Chapter 4 4-1 Figure P4-1 shows a simply supported beam and the cross-section at midspan. The beam supports a uniform service (unfactored) ( unfactored) dead load consisting of its own weight plus 1.4 kips/ft and a uniform service (unfactored) live load of 1.5 kip/ft. The concrete strength is 3500 psi, and the yield strength of the reinforcement is 60,000 psi. The concrete is normalweight concrete. Use load and strength reduction factors from ACI Code Sections 9.2 and 9.3. For the midspan section shown in part (b) of Fig. P4-1, compute M n and show show that it exceeds M u .
1. Calculate the dead load of the beam. 24 12 .15 0.3 0.3 kips/ft 0.15 Weight/ft = 144 2. Compute the factored moment, M u . Factored load/ft: wu = 1.2(0.30 + 1.40) + 1.6(1.50) = 4.44 k/ft M u wu
2
8 4.44 20 8 222 kips-ft 2
3. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: A s = 3 No. 9 bars = 3 1.00 1.00 in.2 = 3.00 in.2 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f 3.00 3.00 6000 60000 0 s y 5.04 in. c 1 0.85 3500 3500 12 12 0.85 f ' b 0.85 c
For f c' 3500 psi, 1 0.85 . Therefore, c 5.04 5.93 in. 1 0.85 Check whether tension steel is yielding: 21.5 5.93 5.93 d c 21.5 using Eq.(4-18) 0.003 0.0078 0.00788 8 0.003 s t 5.93 c cu
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Compute the nominal moment strength, using Eq. (4-21): 5.04 3.0 3.00 6000 0000 21.5 21.5 2 M A f d 285 kips-ft n s y 2 12000 0.00788 8 0.005 0.005 the section is clearly tension-controlled and =0.9. Then, Since, t 0.0078 M n 0.9 0.9 285 285 kip-ft 256 kip-ft. Clearly, M n M u
4-1
4-2 A cantilever beam shown in Fig. P4-2. The The beam supports a uniform service (unfactored) dead load of 1 kip/ft plus its own dead load and it supports a concentrated service (unfactored) live load of 12 kips as shown. The concrete is normal-weight concrete with f c' 4000 psi and the steel is Grade 60. Use load and strength-reduction factors form ACI Code Section 9.2 and 9.3. For the end section shown in part (b) of Fig. P4-2, compute M n and show it exceeds M u .
1. Calculate the dead load of the beam. 30 18 0.15 0.56 0.563 3 kips/ft 0.15 Weight/ft = 144 2. Compute the factored moment, M u . Factored distributed load/ft: wu = 1.2(0.563 + 1.0) = 1.88 k/ft 9.2 kips Factored live load is a concentrated load: Pu 1.6 12 19.2
M u wu
2
2
P u
1 1.88 102 2 19.2 9 267 k-ft
3. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: area: As = 6 No. 8 bars = 6 0.79 in.2 =4.74 in.2 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f 4.74 4.74 6000 60000 0 s y c 2.79 in. 1 ' 0.85 0. 85 400 40 0 0 30 0.85 f b c
For f c' 4000 psi, 1 0.85 . Therefore, c 2.79 3.28 in. 1 0.85 Check whether tension steel is yielding: 15.5 3.28 3.28 d c 15.5 using. Eq.(4-18) 0.003 3 0.01 0.011 1> 0.0021 0.00 s t c cu 3.28
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Compute the nominal moment strength, using Eq. (4-21):
M A f d n s y 2
4.74 .74 60000 0000 15.5 5.5 12000
2.79
2
334 kip-ft
0.011 0.005 0.005 the section is clearly tension-controlled and =0.9. Then, Since, t 0.011 M n 0.9 334 301 kip-ft 267 kip-ft. Clearly, M n M u
4-2
(a) Compare M n for singly reinforced rectangular beams having the following
4-3
properties. Use loads and strength reduction factors from ACI Code Sections 9.2 and 9.3.
Beam
b
d
No.
(in.)
(in.)
1 2 3 4 5
12 12 12 12 12
22 22 22 22 33
Bars 3 No. 7 2 No. 9 plus 1 No. 8 3 No. 7 3 No. 7 3 No. 7
f c'
f y
(psi) 3,000 3,000 3,000 4,500 3,000
(psi) 60,000 60,000 40,000 60,000 60,000
Beam No.1 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. A f 3 0.60 60000 s y c 3.53 in. 1 0.85 f ' b 0.85 3000 12 c For f c' 3000 psi, 1 0.85 . Therefore, c 3.53 4.15 in. 1 0.85
d c 22 4.15 0.003 0.013 s t c cu 4.15
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Since, t 0.005 the section is clearly tension-controlled and =0.9. M A f d n s y 2
0.9 3 0.60 60000 22
3.53
12000
2
164 kip-ft
For Beam 1, M 164 kip-ft n
Beam No.2 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. A f 2 1.00 0.79 60000 s y 5.47 in. c 1 ' 0.85 3000 12 0.85 f b c
For f c' 3000 psi, 1 0.85 . Therefore, c 5.47 6.44 in. 1 0.85
4-3
d c 22 6.44 0.003 0.0072 s t c cu 6.44
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Since, t 0.005 the section is clearly tension-controlled and =0.9. A f d n s y 2 For Beam 2, M 242 kip-ft
0.9 2.79 60000 22
M
5.47
2
12000
242 kip-ft
n
Beam No.3 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. A f 1.8 40000 s y c 2.35 in. 1 ' 0.85 3000 12 0.85 f b c
For f c' 3000 psi, 1 0.85 . Therefore, c 2.35 2.76 in. 1 0.85
d c 22 2.76 0.003 0.021 s t c cu 2.76
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Since, t 0.005 the section is clearly tension-controlled and =0.9. M A f d n s y 2 For Beam 3, M 113 kip-ft
0.9 1.8 40000 22
2.35
2
12000
113 kip-ft
n
Beam No.4 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. A f 1.8 60000 s y c 2.35 in. 1 ' 0.85 4500 12 0.85 f b c For f c' 4500 psi, 1 0.825 . Therefore, c 2.35 2.85 in. 1 0.825
d c 22 2.85 0.003 0.020 s t cu c 2.85
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
4-4
Since, t 0.005 the section is clearly tension-controlled and =0.9. A f d n s y 2 For Beam 4, M 169 kip-ft
0.9 1.8 60000 22
M
2.35
2
12000
169 kip-ft
n
Beam No.5 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. A f 1.8 60000 s y 3.53 in. c 1 ' 0.85 3000 12 0.85 f b c For f c' 3000 psi, 1 0.85 . Therefore, c 3.53 4.15 in. 0.85 1
d c 33 4.15 0.003 0.021 s t c cu 4.15
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Since, t 0.005 the section is clearly tension-controlled and =0.9. A f d n s y 2 For Beam 5, M 253 kip-ft
0.9 1.8 60000 33
M
3.53
2
12000
253 kip-ft
n
(b) Taking beam 1 as the reference point, discuss the effects of changing As , f y , f c' , and d on M n . (Note that each beam has the same properties as beam 1 except for the
italicized quantity.)
Beam
M n
No.
(kips-ft) 164 242 113 169 253
1 2 3 4 5 Effect of As (Beams 1 and 2)
An increase of 55% in As (from 1.80 to 2.79 in. 2) caused on increase of 48% in M n . It is clear that increasing the tension steel area causes a proportional increase in the strength of the section,
4-5
with a loss of ductility. Note that in this case, the strength reduction factor was 0.9 for both sections. Effect of f y (Beams 1 and 3) A decrease of 33% in f y caused a decreased of 31% in M n . A decrease in the steel yield strength has essentially the same effect as decreasing the tension steel area. Effect of f c' (Beams 1 and 4) An increase of 50% in f c' caused an increase of 3% in M n . It is clear that changes in the concrete strength have a much smaller effect on moment strength compared with changes in the tension steel area and steel yield strength. Effect of d (Beams 1 and 5) An increase of 50% in d caused an increase of 54% in M n .It is clear that increasing the effective flexural depth of the section increases the section moment strength (without decreasing the section ductility).
(c) What is the most effective way of increasing M n ? What is the least effective way?
Disregarding any other effects of increasing d , As or f y such as changes in cost, etc., the most effective way to increase M n is the increase the effective flexural depth of the section, d , followed by increasing f y and As . Note that increasing f y and As too much may make the beam over-reinforced and thus will result in a decrease in ductility. The least effective way of increasing M n is to increase f c' .Note that increasing f c' will cause a significant increase in curvature at failure.
4-6
4-4 A 12-ft-long cantilever supports its own dead load plus an additional uniform service (unfactored) dead load of 0.5 kip/ft. The beam is made from normal-weight 4000-psi concrete and has b 16 in., d 15.5 in., and h 18 in. It is reinforced with four No. 7 Grade60 bars. Compute the maximum service (unfactored) concentrated live load that can be applied at 1ft from the free end of the cantilever. Use load and strength – reduction factors from ACI Code Sections 9.2 and 9.3. Also check As ,min .
1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 4 No. 7 bars = 4 0.60 in.2 =2.40 in.2 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f 2.4 60000 s y c 2.65 in. 1 ' 0.85 4000 16 0.85 f b c For f c' 4000 psi, 1 0.85 . Therefore, c
1
2.65
0.85
3.1 in.
Check whether tension steel is yielding: d c 15.5 3.1 using Eq.(4-18) 0.003 0.012 s t c cu 3.1
Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Compute the nominal moment strength, using Eq. (4-21): 2.65 2.4 60000 15.5 2 M A f d 170 kips-ft n s y 2 12000 Since, t 0.012 0.005 the section is clearly tension-controlled and, M n 0.9 170 kips-ft = 153 kips-ft
2. Compute Live Load Set M u M n 153 kips-ft 16 18
0.15 0.3 kips/ft 144 Factored dead load = 1.2 0.3 0.5 0.96 kips/ft Weight/ft of beam =
Factored dead load moment = wl 2 2 0.96 122 2 69.1 kips-ft Therefore the maximum factored live load moment is: 153 kips-ft – 69.1 kip-ft = 83.9 kips-ft Maximum factored load at 1 ft from the tip = 83.9 kips-ft / 11 ft = 7.63 kips Maximum concentrated service live load = 7.63 kips / 1.6 = 4.77 kips
4-7
3. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). Also, 3 f c' is equal to 189 psi, so use 200 psi in the numerator: As ,min
200 f y
bw d
200 60,000
16 15.5 0.82 in.2 < As (o.k.)
4-8
4-5
Compute M n and check As ,min for the beam shown in Fig. P4-5. Use f c' 4500 psi
and f y 60,000 psi.
1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 6 No. 8 bars = 6 0.79 in.2 =4.74 in.2 The tension reinforcement for this section is provided in two layers, where the distance from the tension edge to the centroid of the total tension reinforcement is given as d 19 in. Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the
compression flange, h f and that the tension steel is yielding, s y , using Eq. (4-16):
A f 4.74 60000 s y 1.55 in. h f 6 in. (o.k.) ' 0.85 4500 48 0.85 f b e c
For f c' 4500 psi, 1 0.825 . Therefore, c
1
1.55
0.825
1.88 in.
Comparing the calculated depth to the neutral axis, c , to the values for d and d t , it is clear that the tension steel strain, s , easily exceeds the yield strain (0.00207) and the strain at the level of the extreme layer of tension reinforcement, t , exceeds the limit for tension-controlled sections (0.005). Thus, =0.9 and we can use Eq. (4-21) to calculate M n :
4.74 60000 19
A f d n s y 2 12000 M n 0.9 432 kips-ft = 389 kips-ft
M
1.55
2
432 kips-ft
2. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). Also, 3 f c' is equal to 201 psi, so use 3 f c' in the numerator: As ,min
3 f c' f y
bw d
201 60,000
12 19 0.76 in.2 < As (o.k.)
4-9
4-6
Compute M n and check As ,min for the beam shown in Fig. P4-6. Use f c' 4000 psi
and f y 60,000 psi.
1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 6 No. 8 bars = 6 0.79 in.2 =4.74 in.2 The tension reinforcement for this section is provided in two layers, where the distance from the tension edge to the centroid of the total tension reinforcement is given as d 18.5 in. Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange, h f and that the tension steel is yielding, s y , using Eq. (4-16): A f 4.74 60000 s y 4.18 in. ' 0.85 4000 20 0.85 f b e c
h f 5 in. (o.k.)
For f c' 4000 psi, 1 0.85 . Therefore, c 4.18 4.92 in. 1 0.85 Check whether tension steel is yielding: d c 18.5 4.95 using Eq.(4-18) 0.003 0.0082 s c cu 4.95 Thus, > 0.002 and it is clear that the steel is yielding in both layers of reinforcement. s
It is also clear that the section is tension-controlled ( =0.9), but just for illustration the value of can be calculated as: t d c 19.5 4.92 t 0.003 0.0089 t c cu 4.92 We can use Eq. (4-21) to calculate M n :
4.74 60000 18.5
A f d n s y 2 12000 M n 0.9 389 kips-ft = 350 kips-ft
M
4.18
2
389 kips-ft
2. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we ' should use bw in Eq. (4-11). Also, 3 f c is equal to 190 psi, so use 200 psi in the numerator:
As ,min
200 f y
bw d
200 60,000
12 18.5 0.74 in.2 < As (o.k.)
4-10
4-7
Compute the negative-moment capacity, M n , and check As ,min for the beam shown
in Fig. P4-7. Use f c' 3500 psi and f y 40,000 psi.
1. Calculation of M n This section is subjected to negative bending and tension will develop in the top flange and the compression zone is at the bottom of the section. ACI Code Section 10.6.6 requires that a portion of the tension reinforcement be distributed in the flange, so assuming that the No. 6 bars in the flange are part of the tension reinforcement: As 6 0.44 2.64 in.2
The depth of the Whitney stress block can be calculated using Eq. (4-16) , using b 12 in., since the compression zone is at the bottom of the section:
A f 2.64 40000 s y 2.96 in. ' 0.85 3500 12 0.85 f b e c
For f c' 3500 psi, 1 0.85 . Therefore, c 2.96 3.48 in. 1 0.85 We should confirm that the steel is yielding:
d c 19.5 3.48 3.48 0.003 0.014 c cu is yielding s y 0.00207 and this
using Eq.(4-18) t s Clearly,
the
steel
is
tension-controlled
section t 0.005 . We can use Eq. (4-21) to calculate M n :
2.64 40000 19.5
M A f d n s y 2 12000 M n 0.9 159 kips-ft = 143 kips-ft
2.96
2 159 kips-ft
2. Check of As ,min The beam is subjected to negative bending and since the flanged portion of the beam section is in tension, the value of As ,min will depend on the use of that beam. Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw in Eq. (4-11). Also, 3 f c' is equal to 177 psi, so use 200 psi in the numerator:
4-11
As ,min
200 f y
bw d
200 40,000
12 19.5 1.17 in.2 < As (o.k.)
However, for a statically determinate beam, bw should be replaced by the smaller of 2bw 24 in. or be . Given that be is 48 in. for this beam section, As ,min
200 f y
bw d
200 40,000
24 19.5 2.34 < As (o.k.)
4-12
4-8
For the beam shown in Fig. P4-8, f c' 3500 psi and f y 60,000 psi. (a)
Compute the effective flange width at midspan.
The limits given in ACI Code Section 8.12 for determining the effective compression flange, be , for a flanged section that is part of a continuous floor system are:
4 be bw 2(8h f ) b 2(clear trans. distance)/2 w Assuming that the columns are 18 in. 18 in. , the longitudinal span is approximated as: 18 in. ft 22.5 ft 21 ft 12 in. ft 12 in. 8.5 ft The clear transverse distance for the 9 ft.-6 in. span is: 9.5 ft 12 in. ft 1 12 in. 18 in. 9.75 ft and for the 11 ft. span is: 11 ft 2 12 in. 12 in. ft ft So, the average clear transverse distance is 9.125 ft The effective compression flange can now be computed as:
22.5 ft 12 in./ft 67.5 in. 4 12 in. 2 8 6 in. 108 in. be 12 in. 2 9.125 ft 12 in./ft /2=122 in. The first limit governs for this section, so be 67.5 in. (b)
Compute M n for the positive- and negative-moment regions and check
As ,min for both sections. At the supports, the bottom bars are in one layer; at midspan, the No. 8 bars are in the bottom, the No. 7 bars in a second layer.
Positive moment region 1. Calculation of M n Tension steel area: As = 3 No. 8 bars + 2 No. 7 bars = 3 0.79 2 0.60 3.57 in.2
4-13
The tension reinforcement for this section is provided in two layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. Thus the distance from the top of the section to the extreme layer of tension reinforcement, d t , can be calculated to be: d t 21 in. – 2.5 in. =18.5 in.
The minimum spacing required between layers of reinforcement is 1 in. (ACI Code Section 7.6.2). Thus the spacing between the centers of the layers is approximately 2 in. So the distance from the tension edge to the centroid of the total tension reinforcement is:
3 0.79 2.5 2 0.60 4.5 3.57
3.17 in.
Therefore, the effective flexural depth, d , is: d 21 in. – 3.17 in. =17.8 in.
Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange, h f 6 in. and that the tension steel is yielding, s y , using Eq.(416): A f 3.57 60000 s y 1.07 in. 0.85 f ' b e 0.85 3500 67.5 c
h f 6 in. (o.k.)
1.26 in. For f c' 3500 psi, 1 0.85 . Therefore, c 1.07 0.85 1 Comparing the calculated depth to the neutral axis, c , to the values for d and d t , it is clear that the tension steel strain, s , easily exceeds the yield strain (0.00207) and the strain at the level of the extreme layer of tension reinforcement, t , exceeds the limit for tension-controlled sections (0.005). Thus, =0.9 and we can use Eq. (4-21) to calculate M n :
3.57 60000 17.8
M A f d n s y 2 12000 M n 0.9 308 kips-ft = 277 kips-ft
1.07
2
308 kips-ft
2. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). Also, 3 f c' is equal to 177 psi, so use 200 psi in the numerator:
4-14
As ,min
200 f y
200
bw d
60,000
12 17.8 0.71 in.2 < As (o.k.)
Negative moment region The tension and compression reinforcement for this section is provided in single layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension or compression edge of the section to the centroid of the tension or compression layer of steel is approximately 2.5 in. As = 7 No. 7 bars = 7 0.60 4.2 in.2 , d 18.5 in. ' ' As = 2 No. 8 bars = 2 0.79 1.58 in.2 , d 2.5 in.
Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c. Try c d 4 4.5 in.
c d ' 4.5 2.5 cu 0.003 0.00133 4.5 c f s' E s s' 29,000 ksi 0.00133 38.6 ksi f y
s'
C 's As' f s' 0.85 f c' 1.58 in.2 38.6 ksi 2.98 ksi 56.3 kips Cc 0.85 f c'b 1c 0.85 3.5 ksi 12 in. 0.85 4.5 in.=137 kips T As f y 4.20 in. 60 ksi 252 kips 2
Because T Cc C 's , we should increase c for the second trial. Try c 5.9 in. s' 0.00173
f s 50.2 ksi f y '
C 's 74.6 kips C c 179 kips
T 254 kips Cc C s 254 kips
With section equilibrium established, we must confirm the assumption that the tension steel is yielding. d c 18.5 5.9 using Eq.(4-18) 0.003 0.0064 s c cu 5.9
Thus, the steel is yielding 0.00207 and it is a tension-controlled section t s 0.0102 . s So, using 1c 0.85 5.9 in. 5.0 in. , use Eq. (4-21) to calculate M n . Cc d C 's d d ' 179 kips 16 in. 74.6 kips 16 in. n 2 M 2865 k-in. 1195 k-in 4060 k-in 338 k-ft M
n
4-15
M n 0.9 338 kips-ft = 304 kips-ft
2. Check of As ,min The flanged portion of the beam section is in tension and the value of As ,min will depend on the use of that beam. Since the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw in Eq. (4-11). Also, 3 f c' is equal to 177 psi, so use 200 psi in the numerator: 200 200 As ,min bw d 12 18.5 0.74 in.2 < As (o.k.) 60,000 f y
4-16
4-9
Compute M n and check As ,min for the beam shown in Fig. P4-9. Use f c 3500 psi '
and f y 60,000 psi, and (a)
the reinforcement is six No. 8 bars.
1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 6 No. 8 bars = 6 0.79 in.2 = 4.74 in.2 Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the top flange, 5 in. and that the tension steel is yielding, s y , using Eq. (4-16) with b 30 in. : A f 4.74 60000 s y 3.19 in. h f 5 in. (o.k.) ' 0.85 3500 30 0.85 f b c
For f c' 3500 psi, 1 0.85 . Therefore, c 3.19 3.76 in. 1 0.85 Check whether tension steel is yielding: d c 32.5 3.76 using Eq.(4-18) t 0.003 0.023 s c cu 3.76 Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Since, t 0.005 the section is clearly tension-controlled and =0.9. We can use Eq. (4-21) to calculate M n :
4.74 60000 32.5
A f d n s y 2 12000 M n 0.9 733 kips-ft = 659 kips-ft
M
3.18
2
733 kips-ft
2. Check of As ,min The flanged portion of the beam section is in tension and the value of As ,min will depend on the use of that beam. Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw 2 5 10 in. in Eq. (4-11). Also, 3 f c' is equal to 177 psi, so use 200 psi in the numerator: As ,min
200 f y
bw d
200 60,000
10 32.5 1.08 in.2 < As (o.k.)
4-17
However, for a statically determinate beam, bw should be replaced by the smaller of 2bw 20 in. or be . Given that be is 30 in. for this beam section, As ,min
200
(b)
the reinforcement is nine No. 8 bars.
f y
bw d
200 60,000
20 32.5 2.17 in.2 < As (o.k.)
1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 9 No. 8 bars = 9 0.79 in.2 =7.11 in.2 Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange, h f 5 in. and that the tension steel is yielding, s y , using Eq. (416) with b 30 in. : A f 7.11 60000 s y 4.78 in. h f 5 in. (o.k.) 0.85 f ' b 0.85 3500 30 c
For f c' 3500 psi, 1 0.85 . Therefore, c 4.78 5.62 in. 0.85 1 Check whether tension steel is yielding: d c 32.5 5.62 using Eq.(4-18) t 0.003 0.014 s c cu 5.62 Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Since, t 0.005 the section is clearly tension-controlled and =0.9. We can use Eq. (4-21) to calculate M n :
7.11 60000 32.5
M A f d n s y 2 12000 M n 0.9 1070 kips-ft = 963 kips-ft
2. Check of As ,min
As ,min is the same as in part (a).
4-18
4.78
2
1070 kips-ft
4-10
Compute M n and check As ,min for the beam shown in Fig. P4-10. Use f c' 5000 psi
and f y 60,000 psi, and
1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 8 No. 7 bars = 8 0.60 in.2 =4.8 in.2 Tension will develop in the bottom flange and the compression zone is at the top of the section. Thus, assuming that the tension steel is yielding, s y , in Eq. (4-16) we should use b 2 6 12 in. and we find the depth of the Whitney stress block as: A f 4.8 60000 s y 5.65 in. ' 0.85 5000 12 0.85 f b c
For f c' 5000 psi, 1 0.80 . Therefore, c 5.65 7.06 in. 1 0.80 Check whether tension steel is yielding: d c 23.5 7.06 using Eq.(4-18) t 0.003 0.007 s 7.06 c cu Thus, > 0.002 and the steel is yielding ( f s
f ).
y s Since, t 0.005 the section is tension-controlled and =0.9.
We can use Eq. (4-21) to calculate M n :
4.8 60000 23.5
A f d n s y 2 12000 M n 0.9 496 kips-ft = 446 kips-ft
M
5.65
2
496 kips-ft
2. Check of As ,min The flanged portion of the beam section is in tension and the value of As ,min will depend on the use of that beam. Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw 2 6 12 in. in Eq. (4-11). Also, ' note that 3 f c is equal to 212 psi:
As ,min
212 f y
bw d
212 60,000
12 23.5 1.00 in.2 < As (o.k.)
However, for a statically determined beam,
bw should be replaced by the smaller of
2bw 24 in. or be . Given that be is 42 in. for this beam section, As ,min
212 f y
bw d
212 60,000
24 23.5 1.99 in.2 < As (o.k.)
4-19
4-11
(a) Compute M n for the three beams shown in Fig. P4-11. In each case,
f c' 4000 psi
and f y 60 ksi, b 12 in., d 32.5 in., and h 36 in.
Beam No. 1 Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2 The tension reinforcement for this section is provided in two layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. Thus the distance from the top of the section to the extreme layer of tension reinforcement, d t , can be calculated to be:
d t 36 in. – 2.5 in. =33.5 in. The effective flexural depth, d , is given as : d 32.5 in. Assuming that the tension steel is yielding, s y , using Eq. (4-16): A f 6.00 60000 s y 8.82 in. 0.85 f ' b 0.85 4000 12 c
For f c' 4000 psi, 1 0.85 . Therefore, c 8.82 10.4 in. 1 0.85 We need to check whether tension steel is yielding: d c 32.5 10.4 using Eq.(4-18) 0.003 0.0064 s c cu 10.4 Thus, > 0.002 and the steel is yielding ( f s s
f y ).
Also, clearly t 0.005 , the section is tension-controlled and =0.9. Thus, =0.9 and we can use Eq. (4-21) to calculate M n :
6.00 60000 32.5
M A f d n s y 2 12000 M n 0.9 843 kips-ft = 759 kips-ft
8.82
2 843 kips-ft
Beam No. 2 Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2 Compression steel area: As = 2 No. 9 bars = 2 1.00 in.2 =2.00 in.2 '
As was discussed for beam No. 1, d 32 in., d t 33.5 in. and d is given as d 2.5 in. '
4-20
'
Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c. Try c d 4 8 in.
c d ' 8 2.5 cu 0.003 0.00206 c 8 f s' E s s' 29,000 ksi 0.00206 59.7 ksi f y
s'
C 's As' f s' 0.85 f c' 2.00 in.2 59.7 ksi 3.4 ksi 113 kips Cc 0.85 f c'b 1c 0.85 4 ksi 12 in. 0.85 8 in.=277 kips T As f y 6.00 in. 60 ksi 360 kips 2
Because T Cc C 's , we should decrease c for the second trial. Try c 7.4 in. s' 0.00199
f s 57.7 ksi f y '
C 's 109 kips
C c 257 kips T 360 kips Cc C 's 366 kips
With section equilibrium established, we must confirm the assumption that the tension steel is yielding. d c 32.5 7.4 using Eq.(4-18) 0.003 0.0102 s c cu 7.4
Clearly, the steel is yielding 0.00207 and it is a tension-controlled section t s 0.0102 . s So, using 1c 0.85 7.4 in. 6.3 in. , use Eq. (4-21) to calculate M n . Cc d C 's d d ' 257 kips 29.3 in. 1 08.6 kips 30 in. n 2 M 7545 k-in. 3270 k-in 10800 k-in 901 k-ft M
n M n 0.9 901 kips-ft = 811 kips-ft
Beam No. 3 Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2 Compression steel area: As = 4 No. 9 bars = 4 1.00 in.2 =4.00 in.2 '
As was discussed for beam No. 1, d 32.5 in., and d t 33.5 in. '
The compression reinforcement for this beam section is provided in two layers and d is given as 3.5 in.
4-21
Because this is a doubly reinforced section, we will the same procedure as for beam No. 2 (assuming that the tension steel is yielding). The depth of the neutral axis for this section should be smaller compared with beam section No. 2, since the compression reinforcement is increased for this section. Try c 7 in.
c d ' 7 3.5 s cu 0.003 0.0015 7 c '
f s E s s 29, 000 ksi 0.0015 43.5 ksi f y '
'
C 's As f s 0.85 f c 4.00 in. 43.5 ksi 3.4 ksi 160 kips '
'
'
2
Cc 0.85 f c b 1c 0.85 4 ksi 12 in. 0.85 7 in.=243 kips '
T As f y 6.00 in. 60 ksi 360 kips 2
Because T Cc C 's , we should decrease c for the second trial. Try c 6.3 in. (Note that both layers of the compression steel will actually be in the compression zone) s' 0.00133
f s 38.6 ksi f y '
C 's 141 kips
C c 218 kips T 360 kips Cc C s 359 kips
With section equilibrium established, we must confirm the assumption that the tension steel is yielding.
d c 32.5 6.3 0.003 0.012 6.3 c cu
using Eq.(4-18) s Clearly,
the
steel
is
yielding
s 0.00207
and
it
is
a
section t s 0.012 . So, using 1c 0.85 6.3 in. 5.36 in. , use Eq. (4-21) to calculate M n . Cc d C 's d d ' 218 kips 29.8 in. 141 kips 29 in. n 2 M 6495 k-in. 4090 k-in 10585 k-in 882 k-ft M
n M n 0.9 882 kips-ft = 794 kips-ft
4-22
tension-controlled
(b) From the results of part (a), comment on weather adding compression reinforcement is a cost-effective way of increasing the strength, M n , of a beam.
Comparing the values of M n for the three beams, it is clear that for a given amount of tension reinforcement, the addition of compression steel has little effect on the nominal moment capacity, provided the tension steel yields in the beam without compression reinforcement. As a result, adding compression reinforcement in not a cost effective way of increasing the nominal moment capacity of a beam. However, adding compression reinforcement improves the ductility and might be necessary when large amounts of tension reinforcement are used to change the mode of failure.
4-23
4-12 Compute M n for the beam shown in Fig. P4-12. Use f c' 3500 psi and f y 60,000 psi. Does the steel yield in this beam at nominal strength? As = 6 No. 8 bars = 6 0.79 in.2 =4.74 in.2 , d 25 in. 2.5 in. 22.5 in.
As' = 2 No. 7 bars = 2 0.60 in.2 =1.2 in.2 , d ' 2.5 in. Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c. Try c d 4 5.5 in. For f c' 3500 psi, 1 0.85 . Thus, 0.85 5.5 in. 4.68 in. 5.0 in. Since the depth of the Whitney stress block is less than 5.0 in. , 5.0 in. ,the width of the compression zone is constant and equal to 10 in., i.e. b 10 in. s'
c d ' 5.5 2.5 cu 0.003 0.00164 c 5.5
' ' f E 29, 000 ksi 0.00164 47.6 ksi f s s s y
C 's As' f s' 0.85 f c' 1.2 in.2 47.6 ksi 2.98 ksi 53.5 kips Cc 0.85 f c'b 1c 0.85 3.5 ksi 10 in. 0.85 5.5 in.=139 kips T As f y 4.74 in. 60 ksi 284 kips 2
Because T Cc C 's , we should increase c for the second trial. Try c 6.5 in. and find 0.85 6.5 in. 5.53 in.> 5.0 in. s' 0.00185
f s 53.7 ksi f y '
C 's 60.9 kips
In this case, the width of the compression zone is not constant. Using a similar reasoning as in the case of flanged sections, where the depth of the Whitney stress block is in the web of the section, the compression force can be calculated from the following equations (refer to Fig. S4-12): Ccw 0.85 f c' 10 in. 0.85 3.5 ksi 10 in. 5.53 in.=165 kips Ccf 0.85 f c 20 10 in. 5 in. 0.85 3.5 ksi 10 in. 0.53 in. 15.8 kips '
C c 165 15.8 181 kips T 284 kips > Cc C 's 242 kips , we should increase c for the third trial.
Try c 7.2 in. and find 0.85 7.2 in. 6.12 in.> 5.0 in. s' 0.00196
4-24
f s 56.8 ksi f y '
C 's 64.6 kips C cw 182 kips
C cf 33.3 kips C c 215 kips T 284 kips Cc C 's 276 kips
With section equilibrium established, we must confirm the assumption that the tension steel is yielding.
d c 22.5 6.12 0.003 0.00803 c cu 6.12
using Eq.(4-18) s
Thus, the steel is yielding 0.00207 and it is a tension-controlled section t s 0.012 . s Summing the moments about the level of the tension reinforcement: 5 Ccw d Ccf d 5 C 's d d ' n 2 2 M 182 kips 19.4 in. 33.3 kips 16.9 in. + 64.6 kips 20 in. M
n M 3530 k-in. 563 k-in +1290 k-in 5385 k-in 449 k-ft n M n 0.9 500 kips-ft = 404 kips-ft
4-25
bw 0.85f'c ht
a f's
d
h
fs=fy
b
f
(assumed)
a) total beam section and stress distribution
bw ht
a/2
a
d'
Cs Ccw
d
h
T1 F b b) Part 1: web of section and corresponding internal forces
bw ht
a
(a+ht)/2 Ccf d
h
T2 F b c) Part 2: overhanging flanges and corresponding internal forces
Fig. S4-12.1 Beam section and internal forces for the case of ht .
4-26
