1

CONTENTS

Solutions Manual Chapter

Chapter

2 Properties of Reinforced Concrete

3 Flexural Analysis of Reinforced Concrete Beams

4 Flexural Design of Reinforced Concrete Beams

6 Deflection and Control of Cracking

7 Development Length of Reinforcing Bars

8 Shear and Diagonal Tension

9 One Way Slabs

10 Axially Loaded Columns

11 Members in Compression and Bending

12 Slender Columns

13 Footings

14 Retaining Walls

15 Design for Torsion

16 Continuous Beams and Frames

17 Design of Two-Way Slabs

18 Stairs

19 Introduction to Prestressed Concrete

20 Seismic Design of Reinforced Concrete Structures

21 Beams curved in Plan

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 2.11

a.) 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 = =

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

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) (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 1 0.85 0.85 0.85 0.8 0.85 0.85 0.85 0.85 b 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 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

0.013 0.0100 0.0167 0.0167 0.01414 0.00982 0.01309 0.0084 Ru

0.4227 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 0.205 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

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.m b = 0.02024 ; max = 0.6375b = 0.0129 < max Ok 3.5 c: As = 2827 mm2 = 0.02056 ; max = 0.0129 < ; N.G. , Section does not meet ACI code, reduce steel. Ru(max) = 4.1 Mpa ; Mn = 310 KN.m 3.5 d: = 0.0091 ; b = 0.0214 ; max = 0.01356 ; min < < max Ok Ru = 0.439 ksi ; Mn = 177 k.ft 3.5 e: = 0.02727 ; max = 0.01356 < ; N.G. , Section does not meet ACI code, reduce steel. Ru(max) = 0.615 Ksi ; Mn = 248.1 k.ft Problem 3.6: y = 0.00172 > s = 0.0015 This section is over- reinforced. cb = 11.44 in. ; ab = 9.725 in. ; Asb = 3.97 in.2 c = 12 in. ; a = 10.2 in. ; fs = Ess = 43.5 ksi ; As = 4.783 in.2 ; Mn = 201.3 k.ft max = 0.01628 ; As max = 2.34 in.2 ; a = 5.74 in. ; (max) Mn = 132.8 k.ft (max) Mn allowed by ACI code is less than Mn of the section = 201.3 k.ft = 0.014 < max ; As = 2.016 ; a = 4.941 in. ; Mn = 117.4 k.ft Problem 3.7: a = 7.706 in.; Mn = 356.3 k.ft ; Wl = 3.94 k/ft Problem 3.8: a = 5.537 in ; Mn = 230.1 k.ft ; Pl = 17.6 k Problem 3.9: 3.9 a: (1) At balanced condition: cb = 11.834 in. ; ab = 10.059 in. ; Asb = 5.48 in.2 Asmax = 0.63375Asb = 3.473 in.2 ; As < Asmax (2) T = 188.4 k ; a = 6.618 in ; C1= 81.6 k ; C2= 106.8 k ; Mn = 227.83 k.ft 3.9 b: (1) At balanced condition: cb = 9.467 in. ; ab = 8.047 in. ; Asb = 3.89 in.2 Asmax = 2.465 in.2 ; As > Asmax , N.G. ; Section does not meet ACI code, reduce steel. (2) Use As = As max = 2.465 in.2 ; T = 147.9 k ; a = 5.96 in. ; C1= 68 k ; C2= 79.97 k Mn = 137.5 K.ft

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 < max ; Mn = 119 k.ft 3.10 b: = 0.0223 ; = 0.00556 ; - = 0.0167 > k = 0.0139, Compression steel yields. - < max = 0.0203 ; Ok ; a = 4.73 in. ; Mn = 225.6 K.ft Problem 3.11: (1) At balanced condition: cb = 13.02 in. ; ab = 11.07 in. fs = 70.3 > 60 ; Compression steel yields, fs = 60 ksi C1 = 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 < max ; Mn = 280.7 k.ft WD = 2.208 k/ft ; WL = 1.25 k /ft ; Mu = 232.5 k.ft ; the section is adequate. Problem 3.13: (1) be = 54 in. (2) a = 0.98 in. < t ; Rectangular Section (3) Mn = 236.4 k.ft ; Asmin = 0.6 in2 < As< Asmax = 13.23 in.2 Ok Problem 3.14: (1) a = 3.69 in. > 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

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

Bars rows

5#9 one

6#9 two

5#10 two

7#9 two

h (in.) 21 29 33 28

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 88.86 107.02As1(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

Bars rows

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

9

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 0.804 AS (in2) 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

10

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 Bars rows

8#9 two

6#10 two

6#9 two

7#9 two

6#9 two

5#10 one

6#9 two

7#9 two

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

11

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 in 2 ; 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 in 2 )

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 in2 ; 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 in2 ; 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 in2 ; 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; OK A S'= 0.98 in.2 ; AS(max)= 5.27 in2 > As = 4.02 in2 ; h = 23 in. OK.

12

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.ft For Mu1 = 195 k.ft ; d = 19.5 in. and AS1 = 2.64 in2 ;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 i n . 2 Use 5 # 8 c) Mu = 260 k.ft 1) For two rows of steel h = 26 in, d =22.5 in and dt = 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, ( A S = 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 i n . 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 in2 ) 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.ft Design beam BC for positive moment, AS = 8.86 in2 Use 9#9 bars (two rows, As = 9.0 in2 ) ASmax = 26.65 in.2 ; AS(min)= mi n b 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 i n . 2 Use 8 # 9 in two rows, ( A S = 8 in2 ) A S'= 1.99 in.2 Use 2 # 9 in one row, ( A S = 2 in2 )

13

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 0.2024 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/lf WLL = 1.8 x 7.6 kips/lf = 13.6 kip/ft Strut And Tie Model: P1 = 1195 kip Lclr / 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 Tension Member 3 (Strut) C3 = 1570 kip Compression Member 4 (Strut) C4 = 1195 kip Compression Member 5 (Strut) C5 = 910 kip Compression Member 6, 7 and 8 (Tie) T6 = T7 =T8 = 0 kip

14

Support Reaction D1x = -1016 kip, D1y = 1195 kip Support Reaction D2x = 0 kip, D2y = 1195 kip Support 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 ksi Nodal 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 ksi Nodal 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 in Nodal 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 in Tie 2 width w2 = 6.5 * (1020kip / 910kip) = 5.6 in Strut 5 width w5 = 7.45 in Nodal 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 Good Design of vertical and horizontal reinforcement: Vertical Bars Use #6 @ 12 4 legged stirrups, As = 4 * 0.44 = 1.76 in^2

Sin 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 = 0.00156 Design 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 2s Anchorage Length Ldh = 17 in

15

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 (in2.)

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

16

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

17

Problem 6.3: 1.) d = 23.27 in. ; As = 5.0 in.2 use 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 gI 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 gI 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 in2 /ft./side.; Max. spacing = 6.75 in., say 6.5 in. As/side = (0.12640.5) / (212) = 0.212 in2 ; Choose 3 #3 bars each side.

18

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

1.0 0.75 1.0 1.0 1.0

1.0

1.0 1.0

t e 1.7

Y

Y

Y Y Y Y Y

Y

Y N

'cf

19

Problem 7.3: a.) ld = 69.56 in., use 70 in. b.) Bars with 90 hook, db = 1.128 in.; lhb = 21.4 in. Use 22 in. c.) Bars with 180 hook, db = 1.128 in.; lhb = 22 in. Problem 7.4: a.) ld = 0.875(61.66) = 53.96 in., use 54 in. b.) Bars with 90 hook, db = 0.875 in.; lhb = 16.6 in. Use 17 in. > 8 db c.) Bars with 180 hook, db = 1.128 in.; Basic lhb = 17 in. > 8 db Problem 7.5: a.) ld = 1.128(47.47) = 53.54 in., use 54 in. b.) Bars with 90 hook, db = 1.128 in.; lhb = 16.48 in. Use 17 in. c.) Bars with 180 hook, db = 1.128 in.; Basic lhb = 17 in. Problem 7.6: a.) ld = 1.27(61.66) = 78.3 in., use 79 in. b.) Bars with 90 hook, db = 1.27 in.; lhb = 24.09 in. Use 25 in. c.) Bars with 180 hook, db = 1.27 in.; Basic lhb = 24 in. Problem 7.7: a.) When 50% of bars are spliced; ld = 80.3 in., use 81 in., lst = ldc = 81 in. Class A splice. b.) When 75% of bars are spliced; Class B splice, lst = 104.3 in., use 105 in. c.) All bars are spliced, Rs = 2.0, Class B splice, lst = 105 in. d.) When all bars are spliced, Rs = 1.3, Class B splice, lst = 105 in. Problem 7.8: a.) When 50% of bars are spliced, ld = 92.7 in., use 93 in. Class A splice, lst = ldc = 93 in. b.) When 75% of bars are spliced; Class B splice, ld = 93 in., lst = 121 in. c.) All bars are spliced, Rs = 2.0, Class B splice, lst = 121 in. d.) When all bars are spliced, Rs = 1.3, Class B splice, lst = 121 in.> 12 in. Problem 7.9: ldc = 19.14 in. 3.84 in. ; Use lsc = ldc = 34 in. Controls Problem 7.10: ldc = 23.9 in. 42.3 in. ; Use lsc = ldc = 43 in. Controls Problem 7.11: ldc = 25.52 in. 54.14 in; lsc = 54.14 in., use lsc = 55 in. Controls Problem 7.12: ldc = 21.4 in. 33.84 in.; Use lsc = 34 in. Controls

20

Problem 7.13: 1.) Let d = 18 in; Development length of X1 = 62 in. 2.) X2 = 18 in. 3.) 4#8 bars are extended beyond point of inflection. Total length = 66 in. 4.) X3 = 18 in. 5.) X4 = 4 ft = 48 in. 6.) X5 = 19 in. 7.) X6 = 6 ft. = 48 in. ; X7 = 18 in. Problem 7.14: As max = 3.9 in.2 Use 5#8 bars in two rows, (As = 3.93 in.2) Use h = 22 in. total length; Actual d = 18.5 in. Mur (one bar) = 663 K.in. = 55.3 K.ft. Extend bars 1 and 2 20 ft., 1/3 of total bars, which meets code requirement. Length of bar 3 = 21.5 ft, i.e. span length = 20ft. Length of bar 4 = 17.8 ft. Length of bar 5 = 15.1 ft. (for #8 bars, ld = 48 in.) Problem 7.15: d = 33.6 in.; As = 4.314 in.2 Use 6#8 bars in two rows, (As = 4.72 in.2) Use h = 38 in.; Actual d = 34.5 in. Mur (one bar) = 107.6 K.ft.; For 6 bars, Mu = 645 K.ft. Length of bar # 1 = 7.5 + 3 = 10.5 ft. Length of bar # 2 = 5.5 + 3 = 8.5 ft. Length of bar # 3 = 3.75 + 3 = 6.75 ft. ldh = 22 in. ld = 55 in., For top bars, ld = 1.3(55) = 71.5 in., or 72 in. = 6 ft.

Problem 7.16: 1.) Maximum positive moment Mu = 574 K.ft. Maximum negative moment Mu = 435.2 K.ft. 2.) Section at B: d = 23.04 in.; As max = 4.99 in.2 Use 5#9 bars in two rows, (As = 5 in.2) h = 26.54 ; use 27 in. total length; Actual d = 23.5 in. Mur (one bar) = 88.9 K.ft./ bar; For 5 bars, Mu = 444 K.ft. 3.) Section within AB, (positive moment): Mu = 574 > 44, need compression steel Use 2 # 9 bars; As = 2.0 in.2 Use 7 # 9 bars; As = 7.0 in.2 Actual Mu = 633.4 K.ft. 4.) ld (bottom bars) = 70 in. = X1 ld (top bars) = 54 in. = X2

21

CHAPTER 8 SHEAR AND DIAGONAL TENSION

Problem 8.1. a) Use #3 stirrups spaced at 8.5 in. b) Use #3 @ 5.0 in. c) Use #4 @ 6.5 in.

Problem 8.2. a) Use #3 @ 11.5 in. b) Use #3 @11 in. c) Use #3 @6.5 in. Problem 8.3. a) Use #3 @13.5 in. b) Use #4 @ 7 in. c) Use #4 @ 4.5 in.

Problem 8.4) 1. Vu = 83.25 kip; Vu @ d = 64.29 kip 2. Vc = 0.7535.93 = 26.95 kip < Vu @ d.; Vs = 49.79 kip < Vc1 , 3. Use #3 @ 6 in. 4. Vs = 20.3 k.

6.29 ft. 75 in.X = = 5. Distribution of stirrups 1st stirrup @ S/2 =3.0 in. 6 stirrup @6 in. = 39 in. 39 in. > 38.9 in. 4 stirrup @10 in. = 44 in. 79 in. > 75 in. Problem 8.5) 1. Wu = 7.2 k/ft.; Vu = 64.8 kip; Vu @ d = 50.4 kip 2. Nominal shear provided by concrete Vc = 23.66 kip < Vu @ d; Vs = 35.65 kip 3. Use #3 @ 7 in. 4. For Smax = 12 in., Vs = 13.2 in. X at Vc/2 = 88 in. 5. Distribution of stirrups: Use 1 stirrup @ 2.5 in. 6 stirrup @ 7.0 in. = 42 in. 45.5 > 41.07 in. 4 stirrup @ 12 in. = 48 in. 93.5 > 88.22 in.

22

Problem 8.6) 1. Vu at support = 82 kip, Vu @ d = 68.25 kip 2. Vc = 21.91 k.; Vs = 61.78 kip; Then S2 = d/2 = 8.25 in. 3. Use #3 stirrup @ 3.5 in. 4. For Smax = 4 in.; Vs = 34.04 k.; X = 72.77 in. 5. Distribution of stirrups: 1 stirrup@ 2 in = 2 in. 8-stirrup @ 3.5 in. = 28 in 30 in > 26.7 in. 12-stirrup @ 4 in. = 48 in. 74.7 in.>72.77 in.

Problem 8.7) 1. Mu = 168.5 k-ft ; Use d = 18.6 in. As = 2.33 in2; Use 3 #9, As = 3.0 in2 and h = 21.5 in 2. Design for shear Vu = 51.02 kip; Vu @ d = 46.27 kip; Vc = 18.75 kip; Vs = 36.69 kip 3. Use #3 @ 5.5 in. 4. Vu = 13.2 k. < Vc = 17.25 k. but > Vc/2 = 8.63 k., Vs = 0 Therefore, use #3 stirrup, 2 legs, at maximum spacing of 6.5 in.; X (at Vc/2) = 72.47 in. 5. Distribution of stirrups (from A): 1st stirrup @ S/2 = 3.0 in. 6 stirrup @ 6.5 in. = 39 in. 42 in. > 36 in. 4 stirrup @ 9.5 in. = 38 in. 80.0 in. > 72.47 in. Problem 8.8) 1. Mu = 91.5 k-ft; Use d = 12 in.; Use 2 #9, As = 2 in2 and h = 14.5 in (15 in.) 2. Wu = 5.5 k/ft.; Vu = 30.25 kip, Vu @ d = 24.75 kip; Vc = 13.15 kip ; Vs = 19.9 kip Therefore, use Smax = 6.0 in. all over with first stirrup at 3 in. from the face of the support.

Problem 8.9) 1. Use #3 stirrups, 2 legs, Av = 0.22 in2; let S = S3 = 11 in. 2. h = 37.9 in., say 38 in., d = 35.5 in. Use 16 38 in. section, with #3 stirrups @ 11 in. 3. Vc = 53.89 k., Vs = 21.4 k.

Problem 8.10) 1. Wu = 7.2 k/ft.; Vu = 64.8 kip; Vu @ d = 50.4 kip 2. Vc = 23.66 kip < Vu @ d; Vs = 35.65 kip Use stirrup #3, Av = 0.22 in.2; s = 8.5 in. 3. Use #3 @ 8.5 in. 4. For Smax =12 in.; Vs = 19.8 k.; X = 88 in. at Vc /2 5. Distribution of stirrup: 1st stirrup @ 4 in. = 4.0 in. 4 stirrup @ 8.5 in. = 38 in. 42 in. > 35.5 in. 5 stirrup @ 12 in. = 48 in. 90 in. > 88 in.

23

Problem 8.11) 1. Vu at support = 66 kip; Vu @C.L. = 12 kip; Vu @ d = 55.4 kip 2. Vc = 21.91 k., Vs = 54.65 kip 3. Use #4 Stirrups; Smax = 4 in. (Controls) 4. Vs = 27.2 K. for Smax = 8 in.; X1 = 28 in. 5. Distribution of stirrup: Use 1-stirrup @ 2 in. = 2 in. 17-stirrup @ 4 in. = 68 in. 70 in. > 70 in. Problem 8.12) 1. Mmax = 1296 k-in.; As = 2.22 in.2 Use 3 #8, As = 2.35 in.2 ; Use h = 15.5 in., d = 13 in. 2. Check at 2 ft. from support: W2 = 6.22 k/ft, Mu= 609.8 k-in.; As = 2.08 in.2, use 3 #8 3. ld = 48 in = 4 ft. from the support. 4. a) At support: Vu = 288 psi b) At free end Vu = 0 c) Max d = 13 in., for d = 12.5 in.; Vu = 240 psi d) At midspan Vu = 97 psi e) Vus = 161.5 psi f) Choose #3 stirrup@ 5 in. g) Distribution of stirrups from the support first stirrup at 2 in. = 2 in 21-stirrup at 5.0 in. = 105 in

Total = 107 in. < 108 in.

24

CHAPTER 9 ONE-WAY SLABS

Problems 9.1(a) - 9.1(j):

(a)

(b)

(c) (d) (e) (f) (g)

(h)

(i) (j)

fc (ksi)

3

3

3 3 4 4 4

4

5 5

h (in.)

5

6

7 8 5.5 6 7.5

8

5 6

d (in.)

4

4.9

5.9 6.75 4.44 4.8 6.38

6.75

3.94 4.94

As (in2)

0.39

0.46

0.59 0.78 0.37 0.6 0.88

0.79

0.37 0.46

a (in.)

0.76

0.91

1.16 1.53 0.78 1.17 1.72

1.55

0.73 0.90

Mn (k.ft.)

8.03

9.3

11.46 14.51 7.55 11.65 16.0

14.66

7.55 9.21

Problems 9.2(a) - 9.2(j):

(a)

(b)

(c) (d) (e) (f) (g)

(h)

(i) (j)

fc (ksi)

3.0

3.0

3.0 3.0 4.0 4.0 4.0

4.0

5.0 5.0

Mu (k.ft.)

5.4

13.8

24.4 8.1 22.6 13.9 13.0

11.2

20.0 10.6

h (in.)

6.0

7.5

9.0 5.0 7.5 8.5 6.0

7.5

9.0 6.0

d (in.)

4.9

6.375

6.7 3.87 6.3 7.375 4.875

4.5

7.8 4.875

(%)

0.43

0.685

0.85 1.11 1.18 0.495 1.10

0.51

0.637 0.9

As (in2)

0.252

0.524

0.95 0.526 0.89 0.438 0.67

0.64

0.59 0.52

Bars No.

4

6

8 5 7 6 6

5

7 6

Spacing (in.)

9

10

9 7 8 12 7.5

5.5

12 10

As (Tran.)*

0.13

0.162

0.194 0.108 0.162 0.184 0.13

0.162

0.168 0.13

Bars No.

3

3

4 3 3 3 3

3

4 3

Spacing (in.)

10

8

12 12 8 7 10

8

8 10

* Transverse bars (shrinkage and temperature steel).

25

Problem 9.3: 1. M(D.L.) = 9.2 K.ft.; = 0.015 2. Mu = 24.25 K.ft.; WL = 0.258 K/ft. = 258 psf. Problem 9.4: 1. Mu = 170.2 K.ft. 2. d = 8 in.; As = 1.44 in.2 ;Use #7 bars spaced at 5 in. 3. Vu at support = 5.24 K. ; Vu at distance d = 5.09K. 4. Minimum d for deflection checking = L / 10 = 12 in. h used = 10 in. < 12 in., therefore deflection should be checked. 5. Use #3 bars spaced at 6 in. (As = 0.22 in.2) for shear Problem 9.5: 1. Mu = 6.41 K.ft. /ft.; Let d = 5 in. 2. As = 0.36 in.2;Use # 4 bars spaced at 6 in. (As = 0.39 in.2) 3. Shrinkage reinforcement: As = 0.144 in.2;Use #3 bars spaced at 9 in. (As = 0.15 in.2) Problem 9.6: 1. Mu (at support) = 42.2 K.ft.; As = 1.44 in.2 2. Check section at midspan: External Mu = 18.55 K.ft. Internal moment = Moment capacity = Mu = 30.8K.ft. Use 10 in. depth at fixed end, 4 in. at free end and # 7 bars spaced at 5in. at the top of slab. Problem 9.7: 1. Minimum depth = L / 30 = 5.2 in For interior spans, minimum depth = L/35 = 4.5 in. Assume a uniform thickness of 5.0 in. 2. Mu = 5.81 K.ft.; U = 0.344 K/f t . 3. Assume = 0.014 < max = 0.023; d = 3.6 in. As = 0.60 in.2, choose #5 bars. h = 4.66 in.; Use h = 5 in., d = 3.94 in. 4. Moments and As required at other locations:

Location (See example 9.5)

Moment Coeff.

Mu K.ft.

Ru psi

% As(in.2) # 5 bars spaced at (in.)

A - 1/24 2.42 156 0.45 0.21 12 B + 1/14 4.15 267 0.80 0.38 9 C - 1/10 5.81 374 1.15 0.54 6 D - 1/11 5.29 341 1.03 0.49 6 E + 1/16 3.63 234 0.7 0.33 9

5. Shear is adequate.

26

Problem 9.8: L/28 = 5.57 in., use 5 in. and check deflection, or use 5.5 in. 1. Mu = 5.81 K.ft., U = 0.344 K/ft, Use h = 5 in., and assuming #4 bars are used. Then d = 4.0 in. 2. Moments, As and bars. Location Moment Mu Ru # 4 bars Coeff. K.ft. psi % As(in.2) spaced at (in.) A - 1/24 2.42 151 0.33 0.16 12 B + 1/14 4.15 259 0.51 0.25 9 C - 1/10 5.81 363 0.74 0.36 6 D - 1/11 5.29 331 0.68 0.33 6 E + 1/16 3.63 227 0.45 0.22 10 3. Shear is adequate. Problem 9.9: 1. Mu = 5.81 K.ft.; U = 0.344 K/ft. Use h = 5 in. and assume # 4 bars are used then d = 4.0 in. 2. Moments, As, and bars.

Location (See example 9.5)

Mu K.ft.

Ru Psi

%

As (in.2)

#4 bars spacing (in.)

A 2.42 151 0.33 0.16 12 B 4.15 259 0.50 0.24 9 C 5.81 363 0.73 0.36 6 D 5.29 331 0.66 0.33 6 E 3.63 227 0.45 0.22 10

3. Shear is adequate. Problem 9.10: 1.a. Design of slab: Assume top slab thickness = 2 in. Own weight of slab = 25 psf.; D.L. = 55 psf, U = 226 psf. Mu = 1.41 K.in. 1.b. Assuming that moment is resisted by plain concrete only, M = 1.42 K.in. > Applied moment of 1.41K.in. 1.c. Shrinkage reinforcement, As = 0.0432 in.2 Use #3 bars spaced at 12 in. laid normal to the direction of ribs. Similar bars are used parallel to ribs. 2. Assume h = 12 in. (including 2 in. slab) U = 713 lb/ft.; Mu = 346.5 K.in. 3. Mu (flange) = 1541 K.in.; Choose 2#5 bars (As = 0.62 in. 2) 4. Shear in rib; Vu (at d distance) = 5771 lbs. < Vc Use min. stirrups, # 3 spaced at 5 in.

27

Problem 9.11: 1. 2 in. slab reinforced with #3 bars spaced at 12 in. 2. Use h = 12 in., rib height = 10 in. U = 525 lb/ft.; Mu = 255 K.in. Choose 2#5 bars (As = 0.62 in. 2) 3. Shear is adequate Problem 9.12: 1. Design of slab: same as in Problem 9.10 2. Choose h = 12 in., U = 713 lb/ft. Calculate Mu and As: For positive moment b = 34 in., and for negative moment, b = 4 in., d = 10.875 in. Location Moment Mu Ru % As bars Coeff. K.ft. psi (in) 2 A - 1/24 9.62 244 0.50 0.22 2#3 B + 1/4 16.50 49 - 0.35 2#4 C - 1/10 23.10 586 1.30 0.57 2#5 D + 1/16 14.43 43 - 0.35 2#4 Use 2#4 bars at the bottom of ribs to resist the positive moments, and 2#5 bars at the top of ribs to resist the negative moments at the interior supports. 2 # 3 bars are used at the top of the ribs only at the exterior supports.

28

CHAPTER 10 AXIALLY LOADED COLUMNS

Problems 10.1(a)-10.1(j): Given: = 0.65, K = 0.8, g (max.) = 8%, g (min.) = 1%

(a)

(b)

(c) (d) (e) (f) (g)

(h)

(i) (j)

fc(ksi.) 4

4

4 4 5 5 5

5

6 6

As (in.2)

8

25

6.28 15.2 10 5.06 15.2

12.5

10.12 7.59

g (%)

3.13

6.25

4.36 5.28 5.1 1.98 4.18

2.17

3.95 3.16

Pn(k.) 688 1442 439 955 722 712 1244 1634 968 852

Problems 10.2(a)-10.2(e): Given: = 0.75, K = 0.85, g (max.) = 8%, g (min.) = 1%

(a) (b) (c)

(d) (e)

fc (ksi.) 4 4 5

5 6

Ag (in.2)

153.9 201.1 254.5

314.2 176.7

As (in.2)

8 7.59 10.12

15.2 8

g (%)

5.2 3.77 3.97

4.84 4.53

Pn(k.) 622 710 1049 1391 855

Problem 10.3(a) - 10.3(h): Square and Rectangular Columns.

(a)

(b) (c) (d) (e) (f)

(g) (h)

fc(ksi.)

4

4 4 5 4 4

4 5

Pu (k.)

560 1546 528 764 500 840 492 1564

Ag (in.2)

190 552 138 249 213 272 186 561

Section(in.in.)

14x14

24x24 12x12 16x16 12x18 14x20

12x16 18x32

Ast (in.2)

7.25 17.93 9.3 6.84 4.01 11.72 5.18 10.1

Bars

8#9

16#10 8#10 8#9 6#8 10#10

6#9 8#10

Ties #

3

3 3 3 3 3

3 3

Spacing (in.)

14

18 12 16 12 14

12 18

Add. Ties

NO

#3@18 NO NO NO #3@14

NO #3@18

29

Problem 10.3(i) 10.3(L): Spiral Columns.

(i) (j) (k) (L)

fc(ksi.)

4 4 4 5

Pu (k.)

628 922 896 662

Ag (in.2)

174 276 226 157

Diam (in.)

16 20 18 15

Ast (in.2)

5.33 6.68 9.54 5.16

Bars

6#9 6#10 8#10 6#9

Spiral #

#3 #3 #3 #4

Pitch (in.)

2 2 2 3

30

CHAPTER 11 MEMBERS IN COMPRESSION AND BENDING

11-1-a to 11-1-l. Balanced condition (12 Problems) Rectangular sections Balanced condtion

problem 11-1-a Prob. 11-1-b Prob. 11-1-c

Pb (k.), Mb (k-ft.) Pb = 372, Mb = 541 162, 146.7 550.9, 922.3

Prob. 11-1-d Prob. 11-1-e Prob. 11-1-f Prob. 11-1-g Prob. 11-1-h 453.5, 781.2 197.9, 252.12 230.1, 310.3 264.1, 330.9 351.1, 365.9

Prob.11-1-i Prob. 11-1-j Prob. 11-1-k Prob. 11-1-l

390.4, 344.5 394.1, 486.7 485.2, 679.2 347.2, 370.4

11-2-a to 11-2-l. Compression controls: e = 6 in (12 Problems); = 0.65 Rectangular sections Compression controls

Prob. 11-2-a refer to prob. 11-1-a

Prob. 11-2-b Refer to 11-1-b

Prob. 11-2-c Refer to 11-1-c

Pn (k.), Mn (k.-ft.) Pn = 775.4, Mn = 387.7 264.6, 132.3 1208.6, 604.3

Prob. 11-2-d Refer to 11-1-d

Prob. 11-2-e Refer to 11-1-e

Prob. 11-2-f Refer to 11-1-f

Prob. 11-2-g Refer to 11-1-g

Prob. 11-2-h Refer to 11-1-h

993.2, 496.6 384.6, 192.3 465, 232.5 524.9, 262.4 604.6, 302.3

Prob. 11-2-i

Refer to11-1-i Prob. 11-2-j

Refer to 11-1-j Prob. 11-2-k

Refer to 11-1-k Prob. 11-2-l

Refer to 11-1-l 551.0, 275.5 741, 370.5 994.6, 497.32 613, 306.6 11-3-a to 11-3-l. Tension controls: e = 24 in (12 Problems) Rectangular sections tension controls

Prob. 11-3-a refer to prob. 11-1-a

Prob. 11-3-b refer to 11-1-b

Prob. 11-3-c Refer to 11-1-c

Pb (k.), Mb (k-ft) Pb = 258.8, Mb = 517.6 61.4, 122.8 452.5, 904.9

Prob. 11-3-d Refer to 11-1-d

Prob. 11-3-e Refer to 11-1-e

Prob. 11-3-f Refer to 11-1-f

Prob. 11-3-g Refer to 11-1-g

Prob. 11-3-h Refer to 11-1-h

384.7, 769.5 116.8, 233.7 146.2, 292.4 152.12, 304.24 152.71, 305.43

Step Prob. 11-3-i Refer to 11-1-i

Prob. 11-3-j Refer to 11-1-j

Prob. 11-3-k Refer to 11-1-k

Prob. 11-3-l Refer to 11-1-l

11 146.26, 292.53 215.09, 430.18 309.47, 618.93 149.20, 298.40

31

Problem 11.4: Balanced condition: cb = 12.72in, ab = 10.82in. Cc = 588.38K; Cs1 = 286.4K; Cs2 = 58.94K Tension zone: = 0.65; s2 = 5.7510-4, fs3 = 16.69Ksi, Pb = 587.73K; Mb = 816.97K.ft, e = 8in., e < eb compression controls. Assume c = 16.43 in., a = 13.96 in. Cc = 783.4 K, Cs1 = 286.4K, Cs2 = 97.17 K, Cs3 = 14.60 K Tension zone: = 0.65; s = 9.2610-4, fs = 26.85Ksi, Pn = 1063 K; Mn = 708.71 K.ft Check: take moment about As e/ = 17.5in; Pn = 1026.7 K very close to Pn O.K. Problem 11.5: Balanced condition: cb = 10.36in., a = 8.81in., Cc = 599.08 K; Cs1 = 358.28 K; Cs2 = 79.02 K; Cs3 =-0.97= Zero Tension zone: s4 = 9.8210-4, fs = 28.47ksi, Pb = 584.27K; Mb = 788.16K.ft; eb = 16.19in. e = 8in., e < eb compression controls. Assume c = 12.77in., a = 10.8in. Cc = 734.4K, Cs1 = 358.28K, Cs2 = 104.19K, Cs3 = 39.29K Tension zone: s4 = 2.310-4, fs = 6.68Ksi, s5 = 0.0011, fs = 32.22Ksi, Pn = 1014.93K, Mn = 676.62K.ft; e/ = 15.5in. Problem 11.6: Balanced condition: cb = 10.36in., a = 8.8in., Pb = 408.68K; Mb = 414.1K.ft, eb = 12.16in. Compression controls: e = 8 < eb comp. controls. Assume c = 12.09in., a = 10.28in. Cc = 489.33K; Cs1/ = 169.8K, Cs2/ = 23.28K Tension zone: s = 0.0013, Pn = 565.62K, Mn = 377.08K.ft; e/ = 15.5in. Problem 11.7: Balanced condition: cb = 12.72in., ab = 10.82in., Cc = 735.76K; Cs1 = 283K; Cs2 = 68.03K; Cs3 = 3.05K Tension zone: s4 = 9.510-4, fs = 27.56Ksi, Pn = 734.27K, Mb = 914.52K.ft, eb = 14.95in. Compression controls: e = 8 < eb comp. controls. Assume c = 15.98in., a = 13.58in. Cc = 923.44K; Cs1 = 283K, Cs2 = 88.26K, Cs3 = 36.54K Tension zone: s4 = 1.4310-4, fs = 4.19Ksi, s5 = 0.001, fs = 30.05Ksi Pn = 1172.6K, Mn = 781.73K.ft

32

Problem 11.8: a: Choose #4 ties spaced at 18in. Pn = 821.23K; Mn = 403.1K.ft O.K. b:As/ = 3.68in.2 use 5#8 (As/ = 3.93in.2) Pn = 643.1K, Mn = 669.65K.ft c: As = As/ = 7.56in.2, use 6#10 Pn = 756.25K, Mn = 1608K.ft d: As = As/ = 4.73in.2, use 6#9 Pn = 682.7K, Pn = 443.8K > Pu, O.K. e: As = As/ = 7.59 in2, use 6#10 Pn = 1 753.93 K f: As = As/ = 0.031824/2 = 6.48in.2, Use 5#10 Pn = 1114.26 K, Pn = 724.3 K > Pn O.K. g: As = As/ = 0.021420/2 = 2.8in.2, Use 3#9 Pn = 453.12K, Pn = 317.2K; O.K. section is adequate. h: As = As/ = 6.57in.2, use 6#10 Pn = 1617.7K, Pn = 1051.5 K > Pu, O.K. Section is adequate. i: As = As/ = 0.022014/2 = 2.8in.2, Use 2#10 Pn = 923.93K, Pn = 600.5 K > Pu ; O.K. section is adequate. j: As = As/ = 2.64in2, Use 4#9 Pn = 1037.86 K, Pn = 674.6 K > Pu, O.K. section is adequate. Problem 11.9: a: Pn = 265K compared to 265 K by calculation. b: Pn = 991K compared to 993.2 K by calculation. c: Pn = 462.9K compared to 465 K by calculation. d: As = As/ = 5.8in2, Use 5#10, As = 6.33in.2 e: As = As/ = 6.48in.2, Use 6#10, As = 7.59in.2 f: As = As/ = 8.4in.2, Use 7#10, As = 8.86in.2 Problem 11.10: Pb = 67.11K, Pb = 43.62K Mb = 86.9K.ft, Mb = 56.5K.ft, e = 15.54in. Problem 11.11: Pb = 222.33K, Pb = 144.5 K Mb = 229.61K.ft; Mb = 149.3K.ft, eb = 10.5 in. Problem 11.12: Pb = 427.85K, Pb = 278.1K Mb = 420K.ft; Mb = 273K.ft, eb = 11.78in.

33

Problem 11.13: Pb = 677.6K, Pb = 440.4K Mb = 655.7K.ft, Mb = 425.8K.ft, eb = 11.6in. Problem 11.14: a: Pn = 108K (tied); Mn = 53.9K.ft (tied) b: Pn = 271.4K (tied); Mn = 135.5K.ft (tied) c: Pn = 495K (tied); Mn = 247.7K.ft (tied) d: Pn = 731.8K (tied); Mn = 366.3K.ft (tied) Problem 11.15: a: Pbx = 572.5K, Mbx = 790.9K.ft; eb = 16.58in. Pby = 536 K, Mby = 482 K.ft; eby = 10.79in. Pny = 833.55K, Mny = 416.77K.ft Pn = 559 K, Pn = 363.3 K b: Pbx = 576.88K; Mbx = 740.55K.ft; ebx = 15.4in. Pnx = 929.82K, Mnx = 619.88K.ft Pny = 1107 K, Mny = 553.6K.ft Pn = 633 K, Pn = 411.5 K c: Pbx = 408.68K; Mbx = 414.1K.ft; eb = 12.16in. Pby = 368.7K; Mby = 260.3K.ft; eb = 8.47in. Pny = 496.8K, Mny = 248.4K.ft Pn = 323 K., Pn = 210 K. d: Pbx = 718.7K; Mbx = 865.6K.ft, ebx = 14.5in. Pnx = 1 093.6K, Mnx = 729.1K.ft Pby = 701.9K; Mby = 699.6K.ft, eby = 12 in. Pny = 1145.3K, Mny = 572.65K.ft Pn = 717.6 K., Pn = 466 K. Problem 11.16: a: Pn = 566.34K, Pn = 368 K b: Pn = 674.3K., Pn = 438.3K c: Pn = 335 K., Pn = 218 K d: Pn = 732.24 K, Pn = 476 K Problem 11.17: a: Pn = 645 K., Pn = 419 K b: Pn = 749 K., Pn = 487 K c: Pn = 355 K., Pn = 231 K d:Pn = 817 K., Pn = 531 K

34

Problem 11.18: X-axis: Pn = 1113.4Kk, Pu = 723.5K; Mn = 556.7K.ft, Mu = 362K.ft Y-axis: Pn = 833.65Kk, Pu = 542K; Mn = 416.8K.ft, Mu = 271K.ft Biaxial load: Bresler method; Pu = 397K, Pn = 611.3K Hsu method; Pu = 455K, Pn = 701K Problem 11.19: X-axis: Pn = 1107K, Pu = 720K; Mn = 553.56K.ft, Mu = 360K.ft Y-axis: Pn = 1107K, Pu = 720K; Mn = 553.56K.ft, Mu = 360K.ft Biaxial load: Bresler method: Pu = 462K, Pn = 710.6K Hsu method: Pu = 549 K, Pn = 845K Problem 11.20: X-axis: Pn = 674.1K, Pu = 438K; Mn = 337.1K.ft, Mu = 219K.ft Y-axis: Pn = 494.9K, Pu = 322K; Mn = 247.45K.ft, Mu = 160.8K.ft Biaxial load: Bresler method: Pu = 232.7K, Pn = 358.1K Hsu method: Pu = 262K, Pn = 403K Problem 11.21: X-axis: Pn = 1289.7K, Pu = 838.3K; Mn = 644.86K.ft, Mu = 419.1K.ft Y-axis: Pn = 1145.2K, Pu = 744.4K; Mn = 572.6K.ft, Mu = 372.2K.ft Biaxial load: Bresler method: Pu = 518.1K, Pn = 797.13K Hsu method: Pu = 587K, Pn = 903K

35

CHAPTER 12 SLENDER COLUMNS

Problem 12.1: 1.) Pu = 306 kips; Mu = 300.4 k-ft.; e = 11.8 in. 2.) (K lu) / r = 37.5 34-12(M1 / M2) = 22; NO; Consider the slenderness effect. 3.) EI = 15341171 k- in.2 4.) Pc = 3245.269 kips 5.) Cm = 1.0 6.) ns = 1.144 7.) Pn = 470.8 kips; Mn = 462.2 k-ft.; Mc = 528.76 k-ft.; e = 13.48 in. 8.) Pn = 345.787 kips > Pu = 306 Kips

Problem 12.2: 1.) Pu = 306 kips, Mu = 300.4 k-ft.; e = 11.8 in. 2.) Consider the slenderness effect. 3.) EI = 15341171 k- in.2 4.) Pc = 7301,856 kips 5.) Cm = 1.0 6.) ns = 1.06 7.) Pn = 470.8 kips; Mn = 462.2 k-ft.; Mc = 489.932 k-ft.; e = 12.5 in. 8.) Pn = 366.52 kips > Pu Problem 12.3: 1.) Pu = 306 kips; Mu = 300.4 k-ft.; e = 11.8 in. 2.) Consider the slenderness effect. 3.) EI = 15341171 k- in.2 4.) Pc = 3915.788 kips 5.) Cm = 1.0 6.) ns = 1.116 7.) Pn = 470.8 kips; Mn = 462.2 k-ft.; Mc = 515.82 k-ft.; e = 13.15 in. 8.) Pn = 352.6 kips > Pu Problem 12.4: 1.) Pu = 383 kips, Mu = 256.4 k-ft.; e = 8.03 in. 2.) Consider the slenderness effect. 3.) EI = 38374084.45 k- in.2 4.) Pc = 7085.45 kips 5.) Cm = 0.8667 > 0.40 6.) ns = 0.934 < 1.0 , use 1.0 7.) Pn = 589.23 kips; Mn = 394.462 k-ft.; Mc = 394.462 k-ft.; e = 8.03 in. 8.) Pn = 933.6 kips > Pu

36

Problem 12.5: 1.) Pu = 383 kips, Mu = 256.4 k-ft.; e = 8.03 in. 2.) Consider the slenderness effect. 3.) EI = 38374084.45 k- in.2 4.) Pc = 9254.463 kips 5.) Cm = 0.8667 > 0.40 6.) ns = 0.917 < 1.0 use 1.0 7.) Pn = 589.23 kips, Mn = 394.462 k-ft.; Mc = 394.462 k-ft.; e = 8.03 in. 8.) Pn = 933.6 kips > Pu Problem12.6: 1.) Pu = 383 kips, Mu = 256.4 k-ft.; e = 8.03 in. 2.) Consider the slenderness effect. 3.) EI = 38374084.45 k- in2 4.) Pc = 10273.9 kips 5.) Cm = 1 6.) ns = 1.052 7.) Pn = 589.23 kips, Mn = 394.462 k-ft.; Mc = 414.974 k-ft.; e = 8.45 in. 8.) Pn = 904.283 kips > Pu Problem 12.7: 1.) Pu = 383 kips, Mu = 256.4 k-ft.; e = 8.03 in. 2.) Consider the slenderness effect. 3.) EI = 37.084 106 k- in.2 4.) Pc = 4722.24 kips 5.) Cm = 0.867 6.) s = 1.12 < 1; use 1.0 7.) Pn = 589.23 kips, Mn = 394.462 k-ft.; Mc = 441.797 k-ft., e = 9.0 in. 8.) Pn = 1217.7 kips > Pu Problem 12.8: 1.) Pu = 449.4 kips, Mu = 141 k-ft.; e = 3.765 in. 2.) Consider the slenderness effect. 3.) EI = 9685418 k- in2 4.) Pc = 1659.6 kips 5.) Cm = 1.0 6.) s = 1.565 7.) Pn = 691.385 kips, Mn = 216.923 k-ft; Mc = 339.484 k-ft., e = 5.89 in. 8.) Pn = 512.911 kips > Pu

37

Problem 12.9: 1.) Pu = 449.4 kips, Mu = 141 k-ft.; e = 3.765 in. 2.) Consider the slenderness effect. 3.) EI = 9685418 k- in.2 4.) Pc = 6638.28 kips 5.) Cm = 1.0 6.) s = 1.1 7.) Pn = 691.385 kips, Mn = 216.923 k-ft.; Mc = 238.615 k-ft., e = 4.142 in. 8.) Pn = (0.65) (968.710) = 629.662 kips > Pu

Problem 12.10: 1.) Pu = 449.4 kips, Mu = 141 k-ft.; e = 3.765 in. 2.) Consider the slenderness effect. 3.) EI = 18255749 k- in.2 4.) Pc = 1285.37 kips 5.) Cm = 1.0 6.) s = 1.872 7.) Pn = 953 kips, Mn = 216.923 k-ft.; Mc = 309.332 k-ft., e = 5.369 in. 8.) Pn = 619.54 kips > Pu

38

CHAPTER 13 FOOTINGS

Problem 13.1: Wall Footings:

(a)

(b)

(c) (d) (e) (f) (g) (h)

(i) (j)

h (in.)

19

17

20 27 23 19 19 21

14 16

d (in.)

15.5

13.5

16.5 23.5 19.5 15.5 15.5 17.5

10.5 12.5

L (ft.)

10

7.5

8.5 15.0 11 9 11.5 11.0

6 7

Pu (k.)

45.6

44

59.2 69.6 64 60.8 52.8 65.6

44 51.2

qu (ksf.)

4.56

5.87

6.96 4.8 5.82 6.76 4.59 5.96

7.33 7.31

Vu1 (k.)

14.63

12.47

15.95 23.00 18.67 17.18 17.41 20.63

11.91 13.71

Mu (k.ft.)

46.17

31.00

46.79 111.00 67.96 49.67 59.32 72.09

22.91 31.09

As (in.2)

0.7

0.54

0.66 1.10 0.83 0.77 0.92 0.98

0.54 0.59

Bar No.

7

5

6 8 8 7 8 7

5 6

spacing (in.)

9

9

7.5 8 10 9 9 7

7 8

ld (in.)

48

28

33 55 55 42 55 42

24 29

ldav (in.)

51

36

41 77 55 43 59 53

27 32

Ash (in.2)

0.41

0.367

0.43 0.58 0.5 0.41 0.41 0.45

0.3 0.346

Bar No.

5

5

5 6 6 5 5 5

5 5

spacing (in.)

9

10

8 9 10 9 9 8

12 10

39

Problem 13.2: Square Footings.

(a)

(b)

(c) (d) (e) (f) (g) (h)

(i) (j) AF (ft.2)

61.38

47.9

77.1 77.5 69.1 95.4 61.5 95.1

46.6 60.4

h (in.)

20

19

23 24 21 21 20 22

16 18 d (in.)

15.5

14.5

18.5 19.5 16.5 16.5 15.5 17.5

11.5 13.5

L (ft.)

8

7

9 9 8.5 10 8 10

7 8 Pu (k.) 364 352 548.4 440 424 452 432 546 280 276

qu (ksf.) 5.7 7.2 6.8 5.4 5.9 4.5 6.8 5.5 5.7 4.3 Vu1 (k.) 23.1 77.7 130.1 115.4 114.9 113.1 106.5 148.96 68.2 75.97 Vu2 (k.) 325.5 300 480.8 400.2 388.2 417 382.2 426.3 240.0 249.2

Mu (k.ft.) 253.3 190.6 411.4 388.8 337.2 422.5 287.3 477.4 141.9 191.1 As (in.2) 3.75 3.0 5.1 4.6 4.7 5.9 4.3 6.24 2.9 3.3

Bars 10#6

8#6

13#6 12#6 12#6 11#7 11#6 9#8

8#6 9#6

ld (in.)

33

33

33 33 29 42 29 48

29 33 ldav (in.)

37

30

41 45 41 49 36 47

30 38

N1 (k.) 424.3 537 663 238.6 433.2 565.5 716 884 696.4 334.2 x

4

4

4 7 4 4 4 4

4 4

Ads (in.2) 1.28

1.62

2.0 5.54 0.98 1.28 1.62 2.0

1.57 1.0

Ldd (in.)

22

22 24.7 22 21.4 21.4 19 21.4

21.4 22

40

Problem 13.3: Rectangular Footings:

(a)

(b) (c) (d) (e) (f)

(g) (h) AF (ft.2)

61.4

47.9 77.6 78 69.6 96.7

61.5 95

L (ft.)

10.5

8 10 10 10 12.5

10.5 11 B (ft.)

6

6 8 8 7 8

6 9

h (in.)

23

20 23 24 21 21

23 22 d (in.)

18.5

15.5 18.5 19.5 16.5 16.5

18.5 17.5

Pu (k.) 364 352 548.4 440 424 452 432 546 qu (ksf.) 5.78 7.33 6.86 5.5 6.06 4.52 6.86 5.52 Vu1 (k.) 105.5 56.2 111.5 103.1 93.4 110.6 51 134.1 Vu2 (k.) 316.4 294.7 478.2 402.1 385.1 418.8 368.7 492.6

MuL (k.ft.) 363.7 232.3 476.4 445.5 413.7 563.62 416.7 540.96 MuS (k.ft.) 165.2 148.4 344 336.9 257.8 313.9 182.3 408.2 AsL (in.2) 4.52 3.47 5.92 5.25 5.79 7.9 5.2 7.12

Bars

9#7

10#6 11#7 10#7 9#8 9#9

10#7 8#9 AsS (in.2)

5.22

3.46 4.95 5.18 4.54 5.7

5.22 5.23

Bars

12#6

8#6 11#6 12#6 8#7 10#7

9#7 10#7 AsB (in.2)

3.79

2.98 4.42 4.61 3.7 4.44

3.76 5.28

Bars

9#6

7#6 10#6 11#6 6#7 8#7

7#7 9#7 AsE (in.2)

1.8

0.87 0.8 0.84 1.09 1.6

1.77 0.76

Bars

4#6

4#6 2#6 2#6 2#7 4#7

4#7 2#7 ldL (in.)

48

33 48 48 48 54

42 54

ldaL (in.)

52

36 47 51 50 64

51 53 lds (in.)

28

33 33 33 42 42

42 42

ldas (in.)

25*

24* 35 39 32* 37*

24* 41 N1 (k.) 424.3 537 663 238.6 433.2 565.5 716 884

Ads (in.2))

1.28

1.62 2.0 5.55 1.0 1.28

1.62 2.0 x

4

4 4 7 4 4

4 4

ldd (in.)

22

25 25 22 22 22

19 22

41

Problem 13.4: Rectangular footings.

(a) (b) (c)

(d)

AF (ft.2)

61.4 47.9 77.6

78 L (ft.)

9.5 8.5 11

11

B (ft.)

6.5 5.75 7.25

7.25 h (in.)

20 21 26

24

d (in.)

15.5 16.5 21.5

19.5 Pu (k.) 364 352 548.4 440

qu (ksf.) 5.9 7.2 6.88 5.52 Vu1 (k.) 110.26 91.43 151.72 135.07 Vu2 (k.) 321.05 292.39 466.4 394.6

MuL (k.ft.) 294.15 241.64 505.04 451.48 MuS (k.ft.) 199.29 149.23 331.17 296.48 AsL (in.2) 4.43 3.38 5.44 5.42

Bars

9#7 9#6 10#7

10#7 AsS (in.2)

4.10 3.86 6.18

5.70

Bars

11#6 9#6 10#7

10#7 AsB (in.2)

3.59 3.2 4.9

4.49

Bars

8.#6 8#6 8#7

8#7 AsE (in.2)

1.02 1.0 1.77

1.57

Bars

4#6 4#6 4#7

4#6 ldL (in.)

48 33 48

48

ldaL (in.)

44* 38 51

54 lds (in.)

33 33 48

48

ldas (in.)

29* 24* 33*

35* N1 (k.) 464.3 530.2 636.1 357.5

Ads (in.2))

1.4 1.6 1.92

2.49 x

4 4 4

4

ldd (in.)

22 25 25

22

42

Problem 13.5: Plain concrete wall footings:

(a) (b) (c)

(d)

D.L. (k.)

11 9 14

13

L.L. (k.)

6 7 8

12

B (ft.)

5 4 4

7.5

h (in.)

22 20 22

34

d (in.)

19 17 19

31

Pu (k.) 22.8 22 29.6 34.8

qu (ksf.) 4.56 5.5 7.4 4.64

Mu (k.ft.) 9.12 6.19 7.43 23.27

Ig (in4)

6859 4913 6859

29791

ft (psi.) 152 129 124 145

fta (psi.)

178 178 178

178

Vu (k.) 1.9 0.46 0 2.71

Vc (k.) 18.75 16.76 21.85 35.66

Problem 13.6: 1. Pe = 300 K; Pi = 390 K; L = 18 ft. 2. Use 9 18 ft. 3. Pu = 944 K; Pue= 416 K., Pui= 528 K; qu = 5.83 Ksf; qu= 52.5 K/ft. length. a) Shearing forces = 13.125 K. Shear at right side of Pe = 324.125 K. Shear at right side of Pi = 113.7 K. Shear at left side of Pi = 326.7 K. b) Bending moment: At right of interior column: Mu = 123.04 K.ft.; At left of interior column: Mu = 54.82 K.ft. Max. moment at zero shear; from left hand side, Mmax = 1232.14 K.ft. 4.a) Vu(max) = 366.7 K; VU @ d = 215.14 K. b) heck depth for two-way shear: (a + d) = 45.5 in;b0 = 111 in. 5. Design for bending moment in the longitudinal direction:

Use 12 #9 bars, (As = 12 in.2). Spacing = 9.3 in. 6. Design for beading moment in the short direction:

Use 9 #7 bars-(AS = 5.41 in.2), Spacing = 12.75 in.

43

Problem 13.7: Footing No. 1 2 3 4 5 6 D.L. (K) 130 220 150 180 200 240 L.L. (K) 160 220 210 180 220 200 L.L. / D.L. 1.23 1.00 1.40 1.00 1.10 0.83 Usual Load (K) 170 275 202.5 225 255 290 Area = (Usual Load/2.81) 60.5 97.9 72.0 80.1 90.7 103.2 Max. Soil Pressure (Ksf) 4.8 4.5 5.0 4.5 4.6 4.3 Problem 13.8: 1. P = 360 K; M= 230 K.ft., e = 7.67 in., say 8 in. 2. a)Assume total depth = 2 ft., and assume the weight of the soil is 100 pcf. Net upward pressure = 3.5Ksf; Choose a footing (11 x 10) ft b) Pu = 496 K; qu = 4.51 Ksf. Max. moment in the long direction, Mu = 640.62 K.ft. Max. moment in the short direction, Mu = 502.30 K.ft. 3. Vu = 167.25K 4. Two way shear: b0 = 142 in.; A0 = 1244.25 in.2; Vu = 457.03K. 5. Reinforcement in the long direction: Use 9 #10 (11.39in.2) spaced at 14.25in., provided ld > required 32 in. 6. Reinforcement in the short direction: Use 9 #9 (9 in.2) spaced at 15.75 in., provided ld > required 28 in. 7.Long direction, ld = 53 in. > required ld = 32 in. Short direction, ld = 49 in. > required ld = 25 in. 8. Bearing stress at column base: N1 = 530.1 K > Pu = 496 K Problem 13.9: 1. Assume a total footing depth = 28 in., Allowable upward pressure = q net = 3488 psf. P = 360 K; M = 230 K.ft; e = 7.67 in. Choose a footing 8 x 16 ft. 2. Pu = 496 K; Mu = 320 K.ft. 3. Distance from edge of footing = 5.29 ft.; Vu = 190.3 K. 4. bo = 154 in . ; q at center of column = 3.88 Ksf.; Vu = 456.5 K. 5. Reinforcement in the long direction: Choose 15 #9 bars (As = 15 in.2); Spacing = 6.4 in. 6. Reinforcement in the short direction: Use 20 #8 bars (As = 15.7 in.2) ; Spacing = 9.8 in. 7. Development lengths ld in both directions for #9 and #8 bars are adequate. Problem 13.10:

1. Assume total depth = 27 in; q net = 3488 psf.; P = 360 K, M = 230 K, e = 7.67 in. 2. Choose a square footing 12 x 12 ft. 3. qmax < q (allowable)

44

CHAPTER 14 RETAINING WALLS

Problem 14.1: 1. Using Rankine formula: Ca = 1/3 and Cp = 3.0; Ha = 1833 lbs., acting @3.33 ft. from base. Overturning moment Mo = 6.10 K.ft. 2. Calculate the balancing moment taken about toe O. W1 = 7250 lbs., arm = 2.5 ft., Ml = 18.125 K.ft. -W2 = -1305, arm = 1.0 ft., -M2 = -1.305 K.ft. R = W = 5945 lb., M b = 16.82 K.ft. Factor of safety against overturning = 2.76 > 2.0 3. Force resisting sliding, F = 2.97 K.; Factor of safety against sliding = 1.62 > 1.5 4. q(max) = 2.19 Ksf < 3.5 Ksf; q(min) = 0.19 Ksf. Problem 14.2: 1. Using Rankine formula, Ca = 1/3 and Cp = 3.0; Ha = 4125 lbs., acting @15/3 = 5 ft. from base. Overturning moment Mo = 20.625 K.ft. 2. Calculate the balancing moment taken about the toe O: Weight lb. arm (ft) Moment W1 = 3770 2.00 7.54 W2 = 3770 4.33 16.32 W3 = 2320 4.00 9.28 W4 = 2860 5.67 16.22 W5 = 1430 7.5 10.72 Total W = R = 14.150 K. Mb = 60.08 Factor of safety against overturning = 2.91 > 2.0 3. Force resisting sliding: F = 7.08 K; Factor of safety against sliding = 1.72 > 1.5 4. q(max) = 3.37 Ksf < 3.5 Ksf; q(min) = 0.164 Ksf. Problem 14.3: (a) to (j) BALANCED FORCES AND MOMENTS: ITEM WEIGHT ARM MOMENT

(K) (FT) (K.FT) RESULTANT 7.540 4.24 32.000 EARTH PRESSURE FORCES AND MOMENTS:

FORCE ARM MOMENT (K) (FT) (K. FT) TOTAL 2.146 4.00 8.585 FACTOR OF SAFETY AGAINST OVERTURNING = 3.73 FACTOR OF SAFETY AGAINST SLIDING = 1.76

45

Problem 14.3 (e): BALANCED FORCES AND MOMENTS: ITEM WEIGHT ARM MOMENT (K) (FT) (K. FT) RESULTANT 12.604 5.72 72.139 EARTH PRESSURE FORCES AND MOMENTS:

FORCE ARM MOMENT (K) (FT) (K. FT) TOTAL 4.307 5.67 24.409 FACTOR OF SAFETY AGAINST OVERTURNING = 2.96 FACTOR OF SAFETY AGAINST SLIDING = 1.63 Problem 14.4 (e) : 1. Balanced forces and Moments: W1 = 2.325 K arm = 4 ft M = 9.3K.ft. W2 = 0.581 arm = 3.33 M = 1 .938 W3 = 2.025 arm = 4.5 M = 9.113 W4 = 7.673 arm = 6.75 M = 51.79 W6 = 1.35 arm = 6.75 M = 9.113 Total W = 13.964 Total M = 81.25 2. Earth pressure forces and moments: PAH = 4.307 K arm = 5.67 ft. M = 24.41 K.ft. PH2 = 1.382 arm = 8.5 ft. M = 11.76 K.ft. Total P = 5.69 Total M = 36.16 Overturning F.S. = 81.25/ 36.16 = 2.24 OK 3. Sliding: F.S.= 1.23 less than 1.5, NO GOOD; Key 1.5 x 1.5 ft is not adequate. Increase length of footing L to 11 ft. Shear in toe is not adequate and depth of footing hf to 21. Problem 14.5 (e): BALANCED FORCES AND MOMENTS: ITEM WEIGHT ARM MOMENT

(K) (FT) (K. FT) RESULTANT 12.800 5.75 73.612 EARTH PRESSURE FORCES AND MOMENTS:

FORCE ARM MOMENT (K) (FT) (K. FT)

TOTAL: 4.906 5.93 29.100 FACTOR OF SAFETY AGAINST OVERTURNING = 2.53 FACTOR OF SAFETY AGAINST SLIDING (Pp INCLUDED)= 2.05

46

Problem 14.6 (e) to (h): All the steel areas are given at the last part of each problem of 14.3 (e) to (h). Problem 14.7: 1. For = 33, Ca = 0.295 Ha = 7.08 K; Mo = 47.2 K.ft. Weight arm (ft) Moment (k.ft.) W1 = 2.4 4.50 10.8 W2 = 1.2 3.67 4.4 W3 = 3.6 6.00 21.6 W4 = 13.44 8.50 114.2 W = R = 20.64 M = 151.0 Factor of safety against overturning = 2.3 > 2.0 2. Check sliding condition: F = 9.29; F.S. = 1.31 < 1.5 Not good, use Key 1.5 x 1.5ft F.S. (sliding) = 1.56 > 1.5 3. q1 =0.83 = 2 .55 ksf; q2 = 0.89 ksf F = 11.31 K; F.S.(s l iding) = 1.6 > 1.5 Problem 14.11: 1. For = 30, Ca = 0.333, hs(Surcharge) = 3.33 ft. pa (soil) = 0.5 Ksf, Ha = 3.125 K. pw (water) = 0.78 Ksf., Hw = 4.875 K. For intermittently wet ground, Hw = 2.44 K. Ps (surcharge) = 0.133 Ksf., Hs= 1.66 K. 2. Mu = 18.23 K.ft; R (top) = 2.78 K; R (bottom) = 8.8 K. Mu(positive) = 7.77 K.ft. 3. Use #6 bars spaced at 6 in. For Mu = 7.77 K.ft; Use #5 bars spaced at 12 in. For longitudinal reinforcement: use #4 bars spaced at 12 in. (As = 0.20 in.2)

47

CHAPTER 15 DESIGN FOR TORSION

Problem 15.1:

Problem 15.2:

Problem 15.3: Acp = 428in.2 ;Pcp = 138 in.

Problem 15.4: Acp = 608in.2, Pcp = 156in.

Problem 15.5: Acp = 336 in.2; Pcp = 76in.2 If flanges are included, then Tcr = 390.5 k.ft and Tn = 97.7 k.ft. Problem 15.6: Acp = 504in.2, Pcp = 132 in. Problem 15.7:

Pcp = 72 in. Design for torsion: Assume 1.5in. cover concrete cover and #4 stirrups Aoh = 194.25 in.2; Ao = 165.11 in.2, Pn = 58 in. Use #4 stirrups, use S = 7 in. Problem 15.8: Total area of closed stirrups: For one leg: Avt/S = 0.015+0.004 = 0.019in2 (per one leg) Use #4 stirrups, Use S = 7in Problem 15.9: Vc = 22.2 K < Vu shear reinforcement required. Acp = 264 in2, Pcp = 68 in, Ta = 48.6K.in; Ta LHS O.K., Tn = 300/0.75 = 400 K.in Distribution of longitudinal bars: Total = 1.35in2, Al/3 = 0.45

349.8 . .crT K in =

170.8 . .crT K in =

251.9 . .crT K in = 63 . .nT K in =

449.6 . .crT K in =

25.9

362

c

cu

V KVV

== f

48

Problem 15.10: Ta = 61.5K.in torsional reinforcement required. Vc = 22.2 K shear reinforcement required Design for shear: Av/S = 50.4/(6019.5) = 0.043in2/in (two legs) Av/2S = 0.0215 in2/in (one leg) Design for torsion: Aoh = 197.3in2; Ao = 167.7 in2; Ph = 91 in. LHS = 485, RHS = 474.3 < LHS, section is not adequate. Problem 15.11: Design for moment: assume a = t = 6in Mu = 3000Kips.in Vc/2 shear reinforcement required. Ta = 76.8K.in. Torsional reinforcement required. Design for shear: Av/S = 0.006 in2/in(2 legs); Av/2S = 0.003 in2/in(one leg) Design for torsion: LHS = 375.7, RHS = 474.4>LHS O.K., Tn = 300/0.75 = 400 K.in Use #3stirrups @ S = 5.5in Problem 15.12: Design for moment: Mu = 4800K.in Vc/2 shear reinforcement is required. Pcp = 124in. Ta = 77K.in; Torsional reinforcement is required. Design for shear: Av/S = 0.019in2/in (2 legs), Av/2S = 0.0095in2/in (one leg) Design for torsion: Ao = 206.8in2, Ph = 120in LHS = 348.6, RHS = 474.2 > LHS O.K., Tn = 480K.in At/S = 0.019in2 (per one leg) Longitudinal reinforcement: Use #4 stirrups @ S = 8in. Distribution of longitudinal bars: Total 2.28in2, Al/3 = 0.76in2 Problem 15.13: Mu = 3000K.in Vc/2 shear reinforcement is required. Acp = 448 in2, Pcp = 124in, Ta = 66.5K.in. Torsional reinforcement is required Design for shear: Av/S = 0.01 in2/in (two legs); Av/2S = 0.005 in2/in (one leg) Design for torsion: Aoh = 243.35in2, Ao = 206.8in2, Pn = 120 in. LHS = 375.7, RHS = 410.7 > LHS O.K., Tn = 400 Kin, At/S = 0.016in2 (one leg) Total area of closed stirrups: Use #3 stirrups @ S = 6 in. Distribution of longitudinal bars: Total = 1.92 in2, Al/3 = 0.64in2

49

Problem 15.14: As = 3.3in2 use 4#9; Shear and torsional reinforcement are required. Design for shear: Av/2S = 0.003 in2/in (one leg) Design for torsion: Aoh = 243.25in2, Ao = 206.8in2 Pn = 120in, Acp = 448in2, Pcp = 124in; Tcr = 307K.in < Tu = 300K.in; At/S = 0.016 in2(one leg) Total area of stirrups: For one leg: Use #3stirrups @ S = 6in Distribution of longitudinal bars: Total = 1.92in2, Avt/3 = 0.64in2 Problem 15.15: Design for moment: As = 3.35in2, Use 4#9 bars Design for shear: Ao/2S = 0.005in2/in (one leg) Design for torsion: Aoh = 243.25in2, Ao = 206.8in2, Pn = 120in; Acp = 448in2, Pcp = 124in Tcr = 226K.in < Tu, Use Tu = 266K.in; At/S = 0.014in2 (one leg); Al = 0.014120 = 1.7in2 Total area of closed stirrupss: For one leg: Use #3 stirrups; Use S = 6in Distribution of longitudinal bars: Total = 1.7 in2, Al/3 = 0.57 in2 Problem 15.16: Design for shear: Av/2S = 0.0058in2/in (one leg) Design for torsion: Aoh = 194.25, Ao = 165.11, Pn = 58, Acp = 308, Pcp = 72 Tcr = 216.5K.in < Tu = 216.5 K; Use #4 stirrups At 7in. Problem 15.17: 1. Moy = 334 k.ft., Mox = 248.5 k.ft.; mAsx = 4.86 in.2 , Asy = 5.04 in.2 2. Design for torsion and shear: Tu = 576 k.in., Vu = 12 k. a.) d = 17.5 in., fc = 4 ksi., fy = 60 ksi,

b.) Acp = 320 in.2, Pcp = 72 in.; Tu = 576 > Ta = 67.5 K.in. torsional reinforcement is needed. 3. Design for torsion: Assume 1.5 in. cover concrete cover and #4 stirrups. a.) Aoh = 206.25 in.2, Ph = 58 in.2, Ao = 175.3 in.2 b.) LHS = 464.psi., RHS = 474.5 psi, RHS > LHS o.k. Section is adequate. c.) Al (min.) is not critical d.) For one-leg Use #4 stirrups @ 6 in. e.) Longitudinal bars: Use Al/4 = on all sides of beam = 0.53 in.2 f.) Choice of steel bars: Use 6 #9 bars (As = 9 in.2) Problem 15.18: 1. Assume own weight of beam = 0.3 k/ft. Uniform dead load = 1.1 k./ft.; U = 2.34 k./ft. 2. Bending moment at A, Mu = 49.9 k.ft; Tu (at A) = 63.52 k.ft., Vu (at A) = 18.7 k. 3. The section is L-section, but since Mu is small, rect. section is assumed to obtain min. steel. 4. U = 0.25 k./ft., Mu = 8.0 k.ft.; Let t = 5.0 in. slab. 5. Design for shear and torsion: Tu = 591 k.in., d = 21.5 in.; Vu = 14 < Vc /2 Acp = 336 in.2, Pcp = 76 in.; Tu > Ta torsional reinforcement is needed. LHS = 467.4 psi, RHS = 474.5 psi., RHS > LHS o.k. section is adequate. Total area of closed stirrups: for one-leg: Use #4 stirrups, use S = 5 in.

2 ' 26.6 .c cV f bd k = =

50

CHAPTER 16 CONTINUOUS BEAMS AND FRAMES

Problem 16.1: 1. One way slab; Wu = 268 psf 2. Loads on beam: Wu = 2845 lb/ft, Say 2.9 K.ft. 3. Design of moment: Support: A B C Steel: 3#7 (1.8 in2) 4#7 (2.41 in2) 4#7 (2.41 in2) 4. Design for shear: Use #3 stirrups spaced at 6 in. 5. Deflection: = 0.06 in.; / L = 1/3990 which is very small. Problem 16.2: 1.One way slab: Wu = 268 psf. ; Load on beam: Wu = 2.92 K/f t. 2. Design for moments: Section Support A Support B Support C Midspan AB Midspan BC or D Bars 2#7 3#7 3#7 2#7 3#7 3. Design for shear: Use #3 stirrups spaced at 8.5 in. Problem 16.3: MA=MC= 0; MB = -387.2 K.ft. RA = 55.87 K.; RB = 192.33 K.; RC = 55.8 K Section AB B BC Bars 5#9 4#9+3#8(top) 3#9 2#9(bottom) 5. Design for shear: Use #4 stirrups spaced at 3.5 in. Problem 16.4: Section AB B BC Bars 3#9 6#9 2#9+2#8 Design for shear: Use #4 stirrups at 3.5 in.

Problem 16.5: 1.) Assume beam 16 x 34 in., and column 16 x 30 in. Section Supports B,C , Midspan E Bars 2#10 + 6#9(8.53 in2) (5#10 + 3#9)(As=9.33 in2) Design for shear: Use #4 stirrups at 4 in., max spacing = 15 in. 2. Design of columns AB, DC; a) Section B: Provided Pn = 230,9 K > 186.1 required b) Section at midheight of column: Provided Pn = 332.1 K > 192 K required. c) Ties: choose #4 ties spaced at S = 16 in. 3. Use crossing bars 5#9 and 4#7; Use 4#4 ties within a length = a = 9 in. 4. Rectangular footing: use 7.5 x 4 ft. Longitudinal steel: Use 5#6 bars in the longitudinal direction (As = 2.21 in.2) Transverse steel: Use 7#7 bars in the short direction.

51

Problem 16.7: Mp = 494.2 K.ft. Problem 16.8: Mp = 300 K.ft. Problem 16.9: Mu = 171 K.ft. Design critical sections: use 3#9 bars (As = 3.0 in.2) Let h = 19 in., d = 16.5 in. Fixed end moments at fixed ends using ultimate loads: M FA = MFB = -228 K.ft.; MA = 171 K.ft. Rotation capacity provided: A = 0.0101 greater than the required of 0.0062. Deflection: = 0.266 in.; /L = 1/902 which very small. Shear: Use #4 stirrups spaced at 8 in. Problem 16.10: Final moments: Section D.L. L.L. L.L. D.L.+L.L D.L.+L.L. Support moments max (-) max (+) max (-) max (+) B -311 -302.4 52.8 -613.4* -258.2 C -311 -302.4 52.8 -613.4* -258.2 Midspan AB 276.6 -79.2 280.8 197.4 557.4* BC 121.0 -158.4 201.6 -37.4 322.6* CD 276.6 -79.2 280.8 197.4 557.4 *= Design Moments Problem 16.11: Final moments: Moment due to DL L.L. L.L. Col.1+Co1.2 Col. 1+Co1. 3 Section max ve max +ve max -ve max +ve Support A 0 0 0 0 0 B -333.3 -312.5 - -645.8* -333.3 C -222.2 -277.7 - -499.9* -222.2 D -333.3 -312.5 - -645.8* -333.3 E 0 0 0 0 0 Midspan AB +265.4 -84.9 +276.4 +180.5 +541.8* BC +154.3 -141.5 +218.6 +12.8 +372.9* CD +154.3 -141.5 +218.6 +12.8 +372.9* DE +265.4 -84.9 +276.4 +180.5 +541.8*

52

CHAPTER 17 DESIGN OF TWO-WAY SLABS

Problem 17.1:

Prob.#

ln1

ln2

h = ln/30 h = ln/33 h (in.) Totol floor

a

18

18

7.27.5 6.547.0 7.5

b

22

22

8.89.0 8.0 9.0

c

24

24

9.610 8.739.0 10.0

d

18

14

7.27.5 6.547.0 7.5

e

22

18

8.89.0 8.0 9.0

f

24

20

9.610 8.739.0 10.0

g

28

22

11.211.5 10.1810.5 11.5

h

28

28

11.211.5 10.1810.5 11.5 Problem 17.2 (a):

Strip

Column strip Middle Strip

Moment sign

Neg. Pos. Neg.

Pos.

Mu (k.ft.)

- 150.1 64.6 - 50.0

43.1

As = bd (in.2)

5.46 2.6 2.6

2.6

Min. As = 0.0018bhs (in.2)

1.73 1.73 1.73

1.73

Straight bars

18#5 9#5 9#5

9#5

Spacing = b/No.2hs18 in.

6.7 13 13

13 Problem 17.2(b.):

Strip

Column strip Middle Strip Moment sign

Neg. Pos. Neg.

Pos.

Mu (k.ft.)

- 317.8 136.9 - 105.9

91.3 As = bd (in.2)

7.02 5.05 5.05

5.05

Min. As = 0.0018bhs (in.2)

3.11 3.11 3.11

3.11 Straight bars

12#7 12#6 12#6

12#6

Spacing = b/No.2hs 18 in.

12 12 12

12

53

Problem 17.2(c):

Strip

Column strip Middle Strip Moment sign

Neg. Pos. Neg.

Pos.

Mu (k.ft.)

- 411 177 - 137

118 As = bd (in.2)

8.49 5.66 5.66

5.66

Min. As = 0.0018bhs (in.2)

3.51 3.51 3.51

3.51 Straight bars

14#7 13#6 13#6

13#6

Spacing = b/No.2hs 18 in.

11 12 12

12 Problem 17.3(a):

Long Direction

Column strip Middle strip Ext.(a) Pos.(b) Int.(c) Ext.(a)

Pos.(b)

Int.(c)

Mu (k.ft.)

- 80 96 161.6 0

64

- 53.9 As = bd2 (in.2)

2.80 3.35 5.90 0

2.57

2.57

Min. As = 0.0018bhs (in.2)

1.73 1.73 1.73 1.73

1.73

1.73 Straight bars

9#5 12#5 10#7 8#5

9#5

9#5

Spacing = b/No.2hs = 18 in.

13 12 12 15

13

13 Problem 17.3(b):

Long Direction

Column strip Middle strip Ext.(a) Pos.(b) Int.(c) Ext.(a)

Pos.(b)

Int.(c)

Mu (k.ft.)

169.51 203.41 342.27 0

135.61

114.09 As = bd2 (in.2)

5.2 5.2 7.33 0

5.2

5.2

Min. As = 0.0018bhs (in.2)

3.11 3.11 3.11 3.11

3.11

3.11 Straight bars

12#6 12#6 10#8 10#6

12#6

12#6

Spacing = b/No.2hs = 18 in.

12 12 14.4 14.4

12

12 Problem 17.4:

Prob.#

ln1 (ft.)

ln2

(ft.)

h (in.)

Exterior

h (in.)

Interior h (in.)

Totol floor drop panel

(in.) Total depth

drop panel(in.)

Length of drop panel

(ft.)

a 18

18

6.547

6.0 7.0 2 9

6.676.67

b

22

22

8.0

7.338 8.0 2 10

88

c

24

24

8.29

7.278 9.0 2.25 11.5

8.678.67

d

18

14

6.547

5.455.5 7.0 1.75 9

6.675.33

e

22

18

8.0

6.677 8.0 2 10

86.67

54

f 24 20 8.729 8.0 9.0 2.25 11 8.667.33

g 28

22

10.1810.5

8.5 10.5 2.6 13

108

h

28

28

10.1811.5

10.5 10.5 2.6 13

1010

Problem 17.5(b):

Mo = 572 k.ft.

Long & Short Direction

Column strip Middle strip Neg. Pos. Neg.

Pos.

Mu (k.ft.) 280.3 120.1 91.5 80.1 As = bd2 (in.2) 7.76 3.1 3.1 3.1

Min. As = 0.0018bhs (in.2)

3.11 3.11 3.11

3.11 Straight bars

14#7 11#5 11#5

11#5

Problem 17.6(b):

Mo = 572 k.ft.

Long & Short Direction

Column strip Middle strip Ext. Pos. Int. Ext.

Pos.

Int.

Mu (k.ft.) 149 178.5 300 119 100 As = bd2 (in.2)

4.04 4.89 8.3 3.1 4.2 3.6

Min. As = 0.0018bhs (in.2)

2.59 2.59 2.59 2.07

2.07

2.07 Straight bars

12#7 16#5 14#7 8#6

16#5

8#6

Problem 17.7(a):

Long & Short Direction

Column strip Middle strip Neg. Pos. Neg.

Pos.

Mu (k.ft.) -20.6 11.1 -45.8 24.7 As = bd (in.2)

low low 2.34

1.20

Steel bars

10#4 10#4 12#4

10#4 Spacing (in.)2hs 14 in.

12 12 10

12

Problem 17.8(a):

Strip

Column strip Middle strip Ext.(-) Pos. Int.(-) Ext.(-)

Pos.

Int.(-)

Mu (k.ft.) -5.62 18.1 -22.2 -7.7 40.2 49.4 As = bd2 (in.2)

low low low low

2.10

2.40

Min. As = 0.0018bhs (in.2)

1.3 1.3 1.3 1.3

1.3

1.3 As ( min.=0.0033)

1.9 1.9 1.9 1.9

1.9

1.9

Bars selected

10#4 10#4 10#4 10#4

14#4

14#4 Spacing 15 in. 12 12 12 12 8.5 8.5

55

Problem 17.9:

Waffle slab (Interior panel)

Problem 17.10:

Waffle slab (Exterior panel) Exterior Slab: Mni = 1370.74 K.ft.; Mne = 509.12 K.ft.; Mp= 1018.3 K.ft.

Strip

Column strip Middle strip Ext.(-) Int.(-) Pos. Ext.(-)

Int.(-)

Pos.

Mu (k.ft.) 509.12 1028.1 610.98 0 342.64 407.32 As = bd2 (in.2) 9.3 12.8 9.3 9.3 9.3 9.3

Min. As = 0.0035bh (in.2) 9.3 9.3 9.3 9.3 9.3 9.3 Bars selected #8 12 18 12 12 12 12 Bars per rib #8 2 3 2 0 1 2

Interior Slab: Mn = -1272.83 K.ft.; Mp= 685.37 K.ft.

Strip

Column strip Middle strip negative Positive Negative

Positive

Mu (k.ft.) -954.6 411.22 -318.23 274.15 As = bd2 (in.2) 12 9.3 9.3 9.3

Min. As = 0.0035bh (in.2) 9.3 9.3 9.3 9.3 Bars selected #8 16 12 12 12

Spacing 3 2 1 1

56

CHAPTER 18 STAIRS

Problem 18.1: 1.) Wu (on stairs) = 386 lb./ft. Wu (on landing) = 326 lb./ft. 2.) Calculate the maximum bending moment and steel reinforcement a) Mu = 13.37 k.ft.; Use #5 bars @ 7 in.; For 5.5 in. width stairs, use 10#5 bars b) Transverse reinforcement for shrinkage: use #4 bars @ 12 in. (As = 0.2 in.2) 4.) Design of landing: Use #4 bars @ 12 in. (As = 0.2 in.2) 5.) Check shear (stairs): Vu = 2.4 K < Vc/2 = 3.45 k., no shear reinforcement required . 6.) Check shear at loading: Vud = 1.2 K < Vc/2 O.K. Problem 18.4: 1.) Wu (on stairs) = 252.2 lb./ft. Wu (on landing) = 194 lb./ft. 2.) Maximum bending moment and steel reinforcement: Mu = 0.80 k.ft.; Use #3 bars @ 12 in. (As = 0.11 in.2) 3.) Since Vu = 0.567 K < Vc/2, no shear reinforcement is required. But it is recommended to use stirrups #3 @ 4 in. to hold the main reinforcement. 4.) Design of supporting beam: As = 1.5 in.2, use 3#7 (As = 1.8 in.2) 5.) Check beam A for torsion when L.L. acts on one side of stairs. Ta = 17.75 k.in. < Tu = 38.4 k.in.; Torsional reinforcement is needed and section is not adequate. Use #3 closed stirrups @ 4 in.

57

CHAPTER 19 INTRODUCTION TO PRESTRESSED CONCRETE

Problem 19.1: 1. For the given section: top= 564 psi (Compression) ; bottom= +624 psi (tension) Both stresses are less than the ACI allowable stresses at transfer. Both stresses are less than the allowable stresses after all losses. 2. Allowable uniform live load after all losses: a) based on top fibers stress: WL = 1.94 K/ft. b) based on bottom fibers stress: WL = 1.35 K/ft., controls. Allowable live load = 1.35 K/ft based on tensile stress controls. Problem 19.2: Loss from stress ksi percentage Elastic shortening 12.3 7.26 Shrinkage 8.4 4.96 Creep of concrete 15.29 9.02 Relaxation of steel 4.24 2.50 Total losses = 40.23 23.74 Additional losses 3.39 2.00 Total = 43.62 25.74 Total F = 300.76 K Problem 19.3: Loss from Stress Ksi Percentage Elastic shortening 8.52 4.87 Shrinkage 5.60 3.36 Creep of Concrete 8.02 4.80 Relaxation of steel 4.20 2.50 friction 13.2 7.93 Total losses = 39.54 23.46 Total F = 415 K. Problem 19.4: a) Total moment, MT = 1339.5 K.ft. MD = 320.26 k.ft, and MT = 1339.5 k.ft to get critical e's. emax= 22.45 in. and emin = 15.78 in. e (used) = 17.24 in. b) Section at 12 ft. from midspan, (16 ft. from support) : MD = 261.44 K.ft., ML = 512 K.ft., MT = 1093. 44 k.ft e max = 21.15 in. , e min = 8.66 in. , e(used) = 14.1 in. c) Section at 10 ft. from the support (18 ft. from midspan): MD = 187.91K.ft., ML = 368 K.ft., MT =785.9 K.ft.; e(used) = 10.12 in. d) Section at 3 ft. from the support (25 ft. from midspan): MD = 64.95 K.ft., ML = 127.2 K.ft., MT = 271.65 K.ft.

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Problem 19.5: MD = 712.3 k.ft (total D.L.) and ML= 627.2 k.ft.; Fi used = 544 K which is adequate. If MD =320.26 k.ft (self-weight) is used, then ML = MT - MD = 1019.24 K.ft. Fi (min.) = 349.5 k and Fi (max.) = 1616.6 k. Problem 19.6: 1. MD (self-weight) = 349.7 K.ft. = 4196 K.in. 2. Estimate prestress losses: F = 140.9 Ksi, = 0.839 3. Limits of eccentricity at midspan: MD (self-weight) = 349.7 K.ft. = 4196 K.in. ; F = 487 K. e max = 19.7 in. , e min = 13.7 in. 4. Limits of eccentricity at 22 ft. from support: MD = 3787 K.in., Ma = 9979 K.in., MT = 13766 K.in. 5. Limits of e at 11 ft. from support. MD = 2389 K.in., Ma = 6296 K.in., MT = 8685 K.in. 6. Limits of e at 3 ft. from support: MD = 750 K.in., Ma = 1976 K. in . , M T = 2726 K.in. 7. Fi used = 580.6 Ksi is adequate.

Problem 19.7: 1. Stresses at level of tendon due to D.L. Fi = 165.16 Ksi 2. Loss due to shrinkage = 8.4 Ksi 3. Loss due to creep = 1.5; Elastic strain = 0.002414; Creep loss = 10.14 Ksi. 4. Loss due to relaxation of steel = 7 Ksi = 0.845 and F = 482.5 K. Total Mn = 2591.6 K.ft., Mu = 2332.5 K.f t. 1.2 Mcr = 1825.5 K.f t. < Mn. Section is adequate. Problem 19.8: a) Camber at transfer = -1.089 + 0.424 = -0.665 in. (upward). b) Deflection at service load = 0.553 in. (downward) Problem 19.9: Wu = 3.22 K/ft. Vu (at h/2 from support) = 96.6 K Mu (at h/2 from support) = 199.64 e at midspan = 18.5 in., e @ 2 ft. from the support = 1.7 in Use #3 stirrups spaced at 14.5 in. all over. Problem 19.10: Beam acts as a T-section. Mn = 1950.8 K.ft.; Mn = 1755.7 K.ft.

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CHAPTER 20 SEISMIC DESIGN OF REINFORCED CONCRETE STRUCTURES

Problem 20.1: SMS = 1.64 g; SM1 = 0.496 g; SDS = 1.09 g; SD1 = 0.33 g Seismic design category is D. Problem 20.2: SMS = 1.3 g; SM1 = 0.6 g; SDS = 0.87 g; SD1 = 0.40 g SDC = D; V = 13.16 k Lateral seismic forces are: F1 = 5.98 k; F2 = 7.18 k Problem 20.3: SMS = 0.825 g; SM1 = 0.32 g; SDS = 0.55 g; SD1 = 0.21 g, SDC = D; CS = 0.066 N Wi [k] h (ft) wihi Cvx Fx [k] Vx [k] 5 1000 50 50000 0.333 110 110 4 1000 40 40000 0.267 88 198 3 1000 30 30000 0.2 66 264 2 1000 20 20000 0.133 44 308 1 1000 10 10000 0.067 22 330 150000 330 Problem 20.4: Location Mu (kft) As (in2) Reinforcement Mn (kft) Support -265.5 3.6 6No.7 322.6

131 1.8 3No.7 168 Midspan 82.8 1.2 2No.7 113 Problem 20.5: For 5 No.8 bars; As = 3.95 in2; Wu = 4.35 k/ft Design shear = 105 k; Earthquake induced force = 52.8 k Use No. 3 hoops A 11 in spacing started at 62 in from the face of support will be sufficient. Problem 20.6: Ash = 0.636 in2: Use 4 No.4 ties Shear strength: Vu =120 k; Vc = 90 k; Vs = 252 k Problem 20.7: e = 115 in; Mn = 76923 kft; Pn = 8000 k Rm = 0.132; Mn = 1995840 k ft > 76923 kft (OK) Special boundary elements are needed Transverse reinforcement of the boundary element: Use no.4 hoops and crossties; Smax = 6 in; Ash = 0.954 in2: Use 5 No.4 crossties

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CHAPTER 21 BEAMS CURVED IN PLAN

Problem 21.1: Moment For the section at support, Mu = 187 k-ft., 2in 11.3=sA For the section at midspan, ft.-k 52.94=uM , 2in 47.1=sA Maximum torsional moment, ft.-k 90.18=uT Shear at the point of maximum torsional Moment = 60.55 k Stirrups #4 @ 5 in c/c Longitudinal Bars, 2in. 81.0=lA Sectional Details

a) The total area of top bars is 3.11 + 0.27 = 3.38 in.2, use 4 # 9 bars (negative moment) b) The total are of bottom bars is 1.47 + 0.27 = 1.74 2in. , use 2 # 9 bars at the corner c) At middepth, use 2 # 4 bars (0.4 in.2)

Problem 21.2:

AV = 1.57 Wu r = (1.57)(10.7)(6) = 100.8 k. AM = -Wu r

2 = -(10.7)(36)2 = -385.2 k.-ft, 2in 3.3=sA (Top Steel) AT = -0.3 Wu r

2 = -0.3(10.7)(36)2 = 115.56 k.-ft

Mc= 0.273 Wu r2 = (0.273)(10.7)(36)2 = 105.2 k.-ft CT = CV = 0

# 4 Stirrups, 2 braches, s = 3.89 in., use #4 @ 3.5 in. Distribution bars:

Bottom bars = 3.15/3 = 1.05 2in. , use 2 # 7 (1.21 2in. ) Middepth bars , use 2 # 7 bars Top bars = 3.3 + 1.05 = 4.35 in.2, use 6 # 8 bars in one row (4.71 2in. )

Problem 21.3:

CM =134.11 k-ft, CT =0

DM = -58.20 k-ft, DT = 52.46 k-ft Problem 21.4: Design for AM = -601.77 k-ft. 2in 2.4=sA

k.ft 1.281 k.ft., 6.103 == uu TV # 4 Stirrups, 4 braches, 2in 4.0)2.0(2 ==vA , Spacing; s = 6.31 in., use #4 @ 6 in. Longitudinal reinforcement : 2in. 19.6=lA Maximum spacing = in. 5.138/1088/ ==hP

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Distribution of bars: Bottom bars = 4 # 7 ( sA = 2.41

2in. ), Middepth bars , use 2 # 9 bars, Top bars = 6.3 in.2, use 5 # 10 bars in one row ( sA = 6.33

2in. ) Problem 21.5:

CM = 55.09 k-ft , 2in 86.05.21120033.0 ==sA

AM = - 296.91 k-ft., 2in 2.45.21120150.0 ==sA AT = 31.81 k-ft CT = 31.81 k-ft

Use # 4 Closed Stirrups, s = 0.2/0.0453 = 4.4 in., use #4 @ 4 in. Longitudinal reinforcement : 22 in. 50.03/,in. 47.1 == ll AA For section at support;

Bottom bars, sA = 0.5 2in. , use 2 # 5 bars ( sA = 0.61

2in. ) Middepth bars, sA = 0.5

2in. , use 2 # 5 bars Top bars = 3.87 + 0.5 = 4.37 in.2, use 6 #8 bars in two row ( sA = 4.71

2in. ) For section at mid-span

Bottom bars, = 0.86 + 0.5 = 1.36 in.2, use 2 #8 bars ( sA = 1.57 2in. )

Middepth bars, use 2 # 5 bars Top bars, use 2 # 5 bars