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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 nM and show that it

exceeds uM .

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 28 4.44 20 8 222

u uM w kips-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 5.930.003 0.00788

5.93

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

kips-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 ' 4000cf 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

nM and show it exceeds uM .

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 9

u u uM w P k-ft


Tension steel area: As = 6 No. 8 bars = 6 0.79 in.2 =4.74 in.2



4.74 600002.79

1 ' 0.85 4000 300.85

A fs y

cf bc

in.


2.79 3.280.85

c

in.


using. Eq.(4-18) 15.5 3.280.003 0.011

3.28

d c

s t cuc

> 0.0021

Thus, s

> 0.002 and the steel is yielding ( s yf f ).


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


4-3

4-3 (a) Compare nM for singly reinforced rectangular beams having the following

properties. Use loads and 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 3,000 60,000

2 12 22 2 No. 9 plus 1 No. 8 3,000 60,000

3 12 22 3 No. 7 3,000 40,000

4 12 22 3 No. 7 4,500 60,000

5 12 33 3 No. 7 3,000 60,000

Beam No.1


yielding.

3 0.60 600003.53

1 ' 0.85 3000 120.85

A fs y

cf bc

in.


3.53 4.150.85

c

in.

22 4.150.003 0.013

4.15

d c

s t cuc

Thus, s


Since, 0.005t the section is clearly tension-controlled and =0.9.

3.530.9 3 0.60 60000 22

2164

2 12000M A f d

n s y

kip-ft

For Beam 1, 164Mn

kip-ft

Beam No.2


yielding.

2 1.00 0.79 600005.47

1 ' 0.85 3000 120.85

A fs y

cf bc

in.


5.47 6.440.85

c

in.


4-4

22 6.440.003 0.0072

6.44

d c

s t cuc

Thus, s



5.470.9 2.79 60000 22

2242

2 12000M A f d

n s y

kip-ft

For Beam 2, 242Mn

kip-ft

Beam No.3


yielding.

1.8 400002.35

1 ' 0.85 3000 120.85

A fs y

cf bc

in.


2.35 2.760.85

c

in.

22 2.760.003 0.021

2.76

d c

s t cuc

Thus, s



2.350.9 1.8 40000 22

2113

2 12000M A f d

n s y

kip-ft

For Beam 3, 113Mn

kip-ft

Beam No.4


yielding.

1.8 600002.35

1 ' 0.85 4500 120.85

A fs y

cf bc

in.


2.35 2.850.825

c

in.

22 2.850.003 0.020

2.85

d c

s t cuc

Thus, s



4-5


2.350.9 1.8 60000 22

2169

2 12000M A f d

n s y

kip-ft

For Beam 4, 169Mn

kip-ft

Beam No.5


yielding.

1.8 600003.53

1 ' 0.85 3000 120.85

A fs y

cf bc

in.


3.53 4.150.85

c

in.

33 4.150.003 0.021

4.15

d c

s t cuc

Thus, s



3.530.9 1.8 60000 33

2253

2 12000M A f d

n s y

kip-ft

For Beam 5, 253Mn

kip-ft

(b) Taking beam 1 as the reference point, discuss the effects of changing sA , yf ,

' ,cf and d on nM . (Note that each beam has the same properties as beam 1 except for the

italicized quantity.)

Beam

No.

Mn

(kips-ft)

1 164

2 242

3 113

4 169

5 253

Effect of sA (Beams 1 and 2)

An increase of 55% in sA (from 1.80 to 2.79 in.2) caused on increase of 48% in nM . It is clear

that increasing the tension steel area causes a proportional increase in the strength of the section,


4-6

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)

A decrease of 33% in yf caused a decreased of 31% in nM . A decrease in the steel yield

strength has essentially the same effect as decreasing the tension steel area.

Effect of 'cf (Beams 1 and 4)

An increase of 50% in 'cf caused an increase of 3% in

nM . 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 nM .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 nM ? 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 the 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 16b in., 15.5d in., and 18h 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 ,minsA .


Tension steel area: sA = 4 No. 7 bars = 4 0.60 in.2 =2.40 in.2



2.4 600002.65

1 ' 0.85 4000 160.85

A fs y

cf bc

in.


2.65 3.10.85

c

in.


using Eq.(4-18) 15.5 3.10.003 0.012

3.1

d c

s t cuc

Thus, s



2.652.4 60000 15.5

2170

2 12000M A f d

n s y

kips-ft

Since, 0.012 0.005t the section is clearly tension-controlled and,

0.9 170nM kips-ft = 153 kips-ft

2. Compute Live Load

Set 153u nM M kips-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 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


4-8

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 nM and check ,minsA for the beam shown in Fig. P4-5. Use ' 4500cf psi

and 60,000yf psi.



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.)


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

kips-ft


2. Check of ,minsA


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


and 60,000yf psi.


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.


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.)


4.18 4.920.85

c

in.


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

oft 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

kips-ft


2. Check of ,minsA



,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, nM , and check ,minsA for the beam shown

in Fig. P4-7. Use ' 3500cf psi and 40,000yf psi.

1. Calculation of nM

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:

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:

2.64 400002.96

' 0.85 3500 120.85 e

A fs y

f bc

in.


2.96 3.480.85

c

in.

We should confirm that the steel is yielding:

using Eq.(4-18) 19.5 3.48

0.003 0.0143.48

t

d c

s cuc

Clearly, the steel is yielding 0.00207s y and this is tension-controlled

section 0.005t .


2.962.64 40000 19.5

2159

2 12000M A f d

n s y

kips-ft


2. Check of ,minsA

The beam is subjected to negative bending and since the flanged portion of the beam section is in

tension, 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 wb in Eq. (4-11). Also, '3 cf is equal

to 177 psi, so use 200 psi in the numerator:


4-12

,min

200 20012 19.5 1.17

40,000s 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 24 in.wb or eb . Given that eb is 48 in. for this beam section,

,min

200 20024 19.5 2.34

40,000s w

y

A b df

< sA (o.k.)


4-13

4-8 For the beam shown in Fig. P4-8, ' 3500cf psi and 60,000yf 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 nM for the positive- and negative-moment regions and check

,minsA 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


4-14

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.


compression flange, 6 in.fh and that the tension steel is yielding, s y , using Eq.(4-

16):

3.57 60000

1.07' 0.85 3500 67.50.85 e

A fs y

f bc

in. 6fh in. (o.k.)


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

kips-ft


2. Check of ,minsA




4-15

,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 0.001334.5

s 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


4-16


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. 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.)


4-17

4-9 Compute nM and check ,minsA for the beam shown in Fig. P4-9. Use

'3500

cf psi

and 60,000y

f psi, and

(a) the reinforcement is six No. 8 bars.


Tension steel area: sA = 6 No. 8 bars = 6 0.79 in.2 = 4.74 in.2


top flange, 5 in. and that the tension steel is yielding, s y , using Eq. (4-16) with

30 in.b :

4.74 60000

3.19' 0.85 3500 300.85

A fs y

f bc

in. 5fh in. (o.k.)


3.19 3.760.85

c

in.


using Eq.(4-18) 32.5 3.76

0.003 0.0233.76

t

d c

s cuc

Thus, s




3.184.74 60000 32.5

2733

2 12000M A f d

n s y

kips-ft


2. Check of ,minsA


use of that beam.


minimum tension reinforcement should be calculated using 2 5 10 in.wb in Eq. (4-11). Also,

'3 cf is equal to 177 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.)


4-18

However, for a statically determinate beam, wb should be replaced by the smaller of


,min

200 20020 32.5 2.17

60,000s w

y

A b df

in.2 < sA (o.k.)

(b) the reinforcement is nine No. 8 bars.




compression flange, 5 in.fh and that the tension steel is yielding, s y , using Eq. (4-

16) with 30 in.b :

7.11 60000

4.78' 0.85 3500 300.85

A fs y

f bc

in. 5fh in. (o.k.)


4.78 5.620.85

c

in.


using Eq.(4-18) 32.5 5.62

0.003 0.0145.62

t

d c

s cuc

Thus, s




4.787.11 60000 32.5

21070

2 12000M A f d

n s y

kips-ft


2. Check of ,minsA

,minsA is the same as in part (a).


4-19


and 60,000yf psi, and



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 60000

5.65' 0.85 5000 120.85

A fs y

f bc

in.


5.65 7.060.80

c

in.


using Eq.(4-18) 23.5 7.06

0.003 0.0077.06

t

d c

s cuc

Thus, s


Since, 0.005t the section is tension-controlled and =0.9.


5.654.8 60000 23.5

2496

2 12000M A f d

n s y

kips-ft


2. Check of ,minsA


use of that beam.


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


,min

212 21224 23.5 1.99

60,000s w

y

A b df

in.2 < sA (o.k.)


4-20

4-11 (a) Compute nM for the three beams shown in Fig. P4-11. In each case,

' 4000cf psi and 60yf ksi, 12 in., 32.5 in., and 36 in.b d h

Beam No. 1


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):

6.00 60000

8.82' 0.85 4000 120.85

A fs y

f bc

in.


8.82 10.40.85

c

in.

We need to check whether tension steel is yielding:

using Eq.(4-18) 32.5 10.4

0.003 0.006410.4

d c

s cuc

Thus, s


Also, clearly 0.005t , the section is tension-controlled and =0.9.

Thus, =0.9 and we can use Eq. (4-21) to calculate nM :

8.826.00 60000 32.5

2843

2 12000M A f d

n s y

kips-ft


Beam No. 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.


4-21



Try 4 8 in.c d

'' 8 2.5

0.003 0.002068

s cu

c d

c

' ' 29,000 ksi 0.00206 59.7 ksis s s yf E f

' ' ' 2' 0.85 2.00 in. 59.7 ksi 3.4 ksi 113 kipss s s cC A f f

'10.85 0.85 4 ksi 12 in. 0.85 8 in.=277 kipsc cC f b c


Because 'c sT C C , we should decrease c for the second trial.

Try 7.4 in.c ' 0.00199s

' 57.7 ksis yf f

' 109 kipssC

257 kipscC

360 kips ' 366 kipsc sT C C


yielding.

using Eq.(4-18) 32.5 7.4

0.003 0.01027.4

d c

s cuc

Clearly, the steel is yielding 0.00207s and it is a tension-controlled section 0.0102t s .


'' 257 kips 29.3 in. 108.6 kips 30 in.2

7545 k-in. 3270 k-in 10800 k-in 901 k-ft

c sM C d C d dn

Mn


Beam No. 3


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.


4-22

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 7 in.c '

' 7 3.50.003 0.0015

7s cu

c d

c

' '29,000 ksi 0.0015 43.5 ksi

s s s yf E f

' ' ' 2' 0.85 4.00 in. 43.5 ksi 3.4 ksi 160 kipss s s c

C A f f

'

10.85 0.85 4 ksi 12 in. 0.85 7 in.=243 kips

c cC f b c

26.00 in. 60 ksi 360 kips

s yT A f

Because 'c s

T C C , we should decrease c for the second trial.

Try 6.3 in.c (Note that both layers of the compression steel will actually be in the compression zone) ' 0.00133s

' 38.6 ksis yf f

' 141 kipssC

218 kipscC

360 kips 359 kipsc sT C C


yielding.

using Eq.(4-18) 32.5 6.3

0.003 0.0126.3

d c

s cuc

Clearly, the steel is yielding 0.00207s and it is a tension-controlled

section 0.012t s .


'' 218 kips 29.8 in. 141 kips 29 in.2

6495 k-in. 4090 k-in 10585 k-in 882 k-ft

c sM C d C d dn

Mn



4-23

(b) From the results of part (a), comment on weather adding compression

reinforcement is a cost-effective way of increasing the strength, nM , 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,

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.


4-24

4-12 Compute nM for the beam shown in Fig. P4-12. Use ' 3500cf psi and

60,000yf psi. Does the 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



Try 4 5.5 in.c d

For ' 3500cf 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. 10 in.b

'

' 5.5 2.50.003 0.00164

5.5s cu

c d

c

' ' 29,000 ksi 0.00164 47.6 ksif E fs s s y

' ' ' 2' 0.85 1.2 in. 47.6 ksi 2.98 ksi 53.5 kipss s s cC A f f

'10.85 0.85 3.5 ksi 10 in. 0.85 5.5 in.=139 kipsc cC f b c


Because 'c sT C C , we should increase c for the second trial.

Try 6.5 in.c

and find 0.85 6.5 in. 5.53 in.> 5.0 in. ' 0.00185s

' 53.7 ksis yf f

' 60.9 kipssC

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):

'0.85 10 in. 0.85 3.5 ksi 10 in. 5.53 in.=165 kipscw cC f

'0.85 20 10 in. 5 in. 0.85 3.5 ksi 10 in. 0.53 in. 15.8 kipscf cC f

165 15.8 181 kipscC

284 kips > ' 242 kipsc sT C C , we should increase c for the third trial.

Try 7.2 in.c

and find 0.85 7.2 in. 6.12 in.> 5.0 in. ' 0.00196s


4-25

' 56.8 ksis yf f

' 64.6 kipssC

182 kipscwC

33.3 kipscfC

215 kipscC

284 kips ' 276 kipsc sT C C


yielding.

using Eq.(4-18) 22.5 6.12

0.003 0.008036.12

d c

s cuc

Thus, the steel is yielding 0.00207s and it is a tension-controlled section 0.012t s .

Summing the moments about the level of the tension reinforcement:

'5

5 '2 2

182 kips 19.4 in. 33.3 kips 16.9 in. + 64.6 kips 20 in.

3530 k-in. 563 k-in +1290 k-in 5385 k-in 449 k-ft

cw cf sM C d C d C d dn

Mn

Mn



4-26

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 .

Documents

Full file at ://fratstock.eu/sample/Solutions-Manual... · kip-ft For Beam 4, M 169 n I kip-ft Beam No.5 Compute the depth of the equivalent rectangular stress block, D, assuming