reinforced concrete manual
DESCRIPTION
TRANSCRIPT
INSTRUCTOR’S SOLUTIONS MANUAL
REINFORCEDCONCRETEA FUNDAMENTAL APPROACH
S I X T H E D I T I O N
EDWARD G. NAWY
Pearson Education International
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10 9 8 7 6 5 4 3 2 1
ISBN-13: 978-0-13-136170-6ISBN-10: 0-13-136170-8
CONTENTS
Please note that there are no solutions for Chapters 1 through 4. Solutions begin with Chapter 5.
Chapter 5 Flexure in Beams, 1–41
Chapter 6 Shear and Diagonal Tension in Beams, 42–82
Chapter 7 Torsion, 83–111
Chapter 8 Serviceability of Beams and One-Way Slabs, 112–143
Chapter 9 Combined Compression and Bending: Columns, 144–205
Chapter 10 Bond Development of Reinforcing Bars, 206–221
Chapter 11 Design of Two-Way Slabs and Plates, 222–262
Chapter 12 Footings, 263–281
Chapter 13 Continuous Reinforced Concrete Structures, 282–312
Chapter 14 Introduction to Prestressed Concrete, 313–329
Chapter 15 LRFD AASHTO Design of Concrete Bridge Structures, 330–368
Chapter 16 Seismic Design of Concrete Structures, 369–395
Chapter 17 Strength Design of Masonry Structures, 396–421
iiiv
5.1. For the beam cross-section shown in Fig. 5.33 determine whether the failure of the beam will be initiated bycrushing of concrete or yielding of steel. Given:
Also determine whether the section satisfies ACI Code requirements.
fy � 60,000 psi 1414 MPa2
f ¿c � 6000 psi 141.4 MPa2 for case 1b2, As � 6 in.2 f ¿c � 3000 psi 120.7 MPa2 for case 1a2, As � 1 in.2
1
Figure 5.33
Solution:
(a) The following information is given:
b = 8 in. section width
d = 18 in. section depth
dt = 16 in. depth to reinforcement
cf = 3000 psi required compression strength
fy = 60,000 psi steel strength
As = 1 in2 steel area
First, determine the value for 1 using equation 5.9.
1 = 0.85 ( )40002500 < cf
Then calculate the depth of the compression block.
a = bf
fA
c
ys
85.0
= )8)(3000(85.0
)000,60)(1(
= 2.94 in
1
2
Calculate the depth to the neutral axis using 1 and a.
c = 1
a
= 85.0
94.2
= 3.46 in
Then find the ratio of c and dt.
td
c =
16
46.3
= 0.216
Since this value is less than 0.375, the flexure is tension controlled and the steel yields before the
concrete crushes.
To determine if the section meets ACI Code requirements, calculate the reinforcement ratio.
= bd
As
= )16)(8(
1
= 0.0078
This value must be greater than the larger of y
c
f
f3 and
yf
200.
000,60
30003 = 0.0027
000,60
200 = 0.0033
Since 0.0078 > 0.0033, the section satisfies the ACI Code.
(b) The following information is given:
b = 8 in. section width
d = 18 in. section depth
dt = 16 in. depth to reinforcement
cf = 6000 psi required compression strength
fy = 60,000 psi steel strength
As = 6 in.2 steel area
2
3
First, determine the value for 1 using equation 5.9.
1 = 1000
400005.085.0 cf
, ( )80004000 < cf
= 1000
4000600005.085.0
= 0.75
Then calculate the depth of the compression block.
a = bf
fA
c
ys
85.0
= )8)(6000(85.0
)000,60)(6(
= 8.82 in
Calculate the depth to the neutral axis using 1 and a.
c = 1
a
= 75.0
82.8
= 11.76 in
Then find the ratio of c and dt.
td
c =
16
76.11
= 0.735
Since this value is greater than 0.6, the flexure is compression controlled and the concrete crushes
before the steel yields.
3
4
5
6
7
5.4. Design a one-way slab to carry a live load of 100 psf and an external dead load of 50 psf. The slab is simplysupported over a span of 12 ft. Given:
fy � 60,000 psi 1414 MPa2
f ¿c � 4000 psi 127.6 MPa2, normal-weight concrete
4
Solution:
Design as a 1 ft wide, singly reinforced section.
The minimum depth for deflection is L
20= 12 12
20= 7.2 in, so try a depth of 8 in. Assume for
flexure an effective depth d = 7 in. Calculate the self weight.
self weight of a 12 in. strip = 150144
128
= 100 lb/ft
Then calculate the factored load.
factored external load wu = )load live(6.1)load dead weightself(2.1 ++
= )100(6.1)50100(2.1 ++
= 340 lb/ft
factored external moment Mu = 8
)( 2Lwu
= 8
)12)(340( 2
= 6120 ft-lb = 73,440 in-lb
The required nominal strength for the slab is Mn = 73,440
0.9= 81,600 in-lb .
Assume a moment arm of in 3.679.090.0 ==d and calculate the area of steel per 12 in. strip.
nM = )armmoment (ys fA
sA = )3.6)(000,60(
600,81
= 0.1889 in2
Make sure the area is large enough to meet the minimum reinforcement ratio.
8
5
1 = 0031.0000,60
400033==
y
c
f
f
2 = 0033.0000,60
200200 ==yf
min sA = 2in 28.0)7)(12)(0033.0( =
Determine the value for 1.
1 = 0.85 ( )40002500 < cf
Calculate the depth of the compression block.
a = bf
fA
c
ys
85.0
= )12)(4000(85.0
)000,60)(28.0(
= 0.4118 in
Calculate the depth to the neutral axis using 1 and a.
c = 1
a
= 85.0
4118.0
= 0.4844 in
Then find the ratio of c and dt.
td
c =
7
4844.0
= 0.0692
Verify the strength is sufficient (at least 73,440 in-lb)
nM = 2
adfA ys
= 2
4118.07)000,60)(28.0(
= 114,141 in-lb
9
6
The strength is sufficient, so accept the design. Now design the shrinkage and temperature
reinforcement.
The minimum required steel fraction is 0.0018.
Area = )8)(12(0018.0
= 0.1728 in2
The maximum spacing is the smaller of 5 times the depth and 18 inches. In this case, 5 times the
depth is 40)8(5 = , so the maximum spacing is 18 in.
For a slab with a depth of 8 inches and using #4 bars, the maximum spacing is in 57.8)12(28.0
20.0 =
c-c for the main reinforcement and in 89.13)12(1728.0
20.0 = c-c (for the shrinkage and temperature
reinforcement).
10
5.5. Design the simply supported beams shown in Fig. 5.36 as rectangular sections. Given:
fy � 60,000 psi 1414 MPa2
f ¿c � 6000 psi 141.4 MPa2, normal-weight concrete
7
span = 20 ft (a) Distributed dead load (including self weight) = 600 lb/ft. Distributed live load of
1500 lb/ft. (b) Point load at mid span of 15,000 lb
(c) Point loads at 5 and 15 ft of 7500 lb
Solution:
(a) Calculate the factored load.
factored external load wu = LD WW 6.12.1 +
= )1500(6.1)600(2.1 +
= 3120 lb/ft
factored external moment Mu = 8
)( 2Lwu
= 8
)20)(3120( 2
= 156,000 ft-lb = 1,872,000 in-lb
The required nominal strength for the beam is Mn = 1,872,000
0.9= 2,080,000 in-lb .
Figure 5.36
(a)
(b) (c)
1500 21.9
6008.7
15,000(66.7 kN)
7500 750033.4 33.4
11
8
Make sure the steel area meets the required minimum reinforcement ratio.
1 = 3 f c
fy
= 3 6000
60,000= 0.0039
2 = 0033.0000,60
200200 ==yf
min sA = 2in 488.0)14)(9)(0039.0( =
The code is satisfied because 2.58 in2 > 0.488 in2. The area can be provided by 3 #9 bars (the area
is 3(1.0) = 3.0 in2). Now check the design using the actual steel area.
a = bf
fA
c
ys
85.0
= )9)(6000(85.0
)000,60)(3(
= 3.92 in
Calculate the depth to the neutral axis using 1 and a.
c = 1
a
= 75.0
92.3
= 5.23 in
Then find the ratio of c and d.
d
c =
15
23.5
= 0.348
Determine the minimum depth from the ACI Code: min h = L
16= 20(12)
16=15 in. Try h = 18 in.,
b = 0.5 h = 9 in., and d = 15 in. Assume c/d = 0.30.
Calculate the required steel area.
c = (0.30)(15) = 4.5 in.
1 = 1000
4000600005.085.0 = 0.75
a = 1 c = (0.75)(4.5) = 3.375 in.
Areq = y
c
f
baf)85.0( =
000,60
)375.3)(9)(6000)(85.0( = 2.58 in2
12
9
The strength is sufficient, so accept the design.
(b) Assume some dimensions in order to calculate the self weight.
Determine the minimum depth from the ACI Code: min h = L
16= 20(12)
16=15 in. Try h = 16 in.,
b = 0.5 h = 8 in., and d = 14 in. Assume c/d = 0.30.
Calculate the factored external moment. The self weight is 144
)16)(8)(150( = 133 lb/ft.
Dead load Mu = 8
)20)(133)(2.1( 2
= 8000 ft-lb
Live load Mu = 2
)10)(000,15)(6.1( = 120,000 ft-lb
Mu = 128,000 ft-lb = 1,536,000 in-lb
The required nominal strength for the beam is 9.0
000,536,1=nM = 1,706,667 in-lb.
Calculate the required steel area.
c = (0.30)(14) = 4.2 in
1 = 1000
4000600005.085.0 = 0.75
a = 1 c = (0.75)(4.5) = 3.15 in
Areq = y
c
f
baf)85.0( =
000,60
)15.3)(8)(6000)(85.0( = 2.142 in2
0.348 < 0.375, so the section is tension controlled. Verify the strength is sufficient (at least
2,080,000 in-lb)
nM = 2
adfA ys
= 2
92.314)000,60)(3(
= 2,347,058 in-lb
13
10
a = bf
fA
c
ys
85.0 =
)8)(6000(85.0
)000,60)(37.2( = 3.49 in.
c = 1
a =
75.0
49.3 = 4.65 in.
d
c =
14
65.4 = 0.332
0.332 < 0.375, so the section is tension controlled. Verify the strength is sufficient (at least
1,706,667 in-lb)
nM = 2
adfA ys
= 2
49.314)000,60)(37.2(
= 1,742,995 in-lb
The strength is sufficient, so accept the design.
(c) Assume the same dimensions as in part (b).
Calculate the factored external moment. The self weight is 144
)16)(8)(150( = 133 lb/ft.
Dead load Mu = 8
)20)(133)(2.1( 2
= 8000 ft-lb
Live load Mu = )5)(7500)(6.1( = 60,000 ft-lb
Mu = 68,000 ft-lb = 816,000 in-lb
The required nominal strength for the beam is 9.0
000,536,1=nM = 906,667 in-lb.
From part (b) the strength of the beam is 1,742,995 in-lb, so the design is sufficient.
Make sure the steel area meets the required minimum reinforcement ratio.
1 = 0039.0000,60
600033==
y
c
f
f
2 = 0033.0000,60
200200 ==yf
min sA = )14)(8)(0039.0( = 0.434 in2
The code is satisfied because 2.142 in2 > 0.434 in2. The area can be provided by 3 #8 bars (area is
3(0.79) = 2.37 in2). Now check the design using the actual steel area.
14
15
16
17
18
5.7. Compute the stresses in the compression steel, , for the cross sections shown in Fig. 5.38. Also compute thenominal moment strength for the section in part (b). Given:
fy � 60,000 psi 1414 MPa2
f ¿c � 7000 psi 148.3 MPa2, normal-weight concrete
f ¿s
11
2.050.8
10
18(4
57.2
mm
)9 in.
(228.6 mm)2.0 in.
(50.8 mm)15 in.
(381 mm)
10
30 in
.(7
62 m
m)
Figure 5.38
(a) Calculate the steel areas.
sA = (3)(1.27) = 3.81 in2
sA = (2)(0.31) = 0.62 in2
Calculate 1 = 1000
4000700005.085.0 = 0.7.
Calculate the strain assuming the compression steel has yielded.
a = bf
fAA
c
yss
85.0
)( =
)9)(7000(85.0
)000,60)(62.081.3( = 3.57 in.
c = 1
a =
7.0
57.3 = 5.11 in.
s = c
dc003.0 =
11.5
211.5003.0 = 0.00182 in./in.
Check if the compression steel has actually yielded.
y = s
y
E
f =
61029
000,60 = 0.002 in./in.
19
12
s = c
dc003.0 =
22.5
222.5003.0 = 0.00185 in./in.
Repeat this calculation until the computed values converge. After several trials, sf = 53,605 psi,
a = 3.65 in., c = 5.21 in., and s = 0.00185 in./in.
(b) Calculate the steel areas
sA = (4)(1.27) = 5.08 in2
sA = (2)(0.79) = 1.58 in2
Calculate 1 = 1000
4000700005.085.0 = 0.7.
Calculate the strain assuming the compression steel has yielded.
a = bf
fAA
c
yss
85.0
)( =
)15)(7000(85.0
)000,60)(58.108.5( = 2.35 in.
c = 1
a =
7.0
35.2 = 3.36 in.
s = c
dc003.0 =
36.3
5.236.3003.0 = 0.000769 in./in.
Check if the compression steel has actually yielded.
y = s
y
E
f =
61029
000,60 = 0.002 in./in.
0.002 in./in. > 0.000769 in./in. so the compression steel has not yielded. Calculate the strain again
using the actual compression steel stress found from the first calculation.
0.002 in./in. > 0.00182 in./in. so the compression steel has not yielded. Calculate the strain again
using the actual compression steel stress found from the first calculation.
sf = ss E = (0.00182)(29 106) = 52,923 psi
a = bf
fAfA
c
ssys
85.0 =
)9)(7000(85.0
)923,52)(62.0()000,60)(81.3( = 3.66 in.
c = 1
a =
7.0
66.3 = 5.22 in.
20
13
sf = ss E = (0.000769)(29 106) = 22,294 psi
a = bf
fAfA
c
ssys
85.0 =
)15)(7000(85.0
)294,22)(58.1()000,60)(08.5( = 3.02 in.
c = 1
a =
7.0
02.3 = 4.31 in.
s = c
dc003.0 =
31.4
5.231.4003.0 = 0.00126 in./in.
Repeat this calculation until the computed values converge. After several trials, sf = 33,149 psi,
a = 2.83 in., c = 4.04 in., and s = 0.00114 in./in.
To compute the nominal strength, first verify that the section is tension controlled.
t = c
cdt003.0 = 04.4
04.45.27003.0 = 0.0174 in./in.
The strain 0.0174 in./in. > 0.005 in./in. so the section is tension controlled and = 0.9. Now
verify that the section meets code by checking the minimum reinforcement ratio.
= )5.27)(15(
08.5=t
s
bd
A = 0.0123
1 = 0042.0000,60
700033==
y
c
f
f
2 = 0033.0000,60
200200 ==yf
0.0123 > max(0.0042,0.0033) so the ACI Code is satisfied.
Calculate the nominal strength of the section.
nM = )(2
)( ddfAa
dfAfA ssssys +
= (5.08)(60,000) (1.58)(33,149)[ ] 27.5 2.83
2
+ (1.58)(33,149)(27.5 2.5)
= 7,894,097 in-lb
M = Mn = (0.9)(7,894,097) = 7,104,867 in-lb
21
22
23
24
25
26
27
28
5.10. At failure, determine whether the precast sections shown in Fig. 5.39 will act similarly to rectangular sectionsor as flanged sections. Given:
fy � 60,000 psi 1414 MPa2
f ¿c � 4000 psi 127.6 MPa2, normal-weight concrete
14
20 508
2 in.(50.8 mm)
12 in
.(3
04.8
mm
)
8
8 in.(203.2 mm)
20 5084 in.
(101.6 mm)
9
8
30 in
.(7
63 m
m)
Figure 5.39
(a) Calculate the compression block dimensions assuming the sections behaves as a rectangular
section with a width equal to the flange width.
1 = 0.85
sA = (3)(0.79) = 2.37 in2
a = bf
fA
c
ys
85.0 =
)20)(4000(85.0
)000,60)(37.2( = 2.09 in.
c = 1
a =
85.0
09.2 = 2.46 in.
Since both a and c are greater than the flange thickness, the section must be treated as a T-section.
(b) Calculate the compression block dimensions assuming the sections behaves as a rectangular
section with a width equal to the flange width.
29
15
1 = 0.85
sA = (5)(0.79) = 3.95 in2
a = bf
fA
c
ys
85.0 =
)30)(4000(85.0
)000,60)(95.3( = 2.32 in.
c = 1
a =
85.0
32.2 = 2.73 in.
Since both a and c are less than the flange thickness, the section can be treated as a rectangular
section.
(c) Calculate the compression block dimensions assuming the sections behaves as a rectangular
section with a width equal to the flange width.
1 = 0.85
sA = (4)(1) = 4 in2
a = bf
fA
c
ys
85.0 =
)20)(4000(85.0
)000,60)(4( = 3.53 in.
c = 1
a =
85.0
53.3 = 4.15 in.
Since a is less than the flange thickness and c is greater than the flange thickness, the section can
be treated as either a rectangular section or an L-section.
30
31
32
33
34
35
36
37
38
39
40
41
6.1. A simply supported beam has a clear span ln = 20 ft (6.10 m) and is subjected to an external uniform service deadload WD = 1000 lb per ft (14.6 kN/m) and live load wL = 1500 lb per ft (21.9 kN/m). Determine the maximum fac-tored vertical shear Vu at the critical section. Determine the nominal shear resistance Vc by both the short methodand by the more refined method of taking the contribution of the flexural steel into account. Design the size andspacing of the diagonal tension reinforcement. Given:
Assume that no torsion exists.
fy � 60,000 psi 1413.7 MPa2
f¿c � 4000 psi 127.6 MPa2, normal-weight concrete
As � 6.0 in.2 13780 mm22
h � 20 in. 1508 mm2
d � 18 in. 1457.2 mm2
bw � 12 in. 1305 mm2
1
Calculate the factored loads.
self-weight = 144
)20)(12(150 = 250 lb/ft
wu = 1.2(250 + 1000) + 1.6(1500) = 3900 lb/ft
Vu at support = 2
)20)(3900( = 39,000 lb
Vu at d = 39,000 3900 (18 / 12) = 33,150 lb
Mu at d = 2
)12/18)(3900()12/18(000,39
2
= 54,112.5 ft-lb
= 649,350 in.-lb
Use the simplified method to calculate the shear loads.
Vc = bdfc2 = )18)(12(4000)0.1(2 = 27,322 lb
Vn = uV
= 75.0
150,33 = 44,200 lb
Since Vn > Vc / 2, shear reinforcement is necessary. The required shear support is
Vs = Vn Vc = 44,200 27,322 = 16,878 lb
Choose a size for the stirrup steel. Try No. 3 bars, Av = 2(0.11) = 0.22 in2.
Calculate the minimum spacing based on the steel strength.
s = s
yV
V
dfA =
878,16
)18)(000,60)(22.0( = 14.08 in.
Since Vs < 4 f c bd = 4 4000(12)(18) = 54,644, use d/2. The minimum required spacing is the
minimum of d/2 = 9 in. and 14.08 in.
Therefore, use No. 3 stirrups at 9 in. c-c spacing. 42
43
44
45
46
47
48
49
2
The minimum thickness required for deflection is 16
)12(18
16=L
= 13.5 in., which is quite small.
Assume beam dimensions of h = 24 in., d = 21 in., and b = 12 in.
Now calculate the factored loads.
self-weight = 144
)24)(12(150 = 300 lb/ft
wu = 1.2(300 + 400) + 1.6(800) = 2120 lb/ft
Pu = 1.2(25,000) + 1.6(30,000) = 78,000 lb
Then analyze the beam to find the shear and moment distribution. Split the loading into two
separate problems.
wu = 2120 lb/ft
Combine the two beams.
6.5. A continuous beam has two equal spans ln = 20 ft (6.10 m) and is subjected to an external service dead load wD
of 400 lb per ft. (5.8 kN/m) and a service live load wL of 800 lb per ft. (11.7 kN/m). In addition, an external ser-vice concentrated dead load PD of 25,000 lb and an external service concentrated live load PL of 30,000 lb (133kN) are applied to one midspan only. Design the diagonal tension reinforcement necessary. Given:
fy � 60,000 psi 1413.7 MPa2
f¿c � 5000 psi 134.5 MPa2, normal-weight concrete
50
3
Design the longitudinal support for the largest positive moment, M = 4,438,500 in.-lb. Try using
4 #10 bars at the bottom, As = (4)(1.27) = 5.08 in2.
a = bf
fA
c
ys
85.0 =
)12)(5000(85.0
)000,60)(08.5( = 5.98 in.
c = 1
a =
8.0
98.5 = 7.47 in.
d
c =
21
47.7 = 0.3557 < 0.375 tension controlled, = 0.90
Check the minimum reinforcement area, >yy
cs
ff
f
bd
A 200,
3max , and the strength,
=2
adfAM ysn .
0202.0)21)(12(
08.5 = > == 0033.0000,60
200,0035.0
000,60
50003max
2
98.521)000,60)(08.5(9.0 = 4,940,987 in.-lb > 4,438,500 in.-lb
Design the longitudinal support for the largest negative moment, M = 3,027,000 in.-lb. Try
using 3 #10 bars at the top, As = (3)(1.27) = 3.81 in2.
a = bf
fA
c
ys
85.0 =
)12)(5000(85.0
)000,60)(81.3( = 4.48 in.
c = 1
a =
8.0
48.4 = 5.60 in.
d
c =
21
60.5 = 0.2668 < 0.375 tension controlled, = 0.90
Check the minimum reinforcement area, >yy
cs
ff
f
bd
A 200,
3max , and the strength,
=2
adfAM ysn .
0151.0)21)(12(
81.3 = > == 0033.0000,60
200,0035.0
000,60
50003max
51
4
2
48.421)000,60)(81.3(9.0 = 3,859,440 in.-lb > 3,027,000 in.-lb
So use 4 #10 bars at the bottom and 3 #10 bars at the top of the section. Now design the shear or
diagonal reinforcement. Check to see if stirrups are necessary.
Use the simplified method to determine the shear capacity of the beam.
Vc = bdfc2 = )21)(12(5000)0.1(2 = 35,638 lb
Determine the minimum stirrup spacing at three points: the left end, at the point load, and at the
center support. Choose a size for the stirrup steel and then determine the required spacing. Try
No. 3 bars, Av = 2(0.11) = 0.22 in2.
At the left end, choose a section a distance d from the end. The shear loading at that point is
12
2121205.587,47 = 43,877.5 lbf.
Vn = uV
= 75.0
5.877,43 = 58,503 lb (Vn > Vc / 2, support required)
Vs = Vn Vc = 58,503 35,638 = 22,865 lb
s = s
yV
V
dfA =
865,22
)21)(000,60)(22.0( = 12.12 in.
Since Vs < )21)(12(500044 =bdfc = 71,276 lb, use d/2. The minimum required spacing is the
minimum of d/2 = 10.5 in. and 12.12 in. Therefore, use No. 3 stirrups at 10.5 in. c-c spacing.
Then determine the minimum spacing under the point load. The shear load is 51,612.5 lb.
Vn = uV
= 75.0
5.612,51 = 68,817 lb (Vn > Vc / 2, support required)
Vs = Vn Vc = 68,817 35,638 = 33,178 lb
s = s
yV
V
dfA =
178,33
)21)(000,60)(22.0( = 8.35 in.
The minimum required spacing is the minimum of d/2 = 10.5 in. and 8.35 in. Therefore, use No. 3
stirrups at 8.35 in. c-c spacing.
Finally, determine the minimum spacing at the center support. The shear load is 72,812,5 lb.
52
5
Vn = uV
= 75.0
5.812,72 = 97,083 lb (Vn > Vc / 2, support required)
Vs = Vn Vc = 97,083 35,638 = 61,445 lb
s = s
yV
V
dfA =
445,61
)21)(000,60)(22.0( = 4.51 in.
The minimum required spacing is the minimum of d/2 = 10.5 in. and 4.51 in. Therefore, use No. 3
stirrups at 4.51 in. c-c spacing.
53
6.6. Design the vertical stirrups for a beam having the shear diagram shown in Figure 6.37 assuming that Vc =2 bwd. Given:
fy � 60,000 psi 1414 MPa2
f¿c � 5000 psi 134.5 MPa2, normal-weight concrete
Vu3 � 30,000 lb 1133 kN2
Vu2 � 50,000 lb 1222 kN2
Vu1 � 80,000 lb 1356 kN2
dw � 20 in. 1508 mm2
bw � 12 in. 1304.8 mm2
2f ¿c
6
Figure 6.37
Use the simplified method to determine the shear capacity of the beam.
Vc = bdfc2 = )20)(12(5000)0.1(2 = 33,941 lb
Try No. 4 bars for the stirrups, Av = 2(0.2) = 0.4 in2. Determine the spacing for Vu1. The shear
loading at that point is 80,000 lbf.
Vn = uV
= 75.0
000,80 = 106,667 lb (Vn > Vc / 2, support required)
Vs = Vn Vc = 106,667 33,941 = 72,726 lb
s = s
yV
V
dfA =
726,72
)20)(000,60)(4.0( = 6.60 in.
Since Vs > )20)(12(500044 =bdfc = 67,882 lb, use d/4. The minimum required spacing is the
minimum of d/4 = 5 in and 6.60 in. Therefore, use No. 4 stirrups at 5 in. c-c spacing.
Then determine the spacing for Vu2. The shear load is 50,000 lb.
Vn = uV
= 75.0
000,50 = 66,667 lb (Vn > Vc / 2, support required)
Vs = Vn Vc = 66,667 33,941 = 32,726 lb
s = s
yV
V
dfA =
726,32
)20)(000,60)(4.0( = 14.67 in.
54
7
Since Vs < 67,882 lb, use d/2. The minimum required spacing is the minimum of d/2 = 10 in. and
14.67 in. Therefore, use No. 4 stirrups at 10 in. c-c spacing.
Finally, determine the spacing for Vu3. The shear load is 30,000 lb.
Vn = uV
= 75.0
000,30 = 40,000 lb (Vn > Vc / 2, support required)
Vs = Vn Vc = 40,000 33,941 = 6059 lb
s = s
yV
V
dfA =
6059
)20)(000,60)(4.0( = 79.2 in.
Since Vs < 67,882 lb, use d/2. The minimum required spacing is the minimum of d/2 = 10 in. and
79.2 in. Therefore, use No. 4 stirrups at 10 in. c-c spacing.
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
6.10. Design a bracket to support a concentrated factored load Vu = 100,000 lb (444.8 kN) acting at a lever arm a = 6in. (152.4 mm) from the column face; horizontal factored force Nuc = 25,000 lb (111 kN). Given:
Column size = 12 � 24 in. (305 � 609.6 mm); Corbel width = 24 in.Use both the shear-friction approach and the strut-and-tie method in your solution. Assume that the
bracket was cast after the supporting column cured and that the column surface at the bracket location wasnot roughened before casting the bracket. Detail the reinforcing arrangements for the bracket.
fy � fyt � 60,000 psi 1414 MPa2
f¿c � 5000 psi 134.5 MPa2, normal-weight concrete
b � 24 in. 1609.6 mm2
8
Shear friction approach:
First, choose dimensions for the corbel. Try h = 18 in. and d = 15 in. Then check the vertical
load.
Vn = uV
= 75.0
000,100 = 133,333 lb
V1 = bdfc2.0 = 0.2(5000)(24)(15) = 360,000 lb
V2 = bdfc )08.0480( + = (480 + 0.08 5000)(24)(15) = 316,800 lb
V3 = 1600bd = (1600)(24)(15) = 576,000 lb
Vn is less than all three values, so the design is okay so far. For a corbel cast on an unroughened
hardened column, the value for is 0.6 = 0.6. Calculate the required areas for the steel
reinforcement.
Avf = y
u
f
V =
)6.0)(000,60(
333,133 = 3.70 in2.
Af = )85.0(
)(
df
dhNaV
y
ucu +
= )15)(85.0)(000,60(75.0
)1518)(000,25()6)(333,133( +
= 1.18 in2.
An = y
uc
f
N =
)000,60)(75.0(
000,25 = 0.56 in2.
The primary tension steel area is the maximum of nvf AA +3
2, nf AA + , and bd
f
f
y
c04.0 .
Asc1 = nvf AA +3
2 = 56.0)70.3(
3
2 + = 3.02 in2.
Asc2 = nf AA + = 1.18 + 0.56 = 1.73 in2.
Asc3 = bdf
f
y
c04.0 = )15)(24(000,60
500004.0 = 1.2 in2.
75
9
Therefore, the required steel areas are Asc = 3.02 in2. and Ah = )(5.0 nsc AA = 1.23 in2. Select the
bar sizes.
For the primary steel, use 4 No. 8 bars. For the shear reinforcement, use 4 No. 5 bars spaced 2.5
in. c-c. Also use 4 No. 5 framing bars and 1 No. 5 anchor bar to complete the cage.
The required bearing plate area is )5000)(85.0(7.0
000,100
)85.0(7.0=
c
u
f
V = 33.6 in2. so use a square 6 in.
by 6 in. plate thick enough to be rigid under the load.
Solution using the strut and tie approach:
Assume the following dimensions:
h = 18 in.
d = 15 in.
Points C and D inset 2 in. from the surface
Use statics to calculate the forces.
BCF = d
LV BC
u = 113,333 lb
ABT = uc
BC
BCBC N
L
xF + = 78,333 lb
ACF = AC
ACAB
x
LT = 166,458 lb
ADT = AC
ACL
dF
CEF = BC
BC
AC
ACL
dF
L
dF + = 246,875 lb
CDT = BC
BCBC
AC
ACAC
L
xF
L
xF = 25,000 lb
The required bearing plate area is )5000)(85.0(7.0
000,100
)85.0(75.0=
c
u
f
V = 31.4 in2. so use a square
plate that is 6 in. 6 in. and thick enough to be rigid under the load.
Use the tie loads with a load factor of = 0.75 to design the steel reinforcement cage. The
primary tension steel must support the tension ABT . 76
10
AAB = )000,60(75.0
333,78
75.0=
y
AB
f
T = 1.74 in2.
ACD = )000,60(75.0
000,25
75.0=
y
CD
f
T = 0.56 in2.
A portion of the reinforcement resists the internal friction force.
An = )000,60(75.0
000,25
75.0=
y
uc
f
N = 0.56 in2.
Therefore, the required steel areas are AAB = 1.74 in2., ACD = 0.56 in2., and 2in59.0)(5.0 == nABh AAA . Select the bar sizes.
Use 3 No. 7 bars for the top steel, Area = 3(0.6) = 1.8 in2.
Use 3 No. 4 bars for the bottom steel, Area = 3(0.2) = 0.6 in2.
Use 3 No. 4 closed stirrups for the reinforcement, Area = 3(0.2) = 0.6 in2. The bars should be
spaced to fill (2/3)(15) = 10 in., so space them 3 in. c-c.
Check the shear reinforcement for struts AB and AC.
Strut AC: AC
h
L
d
bs
tieA
)0.3)(24(
)2.0(2sin
/= = 0.002279 < 0.003 so increase the bar size to No. 4 and
use 3 No. 4 closed stirrups spaced 3 in. c-c.
ACL
d
)0.3)(24(
)31.0(2 = 0.003799 > 0.003, OK
In addition, use 3 No. 4 framing bars and 1 No. 4 anchor bar to complete the cage.
Finally, the strut portions need to be checked for adequate strength.
The allowable concrete strength in the nodal zones is fce = 0.85(0.8)(5000) = 3400 psi. So the
required widths for the struts are
Strut AC: )3400)(24(
458,166 = 2.04 in.
Strut CE: )3400)(24(
875,246 = 3.02 in.
Strut BC: )3400)(24(
333,113 = 1.39 in.
They are all within the available space in the corbel, so accept the design.
77
78
79
80
81
82
83
84
7.2. A cantilever beam is subjected to a concentrated service live load of 30,000 lb (133.5 kN) acting at a dis-tance of 3 ft 6 in. (1.07 m) from the wall support. In addition, the beam has to resist an equilibrium fac-tored torsion Tu = 450,000 in.-lb (50.8 kN–m). The beam cross section is 15 in. × 30 in. (381 mm × 762mm) with an effective depth of 27 in. (686 mm). Design the stirrups and the additional longitudinal steelneeded.Given:
As � 4.0 in.2 12580.64 mm22
fy � fyt � 60,000 psi
f ¿c � 4000 psi
1
The length of the beam is not given, so ignore the self weight. Use the following beam
dimensions and loads.
Beam height h = 30 in.
Effective depth d = 27 in.
Beam width b = 15 in.
Force position L = 3.5 ft
Clear cover = 1.5 in.
Factored force uV = LP6.1 = 1.6(30,000) = 48,000 lb
Factored moment nM = LPL6.1 = 1.6(30,000)(3.5 12) = 2,016,000 in-lb
Factored torsion uT = 450,000 in-lb
Determine the minimum steel required for shear support. The load factor for shear is 75.0= ,
and 0.1= for normal weight concrete.
Nominal load nV = uV
= 75.0
000,48 = 64,000 lb
Concrete support cV = bdfc2 = )27)(15(4000)0.1(2 = 51,229 lb
Required support sV = nV cV = 64,000 51,229 = 12,771 lb
Required area s
A =
df
V
yt
s = )27)(000,60(
771,12 = 0.00788 in.2 / in.
Determine the minimum steel required for torsion. The load factor for torsion is also 75.0= .
Assume #4 bars, diameter = 0.5 in., for the stirrups to calculate the dimensions 1x and 1y .
Nominal load nT = uT
= 75.0
000,450 = 600,000 in.-lb
Outer box 0x = b = 15 in.
0y = h = 30 in.
cpA = 00 yx = (15)(30) = 450 in.2
cpp = )(2 00 yx + = 2(15 + 30) = 90 in.
85
2
Strut angle = 45 degrees
Check that torsion support is required for uT = 450,000 in.-lb.
No support if uT < cp
cpc
p
Af 2
= 90
)450(4000)0.1(75.0 2
= 106,727 in.-lb (false)
Then verify the section meets the requirement +<+ cc
oh
huu fbd
V
A
pT
bd
V8
7.1
2
2
2
.
2
2
2
7.1+
oh
huu
A
pT
bd
V =
2
2
2
)75.304(7.1
)76)(000,450(
)27)(15(
000,48 + = 246.92 psi
+ cc f
bd
V8 = + 4000)0.1(8
)27)(15(
229,5175.0 = 474.34 psi
246.92 psi < 474.34 psi, so the section is adequate.
Determine the minimum required area of transverse steel.
Torsion s
At = cot2 0 yt
n
fA
T =
)1)(000,60)(04.259(2
000,600 = 0.0193 in.2 / in.
Torsion and shear s
A =
s
A
s
At +2 = 2(0.0193) + 0.00788 = 0.0465 in.2 / in. (controls)
Minimum s
A = b
f
f
y
c75.0 = )15(
000,60
400075.0 = 0.0119 in.
Determine the minimum required longitudinal reinforcement.
Torsion LA = 2)(cot
y
yt
ht
f
fp
s
A = 2)1)(1)(76)(0193.0( = 1.47 in.2
Minimum s
At s
At ytf
b25 =
000,60
1525 = 0.0063 in. (does not control)
Inner box 1x = 0x 2(clear cover) (stirrup d) = 15 2(1.5) 0.5 = 11.5 in.
1y = 0y 2(clear cover) (stirrup d) = 30 2(1.5) – 0.5 = 26.5 in.
ohA = 11 yx = (11.5)(26.5) = 304.75 in.2
hp = )(2 11 yx + = 2(11.5 + 26.5) = 76 in.
Torsion area 0A = 0.85 ohA = 0.85(304.75) = 259.04 in.2
86
3
Minimum LA y
yt
ht
y
cpc
f
fp
s
A
f
Af5
)1)(76)(0193.0(000,60
)450(4000)0.1(5 = 0.9 in.2
Therefore, the required transverse steel area is 0.0465 in.2 /in., the specified flexural steel is 6.0
in.2, and the required longitudinal reinforcement is 1.47 in.2. Assume that one quarter of the
reinforcement is added to the bottom steel, one quarter goes in the top corners of the stirrups,
and the remaining steel is distributed along the vertical sides of the stirrups. Choose the bar sizes
and spacings.
For the transverse steel, try #4 closed stirrups. The bar area is 0.2 in.2, so the required spacing is
the smallest of in. 5.98
76
8==hp
, 12 in., and 0465.0
)2.0(2 = 4.3 in. (controls)
Therefore, use #4 closed stirrups spaced 4 in. c-c.
For the bottom steel, the required area is 4 + 0.25(1.47) = 4.37 in.2 Use 5 #9 bars (1 in.2 per bar)
for a total area of 4(1) = 5 in.2
For the top steel, the required area is 0.25(1.47) = 0.37 in.2, so use 2 #4 bars in the corners for an
area of 2(0.2) = 0.4 in.2
For the sides, each side requires 0.25(1.47) = 0.37 in.2 of steel distributed in 25 in. Use 2 #4 bars
spaced equally for an area of 2(0.2) = 0.4 in.2 per side. The final design is shown below.
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
4
7.5. Design the rectangular beam shown in Figure 7.34 for bending, shear, and torsion. Assume that thebeam width b = 12 in. (305 mm). Given:
fy � fyt � 60,000 psi 1413.8 MPa2
f ¿c � 4000 psi 127.6 MPa2
10,000 44.5
4 ft (1.22 m)
12 ft (3.66 m)
Figure 7.34
To include the self-weight, a height must be assumed. Try the following dimensions:
Beam height h = 24 in.
Effective depth d = 21 in.
Beam width b = 12 in.
Beam length L = 12 ft
“L” length 2L = 2 ft
Clear cover = 1.5 in.
Start by calculating the support force and moment.
Self weights 1w = 150bh = )150(144
)24)(12( = 300 lb / ft
2w = 1502hL = )150(12
)24)(2( = 600 lb / ft
Factored force AF = LPbwbLw 6.12.1)(2.1 21 ++
= 1.2(300)(12 1) + 1.2(600)(1) + 1.6(10,000)
= 20,680 lb
Factored moment AM = ++2
6.12
2.12
)(2.12
2
1 bLP
bLbw
bLwL
= )5.11)(000,10(6.1)5.11)(1)(600(2.12
)11)(300(2.1 2
++
= 214,060 ft-lb = 2,568,720 in.-lb
Factored torsion AT = 222 6.1
2
2.1LP
bLwL+
= )2)(000,10(6.12
)2)(1)(600(2.1 +
= 32,720 ft-lb = 392,640 in.-lb
102
5
Then calculate the design loads.
Factored shear uV = dwFA 12.1 = 20,680 1.2(300)(21/12) = 20,050 lb
Factored moment nM = AM = 2,568,720 in.-lb
Factored torsion uT = AT = 392,640 in.-lb
Now design the beam for bending. Determine the minimum required area of steel, = 0.9 for
bending.
Nominal load nM = uM
= 9.0
720,568,2 = 2,854,133 in.-lb
Section strength nM 2
adfA ys and a =
bf
fA
c
ys
85.0
Solve for sA sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c
2)21)(12)(4000(85.0
)133,854,2(211
000,60
)21)(12)(4000(85.0
2.48 in.2 (controls)
Minimum area 1 sA bdf
f
y
c3 = )21)(12(
000,60
40003 = 0.8 in.2
Minimum area 2 sA bdf y
200 = )21)(12(
000,60
200 = 0.84 in.2
Determine the minimum steel required for shear support. The load factor for shear is 75.0= ,
and 0.1= for normal weight concrete.
Nominal load nV = uV
= 75.0
050,20 = 26,733 lb
Concrete support cV = bdfc2 = )21)(12(4000)0.1(2 = 31,876 lb
Since cV > nV , the stirrup size will be determined by the torsion design.
Determine the minimum steel required for torsion. The load factor for torsion is also 75.0= .
Assume #4 bars, diameter = 0.5 in., for the stirrups to calculate the dimensions 1x and 1y .
Nominal load nT = uT
= 75.0
640,392 = 523,520 in.-lb
103
6
Torsion area 0A = 0.85 ohA = 0.85(174.25) = 148.11 in.2
Strut angle = 45 degrees
Check that torsion support is required for uT = 392,640 in.-lb.
No support if uT < cp
cpc
p
Af 2
= 72
)288(4000)0.1(75.0 2
= 54,644 in.-lb (false)
Then verify the section meets the requirement +<+ cc
oh
huu fbd
V
A
pT
bd
V8
7.1
2
2
2
.
2
2
2
7.1+
oh
huu
A
pT
bd
V =
2
2
2
)25.174(7.1
)58)(640,392(
)21)(12(
050,20 + = 448.31 psi
+ cc f
bd
V8 = + 4000)0.1(8
)21)(12(
876,3175.0 = 474.34 psi
448.31 psi < 474.34 psi, so the section is adequate.
Determine the minimum required area of transverse steel.
Torsion s
At = cot2 0 yt
n
fA
T =
)1)(000,60)(11.148(2
520,523 = 0.0295 in.
Torsion and shear s
A =
s
A
s
At +2 = 2(0.0295) + 0 = 0.0589 in. (controls)
Minimum s
A = b
f
f
y
c75.0 = )12(
000,60
400075.0 = 0.0095 in.
Outer box 0x = b = 12 in.
0y = h = 24 in.
cpA = 00 yx = (12)(24) = 288 in.2
cpp = )(2 00 yx + = 2(12 + 24) = 72 in.
Inner box 1x = 0x 2(clear cover) (stirrup d) = 12 2(1.5) 0.5 = 8.5 in.
1y = 0y 2(clear cover) (stirrup d) = 24 2(1.5) – 0.5 = 20.5 in.
ohA = 11 yx = (8.5)(20.5) = 174.25 in.2
hp = )(2 11 yx + = 2(8.5 + 20.5) = 58 in.
104
7
)1)(58)(0295.0(000,60
)288(4000)0.1(5 = 0.19 in.2
Therefore, the required transverse steel area is 0.0589 in.2 / in., the required flexural steel is 2.48
in.2, and the required longitudinal reinforcement is 1.71 in.2. Assume that one quarter of the
reinforcement is added to the bottom steel, one quarter goes in the top corners of the stirrups,
and the remaining steel is distributed along the vertical sides of the stirrups. Choose the bar sizes
and spacings.
For the transverse steel, try #5 closed stirrups. The bar area is 0.31 in.2, so the required spacing
is the smallest of in. 25.78
58
8==hp
, 12 in., and 0589.0
)31.0(2 = 5.26 in. (controls). Therefore, use #5
closed stirrups spaced 5 in. c-c.
For the bottom steel, the required area is 2.48 + 0.25(1.71) = 2.91 in.2. Use 4 #8 bars (0.79 in.2
per bar) for a total area of 4(0.79) = 3.16 in.2.
For the top steel, the required area is 0.25(1.71) = 0.43 in.2, so use 2 #5 bars in the corners for an
area of 2(0.31) = 0.62 in.2.
For the sides, each side requires 0.25(1.71) = 0.43 in.2 of steel distributed in 18.88 in. Use 2 #5
bars spaced equally for an area of 2(0.31) = 0.62 in.2 per side. The final design is shown below.
Determine the minimum required longitudinal reinforcement.
Torsion LA = 2)(cot
y
yt
ht
f
fp
s
A = 2)1)(1)(58)(0295.0( = 1.71 in.2
Minimum s
At s
At ytf
b25 =
000,60
1225 = 0.005 in. (does not control)
Minimum LA y
yt
ht
y
cpc
f
fp
s
A
f
Af5
105
106
107
108
109
110
111
112
113
114
8.2. Calculate the maximum immediate and long-term deflection for a 8-in.-thick slab on simple supports spanningover 15 ft. The service dead and live loads are 60 psf (28.7 kPa) and 150 psf (71.8 kPa), respectively. The rein-forcement consists of No. 5 bars (16-mm diameter) at 6 in. center to center (154 mm center to center). Alsocheck which limitations, if any, need to be placed on its usage. Assume that 50% of the live load is sustainedover a 30-month period. Given:
Es � 29 � 106 psi 1200,000 MPa2
fy � 60,000 psi 1414 MPa2
f ¿c � 5000 psi 134.5 MPa2
1
Design the one-way slab as a rectangular beam with a width of 12 in. Assume the depth of the
steel is 6 in.
Beam height h = 8 in.
Effective depth d = 6 in.
Beam width b = 12 in.
Span L = 15 ft
First, estimate the properties of the concrete.
Weight per foot w = 150144
)12)(8( = 100 lb / ft
Modulus cE = cf000,57 = 5000000,57 = 4,030,509 psi
Tensile rupture rf = cf5.7 = 50005.7 = 530 psi
Stiffness ratio n = c
s
E
E =
509,030,4
1029 6
= 7.19
Then calculate the properties of the elastic beam.
Center of gravity y = 2
h =
2
8 = 4 in.
Moment of inertia gI = 3
12
1bh = 3)8)(12(
12
1 = 512 in.4
Strength crM = y
fI rg
r = 4
)530)(512( = 67,882 in-lb
Calculate the properties of the beam during the first phase of cracking.
Center of gravity equation dnAcnAcb
ss+2
2 = 0
Solve for c c = 1.7728 in.
Moment of inertia crI = 23 )(
3
1cdnAbc s+
= 23 )77.16)(31.02)(19.7()77.1)(12(
3
1 +
= 102 in.4
Calculate the load moments. 115
2
Dead load DM = 8
)( 2LwbLD + =
[ ]8
)1215(100)12)(60( 2+ = 54,000 in-lb
Live load LM = 8
2bLLL = 8
)1215()12)(150( 2
= 50,625 in-lb
Now calculate the moments for the deflections.
Dead M = DM = 54,000 in-lb
Dead plus 50% live M = DM + 0.5 LM = 54,000 + 0.5(50,625) = 79,312.5 in-lb
Dead plus live M = DM + LM = 54,000 + 50,625 = 104,625 in-lb
Calculate the effective moment of inertia for each case.
Equation eI = crcrgcr III
M
M+)(
3
if M > crM
Dead eI = gI = 512 in.4 ( M < crM )
Dead plus 50% live eI = 102)102512(5.312,79
882,673
+ = 359 in.4
Dead plus live eI = 102)102512(625,104
882,673
+ = 214 in.4
Calculate the deflections, use t = 1.75 and = 2.
Equation = eEI
ML
48
5 2
(deflection of simply supported beam)
Immediate DL D = )512)(1003.4(48
)1215)(000,54(56
2
= 0.0883 in.
Immediate LL L = )214)(1003.4(48
)1215)(625,104(56
2
D = 0.3211 in.
Immediate 50% LL LS = )359)(1003.4(48
)1215)(5.312,79(56
2
D = 0.0967 in.
Long term deflection LT = LStDL ++
= (0.3211) + (2)(0.0883) + (1.75)(0.0967) = 0.6669 in.
Now check the deflection requirements.
Roof with no supported or attached nonstructural elements
likely to be damaged by large deflections 180
L =
180
)12(15 = 1 in. > L
116
3
Floor with no supported or attached nonstructural elements
likely to be damaged by large deflections 360
L =
360
)12(15 = 0.5 in. > L
Roof or floor with supported or attached nonstructural
elements likely to be damaged by large deflections 480
L =
480
)12(15 = 0.375 in. < LT
Roof or floor with supported or attached nonstructural
elements not likely to be damaged by large deflections 480
L =
240
)12(15 = 0. 75 in. > LT
Therefore, the slab should be restricted to roofs or floors without attached nonstructural
elements, or with attached nonstructural elements not likely to be damaged by large deflections.
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
8.7. A rectangular beam under simple bending has the dimensions shown in Fig. 8.21. It is subjected to an aggres-sive chemical environment. Calculate the maximum expected flexural crack width and whether the beam sat-isfies the serviceability criteria for crack control. Given:
minimum clear cover � 112
in. 138.1 mm2
fy � 60,000 psi 1414 MPa2
f ¿c � 4500 psi 131.0 MPa2
4
24 in.
8
b = 10 in.
d =
20
in.
t = 4 in.
Figure 8.21 Beam geometry.
Beam height h = 24 in.
Effective depth d = 20 in.
Beam width b = 10 in.
First, estimate the properties of the concrete.
Modulus cE = cf000,57 = 4500000,57 = 3.824 106 psi
Tensile rupture rf = cf5.7 = 45005.7 = 503 psi
Stiffness ratio n = c
s
E
E =
6
6
10824.3
1029 = 7.58
Then calculate the properties of the elastic beam.
Center of gravity y = 2
h =
2
24 = 12 in.
Moment of inertia gI = 3
12
1bh = 3)12)(10(
12
1 = 11,520 in.4
Strength crM = y
fI rg
r = 12
)503)(520,11( = 482,991 in-lb
Calculate the center of gravity for the beam during the first phase of cracking.
Center of gravity equation b
2c 2 + nAs + (n 1) A s[ ] c nAsd + (n 1) A s d [ ] = 0
Solve for c c = 7.54 in.
140
5
maxw = 3076.0 Adf cs
= )20)(5.2()36)(32.1(076.0 = 13.3 mils = 0.0133 in.
The maximum tolerable crack for an aggressive chemical environment is 0.007 in., so the beam
does not satisfy the ACI code for crack control tolerance.
Then calculate the values for the crack equation, 3076.0 Adf cs .
Depth factor = cd
ch =
54.720
54.724 = 1.32
Steel stress sf = 0.6 yf = 0.6(60,000) = 36,000 psi = 36 ksi
Number of bars bc = 4
Tension area A = bc
bdh )(2 =
4
10)2024(2 = 20 in.2
First layer depth cd = (clear cover) + (stirrup d) + 0.5(bar diameter)
= 1.5 + 0.5 + 0.5(1) = 2.5 in.
Finally, calculate the maximum crack size.
141
142
8.9. Find the maximum web of a beam reinforced with bundled bars to satisfy the crack-control criteria for interiorexposure conditions. Given:
No. 4 stirrups used 113-mm diameter2
As � three bundles of three No. 9 bars each 1three bars of 28.6-mm diameter each in a bundle2
fy � 60,000 psi 1414 MPa2
f ¿c � 5000 psi 134.5 MPa2
6
Since the beam dimensions are unknown, assume = 1.2, sf = 0.6 yf , and a clear cover of 1.5
in. The crack equation for a beam with bundles bars is
maxw = 3076.0 Adf cs
= 1.2
sf = 0.6 yf = 0.6(60,000 psi) = 36 ksi
Calculate the center of gravity of the bundles from the bottom of the beam.
cd = (clear cover) + (stirrup d) + (bundle CG)
= )1(6
335.05.1
+++ = 2.890 in.
Write the area of concrete in tension in terms of the web width b.
Number of bars bc = 9
Effective number bc = 0.650 bc = 0.650(9) = 5.85
Effective area A = bc
cbd2 = b
85.5
)890.2(2 = 0.9879b
For a beam in interior exposure, the maximum tolerable crack width is 0.016 in. Substitute the
known values into the equation.
Substitute values 016.0 = 1000
)9879.0)(890.2()36)(2.1(076.0 3 b
Solve for b b = )9879.0)(890.2(
1
)36)(2.1)(076.0(
)016.0)(1000(3
= 40.54 in.
Therefore, the maximum beam web width is 40.5 in.
143
9.1. Calculate the axial load strength Pn for columns having the cross-sections shown in Figure 9.46. Assume zeroeccentricity for all cases. Cases (a), (b), (c), and (d) are tied columns; case (e) is spirally reinforced.
1
9
8000
8 in.(203.2 mm)
12 in
.(3
04.8
mm
)Figure 9.46 Column sections.
144
2
Beam area gA = bh = (8)(12) = 96 in.2
Steel area stA = 6(1) = 6 in.2
Since the column is compression controlled (axial load), = 0.65. For a tied column:
Nominal strength nP = (0.80) 0.85 f c (Ag Ast ) + Ast fy[ ]
= )]000,60)(6()696)(8000(85.0)[80.0)(65.0( +
= 505,400 lb
145
146
147
148
149
150
151
152
153
154
155
156
9.4. For the cross section shown in Figure 9.46c of Problem 9.1, determine the safe eccentricity e ifPu = 400,000 lb. Use the trial-and-adjustment method satisfying the compatibility of strains.
3
Beam area gA = bh = (8)(12) = 96 in.2
Top steel area stA = 3(1) = 3 in.2
Bottom steel area stA = 3(1) = 3 in.2
For the column = 0.65 and for cf = 8000 psi, 1 = 0.65. Guess that the neutral axis passes
through the center of the section, c = 6 in.
Compression block a = c1 = (0.65)(6) = 3.9 in.
Concrete force CF = bafc85.0 = (0.85)(8000)(8)(3.9) = 212,160 lb
Top steel strain s = c
dc003.0 =
6
36003.0 = 0.0015 (compression)
Top steel stress sf = ssE = (29 106)(0.0015) = 43,500 psi < 60,000 psi OK
Top steel force SF = ss fA = (3)(43,500) = 130,500 lb
Bottom steel strain s = c
cd003.0 =
6
69003.0 = 0.0015 (tension)
Bottom steel stress sf = ssE = (29 106)(0.0015) = 43,500 psi < 60,000 psi OK
Bottom steel force SF = ss fA = (3)(43,500) = 130,500 lb
Nominal strength nP = CF + SF SF = 212,160 lb
Strength uP = nP = 0.65(212,160) = 137,904 lb
Since the strength is less than the specified load of 400,000 lb, the size of the compression block
must be increased. This can be achieved by increasing c and iterating until the calculated strength
is equal to the load. After several iterations, c = 11.0 in.
Compression block a = c1 = (0.65)(11) = 7.15 in.
Concrete force CF = bafc85.0 = (0.85)(8000)(8)(7.15) = 388,960 lb
Top steel strain s = c
dc003.0 =
6
311003.0 = 0.002182 (compression)
Top steel stress sf = ssE = (29 106)(0.002182) = 63,273 psi > 60,000 psi
= 60,000 psi
Top steel force SF = ss fA = (3)(60,000) = 180,000 lb
Bottom steel strain s = c
dc003.0 =
6
911003.0 = 0.000545 (compression)
Bottom steel stress sf = ssE = (29 106)(0.000545) = 15,818 psi < 60,000 psi OK
Bottom steel force SF = ss fA = (3)(15,818) = 47,455 lb
Nominal strength nP = CF + SF SF = 616,415 lb
157
4
Strength uP = nP = 0.65(616,415) = 400,669 lb
The strength is sufficient, so calculate the eccentricity.
Nominal moment nM = )()(2
dyFdyFa
yF ssC ++
= (388,960)(6 0.5 7.15) + (180,000)(6 3) + (47,455)(6 9)
= 1,340,864 in.-lb
Eccentricity e = n
n
P
M =
415,616
864,340,1 = 2.18 in.
158
159
160
161
162
163
164
165
166
167
168
9.8. Design the reinforcement for a nonslender 12 in. × 15 in. column to carry the following loading. The factoredultimate axial force Pu = 200,000 lb. The eccentricity e to geometric centroid = 8 in. Given:
fy � 60,000 psi
f ¿c � 4000 psi
5
Use the following design parameters.
Force uP = 200,000 lb
Eccentricity e = 7 in.
Moment uM = ePu = (200,000)(7) = 1,400,000 in.-lb
Top steel depth d = 3 in.
Bottom steel depth d = h 3 = 15 – 3 = 12 in.
Reinforcement ratios = = 0.015
Choose the longitudinal reinforcement (the same for both top and bottom).
Required area stA = bd = (0.015)(12)(12) = 2.16 in.2
Use 3 No. 8 bars stA = (3)(0.79) = 2.37 in.2
Check for compression failure using = 0.65 and for cf = 4000 psi, 1 = 0.85.
Neutral axis depth c = 0.6 d = 0.6(12) = 7.2 in.
Compression block a = c1 = (0.85)(7.2) = 6.12 in.
Concrete force CF = bafc85.0 = (0.85)(4000)(12)(6.12) = 249,696 lb
Top steel strain s = c
dc003.0 =
2.7
32.7003.0 = 0.00175 (compression)
Top steel stress sf = ssE = (29 106)(0.00175) = 50,750 psi < 60,000 psi OK
Top steel force SF = ss fA = (2.37)(50,750) = 120,278 lb
Bottom steel strain s = c
cd003.0 =
2.7
2.712003.0 = 0.002 (tension)
Bottom steel stress sf = ssE = (29 106)(0.002) = 58,000 psi < 60,000 psi OK
Bottom steel force SF = ss fA = (2.37)(58,000) = 137,460 lb
Nominal strength nP = CF + SF SF = 232,514 lb
Strength uP = nP = 0.65(232,514) = 151,134 lb
Nominal moment nM = )()(
2dyFdyF
ayF ssC ++
= )125.7)(460,137()35.7)(278,120(2
12.615)696,249( +
= 2,268,469 in-lb
Eccentricity e = n
n
P
M =
514,232
469,268,2 = 9.76 in.
169
6
Since 9.76 in. > 7 in., the column is compression controlled. Choose a larger ratio and iterate until
the calculated eccentricity is 7 in. After a few iterations, the ratio is about 0.673.
Neutral axis depth c = 0.64 d = 0.673(12) = 8.08 in.
Compression block a = c1 = (0.85)(8.08) = 6.87 in.
Concrete force CF = bafc85.0 = (0.85)(4000)(12)(6.87) = 280,076 lb
Top steel strain s = c
dc003.0 =
08.8
308.8003.0 = 0.00189 (compression)
Top steel stress sf = ssE = (29 106)(0.00189) = 54,682 psi < 60,000 psi OK
Top steel force SF = ss fA = (2.37)(54,682) = 129,596 lb
Bottom steel strain s = c
cd003.0 =
08.8
08.812003.0 = 0.00146 (tension)
Bottom steel stress sf = ssE = (29 106)(0.00146) = 42,272 psi < 60,000 psi OK
Bottom steel force SF = ss fA = (2.37)(42,272) = 100,184 lb
Nominal strength nP = CF + SF SF = 309,488 lb
Strength uP = nP = 0.65(309,488) = 201,167 lb
Nominal moment nM = )()(
2dyFdyF
ayF ssC ++
= )125.7)(184,100()35.7)(596,129(2
87.615)076,280( +
= 2,173,277 in.-lb
Eccentricity e = n
n
P
M =
488,309
277,173,2 = 7.02 in.
The strength of the column is sufficient (201,167 lb > 200,000 lb and 2,173,277 in.-lb >
1,400,000 in.-lb). Now design the ties. Since the longitudinal bars are No. 8 bars, use No. 3 ties.
The spacing is the smallest of
1. 48 times No. 3 bar diameter = 48(0.375) = 18 in.
2. 16 times No. 8 bar diameter = 16(1) = 16 in.
3. Smaller column dimension = 12 in.
Therefore, use No. 3 ties spaced 12 in. c-c.
170
171
172
173
174
175
176
177
178
9.12. A rectangular braced column of a multistory frame building has a floor height lu = 30 ft. It is subjected to ser-vice dead-load moments M2 = 4,000,000 on top and M1 = 2,000,000 in.-lb at the bottom. The service live-loadmoments are 80% of the dead-load moments. The column carries a service axial dead load PD = 200,000 lb anda service live load PL = 400,000. Design the cross-section size and reinforcement for this column. Given:
d¿ � 3 in.
�A � 1.3, �B � 0.9
fy � 60,000 psi
f ¿c � 8000 psi
7
First, calculate the factored loads on the column.
Force uP = 1.2(200,000) + 1.6(400,000) = 880,000
Moment 1 uM1 = 1.2(3,000,000) + 1.6(0.8)(3,000,000) = 7,440,000 in.-lb
Moment 2 uM1 = 1.2(4,000,000) + 1.6(0.8)(4,000,000) = 9,920,000 in.-lb
Assume the column dimensions.
Section height h = 25 in.
Section width b = 25 in.
Check if slenderness is an issue.
Effective length 1 k )(05.07.0 BA ++ = 0.7 + 0.05(1.3 + 0.9) = 0.81
Effective length 2 k min05.085.0 + = 0.85 + 0.05(0.9) = 0.895
Effective length k = 0.81
Radius of gyration r = h3.0 = 0.3(25) = 7.5 in.
Slenderness ratio r
kL =
5.7
)1230)(81.0( = 38.88
Max slenderness ratio r
kL
2
11234M
M =
000,920,9
000,440,71234 = 25
Since 38.88 > 25, slenderness needs to be considered. The frame is braced, so there is no side
sway. Check that the moment is greater than the minimum required.
Minimum moment 2M = )03.06.0( hPu + = 880,000(0.6 + 0.03 25) = 1,188,000 in.-
2M = 9,920,000 in.-lb
Now calculate the magnified moment for a non-sway frame.
Concrete strength cE = cfw 5.133 = 8000)150(33 5.1 = 5.422 106 psi
Moment of inertia gI = 12
3bh =
12
)25(25 3
= 32,552 in.4
179
8
EI EI = dns
gc IE
+1
40.0 =
5.01
)552,32)(10422.5(40.0 6
+ = 4.707 1010 in.2-lb
Euler load cP = 2
2
)(kL
EI =
2
102
)123081.0(
)10707.4( = 5,463,464 lb
mC = +2
14.06.0M
M = +
000,920,9
000,440,74.06.0 = 0.9
Magnification factor ns = )75.0/(1 cu
m
PP
C =
)464,463,575.0/(000,8801
mC = 1.146
Design moment uM = 2Mns = 1.146(9,920,000) = 11,369,772 in.-lb
Therefore, design the section as a non-slender column with uP = 880,000 lb, uM = 11,369,772
in.-lb, and e = u
u
P
M = 12.92 in.
Top steel depth d = 3 in.
Bottom steel depth d = h 3 = 25 – 3 = 22 in.
Reinforcement ratios = = 0.015
Choose the longitudinal reinforcement (the same for both top and bottom).
Required area stA = bd = (0.015)(25)(22) = 8.25 in.2
Use 7 No. 10 bars stA = (7)(1.27) = 8.89 in.2
Check for compression failure using = 0.65 and for cf = 8000 psi, 1 = 0.65.
Neutral axis depth c = 0.6 d = 0.6(22) = 13.2 in.
Compression block a = c1 = (0.65)(13.2) = 8.58 in.
Concrete force CF = bafc85.0 = (0.85)(8000)(25)(8.58) = 1,458,600 lb
Top steel strain s = c
dc003.0 =
2.13
32.13003.0 = 0.00232 (compression)
Top steel stress sf = ssE = (29 106)(0.00232) = 67,227 psi > 60,000 psi
sf = 60,000 psi
Top steel force SF = ss fA = (8.89)(60,000) = 533,400 lb
Bottom steel strain s = c
cd003.0 =
2.13
2.1312003.0 = 0.002 (tension)
Bottom steel stress sf = ssE = (29 106)(0.002) = 58,000 psi < 60,000 psi OK
Bottom steel force SF = ss fA = (8.89)(58,000) = 515,620 lb
Nominal strength nP = CF + SF SF = 1,476,380 lb
Strength uP = nP = 0.65(1,476,380) = 959,647 lb
180
9
Nominal moment nM = )()(
2dyFdyF
ayF ssC ++
= )35.12)(400,533(2
58.825)600,458,1( +
)225.12)(620,515(
= 21,940,796 in.-lb
Eccentricity e = n
n
P
M =
380,476,1
796,940,21 = 14.86 in.
Since 14.86 in. > 12.92 in., the column is compression controlled. Choose a larger ratio and iterate
until the calculated eccentricity is 12.92 in. After a few iterations, the ratio is about 0.641.
Neutral axis depth c = 0.641 d = 0.641(22) = 14.10 in.
Compression block a = c1 = (0.65)(14.10) = 9.17 in.
Concrete force CF = bafc85.0 = (0.85)(8000)(25)(9.17) = 1,558,271 lb
Top steel strain s = c
dc003.0 =
10.14
310.14003.0 = 0.00236 (compression)
Top steel stress sf = ssE = (29 106)(0.00236) = 68,492 psi > 60,000 psi
sf = 60,000 psi
Top steel force SF = ss fA = (8.89)(60,000) = 533,400 lb
Bottom steel strain s = c
cd003.0 =
10.14
10.1422003.0 = 0.00168 (tension)
Bottom steel stress sf = ssE = (29 106)(0.00168) = 48,725 psi < 60,000 psi OK
Bottom steel force SF = ss fA = (8.89)(48,725) = 433,169 lb
Nominal strength nP = CF + SF SF = 1,658,502 lb
Strength uP = nP = 0.65(1,658,502) = 1,078,026 lb
Nominal moment nM = )()(
2dyFdyF
ayF ssC ++
= )35.12)(400,533(2
17.925)271,558,1( +
)225.12)(169,433(
= 21,519,003 in.-lb
Strength uM = nM = 0.65(21.519,003) = 13,987,353 in.-lb
Eccentricity e = n
n
P
M =
502,658,1
003,519,21 = 12.97 in.
The strength of the column is sufficient (1,078,026 lb > 880,000 lb and 13,987,353 in-lb >
11,369,772 in-lb). Now design the ties. Since the longitudinal bars are No. 10 bars, use No. 4 ties.
The spacing is the smallest of
1. 48 times No. 4 bar diameter = 48(0.5) = 24 in.
181
10
2. 16 times No. 8 bar diameter = 16(1.25) = 20.32 in.
3. Smaller column dimension = 25 in.
Therefore, use No. 4 ties spaced 20 in. c-c.
182
183
184
185
186
187
188
189
190
191
192
193
194
9.16. A nonslender square corner column is subjected to biaxial bending about its x and y axes. It supports a fac-tored load Pu = 500,000 lb acting at eccentricities ex = ey = 6 in. Design the column size and reinforcementneeded to resist the applied stresses. Given:
Solve by using all the three methods.
d¿ � 2.5 in.
gross reinforcement percentage �g � 0.03
fy � 60,000 psi
f ¿c � 6000 psi
11
Determine the design loads.
Factored load uP = 500,000 lb
Nominal load nP = uP
= 65.0
000,500 = 769,231 lb
Factored x moment uxM = yueP = (500,000)(4) = 2,000,000 in.-lb
Nominal x moment nxM = uxM
= 65.0
000,000,2 = 3,076,923 in.-lb
Factored y moment uyM = xueP = (500,000)(4) = 2,000,000 in.-lb
Nominal y moment nyM = uyM
= 65.0
000,000,2 = 3,076,923 in.-lb
Then make a reasonable guess about the required dimensions of the column. Choose a total
reinforcement ratio of g = 0.03 and try 4 bars per side. Since the moments are equal, use b
h = 1.
Axial column area axialA = )(45.0 gyc
n
ff
P
+ =
)03.0000,606000(45.0
231,769
+ = 219 in.2
Twice the area gA = 2(219) = 438 in.2
Choose a section with h = b = 22 in. Try using 5 bars per side for a total of 16 bars. For the
chosen reinforcement ratio, the steel area required per bar is 0.03 22 22 / 16 = 0.90 in.2, so try 5
No. 9 bars per side. Each side has 5(1.0) = 5.0 in.2 of reinforcement steel.
Each method requires the nominal load strength and bending moment of the section. Calculate the
load and moment strength for bending along the x-axis. First check for compression failure.
Neutral axis depth c = 0.6 d = 11.7 in.
Compression block a = c1 = 8.775 in.
Concrete force CF = bafc85.0 = 984,555 lb
Top steel strain s = c
dc003.0 = 0.00236 (compression)
Top steel stress sf = ssE = 68,410 psi > 60,000 psi so use 60,000 psi
Top steel force SF = ss fA = 300,000 lb
195
12
Bottom steel force SF = ss fA = 290,000 lb
Nominal strength nP = CF + SF SF = 994,555 lb
Nominal moment nM = )()(
2dyFdyF
ayF ssC ++ = 11,525,370 in-lb
Eccentricity e = n
n
P
M = 11.59 in.
Since 11.59 in. > 6 in., the column is compression controlled for bending along the x-axis. Choose
a larger ratio and iterate until the calculated eccentricity is 6 in. After a few iterations, the ratio is
about 0.857.
Neutral axis depth c = 0.857 d = 16.72 in.
Compression block a = c1 = 12.54 in.
Concrete force CF = bafc85.0 = 280,076 lb
Top steel strain s = c
dc003.0 = 0.002551 (compression)
Top steel stress sf = ssE = 73,991 psi > 60,000 psi so use 60,000 psi
Top steel force SF = ss fA = 300,000 lb
Bottom steel strain s = c
cd003.0 = 0.0005 (tension)
Bottom steel stress sf = ssE = 14,467 psi < 60,000 psi OK
Bottom steel force SF = ss fA = 72,335 lb
Nominal strength nP = CF + SF SF = 1,634,630 lb
Nominal moment nM = )()(
2dyFdyF
ayF ssC ++ = 9,819,937 in.-lb
Eccentricity e = n
n
P
M = 6.01 in.
In this case the column is square, so the same calculations apply to bending about the y-axis.
Now check the design.
Load Contour Method
Since the loads are equal, use oxM , where b = h = 22 in., nynx MM = = 4,615,385 in.-lb, and
start with = 0.5.
Required strength, x oxM = + 1
h
bMM nynx = 9,230,769 in.-lb
Bottom steel strain s = c
cd003.0 = 0.002 (tension)
Bottom steel stress sf = ssE 58,000 psi < 60,000 psi
196
13
From the section calculations, oxnM = 9,819,937 in.-lb > 9,230,769 in.-lb, so the strength of the
section is sufficient. Now check nyM . Using oxn
nx
M
M = 0.5 and = 0.5 in the figure,
oyn
ny
M
M = 0.5.
Therefore, nyM = 0.5 oynM , which is 4,909,968 in.-lb. Since nyM = 3,076,923 in.-lb, the strength
is sufficient about this axis as well.
Reciprocal Load Method
Assume the same design as the previous method. Therefore, lb 630,634,1== nynx PP . Calculate
the strength of the column for an axial load, where Ast = 16(1.0) = 16 in.2
Axial strength 0nP = yststgc fAAAf +)(85.0 = 3,346,800 lb
Then calculate the nominal load strength.
Axial strength nP
1 =
0
111
nnynx PPP+
nP = 1,081,401 lb > 769,230 lb
Modified Load Contour Method
Assume the same design and calculate the value of
5.15.1
0
++nby
ny
nbx
nx
nbn
nbn
M
M
M
M
PP
PP, where
nbP = 994,555 lb, nbxM = nbyM = 11,525,370 in.-lb, nxM = nyM = 4,615,385 in.-lb, and 0nP is
the same as the value found for the Reciprocal Load Method.
5.15.1
0
++nby
ny
nbx
nx
nbn
nbn
M
M
M
M
PP
PP = 0.411, which is significantly less than 1, which indicates
that the section is stronger than it needs to be.
197
198
199
200
201
202
203
204
205
10.1. Calculate the basic development lengths in tension for the following deformed bars embedded in normal-weight concrete.(a) No. 4, No. 10. Given:
(b) No. 14, No. 18. Given:
fy � 80,000 psi
fy � 60,000 psi
f ¿c � 4000 psi
fy � 60,000 psi 1413.7 MPa2
f ¿c � 6000 psi 141.4 MPa2
1
For a No. 4 bar:
Bar diameter bd = 0.5 in.
Bar area bA = 0.2 in.2
Assume the clear cover is 1.5 in. and the clear spacing between the bars is 2db. Assume the bars
are not being used for top reinforcement, and that they are not coated. Therefore, 1== et .
For a No. 4 bar, 8.0=s . Use Ktr = 0. Determine the value for c.
c = 0.5db + (clear cover) = 0.5(0.5) + 1.5 = 1.75 in.
c = 0.5(clear spacing + db) = 0.5(2 0.5 + 0.5) = 0.75 in. (controls)
Check the values.
cf = 6000 = 77.5 psi < 100 psi OK
b
tr
d
Kc + =
5.0
075.0 + = 1.5 OK
Calculate the development length.
dL = +
b
tr
ste
c
y
b
d
Kcf
fd
40
3 =
)5.1)(5.77(40
)8.0)(0.1)(0.1)(000,60(3)5.0(
= 15.49 in.
For a No. 10 bar:
Bar diameter bd = 1.27 in.
Bar area bA = 1.27 in.2
206
(a)
Assume the clear cover is 1.5 in. and the clear spacing between the bars is 2db. Assume the bars
are not being used for top reinforcement, and that they are not coated. Therefore, 1== et .
For a No. 10 bar, 0.1=s . Use Ktr = 0. Determine the value for c.
c = 0.5db + (clear cover) = 0.5(1.27) + 1.5 = 2.14 in.
c = 0.5(clear spacing + db) = 0.5(3 1.27) = 1.91 in. (controls)
Check the values.
cf = 6000 = 77.5 psi < 100 psi OK
b
tr
d
Kc + =
27.1
091.1 + = 1.5 OK
Calculate the development length.
dL = +
b
tr
ste
c
y
b
d
Kcf
fd
40
3 =
)5.1)(5.77(40
)0.1)(0.1)(0.1)(000,60(3)27.1(
= 49.19 in.
207
208
209
210
211
212
213
214
10.5. An 15-ft (4.57-m) normal-weight concrete cantilever beam is subjected to a factored Mu = 4,000,000 in-lb (452kN-m) and a factored shear Vu = 50,000 lb (222 kN) at the face of the support. Design the top reinforcementand the appropriate embedment of 90° hook into the concrete wall to sustain the external shear and moment.Given:
fy � 60,000 psi
f ¿c � 5000 psi
4
For a cantilever beam, the minimum depth for deflection is L / 8 or (12 15) / 8 = 22.5 in.
Beam height h = 24 in.
Effective depth d = 21 in.
Beam width b = 15 in.
Beam length L = 15 ft
Design the beam for flexure.
Nominal moment nM =
Mu = 4,000,000
0.9= 4,444,444 in.-lb
Section strength nM 2
adfA ys and a =
bf
fA
c
ys
85.0
Solve for sA sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c = 3.86 in.2 (controls)
Minimum area 1 sA bdf
f
y
c3 = 1.11 in.2
Minimum area 2 sA bdf y
200 = 1.05 in.2
The required area is 3.86 in.2 Use 4 #9 bars. Verify that the strength is sufficient.
Steel area sA = (4)(1) = 4 in.2
Compression block a = bf
fA
c
ys
85.0 = 3.76 in.
Nominal strength nM = 2
adfA ys = 4,129,411 in.-lb > 4,000,000 in.-lb.
Design the anchor.
Bar diameter bd = 1.128 in.
Multiplier e = 1.0 for uncoated bars
Aggregate factor = 1.0 for normal concrete
Basic length hbL = b
c
yed
f
f02.0 = 19.14 in.
215
5
Multiplier d = st
req
A
A = 0.9653
Development length dhL = hbd L = 18.48 in. (> 8 bd and > 6 in. so OK)
Hook length hookL = 12 bd = 13.54 in.
Therefore, use a hook with dhL = 18.5 in and hookL = 14 in.
216
217
218
219
220
221
11.1 An end panel of a floor system supported by beams on all sides carries a uniform service live load wL = 80 psfand an external dead load wD = 20 psf in addition to its self-weight. The center-line dimensions of the panel are20 ft × 30 ft (the dimension of the discontinuous side is 20 ft). Design the panel and the size and spacing of the re-inforcement using the ACI Code direct design method. Given:
= 4000 psi, normal-weight concrete
fy = 60,000 psi
column sizes 24 in. × 24 in.
width of the supporting beam webs = 15 in.
f ¿c
1
Assume the following dimensions.
Slab thickness sh = 8 in.
Beam section height bh = 27 in.
Use a depth of 24 in. for the beam steel, a depth of 7 in. for the East-West slab steel, and a depth
of 6.375 in. for the North-South slab steel. Check the conditions for using the Direct Design
Method.
1. The maximum length ratio is 20
30 = 1.5 which is less than 2.
2. Assume there are more than three panels in each direction.
3. Check the live load to dead load ratio.
dw = 150(8/12) + 20 = 120 lb / ft2
lw = 80 lb/ft2 < 2 dw
Now calculate the stiffness properties. The beams on the north, south, and east sides are all the
same, which the beam on the west side is L shaped. because it is the edge of the panel.
Calculate the moments of inertia for each beam.
T section y = )53)(8()19)(15(
)2/819)(53)(8()2/19)(19)(15(
+++
= 17.57 in. (from bottom)
1bI = 23
23
)57.1723)(53)(8(12
)8)(53()57.175.9)(19)(15(
12
)19)(15( +++
= 41,897 in.4
222
2
L section y = )34)(8()19)(15(
)2/819)(34)(8()2/19)(19)(15(
+++
= 16.09 in.
2bI = 23
23
)09.1623)(34)(8(12
)8)(34()09.165.9)(19)(15(
12
)19)(15( +++
= 35,389 in.4
Then calculate the moments of inertia for the slabs.
Strip Width Height Moment of Inertia
N-S strip on west edge 2
24
2
1230 + = 192 in. 8 in. 12
)8)(192( 3
= 8192 in.4
N-S strip on east edge 1230 = 360 in. 8 in. 12
)8)(360( 3
= 15,360 in.4
E-W strips 1220 = 240 in. 8 in. 12
)8)(240( 3
= 10,240 in.4
Use these values to find the ratios of the beam to slab stiffness along each edge of the panel.
Edge Beam Slab Ratio
North 41,897 in.4 10,240 in.4 = 4.09
South 41,897 in.4 10,240 in.4 = 4.09
East 41,897 in.4 15,360 in.4
= 2.73
West 35,389 in.4 8192 in.4 = 4.32
Average m = 3.81
Now check the minimum slab thickness, using the appropriate equation for m > 2.5.
Length ratio = 241220
241230 = 1.56
Minimum h = 936
)000,200/8.0(
++ yfL
= 7.39 in.
The slab thickness 8 in. > 7.39 in. Check the shear capacity. Use the depth for the steel in the N-S direction since the shortest span is in the N-S direction.
Factored load uw = ld ww 6.12.1 + = 272 lb/ft2
Factored shear uV = 2
15.1 minLwu = 2815.2 lb/ft
Nominal shear nV = 75.0
uV = 3753.6 lb/ft
Shear capacity cV = dfc2 = )7)(5000(2 = 11,879 lb/ft
223
3
cV > nV , so the capacity is sufficient with the given thickness. Now calculate and distribute the
moments.
East-West Direction
Column width cb 24 in.
Span 1S 30 ft
Perpendicular span 2S 20 ft
Strip length 1L = cbS1 336 in.
Minimum strip length 1L = 165.0 S 234 in.
Static moment oM = 8
2
12LSwu 6,397,440
in-lb
Exterior negative moment uM = oM16.0 1,023,590 in-lb
Interior negative moment uM = oM7.0 4,478,208 in-lb
Positive moment uM+ = oM57.0 3,646,541 in-lb
Column strip
Exterior negative moment cuM = )(85.0 uM 870,052 in-lb
Interior negative moment cuM = )(85.0 uM 3,806,477 in-lb
Positive moment cuM+ = )(85.0 uM+ 3,099,560 in-lb
Positive design moment cnM+ = cuM+
3,443,955
in-lb
Negative design moment cnM = )max( cuM
4,229,419
in-lb
Strip width cw = 25.0 S 10 ft
Beam width cbw 53 in.
Slab width csw = cbc ww 67 in.
Positive beam moment )(85.0 cnM+ 2,927,362 in-lb
Negative beam moment )(85.0 cnM 3,595,006 in-lb
Positive slab moment / foot cs
cn
w
M )(15.0 + 92,524
in-lb/ft
Negative slab moment / foot cs
cn
w
M )(15.0 113,626
in-lb/ft
Middle strip
Exterior negative moment muM = )(15.0 uM 153,538 in-lb
Interior negative moment muM = )(15.0 uM 671,731 in-lb
Positive moment muM+ = )(15.0 uM+ 546,981 in-lb
Strip width msw = cwS2 10 ft
Positive design moment / foot mnM+ = ms
mu
w
M+ 60,776
in-lb/ft
224
4
Negative design moment / foot mnM = ms
mu
w
M )max( 74,637
in-lb/ft
Calculate the required reinforcement.
Beam: uM+ = 2,927,362 in-lb, uM = 3,595,006 in-lb, d = 24 in., b = 15 in.
Minimum 1 sA y
c
f
bdf3 1.27
in.2
Minimum 2 sA yf
bd200 1.2
in.2
Bottom sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c 2.12
in.2
Top sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c 2.63
in.2
Use 3 No. 9 bars (3.0 in.2) for both the top and bottom steel of the E-W beams.
Column strip: uM+ = 92,524 in-lb, uM = 113,626 in-lb, d = 7 in., b = 12 in., h = 8 in.
Minimum sA bh0018.0 0.1728 in.2/ft
Bottom sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.2476
in.2/ft
Top sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.3057
in.2/ft
Using No. 5 bars, the required spacings are 15.03 in., and 12.17 in., but the maximum spacing is
12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the E-W column strips.
Middle strip: uM+ = 60,776 in-lb, uM = 74,637 in-lb, d = 7 in., b = 12 in., h = 8 in.
Minimum sA bh0018.0 0.1728 in.2/ft
Bottom sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.1613
in.2/ft
Top sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.1988
in.2/ft
Using No. 5 bars, the required spacings are 21.53 in., and 18.71 in., but the maximum spacing is
12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the E-W middle strips.
225
5
North-South Direction
Column width cb 24 in.
Span 1S 20 ft
Perpendicular span 2S 30 ft
Strip length 1L = cbS1 216 in.
Minimum strip length 1L = 165.0 S 156 in.
Static moment oM = 8
2
12LSwu 3,965,760
in-lb
Interior negative moment uM = oM65.0 2,577,744 in-lb
Positive moment uM+ = oM35.0 1,388,016 in-lb
Column strip
Interior negative moment cuM = )(6.0 uM 1,546,646 in-lb
Positive moment cuM+ = )(6.0 uM+ 832,810 in-lb
Positive design moment cnM+ = cuM+
925,344
in-lb
Negative design moment cnM = )max( cuM
1,718,496
in-lb
Strip width cw = 25.0 S 10 ft
Beam width cbw 53 in.
Slab width csw = cbc ww 67 in.
Positive beam moment )(85.0 cnM+ 786,542 in-lb
Negative beam moment )(85.0 cnM 1,460,722 in-lb
Positive slab moment / foot cs
cn
w
M )(15.0 + 24,860
in-lb/ft
Negative slab moment / foot cs
cn
w
M )(15.0 46,169
in-lb/ft
Middle strip
Interior negative moment muM = )(4.0 uM 1,031,098 in-lb
Positive moment muM+ = )(4.0 uM+ 555,206 in-lb
Strip width msw = cwS2 10 ft
Positive design moment / foot mnM+ = ms
mu
w
M+ 30,845
in-lb/ft
Negative design moment / foot mnM = ms
mu
w
M )max( 57,283
in-lb/ft
226
6
Calculate the required reinforcement.
Beam: uM+ =786,542 in-lb, uM = 1,460,722 in-lb, d = 24 in., b = 15 in.
Minimum 1 sA y
c
f
bdf3 1.27
in.2
Minimum 2 sA yf
bd200 1.2
in.2
Bottom sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c 0.55
in.2
Top sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c 1.03
in.2
Use 2 No. 9 bars (2.0 in.2) for both the top and bottom steel of the N-S beams.
Column strip: uM+ = 24,860 in-lb, uM = 46,169 in-lb, d = 7 in., b = 12 in., h = 8 in.
Minimum sA bh0018.0 0.1728 in.2/ft
Bottom sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.0595
in.2/ft
Top sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.1110
in.2/ft
Using No. 5 bars, the required spacings are 21.53 in., and 21.53 in., but the maximum spacing is
12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the N-S column strips.
Middle strip: uM+ = 30,845 in-lb, uM = 57,283 in-lb, d = 7 in., b = 12 in., h = 8 in.
Minimum sA bh0018.0 0.1728 in.2/ft
Bottom sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.0739
in.2/ft
Top sA 285.0
211
85.0
df
M
f
bdf
c
n
y
c 0.1380
in.2/ft
Using No. 5 bars, the required spacings are 21.53 in., and 21.53 in., but the maximum spacing is
12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the N-S middle strips.
227
7
Therefore, the reinforcement design is as shown below.
East-West Reinforcement
North-South Reinforcement
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
8
Slab thickness h = 16 in.
Steel depth d = 14 in.
Hinge field radius r = 24 in.
Calculate the load and effective circle.
Dead load dw = 150(16/12) + 20 = 220 lb / ft2
Live load lw = 80 lb/ft2
Factored load uw = 1.2wd +1.6wl = 392 lb/ft 2
Panel area A = (20)(30) = 600 ft2
Panel load P = Awu = 235,200 lb
Radius = A
4 = 27.63 ft
Assume MM = and calculate the required steel for a 12 in. strip (b = 12 in.)
Factored moment uM = r
P
3
21
4 = 1,844,675 in-lb / ft
Nominal moment nM = uM
= 2,049,639 in-lb / ft
Section strength nM 2
adfA ys and a =
bf
fA
c
ys
85.0
Solve for sA sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c = 2.87 in.2
Choose 3 No. 9 bars for 3.0 in.2 Check the strength and failure mode.
Compression block a = bf
fA
c
ys
85.0 = 4.41 in.
Neutral axis depth c = 1
a = 5.19 in.
Ratio d
c = 0.370 < 0.375 Tension controlled, OK
Strength nM = 2
adfA ys = 2,122,941 in-lb > 2,049,639 in-lb, OK
11.7 Use the yield-line theory to evaluate the slab thickness needed in the column zone of the flat plate in Problem11.5 for flexure, assuming that the hinge field would have a radius of 24 in.
256
257
258
259
260
261
11.10 Calculate the maximum crack width in a two-way interior panel of a reinforced concrete floor system. Theslab thickness is 7 in. (177.8 mm) and the panel size is 24 ft × 30 ft (7.32 m × 9.14 m). Also design the size andspacing of the reinforcement necessary for crack control assuming that (a) the floor is exposed to normal en-vironment; (b) the floor is part of a parking garage. Given: fy = 60.0 ksi (414 MPa).
9
Slab thickness h = 8 in.
Clear cover cc = 0.75 in.
Ratio r = 24
30 = 0.8
For a normal environment, try No. 4 bars ( bd = 0.5 in.)
Max crack width maxw = 0.012 in.
Coefficient K = 2.1 10 5 in.
2 / lb
Standard value = 1.2
Steel depth cd = bdcc 5.0+ = 1 in.
Calculate 1G 1G =
2
max
4.0 yfK
w = 393.6 in.2
Assume sss == 21 s = c
b
d
Gd
8
1 = 8.79 in.
Therefore, use No. 4 bars at 8.5 in. c-c for the crack reinforcement.
For a parking garage, try No. 4 bars ( bd = 0.5 in.)
Max crack width maxw = 0.016 in.
Coefficient K = 2.1 10 5 in.
2 / lb
Standard value = 1.2
Steel depth cd = bdcc 5.0+ = 1 in.
Calculate 1G 1G =
2
max
4.0 yfK
w = 699.9 in.2
Assume sss == 21 s = c
b
d
Gd
8
1 = 11.72 in.
Therefore, use No. 4 bars at 11.5 in. c-c for the crack reinforcement.
262
12.1. Design a reinforced concrete, square, isolated footing to support an axial column service live load PL =400,000 lb (1779 kN) and service dead load PD = 500,000 lb (2224 kN). The size of the column is 26 in. × 22 in.(0.66 m × 0.56 m). The soil test borings indicate that it is composed of medium compacted sands and gravelysands, poorly graded. The frost line is assumed to be 3 ft below grade. Given:
surcharge � 90 psf 14.3 kPa2
fy � 60,000 psi 1413.7 MPa2
column f ¿c � 4000 psi 127.6 MPa2
footing f ¿c � 3000 psi 120.7 MPa2
average weight of soil and concrete above the footing, � � 140 pcf 122.0 kN>m32
1
Given: Column dimensions: 26 in. 22 in.
Combined soil and concrete weight: = 140 pcf
Surcharge: 90 psf
Frost line depth: 3 ft
Assume the following dimensions, and try using No. 7 bars throughout.
Slab thickness h = 3 ft
Clear cover at base cc = 3 in.
Bar diameter bd = 0.875 in.
Bar area bA = 0.6 in.2
Average steel depth d = bdcch = 32.125 in.
For the given soil and concrete types:
Aggregate factor = 1
Soil capacity soilP = 2.5 ton/ft2 = 5000 psf
Calculate the required footing area using the service load and the allowable soil pressure.
Allowable soil load nP = surcharge)frost( +hPsoil = 4070 psf
Service load P = DP + LP = 900,000 lbf
Required area fA = nP
P = 221.13 ft2
Choose size (square) b = 15 ft ( fA = 225 ft2)
Calculate the factored load intensity.
Factored load UP = LD PP 6.12.1 + = 1,240,000 lbf
Load intensity Uq = f
U
A
P = 5511.11 psf
263
2
Consider the shear capacity for beam action.
Factored shear UV = )15(12
125.32
122
22
2
15Uq = 322,917 lbf
Nominal shear nV = 75.0
UV = 430,556 lbf
Shear capacity cV = bdfc2 = 633,441 lbf > 430,556 lbf, OK
Consider the shear capacity for two-way action.
Factored shear UV = ++
12
125.3226
12
125.3222)15)(15(Uq = 1,119,597 lbf
Nominal shear nV = 75.0
UV = 1,492,796 lbf
Perimeter 0b = )125.3226125.3222(2 +++ = 224.5 in.
Footing size ratio = 1
Column type factor s = 40 for an interior column
Shear capacity 1 cV = dbfc 0
42 + = 2,370,126 lbf
Shear capacity 2 cV = dbfb
dc
s0
0
2+ = 3,051,075 lbf
Shear capacity 3 cV = dbfc 04 = 1,580,083 lbf (controls) > 1,492,796 lbf, OK
Design the reinforcement for the moment at the critical section. The critical section is plane
parallel to the wider column face.
Factored moment UM =
2
2
22
2
1215)15(
2
1Uq = 21,496,778 in-lb
Nominal moment nM = 9.0
UM = 23,885,309 in-lb
Minimum 1 sA bd0018.0 = 10.4085 in.2
Minimum 2 sA 285.0
211
85.0
bdf
M
f
bdf
c
n
y
c = 12.7211 in.2
That requires 22 No. 7 bars for an area of 13.2 in.2 For a footing width of 15 ft, the required
spacing is 15 12/22 = 8.18 in. so use 22 No. 7 bars spaced 8 in. c-c. for both directions. Check
the development length for the reinforcement bars. Use 1=== est and 0=trK .
264
3
Effective cover factor b
trb
d
Kc + = 3.428 > 2.5 so use 2.5
Strength factor cf = 54.77 < 100, OK
Development length dl = best
c
yd
f
f
5.240
3 = 28.75 in.
Available space 2
26
2
1215 = 79 in. > 28.75 in. OK
Check the bearing strength for the column and the footing.
Column area 1A = 144
2622 = 3.9722 ft
2
Column bearing UP = 1)85.0(7.0 Afc = 1,361,360 lbf > 1,240,000 lbf, OK
Footing similar area 2A =
2
126
1215A = 190.38 ft
2
Area factor 1
2
A
A = 6.923 > 2.0 so use 2.0
Footing bearing UP = 1)85.0)(7.0(0.2 Afc = 2,042,040 lbf > 1,240,000 lbf, OK
Since the bearing strength is adequate, only the minimum dowel area is required.
Minimum area dA = 1005.0 A = 2.86 in.2
Using No. 7 bars, 5 are required. Use 6 No. 7 dowels for symmetry. Space 3 dowels 8.5 in. c-c
along each 26 in. face of the column. Check the development length in the column and the footing
Minimum dl yb fd0003.0 = 15.75 in.
Footing dl = c
yb
f
fd02.0 = 19.17 in. (controls)
Column dl = c
yb
f
fd02.0 = 16.60 in. (controls)
The available space in the footing is bdcch 3 = 30.375 in. > 19.17 in. OK
265
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266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
13.3. Find the moments and shears caused by a wind intensity of 30 psf acting on the structural system in Ex. 13.5using the portal method of analysis. Use a height between floors of 10′ – 6″ (3.20 m).
1
Given:
Floor height: uL = 10.5 ft
Roof parapet: pL = 2.5 ft
Section width: L = 20 ft
Wind load: uw = 30 psf
Calculate the force at each floor and at the roof.
Roof roofF = upu LwLL
+2
7.1 = 7905 lb
Floor floorF = uu LwL7.1 = 10,710 lb
Therefore the wind load on the building is as shown.
The shear in the each floor supports the wind forces for all the floors above, and the interior
columns carry twice the shear as the exterior columns. The moment is the shear multiplied by
half the floor height.
Shear (lb) Moment (ft-lb)
Floor Shear Interior Exterior Interior Exterior
1 3(10,710) + 7905 = 40,035 lb 13,345 6672.5 70,061 35,031
2 2(10,710) + 7905 = 29,325 lb 9775 4887.5 51,319 25,659
3 1(10,710) + 7905 = 18,615 lb 6205 3102.5 32,576 16,288
4 7905 lb 2635 1317.5 13,834 6917
297
2
For the beams, the moment at each beam/column joint must sum to zero, and the unbalanced
moment from the columns is assumed to be split evenly between the beams for the interior joints.
The vertical shear in the beams is the moment divided by half the span between the beams.
Moment (ft-lb) Shear (lb)
Floor Interior Exterior Interior Exterior
2 0.5(70,061 + 51,319) = 60,690 35,031 + 25,659 = 60,690 5057.5 6936
3 41,948 41,948 3495.6 4794
4 23,205 23,205 1933.8 2652
R 6917 6917 576.4 790.5
The shear forces in the columns and beams are shown in the figure below.
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14.1. An AASHTO prestressed simply supported I beam has a span of 40 ft (12.9 m) and is 42 in. (106.7 cm) deep.Its cross section is shown in Figure 14.18. It is subjected to a live-load intensity WL = 4800 plf (58.4 kN/m). De-termine the required �� -in.-diameter, stress-relieved, seven-wire strands to resist the applied gravity load andthe self-weight of the beam, assuming that the tendon eccentricity at midspan is ec = 18.42 in. (467.9 mm).Maximum permissible stresses are as follows:
The section properties, given these stresses, are
WL � 4800 plf WD � 568 plf
St � 4531 in.3 Sb � 5900 in.3 cb � 18.24 in.
r 2 �Ic
Ac� 197.5 in.2
Ig � 107,645 in.4 Ac � 545 in.2
fpe � 145,000 psi 11000 MPa2 fpi � 189,000 psi 11303 MPa2 fpu � 270,000 psi 11862 MPa2
ft � 122f ¿c � 930 psi 16.4 MPa2 � 2700 psi 126.7 MPa2
fc � 0.45f ¿c f ¿c � 6000 psi 141.4 MPa2
1
6
4
85
8
8
2 0
Figure 14.18
Given:
Concrete area: cA = 545 in.2
Strands: 1/2 od, 7 wire strand, sA = 0.153 in.2
Final pretension stress: pef = 145,000 psi
Dead load: Dw = 568 lb/ft
Live load: Lw = 4800 lb/ft
Mid span eccentricity: ce = 18.42 in.
Section center of gravity: c = 18.24 in.
Section depth: h = 42 in.
Section radius of gyration: 2r = 197.5 in.
2
Section bottom modulus: bS = 5900 in.3
Section top modulus: tS = 4531 in.3
Maximum tensile stress: tf = 930 psi
Maximum compressive stress: cf = 2700 psi
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2
Let s be the number of strands needed.
Total steel area psA = sAs = 0.153s in.2
Final pretension eP = peps fA = 21,380.28s lb
Mid span moment TM = 88
22 LwLw LD + = 12,882,500 in-lb
Stress at top tf = t
Ttc
c
e
S
M
r
ce
A
P2
1 = 47.68s 2843 psi
Stress at bottom bf = b
Tbc
c
e
S
M
r
ce
A
P+
21 = 105.98s + 2183.38 psi
Calculate the minimum number of strands.
Strands for top s 68.47
2843+tf = 2.9994
Strands for bottom s 98.105
38.2183cf = 11.8315
Therefore, use 12 1/2” O.D. 7 wire strands for pretensioning.
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16.1. A 3 × 18 panel, ductile, moment-resistant category II, site class B frame building has a ground story 12 ft high(3.66 m) and 10 upper stories of equal height of 10 ft (3.05 m). Calculate the base shear V and the overturningmoment at each story level in terms of the weight Ws of each floor. Use the Equivalent Lateral Force Methodin the solution. Given:
Ws per floor � 2,000,000 lb 18896 kN2
R � 5
S1 � 0.34 sec., Ss � 0.90 sec.
1
Building height bh = 12 + (10)(10) = 112 ft
Site coefficient af = 1.0 for SS = 0.90
Site coefficient vf = 1.0 for 1S = 0.34
Importance factor I = 1 for class II
Spectral response
Maximum response MSS = SaSf = 0.9
1MS = 1Sfv = 0.34
Damped response DSS = MSS3
2 = 0.6
1DS = 13
2MS = 0.2267
Calculate the seismic response coefficient.
Nominal value SC = IR
SDS
/ = 0.15
Period coefficient TC = 0.020 for generic building
Approximate period aT = 4/3
bT hC = 0.6886 s
Limit coefficient UC = 1.37 for 1DS = 0.2267
Maximum period aT = aUTC = 0.9456 s
Building period T = aT2.1 = 1.1348 s
Maximum value SC TIR
SD
)/(
1 = 0.049938 (controls)
Minimum value SC DSS044.0 = 0.0264
Calculate the total shear.
Weight at base W = sW11 = 22,000,000 lb
Shear at base V = WCS = 0.549313 sW = 1,098,625 lb
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2
The coefficient for the force at each floor is vxC =
=
n
i
k
is
k
xs
hW
hW
1
, where ih is the height of floor i
above the ground and k = 1)12(5.05.2
5.0 +T = 1.317376.
The force on each floor is xF = VCvx , the story shear is xV = =
x
i
iFV0
, and the overturning
moment is xM = )( xi
n
xi
i hhF=
. For the top ten stories, = 1.
Denominator for vxC =
n
i
k
ishW1
= 2672.35 sW
Floor xh vxC xF xV xV (lb) xM (ft-lb)
Base 0 0 0 0.549313 sW 1,098,625 89,759,342
1 12 0.009881 0.005428 sW 0.543885 sW 1,087,770 76,575,838
2 22 0.021957 0.012061 sW 0.531824 sW 1,063,647 65,698,137
3 32 0.035971 0.019759 sW 0.512064 sW 1,024,128 55,061,666
4 42 0.051468 0.028272 sW 0.483792 sW 967,584 44,820,384
5 52 0.068191 0.037458 sW 0.446334 sW 892,668 35,144,540
6 62 0.085973 0.047226 sW 0.399108 sW 798,216 26,217,859
7 72 0.104691 0.057508 sW 0.341600 sW 683,200 18,235,695
8 82 0.124256 0.068256 sW 0.273344 sW 546,689 11,403,697
9 92 0.144595 0.079428 sW 0.193917 sW 387,833 5,936,809
10 102 0.165648 0.090993 sW 0.102924 sW 205,848 2,058,477
11 112 0.187368 0.102924 sW 0 0 0
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17.1. Design for flexure and shear an 8 in. thick a grouted CMU masonry lintel having a 20 ft span and a bear-ing length of 20 in. on the supporting wall to support a roof transmitting 350 plf load to the lintelGiven:
Effective t 7.63 in.
Unit weight 80 psf
Wall height to roof 14 ft.
Height of lintel
Parapet height
Service live load on roof 200 plf
Governing factored loading
163 psi
1500 psi
60,000 psify
fm
fr
1.2D � 1.6L
3¿-6–5¿-3–
1
Given:
Lintel width b = 7.63 in.
Lintel height h = 5 -3 = 63 in.
Lintel span L = Clear span + 1/2 bearing = 20 + 10 = 250 in.
Dead load Dw = bh80350 + = 770 lb/ft
Live load Lw = 200 lb/ft
Masonry strength mf = 1500 psi
Rupture strength rf = 163 psi
Steel strength yf = 60,000 psi
Steel modulus sE = 29 106 psi
Calculate the loads.
Factored load Uw = LD ww 6.12.1 + = 1244 lb/ft
Factored shear UV = 2
LwU = 12,958.33 lb
Nominal shear nV = 80.0
UV = 16,197.92 lb
Factored moment UM = 8
2LwU = 809,895.83 in.-lb
Nominal moment nM = 90.0
UM = 899,884.26 in.-lb
Check the design for shear. Assume 3” cover for the steel reinforcement so d = h 3 = 60 in.
Shear plane area nA = bd = 457.8 in.2
Masonry strength mV = mn fA)75.14( = 39,893.66 lb (controls)
Maximum strength nV = mn fA4 = 70,922.07 lb
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Since 39,893.66 lb > 16,197.92 lb, the shear strength is sufficient.
Determine the required reinforcement steel area.
Required area nM = 2
adfA ys and a =
bf
fA
m
ys
80.0
Solve for sA sA = 280.0
211
80.0
bdf
M
f
bdf
m
n
y
m = 0.2535 in.2
Check the cracking strength.
Critical moment 1.3Mcr = rfbh
63.1
2
= 1,069,511 in.-lb > 899,884.26 in.-lb
Required steel area sA = s
n
cr AM
M3.1 = 0.3013 in.2
Therefore, choose 2 No. 4 bars for an area of 2(0.2) = 0.4 in.2 Check the reinforcement ratio.
Actual ratio = bd
As = 0.000874
Max steel strain y = s
y
E
f = 0.002069
Max masonry strain mu = 0.0025 for concrete
Load factor = 1.5
Maximum ratio + ymu
mu
y
m
f
f64.0 = 0.007138 > 0.000874 OK
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17.8. Design a tension anchor for a grouted masonry wall to withstand a tensile force of magnitudeand a shear force without causing any masonry breakout. Consider the di-
rection of the shear force to be along the wall length.Given
fm � 1500 psi
fye � 27,000 psi
fy � 36,000 psi
Pv � 3000 lbPu � 14,000 lb
3
Given:
Bolt axial strength yf = 36,000 psi
Bolt shear strength yvf = 27,000 psi
Masonry strength mf = 1500 psi
Axial load aP = 14,000 lb
Shear load vP = 3000 lb
Try a design using bN = 2 bolts with a diameter of bd = 0.75 in. and an embedment length of bI
= 6.5 in. Space the bolts 13 in. apart to avoid overlap of the anchorage areas.
Bolt area bA = 4
2
bb dN = 0.8836 in.2
Masonry axial area ptA = 2
bb IN = 265.46 in.2
Masonry shear area pvA = 2
2
bb IN = 132.73 in.2
Masonry axial failure anB = mpt fA4)5.0( = 20,562.8 lb (controls) > 14,000 lb OK
Bolt axial failure anB = yb fA)9.0( = 28,627.8 lb
Masonry shear failure vnB = mpv fA4)5.0( = 10,281.4 lb (controls) > 3000 lb OK
Bolt shear failure vnB = yvb fA)6.0)(9.0( = 12,882.5 lb
Combined load vn
v
an
a
B
P
B
P+ = 0.97263 < 1 OK
Therefore, the design is sufficient.
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