1

CHAPTER 3 FLEXURAL ANALYSIS OF REINFORCED CONCRETE BEAMS

Problem 3.1 a ( ρmax = 0.01806, ρ min = 0.00333 ) for all problems. Given: b = 14 in, d = 22.5 in, As = 5.08 in2, fc’ = 4 ksi, fy = 60 ksi Solution: φ = 0.9

a =A ff bs y

c085. ' = 5 08 60

0 85 4 14.

.×

× × = 6.4 in.

φMn = φAsfy (d - a/2) = 0.9 × 5.08 × 60 × (22.5 - 6.4/2) = 5294 k.in. = 441.2 k.ft

ρ = As / bd = 5 0814 22 5

..×

= 0.016

ρ < ρmax = 0.01806, ρ > ρmin = 0.00333 Ok General Equations:

a =A ff bs y

c085. ' ,

ρ = As / bd,φMn = φAsfy (d - a/2) Problem 3.1

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

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

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

ρ 0.0161 0.0149 0.0142 0.0142 0.0162 0.0135 0.0171 0.0063

ρmax<ρ<ρmin Yes Yes Yes Yes Yes Yes Yes Yes

2

Problem 3.2a ( ρmax = 0.01806, ρ min = 0.00333 ) for all problems. Typical Solution, comp. steel yields Given: b = 15 in, d = 22.5 in, As = 8 in2, As’ = 2 in2 , fc’ = 4 ksi, fy = 60 ksi Solution: φ = 0.9

ρ = As/bd = 815 22.5×

= 0.0237

ρ’ = As’/bd = 215 22.5×

= 0.0059

ρ - ρ’ = 0.0237 - 0.0059 = 0.0178

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

= 085 460

2 522 5

8787 60

2. ..

× × ×−

= 0.0172

ρ - ρ’ = 0.0178 > k = 0.0172, Compression steel yields.

a =bffAA

c

yss

'85.0)'( −

=(8 2) 600.85 4 15− ×

× ×= 7.06 in.

φMn = φ[(As - As’) fy (d - a/2)+ As’fy(d-d’)]

= 0.9[( 8 - 2)×60×(22.5 - 7.06/2) + 2×60(22.5-2.5)]= 8306.5 k.in. = 692.2 k.ft

ρmax < ( ρ - ρ’ )< ρ min , OK General Equations:ρ = As/bd, ρ’ = As’/bd,

a =bffAA

c

yss

'85.0)'( −

,

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

A = 0.85 fc’β1b, B = As’ (87 - 0.85 fc’) - As fy, C = -87As’d’ c = [-B ± ACB 42 − ] / 2A fs’= 87( c - d’ )/c ≤ fy, C1 = 0.85 fc’ab, C2 = As’ (fs’- 0.85 fc ’) Comp. steel yields:

a =bffAA

c

yss

'85.0)'( −

,

φMn = φ[(As - As’)fy (d - a/2)+ As’fy(d-d’)]

3

4

3.2 e: Typical solution, Comp. steel does not yield. Given: b = 14 in., d = 20.5 in., As = 7.62 in.2, As’ = 2.54 in.2 , fc’ = 4 ksi, fy = 60 ksi Solution: φ = 0.9

ρ = As/bd = 5.2014

62.7×

= 0.0266

ρ’ = As’/bd = 2 5414 20 5

..×

= 0.0089

ρ - ρ’ = 0.0266 - 0.0089 = 0.0177

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

= 085 460

2 520 5

8787 60

2. ..

× × ×−

= 0.0189

ρ - ρ’ = 0.0177 < k = 0.0189, Compression steel doesn’t yield. A = 0.85 fc’β1b = 0.85 × 4 × 0.85 × 14 = 40.46 B = As’ (87 - 0.85 fc’) - As fy = 2.54(87 - 0.85 ×4) - 7.62 ×60 = - 244.9 C = -87As’d’ = - 87 ×2.54 ×2.5 = - 552.45 c = [-B ± ACB 42 − ] / 2A = 7.8 in. a = β1c = 0.85 ×7.8 = 6.63 in. fs’= 87( c - d’ )/c = 87( 7.8 - 2.5 )/7.8 = 59.1 ksi ≤ fy

Cc = 0.85 fc’ab = 0.85 × 4 × 6.63 × 14 = 315.7 k Cs = As’ (fs’- 0.85 fc ’) = 2.54(59.1 - 0.85 ×4) = 141.48 k As1 = Cc / fy = 315.7/60 = 5.26 in.2

ρ1 = As1 / bd = 5.26/(14 × 20.5) = 0.0183 < ρmax

φMn = φ[ Cc (d - a/2)+ Cs(d-d’)] = 0.9[315.7×(20.5 - 6.63/2) + 141.48×(20.5-2.5)]= 7173.2 k.in. = 597.9 k.ft

5

Problem 3.2

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

b (in.) 15 17 13 10 14 16 20 18

d (in.) 22.5 24.5 22 21.5 20.5 20.5 18 20.5

As (in.2) 8#9 8.00

8#10 10.08

7#9 7.00

4#10 5.08

6#10 7.62

9#9 9.00

12#9 12.00

8#10 10.12

As’(in.2) 2#9 2.0

2#10 2.54

3#7 1.8

2#7 1.2

2#10 2.54

4#9 4.0

6#9 6.0

4#10 5.08

ρ 0.0237 0.0242 0.0245 0.0236 0.0266 0.0274 0.0333 0.0275

ρ’ 0.0059 0.0061 0.0063 0.0056 0.0089 0.0122 0.0167 0.0138

ρ- ρ’ 0.0178 0.0181 0.0182 0.0180 0.0177 0.0152 0.0167 0.0137

k 0.0172 0.0158 0.0176 0.0180 0.0189 0.0189 0.0216 0.0189

a ( in.) 7.06 7.83 7.06 6.85

φ Mn 692.2 950 590.2 418.2

A 40.46 46.24 57.8 52.6

B -244.9 -205.6 -218.4 -184.9

C -552.45 -870 -1305 -1100.49

c 7.8 7.097 7.0 6.7

a 6.63 6.033 5.95 5.71

fs’ 59.1 56.36 55.94 54.62

C1 315.7 328.2 504.8 349.4

C2 141.5 211.8 315.2 260.2

As1 5.26 5.47 6.75 5.82

ρ1 0.0183 0.0167 0.0187 0.0158

ρ1≤ρmax Yes Yes Yes Yes

φ Mn 597.9 716.3 822.5 813.7

6

Problem 3.3 a: Typical solution of T-sections, Rect. Section analysis. Given: bw = 14 in., b = 54 in., t = 3 in., d = 17.5 in., As = 5.08 in.2,

fc’ = 3 ksi, fy = 60 ksi Solution: φ = 0.9

a =A ff bs y

c085. ' =

54385.06008.5××

× = 2.213 in. < t

c = 2.6 in. dt = 17.5 in. εt = 0.003(dt – c)/c = 0.0172 > 0.005 Rectangular section analysis φMn = φAsfy ( d - a/2 ) = 0.9 × 5.08 × 60 × (17.5 – 2.21/2) = 4497 k.in. = 374.7 k.ft Asmax = 0.0425 [t (be-bw) + 0.319bwd] = 0.0425 [3(54-14) + 0.319×54×17.5]) = 8.42 in.2

Asmin = ρminbwd = 0.00333 ×14×17.5 = 0.81 in.2

Asmin < As< Asmax Ok Problem 3.3 d: Typical solution of T-sections analysis. Given: bw = 16 in., b = 32 in., t = 3 in., d = 15.5 in., As = 6 in.2, fc’ = 3 ksi, fy = 60 ksi Solution:

a =A ff bs y

c085. ' =

32385.0606

××× = 4.41 in. > t

T-section Asmax = 0.0425 [t (be-bw) + 0.319bwd] = 0.0425 [3(32-16) + 0.319×16×15.5]

= 5.40 in.2

As = 6 in2 > Asmax N.G. Section doesn’t meet ACI Code. If ρmax is used: Asf = 0.85 fc’t(b-bw)/fy = 0.85×3×3× (32-16) /60 =2.04 in.2

Mflange = φAsffy (d - t/2) = 0.9 × 2.04 × 60 (15.5 – 3/2) = 1542.24 k.in. = 128.52 k.ft As1 (web) = 5.40 – 2.04 = 3.36 in.2

a = 3.36 600.85 3 16

×× ×

= 4.94 in.; c = 5.81 in.

dt = 15.5 + 1 = 16.5 in. εt = 0.003(dt – c)/c = 0.0055 > 0.005 φMn (web)= φAs1fy ( d - a/2 ) = 0.9 × 3.36× 60 × (15.5 – 4.94/2)/12 = 197.01 k.ft Total φMn = 128.52 + 197.01 = 325.53 k.ft

7

Problem 3.3 e: Typical solution of T-sections analysis. Given: bw = 12 in., b = 44 in., t = 4 in., d = 20.5 in., As = 8 in.2, fc’ = 3 ksi, fy = 60 ksi Solution: φ = 0.9

a =bf

fA

c

ys

'85.0=

44385.0608

××× = 4.27 in. > t

T-section Asf = 0.85 fc’t(b-bw)/fy

= 0.85×3×4× (44-12) /60 = 5.44 in.2

a =wc

ysfs

bffAA

'85.0)( −

=12385.060)44.58(

×××− = 5.02 in.

c = 5.906 in. dt = 21.5 in. εt = 0.003(dt – c)/c = 0.0079 > 0.005 φMn = φ[(As - Asf ) fy ( d - a/2 ) + Asf fy ( d - t/2 ) = 0.9[(8 – 5.44)×60(20.5 – 5.02/2) + 5.44 × 60(20.5 – 4/2) = 7921 k.in. = 660.1 k.ft Asmax = 0.0425 [t (be-bw) + 0.319bwd] = 0.0425 [4(44-12) + 0.319×12×20.5]

= 8.77 in.2

Asmin = ρminbwd = 0.00333 ×12×20.5 = 0.82 in.2

Asmin < As< Asmax Ok Problem 3.3

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

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 φMn (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 3.83a ( in. ) 4.94 5.02 4.05 4.84 4.25

φMn (k.ft) 293.41 660.1 466.8 415 425.9

Asmin< As < Asmax No Yes Yes Yes Yes

8

9

3.4 a: Typical Solution: Given: b = 12 in., d = 20 in., As = 3.14 in.2, fc’ = 3 ksi, fy = 40 ksi Solution: ρb = 0.85β1(fc’/fy)[87/( 87+ fy)] = 0.85 ×0.85( 3/40 ) × [87/(87+40)] = 0.0371 ρmax = 0.5474ρb = 0.5474 × 0.0371 = 0.0203

Ru(max) = φρmax fy ( )'7.1

1 max

c

y

ffρ

−

= 0.9 ×0.0203×40 ( )0.0203 4011.7 3

×−

×= 0.615 ksi

ρ = As / bd = 3.14/ (12×20) = 0.01308

Ru = φρ fy ( )'7.1

1c

y

ffρ

−

= 0.9 ×0.01308×40 ( )37.1

4001308.01×

×− = 0.4226 ksi

a / d ratio = '85.0 c

y

ffρ

=385.0

4001308.0×× = 0.2052

(a / d) max = '85.0

max

c

y

ffρ

= 0.0203 400.85 3

××

= 0.319

Problem 3.4 e: Typical Solution:

Given: b = 300 mm, d = 500mm, As = 3× ( )430 2π = 2121mm2, fc’ = 30Mpa, fy = 400Mpa

Solution: ρb = 0.85β1(fc’/fy)[600/( 600+ fy)] = 0.85 ×0.85(30/400) × [600/ (600+400)] = 0.0325 ρmax = 0.625ρb = 0.625× 0.0325 = 0.0203

Ru(max) = φρmax fy ( )'7.1

1 max

c

y

ffρ

−

= 0.9 ×0.0203×400 ( )0.0203 40011.7 30

×−

×= 6.14 Mpa

ρ = As / bd = 2121/(300×500) = 0.01414

Ru = φρ fy ( )'7.1

1c

y

ffρ

− = 0.9 ×0.01414×400 ( )307.1

40001414.01×

×− = 4.53 Mpa

a / d ratio = '85.0 c

y

ffρ

=3085.0

40001414.0×× = 0.2218

(a/ d) max = '85.0

max

c

y

ffρ

= 0.0203 4000.85 30

××

= 0.3184

10

Problem 3.4

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

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)(mm2)

4#8

3.14

4#7

2.41

4#9

4.00

4#9

4.00

3*30 mm

2121mm2

3*25 mm

1473

4*25 mm

1963

4*20 mm

1257

β1 0.85 0.85 0.85 0.8 0.85 0.85 0.85 0.85

ρb 0.0371 0.0285 0.0207 0.0335 0.0325 0.0321 0.0236 0.04

ρmax 0.0203 0.0181 0.0145 0.0212 0.0203 0.01805 0.0162 0.0225

Rumax 0.614

ksi 0.821

ksi 0.822

ksi 0.973

ksi 6.14 Mpa

4.11 Mpa

6.13 Mpa

5.11 Mpa

ρ 0.0131 0.01004 0.01667 0.01667 0.01414 0.00982 0.01309 0.0084

Ru0.4227

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 0.2052 0.1772 0.3676 0.2353 0.2218 0.1732 0.2567 0.118

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

ρmin<ρ<ρmax Yes Yes No Yes Yes Yes Yes Yes

3.5 a: Given: b = 250 mm, d = 550mm, As = 4×20mm, fc’ = 20 Mpa, fy = 420 Mpa Solution:

As = 4× ( )420 2π = 1256.6mm2

ρ = As / bd = 1256.6/(250×550) = 0.00914

Ru = φρ fy ( )'7.1

1c

y

ffρ

−

= 0.9 ×0.00914×420 ( )207.1

42000914.01×

×− = 3.6 Mpa

φMn = Ru bd2 =3.6×106×0.25×0.552 = 272.3 KN.m ρb = 0.85β1(fc’/fy)[600/( 600+ fy)] = 0.85 ×0.85(20/420) × [600/ (600+420)] = 0.02024 ρmax = 0.6375ρb = 0.6375× 0.02024 = 0.0129 Ok

11

3.5 b: Given: b = 250 mm, d = 550mm, As = 3×25mm, fc’ = 20 Mpa, fy = 420 Mpa Solution:

As = 3× ( )425 2π = 1472 mm2

ρ = As / bd = 1472 / (250×550) = 0.0107 ρb = 0.85β1(fc’/fy)[600/( 600+ fy)] = 0.02024 ρmax = 0.6375ρb = 0.0129 ρ < ρmax Ok

Ru = φρ fy ( )'7.1

1c

y

ffρ

−

= 0.9 ×0.0107×420 ( )0.0107 42011.7 20

×−

×= 3.51 Mpa

φMn = Ru bd2 =3.51×0.25×0.552

= 265.4 KN.m 3.5 c: Given: b = 250 mm, d = 550mm, As = 4×30 mm, fc’ = 20 Mpa, fy = 420 Mpa Solution:

As = 4× ( )430 2π = 2827 mm2

ρ = As / bd = 2827/ (250×550 ) = 0.02056 ρmax = 0.0129 ρ> ρmax N.G. , Section does not meet ACI code, reduce steel.

Ru(max) = φρmax fy ( )'7.1

1 max

c

y

ffρ

−

= 0.9 ×0.0129×420 ( )0.0129 42011.7 20

×−

×= 4.1 Mpa

φMn = Ru (max) bd2 = 4.1×0.25×0.552

= 310 KN.m

12

3.5 d: Given: b = 10 in., d = 22 in., As = 2.0 in.2, fc’ = 3 ksi, fy = 60 ksi Solution: ρ = As / bd = 2/(10×22) = 0.0091 ρb = 0.85β1(fc’/fy)[87/( 87+ fy)] = 0.85 ×0.85(3/60) × [87/ (87+60)] = 0.0214 ρmax = 0.01356 ρmin < ρ < ρmax Ok

Ru = φρ fy ( )'7.1

1c

y

ffρ

−

= 0.9 ×0.0091×60 ( )0.0091 6011.7 3

×−

×= 0.439 ksi

φMn = Ru bd2 = 0.439×10×222 = 2124 k.in. = 177 k.ft

3.5 e: Given: b = 10 in., d = 22 in., As = 6.0 in.2, fc’ = 3 ksi, fy = 60 ksi Solution: ρ = As / bd = 6/(10×22) = 0.02727 ρmax = 0.01356 < ρ N.G. , Section does not meet ACI code, reduce steel.

Ru(max) = φρmax fy ( )'7.1

1 max

c

y

ffρ

−

= 0.9 ×0.01356×60 ( )0.01356 6011.7 3

×−

×= 0.615 Ksi

φMn = Ru(max) bd2 =0.615×10×222 = 2976.6 k.in. = 248.1 k.ft

13

3.6: Given: b = 8 in., d = 18 in., fc’ = 3 ksi, fy = 50 ksi Solution:

(a) εy =Ef y =

2900050 = 0.00172, εc = 0.003

εy > εs = 0.0015 This section is over- reinforced.

(b) dcb =

00172.0003.0003.0

+

cb =00172.0003.0

003.0+

× 18 = 11.44 in.

ab = β1cb = 0.85×11.44 = 9.725 in. 0.85fc’ab b = Asbfy

Asb =y

bc

fbaf '85.0

=50

8725.9385.0 ××× = 3.97 in.2

(c) dc =

0015.0003.0003.0+

c =0015.0003.0

003.0+

× 18 = 12 in.

a = β1c = 0.85×12 = 10.2 in. fs = Esεs = 29000×0.0015 = 43.5 ksi 0.85fc’a b = Asfs

As = 0.85 'cs

f a bf

=5.43

82.10385.0 ××× = 4.783 in.2

φMn = φAsfs (d - a/2) = 0.9×4.783×43.5(18 – 10.2/2) = 2416 k.in. = 201.3 k.ft From tables: ρmax = 0.01628 As max = ρmax bd = 0.01628×8×18 = 2.34 in.2

a = max

0.85 's y

c

A ff b

= 2.34 500.85 3 8

×× ×

= 5.74 in.

(max) φMn = φAsmaxfy (d - a/2 ) = 0.9×2.34×50(18 – 5.74/2) = 1593.2 k.in. = 132.8 k.ft (max) φMn allowed by ACI code is less than φMn of the section = 201.3 k.ft (d) ρ = 0.014 < ρmax As = ρbd = 0.014×8×18 = 2.016

a =bf

fA

c

ys

'85.0=

8385.050016.2××

× = 4.941 in.

φMn = φAsfy (d - a/2) = 0.9×2.016×50(18 – 4.941/2) = 1408.8 k.in. = 117.4 k.ft

14

15

Problem 3.7: Given: b = 12 in., d = 24 in., L = 10 ft, As = 3.93 in.2, fc’ = 3 ksi, fy = 60 ksi Solution:

a =bf

fA

c

ys

'85.0=

12385.06093.3××

× = 7.706 in.

φMn = φAsfy (d - a/2) = 0.9×3.93×60(24 – 7.706/2) = 4275.6 k.in. = 356.3 k.ft

Mu = 1.2(2

2lWD ) + 1.6(2

2lWL ) = φMn

1.2(2

10685.0 2× ) + 1.6(2102×lW ) = 356.3

Wl = 3.94 k/ft

Problem 3.8: Given: b = 10 in. h = 28 in., d = 24.5 in., L = 17 ft, As = 3.53 in.2, fc’ = 3 ksi, fy = 40 ksi Solution:

a =bf

fA

c

ys

'85.0=

10385.04053.3××

× = 5.537 in

φMn = φAsfy (d - a/2) = 0.9×3.53×40(24.5 – 5.537/2) = 2761.6 k.in = 230.1 k.ft

Mu = 1.2(8

2lWD ) + 1.6(4lpL ) = φMn

1.2(8

1755.2 2× ) + 1.6(417×Lp ) = 230.1

Pl = 17.6 k

Problem 3.9 a: Given: b = 12 in., d = 20 in., As = 3.14 in.2, fc’ = 4 ksi, fy = 60 ksi Solution: (1) At balanced condition

dcb =

00207.0003.0003.0

+cb = 11.834 in.

ab = β1cb = 0.85×11.834 = 10.059 in. C = 2×0.85fc’×3×4 + 0.85 fc’×12(10.059 – 4) = 328.8 k Asb = C/fy = 328.8/60 = 5.48 in.2

Asmax = 0.63375Asb = 0.63375×5.48 = 3.473 in.2

As < Asmax

16

(2) T = Asfy = 3.14×60 = 188.4 k T = C = 2×0.85 fc’×3×4 + 0.85 fc’×12(a – 4) 188.4 = 2×0.85×4×3×4 + 0.85×4×12(a – 4) a = 6.618 in C1= 2×0.85fc’×3×4 = 81.6 kC2= 0.85 fc’×12(a – 4) = 0.85×4×12(6.618 – 4) = 106.8 k

φMn = φ[C1( d - 4/2 )+ C2( d – 4 -2

4−a )]

= 0.9[81.6( 20 - 4/2 )+ 106.8( 20 – 4 - 2

4618.6 − )]

= 2734 k.in = 227.83 k.ft

Problem 3.9 b: Given: b = 12 in., d = 16 in., As = 3.93 in.2, fc’ = 4 ksi, fy = 60 ksi Solution: (1) At balanced condition

dcb =

00207.0003.0003.0

+cb = 9.467 in.

ab = β1cb = 0.85×9.467 = 8.047 in. C = 0.85 fc’×5×4 + 0.85 fc’×12(ab – 4 ) = 233.1 kAsb = C/fy = 233.1/60 = 3.89 in.2

Asmax = 0.63375Asb = 0.63375×3.89 = 2.465 in.2

As = 3.93> Asmax , N.G. Section does not meet ACI code, reduce steel. (2) Use As = As max = 2.465 in.2

T = Asfy = 2.465×60 = 147.9 kT = C = 0.85 fc’×5×4 + 0.85 fc’×12(a – 4) 174.85 = 0.85×4×5×4 + 0.85×4×12(a – 4) a = 5.96 in. C1= 0.85fc’×5×4 = 68 kC2= 0.85 fc’×12(a – 4) = 0.85×4×12(5.96 – 4) = 79.97 k

φMn = φ[C1( d – 4/2 )+ C2( d – 4 –2

4−a )]

= 137.5 K.ft

17

Problem 3.10 a: Given: b = 12 in., d = 18in., d’ = 2.5in., As = 2.41 in.2, As’ = 1.2 in.2, fc’ = 3ksi, fy = 40ksi Solution:

ρ = As/bd = 1812

41.2×

= 0.011157

ρ’ = As’/bd = 1812

2.1×

= 0.00556

ρ - ρ’ = 0.005597 < ρmax = 0.0203

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

=4087

8718

5.240385.0 2

−××× = 0.0139

ρ - ρ’ < k = 0.0139, Compression steel doesn’t yield. A = 0.85 fc’β1b = 0.85 × 3 × 0.85 × 12 = 26.01 B = As’ ( 87 - 0.85 fc’) - As fy = 1.2( 87 - 0.85 ×3) – 2.41 ×40 = 4.94 C = -87 As’d’ = - 87 ×1.2 ×2.5 = - 261 c = [-B ± ]B AC2 4− / 2A = 3.07 in. a = β1c = 0.85 ×3.07 = 6.61 in. fs’= 87( c - d’ )/c = 87( 3.07 - 2.5 )/3.07 = 16.15 ksi ≤ fyC1 = 0.85 fc’ab = 0.85 × 3 × 2.61 × 12 = 79.85 k C2 = As’ ( fs’- 0.85 fc ’) = 1.2( 16.15 - 0.85 ×3 ) = 16.32 k As1 = C1 / fy = 79.85/40 = 1.996 in.2

ρ1 = As1 / bd = 1.996/(12 × 18) = 0.009242 < ρmax φMn = φ[ C1(d - a/2 )+ C2(d-d’)]

= 0.9[79.85×(18 – 2.61/2) + 16.32×(18-2.5)] = 1428 k.in. = 119 k.ft Problem 3.10 b: Given: b = 12 in., d = 18 in., As = 4.82 in.2, As’ = 1.2 in.2, fc’ = 3 ksi, fy = 40 ksi Solution:

ρ = As/bd = 4.8212 18×

= 0.0223

ρ’ = As’/bd = 1812

2.1×

= 0.00556

ρ - ρ’ = 0.0167

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

=4087

8718

5.240385.0 2

−××× = 0.0139

ρ - ρ’ > k = 0.0139, Compression steel yields. ρ - ρ’< ρmax = 0.0203 Ok

a =bffAA

c

yss

'85.0)'( −

= (4.82 1.2) 400.85 3 12

− ×× ×

= 4.73 in.

φMn = φ[(As-As’)fy ( d - a/2 ) + As’fy ( d – d’ ) = 0.9[(4.82-1.2)×40(18 – 4.73/2) +1.2×40(18 – 2.5)] = 2707 k.in. = 225.6 K.ft

18

Problem 3.11: Given: b = 12 in., d = 22 in., d’ = 2.5 in., As’= 1.32 in.2, fc’ = 4 ksi, fy = 60 ksi, l = 16 ft Solution: (1) At balanced condition:

dcb = 0.003

0.003 0.00207+cb = 13.02 in.

ab = β1cb = 0.85×13.02 = 11.07 in.

fs’ = )5.2

(87b

b

cc −

= 13.02 2.587( )13.02

− = 70.3 > 60

Compression steel yields, fs’ = 60 ksi C1 = 0.85 fc’abb = 0.85×4×12×11.07 = 451.66 k C2 = As’ fy = 1.32×60 = 79.2 k T = C1 + C2 = 530.86 k As1b = C1/fy =451.66/60 = 7.53 in.2

As max = 0.63375As1b + As’= 0.63375×7.53+1.32 = 6.09 in.2

(2) ρ = As / bd = 6.0912 22×

= 0.02307

ρ’ = As’/bd = 1.3212 22×

= 0.005

ρ - ρ’ = 0.01807

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

= 2 4 2.5 870.8560 22 87 60

× × ×−

= 0.0176

ρ - ρ’ > k, Compression steel yields. ρ - ρ’= ρmax Ok

a =bffAA

c

yss

'85.0)'( −

= (6.09 1.32) 600.85 4 12

− ×× ×

= 7.01 in.

φMn = φ[(As-As’)fy ( d - a/2 ) + As’fy ( d – d’ ) = 0.9[(6.09-1.32)×60(22 – 7.01/2) + 1.32×60(22 – 2.5)] = 6153.9 k.in. = 512.8 k.ft Mu = 1.2 MD + 1.6ML = φMn

1.2(8162 2× )+1.6(

2168

LW × )= 512.8

Wl = 8.516 k/ft

19

Problem 3.12 Given: b = 10 in., d = 20 in., d’ = 2.5 in., As = 3.61 in.2, As’ = 0.62 in.2, fc’ = 4 ksi, fy = 60 ksi, l = 10 ft, Wl = 1.25 k/ft, WD = 2 k/ft (excluding self weight) Solution:

ρ = As/bd = 2010

61.3×

= 0.01805

ρ’ = As’/bd = 0.6210 20×

= 0.0031

ρ - ρ’ = 0.015 < ρmax = 0.01806

k = 085 8787

2.' '× × ×

−ff

dd f

c

y y

=6087

8720

5.260485.0 2

−××× = 0.0194

ρ - ρ’ < k , Compression steel doesn’t yield. A = 0.85 fc’β1b = 0.85 × 4 × 0.85 × 10 = 28.9 B = As’ (87 - 0.85 fc’) - As fy = 0.62(87 - 0.85 ×4) – 3.61 × 60 = -165 C = -87 As’d’ = - 87 ×0.62 ×2.5 = - 134.85 c = [-B ± ]B AC2 4− / 2A = 6.44 in. a = β1c = 0.85 ×6.44 = 5.47 in. fs’= 87( c - d’ )/c = 87( 6.44 - 2.5 )/6.44 = 53.23 ksi < fyC1 = 0.85 fc’ab = 0.85 × 4 × 5.47 × 10 = 186 k C2 = As’ ( fs’- 0.85 fc ’) = 0.62( 53.23 - 0.85 ×4 ) = 30.4 k As1 = C1 / fy = 186/60 = 3.1 in.2

ρ1 = As1 / bd = 3.1/(10 × 20) = 0.0155 < ρmax

φMn = φ[ C1 (d - a/2)+ C2(d-d’)] = 0.9[186×(20 – 5.47/2) + 30.4×(20-2.5)]= 3369 k.in. = 280.7 k.ft

WD = 2 + 15.0144

2010 ×× = 2.208 k/ft

WL = 1.25 k /ft

Mu = 1.2 MD + 1.6ML = 1.2(2

2lWD ) + 1.6(2

2lWL ) = φMn

1.2(2

10208.2 2× ) + 1.6(2

1025.1 2× ) = 232.5 k.ft

The internal design moment of 280.7 k.ft is great than the external ultimate moment of 232.5 k.ft, the section is adequate.

20

Problem 3.13: Given: bw= 10 in., t = 4 in., d =18 in., As = 3 in.2, l = 18 ft, b = 9 ft, fc’ = 4 ksi, fy = 60 ksi Solution: (1) Effective width, be= b = 108 in. be= l/4 = 18×12/4 = 54 in. be= 16t+bw = 16×4 +10 = 74 in, use be = 54 in.

(2) a =bf

fA

c

ys

'85.0=

54485.0603

××× = 0.98 in. < t

Rectangular Section (3) φMn = φAsfy (d - a/2) = 0.9×3×60(18 – 0.98/2) = 2836.6 k.in. = 236.4 k.ft Asmin = ρminbwd = 0.00333×10×18 = 0.6 in2

Asmax = 0.0567 [t (be-bw) + 0.319bwd] = 0.0567 [4(54-10) + 0.319×10×18]

= 13.23 in.2

Asmin < As< Asmax Ok Problem 3.14: Given: bw= 10 in., t = 3 in., d =18 in., As = 4.71 in.2, b = 30 in., fc’ = 3 ksi, fy = 60 ksi Solution:

(1) a =bf

fA

c

ys

'85.0=

30385.06071.4××

× = 3.69 in. > t

T section Asmax = 0.0425 [t (be-bw) + 0.319bwd] = 0.0425 [3(30-10) + 0.319×10×18]

= 4.99 in.2 > As Ok (2) Asf = 0.85 fc’t(b-bw)/fy = 0.85×3×3( 30 – 10)/60 = 2.55 in.2

a =( ')

0.85 's s f y

c

A A ff b

−=

10385.060)55.271.4(

×××− = 5.08 in.

φMn = φ[(As-As’) fy ( d - a/2 ) + Asf fy ( d – d’ )] = 0.9[(4.71-2.55)×60(18 – 5.08/2)+2.55×60(18 – 3/2)] = 4075.3 k.in. = 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