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
1
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; f r : Density
f c΄
Ec
n
f r r
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.
3
35 MPa
29,910 Mpa
6.69
3.668 MPa
3
30 MPa
25,980 Mpa
7.70
3.396 MPa
3
25 MPa
20,690 Mpa
9.67
3.10 MPa
2400 kg/m 2300 kg/m 2100 kg/m 2.11
a.) See figure below. b.) Secant modulus (at f c΄/2 = 1910 psi.) -4 Ec = 1910 / 6.10× 6.10×10 = 3130 ksi. -4 Approximate Initial Modulus = 2.6(ksi.) / 5.45× 5.45×10 = 4771 ksi. (Possible range 4600 – 5200) c.) Ec (ACI formula) =
57000 fc ' = 57000 3820 = 3523 ksi.
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; f r : Density
f c΄
Ec
n
f r r
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.
3
35 MPa
29,910 Mpa
6.69
3.668 MPa
3
30 MPa
25,980 Mpa
7.70
3.396 MPa
3
25 MPa
20,690 Mpa
9.67
3.10 MPa
2400 kg/m 2300 kg/m 2100 kg/m 2.11
a.) See figure below. b.) Secant modulus (at f c΄/2 = 1910 psi.) -4 Ec = 1910 / 6.10× 6.10×10 = 3130 ksi. -4 Approximate Initial Modulus = 2.6(ksi.) / 5.45× 5.45×10 = 4771 ksi. (Possible range 4600 – 5200) c.) Ec (ACI formula) =
57000 fc ' = 57000 3820 = 3523 ksi.
2
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
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
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
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
2
As (in. ) a(in.)
ρmax<ρ<ρmin Problem 3.2
DIM.
No.a
No.b
No.c
No.d
No.e
No.f
No.g
No.h
b (in.) d (in.)
15 22.5 8#9 8.00 2#9 2.0 0.0237 0.0059 0.0178 0.0172 7.06 692.2
17 24.5 8#10 10.08 2#10 2.54 0.0242 0.0061 0.0181 0.0158 7.83 950
13 22 7#9 7.00 3#7 1.8 0.0245 0.0063 0.0182 0.0176 7.06 590.2
10 21.5 4#10 5.08 2#7 1.2 0.0236 0.0056 0.0180 0.0180 6.85 418.2
14 20.5 6#10 7.62 2#10 2.54 0.0266 0.0089 0.0177 0.0189
16 20.5 9#9 9.00 4#9 4.0 0.0274 0.0122 0.0152 0.0189
20 18 12#9 12.00 6#9 6.0 0.0333 0.0167 0.0167 0.0216
18 20.5 8#10 10.12 4#10 5.08 0.0275 0.0138 0.0137 0.0189
40.46 -244.9 -552.45 7.8 6.63 59.1 315.7 141.5 5.26 0.0183 Yes 597.9
46.24 -205.6 -870 7.097 6.033 56.36 328.2 211.8 5.47 0.0167 Yes 716.3
57.8 -218.4 -1305 7.0 5.95 55.94 504.8 315.2 6.75 0.0187 Yes 822.5
52.6 -184.9 -1100.5 6.7 5.71 54.62 349.4 260.2 5.82 0.0158 Yes 813.7
2
As (in. ) 2
As’(in. ) ρ ρ’ ρ- ρ’ k a ( in.) φ Mn A B C c a f s’ C1 C2 As1 ρ1 ρ1≤ρmax φ Mn
3
Problem 3.3
DIM.
No.a
No.b
No.c
No.d
.No.e
No.f
No.g
No.h
b ( in.) bw ( in.) t ( in. ) d ( in. )
54 14 3 17.5 4#10 5.08 2.213 R 374.7 Yes
48 14 4 16.5 4#9 4.00 1.961 R 279.4 Yes
72 16 4 18.5 8#10 10.16 3.32 R 769.9 Yes
32 16 3 15.5 6#9 6.00 4.41 T
44 12 4 20.5 8#9 8.00 4.27 T
50 14 3 16.5 7#9 7.00 3.29 T
40 16 3 16.5 5#10 6.35 3.74 T
42 12 3 17.5 6#9 6.00 3.36 T
2.04 4.94 293.4 No
5.44 5.02 660.1 Yes
4.59 4.05 466.8 Yes
3.06 4.84 415 Yes
3.83 4.25 425.9 Yes
2
As (in. ) a ( in. ) Rect. or T φMn (k.ft) Asmin< As < Asmax 2 Asf (in. ) a ( in. ) φMn (k.ft) Asmin< As < Asmax Problem 3.4
DIM.
No.a
No.b
No.c
No.d
No.e
No.f
No.g
No.h
fc’ fy b d As 2 (in. ) 2 (mm ) β1 ρ ρmax
3 ksi 40 ksi 12 in. 20 in. 4#8
4 ksi 60 ksi 12 in. 20 in. 4#7
4 ksi 75 ksi 12 in. 20 in. 4#9
5 ksi 60 ksi 12 in. 20 in. 4#9
30 Mpa 400 Mpa 300 mm 500 mm 3*30 mm
20 Mpa 300 Mpa 300 mm 500 mm 3*25 mm
30 Mpa 500 Mpa 300 mm 500 mm 4*25 mm
25 Mpa 300 Mpa 300 mm 500 mm 4*20 mm
3.14 0.85 0.037 0.020 0.614 ksi 0.013 0.422 7 ksi 0.205 0.318 Yes
2.41 0.85 0.0285 0.0181 0.821 ksi 0.0100 0.4942 ksi 0.1772 0.3194 Yes
4.00 0.85 0.0207 0.0145 0.822 ksi 0.0167 0.9182 ksi 0.3676 0.3199 No
4.00 0.8 0.0335 0.0212 0.973 ksi 0.0167 0.7941 ksi 0.2353 0.2993 Yes
2121mm 0.85 0.0325 0.0203 6.14 Mpa 0.01414 4.52 Mpa 0.2218 0.3184 Yes
2
1473 0.85 0.0321 0.01805 4.11 Mpa 0.00982 2.42 Mpa 0.1732 0.3194 Yes
1963 0.85 0.0236 0.0162 6.13 Mpa 0.01309 5.13 Mpa 0.2567 0.3176 Yes
1257 0.85 0.04 0.0225 5.11 Mpa 0.0084 2.13 Mpa 0.118 0.3176 Yes
R umax ρ R u a/d ratio (a/d) max ρmin<ρ<ρmax
4
Problem 3.5: 3.5 a: 2 As = 1256.6mm ; ρ = 0.00914 ; R u = 3.6 Mpa ; φMn = 272.3 KN.m ρ b = 0.02024 ; ρmax = 0.6375ρ b = 0.0129 Ok 3.5 b: 2 As = 1472 mm ; ρ = 0.0107 ; R u = 3.51 Mpa ; φMn = 265.4 KN.m ρ b = 0.02024 ; ρmax = 0.6375ρ b = 0.0129 ρ < ρmax Ok 3.5 c: 2 As = 2827 mm ρ = 0.02056 ; ρmax = 0.0129 < ρ ; N.G. , Section does not meet ACI code, reduce steel. R u(max) = 4.1 Mpa ; φMn = 310 KN.m 3.5 d: ρ = 0.0091 ; ρ b = 0.0214 ; ρmax = 0.01356 ; ρmin < ρ < ρmax Ok R u = 0.439 ksi ; φMn = 177 k.ft 3.5 e: ρ = 0.02727 ; ρmax = 0.01356 < ρ ; N.G. , Section does not meet ACI code, reduce steel. R u(max) = 0.615 Ksi ; φMn = 248.1 k.ft Problem 3.6: εy = 0.00172 > εs = 0.0015
This section is over- reinforced.
c b = 11.44 in. ; a b = 9.725 in. ; Asb = 3.97 in.
2
c = 12 in. ; a = 10.2 in. ; f s = Esεs = 43.5 ksi ; As = 4.783 in. ; φMn = 201.3 k.ft 2 ρmax = 0.01628 ; As max = 2.34 in. ; a = 5.74 in. ; (max) φMn = 132.8 k.ft (max) φMn allowed by ACI code is less than φMn of the section = 201.3 k.ft 2
ρ = 0.014 <
ρmax ; As = 2.016 ; a = 4.941 in. ; φMn = 117.4 k.ft
Problem 3.7: a = 7.706 in.; φMn = 356.3 k.ft ; Wl = 3.94 k/ft Problem 3.8: a = 5.537 in ; φMn = 230.1 k.ft ; Pl = 17.6 k Problem 3.9: 3.9 a:
(1) At balanced condition: c b = 11.834 in. ; a b = 10.059 in. ; As b = 5.48 in. 2 Asmax = 0.63375Asb = 3.473 in. ; As < Asmax (2) T = 188.4 k ; a = 6.618 in ; C1= 81.6 k ; C2= 106.8 k ; φMn = 227.83 k.ft 3.9 b: 2
(1) At balanced condition: c b = 9.467 in. ; a b = 8.047 in. ; As b = 3.89 in. 2 Asmax = 2.465 in. ; As > Asmax , N.G. ; Section does not meet ACI code, reduce steel. 2 (2) Use As = As max = 2.465 in. ; T = 147.9 k ; a = 5.96 in. ; C1= 68 k ; C2= 79.97 k φMn = 137.5 K.ft 2
5
Problem 3.10 3.10 a: ρ = 0.011157 ; ρ’ = 0.00556 ; ρ - ρ’ = 0.005597 < ρmax = 0.0203 ρ - ρ’ < k = 0.0139, Compression steel doesn’t yield. c = 3.07 in. ; a = β1c = 6.61 in. ; f s’= 16.15 ksi ≤ f y 2 C1 = 79.85 k ; C2 = 16.32 k ; As1 = 1.996 in. ; ρ1 = 0.009242 < ρmax ; φMn = 119 k.ft 3.10 b: ρ = 0.0223 ; ρ’ = 0.00556 ; ρ - ρ’ = 0.0167 > k = 0.0139, Compression steel yields. ρ - ρ’< ρmax = 0.0203 ; Ok ; a = 4.73 in. ; φMn = 225.6 K.ft Problem 3.11:
(1) At balanced condition: c b = 13.02 in. ; a b = 11.07 in. f s’ = 70.3 > 60 ; Compression steel yields, fs’ = 60 ksi 2 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) ρ = 0.02307 ; ρ’ = 0.005 ; ρ - ρ’ = 0.01807 > k, Compression steel yields. ρ - ρ’= ρmax ; Ok ; a = 7.01 in. ; φMn = 512.8 k.ft ; Wl = 8.516 k/ft Problem 3.12: ρ = 0.01805 ; ρ’ = 0.0031 ; ρ - ρ’ = 0.015 < ρmax = 0.01806 ρ - ρ’ < k = 0.0194 , Compression steel doesn’t yield. c = 6.44 in. ; a = 5.47 in. ; f s’= 53.23 ksi < f y 2 C1 = 186 k ; C2 = 30.4 k ; As1 = 3.1 in. ; ρ1 = 0.0155 < ρmax ; φMn = 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 2 2 (3) φMn = 236.4 k.ft ; Asmin = 0.6 in < As< Asmax = 13.23 in.
Problem 3.14: 2 (1) a = 3.69 in. > t ; T section ; A smax = 4.99 in. > As 2 (2) Asf = 2.55 in. ; a = 5.08 in. ; φMn = 339.6 k.ft
Ok
Ok
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
6
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
R u 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
3.013
6.947
8.029
6.36
6.0
2.39
5.48
6.42
Bars rows h (in)
5#7 One 24
7#9 two 36
7#10 two 32
5#10 one 29
6#9 two 28
6#6 one 24
6#9 two 25
8#9 two 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
ρ (%)
1.77
1.81
1.8
1.8
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
%
ρ ( ) 2
AS (in )
AS (in)
2
7
Problem 4.2: Assume c/dt = 0.375 for f y = 60Ksi,(all problem) dt = (d + 1)in. for 2 rows of bars; d’ = 2.5 in.
DIM.
Prob. Prob. Prob. Prob. Prob. Prob. Prob. Prob. Prob. No.a No.b No.c No.d No.e No.f No.g No.h No.i
Prob. Prob. No.j 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
4.606
5.56
Mu1(k.ft)
402
656.3 280.6 344.6 459.5 595.3 374.2 223.2 297.7 223.2 201.1
293.0
Mu2(k.ft)
152
133.7 167.4 175.4 305.5 259.7 180.8 76.76 102.3 56.76 88.86 107.02
5.44
7.37
3.996 4.428 5.904 7.128 5.328 3.564
4.75
3.576 3.654
4.41
1.68
1.13
2.324 2.165
3.77
1.62
1.35
1.645
1.585
As (in )
7.12
8.50
6.38
6.66
9.674 10.09 7.839 4.784 6.376 4.925 5.299
5.995
f’s (ksi)
60
60
55.52
58.6
58.6
60
46.8
53.74
A’s (in )
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
2#8
2#10
2#10
9#6
10#5
6#10
5#5
10#4
2#7
7#5
4#6
Two
Two
Two
Two
Two
Two
Two
Two
Two
Two
Two
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
28
22
24
24
26
22
20
20
20
18
21
5.24
5.241
5.26
2
As1(in ) 2
As2(in )
2.959 2.511
1.22
2
2
h (in)
24
8
20.24 51.73 51.73 51.84
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
φMn(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
R u (psi)
288
421
303
464
375
456
0.0063
0.769
0.646
0.558
0.76
0.804
AS (in )
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
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
244.8
244.8
ASf (in )
4.08
4.08
Muf (ft)
266.2
247.9
Muw (ft)
121.8
132.1
R u (web) (psi)
476
503
ρw(%)
1.0
1.07
1.92
2.25
6.0
6.33
ρ (%) 2
c/ dt ≤ 0.375 Cf (k) 2
2
ASw (in ) 2
AS (in ) 2
AS (max) (in )
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
9
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
φMn(k.ft)
516.4
495.7
340.8
401.6
T
T
T
T
R
R
R
T
379
452
280
R or_T R u (psi) 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 (in )
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
R u (web) (psi)
436.12
407
485
413
ρw(%)
0.9
0.84
1.02
0.85
2
2.38
2.35
2.3
1.84
AS (in )
7.99
7.2
6.0
6.43
AS (max) (in )
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
2
ASw (in ) 2
2
10
Problem 4.4: 2 2 (a) ρ = ρmax , d = 18.4 in. ; AS = 3.31 in. ; Use 4#9 bars ( two rows. AS = 4.0 in. ); h = 22 in. ; a = 5.84 in.; c = 6.87 in.; dt = 19.5 in. => c / dt =0.352 < 0.375 ; OK 2 2 (b) ρ=0.016, d = 19.4 in. ; AS = 3.1 in. ; Use 4#8 bars ( two rows. A S = 3.16 in. ); h = 23 in. ; a = 5.47 in.; c = 6.43 in.; dt = 20.5 in. => c / dt =0.314 < 0.375 ; OK 2 2 (c) ρ= 0.012, d = 21.9 in. ; AS = 2.63 in. ; Use 3#9 bars ( AS = 3 in. ); 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: 2 2 R u = 465 psi ; ρ=0.0146 ; As = 3.5 in ; Use 3# 10 bars (As = 3.79 in ) Problem 4.6: 2 2 ρ = 0.0094 ; R u = 465 psi ; As = 2.27 in ; Use 2#10 bars ( AS = 2.53 in ) Problem 4.7: Assume a total depth = L / 15 = 20×12 / 15 =16 in. Mu = 132 k.ft 2 (a) Using ρmax = 0.01806, R u(max) = 820 psi ; d = 12.7 in. ; AS = 2.75 in ; Use 3#9 ; h =16 in ; ρmin < ρ = ρmax , εt = 0.005 2 2 (b) Using ρs = 0.014, R u = 662 psi ; d = 14.2 in. ; AS = 2.38 in ; Use 4#7 bars (A s = 2.41 in ). h= 17 in. ; ρmin < ρ < ρmax , εt > 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 ρ=0.015, R u = 700 psi, d = 20.88 in. 2 (3) AS = 3.76 in ; 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 2 ρ = 0.00674; ρmin < ρ < ρmax , εt > 0.005 ; AS = 1.74 in ; Use 2#9 bars or 3#7 Problem 4.9: Mu = 317 k.ft ; R u = 622 psi, ρmin < ρ < ρmax , εt > 0.005 ; d = 20.27 in. 2 2 AS= 3.97 in ; Use 4#9 bars ( A S = 4.0 in ) in one row. ; h = 23 in. Problem 4.10: Mu = 151.6 k.ft ; R u = 476 psi, ρmin < ρ < ρmax , εt > 0.005 ; d = 17.8 in. 2 2 AS= 3.2 in ; Use 3#10 bars ( AS = 3.79 in ) in one row. ; h = 21 in. Problem 4.11: 2 φMn = R umax bd = 222 k.ft < Mu ; Compression steel is needed. 2 2 2 Mu1 = 222 k.ft ; Mu2 = 68 k.ft ; AS1= 3.10 in. ; AS2 = 0.92 in. ; AS = 4.02 i n . ; Use 4 # 9 a = 6.08 in. ; c = 7.15 in.; dt = 20.5 in.; εt =0.0056 > 0.005; OK 2 2 2 A S' = 0.98 in. ; AS(max)= 5.27 in > As = 4.02 in ; h = 23 in. OK.
11
Problem 4.12: 2 2 a)d = 22.5 in. and AS = 3.05 in ;Use 3#9 bars in one row (AS = 3.0 in ) ; h =26 in Mu = 269.6 k.ft.; a = 7.06 in.; c = 8.3 in.; dt = 23.5 in.; εt =0.0055>0.005; b) Mu1 =195 k.ft, M u 2 = 65 k.ft 2 For Mu1 = 195 k.ft ; d = 19.5 in. and AS1 = 2.64 in ;Let h = 23 in. and d’ = 2.5 in. 2 2 For M u 2 = 65 k.ft. ; As’= AS2 = 0.85 in. ; AS = 3.49 in . Use 5 # 8 c) Mu = 260 k.ft 1) For two rows of steel h = 26 in, d =22.5 in and d t = 23.5 in. 2) Section behaves as a T-section. 2 2 ASf = 1.28 in. ; M u2 =120.5 k.ft ; Mu1 = 139.5 k.ft ; AS1 = 1.53 in. 2 2 2 AS(max)= 4.33 in > Total As = 2.81 in ; OK ; Choose 3#9 bars, ( AS = 3.0 in ) Problem 4.13: 2 2 d = 19.8 in.; As = 6.8 in. ; Use 7#9 bars (two rows. A S = 7.0 in. ) if d = 22 in. is used; a=3.53 in. < 4 in. ; h = 26in. 2 2 AS(min) = 0.86 in. < AS < AS(max) = 7.78 in. OK Problem 4.14: Ultimate load W = 7.9 k/ft MB( maximum negative) = -395 K.ft M Dmax = 270.8 k.ft b) φMn = 344.6 k.ft < Mu Compression steel is needed. 2 AS = 5.04 in . Use 7 # 8 ( two rows) 2 2 2 A S' = 0.64 in. ; AS1 = 3.12 in. ;Use 4#8 bars in one row, ( A S = 3.16 in ) Problem 4.15: a) W = 8.56 k/ft ; H A = HD = 34.24 k ; R A = R D = 154.08 k M B = MC = - 616.32 k.ft. ; MBC (midspan) = 770.4 k.ft Design beam BC for positive moment , 2 2 AS = 8.86 in Use 9#9 bars (two rows, As = 9.0 in ) 2 2 ASmax = 26.65 in. ; AS(min)= ρ mi n b w d = 1 . 0 9 in AS(min) < AS < AS(max) OK Design beam BC for negative moment , Mu =616.32 k.ft Compression steel is needed. 2 2 AS = 7.93 in . Use 8 # 9 in two rows, ( A S = 8 in ) 2 2 A S'= 1.99 in. Use 2 # 9 in one row, ( A S = 2 in )
12
CHAPTER 5 ALTERNATE DESIGN METHODS Extra Problems
Problem E5.1 Tie AB F u = 58.75kips Ats = 1.31in 2 Provide 5 No. 5 Bar Ats = 1.55in 2 Tie CD F u = 15.0kips Ats = 0.33in 2 Provide 1 No. 4 Bar, 2 legs Ats = 0.40in 2 Tie BD & DF F u = 68.21kips Ats = 1.52in 2 Provide 5 No. 5 Bar Ats = 1.55in 2 Calculate strut widths β s = 0.75 φ f ce = 0.75 × 3825 = 2868 psi Strut AC, P u = 86.83kips , w = 1.26in Strut BC, P u = 80.93kips , w = 1.18in Strut CE, P u = 130.67 kips , w = 1.90in Strut DE, P u = 16kips , w = 1.26in 2 0.20 sin 0.0032 0.003 24 4.5 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 Ө = 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 ФVn = 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 13
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 Φfns = Φ*fce * Acs = 1751 kips > 1570 Good Capacity of Strut 4 Φfns = Φ*fce * Acs = 1195 kips = 1195 Good Capacity of Strut 5 Φfns = Φ*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 γ = 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 γ = 0.00156 Design of horizontal tie 1 and 2: Fu = Ф*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 2’s Anchorage Length Ldh = 17 in
14
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
5
7.59
4.71
7.59
9.37
8
As΄ (in .)
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 (in .)
9333.3
45000
12167
20736
23434.7
16128
x (in.)
7.543
10.238
8.363
0
0
0
5968.5
25247.2
7013.2
0
0
0
0
0
0
8.475
10.017
9.377
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 (in .)
5989.6
25468.5
7046.3
12971.1
18055.8
12436.6
ΔI (in.)
0.667
0.416
0.638
0.616
0.551
0.62
2
2
2
1.57
1.48
1.48
ΔI (in.)
0.397
0.289
0.447
0.478
0.352
0.384
Δa (in.)
0.794
0.577
0.893
0.751
0.521
0.568
ΔT = ΔI + Δa (in.)
1.46
0.99
1.53
1.37
1.07
1.19
2
As (in .) 2
4
4
Icr (in .) x (in.) 4
Icr (in .)
4
λ
15
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
8
7.59
6.28
8
As΄ (in .)
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 (in .)
4
17280
26364
12167
16128
x (in.)
9.175
9.401
8.822
9.377
12693.9
1540.3
8914
12427.9
0
0
0
0
0
0
0
0
Mcr (k-ft.)
56.92
80.16
41.82
53.13
Ma (k-ft.)
396
381
288
393
Ie (in .)
12707.5
15505.9
8923.9
12437.1
ΔI (in.)
0.538
0.457
0.558
0.577
λ
1.51
2
1.5
1.48
ΔI (in.)
0.342
0.29
0.33
0.42
Δa (in.)
0.516
0.58
0.5
0.63
ΔT = ΔI + Δa (in.)
1.05
1.042
1.06
1.203
2
As (in .) 2
4
Icr (in .) x (in.) 4
Icr (in .)
4
16
Problem 6.3: 2 1.) d = 23.27 in. ; As = 5.0 in. use 5 #9 ; say h = 27 in. 4 2.) Ig = 19683 in. ; Ma = 3916 k.in. ; Mcr = 691.1 k.in.
I cr = 11268 in.4 ; I e = 11217 in.4 < I g
Δ = 1.12 in.
Problem 6.4: 1.) Mu = 444 k.ft. 2 2 Use 6 #9, As = 6.0 in. d = 20.82 in. ; Total As = 5.57 in. Compression steel, fs΄ = 59 ksi. 2 h = 24.32 in., say 25 in. ; d = 21.5 in. ; For comp. steel, use 2 #7, As = 1.2 in. 4 2.) Ig = 15625 in. ; Ma = 3916 k.in. ; Mcr = 592.93 k.in. ΄
I cr = 10750 in.4 ; I e = 10767in.4 < I g
Δ = 1.17 in. 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 = (0.126×40.5) / (2×12) = 0.212 in2 ; Choose 3 #3 bars each side.
17
CHAPTER 7 DEVELOPMENT LENGTH OF REINFORCING BARS Problems 7.1(a) - 7.1(j)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
f c΄(ksi.)
3
4
5
3
4
5
5
3
4
4
f y (ksi.)
60
60
60
40
60
60
60
40
60
60
Bar No.
5
6
7
8
9
10
11
9
8
6
d b (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
ψt
1.0
1.0
1.0
1.3
1.0
1.0
1.0
1.0
1.0
1.3
ψe
1.0
1.0
1.5
1.0
1.0
1.0
1.0
1.5
1.0
1.5
Λ
1.0
0.75
1.0
0.75
1.0
1.0
1.0
1.0
1.0
1.0
ψ t ψ e ≤1.7
Y
Y
Y
Y
Y
Y
Y
Y
Y
N
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
28
38
56
64
81
54
60
93
48
49
f c '
<100
l d (in.)
Problem 7.2 (a): ldc = 21.9 in. ≥ 18 in.; Use l d = 22 in. (b): ldc = 21.4 in. ≥ 20.3 in.; Use l d = 22 in. (c): l dc = 16.06 in. ≥15.24 in.,use l d = 17 in. (d): l dc = 23.93 in. ≤ 25.38 in., l d = 0.8(25.38) = 20.3 in., use 21 in. (e): l dc = 13.55 in. ≤ 15.75 in., l d = 0.9(15.75) = 14.12 in., use 15 in. (f): l dc = 19.1 in. ≤ 20.25 in., l d = 0.75(20.25) = 15.2 in., use 16 in.
18
Problem 7.3: a.) l d = 69.56 in., use 70 in. b.) Bars with 90° hook, d b = 1.128 in.; lhb = 21.4 in. Use 22 in. c.) Bars with 180° hook, d b = 1.128 in.; l hb = 22 in. Problem 7.4: a.) l d = 0.875(61.66) = 53.96 in., use 54 in. b.) Bars with 90° hook, d b = 0.875 in.; lhb = 16.6 in. Use 17 in. > 8 d b c.) Bars with 180° hook, d b = 1.128 in.; Basic l hb = 17 in. > 8 d b Problem 7.5: a.) l d = 1.128(47.47) = 53.54 in., use 54 in. b.) Bars with 90° hook, d b = 1.128 in.; lhb = 16.48 in. Use 17 in. c.) Bars with 180° hook, d b = 1.128 in.; Basic l hb = 17 in. Problem 7.6: a.) l d = 1.27(61.66) = 78.3 in., use 79 in. b.) Bars with 90° hook, d b = 1.27 in.; lhb = 24.09 in. Use 25 in. c.) Bars with 180° hook, d b = 1.27 in.; Basic l hb = 24 in. Problem 7.7: a.) When 50% of bars are spliced; l d = 80.3 in., use 81 in., l st = l dc = 81 in. Class A splice. b.) When 75% of bars are spliced; Class B splice, l st = 104.3 in., use 105 in. c.) All bars are spliced, R s = 2.0, Class B splice, l st = 105 in. d.) When all bars are spliced, R s = 1.3, Class B splice, l st = 105 in. Problem 7.8: a.) When 50% of bars are spliced, l d = 92.7 in., use 93 in. Class A splice, l st = l dc = 93 in. b.) When 75% of bars are spliced; Class B splice, l d = 93 in., l st = 121 in. c.) All bars are spliced, R s = 2.0, Class B splice, l st = 121 in. d.) When all bars are spliced, R s = 1.3, Class B splice, l st = 121 in.> 12 in. Problem 7.9: l dc = 19.14 in. ≥ 3.84 in. ; Use lsc = l dc = 34 in. Controls Problem 7.10: l dc = 23.9 in. ≥ 42.3 in. ; Use l sc = l dc = 43 in. Controls Problem 7.11: l dc = 25.52 in. ≥ 54.14 in; l sc = 54.14 in., use Problem 7.12: l dc = 21.4 in. ≥ 33.84 in.; Use
l sc =
l sc =
34 in. Controls
19
55 in. Controls
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: 2 2 As max = 3.9 in. Use 5#8 bars in two rows, (As = 3.93 in. ) 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: 2 2 d = 33.6 in.; As = 4.314 in. Use 6#8 bars in two rows, (As = 4.72 in. ) 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 M u = 574 K.ft. Maximum negative moment Mu = 435.2 K.ft. 2.) Section at B: 2 2 d = 23.04 in.; As max = 4.99 in. Use 5#9 bars in two rows, (As = 5 in. ) 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 2 Use 2 # 9 bars; As = 2.0 in. 2 Use 7 # 9 bars; As = 7.0 in.
Actual Mu = 633.4 K.ft. 4.) ld (bottom bars) = 70 in. = X 1 ld (top bars) = 54 in. = X2
20
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. V u = 83.25 kip; V u @ d = 64.29 kip 2. φV c = 0.75×35.93 = 26.95 kip < V u @ d.; V s = 49.79 kip < V c1 , 3. Use #3 @ 6 in. 4. φV s = 20.3 k. X = 6.29 ft. = 75 in. 5. Distribution of stirrups st 1 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. W u = 7.2 k/ft.; V u = 64.8 kip; V u @ d = 50.4 kip 2. Nominal shear provided by concrete φV c = 23.66 kip < V u @ d; V s = 35.65 kip 3. Use #3 @ 7 in. 4. For Smax = 12 in., φV s = 13.2 in. X at φV c/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.
21
Problem 8.6) 1. Vu at support = 82 kip, V u @ d = 68.25 kip 2. φV c = 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.; φV s = 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. M u = 168.5 k-ft ; Use d = 18.6 in. 2 2 As = 2.33 in ; Use 3 #9, As = 3.0 in and h = 21.5 in 2. Design for shear V u = 51.02 kip; V u @ d = 46.27 kip; φV c = 18.75 kip; V s = 36.69 kip 3. Use #3 @ 5.5 in. 4. V u = 13.2 k. < φV c = 17.25 k. but > φVc/2 = 8.63 k., Vs = 0 Therefore, use #3 stirrup, 2 legs, at maximum spacing of 6.5 in.; X (at φVc/2) = 72.47 in. 5. Distribution of stirrups (from A): st 1 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) 2 1. M u = 91.5 k-ft; Use d = 12 in.; Use 2 #9, As = 2 in and h = 14.5 in (15 in.) 2. W u = 5.5 k/ft.; V u = 30.25 kip, V u @ d = 24.75 kip; φV c = 13.15 kip ; V s = 19.9 kip Therefore, use S max = 6.0 in. all over with first stirrup at 3 in. from the face of the support. Problem 8.9) 2 1. Use #3 stirrups, 2 legs, Av = 0.22 in ; let S = S 3 = 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. φV c = 53.89 k., φV s = 21.4 k. Problem 8.10) 1. W u = 7.2 k/ft.; V u = 64.8 kip; V u @ d = 50.4 kip 2. φV c = 23.66 kip < V u @ d ; V s = 35.65 kip 2 Use stirrup #3, Av = 0.22 in. ; s = 8.5 in. 3. Use #3 @ 8.5 in. 4. For S max =12 in.; φV s = 19.8 k.; X = 88 in. at φV c /2 5. Distribution of stirrup: st 1 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.
22
Problem 8.11) 1. Vu at support = 66 kip; V u @C.L. = 12 kip; V u @ d = 55.4 kip 2. φV c = 21.91 k., V s = 54.65 kip 3. Use #4 Stirrups; Smax = 4 in. (Controls) 4. V s = 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) 2 1. M max = 1296 k-in.; As = 2.22 in. 2 Use 3 #8, As = 2.35 in. ; Use h = 15.5 in., d = 13 in. 2 2. Check at 2 ft. from support: W2 = 6.22 k/ft, Mu= 609.8 k-in.; As = 2.08 in. , use 3 #8 3. ld = 48 in = 4 ft. from the support. 4. a) At support: V u = 288 psi b) At free end V u = 0 c) Max d = 13 in., for d = 12.5 in.; V u = 240 psi d) At midspan V u = 97 psi e) V us = 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.
23
CHAPTER 9 ONE-WAY SLABS
Problems 9.1(a) - 9.1(j): (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
f c΄ (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
A s (in )
2
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
8.03
9.3
11.46
14.51
7.55
11.65
16.0
14.66
7.55
9.21
φ M n
(k.ft.)
Problems 9.2(a) - 9.2(j): (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
f c΄ (ksi)
3.0
3.0
3.0
3.0
4.0
4.0
4.0
4.0
5.0
5.0
M u (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
ρ (%)
0.43
0.685
0.85
1.11
1.18
0.495
1.10
0.51
0.637
0.9
A s (in )
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
A s (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
2
* Transverse bars (shrinkage and temperature steel).
24
Problem 9.3: 1. M(D.L.) = 9.2 K.ft.; ρ = 0.015 2. Mu = 24.25 K.ft.; WL = 0.258 K/ft. = 258 psf. Problem 9.4: 1. M u = 170.2 K.ft. 2 2. d = 8 in.; As = 1.44 in. ;Use #7 bars spaced at 5 in. 3. Vu at support = 5.24 K. ; V u 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. 2 5. Use #3 bars spaced at 6 in. (As = 0.22 in. ) for shear Problem 9.5: 1. Mu = 6.41 K.ft. /ft.; Let d = 5 in. 2 2 2. A s = 0.36 in. ;Use # 4 bars spaced at 6 in. (A s = 0.39 in. ) 2 3. Shrinkage reinforcement: As = 0.144 in. ;Use #3 bars spaced at 9 in. (A s = 0.15 2 in . ) Problem 9.6: 2 1. M u (at support) = 42.2 K.ft.; A s = 1.44 in. 2. Check section at midspan: External Mu = 1 8.55 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 ρ = 0.014 < ρmax = 0.023; d = 3.6 in. 2 As = 0.60 in. , choose #5 bars. h = 4.66 in.; Use h = 5 in., d = 3.94 in. 4. Moments and As required at other locations: Location Moment Mu R u (See example 9.5) Coeff. K.ft. psi A B C D E 5. Shear is adequate.
- 1/24 + 1/14 - 1/10 - 1/11 + 1/16
2.42 4.15 5.81 5.29 3.63
156 267 374 341 234
25
2
ρ%
As(in. )
# 5 bars spaced at (in.)
0.45 0.80 1.15 1.03 0.7
0.21 0.38 0.54 0.49 0.33
12 9 6 6 9
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 Coeff. A - 1/24 B + 1/14 C - 1/10 D - 1/11 E + 1/16 3. Shear is adequate.
Mu K.ft. 2.42 4.15 5.81 5.29 3.63
R u psi 151 259 363 331 227
ρ% 0.33 0.51 0.74 0.68 0.45
2
As(in. ) 0.16 0.25 0.36 0.33 0.22
# 4 bars spaced at (in.) 12 9 6 6 10
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, A s , and bars. Location Mu (See example 9.5) K.ft.
3.
R u Psi
ρ%
As 2 (in. )
#4 bars spacing (in.)
A B C
2.42 4.15 5.81
151 259 363
0.33 0.50 0.73
0.16 0.24 0.36
12 9 6
D E
5.29 3.63
331 227
0.66 0.45
0.33 0.22
6 10
Shear is adequate.
Problem 9.1 0: 1.a. Design of slab: Assume top slab thickness = 2 in. Own weight of slab = 25 psf.; D.L. = 55 psf, U = 226 psf. M u = 1.41 K.in. 1.b. Assuming that moment is resisted by plain co ncrete only, M = 1.42 K.in. > Applied moment of 1.41K.in. 2 1.c. Shrinkage reinforcement, As = 0.0432 in. 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. 2 3. Mu (flange) = 1541 K.in.; Choose 2#5 bars (As = 0.62 in. ) 4. Shear in rib; Vu (at d distance) = 5771 lbs. < V c Use min. stirrups, # 3 spaced at 5 in.
26
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 . 2 Choose 2#5 bars (As = 0.62 in. ) 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 M u and As: For positive moment b = 34 in., and for negative moment, b = 4 in., d = 10.875 in.
Location
A B C D
Moment Coeff.
Mu K.ft.
R u psi
ρ%
As 2 (in)
bars
- 1/24 + 1/4 - 1/10 + 1/16
9.62 16.50 23.10 14.43
244 49 586 43
0.50 1.30 -
0.22 0.35 0.57 0.35
2#3 2#4 2#5 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.
27
CHAPTER 10 AXIALLY LOADED COLUMNS Problems 10.1(a)-10.1(j): Given: φ = 0.65, K = 0.8, ρg (max.) = 8%, ρg (min.) = 1%
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
4
4
4
4
5
5
5
5
6
6
As (in. )
8
25
6.28
15.2
10
5.06
15.2
12.5
10.12
7.59
ρg (%)
3.13
6.25
4.36
5.28
5.1
1.98
4.18
2.17
3.95
3.16
φPn(k.)
688
1442
439
955
722
712
1244
1634
968
852
f c΄(ksi.) 2
Problems 10.2(a)-10.2(e): Given: φ = 0.75, K = 0.85, ρg (max.) = 8%, ρg (min.) = 1%
(a)
(b)
(c)
(d)
(e)
4
4
5
5
6
2
153.9
201.1
254.5
314.2
176.7
As (in. )
2
8
7.59
10.12
15.2
8
ρg (%)
5.2
3.77
3.97
4.84
4.53
φPn(k.)
622
710
1049
1391
855
f c΄ (ksi.) Ag (in. )
Problem 10.3(a) - 10.3(h): Square and Rectangular Columns .
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
f c΄(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. )
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
2
28
Problem 10.3(i) – 10.3(L): Spiral Columns .
(i)
(j)
(k)
(L)
f c΄(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. )
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
2
29
CHAPTER 11 MEMBERS IN COMPRESSION AND BENDING 11-1-a to 11-1-l. Balanced condition (12 Problems) Rectangular sections problem 11-1-a Balanced condtion φ P b (k.), φ M b (k-ft.) φ P b = 372, φ M b = 541
Prob. 11-1-d 453.5, 781.2
Prob. 11-1-e 197.9, 252.12
Prob.11-1-i
Prob. 11-1-f 230.1, 310.3
Prob. 11-1-c
162, 146.7
550.9, 922.3
Prob. 11-1-g
Prob. 11-1-h
264.1, 330.9
Prob. 11-1-j
390.4, 344.5
Prob. 11-1-b
351.1, 365.9
Prob. 11-1-k
394.1, 486.7
485.2, 679.2
Prob. 11-1-l 347.2, 370.4
e = 6 in 11-2-a to 11-2-l. Compression controls: Rectangular sections Prob. 11-2-a Compression controls refer to prob. 11-1-a
(12 Problems); φ = 0.65 Prob. 11-2-b Prob. 11-2-c Refer to 11-1-b Refer to 11-1-c
φPn (k.), φMn (k.-ft.)
264.6, 132.3
1208.6, 604.3
Prob. 11-2-g Refer to 11-1-g
Prob. 11-2-h Refer to 11-1-h
524.9, 262.4
604.6, 302.3
Prob. 11-2-d Refer to 11-1-d 993.2, 496.6 Prob. 11-2-i Refer to11-1-i 551.0, 275.5
φPn = 775.4, φMn = 387.7
Prob. 11-2-e Refer to 11-1-e 384.6, 192.3
Prob. 11-2-f Refer to 11-1-f 465, 232.5
Prob. 11-2-j Refer to 11-1-j 741, 370.5
Prob. 11-2-k Refer to 11-1-k 994.6, 497.32
Prob. 11-2-l Refer to 11-1-l 613, 306.6
11-3-a to 11-3-l. Tension controls: e = 24 in (12 Problems) Rectangular sections Prob. 11-3-a Prob. 11-3-b tension controls refer to prob. 11-1-a refer to 11-1-b 61.4, 122.8 φP (k.), φM (k-ft) φP = 258.8, φM = 517.6
Prob. 11-3-d Refer to 11-1-d 384.7, 769.5 Step
11
Prob. 11-3-e Refer to 11-1-e 116.8, 233.7
Prob. 11-3-i Refer to 11-1-i 146.26, 292.53
Prob. 11-3-f Refer to 11-1-f 146.2, 292.4
Prob. 11-3-j Refer to 11-1-j 215.09, 430.18
30
Prob. 11-3-g Refer to 11-1-g 152.12, 304.24
Prob. 11-3-k Refer to 11-1-k 309.47, 618.93
Prob. 11-3-c Refer to 11-1-c 452.5, 904.9 Prob. 11-3-h Refer to 11-1-h 152.71, 305.43
Prob. 11-3-l Refer to 11-1-l 149.20, 298.40
Problem 11.4: Balanced condition : c b = 12.72in, a b = 10.82in. Cc = 588.38K; Cs1 = 286.4K; Cs2 = 58.94K -4 Tension zone : φ = 0.65; ε s2 = 5.75×10 , f s3 = 16.69Ksi, P b = 587.73K; M b = 816.97K.ft, e = 8in., e < e b 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 -4 Tension zone : φ = 0.65; ε s = 9.26×10 , f s = 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 : c b = 10.36in., a = 8.81in., Cc = 599.08 K; Cs1 = 358.28 K; Cs2 = 79.02 K; Cs3 =”-0.97“= Zero -4 Tension zone: εs4 = 9.82×10 , f s = 28.47ksi, P b = 584.27K; M b = 788.16K.ft; e b = 16.19in. e = 8in., e < e b compression controls. Assume c = 12.77in., a = 10.8in. Cc = 734.4K, Cs1 = 358.28K, Cs2 = 104.19K, Cs3 = 39.29K -4 Tension zone: εs4 = 2.3×10 , f s = 6.68Ksi, εs5 = 0.0011, f s = 32.22Ksi, / Pn = 1014.93K, Mn = 676.62K.ft; e = 15.5in. Problem 11.6: Balanced condition : c b = 10.36in., a = 8.8in., P b = 408.68K; M b = 414.1K.ft, e b = 12.16in. Compression controls: e = 8 < e b comp. controls. Assume c = 12.09in., a = 10.28in. / / Cc = 489.33K; Cs1 = 169.8K, Cs2 = 23.28K / Tension zone: ε s = 0.0013, Pn = 565.62K, Mn = 377.08K.ft; e = 15.5in. Problem 11.7: Balanced condition: c b = 12.72in., a b = 10.82in., Cc = 735.76K; Cs1 = 283K; Cs2 = 68.03K; Cs3 = 3.05K -4 Tension zone: ε s4 = 9.5×10 , f s = 27.56Ksi, Pn = 734.27K, M b = 914.52K.ft, e b = 14.95in. Compression controls: e = 8 < e b comp. controls. Assume c = 15.98in., a = 13.58in. Cc = 923.44K; Cs1 = 283K, Cs2 = 88.26K, Cs3 = 36.54K -4 Tension zone: ε s4 = 1.43×10 , f s = 4.19Ksi, ε s5 = 0.001, f s = 30.05Ksi Pn = 1172.6K, Mn = 781.73K.ft
31
Problem 11.8: a: Choose #4 ties spaced at 18in. Pn = 821.23K; Mn = 403.1K.ft O.K. / 2 / 2 b:As = 3.68in. use 5#8 (As = 3.93in. ) Pn = 643.1K, Mn = 669.65K.ft / 2 c: As = As = 7.56in. , use 6#10 Pn = 756.25K, Mn = 1608K.ft / 2 d: As = As = 4.73in. , use 6#9 Pn = 682.7K, φ Pn = 443.8K > Pu, O.K. / 2 e: As = As = 7.59 in , use 6#10 Pn = 1 753.93 K / 2 f: As = As = 0.03×18×24/2 = 6.48in. , Use 5#10 Pn = 1114.26 K, φ Pn = 724.3 K > Pn O.K. / 2 g: As = As = 0.02×14×20/2 = 2.8in. , Use 3#9 Pn = 453.12K, φ Pn = 317.2K; O.K. section is adequate. / 2 h: As = As = 6.57in. , use 6#10 Pn = 1617.7K, φ Pn = 1051.5 K > Pu, O.K. Section is adequate. / 2 i: As = As = 0.02×20×14/2 = 2.8in. , Use 2#10 Pn = 923.93K, φ Pn = 600.5 K > Pu ; O.K. section is adequate. / 2 j: As = As = 2.64in , Use 4#9 Pn = 1037.86 K, φ Pn = 674.6 K > Pu, O.K. section is adequate. Problem 11.9: a: φPn = 265K compared to 265 K by calculation. b: φPn = 991K compared to 993.2 K by calculation. c: φPn = 462.9K compared to 465 K by calculation. / 2 2 d: As = As = 5.8in , Use 5#10, As = 6.33in. / 2 2 e: As = As = 6.48in. , Use 6#10, As = 7.59in. / 2 2 f: As = As = 8.4in. , Use 7#10, As = 8.86in. Problem 11.10: P b = 67.11K, φ P b = 43.62K M b = 86.9K.ft, φ M b = 56.5K.ft, e = 15.54in. Problem 11.11: P b = 222.33K, φ P b = 144.5 K M b = 229.61K.ft; φ M b = 149.3K.ft, e b = 10.5 in. Problem 11.12: P b = 427.85K, φ P b = 278.1K
M b = 420K.ft; φ M b = 273K.ft, e b = 11.78in.
32
Problem 11.13: P b = 677.6K, φ P b = 440.4K M b = 655.7K.ft, φ M b = 425.8K.ft, e b = 11.6in. Problem 11.14: a: φ Pn = 108K (tied); φ Mn = 53.9K.ft (tied) b: φ Pn = 271.4K (tied); φ Mn = 135.5K.ft (tied) c: φ Pn = 495K (tied); φ Mn = 247.7K.ft (tied) d: φ Pn = 731.8K (tied); φ Mn = 366.3K.ft (tied) Problem 11.15: a: P bx = 572.5K, M bx = 790.9K.ft; e b = 16.58in. P by = 536 K, M by = 482 K.ft; e by = 10.79in. Pny = 833.55K, Mny = 416.77K.ft Pn = 559 K, φ Pn = 363.3 K b: P bx = 576.88K; M bx = 740.55K.ft; e bx = 15.4in. Pnx = 929.82K, Mnx = 619.88K.ft Pny = 1107 K, Mny = 553.6K.ft Pn = 633 K, φ Pn = 411.5 K c: P bx = 408.68K; M bx = 414.1K.ft; e b = 12.16in. P by = 368.7K; M by = 260.3K.ft; e b = 8.47in. Pny = 496.8K, Mny = 248.4K.ft Pn = 323 K., φPn = 210 K. d: P bx = 718.7K; M bx = 865.6K.ft, e bx = 14.5in. Pnx = 1 093.6K, Mnx = 729.1K.ft P by = 701.9K; M by = 699.6K.ft, e by = 12 in. Pny = 1145.3K, Mny = 572.65K.ft Pn = 717.6 K., φ Pn = 466 K. Problem 11.16: a: Pn = 566.34K, φPn = 368 K b: Pn = 674.3K., φPn = 438.3K c: Pn = 335 K., φPn = 218 K d: Pn = 732.24 K, φPn = 476 K Problem 11.17: a: Pn = 645 K., φPn = 419 K b: Pn = 749 K., φPn = 487 K c: Pn = 355 K., φPn = 231 K d:Pn = 817 K., φPn = 531 K
33
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
34
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.) δns = 1.144 7.) Pn = 470.8 kips; Mn = 462.2 k-ft.; Mc = 528.76 k-ft.; e = 13.48 in. 8.) φPn = 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.) δns = 1.06 7.) Pn = 470.8 kips; Mn = 462.2 k-ft.; Mc = 489.932 k-ft.; e = 12.5 in. 8.) φPn = 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.) δns = 1.116 7.) Pn = 470.8 kips; Mn = 462.2 k-ft.; Mc = 515.82 k-ft.; e = 13.15 in. 8.) φPn = 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.) δns = 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.) φPn = 933.6 kips > Pu
35
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.) δns = 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.) φPn = 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.) δns = 1.052 7.) Pn = 589.23 kips, Mn = 394.462 k-ft.; Mc = 414.974 k-ft.; e = 8.45 in. 8.) φPn = 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.) δs = 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.) φPn = 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.) δs = 1.565 7.) Pn = 691.385 kips, Mn = 216.923 k-ft; Mc = 339.484 k-ft., e = 5.89 in. 8.) φPn = 512.911 kips > Pu
36
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.) δs = 1.1 7.) Pn = 691.385 kips, Mn = 216.923 k-ft.; Mc = 238.615 k-ft., e = 4.142 in. 8.) φPn = (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.) δs = 1.872 7.) Pn = 953 kips, Mn = 216.923 k-ft.; Mc = 309.332 k-ft., e = 5.369 in. 8.) φPn = 619.54 kips > Pu
37
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
P u (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
V u1 (k.)
14.63
12.47
15.95
23.00
18.67
17.18 17.41 20.63
11.91
13.71
M u (k.ft.)
46.17
31.00
46.79
111.00
67.96
49.67 59.32 72.09
22.91
31.09
A s (in. )
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
l d (in.)
48
28
33
55
55
42
55
42
24
29
l dav (in.)
51
36
41
77
55
43
59
53
27
32
A sh (in. )
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
2
2
38
Problem 13.2: Square Footings .
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i) (i )
(j)
A F (ft. )
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.) P u (k.)
8 364
7 352
9 548.4
9 440
8.5 424
10 452
8 432
10 546
7 280
8 276
qu (ksf.)
5.7
7.2
6.8
5.4
5.9
4.5
6.8
5.5
5.7
4.3
V u1 u1 (k.)
23.1
77.7
130.1
115.4
114.9
113.1
106.5
148.96
68.2
75.97
V u2 u2 (k.)
325.5
300
480.8
400.2
388.2
417
382.2
426.3
240.0
249.2
M u (k.ft.)
253.3
190.6
411.4
388.8
337.2
422.5
287.3
477.4
141.9
191.1
A s (in. )
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
l d d (in.)
33
33
33
33
29
42
29
48
29
33
l dav dav (in.) N 1 (k.)
37 424.3
30 537
41 663
45 238.6
41 433.2
49 565.5
36 716
47 884
30 696.4
38 334.2
x
4
4
4
7
4
4
4
4
4
4
Ads (in. )
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
2
2
2
39
Problem 13.3: Rectangular Footings:
A F (ft. ) L (ft.) B (ft.) h (in.) d (in.) P u (k.) qu (ksf.)
(a) 61.4 10.5 6 23 18.5 364 5.78
(b) 47.9 8 6 20 15.5 352 7.33
(c) 77.6 10 8 23 18.5 548.4 6.86
(d) 78 10 8 24 19.5 440 5.5
(e) 69.6 10 7 21 16.5 424 6.06
(f) 96.7 12.5 8 21 16.5 452 4.52
(g) 61.5 10.5 6 23 18.5 432 6.86
(h) 95 11 9 22 17.5 546 5.52
V u1 u1 (k.) V u2 u2 (k.) M uL uL (k.ft.) M uS uS (k.ft.)
105.5 316.4 363.7 165.2
56.2 294.7 232.3 148.4
111.5 111.5 478.2 476.4 344
103.1 402.1 445.5 336.9
93.4 385.1 413.7 257.8
110.6 418.8 563.62 313.9
51 368.7 416.7 182.3
134.1 492.6 540.96 408.2
A sL (in.2) Bars A sS (in.2) Bars A sB (in.2) Bars A sE (in.2) Bars
4.52
3.47
5.92
5.25
5.79
7.9
5.2
7.12
9#7 5.22 12#6 3.79 9#6 1.8 4#6
10#6 3.46 8#6 2.98 7#6 0.87 4#6
11#7 4.95 11#6 4.42 10#6 0.8 2#6
10#7 5.18 12#6 4.61 11#6 0.84 2#6
9#8 4.54 8#7 3.7 6#7 1.09 2#7
9#9 5.7 10#7 4.44 8#7 1.6 4#7
10#7 5.22 9#7 3.76 7#7 1.77 4#7
8#9 5.23 10#7 5.28 9#7 0.76 2#7
l dL dL (in.) l daL daL (in.) l ds ds (in.) l das das (in.)
48 52 28 25*
33 36 33 24*
48 47 33 35
48 51 33 39
48 50 42 32*
54 64 42 37*
42 51 42 24*
54 53 42 41
N 1 (k.) Ads (in.2)) x l dd dd (in.)
424.3
537
663
238.6
433.2
565.5
716
884
1.28 4 22
1.62 4 25
2.0 4 25
5.55 7 22
1.0 4 22
1.28 4 22
1.62 4 19
2.0 4 22
2
40
Problem 13.4: Rectangular footings.
(a)
(b)
(c)
(d)
A F (ft. ) L (ft.) B (ft.) h (in.) d (in.) d (in.) P u (k.) qu (ksf.)
61.4 9.5 6.5 20 15.5 364 5.9
47.9 8.5 5.75 21 16.5 352 7.2
77.6 11 7.25 26 21.5 548.4 6.88
78 11 7.25 24 19.5 440 5.52
V u1 u1 (k.) V u2 u2 (k.) M uL uL (k.ft.) M uS uS (k.ft.) 2 A sL (in. ) Bars 2 A sS (in. (in. ) Bars 2 A sB (in. ) Bars 2 A sE (in. ) Bars
110.26 321.05 294.15 199.29 4.43
91.43 292.39 241.64 149.23 3.38
151.72 466.4 505.04 331.17 5.44
135.07 394.6 451.48 296.48 5.42
9#7 4.10 11#6 3.59 8.#6 1.02 4#6
9#6 3.86 9#6 3.2 8#6 1.0 4#6
10#7 6.18 10#7 4.9 8#7 1.77 4#7
10#7 5.70 10#7 4.49 8#7 1.57 4#6
l dL dL (in.) l daL daL (in.) l ds ds (in.) l das das (in.)
48 44* 33 29* 464.3
33 38 33 24* 530.2
48 51 48 33* 636.1
48 54 48 35* 357.5
1.4 4 22
1.6 4 25
1.92 4 25
2.49 4 22
2
N 1 (k.) 2 Ads (in. )) x l dd dd (in.)
41
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
22.8
22
29.6
34.8
4.56
5.5
7.4
4.64
9.12
6.19
7.43
23.27
6859
4913
6859
29791
152
129
124
145
178
178
178
178
1.9
0.46
0
2.71
18.75
16.76
21.85
35.66
P u (k.) qu (ksf.) M u (k.ft.) 4
I g (in ) f t (psi.) f ta (psi.) V u (k.) φV c (k.)
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; q u = 5.83 Ksf; q u’= 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 = 12 3.0 4 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: 2 Use 12 #9 bars, (As = 12 in. ). Spacing = 9.3 in. 6. Design for beading moment in the short direction: 2 Use 9 #7 bars (AS = 5.41 in. ), Spacing = 12.75 in.
42
Problem 13.7: Footing No. 1 D.L. (K) 130 L.L. (K) 160 L.L. / D.L. 1.23 Usual Load (K) 170 Area = (Usual Load/2.81) 60.5 Max. Soil Pressure (Ksf) 4.8
2 220 220 1.00 275 97.9 4.5
3 150 210 1.40 202.5 72.0 5.0
4 180 180 1.00 225 80.1 4.5
5 200 220 1.10 255 90.7 4.6
6 240 200 0.83 290 103.2 4.3
Problem 13.8: 1. P = 360 K; M= 230 K.ft., e = 7.67 in., say 8 in. 2. a)Assume total dep th = 2 ft., a nd assume the weight of the soil is 100 pcf. Ne t up wa rd pre ss ur e = 3.5 Ks f; Ch oo se a fo oti ng (1 1 x 10 ) ft b) P u = 496 K; q u = 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. V u = 167.25K 2 4. Two way shear: b0 = 142 in.; A0 = 1244.25 in. ; V u = 457.03K. 5. Reinforcement in the long direction: 2 Use 9 #10 (11.39in. ) spaced at 14.25in., provided l d > required 32 in. 6. Reinforcement in the short direction: 2 Use 9 #9 (9 in. ) 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: 2 Choose 15 #9 bars (As = 15 in. ); Spacing = 6.4 in. 6. Reinforcement in the short direction: 2 Use 20 #8 bars (As = 15.7 in. ) ; 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. q max < q (allowable)
43
CHAPTER 14 RETAINING WALLS Problem 14.1: 1. Using Rankine formula: Ca = 1/3 and C p = 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 = ΣW = 5945 lb., ΣM 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 C p = 3.0; Ha = 4125 lbs., acting @15/3 = 5 ft. from base. Overturning moment M o = 20.625 K.ft. 2. Calculate the balancing moment taken about the toe O:
Weight lb. W1 = 3770 W2 = 3770 W3 = 2320 W4 = 2860 W5 = 1430 Total W = R = 14.150 K.
arm (ft) 2.00 4.33 4.00 5.67 7.5
Moment 7.54 16.32 9.28 16.22 10.72 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 (K) (FT) RESULTANT 7.540 4.24
EARTH PRESSURE FORCES AND MOMENTS: FORCE ARM (K) (FT) TOTAL 2.146 4.00 FACTOR OF SAFETY AGAINST OVERTURNING = 3.73 FACTOR OF SAFETY AGAINST SLIDING =
44
MOMENT (K.FT) 32.000
MOMENT (K. FT) 8.585
1.76
Problem 14.3 (e) : BALANCED FORCES AND MOMENTS: ITEM WEIGHT (K) RESULTANT 12.604
ARM (FT) 5.72
MOMENT (K. FT) 72.139
EARTH PRESSURE FORCES AND MOMENTS: FORCE ARM (K) (FT) TOTAL 4.307 5.67 FACTOR OF SAFETY AGAINST OVERTURNING FACTOR OF SAFETY AGAINST SLIDING Problem 14.4 (e) : 1. Balanced forces and Moments: W1 = 2.325 K arm W2 = 0.581 arm W3 = 2.025 arm W4 = 7.673 arm W6 = 1.35 arm Total W = 13.964
= 4 ft = 3.33 = 4.5 = 6.75 = 6.75
2. Earth pressure forces and moments: PAH = 4.307 K arm = 5.67 ft. PH2 = 1.382 arm = 8.5 ft. Total P = 5.69 Overturning F.S. = 81.25/ 36.16 = 2.24
MOMENT (K. FT) 24.409 = =
2.96 1.63
M = 9.3K.ft. M = 1 .938 M = 9.113 M = 51.79 M = 9.113 Total M = 81.25
M = 24.41 K.ft. M = 11.76 K.ft. Total M = 36.16 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
45
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 φ = 33°, Ca = 0.295 Ha = 7.08 K ; Mo = 47.2 K.ft.
Weight arm (ft) W 1 = 2.4 4.50 W 2 = 1.2 3.67 W 3 = 3.6 6.00 W 4 = 13.44 8.50 W = R = 20.64 Factor of safety against overturning = 2.3 > 2. 0 2. Check sliding condition: F = 9.2 9; F.S. = 1.31 < 1.5 F.S. (sliding) = 1.56 > 1.5 3. q 1 =0.83 = 2 .55 ksf; q2 = 0.89 ksf F = 11.31 K; F. S.(slidin g) = 1.6 > 1.5
Moment (k.ft.) 10.8 4.4 21.6 114.2 M = 151.0 Not good, use Key 1.5 x 1. 5ft
Problem 14.11: 1. For φ = 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. 2 For longitudinal reinforcement: use #4 bars spaced at 12 in. (As = 0.20 in. )
46
CHAPTER 15 DESIGN FOR TORSION Problem 15.1:
φ Tcr = 349.8 K .in. Problem 15.2:
φ Tcr = 170.8 K .in. Problem 15.3: 2 Acp = 428in. ;Pcp = 138 in.
φ Tn = 63 K .in.
φ Tcr = 251.9 K .in. Problem 15.4: 2 Acp = 608in. , Pcp = 156in.
φ Tcr = 449.6 K .in. Problem 15.5: 2 2 Acp = 336 in. ; Pcp = 76in. If flanges are included, then φTcr = 390.5 k.ft and φTn = 97.7 k.ft. Problem 15.6: 2 Acp = 504in. , Pcp = 132 in. Problem 15.7: φ Vc = 25.9 K
V u = 36 f
φ V c 2
Pcp = 72 in. Design for torsion: Assume 1.5in. cover concrete cover and #4 stirrups 2 2 Aoh = 194.25 in. ; Ao = 165.11 in. , Pn = 58 in. Use #4 stirrups, use S = 7 in. Problem 15.8: Total area of closed stirrups: 2 For one leg: Avt/S = 0.015+0.004 = 0.019in (per one leg) Use #4 stirrups, Use S = 7in Problem 15.9: φ Vc = 22.2 K < Vu shear reinforcement required. 2 Acp = 264 in , Pcp = 68 in, Ta = 48.6K.in; Ta LHS O.K., Tn = 300/0.75 = 400 K.in 2 Distribution of longitudinal bars: Total = 1.35in , Al/3 = 0.45
47
Problem 15.10: Ta = 61.5K.in torsional reinforcement required. φ Vc = 22.2 K shear reinforcement required Design for shear: 2 Av/S = 50.4/(60×19.5) = 0.043in /in (two legs) 2 Av/2S = 0.0215 in /in (one leg) Design for torsion: 2 2 Aoh = 197.3in ; Ao = 167.7 in ; 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 φ Vc/2 shear reinforcement required. Ta = 76.8K.in. Torsional reinforcement required. Design for shear: 2 2 Av/S = 0.006 in /in(2 legs); Av/2S = 0.003 in /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 φ Vc/2 shear reinforcement is required. Pcp = 124in. Ta = 77K.in; Torsional reinforcement is required. 2 2 Design for shear: Av/S = 0.019in /in (2 legs), Av/2S = 0.0095in /in (one leg) 2 Design for torsion: Ao = 206.8in , Ph = 120in LHS = 348.6, RHS = 474.2 > LHS O.K., Tn = 480K.in 2 At/S = 0.019in (per one leg) Longitudinal reinforcement: Use #4 stirrups @ S = 8in. 2 2 Distribution of longitudinal bars: Total 2.28in , Al/3 = 0.76in Problem 15.13: Mu = 3000K.in φ Vc/2 shear reinforcement is required. 2 Acp = 448 in , Pcp = 124in, Ta = 66.5K.in. Torsional reinforcement is required 2 2 Design for shear: Av/S = 0.01 in /in (two legs); Av/2S = 0.005 in /in (one leg) 2 2 Design for torsion: Aoh = 243.35in , Ao = 206.8in , Pn = 120 in. 2 LHS = 375.7, RHS = 410.7 > LHS O.K., Tn = 400 Kin, At/S = 0.016in (one leg) Total area of closed stirrups: Use #3 stirrups @ S = 6 in. 2 2 Distribution of longitudinal bars: Total = 1.92 in , Al/3 = 0.64in
48
Problem 15.14: 2 As = 3.3in use 4#9; Shear and torsional reinforcement are required. 2 Design for shear: Av/2S = 0.003 in /in (one leg) 2 2 Design for torsion: Aoh = 243.25in , Ao = 206.8in 2 2 Pn = 120in, Acp = 448in , Pcp = 124in; φ Tcr = 307K.in < Tu = 300K.in; At/S = 0.016 in (one leg) Total area of stirrups: For one leg: Use #3stirrups @ S = 6in 2 2 Distribution of longitudinal bars: Total = 1.92in , Avt/3 = 0.64in Problem 15.15: 2 Design for moment: As = 3.35in , Use 4#9 bars 2 Design for shear: Ao/2S = 0.005in /in (one leg) 2 2 2 Design for torsion: Aoh = 243.25in , Ao = 206.8in , Pn = 120in; Acp = 448in , Pcp = 124in 2 2 φ Tcr = 226K.in < Tu, Use Tu = 266K.in; At/S = 0.014in (one leg); Al = 0.014×120 = 1.7in Total area of closed stirrupss: For one leg: Use #3 stirrups; Use S = 6in 2 2 Distribution of longitudinal bars: Total = 1.7 in , Al/3 = 0.57 in Problem 15.16: 2 Design for shear: Av/2S = 0.0058in /in (one leg) Design for torsion: Aoh = 194.25, Ao = 165.11, Pn = 58, Acp = 308, Pcp = 72 φ Tcr = 216.5K.in < Tu = 216.5 K; Use #4 stirrups At 7in. Problem 15.17: 2 2 1. Moy = 334 k.ft., Mox = 248.5 k.ft.; mAsx = 4.86 in. , Asy = 5.04 in. 2. Design for torsion and shear: Tu = 576 k.in., Vu = 12 k. a.) d = 17.5 in., f c’ = 4 ksi., f y = 60 ksi,
φVc = φ 2 fc 'bd = 26.6 k . 2
b.) Acp = 320 in. , 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. 2 2 2 a.) Aoh = 206.25 in. , Ph = 58 in. , Ao = 175.3 in. 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. 2 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. ) 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 < φ Vc /2 2 Acp = 336 in. , 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.
49
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 2 2 2 Steel: 3#7 (1.8 in ) 4#7 (2.41 in ) 4#7 (2.41 in ) 4. Design for shear: Use #3 stirrups spaced at 6 in. 5. Deflection: Δ = 0.06 in.; Δ / L = 1/3990 which is very small. Problem 16.2: 1.One way slab: Wu = 268 psf. ; Load o n 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. R A = 55.87 K.; R B = 192.33 K.; R C = 55.8 K Section AB B Bars 5#9 4#9+3#8(top) 2#9(bottom) 5. Design for shear: Use #4 stirrups spaced at 3.5 in. Problem 16.4: Section AB B Bars 3#9 Design for shear: Use #4 stirrups at 3.5 in.
BC 3#9
BC 6#9
2#9+2#8
Problem 16.5: 1.) Assume beam 16 x 34 in., and column 16 x 30 in. Section Supports B,C , Midspan E 2 2 Bars 2#10 + 6#9(8.53 in ) (5#10 + 3#9)(As=9.33 in ) Design for shear: Use #4 stirr ups at 4 in., max spac ing = 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: Pr ov id ed 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. Rect angu lar footi ng: use 7.5 x 4 ft. 2 Longitudinal steel: Use 5#6 bars in the longitudinal direction (As = 2.21 in. ) Transverse steel: Use 7#7 bars in the short direction.
50
Problem 16.7: M p = 494.2 K.ft. Problem 16.8: M p = 300 K.ft. Problem 16.9: Mu = 171 K.ft. 2 Design critical sections: use 3#9 bars (As = 3.0 in. ) 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: θA = 0.0101 greater than the required θ of 0.0062. Deflection: Δ = 0.266 in.; Δ/L = 1/902 which very small. Shear: Use #4 stirrups spaced at 8 in. Problem 16.10: Final moments: Section Support
D.L. moments
B C Midspan AB BC CD *= Design Moments Problem 16.11: Final moments: Moment due to Section
Support
A B C D E
Midspan AB BC CD DE
L.L. max (-)
L.L. max (+)
D.L.+L.L max (-)
D.L.+L.L. max (+) -258.2 -258.2
-311 -311
-302.4 -302.4
52.8 52.8
-613.4* -613.4*
276.6 121.0 276.6
-79.2 -158.4 -79.2
280.8 201.6 280.8
197.4 -37.4 197.4
DL
L.L. max –ve
L.L. max +ve
0 -333.3 -222.2 -333.3 0
0 -312.5 -277.7 -312.5 0
0 0
+265.4 +154.3 +154.3 +265.4
-84.9 -141.5 -141.5 -84.9
+276.4 +218.6 +218.6 +276.4
51
Col.1+Co1.2 max -ve 0 -645.8* * -499.9 * -645.8 0 +180.5 +12.8 +12.8 +180.5
557.4* 322.6* 557.4
Col. 1+Co1. 3 max +ve 0 -333.3 -222.2 -333.3 0 +541.8* +372.9* +372.9* +541.8*
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
7.2→7.5
6.54→7.0
7.5
b
22
22
8.8→9.0
8.0
9.0
c
24
24
9.6→10
8.73→9.0
10.0
d
18
14
7.2→7.5
6.54→7.0
7.5
e
22
18
8.8→9.0
8.0
9.0
f
24
20
9.6→10
8.73→9.0
10.0
g
28
22
11.2→11.5
10.18→10.5
11.5
h
28
28
11.2→11.5
10.18→10.5
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
5.46
2.6
2.6
2.6
Min. As = 0.0018bhs (in. )
1.73
1.73
1.73
1.73
Straight bars
18#5
9#5
9#5
9#5
Spacing = b/No.≤2hs≤18 in.
6.7
13
13
13
2
As = ρ bd (in. ) 2
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
7.02
5.05
5.05
5.05
Min. As = 0.0018bhs (in. )
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
2
As = ρ bd (in. ) 2
52
Problem 17.2(c) : Strip
Column strip
Middle Strip
Moment sign
Neg.
Pos.
Neg.
Pos.
Mu (k.ft.)
- 411
177
- 137
118
8.49
5.66
5.66
5.66
Min. As = 0.0018bhs (in. )
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
2
As = ρ bd (in. ) 2
Problem 17.3(a) : Long Direction
Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0018bhs (in. ) Straight bars Spacing = b/No.≤2hs = 18 in. Problem 17.3(b): Long Direction
Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0018bhs (in. ) Straight bars Spacing = b/No.≤2hs = 18 in.
Column strip Ext.(a) Pos.(b) - 80 96 2.80 3.35 1.73 1.73 9#5 12#5 13 12
Int.(c) 161.6 5.90 1.73 10#7 12
Middle strip Ext.(a) Pos.(b) Int.(c) 0 64 - 53.9 0 2.57 2.57 1.73 1.73 1.73 8#5 9#5 9#5 15 13 13
Column strip Ext.(a) Pos.(b) Int.(c) 169.51 203.41 342.27 5.2 5.2 7.33 3.11 3.11 3.11 12#6 12#6 10#8 12 12 14.4
Middle strip Ext.(a) Pos.(b) Int.(c) 0 135.61 114.09 0 5.2 5.2 3.11 3.11 3.11 10#6 12#6 12#6 14.4 12 12
Problem 17.4: Pro b.#
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
6.54→7
6.0
7.0
2
9
6.67×6.67
b
22
22
8.0
7.33→8
8.0
2
10
8×8
c
24
24
8.2→9
7.27→8
9.0
2.25
11.5
8.67×8.67
d
18
14
6.54→7
5.45→5.5
7.0
1.75
9
6.67×5.33
e
22
18
8.0
6.67→7
8.0
2
10
8×6.67
53
f
24
20
8.72→9
8.0
9.0
2.25
11
8.66×7.33
g
28
22
10.18→10.5
8.5
10.5
2.6
13
10×8
h
28
28
10.18→11.5
10.5
10.5
2.6
13
10×10
Problem 17.5(b): Mo = 572 k.ft.
Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0018bhs (in. ) Straight bars
Long & Short Direction Column strip Middle strip Neg. Pos. Neg. Pos. 280.3 120.1 91.5 80.1 7.76 3.1 3.1 3.1
3.11 14#7
3.11 11#5
3.11 11#5
3.11 11#5
Problem 17.6(b): Mo = 572 k.ft.
Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0018bhs (in. ) Straight bars
Long & Short Direction Column strip Middle strip Ext. Pos. Int. Ext. Pos. 149 178.5 300 119 8.3 3.1 4.2 4.04 4.89 2.59 2.59 2.59 2.07 2.07 12#7 16#5 14#7 8#6 16#5
Int. 100 3.6 2.07 8#6
Problem 17.7(a) : Long & Short Direction
Column strip Neg. Pos. -20.6 11.1
Mu (k.ft.) 2
Middle strip Neg. Pos. -45.8 24.7
As = ρ bd (in. )
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
Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0018bhs (in. ) As (ρ min.=0.0033) Bars selected Spacing ≤15 in.
Column strip Ext.(-) Pos. Int.(-) -5.62 18.1 -22.2 low 1.3 1.9 10#4 12
low 1.3 1.9 10#4 12 54
low 1.3 1.9 10#4 12
Middle strip Ext.(-) Pos. Int.(-) -7.7 40.2 49.4 low 1.3 1.9 10#4 12
2.10 1.3 1.9 14#4 8.5
2.40 1.3 1.9 14#4 8.5
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.; M p= 1018.3 K.ft.
Strip Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0035bh (in. ) Bars selected #8 Bars per rib #8
Column strip Ext.(-) Int.(-) Pos. 509.12 1028.1 610.98 9.3 12.8 9.3 9.3 9.3 9.3 12 18 12 2 3 2
Middle strip Ext.(-) Int.(-) Pos. 0 342.64 407.32 9.3 9.3 9.3 9.3 9.3 9.3 12 12 12 0 1 2
Interior Slab: Mn = -1272.83 K.ft.; M p= 685.37 K.ft.
Strip Mu (k.ft.) 2 2 As = ρ bd (in. ) 2 Min. As = 0.0035bh (in. ) Bars selected #8 Spacing
Column strip negative -954.6 12 9.3 16 3
Middle strip Positive 411.22 9.3 9.3 12 2
55
Negative -318.23 9.3 9.3 12 1
Positive 274.15 9.3 9.3 12 1
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 2 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. ) 5.) Check shear (stairs): Vu = 2.4 K < φVc/2 = 3.45 k., no shear reinforcement required . 6.) Check shear at loading: Vud = 1.2 K < φVc/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: 2 Mu = 0.80 k.ft.; Use #3 bars @ 12 in. (As = 0.11 in. ) 3.) Since Vu = 0.567 K < φVc/2, no shear reinforcement is required. But it is recommended to use stirrups #3 @ 4 in. to hold the main reinforcement. 2 2 4.) Design of supporting beam: As = 1.5 in. , use 3#7 (As = 1.8 in. ) 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.
56
CHAPTER 19 INTRODUCTION TO PRESTRESSED CONCRETE Problem 19.1: 1. For the given section: σtop= 564 psi (Compression) ; σ 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 Elastic shortening Shrinkage Creep of concrete Relaxation of steel Total losses = Additional losses Total = Total F = 300.76 K Problem 19.3: Loss from Elastic shortening Shrinkage Creep of Concrete Relaxation of steel friction Total losses = Total F = 415 K.
stress ksi 12.3 8.4 15.29 4.24 40.23 3.39 43.62
Stress Ksi 8.52 5.60 8.02 4.20 13.2 39.54
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.
57
percentage 7.26 4.96 9.02 2.50 23.74 2.00 25.74
Percentage 4.87 3.36 4.80 2.50 7.93 23.46
Problem 19.5: M D = 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, η = 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 η = 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. < φMn. 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.; φMn = 1755.7 K.ft.
58
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] 5 1000 50 50000 0.333 110 4 1000 40 40000 0.267 88 3 1000 30 30000 0.2 66 2 1000 20 20000 0.133 44 1 1000 10 10000 0.067 22 150000 330 Problem 20.4: Location Mu (kft) Support -265.5 131 Midspan 82.8
2
As (in ) 3.6 1.8 1.2
Reinforcement 6No.7 3No.7 2No.7
Vx [k] 110 198 264 308 330
ØMn (kft)
322.6 168 113
Problem 20.5: 2 For 5 No.8 bars; As = 3.95 in ; 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: 2 Ash = 0.636 in : 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: 2 Use no.4 hoops and crossties; Smax = 6 in; Ash = 0.954 in : Use 5 No.4 crossties
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CHAPTER 21 BEAMS CURVED IN PLAN Problem 21.1:
Moment For the section at support, M u = 187 k-ft., A s = 3.11 in 2 For the section at midspan, M u = 94.52 k - ft. , A s = 1.47 in 2 Maximum torsional moment, T u = 18.90 k - ft. Shear at the point of maximum torsional Moment = 60.55 k Stirrups #4 @ 5 in c/c Longitudinal Bars, Al = 0.81 in.2 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 in. 2 , use 2 # 9 bars at the corner 2 c) At middepth, use 2 # 4 bars (0.4 in. ) Problem 21.2:
V A = 1.57 W u r = (1.57)(10.7)(6) = 100.8 k. M A = -W u r 2 = -(10.7)(36)2 = -385.2 k.-ft, A s = 3.3 in 2 (Top Steel) T A = -0.3 W u r 2 = -0.3(10.7)(36)2 = 115.56 k.-ft Mc = 0.273 W u r 2 = (0.273)(10.7)(36)2 = 105.2 k.-ft T C = V C = 0 # 4 Stirrups, 2 braches, s = 3.89 in., use #4 @ 3.5 in. Distribution bars: Bottom bars = 3.15/3 = 1.05 in. 2 , use 2 # 7 (1.21 in. 2 ) 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 in. 2 ) Problem 21.3: M C =134.11 k-ft, T C =0
M D = -58.20 k-ft, T D = 52.46 k-ft Problem 21.4:
Design for M A = -601.77 k-ft. A s = 4.2 in 2 V u = 103.6 k.ft., T u = 281.1 k.ft
# 4 Stirrups, 4 braches, Av = 2(0.2) = 0.4 in 2 , Spacing; s = 6.31 in., use #4 @ 6 in. Longitudinal reinforcement : Al = 6.19 in.2 Maximum spacing = P h / 8 = 108 / 8 = 13.5 in.
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