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-
4-1
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) 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 normal-weight 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 and show that it exceeds .
1. Calculate the dead load of the beam.
Weight/ft = 24 12
0.15 0.3144
kips/ft
2. Compute the factored moment,u
M :
Factored load/ft: uw = 1.2(0.30 + 1.40) + 1.6(1.50) = 4.44 k/ft 2 2
8 4.44 20 8 222u u
M w kip-ft
3. Compute the nominal moment capacity of the beam, nM and the strength reduction factor, .
Tension steel area: As = 3 No. 9 bars = 3 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)):
3.00 600005.04
1 ' 0.85 3500 120.85
A fs y
cf bc
in.
For ' 3500cf psi, 1 0.85 . Therefore,
1
5.04 5.930.85
c
in.
Check whether tension steel is yielding:
Using Eq.(4-18) 21.5 5.93 0.003 0.007885.93
d c
s t cuc
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Compute the nominal moment strength, using Eq. (4-21):
5.043.00 60000 21.5
2285
2 12000M A f d
n s y
kip-ft
Since, 0.00788 0.005t the section is clearly tension-controlled and =0.9. Then,
0.9 285nM kip-ft 256 kip-ft. Clearly, n uM M
-
4-2
4-2 A cantilever beam shown in Fig. P4-2. 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 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 and show it exceeds .
1. Calculate the dead load of the beam.
Weight/ft = 30 18
0.15 0.563144
kips/ft
2. Compute the factored moment,u
M .
Factored distributed load/ft: uw = 1.2(0.563 + 1.0) = 1.88 k/ft
Factored live load is a concentrated load: 1.6 12 19.2uP kips
2 22 1.88 10 2 2671 19.2 9u u uM w P kip-ft
3. Compute the nominal moment capacity of the beam, nM and the strength reduction factor, .
Tension steel 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)):
4.74 600002.79
1 ' 0.85 4000 300.85
A fs y
cf bc
in.
For ' 4000cf psi, 1 0.85 . Therefore,
1
2.79 3.280.85
c
in.
Check whether tension steel is yielding:
Using. Eq.(4-18) 15.5 3.28 0.003 0.0113.28
d c
s t cuc
> 0.0021
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Compute the nominal moment strength, using Eq. (4-21):
2.794.74 60000 15.5
2334
2 12000M A f d
n s y
kip-ft
Since, 0.011 0.005t the section is clearly tension-controlled and =0.9. Then,
0.9 334 301nM kip-ft 267 kip-ft. Clearly, n uM M
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4-3
4-3 (a) Compare for singly reinforced rectangular beams having the following properties. Use strength reduction factors from ACI Code Sections 9.2 and
9.3.
Beam
No.
b
(in.)
d
(in.)
Bars
'cf
(psi)
yf
(psi)
1 12 22 3 No. 7 4,000 60,000
2 12 22 2 No. 9 plus 1 No. 8 4,000 60,000
3 12 22 3 No. 7 4,000 80,000
4 12 22 3 No. 7 6,000 60,000
5 12 33 3 No. 7 4,000 60,000
Beam No.1
Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding.
For ,
1 0.85 . Therefore,
( ) (
) (
)
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is tension-controlled and =0.9.
(
)
( )
For Beam 1, -
Beam No.2
Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding.
( )
For , 1 0.85 . Therefore,
-
4-4
( ) (
) (
)
Thus,
s > 0.002 and the steel is yielding ( ).
Since, 0.005t the section is clearly tension-controlled and =0.9.
(
)
( ) ( )
For Beam 2, -
Beam No.3
Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding.
For , 1 0.85 . Therefore,
( ) (
) (
)
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is clearly tension-controlled and =0.9.
(
)
( )
For Beam 3, -
Beam No.4
Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding.
For , . Therefore,
( ) (
) (
)
s yf f
-
4-5
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is tension-controlled and =0.9.
(
)
( )
For Beam 4, -
Beam No.5
Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding.
For , 1 0.85 . Therefore,
( ) (
) (
)
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is tension-controlled and =0.9.
(
)
( )
For Beam 5, -
(b) Taking beam 1 as the reference point, discuss the effects of changing
and d on . (Note that each beam has the same properties as
beam 1 except for the italicized quantity.)
Beam
No.
Mn
(kip-ft)
1 167
2 250
3 219
4 171
5 257
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4-6
Effect of sA (Beams 1 and 2)
An increase of 55% in sA (from 1.80 to 2.79 in.
2) caused an increase of 50% in nM . Increasing
the tension steel area causes a proportional increase in the strength of the section, with a loss of
ductility. Note that in this case, the strength reduction factor was 0.9 for both sections.
Effect of yf (Beams 1 and 3)
An increase of 33% in yf caused an increase of 31% in nM . Increasing the steel yield strength
has essentially the same effect as increasing the tension steel area.
Effect of '
cf (Beams 1 and 4)
An increase of 50% in '
cf caused an increase of 2% in nM . Changing the concrete strength has
approximately no impact on moment strength, relative to 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 nM . 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 ? What is the least effective way?
Disregarding any other effects of increasing , sd A or yf such as changes in cost, etc., the most
effective way to increase nM is to increase the effective flexural depth of the section, d ,
followed by increasing yf and sA . Note that increasing yf and sA too much may make the beam
over-reinforced and thus will result in a decrease in ductility.
The least effective way of increasing nM is to increase '
cf . Note that increasing '
cf will cause a
significant increase in curvature at failure.
-
4-7
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 in., in., and in. It is reinforced with four No. 7 Grade-60 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 .
1. Compute the nominal moment capacity of the beam, nM and the strength reduction factor, .
Tension steel area: sA = 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)):
2.4 600002.65
1 ' 0.85 4000 160.85
A fs y
cf bc
in.
For ' 4000cf psi, 1 0.85 . Therefore,
1
2.65 3.10.85
c
in.
Check whether tension steel is yielding:
Using Eq.(4-18) 15.5 3.1 0.003 0.0123.1
d c
s t cuc
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Compute the nominal moment strength, using Eq. (4-21):
2.652.4 60000 15.5
2170
2 12000M A f d
n s y
kip-ft
Since, 0.012 0.005t the section is clearly tension-controlled and,
0.9 170nM kip-ft = 153 kip-ft
2. Compute Live Load
Set 153u nM M kip-ft
Weight/ft of beam = 16 18
0.15 0.3144
kips/ft
Factored dead load = 1.2 0.3 0.5 0.96 kips/ft
Factored dead load moment = 2 22 0.96 12 2 69.1wl kip-ft
Therefore the maximum factored live load moment is: 153 kip-ft 69.1 kip-ft = 83.9 kip-ft
Maximum factored load at 1 ft from the tip = 83.9 kip-ft / 11 ft = 7.63 kips
Maximum concentrated service live load = 7.63 kips / 1.6 = 4.77 kips
-
4-8
3. Check of ,minsA
The section is subjected to positive bending and tension is at the bottom of this section, so we
should use wb in Eq. (4-11). Also, '3 cf is equal to 189 psi, so use 200 psi in the numerator:
,min
200 20016 15.5 0.82
60,000s w
y
A b df
in.2 <
sA (o.k.)
-
4-9
4-5 Compute and check for the beam shown in Fig. P4-5. Use
psi and psi.
1. Compute the nominal moment capacity of the beam, nM 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, fh and that the tension steel is yielding, s y , using Eq. (4-16):
4.74 60000
1.55' 0.85 4500 480.85 e
A fs y
f bc
in. 6fh in. (o.k.)
For ' 4500cf psi, 1 0.825 . Therefore,
1
1.55 1.880.825
c
in.
Comparing the calculated depth to the neutral axis, c , to the values for d and td , 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 nM :
1.554.74 60000 19
2432
2 12000M A f d
n s y
kip-ft
0.9 432nM kip-ft = 389 kip-ft
2. Check of ,minsA
The section is subjected to positive bending and tension is at the bottom of this section, so we
should use wb in Eq. (4-11). Also, '3 cf is equal to 201 psi, so use
'3 cf in the numerator:
'
,min
3 20112 19 0.76
60,000
c
s w
y
fA b d
f in.2 < sA (o.k.)
-
4-10
4-6 Compute and check for the beam shown in Fig. P4-6. Use
psi and psi.
1. Compute the nominal moment capacity of the beam, nM and the strength reduction factor, .
Tension steel area:
sA = 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, fh and that the tension steel is yielding, s y , using Eq. (4-16):
4.74 60000
4.18' 0.85 4000 200.85 e
A fs y
f bc
in. 5fh in. (o.k.)
For ' 4000cf psi, 1 0.85 . Therefore,
1
4.18 4.920.85
c
in.
Check whether tension steel is yielding:
Using Eq.(4-18) 18.5 4.95
0.003 0.00824.95
d c
s cuc
Thus, s
> 0.002 and it is clear that the steel is yielding in both layers of reinforcement.
It is also clear that the section is tension-controlled ( =0.9), but just for illustration the value of
t can be calculated as:
19.5 4.920.003 0.0089
4.92
td c
t cuc
We can use Eq. (4-21) to calculate nM :
4.184.74 60000 18.5
2389
2 12000M A f d
n s y
kip-ft
0.9 389nM kip-ft = 350 kip-ft
2. Check of ,minsA
The section is subjected to positive bending and tension is at the bottom of this section, so we
should use wb in Eq. (4-11). Also, '3 cf is equal to 190 psi, so use 200 psi in the numerator:
,min
200 20012 18.5 0.74
60,000s w
y
A b df
in.2 < sA (o.k.)
-
4-11
4-7 Compute the negative-moment capacity, , and check for the beam shown in Fig. P4-7. Use
psi and psi.
1. Calculation of
nM
This section is subjected to negative bending, so 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: 6 0.44 2.64sA in.2
The depth of the Whitney stress block can be calculated using Eq. (4-16), using 12b in., since the compression zone is at the bottom of the section:
For , 1 0.85 . Therefore,
(
) (
)
The steel is yielding 0.00207s y and it is tension-controlled 0.005t so = 0.9.
(
)
( )
2. Check of ,minsA
The flanged portion of the beam section is in tension because the beam is subjected to negative
bending. Therefore, the value of ,minsA will depend on whether the beam is statically determinate.
Assuming that the beam is part of a continuous, statically indeterminate floor system, the
minimum tension reinforcement should be calculated using wb in Eq. (4-11). Also, '3 cf is
equal to 189 psi, so use 200 psi in the numerator:
( )
However, for a statically determinate beam, wb should be replaced by the smaller of
2 24 in.wb or eb . Given that eb is 48 in. for this beam section,
( )
-
4-12
4-8 For the beam shown in Fig. P4-8, psi and psi.
(a) Compute the effective flange width at midspan.
The limits given in ACI Code Section 8.12 for determining the effective compression flange, eb ,
for a flanged section that is part of a continuous floor system are:
4
2(8 )
2(clear trans. distance)/2
e w f
w
b b h
b
Assuming that the columns are 18 in. 18 in. , the longitudinal span is approximated as:
18 in.21 ft ft 22.5 ft
in.12ft
The clear transverse distance for the 9 ft.-6 in. span is: 12 in.
9.5 ft 8.5 ftin.12
ft
and for the 11 ft. span is: 1 12 in. 18 in.
11 ft 9.75 ftin. in.2 12 12
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./ft67.5 in.
4
12 in. 2 8 6 in. 108 in.
12 in. 2 9.125 ft 12 in./ft /2=122 in.
eb
The first limit governs for this section, so 67.5 in.eb
(b) Compute for the positive- and negative-moment regions and check 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 nM
Tension steel area: sA = 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, td ,
can be calculated to be:
td 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 4.53.17
3.57
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, 6 in.fh and that the tension steel is yielding, s y ; using Eq.(4-16) we have:
3.57 60000
1.07' 0.85 3500 67.50.85 e
A fs y
f bc
in. 6fh in. (o.k.)
For ' 3500cf psi, 1 0.85 . Therefore,
1
1.07 1.260.85
c
in.
Comparing the calculated depth to the neutral axis, c , to the values for d and td , 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 nM :
1.073.57 60000 17.8
2308
2 12000M A f d
n s y
kip-ft
0.9 308nM kip-ft = 277 kip-ft
Check of ,minsA : The section is subjected to positive bending and tension is at the bottom of this
section, so we should use wb in Eq. (4-11). '3 cf is equal to 177 psi, so use 200 psi in the
numerator.
-
4-14
,min
200 20012 17.8 0.71
60,000s w
y
A b df
in.2 < sA (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.
sA = 7 No. 7 bars = 7 0.60 4.2 in.2 , d 18.5 in.
'sA = 2 No. 8 bars = 2 0.79 1.58 in.
2 , ' 2.5d 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 4 4.5 in.c d
'' 4.5 2.5 0.003 0.00133
4.5s cu
c d
c
' ' 29,000 ksi 0.00133 38.6 ksis s s yf E f
' ' ' 2' 0.85 1.58 in. 38.6 ksi 2.98 ksi 56.3 kipss s s cC A f f '
10.85 0.85 3.5 ksi 12 in. 0.85 4.5 in.=137 kipsc cC f b c 24.20 in. 60 ksi 252 kipss yT A f
Because 'c sT C C , we should increase c for the second trial.
Try 5.9 in.c ' 0.00173s
' 50.2 ksis yf f ' 74.6 kipssC
179 kipscC
254 kips 254 kipsc sT C C
With section equilibrium established, we must confirm the assumption that the tension steel is
yielding.
using Eq.(4-18) 18.5 5.9
0.003 0.00645.9
d c
s cuc
Thus, the steel is yielding 0.00207s and it is a tension-controlled section 0.0102t s . So, using 1 0.85 5.9 in. 5.0 in.c , use Eq. (4-21) to calculate nM .
'' 179 kips 16 in. 74.6 kips 16 in.2
2865 k-in. 1195 k-in 4060 k-in 338 k-ft
c sM C d C d dn
Mn
-
4-15
0.9 338nM kip-ft = 304 kip-ft
Check of ,minsA : The flanged portion of the beam section is in tension and the value of ,minsA 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 wb in Eq. (4-11).
Also, '3 cf is equal to 177 psi, so use 200 psi in the numerator.
,min
200 20012 18.5 0.74
60,000s w
y
A b df
in.2 <
sA (o.k.)
-
4-16
4-9 Compute and check for the beam shown in Fig. P4-9. Use
psi and psi, and
(a) the reinforcement is six No. 8 bars.
1. Compute the nominal moment capacity of the beam, nM and the strength reduction factor, .
Tension steel area:
sA = 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 30 in.b :
( )
For , 1 0.85 . Therefore,
( ) (
) (
)
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is tension-controlled and =0.9.
We can use Eq. (4-21) to calculate nM :
(
)
( )
2. Check of ,minsA
The flanged portion of the beam section is in tension and the value of ,minsA 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 2 5 10 in.wb in Eq.
(4-11). Also, '3 cf is equal to 189 psi, so use 200 psi in the numerator:
,min
200 20010 32.5 1.08
60,000s w
y
A b df
in.2 < sA (o.k.)
However, for a statically determinate beam, wb should be replaced by the smaller of
2 20 in.wb or eb . Given that eb is 30 in. for this beam section,
,min
200 20020 32.5 2.17
60,000s w
y
A b df
in.2 < sA (o.k.)
-
4-17
(b) the reinforcement is nine No. 8 bars.
1. Compute the nominal moment capacity of the beam, nM 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, 5 in.fh and that the tension steel is yielding, s y , using Eq. (4-16) with 30 in.b :
( )
For , 1 0.85 . Therefore,
( ) (
) (
)
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is tension-controlled and =0.9.
We can use Eq. (4-21) to calculate nM :
(
)
( )
2. Check of ,minsA
,minsA is the same as in part (a).
-
4-18
4-10 Compute and check for the beam shown in Fig. P4-10. Use
psi and psi.
1. Compute the nominal moment capacity of the beam, nM 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 2 6 12 in.b and we find the depth of the Whitney stress block as:
4.8 600005.65
' 0.85 5000 120.85
A fs y
f bc
in.
For ' 5000cf psi, 1 0.80 . Therefore,
1
5.65 7.060.80
c
in.
Check whether tension steel is yielding:
using Eq.(4-18) 23.5 7.06
0.003 0.0077.06
t
d c
s cuc
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Since, 0.005t the section is tension-controlled and =0.9.
We can use Eq. (4-21) to calculate nM :
5.654.8 60000 23.5
2496
2 12000M A f d
n s y
kip-ft
0.9 496nM kip-ft = 446 kip-ft
2. Check of ,minsA
The flanged portion of the beam section is in tension and the value of ,minsA 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 2 6 12 in.wb in Eq. (4-11). Also,
note that '3 cf is equal to 212 psi:
,min
212 21212 23.5 1.00
60,000s w
y
A b df
in.2 < sA (o.k.)
However, for a statically determined beam, wb should be replaced by the smaller of
2 24 in.wb or eb . Given that eb is 42 in. for this beam section,
,min
212 21224 23.5 1.99
60,000s w
y
A b df
in.2 < sA (o.k.)
-
4-19
4-11 (a) Compute for the three beams shown in Fig. P4-11. In each case,
psi and ksi,
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, td ,
can be calculated to be:
td 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):
For , . Therefore,
(
) (
)
Thus, s
> 0.002 and the steel is yielding ( s yf f ).
Also, 0.005t , the section is tension-controlled and =0.9.
We can use Eq. (4-21) to calculate nM :
(
)
( )
Beam No. 2
Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2
Compression steel area: '
sA = 2 No. 9 bars = 2 1.00 in.2 =2.00 in.
2
As was discussed for beam No. 1, d 32 in., td 33.5 in. and 'd is given as ' 2.5d 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.
-
4-20
Try 4 8 in.c d
(
) (
)
( )
(
) ( )
Because 'c sT C C , we should decrease c for the second trial.
Try
( )
With section equilibrium established, we must confirm the assumption that the tension steel is
yielding.
(
) (
)
Clearly, the steel is yielding 0.00207s and it is a tension-controlled section. So, using , use Eq. (4-21) to calculate nM .
* (
)
( )+ [ (
) ( )]
Beam No. 3
Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2
Compression steel area: '
sA = 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 td 33.5 in.
The compression reinforcement for this beam section is provided in two layers and 'd is given as
3.5 in.
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 (Note that both layers of the compression steel will be in the compression zone)
( )
-
4-21
Because 'c s
T C C , we should decrease c for the second trial.
Try
( )
With section equilibrium established, we must confirm the assumption that the tension steel is
yielding.
(
) (
)
Clearly, the steel is yielding 0.00207s and it is a tension-controlled section. So, using , use Eq. (4-21) to calculate nM .
* (
)
( )+ [ (
) ( )]
(b) From the results of part (a), comment on whether adding compression
reinforcement is a cost-effective way of increasing the strength, , of a beam.
Comparing the values of nM 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,
as long as 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 behavior
from compression controlled to tension controlled.
-
4-22
4-12 Compute for the beam shown in Fig. P4-12. Use psi and
psi. Does the compression steel yield in this beam at nominal strength?
sA = 6 No. 8 bars = 6 0.79 in.2 = 4.74 in.
2 , 25 in. 2.5 in. 22.5 in.d
'
sA = 2 No. 7 bars = 2 0.60 in.2 =1.2 in.
2 , ' 2.5 in.d
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 For
psi, . Thus,
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. 10 in.b
(
) (
)
( )
(
) ( )
Because , we should increase c for the second trial.
Try
(
) (
)
( )
(
) ( )
Since , 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):
( )( )
With section equilibrium established, we must confirm the assumption that the tension steel is
yielding.
(
) (
)
Thus, the tension steel is yielding 0.00207s and it is a tension-controlled section.
-
4-23
Summing the moments about the level of the tension reinforcement:
[ (
) (
)
( )]
[ (
) (
) ( )]
The strain in the compression steel at nominal moment capacity is 0.00185, the compression steel
has not yielded at nominal strength.
-
4-24
0.85f'c
a
fs=fy
dh
dh
f's
dh
a/2
Ccw
Csd'
T1
bw
b
bw
b
bw
b
Ccf
T2
ht
ht
ht (a+ht)/2
a) total beam section and stress distribution
b) Part 1: web of section and corresponding internal forces
c) Part 2: overhanging flanges and corresponding internal forces
f
F
F
a
a
(assumed)
Fig. S4-12.1 Beam section and internal forces for the case of th .
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