dt =0.352 < 0.375 ; OK

(b) =0.016, d = 19.4 in. ; AS = 3.1 in.2

; Use 4#8 bars ( two rows. AS = 3.16 in.2

);h = 23 in. ; a = 5.47 in.; c = 6.43 in.; dt = 20.5 in. => c / dt =0.314 < 0.375 ; OK

(c) = 0.012, d = 21.9 in. ; AS = 2.63 in.2

; Use 3#9 bars ( AS = 3 in.2

);h = 25 in. ; a = 4.64 in.; c = 5.46 in.; dt = 22.5 in. => c / dt =0.243 < 0.375 ; OK

Problem 4.5:

Ru = 465 psi ; =0.0146 ; As = 3.5 in2

; Use 3# 10 bars (As = 3.79 in2

)

Problem 4.6:

= 0.0094 ; Ru = 465 psi ; As = 2.27 in2

; Use 2#10 bars ( AS = 2.53 in2)

Problem 4.7:

Assume a total depth = L / 15 = 2012 / 15 =16 in.

Mu = 132 k.ft(a) Using max = 0.01806, Ru(max) = 820 psi ; d = 12.7 in. ; AS = 2.75 in

2; Use 3#9 ;

h =16 in ; min < = max , t = 0.005

(b) Using s = 0.014, Ru = 662 psi ; d = 14.2 in. ; AS = 2.38 in2

; Use 4#7 bars (As = 2.41 in2

).

h= 17 in. ; min < < max , t > 0.005 OK

Problem 4.8:

Case 1:

(1) D.L. on ABC, L.L. on AB for maximum positive B.M. ; U = 4.84 k/ft.

(2) Positive ultimate B.M. = 305.15 k.ft; Let =0.015, Ru =700 psi, d = 20.88 in.(3) AS = 3.76 in

2; use 4#9 bars in one row.; bmin= 11.625 in.< 12 in. ; h = 24 in.

Case 2:D.L. on ABC, L.L. on BC only for maximum negative B.M.

d = 21.5 in. ; Mu = 154.88 k.ft ; Ru = 335 psi

= 0.00674; min < < max , t > 0.005 ; AS = 1.74 in2

; Use 2#9 bars or 3#7

Problem 4.9:

Mu = 317 k.ft ; Ru = 622 psi, min < < max , t > 0.005 ; d = 20.27 in.AS= 3.97 in

2; Use 4#9 bars ( AS = 4.0 in

2)

in one row. ; h = 23 in.

Problem 4.10:

Mu = 151.6 k.ft ; Ru = 476 psi, min < < max , t > 0.005 ; d = 17.8 in.

AS= 3.2 in2 ; Use 3#10 bars ( AS = 3.79 in 2 )in one row. ; h = 21 in.

Problem 4.11:

Mn = Rumaxbd2

= 222 k.ft < Mu ; Compression steel is needed.Mu1 = 222 k.ft ; Mu2 = 68 k.ft ; AS1= 3.10 in.

2; AS2 = 0.92 in.

2; AS = 4.02 i n .

2; Use 4 # 9

a = 6.08 in. ; c = 7.15 in.; dt = 20.5 in.; t =0.0056 > 0.005; OKA S' = 0.98 in.

2; AS(max)=5.27 in

2> As = 4.02 in

2; h = 23 in. OK.

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Problem 4.12:

a)d = 22.5 in. and AS = 3.05 in2;Use 3#9 bars in one row (AS = 3.0 in

2); h =26 in

Mu = 269.6 k.ft.; a = 7.06 in.; c = 8.3 in.; dt = 23.5 in.; t =0.0055>0.005;

b) Mu1 =195 k.ft, M u 2 = 65 k.ftFor Mu1 = 195 k.ft ; d = 19.5 in. and AS1 = 2.64 in

2;Let h = 23 in. and d = 2.5 in.

For M u 2 = 65 k.ft. ; As= AS2 = 0.85 in.

2

; AS = 3.49 in .

2

Use 5 # 8c) Mu = 260 k.ft

1) For two rows of steel h = 26 in, d =22.5 in and d t = 23.5 in.2) Section behaves as a T-section.

ASf= 1.28 in.2

; M u2 =120.5 k.ft ; Mu1 = 139.5 k.ft ; AS1 = 1.53 in.2

AS(max)=4.33 in2

> Total As = 2.81 in2

; OK ; Choose 3#9 bars, ( AS = 3.0 in2

)

Problem 4.13:

d = 19.8 in.; As = 6.8 in.2

; Use 7#9 bars (two rows. AS = 7.0 in.2

)if d = 22 in. is used; a=3.53 in. < 4 in. ; h = 26in.

AS(min) = 0.86 in.2

< AS < AS(max) = 7.78 in.2

OK

Problem 4.14:Ultimate load W = 7.9 k/ft

MB( maximum negative) = -395 K.ft

MDmax = 270.8 k.ft

b) Mn = 344.6 k.ft < Mu Compression steel is needed.

AS = 5.04 in .2

Use 7 # 8 ( two rows)A S' = 0.64 in.

2; AS1 = 3.12 in.

2;Use 4#8 bars in one row, ( A S = 3.16 in

2)

Problem 4.15:

a) W = 8.56 k/ft ; HA = HD = 34.24 k ; RA = RD = 154.08 k

MB = MC = - 616.32 k.ft. ; MBC (midspan) = 770.4 k.ftDesign beam BC for positive moment,AS = 8.86 in

2Use 9#9 bars (two rows, As = 9.0 in

2)

ASmax = 26.65 in.2

; AS(min)= mi nb w d = 1 . 0 9 in2

AS(min) < AS < AS(max) OK

Design beam BC for negative moment,

Mu =616.32 k.ft

Compression steel is needed.

AS = 7.93 in .2

Use 8 # 9 in two rows, ( AS = 8 in2

)A S'= 1.99 in.

2Use 2 # 9 in one row, ( A S = 2 in

2)

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CHAPTER 5

ALTERNATE DESIGN METHODS

Extra Problems

Problem E5.1

Tie AB kipsFu 75.58= 231.1 inAts = Provide 5 No. 5 Bar

255.1 inAts =

Tie CD kipsFu 0.15= 233.0 inAts = Provide 1 No. 4 Bar, 2 legs

240.0 inAts =

Tie BD & DF kipsFu 21.68= 252.1 inAts = Provide 5 No. 5 Bar

255.1 inAts =

Calculate strut widths

75.0=s psifce 2868382575.0 ==

Strut AC, inwkipsPu 26.1,83.86 ==

Strut BC, inwkipsPu 18.1,93.80 ==

Strut CE, inwkipsPu 90.1,67.130 ==

Strut DE, inwkipsPu 26.1,16 ==

sin

2 0.20

24 4.5 0.0032 0.003

Problem E5.2: Deep Beam Design Example: Bridge Bent Cap Using Strut and Tie Method

Load Combination: U = 1.8 [DL + (LL+I)]

PGirder= 1.8 x 580 kips = 1044 kip

WDL = 1.8 x 8.5 kips/lf = 15.3 kip/lfWLL = 1.8 x 7.6 kips/lf = 13.6 kip/ft

Strut And Tie Model:

P1 = 1195 kipLclr / Ds = 1.21 < 4 Bent Cap is considered a deep beam

Angle between strut and tie = 46d > 26d Good

Check Maximum Shear Strength of Beam X-Section: (ACI 318-08 11.7.3)Vu = 1195 kip

Assume d = 0.9 * h = 75.6 in

Vn = 2886 kip > Vu Good

Force Resultants:From Truss Analysis, internal forces are presented below. Since the bent cap is symmetrical, the

left side is designed and reinforcement will be applied symmetrically.

Member 1 and 2 (Tie) T1 = T2= 1020 kip TensionMember 3 (Strut) C3 = 1570 kip Compression

Member 4 (Strut) C4 = 1195 kip Compression

Member 5 (Strut) C5 = 910 kip CompressionMember 6, 7 and 8 (Tie) T6 = T7 =T8 = 0 kip

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Support Reaction D1x = -1016 kip, D1y = 1195 kip

Support Reaction D2x = 0 kip, D2y = 1195 kipSupport Reaction D3x = 627 kip, D3y = 655 kip

Calculate Effective Strength, fce:

Member 3 and Member 5 Bottle-shape struts fce = 3.18 ksi

Member 4 Uniform x-section strut fce = 4.25 ksiNodal Zone A C-C-T fce = 3.4 ksi

Nodal Zone B C-T-T fce = 2.55 ksi

Nodal Zone C C-C-T fce = 3.4 ksiNodal Zone D1 C-C-T fce = 3.4 ksi

Nodal Zone D3 C-C-T fce = 3.4 ksi

Dimensions of nodal zones:Nodal zone A Horizontal width wc = 1570 kip / 0.75 * 3.4ksi *72in = 8.55 in

Tie 1 width w1 = 8.55 * (1020kip / 1570kip) = 5.55 in

Strut 3 width w3 = 10.2 inNodal zone B Horizontal width wc = 1195 kip / 0.75 * 2.55 ksi *72in = 8.68 in

Nodal zone C Horizontal width wc = 910 kip / 0.75 * 3.4ksi *72in = 4.96 inTie 2 width w2 = 6.5 * (1020kip / 910kip) = 5.6 in

Strut 5 width w5 = 7.45 inNodal zone D1 and D3 Assume same widths as Nodal zone A and C

Nodal zone D2 Assume same widths as Nodal zone B

Check Capacity of Struts:Capacity of Strut 3 fns = *fce * Acs = 1751 kips > 1570 Good

Capacity of Strut 4 fns = *fce * Acs = 1195 kips = 1195 Good

Capacity of Strut 5 fns = *fce * Acs = 1280 kips > 910 GoodDesign of vertical and horizontal reinforcement:

Vertical Bars Use #6 @ 12 4 legged stirrups, As = 4 * 0.44 = 1.76 in^2Sin 50d = 0.76, As / (b * s) * sin = 0.0015

Horizontal Bars Use #8 @ 9 2 sides, As = 2 * 0.79 = 1.58 in^2

Sin 40d = 0.64, As / (b * s) * sin = 0.00156Design of horizontal tie 1 and 2:

Fu = *As*fy As = 1020 kip / (0.75 * 60 ksi) = 22.7 in^2

22.7 in^2 / (1 in^2) #9 Bars = 23

Use #9 Tot 24, Vertically Bundled in 2sAnchorage Length Ldh = 17 in

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CHAPTER 6

DEFLECTION AND CONTROL OF CRACKING

Problem 6.1(a)-6.1(f):

Description Prob. (a) Prob. (b) Prob. (c) Prob. (d) Prob. (e) Prob. (f)

b (in.) 14 20 12 18 16 14

d (in.) 17.5 27.5 19.5 20.5 22.5 20.5

h (in.) 20 30 23 24 26 24

As (in2.) 5 7.59 4.71 7.59 9.37 8

As (in

2

.) 0 0 0 2 2.53 2

Wd (k/ft.) 2.2 7 3 6 5 3.8

Wl (k/ft.) 1.8 3.6 1.5 2 3.2 2.8

Pd (k.) 0 0 0 0 12 8

Pl (k.) 0 0 0 0 10 6

Ig (in4.) 9333.3 45000 12167 20736 23434.7 16128

x (in.) 7.543 10.238 8.363 0 0 0

Icr (in4.) 5968.5 25247.2 7013.2 0 0 0

x (in.) 0 0 0 8.475 10.017 9.377

Icr (in4.) 0 0 0 12932.3 18042 12427.9

Mcr (k-ft.) 36.89 118.59 41.82 68.31 71.26 53.13

Ma (k-ft.) 200 530 225 400 520 400

Ie (in4.) 5989.6 25468.5 7046.3 12971.1 18055.8 12436.6

I (in.) 0.667 0.416 0.638 0.616 0.551 0.62

2 2 2 1.57 1.48 1.48

I (in.) 0.397 0.289 0.447 0.478 0.352 0.384

a (in.) 0.794 0.577 0.893 0.751 0.521 0.568

T = I +a (in.) 1.46 0.99 1.53 1.37 1.07 1.19

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Problem 6.2(a)-6.2(d):

Discription Prob. (a) Prob. (b) Prob. (c) Prob. (d)

b (in.) 15 18 12 14

d (in.) 20.5 22.5 19.5 20.5

h (in.) 24 26 23 24

As (in2.) 8 7.59 6.28 8

As (in2.) 2 0 1.57 2

Wd (k/ft.) 3.5 2 2.4 3

Wl (k/ft.) 2 1.5 1.6 1.1

Pd (k.) 0 7.4 0 5.5

Pl (k.) 0 5 0 4

Ig (in4.) 17280 26364 12167 16128

x (in.) 9.175 9.401 8.822 9.377

Icr (in4.) 12693.9 1540.3 8914 12427.9

x (in.) 0 0 0 0

Icr (in4.) 0 0 0 0

Mcr (k-ft.) 56.92 80.16 41.82 53.13

Ma (k-ft.) 396 381 288 393

Ie (in4.) 12707.5 15505.9 8923.9 12437.1

I (in.) 0.538 0.457 0.558 0.577

1.51 2 1.5 1.48

I (in.) 0.342 0.29 0.33 0.42

a (in.) 0.516 0.58 0.5 0.63

T = I +a (in.) 1.05 1.042 1.06 1.203

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Problem 6.3:1.) d = 23.27 in. ; As = 5.0 in.

2use 5 #9 ; say h = 27 in.

2.) Ig = 19683 in.4

; Ma = 3916 k.in. ; Mcr= 691.1 k.in.4 411268 in. ; 11217 in.cr e g I I I= = <

in.12.1=

Problem 6.4:1.) Mu = 444 k.ft.

d = 20.82 in. ; Total As = 5.57 in.2 Use 6 #9, As = 6.0 in.

2

Compression steel, fs = 59 ksi.

h = 24.32 in., say 25 in. ; d = 21.5 in. ; For comp. steel, use 2 #7, As = 1.2 in.2

2.) Ig = 15625 in.4

; Ma = 3916 k.in. ; Mcr= 592.93 k.in.4 410750 in. ; 10767 .cr e g I I in I= = <

in.17.1=

Additional long-term deflection = 1.29 in.Total deflection = 2.46 in.

Problem 6.5:

a) S = 10.31 in less than 12 in.; provided spacing = 3.5 in., OK.W=0.0121 in.

b) W = 0.011 in.

c) W = 0.0103 in.

d) W = 0.011 in.

Problem 6.6:

As (min.) = 0.126 in

2

/ft./side.; Max. spacing = 6.75 in., say 6.5 in.As/side = (0.12640.5) / (212) = 0.212 in

2; Choose 3 #3 bars each side.

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CHAPTER 7

DEVELOPMENT LENGTH OF REINFORCING BARS

Problems 7.1(a) - 7.1(j)

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

fc(ksi.) 3 4 5 3 4 5 5 3 4 4

fy (ksi.) 60 60 60 40 60 60 60 40 60 60

Bar No. 5 6 7 8 9 10 11 9 8 6

db (in.) 0.625 0.75 0.875 1.0 1.128 1.27 1.41 1.128 1.0 0.75

Clear

Cover (in.)2.0 2.0 2.0 2.5 1.5 2.0 3.0 2.0 2.0 1.5

Clear

Spacing(in.)2.25 2.5 2.13 2.3 1.5 2.5 3.0 1.5 1.75 1.65

Cond. met Y Y Y Y N Y Y N N Y

t 1.0 1.0 1.0 1.3 1.0 1.0 1.0 1.0 1.0 1.3

e 1.0 1.0 1.5 1.0 1.0 1.0 1.0 1.5 1.0 1.5

1.0 0.75