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; 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 / 6.10×10-4 = 3130 ksi.
Approximate Initial Modulus = 2.6(ksi.) / 5.45×10-4 = 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
a(in.)
14
22.5
4#10
5.08
6.4
18
28.5
6#10
7.62
7.47
12
23.5
4#9
4.00
5.88
12
18.5
4#8
3.16
4.65
16
24.5
5#10
6.35
7.00
14
26.5
5#9
5.00
6.3
10
17.5
3#9
3.00
5.29
20
31.5
4#9
4.00
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
b (in.)
d (in.)
As (in.2)
ρmax<ρ<ρmin
Problem 3.2
DIM.
No.a
b (in.)
15
d (in.)
22.5
8#9
As (in.2)
8.00
2#9
As’(in.2)
2.0
0.0237
ρ
0.0059
ρ’
0.0178
ρ- ρ’
k
0.0172
a ( in.)
7.06
692.2
φ Mn
A
B
C
c
a
fs’
C1
C2
As1
ρ1
ρ1≤ρmax
φ Mn
No.b
17
24.5
8#10
10.08
2#10
2.54
0.0242
0.0061
0.0181
0.0158
7.83
950
No.c
13
22
7#9
7.00
3#7
1.8
0.0245
0.0063
0.0182
0.0176
7.06
590.2
No.d
10
21.5
4#10
5.08
2#7
1.2
0.0236
0.0056
0.0180
0.0180
6.85
418.2
3
No.e
14
20.5
6#10
7.62
2#10
2.54
0.0266
0.0089
0.0177
0.0189
No.f
16
20.5
9#9
9.00
4#9
4.0
0.0274
0.0122
0.0152
0.0189
No.g
20
18
12#9
12.00
6#9
6.0
0.0333
0.0167
0.0167
0.0216
No.h
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
Problem 3.3
DIM.
b ( in.)
bw ( in.)
t ( in. )
d ( in. )
As (in.2)
a ( in. )
Rect. or T
φMn (k.ft)
Asmin< As < Asmax
Asf (in.2)
a ( in. )
φMn (k.ft)
Asmin< As < Asmax
Problem 3.4
DIM.
fc’
fy
b
d
As
(in.2)
(mm2)
β1
ρb
ρmax
Rumax
ρ
Ru
a/d ratio
(a/d) max
ρmin<ρ<ρmax
No.a
No.b
No.c
No.d
.No.e
No.f
No.g
No.h
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
No.a
3 ksi
40 ksi
12 in.
20 in.
4#8
No.b
4 ksi
60 ksi
12 in.
20 in.
4#7
No.c
4 ksi
75 ksi
12 in.
20 in.
4#9
No.d
5 ksi
60 ksi
12 in.
20 in.
4#9
No.e
30 Mpa
400 Mpa
300 mm
500 mm
3*30 mm
No.f
20 Mpa
300 Mpa
300 mm
500 mm
3*25 mm
No.g
30 Mpa
500 Mpa
300 mm
500 mm
4*25 mm
No.h
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
2121mm2
0.85
0.0325
0.0203
6.14
Mpa
0.01414
4.52
Mpa
0.2218
0.3184
Yes
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
4
Problem 3.5:
3.5 a:
As = 1256.6mm2 ; ρ = 0.00914 ; Ru = 3.6 Mpa ; φMn = 272.3 KN.m
Ok
ρb = 0.02024 ; ρmax = 0.6375ρb = 0.0129
3.5 b:
As = 1472 mm2 ; ρ = 0.0107 ; Ru = 3.51 Mpa ; φMn = 265.4 KN.m
ρb = 0.02024 ; ρmax = 0.6375ρb = 0.0129
ρ < ρmax
Ok
3.5 c:
As = 2827 mm2
ρ = 0.02056 ; ρmax = 0.0129 < ρ ; N.G. , Section does not meet ACI code, reduce steel.
Ru(max) = 4.1 Mpa ; φMn = 310 KN.m
3.5 d:
ρ = 0.0091 ; ρb = 0.0214 ; ρmax = 0.01356 ; ρmin < ρ < ρmax
Ok
Ru = 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.
Ru(max) = 0.615 Ksi ; φMn = 248.1 k.ft
Problem 3.6:
εy = 0.00172 > εs = 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 = Esεs = 43.5 ksi ; As = 4.783 in.2 ; φMn = 201.3 k.ft
ρmax = 0.01628 ; As max = 2.34 in.2 ; 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
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: 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 ; φMn = 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
φMn = 137.5 K.ft
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. ; fs’= 16.15 ksi ≤ fy
C1 = 79.85 k ; C2 = 16.32 k ; As1 = 1.996 in.2 ; 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: 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) 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. ; fs’= 53.23 ksi < fy
C1 = 186 k ; C2 = 30.4 k ; As1 = 3.1 in.2; 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
(3) φMn = 236.4 k.ft ; Asmin = 0.6 in2 < As< Asmax = 13.23 in.2
Problem 3.14:
(1) a = 3.69 in. > t ; T section ; Asmax = 4.99 in.2 > As
(2) Asf = 2.55 in.2 ; 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
Ru psi
589.9
631
780
700
799
660.7
818
708
ρ ( %)
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
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
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
7
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. 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.
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
4.606
5.56
Mu1(k.ft)
402
656.3 280.6 344.6 459.5 595.3 374.2 223.2 297.7 223.2 201.1
293.0
Mu2(k.ft)
152
133.7 167.4 175.4 305.5 259.7 180.8 76.76 102.3 56.76 88.86 107.02
As1(in2)
5.44
7.37
3.996 4.428 5.904 7.128 5.328 3.564
4.75
3.576 3.654
4.41
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 (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
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
2
As2(in )
2
h (in)
24
8
2.959 2.511
5.24
1.22
5.241
5.26
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
Ru (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
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
ρw(%)
1.0
1.07
ASw (in2)
1.92
2.25
AS (in2)
6.0
6.33
ρ (%)
2
c/ dt ≤ 0.375
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
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
452
5#10
one
280
6#9
two
7#9
two
R or_T
Ru (psi)
Bars
rows
8#9
two
6#10
two
6#9
two
7#9
two
379
6#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
ρw(%)
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
10
Problem 4.4:
(a) ρ = ρmax , 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) ρ=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) ρ= 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 ; ρ=0.0146 ; As = 3.5 in 2 ; Use 3# 10 bars (As = 3.79 in2 )
Problem 4.6:
ρ = 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 = 20×12 / 15 =16 in.
Mu = 132 k.ft
(a) Using ρmax = 0.01806, Ru(max) = 820 psi ; d = 12.7 in. ; AS = 2.75 in2 ; Use 3#9 ;
h =16 in ; ρmin < ρ = ρmax , εt = 0.005
(b) Using ρs = 0.014, Ru = 662 psi ; d = 14.2 in. ; AS = 2.38 in2 ; Use 4#7 bars (As = 2.41 in2 ).
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, 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
ρ = 0.00674; ρmin < ρ < ρmax , εt > 0.005 ; AS = 1.74 in2 ; Use 2#9 bars or 3#7
Problem 4.9:
Mu = 317 k.ft ; Ru = 622 psi, ρmin < ρ < ρmax , εt > 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, ρmin < ρ < ρmax , εt > 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:
φMn = 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
OK
a = 6.08 in. ; c = 7.15 in.; dt = 20.5 in.; εt =0.0056 > 0.005;
A S' = 0.98 in.2 ; AS(max)= 5.27 in2 > As = 4.02 in2 ; h = 23 in.
OK.
11
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.; εt =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.
OK
AS(min) = 0.86 in.2 < AS < AS(max) = 7.78 in.2
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.
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 ; H A = H D = 34.24 k ; R A = R D = 154.08 k
M B = M C = - 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)= ρ 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 )
2
A S' = 1.99 in.
Use 2 # 9 in one row, ( A S = 2 in2 )
12
CHAPTER 5
ALTERNATE DESIGN METHODS
Extra Problems
Problem E5.1
Tie AB Fu = 58.75kips Ats = 1.31in 2 Provide 5 No. 5 Bar Ats = 1.55in 2
Tie CD Fu = 15.0kips Ats = 0.33in 2 Provide 1 No. 4 Bar, 2 legs Ats = 0.40in 2
Tie BD & DF Fu = 68.21kips Ats = 1.52in 2 Provide 5 No. 5 Bar Ats = 1.55in 2
Calculate strut widths
β s = 0.75 φf ce = 0.75 × 3825 = 2868 psi
Strut AC, Pu = 86.83kips , w = 1.26in
Strut BC, Pu = 80.93kips , w = 1.18in
Strut CE, Pu = 130.67 kips , w = 1.90in
Strut DE, Pu = 16 kips , 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
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
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
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
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
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
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
Icr (in4.)
16
Problem 6.3:
1.) d = 23.27 in. ; As = 5.0 in.2
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.
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.
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)
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
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
ψt 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
fc '
<100
ld (in.)
Problem 7.2
(a): ldc = 21.9 in. ≥ 18 in.; Use ld = 22 in.
(b): ldc = 21.4 in. ≥ 20.3 in.; Use ld = 22 in.
(c): ldc = 16.06 in. ≥15.24 in.,use ld = 17 in.
(d): ldc = 23.93 in. ≤ 25.38 in., ld = 0.8(25.38) = 20.3 in., use 21 in.
(e): ldc = 13.55 in. ≤ 15.75 in., ld = 0.9(15.75) = 14.12 in., use 15 in.
(f): ldc = 19.1 in. ≤ 20.25 in., ld = 0.75(20.25) = 15.2 in., use 16 in.
18
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
19
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 M u = 574 K.ft.
Maximum negative moment M u = 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
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. Vu = 83.25 kip; Vu @ d = 64.29 kip
2. φVc = 0.75×35.93 = 26.95 kip < Vu @ d.; Vs = 49.79 kip < Vc1 ,
3. Use #3 @ 6 in.
4. φVs = 20.3 k.
X = 6.29 ft. = 75 in.
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
φVc = 23.66 kip < Vu @ d; Vs = 35.65 kip
3. Use #3 @ 7 in.
4. For Smax = 12 in., φVs = 13.2 in.
X at φVc/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, Vu @ d = 68.25 kip
2. φVc = 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.; φVs = 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; φVc = 18.75 kip; Vs = 36.69 kip
3. Use #3 @ 5.5 in.
4. Vu = 13.2 k. < φVc = 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):
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; φVc = 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. φVc = 53.89 k., φVs = 21.4 k.
Problem 8.10)
1. Wu = 7.2 k/ft.; Vu = 64.8 kip; Vu @ d = 50.4 kip
2. φVc = 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.; φVs = 19.8 k.; X = 88 in. at φVc /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.
22
Problem 8.11)
1. Vu at support = 66 kip; Vu @C.L. = 12 kip; Vu @ d = 55.4 kip
2. φVc = 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.
23
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
φMn (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
ρ (%)
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
7
10
8
8
10
Spacing (in.)
10
8
12
12
8
* 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.; W L = 0.258 K/ft. = 258 psf.
Problem 9.4:
1. M u = 170.2 K.ft.
2. d = 8 in.; A s = 1.44 in. 2 ;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.
5. Use #3 bars spaced at 6 in. (A s = 0.22 in.2 ) for shear
Problem 9.5:
1. Mu = 6.41 K.ft. /ft.; Let d = 5 in.
2. A s = 0.36 in. 2 ;Use # 4 bars spaced at 6 in. (A s = 0.39 in. 2 )
3. Shrinkage reinforcement: A s = 0.144 in. 2 ;Use #3 bars spaced at 9 in. (A s = 0.15
in. 2 )
Problem 9.6:
1. M u (at support) = 42.2 K.ft.; A s = 1.44 in. 2
2. Check section at midspan:
External M u = 18.55 K.ft.
Internal moment = Moment capacity = M u = 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.
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
Moment
Mu
K.ft.
(See example 9.5)
Coeff.
A
- 1/24
2.42
B
+ 1/14
4.15
C
- 1/10
5.81
D
- 1/11
5.29
E
+ 1/16
3.63
5. Shear is adequate.
Ru
psi
156
267
374
341
234
25
ρ%
As(in.2)
0.45
0.80
1.15
1.03
0.7
0.21
0.38
0.54
0.49
0.33
# 5 bars
spaced at (in.)
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
Ru
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
K.ft.
(See example 9.5)
3.
A
B
C
D
E
Shear is adequate.
2.42
4.15
5.81
5.29
3.63
Ru
Psi
ρ%
As
(in. 2 )
#4 bars
spacing (in.)
151
259
363
331
227
0.33
0.50
0.73
0.66
0.45
0.16
0.24
0.36
0.33
0.22
12
9
6
6
10
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.
M u = 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, A s = 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.
(As = 0.62 in. 2)
3. Mu (flange) = 1541 K.in.; Choose 2#5 bars
4. Shear in rib; Vu (at d distance) = 5771 lbs. < Vc
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.
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 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.
Ru
psi
ρ%
As
(in) 2
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)
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
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
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)
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
ρg (%)
5.2
3.77
3.97
4.84
4.53
φPn(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
28
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
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 Pb (k.), φ Mb (k-ft.)
φ Pb = 372, φ Mb = 541
Prob. 11-1-d
453.5, 781.2
Prob. 11-1-e
197.9, 252.12
Prob.11-1-i
390.4, 344.5
Prob. 11-1-f
230.1, 310.3
Prob. 11-1-j
993.2, 496.6
Prob. 11-2-i
Refer to11-1-i
551.0, 275.5
384.6, 192.3
162, 146.7
550.9, 922.3
Prob. 11-1-g
Prob. 11-1-h
351.1, 365.9
Prob. 11-1-k
394.1, 486.7
Prob. 11-2-e
Refer to 11-1-e
Prob. 11-1-c
264.1, 330.9
485.2, 679.2
11-2-a to 11-2-l. Compression controls:
e = 6 in
Rectangular sections
Prob. 11-2-a
Compression controls
refer to prob. 11-1-a
φPn (k.), φMn (k.-ft.)
φPn = 775.4, φMn = 387.7
Prob. 11-2-d
Refer to 11-1-d
Prob. 11-1-b
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. 11-1-l
347.2, 370.4
(12 Problems); φ = 0.65
Prob. 11-2-b
Prob. 11-2-c
Refer to 11-1-b Refer to 11-1-c
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
Prob. 11-2-k
Refer to 11-1-k
994.6, 497.32
604.6, 302.3
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
φPb (k.), φMb (k-ft)
φPb = 258.8, φMb = 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: cb = 12.72in, ab = 10.82in.
Cc = 588.38K; Cs1 = 286.4K; Cs2 = 58.94K
Tension zone: φ = 0.65; ε s2 = 5.75×10-4, 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: φ = 0.65; ε s = 9.26×10-4, 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: εs4 = 9.82×10-4, 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: εs4 = 2.3×10-4, fs = 6.68Ksi, εs5 = 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: ε 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: ε s4 = 9.5×10-4, 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: ε s4 = 1.43×10-4, fs = 4.19Ksi, ε s5 = 0.001, fs = 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.
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, φ Pn = 443.8K > Pu, O.K.
e: As = As/ = 7.59 in2, use 6#10
Pn = 1 753.93 K
f: As = As/ = 0.03×18×24/2 = 6.48in.2, Use 5#10
Pn = 1114.26 K, φ Pn = 724.3 K > Pn O.K.
g: As = As/ = 0.02×14×20/2 = 2.8in.2, Use 3#9
Pn = 453.12K, φ Pn = 317.2K; O.K. section is adequate.
h: As = As/ = 6.57in.2, use 6#10
Pn = 1617.7K, φ Pn = 1051.5 K > Pu, O.K. Section is adequate.
i: As = As/ = 0.02×20×14/2 = 2.8in.2, Use 2#10
Pn = 923.93K, φ Pn = 600.5 K > Pu ; O.K. section is adequate.
j: As = As/ = 2.64in2, 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.
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, φ Pb = 43.62K
Mb = 86.9K.ft, φ Mb = 56.5K.ft, e = 15.54in.
Problem 11.11:
Pb = 222.33K, φ Pb = 144.5 K
Mb = 229.61K.ft; φ Mb = 149.3K.ft, eb = 10.5 in.
Problem 11.12:
Pb = 427.85K, φ Pb = 278.1K
Mb = 420K.ft; φ Mb = 273K.ft, eb = 11.78in.
32
Problem 11.13:
Pb = 677.6K, φ Pb = 440.4K
Mb = 655.7K.ft, φ Mb = 425.8K.ft, eb = 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: 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, φ 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, φ 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., φPn = 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., 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
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
38
Problem 13.2: Square Footings.
2
AF (ft. )
h (in.)
d (in.)
L (ft.)
Pu (k.)
qu (ksf.)
Vu1 (k.)
Vu2 (k.)
Mu (k.ft.)
As (in.2)
Bars
ld (in.)
ldav (in.)
N1 (k.)
x
Ads (in.2)
Ldd (in.)
(a)
61.38
20
15.5
8
364
5.7
23.1
325.5
253.3
3.75
(b)
47.9
19
14.5
7
352
7.2
77.7
300
190.6
3.0
(c)
77.1
23
18.5
9
548.4
6.8
130.1
480.8
411.4
5.1
(d)
77.5
24
19.5
9
440
5.4
115.4
400.2
388.8
4.6
(e)
69.1
21
16.5
8.5
424
5.9
114.9
388.2
337.2
4.7
(f)
95.4
21
16.5
10
452
4.5
113.1
417
422.5
5.9
(g)
61.5
20
15.5
8
432
6.8
106.5
382.2
287.3
4.3
(h)
95.1
22
17.5
10
546
5.5
148.96
426.3
477.4
6.24
(i)
46.6
16
11.5
7
280
5.7
68.2
240.0
141.9
2.9
(j)
60.4
18
13.5
8
276
4.3
75.97
249.2
191.1
3.3
10#6
33
37
424.3
8#6
33
30
537
13#6
33
41
663
12#6
33
45
238.6
12#6
29
41
433.2
11#7
42
49
565.5
11#6
29
36
716
9#8
48
47
884
8#6
29
30
696.4
9#6
33
38
334.2
4
1.28
22
4
1.62
22
4
2.0
24.7
7
5.54
22
4
0.98
21.4
4
1.28
21.4
4
1.62
19
4
2.0
21.4
4
1.57
21.4
4
1.0
22
39
Problem 13.3: Rectangular Footings:
2
AF (ft. )
L (ft.)
B (ft.)
h (in.)
d (in.)
Pu (k.)
qu (ksf.)
Vu1 (k.)
Vu2 (k.)
MuL (k.ft.)
MuS (k.ft.)
AsL (in.2)
Bars
AsS (in.2)
Bars
AsB (in.2)
Bars
AsE (in.2)
Bars
ldL (in.)
ldaL (in.)
lds (in.)
ldas (in.)
N1 (k.)
Ads (in.2))
x
ldd (in.)
(a)
61.4
10.5
6
23
18.5
364
5.78
105.5
316.4
363.7
165.2
4.52
9#7
5.22
12#6
3.79
9#6
1.8
4#6
48
52
28
25*
424.3
1.28
4
22
(b)
47.9
8
6
20
15.5
352
7.33
56.2
294.7
232.3
148.4
3.47
10#6
3.46
8#6
2.98
7#6
0.87
4#6
33
36
33
24*
537
1.62
4
25
(c)
77.6
10
8
23
18.5
548.4
6.86
111.5
478.2
476.4
344
5.92
11#7
4.95
11#6
4.42
10#6
0.8
2#6
48
47
33
35
663
2.0
4
25
(d)
78
10
8
24
19.5
440
5.5
103.1
402.1
445.5
336.9
5.25
10#7
5.18
12#6
4.61
11#6
0.84
2#6
48
51
33
39
238.6
5.55
7
22
40
(e)
69.6
10
7
21
16.5
424
6.06
93.4
385.1
413.7
257.8
5.79
9#8
4.54
8#7
3.7
6#7
1.09
2#7
48
50
42
32*
433.2
1.0
4
22
(f)
96.7
12.5
8
21
16.5
452
4.52
110.6
418.8
563.62
313.9
7.9
9#9
5.7
10#7
4.44
8#7
1.6
4#7
54
64
42
37*
565.5
1.28
4
22
(g)
61.5
10.5
6
23
18.5
432
6.86
51
368.7
416.7
182.3
5.2
10#7
5.22
9#7
3.76
7#7
1.77
4#7
42
51
42
24*
716
1.62
4
19
(h)
95
11
9
22
17.5
546
5.52
134.1
492.6
540.96
408.2
7.12
8#9
5.23
10#7
5.28
9#7
0.76
2#7
54
53
42
41
884
2.0
4
22
Problem 13.4: Rectangular footings.
2
AF (ft. )
L (ft.)
B (ft.)
h (in.)
d (in.)
Pu (k.)
qu (ksf.)
Vu1 (k.)
Vu2 (k.)
MuL (k.ft.)
MuS (k.ft.)
AsL (in.2)
Bars
AsS (in.2)
Bars
AsB (in.2)
Bars
AsE (in.2)
Bars
ldL (in.)
ldaL (in.)
lds (in.)
ldas (in.)
N1 (k.)
Ads (in.2))
x
ldd (in.)
(a)
(b)
(c)
(d)
61.4
9.5
6.5
20
15.5
364
5.9
110.26
321.05
294.15
199.29
4.43
9#7
4.10
11#6
3.59
8.#6
1.02
4#6
48
44*
33
29*
464.3
1.4
4
22
47.9
8.5
5.75
21
16.5
352
7.2
91.43
292.39
241.64
149.23
3.38
9#6
3.86
9#6
3.2
8#6
1.0
4#6
33
38
33
24*
530.2
1.6
4
25
77.6
11
7.25
26
21.5
548.4
6.88
151.72
466.4
505.04
331.17
5.44
10#7
6.18
10#7
4.9
8#7
1.77
4#7
48
51
48
33*
636.1
1.92
4
25
78
11
7.25
24
19.5
440
5.52
135.07
394.6
451.48
296.48
5.42
10#7
5.70
10#7
4.49
8#7
1.57
4#6
48
54
48
35*
357.5
2.49
4
22
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
22.8
17
22
19
29.6
31
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
152
4913
129
6859
124
29791
145
178
1.9
178
0.46
178
0
178
2.71
18.75
16.76
21.85
35.66
Pu (k.)
ft (psi.)
fta (psi.)
Vu (k.)
φVc (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 = 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.
42
Problem 13.7:
Footing No.
D.L. (K)
L.L. (K)
L.L. / D.L.
Usual Load (K)
Area = (Usual Load/2.81)
Max. Soil Pressure (Ksf)
1
130
160
1.23
170
60.5
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 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) P u = 496 K; q u = 4.51 Ksf.
Max. moment in the long direction, M u = 640.62 K.ft.
Max. moment in the short direction, M u = 502.30 K.ft.
3. V u = 167.25K
4. Two way shear: b0 = 142 in.; A0 = 1244.25 in.2; V u = 457.03K.
5. Reinforcement in the long direction:
Use 9 #10 (11.39in. 2 ) spaced at 14.25in., provided l d > required 32 in.
6. Reinforcement in the short direction:
Use 9 #9 (9 in. 2 ) spaced at 15.75 in., provided l d > 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.; V u = 190.3 K.
4. bo = 154 in . ; q at center of column = 3.88 Ksf.; V u = 456.5 K.
5. Reinforcement in the long direction:
Choose 15 #9 bars (A s = 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)
43
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 = 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 Cp = 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
ΣMb = 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
ARM
(K)
(FT)
RESULTANT
12.604
5.72
MOMENT
(K. FT)
72.139
EARTH PRESSURE FORCES AND MOMENTS:
FORCE
ARM
(K)
(FT)
TOTAL
4.307
5.67
MOMENT
(K. FT)
24.409
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
=
=
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°, C a = 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
6.00
W 3 = 3.6
W 4 = 13.44
8.50
ΣW = R = 20.64
Factor of safety against overturning = 2.3 > 2.0
2. Check sliding condition: F = 9.29; F.S. = 1.31 < 1.5
F.S. (sliding) = 1.56 > 1.5
3. q 1 =0.83 = 2 .55 ksf; q 2 = 0.89 ksf
F = 1 1 . 3 1 K ; F . S . ( s l i d i n g ) = 1.6 > 1.5
Moment (k.ft.)
10.8
4.4
21.6
114.2
ΣM = 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.
Hw = 4.875 K.
pw (water) = 0.78 Ksf.,
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)
46
CHAPTER 15
DESIGN FOR TORSION
Problem 15.1:
φTcr = 349.8 K .in.
Problem 15.2:
φTcr = 170.8 K .in.
Problem 15.3:
Acp = 428in.2 ;Pcp = 138 in.
φTn = 63 K .in.
φTcr = 251.9 K .in.
Problem 15.4:
Acp = 608in.2, Pcp = 156in.
φTcr = 449.6 K .in.
Problem 15.5:
Acp = 336 in.2; Pcp = 76in.2
If flanges are included, then φTcr = 390.5 k.ft and φTn = 97.7 k.ft.
Problem 15.6:
Acp = 504in.2, Pcp = 132 in.
Problem 15.7:
φ Vc = 25.9 K
Vu = 36 f
φVc
2
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:
φ Vc = 22.2 K < Vu shear reinforcement required.
Acp = 264 in2, Pcp = 68 in, Ta = 48.6K.in; Ta<Tu torsional reinforcement required.
Design for shear: Av/S = 0.043 in2/in (2legs); Av/2S = 0.043/2 = 0.0215 in2/in (leg)
Design for torsion: Assume 1.5in cover and use # 4 stirrups @ S = 4 in.
LHS = 463Psi, RHS = 474.3Psi > LHS O.K., Tn = 300/0.75 = 400 K.in
Distribution of longitudinal bars: Total = 1.35in2, 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:
Av/S = 50.4/(60×19.5) = 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<Muft = 11181 K.in., Then rectangular section
Use 4#9 bars; use one raw of bars
Vc = 31Kips; Vu = 28K> φ 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<Muft then rectangular section; Use 5#9
Vc = 29.22K; Vu = 36> φ 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<Muft = 8386 K.in. then rectangular section.
Use 4#9 bars in one row, bmin = 11.6in
Vc = 26.8K; Vu = 28K > φ 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
48
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; φ 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
φ Tcr = 226K.in < Tu, Use Tu = 266K.in; At/S = 0.014in2 (one leg); Al = 0.014×120 = 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
φ 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,
φVc = φ 2 f c 'bd = 26.6 k .
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 < φ 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.
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
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: Δ = 0.06 in.; Δ / 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
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
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 P n = 230,9 K > 186.1 required
b) Section at midheight of column: Provided P n = 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.
50
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: 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 (+)
-311
-311
-302.4
-302.4
52.8
52.8
-613.4*
-613.4*
-258.2
-258.2
276.6
121.0
276.6
-79.2
-158.4
-79.2
280.8
201.6
280.8
197.4
-37.4
197.4
557.4*
322.6*
557.4
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
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
As = ρbd (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.≤2hs≤18 in.
6.7
13
13
13
Problem 17.2(b.):
Strip
Moment sign
Mu (k.ft.)
As = ρbd (in.2)
Min. As = 0.0018bhs (in.2)
Straight bars
Spacing = b/No.≤2hs ≤18 in.
Column strip
Neg.
Pos.
- 317.8
136.9
7.02
5.05
3.11
3.11
12#7
12#6
12
12
52
Middle Strip
Neg.
Pos.
- 105.9
91.3
5.05
5.05
3.11
3.11
12#6
12#6
12
12
Problem 17.2(c):
Strip
Moment sign
Mu (k.ft.)
As = ρbd (in.2)
Min. As = 0.0018bhs (in.2)
Straight bars
Spacing = b/No.≤2hs ≤18 in.
Problem 17.3(a):
Long Direction
Mu (k.ft.)
As = ρbd2 (in.2)
Min. As = 0.0018bhs (in.2)
Straight bars
Spacing = b/No.≤2hs = 18 in.
Problem 17.3(b):
Long Direction
Mu (k.ft.)
As = ρbd2 (in.2)
Min. As = 0.0018bhs (in.2)
Straight bars
Spacing = b/No.≤2hs = 18 in.
Column strip
Neg.
Pos.
- 411
177
8.49
5.66
3.51
3.51
14#7
13#6
11
12
Column strip
Ext.(a) Pos.(b)
- 80
96
2.80
3.35
1.73
1.73
9#5
12#5
13
12
Middle Strip
Neg.
Pos.
- 137
118
5.66
5.66
3.51
3.51
13#6
13#6
12
12
Int.(c)
161.6
5.90
1.73
10#7
12
Middle strip
Ext.(a) Pos.(b)
0
64
0
2.57
1.73
1.73
8#5
9#5
15
13
Int.(c)
- 53.9
2.57
1.73
9#5
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)
0
135.61
0
5.2
3.11
3.11
10#6
12#6
14.4
12
Int.(c)
114.09
5.2
3.11
12#6
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.)
As = ρbd2 (in.2)
Min. As = 0.0018bhs (in.2)
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
3.11
3.11
3.11
14#7
11#5
11#5
11#5
Problem 17.6(b):
Mo = 572 k.ft.
Mu (k.ft.)
As = ρbd2 (in.2)
Min. As = 0.0018bhs (in.2)
Straight bars
Ext.
149
4.04
2.59
12#7
Long & Short Direction
Column strip
Middle strip
Pos.
Int.
Ext.
Pos.
178.5
300
119
8.3
3.1
4.2
4.89
2.59
2.59
2.07
2.07
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
Middle strip
Neg.
Pos.
-45.8
24.7
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
Mu (k.ft.)
As = ρbd (in.2)
Problem 17.8(a):
Strip
Mu (k.ft.)
As = ρbd2 (in.2)
Min. As = 0.0018bhs (in.2)
As (ρ min.=0.0033)
Bars selected
Spacing ≤15 in.
Column strip
Ext.(-)
Pos.
Int.(-)
-5.62
18.1
-22.2
low
low
low
1.3
1.3
1.3
1.9
1.9
1.9
10#4
10#4
10#4
12
12
12
54
Middle strip
Ext.(-)
Pos.
Int.(-)
-7.7
40.2
49.4
low
2.10
2.40
1.3
1.3
1.3
1.9
1.9
1.9
10#4
14#4
14#4
12
8.5
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.; Mp= 1018.3 K.ft.
Strip
Mu (k.ft.)
As = ρbd2 (in.2)
Min. As = 0.0035bh (in.2)
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.; Mp= 685.37 K.ft.
Strip
Mu (k.ft.)
As = ρbd2 (in.2)
Min. As = 0.0035bh (in.2)
Bars selected #8
Spacing
Column strip
negative
-954.6
12
9.3
16
3
Middle strip
Positive Negative
411.22 -318.23
9.3
9.3
9.3
9.3
12
12
2
1
55
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
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 < 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:
Mu = 0.80 k.ft.; Use #3 bars @ 12 in. (As = 0.11 in.2)
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.
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.
56
CHAPTER 19
INTRODUCTION TO PRESTRESSED CONCRETE
Problem 19.1:
1. For the given section: σtop= 564 psi (Compression) ; σbottom= +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 . i n . , 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
As (in2)
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:
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
59
CHAPTER 21
BEAMS CURVED IN PLAN
Problem 21.1:
Moment
For the section at support, Mu = 187 k-ft., As = 3.11 in 2
For the section at midspan, M u = 94.52 k - ft. , As = 1.47 in 2
Maximum torsional moment, Tu = 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
c) At middepth, use 2 # 4 bars (0.4 in.2)
Problem 21.2:
VA = 1.57 Wu r = (1.57)(10.7)(6) = 100.8 k.
M A = -Wu r2 = -(10.7)(36)2 = -385.2 k.-ft, As = 3.3 in 2 (Top Steel)
T A = -0.3 Wu r2 = -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
TC = VC = 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, TC =0
M D = -58.20 k-ft, TD = 52.46 k-ft
Problem 21.4:
Design for M A = -601.77 k-ft. As = 4.2 in 2
Vu = 103.6 k.ft., Tu = 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 = Ph / 8 = 108 / 8 = 13.5 in.
60
Distribution of bars: Bottom bars = 4 # 7 ( As = 2.41 in. 2 ), Middepth bars , use 2 # 9 bars, Top
bars = 6.3 in.2, use 5 # 10 bars in one row ( As = 6.33 in. 2 )
Problem 21.5:
M C = 55.09 k-ft , As = 0.0033 × 12 × 21.5 = 0.86 in 2
M A = - 296.91 k-ft., As = 0.0150 × 12 × 21.5 = 4.2 in 2
TA = 31.81 k-ft
TC = 31.81 k-ft
Use # 4 Closed Stirrups, s = 0.2/0.0453 = 4.4 in., use #4 @ 4 in.
Longitudinal reinforcement : Al = 1.47 in.2 , Al / 3 = 0.50 in.2
For section at support;
Bottom bars, As = 0.5 in. 2 , use 2 # 5 bars ( As = 0.61 in. 2 )
Middepth bars, As = 0.5 in. 2 , use 2 # 5 bars
Top bars = 3.87 + 0.5 = 4.37 in.2, use 6 #8 bars in two row ( As = 4.71 in. 2 )
For section at mid-span
Bottom bars, = 0.86 + 0.5 = 1.36 in.2, use 2 #8 bars ( As = 1.57 in. 2 )
Middepth bars, use 2 # 5 bars
Top bars, use 2 # 5 bars
61