solucion cap 4 libro

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


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.


1

2.79 3.280.85

c

in.


Using. Eq.(4-18) 15.5 3.28 0.003 0.0113.28

d c

s t cuc

> 0.0021

Thus, s



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 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


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

(

)

( )

For Beam 1, -

Beam No.2


( )

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



( ) (

) (

)

Thus, s


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

(

)

( )

For Beam 3, -

Beam No.4


For , . Therefore,

( ) (

) (

)

s yf f

4-5

Thus, s



(

)

( )

For Beam 4, -

Beam No.5



( ) (

) (

)

Thus, s



(

)

( )

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

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 .


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.


1

2.65 3.10.85

c

in.


Using Eq.(4-18) 15.5 3.1 0.003 0.0123.1

d c

s t cuc

Thus, s



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.


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.)


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


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


psi and 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.)


1

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 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


2. Check of ,minsA


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:


(

) (

)

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.


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.)


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


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


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


psi and 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 :

( )


( ) (

) (

)

Thus, s




(

)

( )

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


,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.




compression flange, 5 in.fh and that the tension steel is yielding, s y , using Eq. (4-16) with 30 in.b :

( )


( ) (

) (

)

Thus, s




(

)

( )

2. Check of ,minsA

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

4-18


psi and psi.



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.


1

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




5.654.8 60000 23.5

2496

2 12000M A f d

n s y

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


,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


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


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


(

)

( )

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.



4-20

Try 4 8 in.c d

(

) (

)

( )

(

) ( )

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

Try

( )


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


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

( )


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



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):

( )( )


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 .

Text1: 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

solucion cap 4 libro

Documents