Reinforced Concrete Mechanics Design Solutions

3-1

Chapter 3

3-1 What is the significance of the “critical stress”?

(a) with respect to the structure of the concrete?

A continuous pattern of mortar cracks begins to form. As a result there are few undamaged portions

to carry load and the stress-strain curve is highly nonlinear.

(b) with respect to spiral reinforcement?

At the critical stress the lateral strain begins to increase rapidly. This causes the concrete core within

the spiral to expand, stretching the spiral. The tension in the spiral is equilibrated by a radial

compression in the core. This in turn, biaxially compresses the core, and thus strengthens it.

(c) with respect to strength under sustained loads?

When concrete is subjected to sustained loads greater than the critical stress, it will eventually fail.

3-2 A group of 45 tests on a given type of concrete had a mean strength of 4780 psi and a

standard deviation of 525 psi. Does this concrete satisfy the requirements of ACI Code

Section 5.3.2 for 4000-psi concrete?

From Eq. 3-3a:

Using

(for design)

From Eq. 3-3b:

Using

(for design)

Because both of these exceed 4000 psi, the concrete satisfies the requirements of ACI Code

Section 5.3.2 for 4000 psi concrete.

3-2

3-3 The concrete containing Type I cement in a structure is cured for 3 days at 70° F

followed by 6 days at 40° F. Use the maturity concept to estimate its strength as a

fraction of the 28-day strength under standard curing.

Note: 5

329

C F , so 70° F = 21.1° C and 40° F = 4.4° C

From Eq. 3-6:

1

( 10)( )

(21.1 10)(3) (4.4 10)(6) 180 C days

n

i i

i

M T t

From Fig. 3-8 the compressive strength will be between 0.60 and 0.70 times the 28-day strength

under standard curing conditions.

3-4 Use Fig. 3-12a to estimate the compressive strength 2 for bi-axially loaded concrete

subject to:

(a) 1 = 0.0, 2 = fc'

(b) 1 = 0.75 ft' in tension, 2 = 0.5 fc'

(c) 1 = 0.5 fc' in compression, 2 = 1.2 fc'

3-5 The concrete in the core of a spiral is subjected to a uniform confining stress 3 of 750

psi. What will the compressive strength, 1 be? Assume .

From Eq. 3-16:

3-3

3-6 What factors affect the shrinkage of concrete?

(a) Relative humidity. Shrinkage increases as the relative humidity decreases, reaching a

maximum at RH ≤ 40%.

(b) The fraction of the total volume made up of paste. As this fraction increases,

shrinkage increases.

(c) The modulus of elasticity of the aggregate. As this increases, shrinkage decreases.

(d) The water/cement ratio. As the water content increases, the aggregate fraction

decreases, causing an increase in shrinkage.

(e) The fineness of the cement. Shrinkage increases for finely ground cement that has

more surface area to attract and absorb water.

(f) The effective thickness or volume to surface ratio. As this ratio increases, the

shrinkage occurs more slowly and the total shrinkage is likely reduced.

(g) Exposure to carbon dioxide tends to increase shrinkage.

3-7 What factors affect the creep of concrete?

(a) The ratio of sustained stress to the strength of the concrete. The creep coefficient, ,

is roughly constant up to a stress of 0.5 fc', but increases above that value.

(b) The humidity of the environment. The amount of creep decreases as the RH increases

above 40%.

(c) As the effective thickness or volume to surface ratio increases, the rate at which creep

develops decreases.

(d) Concretes with a high paste content creep more that concretes with a large aggregate

fraction because only the paste creeps.

3-4

3-8 A structure is made from concrete containing Type 1 cement. The average ambient

relative humidity is 70 percent. The concrete was moist-cured for 7 days. fc' = 4000 psi.

(a) Compute the unrestrained shrinkage strain of a rectangular beam with cross-

sectional dimensions of 8 in. x 20 in. at 2 years after the concrete was placed.

1. Compute the humidity modification factor from Eq. (3-30a):

2. Use Eq. (3-31) to compute the volume/surface area ratio modification factor:

( ) ( )

⁄ ⁄

3. Use Eq. (3-29) to compute the ultimate shrinkage strain:

( )

4. Use Eq. (3-28) to compute the shrinkage strain after 2 years:

( )

( )

(b) Compute the stress dependent (creep) strain in the concrete of a 20 in. x 20 in. x

12 ft column at age 3 years. A compression load of 400 kips was applied to the

column at 30 days.

1. Compute the ultimate shrinkage strain coefficient, , using Eqs. (3-36)-(3-39).

⁄ ,

Where:

[ ⁄ ]

3-5

2. Compute the creep coefficient for the time since loading, , using Eq. (3-35).

3. Compute the total stress-dependent strain, ( ), using Eqs. (3-5), (3-18), and (3-35).

First, calculate the creep strain since the load was applied:

( ) √ √

( ) ( )

( )

Then, calculate the initial strain when the load is applied:

( )

( )

( ) ( )

( ) √ ( ) √

( ) ( )

( )

Thus,

( ) ( ) ( )

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 28 4.44 20 8 222

u uM 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 5.930.003 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

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 9

u 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 3.280.003 0.011

3.28

d c

s t cuc

> 0.0021



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


yielding.

For ,

1 0.85 . Therefore,

⁄ ⁄

( ) (

) (

)


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

(

)

( )

For Beam 1, -

Beam No.2


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


yielding.


⁄

⁄

( ) (

) (

)


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

(

)

( )

For Beam 3, -

Beam No.4


yielding.

For , . Therefore,

⁄

⁄

( ) (

) (

)

s yf f

4-5



(

)

(

)

For Beam 4, -

Beam No.5


yielding.


⁄

⁄

( ) (

) (

)



(

)

( )

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 3.10.003 0.012

3.1

d c

s t cuc



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 andtd , 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 0.001334.5

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

4-15


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 :

( )


⁄

⁄

( ) (

) (

)




(

)

( )

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.



2 =7.11 in.

2


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

16) with 30 in.b :

( )


⁄

⁄

( ) (

) (

)




(

)

(

)

2. Check of ,minsA

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

4-18


psi and psi.



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.


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




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


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,

⁄

⁄

(

) (

)


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


(

)

(

)

Beam No. 2


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.



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


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

( )


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 .

5-1

Chapter 5

5-1 Give three reasons for the minimum cover requirements in the ACI code.

- To ensure enough concrete is present to develop the reinforcement.

- To protect reinforcement from corrosive agents.

- To insulate reinforcement in case of fire.

Under what circumstances are larger covers used?

- Highly corrosive environments.

- Situations where abrasion to the concrete surface may result in a reduction in cover provided.

Note: See Section 5.3 “Concrete Cover and Bar Spacing” for further discussion.

5-2 Give three reasons for using compression reinforcement in beams.

- To reduce long-term deflections (i.e. creep).

- It tends to lead to a more ductile failure mode.

- Some compression reinforcement is always required for fabrication of rebar cages.

5-2

5-3 Design a rectangular beam section, i.e. select b, d, h, and the required tension

reinforcement, at midspan for a 22 ft-span simply supported rectangular beam that

supports its own dead load, a superimposed service dead load of 1.25 kip/ft, and a

uniform service load of 2 kip/ft. Use the procedure in Section 5.3 for the design of

beam sections where the dimensions are unknown. Use and fy = 60 ksi.

Step 1: Estimate the dead load due to self-weight of the beam:

Method 1: 0.10 to 0.15 ( )DL SDL LL

0.10 to 0.15 (1.25 k/ft 2 k/ft )DL

325 lb/ft to 490 lb/ft DL

Method 2: /18 /12, so estimate 22 in h h

0.8 18 in b h

2 2 3 2 2 3

lb 18 in 22 in lb lb 150 150 413

144 in /ft ft 144 in /ft ft ft

bhDL

Therefore, DL = 410 lb/ft seems like a good first estimate of the weight of the beam.

Step 2: Compute the total factored load and factored design moment, Mu:

From ACI-08 Chapter 9:

1.2 1.6 , 1.4 u D L Dw w w or w

1.2 (0.41 k/ft 1.25 k/ft) 1.6 (2 k/ft), or 1.4 (0.41 k/ft 1.25 k/ft) uw

5.20 k/ft, or 2.35 k/ftuw

For this simply supported span, 2 25.20 k/ft (22 ft)

315 k-ft 3775 k-in8 8

uu

wM

Step 3: Select ρ and the corresponding R-factor:

Assume the desirable strain diagram shown in Fig. 5-27b, which leads to .

From Eq. (5-19):

1 ' 0.825 4,500 psi0.0154

4 4 60,000 psi

c

y

f

f

Note: β1 = 0.825 for f’c = 4,500 psi

From Eq. (5-21):

0.0154 60,000 psi0.205

' 4,500 psi

y

c

f

f

From Eq. (5-22):

' (1 0.59 ) 0.205 4.5 ksi (1 0.59 0.205) 0.811 ksicR f

(Note that this R-factor is reasonable based on values given in Table A-3)

9.0

5-3

Step 4: Select section dimensions, b and h:

From Eq. (5-23):

2 33775 k-in5170 in

0.9 0.811 ksi

uMbd

R

Since no column dimensions are given which control the width of the beam, the designer can

assume any reasonable α value. Here we assume α = 0.7 1 1

3 33775 k-in19.5 in

0.7 0.9 0.811 ksi

uMd

R

2.5 in 22 inh d

0.7 19.5 in 13.65 in 14 inb d

Note that both h and b are rounded up to the nearest even inch value for constructability.

Step 5: Determine As and select the reinforcing bars:

First, go back and recalculate the weight of the beam with the final selected dimensions:

2 2 3 2 2 3

lb 14 in 22 in lb lb 150 150 321


bhDL

So, 5.09 k/ftuw and 308 k-ft 3695 k-inuM

Calculate the required area of steel, assuming 0.9jd d :

From Eq. (5-16):

23695 k-in

3.90 in0.9 60 ksi 0.9 19.5 in

2

us

y

MA

af d

With this estimate, iterate once to have a better estimate of the lever arm jd .

From Eq. (5-17): 23.90 in 60 ksi

4.37 in0.85 ' 0.85 4.5 ksi 14 in

s y

c

A fa

f b

From Eq. (5-16):

23695 k-in 3.95 in

4.37 in0.9 60 ksi 19.5 in -

2 2

us

y

MA

af d

No further iterations are necessary, since the estimated lever arm was very reasonable. Select 4#9

bars as bottom reinforcement at the critical section of the beam. 2 2 2 4 4 1.0 in 4.0 in 3.95 ins bA A OK

Step 6: Required Checks:

1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 14 in is sufficiently

wide for 4 #9 bars to be placed in a single layer. OK

5-4

2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum

required.

For f’c = 4,500 psi, 3 ' 201 200cf so use 3 'cf

2 2

,min

3 ' 3 4500 14 in 19.5 in 0.92 in 4.0 in

60,000 psi

c

s w

y

fA b d

f OK

3) Calculate the strain in the extreme layer of tension steel to verify that assuming is valid.

From Eq. (5-17): 24.0 in 60 ksi

4.48 in0.85 ' 0.85 4.5 ksi 14 in

s y

c

A fa

f b

We know that1 0.825 (see above).

1

4.48 in5.43 in

0.825

ac

From Eq. (4-18):

- 19.5 in - 5.43 in0.003 0.0078 0.005

5.43 int cu

d c

c OK

Therefore the designer is permitted to use for this beam design.

4) Finally, verify that the nominal flexural strength is sufficient for the applied loads.

From Eq. (5-15):

2 4.48 in 0.9 4.0 in 60 ksi 19.5 in 3725 k-in

2 2n s y

aM A f d

310 k-ft 308 k-ftn uM M

Therefore, this design is sufficient without being too conservative.

Note that other combinations of b, h, and As may also be correct if different assumptions were

made by the designer. If all checks listed in Step 6 are satisfied, without being unreasonably

conservative, the design may be considered adequate.

Fig. S5-4 Cross-section of final design at mid-span

9.0

9.0

22in.

14in.

4 #9 bars

2.5in.

5-5

5-4 The rectangular beam shown in Fig. P5-4 carries its own dead load (you must guess

values for b and h) plus an additional uniform service load of 0.5 kip/ft and a uniform

service live load of 1.5 kip/ft. The dead load acts on the entire beam, of course, but

the live load can act on parts of the span. Three possible loading cases are shown in

Fig. P5-4. Use load and strength reduction factors from ACI Code sections 9.2 and

9.3.

a) Draw factored bending-moment diagrams for the three loading cases shown

and superimpose them to draw a bending-moment envelope.

Begin by estimating the dead load due to self-weight of the beam:

Method 1: 0.10 to 0.15 ( )DL SDL LL

0.10 to 0.15 (0.5 k/ft 1.5 k/ft )DL

200 lb/ft to 300 lb/ft DL

Method 2: /18 /12, so estimate 22 in h h

0.8 18 in b h

2 2 3 2 2 3

lb 18 in 22 in lb lb 150 150 413

144 in /ft ft 144in /ft ft ft

bhDL

We select DL = 350 lb/ft as a first estimate of the weight of the beam.

So, using 1.2 1.6u D Lw w w from ACI 318-08, Chapter 9, the bending-moment envelope is as

follows:

Fig. S5-5a Bending moment envelope

5-6

b) Design a rectangular beam section for the maximum positive bending moment

between the supports, selecting b, d, h, and the reinforcing bars. Use the

procedure in Section 5.3 for the design of beam sections where the dimensions

are unknown. Use f’c = 5000 psi and fy = 60 ksi.

It is necessary to design this beam section for both negative and positive bending. The

need for a practical design makes it reasonable to assume that the outer dimensions of the beam

will be constant along the length, and that these dimensions will be controlled by the design of the

section subjected to the largest absolute value of moment. As seen in Part (a), the largest expected

moment is a positive moment of 226 kip-ft (2715 k-in). Therefore, it is reasonable to begin the

beam design by designing the beam at this location.


Assume the strain diagram shown in Fig. 5-27b, which leads to .

From Eq. (5-19):

1 ' 0.80 5,000 psi0.0167

4 4 60,000 psi

c

y

f

f

Note: β1 =0.80 for f’c = 5,000 psi

From Eq. (5-21):

0.0167 60,000 psi0.20

' 5,000 psi

y

c

f

f

From Eq. (5-22):

' (1 0.59 ) 0.20 5 ksi (1 0.59 0.20) 0.882 ksicR f



From Eq. (5-23):

2 32715 k-in3420 in

0.9 0.882 ksi

uMbd

R



3 32715 k-in17.0 in 17.5 in

0.7 0.9 0.882 ksi

uMd

R

2.5 in 20 inh d

0.7 17.5 in 12.5 in 14 inb d


Step 3: Determine As and select the reinforcing bars:

First, go back and recalculate the weight of the beam with the final selected dimensions:

2 2 3 2 2 3

lb 14 in 20 in lb lb 150 150 292


bhDL

So, using pattern loading again, the maximum positive moment is: 222 k-ft 2670 k-inuM

9.0

5-7

Now calculate the required area of steel, assuming 0.9jd d :

From Eq. (5-16):

22670 k-in

3.14 in0.9 60 ksi 0.9 17.5 in

2

us

y

MA

af d

With this estimate, iterate once to have a better estimate of the lever arm jd

From Eq. (5-17): 23.14 in 60 ksi

3.16 in0.85 ' 0.85 5 ksi 14 in

s y

c

A fa

f b

From Eq. (5-16):

22670 k-in 3.11 in

3.16 in0.9 60 ksi 17.5 in -

2 2

us

y

MA

af d



Step 4: Required Checks:




required.

For f’c = 5,000 psi, 3 ' 212 200cf so use 3 'cf

2 2

,min

3 ' 3 5000 14 in 17.5 in 0.87 in 3.16 in

60,000 psi

c

s w

y

fA b d

f OK


From Eq. (5-17): 23.16 in 60 ksi

3.19 in0.85 ' 0.85 5 ksi 14 in

s y

c

A fa

f b


1

3.19 in3.99 in

0.80

ac

From Eq. (4-18):

- 17.5 in - 3.99 in0.003 0.0102 0.005

3.99 int cu

d c

c OK


9.0

9.0

5-8


From Eq. (5-15):

2 3.19 in 0.9 3.16 in 60 ksi 17.5 in 2715 k-in

2 2n s y

aM A f d


Therefore, this design for positive bending is sufficient without being too conservative.

c) Using the beam section from part (b), design flexural reinforcement for the

maximum negative moment over the roller support.

Since the outer dimensions are selected, the design for negative bending follows the method for

designing a rectangular section where the section dimensions are known. The maximum expected

negative moment, considering pattern loading, is - -

Determine As and select the reinforcing bars for negative bending:

Assume :

From Eq. (5-16):

( )

( )


From Eq. (5-17):

From Eq. (5-16):

( )

( )

No further iterations are necessary since the solution has essentially converged. Select 2#9 bars as

top reinforcement at the critical section of the beam.

Required Checks:




required.

For f’c = 5,000 psi, √ so use √

2 2

,min

3 ' 3 5000 14 in 17.5 in 0.87 in 2.0 in

60,000 psi

c

s w

y

fA b d

f

OK

5-9


From Eq. (5-17): 22.0 in 60 ksi

2.02 in0.85 ' 0.85 5 ksi 14 in

s y

c

A fa

f b


1

2.02 in2.53 in

0.80

ac

From Eq. (4-18):

- 17.5 in - 2.53 in0.003 0.0178 0.005

2.53 int cu

d c

c

OK



From Eq. (5-15):

2 2.02 in 0.9 2.0 in 60 ksi 17.5 in 1780 k-in

2 2n s y

aM A f d


The cross-sections of the beam design at maximum positive and negative bending moments is

shown below.

Fig. S5-5b Cross-sections of final designs for positive and negative bending respectively

Note that other combinations of b, h, and As may also be correct if different assumptions were

made by the designer. If all checks listed in Steps 4 and 6 are satisfied, without being unreasonably

conservative, the design may be considered adequate.

9.0

9.0

20in.

14in.

4 #8 bars

2.5in.

20in.

14in.

2 #9 bars2.5in.

5-10

5-5 Design three rectangular beam sections, i.e. select b and d and the tension steel area

As, to resist a factored design moment, Mu = 260 kip-ft. For all three cases select a

section with b = 0.5d and use f’c = 4000 psi and fy = 60 ksi.

a) Start your design by assuming that εt = 0.0075 (as was done in Section 5.3)

The equations presented in section 5.3 initially assumed that εt = 0.0075. So, while no changes

need to be made, their derivation will briefly be shown to easy comparisons with the solutions to

parts (b) and (c).

0.0030.286

0.003 0.0075

cu

cu s

c d d d

1 1 10.85 ' 0.85 ' 0.286 0.24 'c c c cC f bc f b d f bd

Now, enforce equilibrium:

cC T

10.24 'c s yf bd A f

Thus we have an expression for the initial ρ value when εt = 0.0075 is assumed, from Eq. 5-19:

10.24 ' 0.24 0.85 4,000 psi0.0136

60,000 psi

cinitial

y

f

f

From Eq. (5-21):

0.0136 60,000 psi0.204

' 4,000 psi

y

c

f

f

From Eq. (5-22):

' (1 0.59 ) 0.204 4 ksi (1 0.59 0.204) 0.718 ksicR f


From Eq. (5-23):

2 33120 k-in4830 in

0.9 0.718 ksi

uMbd

R

Note that with εt = 0.0075, 0.9

We are told to assume α = 0.5 1 1

3 33120 k-in21.3 in 21.5 in

0.5 0.9 0.718 ksi

uMd

R

0.5 21.5 in 10.8 in 12 inb d

Note that both d and b are rounded up so that h and b both result in even inch values for

constructability.

Now, determine As and select the reinforcing bars, assuming 0.9jd d :

From Eq. (5-16):

23120 k-in

2.99 in0.9 60 ksi 0.9 21.5 in

2

us

y

MA

af d


5-11

From Eq. (5-17): 22.99 in 60 ksi

4.40 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

23120 k-in 3.00 in

4.4 in0.9 60 ksi 21.5 in -

2 2

us

y

MA

af d



Now do the required checks:




required.

For f’c = 4,000 psi, 3 ' 190 200cf so use 200.

2 2

,min

200 psi 200 psi 12 in 21.5 in 0.86 in 3.0 in

60,000 psis w

y

A b df

OK


From Eq. (5-17): 23.0 in 60 ksi

4.41 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

We know that1 0.85 .

1

4.41 in5.19 in

0.85

ac

From Eq. (4-18):

- 21.5 in -5.19 in0.003 0.0094 0.005

5.19 int cu

d c

c OK



From Eq. (5-15):

2 4.41 in 0.9 3.0 in 60 ksi 21.5 in 3125 k-in

2 2n s y

aM A f d



b) Start your design by assuming that εt = 0.005

9.0

9.0

5-12

We need to re-derive an expression for ρ, using the same approach used in part (a).

0.0030.375

0.003 0.005

cu

cu s

c d d d

1 1 10.85 ' 0.85 ' 0.375 0.319 'c c c cC f bc f b d f bd


cC T

10.319 'c s yf bd A f

Thus we have an expression for the initial ρ value when εt = 0.005 is assumed:

10.319 ' 0.319 0.85 4,000 psi0.0181

60,000 psi

cinitial

y

f

f

From Eq. (5-21):

0.0181 60,000 psi0.271

' 4,000 psi

y

c

f

f

From Eq. (5-22):

' (1 0.59 ) 0.271 4 ksi (1 0.59 0.271) 0.911 ksicR f


From Eq. (5-23):

2 33120 k-in3805 in

0.9 0.911 ksi

uMbd

R

Note that with εt = 0.005, 0.9

We are told to assume α = 0.5 1 1

3 33120 k-in19.7 in 19.5 in

0.5 0.9 0.911 ksi

uMd

R

2.5 in 22 inh d

0.5 19.5 in 10 inb d

Note that both h and b are rounded to even inch values for constructability. Normally these values

would be rounded up, but since the estimate for d is so much nearer to 19.5 in than 21.5 in, it is

rounded down in this solution. Adequate strength of the section will still be achieved by selection

of an appropriate As value.

Now, determine As and select the reinforcing bars, assuming 0.9jd d :

From Eq. (5-16):

23120 k-in

3.29 in0.9 60 ksi 0.9 19.5 in

2

us

y

MA

af d


From Eq. (5-17): 23.29 in 60 ksi

5.81 in0.85 ' 0.85 4 ksi 10 in

s y

c

A fa

f b

From Eq. (5-16):

23120 k-in 3.48 in

5.81 in0.9 60 ksi 19.5 in -

2 2

us

y

MA

af d

Iterate once more to have a better estimate of the lever arm jd .

5-13

From Eq. (5-17): 23.48 in 60 ksi

6.14 in0.85 ' 0.85 4 ksi 10 in

s y

c

A fa

f b

From Eq. (5-16):

23120 k-in 3.52 in

6.14 in0.9 60 ksi 19.5 in -

2 2

us

y

MA

af d

No further iterations are necessary, since the solution has converged. Select 3#10 bars as bottom

reinforcement at the critical section of the beam. 2 2 2 3 3 1.27 in 3.81 in 3.52 ins bA A OK



wide for 3 #10 bars to be placed in a single layer. NOT OK

So, either different bars must be selected, or the beam must be widened. Here we choose to widen

the beam so that b = 12 in.


required.


2 2

,min

200 psi 200 psi 12 in 19.5 in 0.78 in 3.81 in

60,000 psis w

y

A b df

OK


From Eq. (5-17): 23.81 in 60 ksi

5.60 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

We know that1 0.85 .

1

5.60 in6.59 in

0.85

ac

From Eq. (4-18):

- 19.5 in - 6.59 in0.003 0.0059 0.005

6.59 int cu

d c

c OK



From Eq. (5-15):

2 5.60 in 0.9 3.81 in 60 ksi 19.5 in 3435 k-in

2 2n s y

aM A f d


9.0

9.0

5-14

Therefore, this design is sufficient. SincenM is 10% larger than

uM , this design is perhaps too

conservative. Options for optimizing this design might be selecting different bars, or slightly

resizing the element. Notice that the design in part (a) was more efficient, so different initial

assumptions can lead to different designs.

c) Start your design by assuming that εt = 0.0035. You will probably need to add

compression reinforcement to make this a tension-controlled section.

We need to re-derive an expression for ρ, using the same approach used in part (a).


cC T

Thus we have an expression for the initial ρ value when εt = 0.0035 is assumed:

From Eq. (5-21):

From Eq. (5-22):

( ) ( )


Before we can estimate the dimensions of the beam with Eq. (5-23), we need to determine .

Although we began by assuming that εt = 0.0035, tension controlled sections are so desirable that

we will ensure that the section is tension controlled. Compression reinforcement might be required.

Set .

From Eq. (5-23):

We are told to assume α = 0.5. Also, since we had difficulty placing all the steel required in part

(b) due to insufficient beam width, and part (c) requires a higher reinforcement ratio to limit the

tensile strains, we will begin by assuming that two layers of reinforcement will be required.

(

)

⁄

(

)

⁄

Recall that we use 3.5 in. instead of 2.5 in. to account for the effect the second layer of steel has on

the location of the centroid of the tension reinforcement.

5-15

Note that both h and b are rounded to even inch values for constructability.

Now, Determine As and select the reinforcing bars, assuming 0.9jd d :

From Eq. (5-16):

( )

( )


From Eq. (5-17):

From Eq. (5-16):

( )

(

)

Iterate once more to have a better estimate of the lever arm jd .

From Eq. (5-17):

From Eq. (5-16):

( )

(

)

No further iterations are necessary, since the solution has converged. Select 5#8 bars as bottom

reinforcement at the critical section of the beam.

OK

Check whether the section is tensioned controlled:

1

6.97 in8.2 in

0.85

ac

Compression steel must be added to have a tension-controlled section.

Try adding 2 #8 bars in the compression zone, which is approximately , with Try

(

) (

)

( )

(

) ( )

5-16

Try

( )





required.


2 2

,min

200 psi 200 psi 10 in 20.5 in 0.68 in 3.95 in

60,000 psis w

y

A b df

OK


From Eq. (4-18):

Ok, .


From Eq. (5-15):

* (

)

( )+ [ (

) ( )]

- -

Therefore, this design is sufficient. Again, sincenM is 10% larger than

uM , this design is perhaps

too conservative. Options for optimizing this design might be selecting different bars, or slightly

resizing the element. Notice that the design in part (a) was more efficient, so different initial

assumptions can lead to different designs.

5-17

d) Compare and discuss your three section designs.

Fig S5-6 Cross sections of section designs from parts (a), (b), and (c), respectively

Note that the area of reinforcement provided is higher in the third design compared to

either of the first two, even though the section size is smaller. The design for part (c) also required

compression steel to ensure a tension controlled section, whereas the sections in parts (a) and (b)

were tension controlled as singly-reinforced sections.

24in.

12in.

3 #9 bars2.5in.

22in.

12in.

3 #10 bars2.5in.

22in.

10in.

5 #8 bars

3.5in.

2 #8 bars

5-18

5-6 You are to design a rectangular beam section to resist a negative bending moment of

275 kip-ft. Architectural requirements will limit your beam dimensions to a width of

12 in. and a total depth of 18 in. Using those maximum permissible dimensions, select

reinforcement to provide the required moment strength following the ACI Code

provisions for the strength reduction factor, . Use and .

Begin by trying to design the section as singly reinforced, with one layer of tension steel.

Set

( )

( )

Either 5 #9 bars of 6 #8. To fit within the 12 in. beam width, 2 layers of bars are required.

Try 5 #9 bars:

Check whether the tension steel has yielded, and whether the section is tension controlled.

Add compression steel so that ; try 3 #8 bars so that

By iteration,

( )

Check whether the section is tension controlled.

Calculate the nominal moment capacity:

* (

)

( )+ [ (

) ( )]

5-19


1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in. is sufficiently



required.

For f’c = 5,000 psi, √ so use 212.

√

Already verified that is OK, and calculated the nominal moment capacity.

5-20

5-7 For column line 2, use the ACI Moment Coefficients given in ACI Code section 8.3.3

to determine the maximum positive and negative factored moments at the support

faces for columns A-2 and B-2, and at the midspan of an exterior span and the

interior span.

First, confirm that the ACI moment coefficients can be used.

a) Two or more spans OK

b) Longer span length within 20% of shorter OK

c) Loads are uniformly distributed OK

d) Unfactored LL load does not exceed 3 times DL OK

e) Members are prismatic OK

Now, estimate the dead load supported by column line 2:

Dead load from beam, per foot of beam:

2 3

2

12 in 18 in lb lb150 225

in ft ft144 ft

Dead load from slab, per foot of beam:

3

12 ft 11 ft6 in

2 lb lb150 863

in ft ft12 ft

Superimposed dead load, per foot of beam:

2

12 ft 11 ft lb lb20 230

2 ft ft

Can the live load be reduced? From Eq. (5-3):

At column A-2 and midspan of exterior beam:

2 215 150.25 0.050 k/ft 0.25 0.041 k/ft

2 11.5 ft 30 ftr

LL T

L LK A

At column B-2:

2 215 150.25 0.050 k/ft 0.25 0.034 k/ft

2 11.5 ft 55 ftr

LL T

L LK A

At midspan of interior span:

2 215 150.25 0.050 k/ft 0.25 0.044 k/ft

2 11.5 ft 25 ftr

LL T

L LK A

Live load, per foot of beam:

At column A-2 and midspan of exterior beam: 2

12 ft 11 ft lb lb41 472

2 ft ft

At column B-2: 2

12 ft 11 ft lb lb34 391

2 ft ft

At midspan of interior span: 2

12 ft 11 ft lb lb44 506

2 ft ft

5-21

So, using 1.2 1.6uw DL LL , the factored loads are:


1.2 0.225 k/ft 0.863 k/ft 0.230 k/ft 1.6 0.472 k/ft 2.34 k/ftuw

At column B-2:




Calculate the clear span length,n

:

At column A-2: 30 ft -16 in 28.67 ftn

At midspan of exterior span: 30 ft -16 in 28.67 ftn

At column B-2 (exterior): 30 ft 25 ft -16 in 26.17 ft2n

At column B-2 (interior): 30 ft 25 ft -16 in 26.17 ft2n

At midspan of interior span: 25 ft -16 in 23.67 ftn

Design Moments using ACI Moment Coefficients from section 8.3.3:

At column A-2:

22 2.34 k/ft 28.67 ft120 k-ft

16 16

u nu

wM

At midspan of exterior span:

22 2.34 k/ft 28.67 ft137 k-ft

14 14

u nu

wM

At column B-2 (exterior):

22 2.21 k/ft 26.17 ft151 k-ft

10 10

u nu

wM

At column B-2 (interior):

22 2.21 k/ft 26.17 ft138 k-ft

11 11

u nu

wM


22 2.39 k/ft 23.67 ft83.7 k-ft

16 16

u nu

wM

5-22

5-8 Repeat problem 5-7, but use structural analysis software to determine the maximum

positive and negative moments described in problem 5-7. The assumed beam, slab

and column dimensions are given in the figure. Assume 12 ft story heights above and

below this floor level. You must use appropriate live load patterns to maximize the

various factored moments. Use a table to compare the answers from Problems 5-7

and 5-8.

From Problem 5-7, the factored loads are:

At column A-2 and midspan of exterior beam: 2.34 k/ftuw

At column B-2: 2.21 k/ftuw

At midspan of interior span: 2.39 k/ftuw

Properties of elements used for model:

Column:

216 in 16 in 256 ingA

4 416 in /12 5,460 ingI

Beam (Use properties of the web as an approximation of the cracked properties):

212 in 24 in 288 ingA

3 412 in 24 in /12 13,800 ingI

We assume story heights of 12 ft above and below the continuous beam being modeled, and

include columns, fixed at their ends, in our model. Then we apply the appropriate load

combinations following Example 5-2, and the following design moments result:

Location ACI Design Moment

(From Problem 5-7)

Design Moment

(From software)

At column A-2: -120 kip-ft -135 kip-ft

At midspan of exterior span: 137 kip-ft 93.0 kip-ft

At column B-2 (exterior): -151 kip-ft -159 kip-ft

At column B-2 (interior): -138 kip-ft -114 kip-ft

At midspan of interior span: 83.7 kip-ft 55.0 kip-ft

5-23

5-9 Repeat problems 5-7 and 5-8 for column line 1.







Now, estimate the dead weight supported by column line 1:

Dead weight from beam, per foot of beam:

2 3

2

12 in 18 in lb lb150 225

in ft ft144 ft

Dead weight from slab, per foot of beam:

3

6 in 6.5 ft lb lb150 488

in ft ft12 ft


2

lb lb6.5 ft 20 130

ft ft



2 215 150.25 0.050 k/ft 0.25 0.052 k/ft

2 6 ft 30 ftr

LL T

L LK A

since Lr ≥ LL, no reduction is possible, so use LL = 0.050 k/ft2

At column B-1:

2 215 150.25 0.050 k/ft 0.25 0.042 k/ft

2 6 ft 55 ftr

LL T

L LK A


2 215 150.25 0.050 k/ft 0.25 0.056 k/ft

2 6 ft 25 ftr

LL T

L LK A

since Lr ≥ LL, no reduction is possible, so use LL = 0.050 k/ft2


At column A-1 and midspan of exterior beam: 2

lb lb6.5 ft 50 325

ft ft

At column B-1: 2

lb lb6.5 ft 42 273

ft ft


lb lb6.5 ft 50 325

ft ft

5-24




At column B-1:





:

At column A-1: 30 ft -16 in 28.67 ftn

At midspan of exterior span: 30 ft -16 in 28.67 ftn

At column B-1 (exterior): 30 ft 25 ft -16 in 26.17 ft2n

At column B-1 (interior): 30 ft 25 ft -16 in 26.17 ft2n

At midspan of interior span: 25 ft -16 in 23.67 ftn


At column A-1:

22 1.53 k/ft 28.67 ft78.6 k-ft

16 16

u nu

wM


22 1.53 k/ft 28.67 ft89.8 k-ft

14 14

u nu

wM

At column B-1 (exterior):

22 1.45 k/ft 26.17 ft99.3 k-ft

10 10

u nu

wM

At column B-1 (interior):

22 1.45 k/ft 26.17 ft90.3 k-ft

11 11

u nu

wM


22 1.53 k/ft 23.67 ft53.6 k-ft

16 16

u nu

wM

Now, assemble a model using structural analysis software. Use the following properties:

Column:

216 in 16 in 256 ingA

4 416 in /12 5,460 ingI


212 in 24 in 288 ingA

3 412 in 24 in /12 13,800 incrI

5-25




Location ACI Design Moment Design Moment

(From software)

At column A-1: -78.6 kip-ft -88.8 kip-ft

At midspan of exterior span: 89.8 kip-ft 60.9 kip-ft

At column B-1 (exterior): -99.3 kip-ft -103 kip-ft

At column B-1 (interior): -90.3 kip-ft -74.7 kip-ft


5-26

5-10 Repeat problems 5-7 and 5-8 for the beam m-n-o-p in Fig. P5-7. Be sure to comment

on the factored design moment at the face of the spandrel beam support at point m.







Now, estimate the dead weight supported by beam m-n-o-p:


2 3

2

12 in 18 in lb lb150 225

in ft ft144 ft


3

6 in 12 ft lb lb150 900

in ft ft12 ft


2

lb lb12 ft 20 240

ft ft


At m and midspan of exterior beam:

2 215 15

0.25 0.050 k/ft 0.25 0.040 k/ft2 12 ft 30 ft

r

LL T

L LK A

At n:

2 215 15

0.25 0.050 k/ft 0.25 0.033 k/ft2 12 ft 55 ft

r

LL T

L LK A


2 215 15

0.25 0.050 k/ft 0.25 0.043 k/ft2 12 ft 25 ft

r

LL T

L LK A


At m and midspan of exterior beam: 2

lb lb12 ft 40 480

ft ft

At n: 2

lb lb12 ft 33 396

ft ft


lb lb12 ft 43 516

ft ft

5-27


At m and midspan of exterior beam:


At n:





:

At m: 30 ft -12 in 29 ftn

At midspan of exterior span: 30 ft -12 in 29 ftn

At n (exterior): 30 ft 25 ft -12 in 26.5 ft2n

At n (interior): 30 ft 25 ft -12 in 26.5 ft2n

At midspan of interior span: 25 ft -12 in 24 ftn


At m:

22 2.41 k/ft 29 ft84.5 k-ft

24 24

u nu

wM


22 2.41 k/ft 29 ft145 k-ft

14 14

u nu

wM

At n (exterior):

22 2.27 k/ft 26.5 ft159 k-ft

10 10

u nu

wM

At n (interior):

22 2.27 k/ft 26.5 ft145 k-ft

11 11

u nu

wM


22 2.46 k/ft 24 ft88.6 k-ft

16 16

u nu

wM

Now, assemble a model using structural analysis software. Use the following properties:


212 in 24 in 288 ingA

3 412 in 24 in /12 13,800 ingI

Here we follow the recommendations from chapter 5 of the text, and assume that the beam is

pinned at m and supported by rollers (which are free to rotate) at n, o, and p. This neglects the

relatively small amount of moment transferred into the supporting beams due to their torsional

rigidity.

5-28

Now apply the appropriate load combinations following Example 5-2, and the following design

moments result:

Location ACI Design Moment Design Moment

(From software)

At m: -84.5 kip-ft -14.6 kip-ft

At midspan of exterior span: 145 kip-ft 184 kip-ft

At n (exterior): -159 kip-ft -195 kip-ft

At n (interior): -145 kip-ft -200 kip-ft


The software model assumes that there is no torsional rigidity supplied by the supporting beams.

Therefore, the moment that is predicted by the software at m is only due to the fact that the beam is

offset from the centerline of the supporting spandrel beam. While neglecting the torsional rigidity

of the spandrel beams is not realistic, it is also unlikely that the spandrel beam is torsionally rigid

enough to result in a moment as high as the ACI Design Moments.

5-29

5-11 Repeat problems 5-7 and 5-8 for the one-way slab strip shown in Fig. P5-7. For this

problem, find the factored design moments at all the points, a through i, indicated in

Fig. P5-7.







Now, estimate the dead load supported by slab strip a-i:

Dead load from slab, per foot of slab:

2 3

2

6 in 12 in lb lb150 75

in ft ft144 ft

Superimposed dead load, per foot of slab:

2

lb lb1 ft 20 20

ft ft

Live load, per foot of slab:

2

lb lb1 ft 50 50

ft ft

This live load cannot be reduced due to the very small influence area of the slab strip.


1.2 0.075 k/ft 0.020 k/ft 1.6 0.050 k/ft 0.194 k/ftuw

Calculate the clear span lengths,n

:

12 ft -12 in 11 ftn

11 ft -12 in 10 ftn

Also, assemble a model using structural analysis software. Use the following cracked properties:

Beam:

212 in 6 in 72 ingA

3 40.5 0.5 12 in 6 in /12 108 incr gI I

Here, we follow the recommendations from chapter 5 of the text, and assume that the slab strip is

pinned at a, and supported by rollers (which are free to rotate) at c, e, g, and i. Note that although i

is a point of geometrical symmetry, it cannot be modeled as fixed, since the pattern loads are not

necessarily symmetrical. Also note that pinning these supports neglects the relatively small

amount of moment transferred into the supporting beams due to their torsional rigidity.

Once the model is constructed, apply the appropriate load combinations following Example 5-2.

5-30

Design Moments using ACI Moment Coefficients from section 8.3.3, compared to design moments

resulting from software model:

Location ACI Moment

Coefficient

ACI Design Moment Design Moment

(From Software)

a -1/24 -0.978 kip-ft 0.423 kip-ft

b 1/14 1.68 kip-ft 2.14 kip-ft

c -1/11 -2.13 kip-ft -2.86 kip-ft

d 1/16 1.47 kip-ft 1.29kip-ft

e -1/11 -1.76 kip-ft -1.99 kip-ft

f 1/16 1.21 kip-ft 1.29 kip-ft

g -1/11 -1.76 kip-ft -1.99 kip-ft

h 1/16 1.21 kip-ft 1.24 kip-ft

i -1/11 -1.76 kip-ft -1.94 kip-ft

5-31

5-12 Use structural analysis software to find the maximum factored moments for the

girder on column line C. Find the maximum factored positive moments at o and y,

and the maximum factored negative moments at columns C-1, C-2, and C-3.

Estimate the loads supported by the beam along column line C:

Distributed dead load from beam, per foot of beam:

2 3

2

12 in 24 in lb lb150 300

in ft ft144 ft

The remaining loads are transferred to the girder as a point load at mid-span. Loads transferring

from the adjacent interior span will transfer directly to the girder, whereas loads from the adjacent

exterior span must be amplified by 15% according to the ACI shear coefficient.

Dead load from beams and slab, applied at the mid-point of the girder between C-1 and C-2:

2 3

2

12 in 18 in 25 ft 6 in 6 in 12 ft 25 ft 6 in lb150 13.5 k

in in inin 2 2 ft12 12 12 144 ft ft ftft

2 3

2

12 in 18 in 30 ft 6 in 6 in 12 ft 30 ft 6 in lb1.15 ×150 18.8 k


18.8 k 13.5 k 32.3 k

Dead load from beams and slab, applied at the mid-point of the girder at the interior spans:

2 3

2

12 in 18 in 25 ft 6 in 6 in 11 ft 25 ft 6 in lb150 12.6 k


2 3

2

12 in 18 in 30 ft 6 in 6 in 11 ft 30 ft 6 in lb1.15 150 17.5 k


12.6 k 17.5 k 30.1 k

Superimposed dead load, applied at the mid-point of the girder at the exterior spans:

2

25 ft lb12 ft 20 3.0 k

2 ft

3

30 ft lb1.15 12 ft 20 4.14 k

2 ft

3.00 k 4.14 k 7.14 k

Superimposed dead load, applied at the mid-point of the girder at the interior spans:

2

25 ft lb11 ft 20 2.75 k

2 ft

3

30 ft lb1.15 11 ft 20 3.80 k

2 ft

5-32

2.75 k 3.80 k 6.55 k

Live load, applied as a point load:

For negative moment at C-1 and positive moment at o:

2 215 150.25 50 lb/ft 0.25 33 lb/ft

24 ft 55 ftr

I

L LA

2 25 ft 30 ft33 lb/ft 12 ft 1.15 12 ft 11.8 k

2 2oLL

For negative moment at C-2:

2 215 150.25 50 lb/ft 0.25 27.4 lb/ft

46 ft 55 ftr

I

L LA

2 25 ft 30 ft27.4 lb/ft 12 ft 1.15 12 ft 9.78 k

2 2oLL

2 25 ft 30 ft27.4 lb/ft 11 ft 1.15 11 ft 8.97 k

2 2yLL

For positive moment at y:

2 215 150.25 50 lb/ft 0.25 34 lb/ft

22 ft 55 ftr

I

L LA

2 25 ft 30 ft34 lb/ft 11 ft 1.15 11 ft 11.1 k

2 2yLL

For negative moment at C-3:

2 215 150.25 50 lb/ft 0.25 27.7 lb/ft

44 ft 55 ftr

I

L LA

2 25 ft 30 ft27.7 lb/ft 11 ft 1.15 11 ft 9.06 k

2 2yLL


:

Between C-1 and C-2: 24 ft -16 in 22.67 ftn

Between C-2 and C-3: 22 ft -16 in 20.67 ftn

Now assemble a model using structural analysis software. Use the following cracked properties:

Column:

216 in 16 in 256 ingA

4 416 in /12 5,460 ingI


212 in 24 in 288 ingA

3 412 in 24 in /12 13,800 incrI

5-33




Location Design Moment

(From Software)

Negative at C-1 -121 kip-ft

Positive at o 236 kip-ft

Negative at C-2 (o-side) -224 kip-ft

Negative at C-2 (y-side) -198 kip-ft

Positive at y 175 kip-ft

Negative at C-3 -164 kip-ft

5-34

5-13 Assume the maximum factored positive moment near midspan of the floor beam

between columns A-2 and B-2 is 60 kip-ft. Using the beam dimensions given in

Fig. P5-7, determine the required area of tension reinforcement to satisfy all the ACI

Code requirements for strength and minimum reinforcement area. Select bars and

provide a sketch of your final section design.

The maximum expected positive moment is given as 60 k-ft 720 k-inuM

Determine the effective flange width:

By ACI code section 8.12:

30 ft / 4 7.5 ft 90 infb

Check that:

2 8 6 in 12 in 108 in

11 ft -12 in2 12 in 132 in

2

fb

OK

Determine As and select the reinforcing bars:

Assume 0.95jd d :

From Eq. (5-16):

2720 k-in

0.65 in0.9 60 ksi 0.95 21.5 in

2

us

y

MA

af d


From Eq. (5-17): 20.65 in 60 ksi

0.13 in0.85 ' 0.85 4 ksi 90 in

s y

c

A fa

f b

From Eq. (5-16):

2720 k-in 0.62 in

0.13 in0.9 60 ksi 21.5 in -

2 2

us

y

MA

af d

This seems like a very small area of required steel, so check minimum steel requirement before

selecting bars.

From ACI-08 Eq. (10-3):

For f’c = 4,000 psi, 3 ' 189 200cf so use 200

2

,min

3 ' 200 12 in 21.5 in 0.86 in

60,000 psi

c

s w

y

fA b d

f

The minimum steel requirement will govern here. Select 3 #5 bars. 2 2 2 3 3 0.31 in 0.93 in 0.86 ins bA A OK

5-35

Required Checks:




required. Since this requirement governed our bar selection, this is satisfied by default. OK


From Eq. (5-17): 20.93 in 60 ksi

0.182 in0.85 ' 0.85 4 ksi 90 in

s y

c

A fa

f b

We know that1 0.85 for f’c = 4,000 psi

1

0.182 in0.214 in

0.85

ac

From Eq. (4-18):

- 21.5 in - 0.214 in0.003 0.297 0.005

0.214 int cu

d c

c OK



From Eq. (5-15):

2 0.182 in 0.9 0.93 in 60 ksi 21.5 in 1075 k-in

2 2n s y

aM A f d

89.6 k-ft 60 k-ftn uM M

Therefore, this design is sufficient. It seems too conservative, but that is because the minimum

area requirement governed bar selection.

Fig. S5-13 Cross-section of final design for positive bending region

Note that other selections As may also be correct. If all checks are satisfied, without being

unreasonably conservative, the design may be considered adequate.

9.0

9.0

24in.

90in.

3 #5 bars2.5in.

5-36

5-14 Assume the maximum factored negative moment at the face of column B-2 for the

floor beam along column line 2 is -120 kip-ft. Using the beam and slab dimensions

given in Fig. P5-7, determine the required area of tension reinforcement to satisfy all

the ACI Code requirements for strength and minimum reinforcement area. Select

bars and provide a sketch of your final section design.

The maximum expected negative moment is given as 120 k-ft 1440 k-inuM

Although this is a T-section, there is no need to determine the effective flange width since the

compression zone is located in the web of the beam.


Assume 0.9jd d :

From Eq. (5-16):

21440 k-in

1.38 in0.9 60 ksi 0.9 21.5 in

2

us

y

MA

af d


From Eq. (5-17): 21.38 in 60 ksi

2.03 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

21440 k-in 1.30 in

2.03 in0.9 60 ksi 21.5 in -

2 2

us

y

MA

af d

Select 3 #5 bars over the web, and 4 #3 bars in the flange. This will result in “part” of the

reinforcement (about 1/3) being spread into the flange, and most (2/3) remaining over the web. 2 2 2 2

1 2 3 4 3 0.31 in 4 0.11 in 1.37 in 1.30 ins b bA A A OK

Required Checks:




required.

From ACI-08 Eq. (10-3):

For f’c = 4,000 psi, 3 ' 190 200cf so use 200 psi

2 2

,min

3 ' 200 psi 12 in 21.5 in 0.86 in 1.37 in

60,000 psi

c

s w

y

fA b d

f OK

5-37


From Eq. (5-17): 21.37 in 60 ksi

2.01 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b


1

2.01 in2.36 in

0.85

ac

From Eq. (4-18):

- 21.5 in - 2.36 in0.003 0.024 0.005

2.36 int cu

d c

c OK



From Eq. (5-15):

2 2.01 in 0.9 1.37 in 60 ksi 21.5 in 1515 k-in

2 2n s y

aM A f d


Therefore, this design is sufficient.

Fig. S5-14 Cross-section of final design for negative bending region

Note that other selections of As may also be correct. If all checks are satisfied, without being

unreasonably conservative, the design may be considered adequate. Also note that all flange

reinforcement that is considered to contribute to the negative bending capacity of this section is

placed within two flange depths of the web.

9.0

9.0

24in.

90in.

12in.

3 #5 bars and 4 #3 bars

5-38

5-15 Assume the maximum factored negative moment at support n of the floor beam

m-n-o-p is -150 kip-ft. Using the design procedure for singly reinforced beam sections

given in Section 5-3 (design of beams when section dimensions are not known),

determine the beam dimensions and select the required area of tension reinforcement

to satisfy all the ACI Code requirements for strength and minimum reinforcement

area. Select bars and provide a sketch of your final section design.






From Eq. (5-19):

1 ' 0.85 4,000 psi0.0142

4 4 60,000 psi

c

y

f

f

Note: β1 =0.85 for f’c = 4,000 psi

From Eq. (5-21):

0.0142 60,000 psi0.2125

' 4,000 psi

y

c

f

f

From Eq. (5-22):

' (1 0.59 ) 0.2125 4 ksi (1 0.59 0.2125) 0.743 ksicR f



From Eq. (5-23):

2 31800 k-in2690 in

0.9 0.743 ksi

uMbd

R



3 31800 k-in17.6 in 17.5 in

0.5 0.9 0.734 ksi

uMd

R

2.5 in 20 inh d

0.5 17.5 in 8.75 in 10 inb d

Note that both h and b are rounded to the nearest even inch value for constructability.


Assume 0.9jd d : From Eq. (5-16):

21800 k-in

2.12 in0.9 60 ksi 0.9 17.5 in

2

us

y

MA

af d

9.0

5-39


From Eq. (5-17): 22.12 in 60 ksi

3.74 in0.85 ' 0.85 4 ksi 10 in

s y

c

A fa

f b

From Eq. (5-16):

21800 k-in 2.13 in

3.74 in0.9 60 ksi 17.5 in -

2 2

us

y

MA

af d

Since I am not satisfied that any of the possible bar combinations that fulfill this requirement are

not too conservative, I will choose to widen the beam slightly to help get a more efficient design.

Select b = 12 in.

Now re-determine As and select the reinforcing bars:

Assume 0.9jd d :

From Eq. (5-16):

21800 k-in

2.12 in0.9 60 ksi 0.9 17.5 in

2

us

y

MA

af d


From Eq. (5-17): 22.12 in 60 ksi

3.12 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

21800 k-in 2.09 in

3.12 in0.9 60 ksi 17.5 in -

2 2

us

y

MA

af d




Required Checks:




required.

From ACI-08 Eq. (10-3):


2 2

,min

3 ' 200 psi 12 in 17.5 in 0.70 in 2.12 in

60,000 psi

c

s w

y

fA b d

f OK

5-40


From Eq. (5-17): 22.12 in 60 ksi

3.12 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b


1

3.12 in3.67 in

0.85

ac

From Eq. (4-18):

- 17.5 in -3.67 in0.003 0.011 0.005

3.67 int cu

d c

c OK



From Eq. (5-15):

2 3.12 in 0.9 2.12 in 60 ksi 17.5 in 1820 k-in

2 2n s y

aM A f d




Note that other selections As may also be correct. If all checks are satisfied, without being

unreasonably conservative, the design may be considered adequate. Also note that all flange



9.0

9.0

20in.

90in.

12in.


12in.

6in.

5-41

5-16 Assume the maximum factored negative moment at the face of column C-2 for the

girder along column line C is -250 kip-ft. Using the design procedure given in

section 5-4 for the design of doubly reinforced sections, determine the beam

dimensions and select the required areas of tension and compression

reinforcement to satisfy all the ACI Code requirements for strength and minimum

reinforcement area. Select bars and provide a sketch of your final section design.




Select ρ and the corresponding R-factor:


From Eq. (5-25):

10.36 ' 0.36 0.85 4,000 psi0.0204

60,000 psi

c

y

f

f

Note: β1 =0.85 for f’c = 4,000 psi

From Eq. (5-21):

0.0204 60,000 psi0.306

' 4,000 psi

y

c

f

f

From Eq. (5-22):

' (1 0.59 ) 0.306 4 ksi (1 0.59 0.306) 1.0 ksicR f


Select section dimensions, b and h:

From Eq. (5-23):

2 33000 k-in3330 in

0.9 1.0 ksi

uMbd

R



3 33000 k-in17.2 in 17.5 in

0.65 0.9 1.0 ksi

uMd

R

2.5 in 20 inh d

0.65 17.5 in 11.4 in 12 inb d

Note that both h and b are rounded to the nearest even inch value for constructability.


Assume 0.9jd d :

From Eq. (5-16):

23000 k-in

3.53 in0.9 60 ksi 0.9 17.5 in

2

us

y

MA

af d

9.0

5-42


From Eq. (5-17): 23.53 in 60 ksi

5.2 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

23000 k-in3.73 in

5.2 in0.9 60 ksi 17.5 in -

2 2

us

y

MA

af d




Also, select compression reinforcement such that the area of compression reinforcement (A’s) is

greater than one half of the tension steel. Here we select 3 #8 bars so that A’s = 2.37 in2.

Required Checks:




required.

From ACI-08 Eq. (10-3):


2 2

,min

3 ' 200 12 in 17.5 in 0.70 in 3.80 in

60,000 psi

c

s w

y

fA b d

f OK


First, an iterative procedure must be used to determine the depth of the neutral axis. Following the

procedure shown in Chapter 4, c = 4.2 in.

From Eq. (4-18):

- 17.5 in - 4.2 in0.003 0.010 0.005

4.2 int cu

d c

c OK



From Eq. (5-15):

As a result of the iterations used to determine c, the following forces were determined: 2 3.8 in 60 ksi 228 ks yT A f

1 0.85 ' 0.85 4.2 in 0.85 4 ksi 12 in 145 kc c wC c f b

2' 4.2 in 2.5 in' 0.003 0.003 29,000 ksi 2.37 in 83 k

4.2 ins s s

c dC E A

c

9.0

9.0

5-43

So, the nominal moment capacity of the section is:

' '2

n c s

aM C d C d d

0.85 4.2 in

0.9 145 kip 17.5 in 83 k 17.5 in 2.5 in2

nM

3170 k-in 264 k-ft 250 k-ftn uM M



Note that other selections of b, h, and As may also be correct. If all checks are satisfied, without

being unreasonably conservative, the design may be considered adequate. Also note that all flange



20in.

90in.

3 #8 bars2.5in.

12in.


12in.

5-44

5-17 For the one-was slab shown in Fig. P5-7, assume the maximum negative moment at

support c is -3.3 kip-ft/ft, and the maximum factored positive moment at midspan

point b is 2.4 kip-ft/ft.

(a) Using the given slab thickness of 6 in, determine the required reinforcement

size and spacing at both of these locations to satisfy ACI Code flexural

strength requirements. Be sure to check the ACI Code requirements for

minimum flexural reinforcement in slabs.

Negative Moment 3.3 k-ft 39.6 k-inuM :

Assume 0.75 in of cover will be provided. This results in a d = 5 in.

From Eq. (5-16):

239.6 k-in

ft ft 0.16 inftft 0.9 60 ksi 0.9 5 in

2

u

s

y

MA

af d


From Eq. (5-17): 20.16 in 60 ksi

0.24 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

239.6 k-in

ft ft in0.15 ft0.24 inft

0.9 60 ksi 5 in -2 2

u

s

y

MA

af d

Minimum reinforcement:

2,min in0.0018 0.0018 12 in 6 in 0.13

ft ftsA

bh

Strength requirements govern here.

Maximum reinforcement spacing is limited to 3h or 18 in, which is the same value for this 6 in deep

slab. Also, we must check reinforcement spacing for crack control. Since fy and cc are the same

here as in Example 5-7, the maximum spacing for crack control is 12 in. This governs.

If we select #3 bars at 8 in, the result is:

2 22 12 in in in0.11 in 0.165 0.15 ft ft ft8 in

sA OK

Check that strength is satisfied:

2 0.24 in 0.9 0.165 in 60 ksi 5 in 43.5 k-in

2 2n s y

aM A f d

k-ft k-ft 3.62 3.3 ft ftn uM M OK

5-45

Remember to calculate the strain in the extreme layer of tension steel to verify that assuming

is valid.

- 5 in - 0.24 / 0.85 in0.003 0.050 0.005

0.24 / 0.85 int cu

d c

c OK


Positive Moment 2.4 k-ft 28.8 k-inuM :

Assume 0.75 in of cover will be provided. This results in a d = 5 in.

From Eq. (5-16):

228.8 k-in

ft ft 0.12 inftft 0.9 60 ksi 0.9 5 in

2

u

s

y

MA

af d


From Eq. (5-17): 20.12 in 60 ksi

0.17 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

228.8 k-in

ft ft in0.11 ft0.17 inft

0.9 60 ksi 5 in -2 2

u

s

y

MA

af d

Minimum reinforcement:

2,min in0.0018 0.0018 12 in 6 in 0.13

ft ftsA

bh

Minimum requirements govern here.

Maximum reinforcement spacing is limited to 3 h or 18 in, which is the same value for this 6 in

deep slab. Also, we must check reinforcement spacing for crack control. Since fy and cc are the

same here as in Example 5-7, the maximum spacing for crack control is 12 in. This governs.

If we select #3 bars at 10 in, the result is:

2 22 12 in in in0.11 in 0.132 0.13 ft ft ft10 in

sA OK

Check that strength is satisfied:

2 0.17 in 0.9 0.132 in 60 ksi 5 in 35.0 k-in

2 2n s y

aM A f d

k-ft k-ft 2.9 2.4 ft ftn uM M OK

9.0

9.0

5-46

Remember to calculate the strain in the extreme layer of tension steel to verify that assuming

is valid.

- 5 in - 0.17 / 0.85 in0.003 0.072 0.005

0.17 / 0.85 int cu

d c

c OK


(b) At both locations, determine the required bar size and spacing to be provided

in the transverse direction to satisfy ACI Code section 7.12.2 requirements for

minimum shrinkage and temperature reinforcement.

Since the positive flexural region was controlled by temperature and shrinkage reinforcement, the

reinforcement specified there would suffice in the transverse direction at all locations.

So, use #3 bars at 10 in. Placement near the top or bottom of the slab makes no difference here, so

specify that bars are to be placed wherever is easiest.

(c) For both locations provide a sketch of the final design of the slab section.

Fig. S5-17 Cross-section of final design for negative and positive bending regions, respectively

Note that other selections of As may also be correct. If all checks are satisfied, without being

unreasonably conservative, the design may be considered adequate.

9.0

9.0

12in.

6in.

#3@8in.

12in.

6in.

#3@10in.

6-1

Chapter 6

6-1 For the rectangular beam shown in Fig P6-1:

(a) Draw a shear force diagram.

7.3

V, kips 0.2

-9.8

16.0

-17.7

10

(b) Assuming the beam is uncracked, show the direction of the principal tensile

stresses at middepth at points A, B, and C.

A B C

(c) On a drawing of the beam sketch the inclined cracks that would develop at

A, B, and C.

A B C

6-2

6-2, 6-3, 6-4 and 6-5 Compute for the cross sections shown in Figs. P6-2, P6-3, P6-4

and P6-5. In each case use and

6-2 ( )

√ √ ⁄

⁄ [( ) ⁄ ] ⁄

( )

6-3 ( )

√ √ ⁄

⁄ [( ) ⁄ ] ⁄

( )

6-4 ( )

√ √ ⁄

⁄ [( ) ⁄ ] ⁄

( )

6-5 ( )

√ √ ( ) ⁄

⁄ [( ) ⁄ ] ⁄

( )

6-6 ACI Sec. 11.4.5.1 sets the maximum spacing of vertical stirrups at ⁄ . Explain

why.

Every inclined crack must be crossed by at least one stirrup. The assumed horizontal

projection of the crack is d so every crack will be crossed by at least one stirrup.

6-3

6-7 Figure P6-7 shows a simply supported beam. The beam has No. 3 Grade 40 double-

leg stirrups with and 4 No.8 Grade 60 longitudinal bars with

. The plastic truss model for the beam is shown in the figure.

Assuming that the stirrups are all loaded to .

(a) Use the method of joints to compute the forces in each panel of the

compression and tension chords and plot them. The force in member

is ⁄ .

Assume all stirrups yield. 8.8s ytA f kips

Joint 11L : Vertical force in stirrup = 8.8 kips

0V at Joint 11L gives vertical component in strut 12 11U L as 8.8

kips downward (compression in 12 11U L ).

Horizontal component in 12 11

12 in.8.8

20 in.U L kips = 5.28 kips acting

to the left on Joint 11L .

Joint 12U : Each inclined strut has a vertical component of 8.8 kips, therefore 6

struts are needed to equilibrate 52.8 kips force at Joint 12U as shown.

Joint 8U : Total downward load is 26.4 kips applied load plus 8.8 kips in stirrup

8 8U L , therefore 4 struts are needed to equilibrate 35.2 kips force at

8U as shown.

Forces in Inclined Struts in Web

Member

Vertical

Component

(kips)

Horizontal

Projection

(in.)

Horizontal

Component

(kips)

12 11U L 8.8 12 5.28

12 10U L 8.8 24 10.56

12 9U L 8.8 36 15.84

11 8U L 8.8 36 15.84

10 7U L 8.8 30 13.20

9 6U L 8.8 24 10.56

8 5U L 8.8 18 7.92

8 4U L 8.8 24 10.56

8 3U L 8.8 30 13.20

8 2U L 8.8 36 15.84

6-4

7 1U L 8.8 36 15.84

Member (cont’d)

Vertical

Component

(kips)

Horizontal

Projection

(in.)

Horizontal

Component

(kips)

6 0U L 8.8 36 15.84

5 0U L 8.8 30 13.20

4 0U L 8.8 24 10.56

3 0U L 8.8 18 7.92

2 0U L 8.8 12 5.28

1 0U L 8.8 6 2.64

Compute Forces in Lower Chord

Maximum moment = 52.8 8 26.4 4 316.8 ft-kips

Force in 11 13

316.8 12190.0

20

ML L

jd

kips (in tension)

H at Joint 11L gives force in 10 11 190.0 5.28 184.7L L kips

Force in Lower Chords

Member Joint

Horizontal

Force in Strut

(kips)

Lower Chord

Force

(kips)

11 13L L - - 190.0 tension

10 11L L 11L 5.28 184.7

9 10L L 10L 10.56 174.2

8 9L L 9L 15.84 158.3

7 8L L 8L 15.84 142.5

6 7L L 7L 13.20 129.3

5 6L L 6L 10.56 118.7

4 5L L 5L 7.92 110.8

3 4L L 4L 10.56 100.2

2 3L L 3L 13.20 87.0

1 2L L 2L 15.84 71.2

0 1L L 1L 15.84 55.4

0L 15.84

+13.20

6-5

+10.56

+7.92

Member (cont’d) Joint

Horizontal

Force in Strut

(kips)

Lower Chord

Force

(kips)

+5.28

+2.64

Total: 55.4

The force in 0 1L L is 55.4 kips. This must be anchored in Joint 0L . The sum of the

horizontal forces in the struts at Joint 0L is 55.4 kips and 0H at Joint 0L .

Forces in Upper Chords:

Joint 1U : There is no force in the compression chord to the right of 1U .

Member Joint

Horizontal Force

in Struts

(kips)

Lower Chord

Force

(kips)

1 2U U 1U 2.64 2.64

compression

2 3U U 2U 5.28 7.92

3 4U U 3U 7.92 15.84

4 5U U 4U 10.56 26.4

5 6U U 5U 13.20 39.8

6 7U U 6U 15.84 55.4

7 8U U 7U 15.84 71.3

8 9U U 8U 15.84

+13.20

+10.56

+7.92 118.8

9 10U U 9U 10.56 129.4

10 11U U 10U 13.20 142.6

11 12U U 11U 15.84 158.4

Compression at Midspan 12U 15.84

+10.56

6-6

+5.28 190.0

Plot of Forces in Upper Chords:

Plot of Forces in Lower Chords:

6-7

(b) Plot ⁄ on the diagram from part (a) and compare the bar forces

from the truss model to those computed from ⁄ .

The bar forces in the lower chords provided by /s sA f M jd appear to be less

than the values provided by the truss model throughout the span except at the

midspan where the two values match.

(c) Compute the compression stress in the diagonal member (see Eq. (6-

11)). The beam width, , is 12 in.

Slope of 1 7L L : tan 20 / 36 0.556 29.05

1 52800 1tan 0.556 518

tan 12 20 0.556cd

w

Vf

b jd

psi

6-8

6-8 The beam shown in Fig. P6-8 supports the unfactored loads shown. The dead load

includes the weight of the beam.

(a) Draw shearing force diagram for

1.2 1.4 1.68Duw k/ft

1.6 1.5 2.4Luw k/ft

(1) factored dead and live load on the entire beam.

At the left support: 1.68 2.4 25

51.0 kips2

uV

51

.0

- 5

1.0

Case (1)

, kipsuV

(2) factored dead load on the entire beam plus factored live load on the

left haft-span.

At the right support:

2 225 12.51.68 2.4

2 2 28.5 kips25

uV

At the right support: 1.68 25 2.4 12.5 28.5 43.5 kipsuV

At the midspan: 43.5 1.68 2.4 12.5 7.5 kipsuV

-28

.5

-7.5

Case (2)

, kipsuV

(3) factored dead load on the entire beam plus factored live load on the

right haft-span.

Due to the asymmetry of loadings Case (3) to loadings Case (2), the shear

diagram of Case (3) is asymmetric to that of Case (2).

6-9

28

.5

- 4

3.5

7.5

Case (3)

, kipsuV

(b) Superimpose the diagram to get a shear force envelope. Compare the shear

at midspan to that from Eq. 6-26.

Eq. 6-26: 2.4 25

(midspan) 7.58 8

Luu

wV

kips

51.0

-7.5

7.5

- 51.0

, kipsuV

(c) Design stirrups. Use and No.3 double-leg stirrups with

.

68

.0

- 1

0.0

10

.0

- 6

8.0

, kipsuV

0.75

Are stirrups required?

2 4500 12 17.5 /1000 28.2 68.0uc

VV

kips

Stirrups are required.

Check stirrups anchorage

Use No. 3 Grade 40 stirrups. ACI code Sec. 12.13.2.1. allows these to be anchored by a

hook around a top bar.

Maximum spacing

d/2 = 8.75 in.

uV

6 82.1c wf b d kips Maximum spacing of / 4d is not required.

0.22 4000014.6

50.3 120.75

v y

c w

A f

f b

in. (note 0.75 50.3 psi 50 psi, use 50.3 psicf )

6-10

Maximum spacing max 8.5s in.

Compute spacing to resist shear forces

At d from supports:

68 10 17.568 61.2

12.5 12

uV

kips

max

0.22 40 17.54.5 in. 8.5 in.

/ 61.4 27.4

v yt

u c

A f ds s

V V

Use s = 4.0 in.

Change this to 6 in. where 0.22 40 17.5

27.4 53.16

uV

kips.

This occurs at 68 53.1

12.5 12 38.568 10

x

in. from end.

Change this to 8 in. where 0.22 40 17.5

27.4 46.78

uV

kips.


12.5 12 55.168 10

x

in. from end.

Terminate stirrups where 28.2

14.12 2

u cV V

kips.


12.5 1268 10

x

= 139 in.

Compute the number of stirrups

From the center of the support, use 1 @ 2 in.

Required number of stirrups at the spacing of 4 in.:38.5 2

9.14

, use 10 @ 4 in.

Required no. of stirrups at the spacing of 6 in.: 55.1 2 10 4

2.26

, use 3 @ 6 in.

Required number of stirrups at the spacing of 8 in.: 139 2 10 4 3 6

9.98

, use

10 @ 8 in.

Provide No.3 U stirrups Grade 40 steel. Starting from the center of the support,

use 1 @ 2 in., 10 @ 4 in., 3 @ 6 in., and 10 @ 8 in. from each end.

6-11

(Note that 0.22 40 17.5

(max) 38.5 kips < 8 113 kips4

s c wV f b d

, OK)

6-9 The beam shown in Fig. P6-9 supports the unfactored loads shown in the figure. The

dead load includes the weight of the beam.

(a) Draw shearing force diagrams for

(1) factored dead and live load on the entire length of beam.

(2) factored dead load on the entire beam plus factored live load

between B and C.

(3) factored dead load on the entire beam plus factored live load

between A and B and between C and D.

Loadings (2) and (3) will give the maximum positive and negative shears at B.

2.4Duw k/ft

2.4Luw k/ft

4.8uw k/ft

6-12

33.0

V, kips

-5.4

28.8

-43.8

21.3

2.1

14.4

-36.3

28.2

-10.2

28.8

-29.4

(a1)

(a2)

(a3)

(b) Draw the factored shear force envelope. The shear at B should be the

factored dead load shear plus or minus the shear from Eq. 6-26.

Eq. 6-26: 2.4 16

(midspan) 4.88

uV

kips

6-13

33

.0

-10

.2

28

.8

-43

.8

2.1

33.0

-4.8-2.7 = -7.5

28.8

-43

.8

4.8-2.7 = 2.1

Envelope from Part (a)

Envelope from Eq. 6-26

Eq. 6-26 is correct for s simple beam without overhangs. It is an approximation for all

other cases. However, we shall assume it is close enough and use it for the rest of this

example.

(c) Design stirrups. Use and .


√ ⁄


Check anchorage

Use No. 3 stirrups. These can be anchored by a hook around a top bar.

Maximum spacing

d/2 = 10.8 in.

√

Maximum spacing of / 4d is not required.

0.22 4000014.7

50 50 12

v y

w

A f

b

in. (note √

)


Compute spacing to resist shear forces – Part AB

6-14

At d from support A:

44 2.8 21.544 34.8

8 12

uV

kips

max

0.22 40 21.586 in. 10.8 in.

34.8 32.6s s

Use s = 10 in.

Terminate stirrups where:

This occurs at:

Part A-B: Provide No.3 stirrups Grade 40 steel. Starting from A use 1 @ 5 in., 7

@ 10 in.

Compute spacing to resist shear forces – Part BC

At d from support C

58.4 10 21.5

58.4 47.68 12

uV

kips

max

0.22 40 21.512.6 in. 10.8 in.

47.6 32.6s s

Use s = 10 in.

Terminate stirrups where:

This occurs at:

Part BC: Provide No.3 stirrups Grade 40 steel. Starting from C use 1 @ 5 in., 9

@ 10 in.

Compute spacing to resist shear forces – Part CD

At d from support C

6-15

Can terminate stirrups where 2

u cV V

. However, place minimum shear reinforcement

throughout part CD.

Part CD: Use No. 3 U stirrups, Grade 40. Starting from C use 1 @ 5 in., 7 @ 10

in.

6-16

6-10 Fig. P6-10 shows an interior span of a continuous beam. The shears at the ends are

⁄ . The shear at midspan is from Eq. (6-26).

(a) Draw a shear force envelope.

1.2 1.5 1.8Duw k/ft

1.6 1.8 2.88Luw k/ft

4.68uw k/ft

End shears = 4.68 22 / 2 51.5 kips

Midspan shears = 2.88 22/8 7.92 kips

68

.7

- 1

0.6

10

.6

- 6

8.7

, kipsuV

0.75

(b) Design stirrups using and .


2 4000 12 17.5 /1000 26.6cV kips


Check anchorage

Use No. 3 stirrups. These can be anchored by a bend around a top bar.

Maximum spacing

d/2 = 8.8 in.

uV

6 79.8c wf b d kip Maximum spacing of / 4d is not required.

0.22 40000

14.750 50 12

v y

w

A f

b

in. (note 0.75 47.4 psi < 50 psi, use 50 psicf )


Computing spacing to resist shear forces

At d from face of column:

68.7 10.6 17.568.7 61.0

11 12

uV

kips

0.22 40 17.5

4.561.0 26.6

s

in.

6-17

Use s = 4 in.

Changing stirrup spacing to 6 in. where 0.22 40 17.5

26.6 52.36

uV

kips

This occurs at 68.7 52.3

11 12 37.3 in.68.7 10.6

x

Changing stirrup spacing to 8 in. where 0.22 40 17.5

26.6 45.98

uV

kips


11 12 51.8 in.68.7 10.6

x

Terminate stirrups when 26.6

13.32 2

u cV V

kips


11 12 126 in.68.7 10.6

x

Use No.3 Grade 40 U stirrups. Starting from face of column at each end, use 1 @

2 in., 9@ 4 in., 3 @ 6in., and 9 @ 8in.

6-18

6-11 Design shear reinforcement for the C1-C2 span of the girder designed in Example 5-

6 (final section given in Fig. 5-32). From the structural analysis discussed in

Example 5-6, the factored design end shears for this girder are 28.9 kips at the face

of column C1 and 39.2 kips at the face of column C2. Use and

.


√ ⁄


Check anchorage

Select No. 3 stirrups. These can be anchored by a bend around a top bar.

Maximum spacing

⁄

√

Maximum spacing of ⁄ is not required.

( √

)

Maximum spacing

Compute the spacing of stirrups required to resist shear forces

It is desirable to have the most efficient design of stirrups possible, so stirrup size and

spacing should be adjusted over the span length whenever possible to account for the

changing shear demand. Likewise, it is advantageous to use the factored shear demand at

from the face of the column rather than directly at the face of the column to design

stirrups. However, for a girder loaded like this one, the majority of shear comes from the

point load at midspan. Therefore, it is reasonable to select only two different stirrup

spacing, one for each half of the girder, based on the factored shear demand at the face of

the columns.

At the exterior column face (C1):

⁄

The maximum spacing controls, so select s = 8 in.

At the interior column face (C2):

⁄

The strength requirement controls, so select s = 6 in.

Use No.3 Grade 40 U stirrups. Starting from face of C2 column, use s = 6 in. until

past the point load, then use s = 8 in.

6-19

6-12 Fig. P6-12 shows a rigid frame and the factored loads acting on the frame. The 7-kip

horizontal load can act from the left or the right. and

.

(a) Design stirrups in the beam

-27.2

-34.9

27.2

34.9

46.5

-3.85

3.85

44.1

38.0

d from face

of column

face of column

5.1

-5.1

CL

CL

Wind from right

Wind from leftCL

, kipsuV

, kipsuV


Assume d = 20 – 2.5 = 17.5 in.

√ ⁄


Maximum spacing

Use No. 3

d/2 = 8.8 in.

√

Maximum spacing of / 4d is not required.

√

( √

)

Maximum spacing maxs = 8.8 in.

6-20

Compute spacing required to resist shear

At d from face of column, 38uV

kip

⁄

Terminate stirrups when:

This occurs at:

Provide No. 3, Grade 40 U stirrups. Starting at face of column from each end, use

1 @ 4 in., 11@ 8 in.

(b) Are stirrups required in the columns? If so, design the stirrups for the

columns.

Maximum shear is 8uV kips in leeward column.

(

)√

( ( )

)√ ( )


Stirrups are not required. Use minimum amount of shear reinforcement.

7-1

Chapter 7

7-1 A cantilever beam 8 ft long and 18 in wide supports its own dead load plus a

concentrated load located 6 in from the end of the beam and 4.5 in away from the

vertical axis of the beam. The concentrated load is 15 kips dead load and 20 kips live

load. Design reinforcement for flexure, shear, and torsion. Use fy = 60,000 psi for all

steel and f’c = 3750 psi.

Design for moment:

Since wide sections are desirable for torsion, estimate that α = 0.8, so 18 in / 0.8 22.5 ind .

Therefore, the dead load due to the weight of the beam is:

3

2 2

(22.5 in 2.5 in) 18 in0.15 k/ft 0.469 k/ft

144 in /ftDL

So, the moment demand due to factored loads is:

2

0.469 k/ft 8 ft1.2 1.2 15 k 1.6 20 k 7.5 ft 393 k-ft 4720 k-in

2uM

Select ρ and corresponding R-factor, assuming the desirable strain diagram shown in Fig. 5-27b,

which leads to 9.0 .

1 ' 0.85 3,750 psi0.0133

4 4 60,000 psi

c

y

f

f

Note: β1 = 0.85 for f’c = 3,750 psi

0.0133 60,000 psi0.213

' 3,750 psi

y

c

f

f

' (1 0.59 ) 0.213 3.75 ksi (1 0.59 0.213) 0.698 ksicR f


Now select h with b = 18 in:

2 34720 k-in7500 in

0.9 0.698 ksi

uMbd

R

Since the width of the beam is given as 18 in, we can directly solve for a reasonable d value.

4720 k-in20.4 in 21.5 in

0.9 18 in 0.698 ksi

uMd

bR

2.5 in 24 inh d

18 inb


Finally, determine As required for resisting this applied moment by first going back and

recalculating the weight of the beam with the final selected dimensions:

2 2 3 2 2 3

k 18 in 24 in k k 0.15 0.15 0.45


bhDL

So 393 k-ft 4720 k-inuM

Calculate the required area of steel, assuming 0.9jd d :

24720 k-in

4.52 in0.9 60 ksi 0.9 21.5 in

2

us

y

MA

af d

7-2

With this estimate, iterate once to have a better estimate of the lever arm jd . 24.52 in 60 ksi

4.73 in0.85 ' 0.85 3.75 ksi 18 in

s y

c

A fa

f b

24720 k-in 4.57 in

4.73 in0.9 60 ksi 21.5 in -

2 2

us

y

MA

af d

No further iterations are necessary, since the estimated lever arm was very reasonable. Use

b = 18 in, h = 24 in, and As = 4.57 in2. Bars will be selected later.

Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum

required.

For f’c = 3,750 psi, 3 ' 184 200cf so use 200 psi

2 2

,min

200 200 psi 18 in 21.5 in 1.29 in 4.57 in

60,000 psis w

y

A b df

OK

Also, calculate the strain in the extreme layer of tension steel to verify that assuming 9.0 is

valid. 24.57 in 60 ksi

4.78 in0.85 ' 0.85 3.75 ksi 18 in

s y

c

A fa

f b

We know that 1 0.85 .

1

4.78 in5.62 in

0.85

ac

- 21.5 in - 5.62 in

0.003 0.0085 0.0055.62 in

t cu

d c

c OK

Therefore the designer is permitted to use 9.0 for this beam design.

Now design for torsion and shear:

At d = 21.5 in away from the face, the factored torsion is:

4.5 in

1.2 15 k 1.6 20 k 18.8 k-ft12 in/ft

uT

At d = 21.5 in away from the face, the factored shear is:

21.5

1.2 15 k 1.2 (8 ft ft) 0.45 k/ft 1.6 20 k 53.4 k12

uV

Should torsion be considered in this design?

218 in 24 in 432 incpA

(18 in 24 in) 2 84 incpP

From Eq. (7-18b):

222 432 in

' 0.75 1.0 3,750 psi 102,000 lb-in 8.5 k-ft84 in

cp

th c

cp

AT f

P

7-3

Since u thT T , torsion must be considered. Also, since we are dealing with a statically determinate

system, we have a case of equilibrium torsion, in which the full factored torsion must be sustained

by our beam. Therefore, the design torsion cannot be reduced.

Check whether the section dimensions are sufficient to withstand the combined stresses due to

shear and torsion.

From Eq. (7-33):

2 2

28 '

1.7

u u h cc

w oh w

V T P Vf

b d A b d

Assuming #4 stirrups,

2 24 in 2 1.5 in 2 0.25 in 18 in 2 1.5 in 2 0.25 in 70 inhP

224 in 2 1.5 in 2 0.25 in 18 in 2 1.5 in 2 0.25 in 297 inohA

22

22

2 3,750 psi 18 in 21.5 in53.4 k 18.8 k-ft 70 in0.75 8 3,750 psi

18 in 21.5 in 18 in 21.5 in1.7 297 in

0.174 ksi 0.459 ksi

The section is sufficiently large.

Now determine the area of stirrups required to resist Vu:

2 ' 2 3,750 psi 18 in 21.5 in 47.4 kc c wV f b d

53.4 k

47.4 k 23.8 k0.75

us c

VV V

223.8 k in

0.01860 ksi 21.5 in in

v s

yt

A V

s f d

Determine the additional area of stirrups required to resist Tu:

18.8 k-ft

25.1 k-ft0.75 0.75

un

TT

From Eq. (7-24), using 0.85o ohA A :

2

2

25.1 k-ft 12 in/ft in0.0099

2 cot in2 0.85 297 in 60 ksi cot 45

t n

o yt

A T

s A f

So, evaluate the total required area of stirrups:

For strength: 2 2 2in in in

0.018 2 0.0099 0.0378in in in

Minimum required: 250 50 18 in in

0.01560,000 psi in

w

yt

b

f

The strength requirement governs here.

7-4

Now determine stirrup size and spacing:

max

12 in

/ 8 70 / 8 8.75 in

/ 2 10.75 in (for shear)

hs P

d

If we select #3 stirrups, 2

2

2 0.11 in5.82 in

in0.0378

in

s


2

2 0.20 in10.6 in

in0.0378

in

s

Although it is a tight spacing, select to use closed #3 stirrups at 5 in spacing.

Determine the need for longitudinal reinforcement resisting torsion:

For strength: 2

2 2incot 0.00990 70 in 1.0 1.0 0.693 in

in

yttl h

yl

fAA P

s f

Minimum required: ,min

5 'c cp yttl h

yl yl

f A fAA P

f s f

Since 2 225in in

0.022 0.0075in in

t w

yt

A b

s f , we must use

2in0.022

in

2 22

,min

5 3,750 psi 432 in in0.022 70 in 1.0 0.66 in

60 ksi inlA

Use 20.693 inlA

Longitudinal bars are required in the corner of each stirrup. Also, longitudinal bars must be spaced

no more than 12 in apart around the perimeter of the section, so a bar is needed in the middle of

each face. So 8 bars are required.

2 2/ 0.693 in / 8 bars 0.0866 in /barArea bar

Select #3 bars along bottom and sides of the cantilevered section.

The reinforcement required along the top of the beam for resisting moment and torsion is:

2 2 24.56 in 3 0.0866 in 4.82 insA

Select 5#8 bars along the top of the beam such that2 2 25 5 1.0 in 5 in 4.82 ins bA A .

7-5

7-2 Explain why the torsion in the edge beam A-B in Fig. 7-21a is called “equilibrium

torsion,” while the torsion in the edge beam A1-B1 in Fig. P7-3 is called

“compatibility torsion.”

If the edge beam A-B in Fig. 7-21c did not resist torsion, the beam would rotate,

uninhibited, about its longitudinal axis and fail to resist the action of load P. Essentially, the

torsional resistance of the beam is required for equilibrium to be satisfied.

On the contrary, if beam A1-B1 in Fig. P7-3 did not resist torsion, the beam would rotate

only slightly before the floor’s weight and superimposed loads would be redistributed to other

elements, thereby satisfying equilibrium through the redundancy of the system. The torsion in A1-

B1 only arises from the need to maintain compatibility of deformations between the ends of the

joists and the twisting of the edge beam.

7-3 The two parts of this problem refer to the floor plan shown in Fig. P7-3. Assume that

the entire floor system is constructed with normal-weight concrete that has a

compressive strength, f’c = 4,500 psi. Also, assume that the longitudinal steel has a

yield strength of fy = 60 ksi and that the transverse steel has a yield strength of

fyt = 40 ksi.

a) Design the spandrel beam between columns B1 and C1 for bending, shear,

and torsion. Check all of the appropriate ACI Code requirements for

strength, minimum reinforcement area, and reinforcement spacing are

satisfied.

Step 1: Determine Mu, Vu, and Tu:


2 3

2

12 in 18 in lb lb150 225

in ft ft144ft

If we assume that the slab dimensions given in the figure are measured center to center of the

beams, we need to account for half the width of the beam in addition the span length given when

calculating loads. Therefore the dead weight from slab, per foot of beam, is:

3

6 in 6.5 ft lb lb150 488

in ft ft12 ft


2

lb lb6.5 ft 20 130

ft ft

To simplify the design, we will not reduce the live load. Therefore, the live load per foot of beam:

2

lb lb6.5 ft 50 325

ft ft

So, using 1.2 1.6uw DL LL , the factored load is, per foot of beam:


7-6

With that, we can determine the design moments using ACI Moment Coefficients:

At columns:

( )

( )

At midspan:

( )

( )

Note that n is taken as the average of the adjoining clear spans for calculating the negative

moment at the columns.

The design shear is:

At midspan:

At d away from the column face:

⁄(

)

⁄(

)

Note that n is taken as the clear span length of the interior span when calculating the shear

acting on the face of the support. Do not reduce this value to the n used for calculating

the negative moment at column B-1.

To calculate the design torsion, we first have to determine the moment and shear that the slab is

applying to the edge of the beam.

1.2 0.075 k/ft 0.020 k/ft 1.6 0.050 k/ft 0.194 k/ftuw

12 ft -12 in 11 ftn

22 0.194 k/ft 11 ft0.978 k-ft

24 24

u nu

wM

0.194 k/ft 11 ft

1.07 k2 2

u nu

wV

Therefore, the torsion applied to our beam by the slab is:

6 in

0.978 k-ft 1.07 k 1.51 k-ft/ft12 in/ft

t

And thus our design torsion, at d away from the ends of our beam, is:

( )

( )

7-7

Step 2: Determining the area of longitudinal steel required for flexure.

At the column:

Refer to Table A-3. For f’c = 4,500 psi, and R = 229 psi, . So, at the

column,

At midspan:

Refer to Table A-3. There are no R values below 190 psi, which corresponds to the

minimum reinforcement ratio. Therefore, √

√

Step 3: Determine whether torsion must be considered in the design of this beam. Begin by

determining the dimensions of the beam section active in torsion, and calculate the threshold

torsion.

height of beam below slab 18 in

4 24 inf

fh

, where f is the length of flange active in torsion.

Therefore,

224 in 12 in 6 in 18 in 396 incpA

24 in 12 in 18 in 18 in 6 in 30 in 108 incpP

222 396 in

' 0.75 1.0 4,500 psi 73,000 lb-in 6.09 k-ft108 in

cp

th c

cp

AT f

P

Since u thT T , torsion must be considered in this design.

Step 4: Since the torsion resisted by this edge beam is not required to maintain equilibrium of the

structure, we have a case of compatibility torsion. Therefore we can likely reduce our Tu to the

following:

2

22

,

396 in4 ' 0.75 4 1.0 4,500 psi 292,000 lb-in 24.4 k-ft

108 in

cp

u comp c

cp

AT f

P

Unfortunately, since ,u u compT T , we cannot reduce our design torsion. Since our design torsion is

not being reduced, no redistribution of design moments is required in the adjoining slab.

Step 5: Determine whether the section is large enough to resist the combined actions of shear and

torsion. First assume that a closed #4 stirrup will be used in the web of this beam.

224 in 2 1.5 in 0.5 in 12 in 2 1.5 in 0.5 in 174 inohA

2 24 in 2 1.5 in 0.5 in 12 in 2 1.5 in 0.5 in 58 inhP

7-8

From Eq. (7-33):

√(

)

(

)

(

√

)

√(

)

(

)

( √

√ )


Step 6: Determine the area of stirrups required to resist Vu:


Step 7: Determine the area of stirrups required to resist Tu:


( )

Step 8: Evaluate the total required area of stirrups, and select spacing:

For strength:

Minimum required: 20.75 4500 50.31 psi 12 in in

0.01540,000 psi in

w

yt

b

f



max

12 in

/ 8 58 in / 8 7.25 in


hs P

d

If we select #3 stirrups,


7-9

Select closed #3 stirrups at 5 in. spacing at the ends of the beam. It is also possible that stirrups are

not required along the full length of this beam. For torsion, determine where /u thT T . By

similar triangles, this occurs at 66.5 in. away from the face of the column. However, torsional

reinforcement must be continued for ( ) (21.5 in 12 in) 33.5 intd b past this theoretical point.

Therefore, no torsional reinforcement is required beyond 100 in. away from the face of the column.

Since / 2u cV V at this point, no shear reinforcement is required beyond this point either.

Final design of transverse reinforcement:

Using closed #3 stirrups,

One stirrup at 2.5 in from either column face, followed by stirrups spaced at 5 in. until beyond 100

in. from the face of the column.

Step 9: Finalize the design of the longitudinal reinforcement:


For strength:

Minimum required:

,min

5 'c cp yttl h

yl yl

f A fAA P

f s f

Since 2 2

,min

25in in0 0.0075

in in

t w

yt

A b

s f

, we must use

2in0.0075

in

2 22

,min

5 4,500 psi 396 in in 20.0075 58 in 1.92 in

60 ksi in 3lA

Use 21.92 inlA



each vertical face. So 6 bars are required.

2 2/ 1.92 in / 6 bars 0.32 in / barArea bar

-At the columns, use #6 bars in the bottom corners and halfway up the vertical face of the beam.

As top reinforcement, ⁄ ( ) ⁄ , so specify 3 #7 bars along the

top of the section.

-At midspan, use #6 bars in the top corners and halfway up the vertical face of the beam. As

bottom reinforcement, ⁄ ( ) ⁄ , so specify 3 #7 bars along the

bottom of the section.

7-10

b) Design the spandrel beam between columns A1 and A2 for bending, shear,

and torsion. Check that all of the appropriate ACI Code requirements for


satisfied.


Dead weight from spandrel beam, per foot of beam:

32

2

12 in 24 in lblb150 300ftin ft144

ft

Superimposed dead load, per foot of spandrel beam:

2lb1 ft 20 20 lb/ft

ft


2lb1 ft 50 50 lb/ft

ft

Dead weight from joist, applied as a point load at midspan:

32

2

12 in 18 in lb150 14.5 ft 3.26 kftin144

ft

Dead weight from slab, applied as a point load at midspan:

3

6 in 12 ft lb150 14.5 ft 13.1 kftin12

ft

Superimposed dead load, applied as a point load at midspan:

2lb12 ft 14.5 ft 20 3.48 k

ft

Since the live load is not reduced, the live load per foot of beam is:

2lb12 ft 14.5 ft 50 8.7 k

ft

So, using 1.2 1.6uw DL LL , the factored load is:

1.2 0.300 k/ft 0.02 k/ft 1.6 0.05 k/ft 0.464 k/ftuw , per foot of beam

, int 1.2 3.26 k 13.1 k 3.48 k 1.6 8.7 k 37.7 ku pow , applied as a point load

24 ft -16 in 22.67 ftn

With that, we can determine the design moments, but structural analysis software must be used

since the ACI Moment Coefficients cannot be applied when not all loads are distributed. Input the

structural model and applied loads using the appropriate pattern loading for the live loads.

At column A-1: 102 k-ftuM

At midspan: 128 k-ftuM

At column A-2: 139 k-ftuM

The design shear at d away from the supports is:

At column A-1:

, int2 0.464 k/ft 22.67 ft 2 21.5 in 37.7 k

23.3 k2 2 2 2

u pou n

u

ww dV

At midspan: 0 kuV

At column A-2: , 11.15 26.8 ku u AV V

7-11

To calculate the design torsion, we first have to determine the moment and shear that the joist is

applying to the edge of the spandrel beam.


30 ft -12 in 29 ftn

22 2.6 k/ft 29 ft91.1 k-ft

24 24

u nu

wM

, int 37.7 ku u poV w

Thus our design torsion, at d away from the ends of our beam, is:

6 in 6 in

91.1 k-ft 37.7 k 110 k-ft12 in/ft 12 in/ft

u u uT M V


At column A-1:

22

102 k-ft 12 in/ft245 psi

0.9 12 in 21.5 in

uMR

bd

Refer to Table A-3. For f’c = 4,500 psi, and R = 245 psi, 0.0042 . So, at

column A-1,20.0042 12 in 21.5 in 1.08 insA bd .

At midspan:

22


0.9 12 in 21.5 in

uMR

bd


midspan,20.0054 12 in 21.5 in 1.39 insA bd .

At column A-2:

22


0.9 12 in 21.5 in

uMR

bd


column A-2, 20.0058 12 in 21.5 in 1.50 insA bd .



torsion.


4 24 inf

fh


Therefore,

224 in 12 in 6 in 18 in 396 incpA


222 396 in

' 0.75 1.0 4,500 psi 73,000 lb-in 6.09 k-ft108 in

cp

th c

cp

AT f

P


7-12



following:

2

22

,

396 in4 ' 0.75 4 1.0 4,500 psi 292,000 lb-in 24.4 k-ft

108 in

cp

u comp c

cp

AT f

P

Since ,u u compT T , we can reduce our design torsion to 24.4 k-ft. Since our design torsion for the

spandrel beam is being reduced, it is necessary to redistribute the design moments for the joist that

frames into the spandrel beam. See chapter 7 for further discussion of why this is required.





From Eq. (7-33): 2 2

28 '

1.7

u u h cc

w oh w

V T P Vf

b d A b d

22

22

2 4,500 psi 12 in 21.5 in25.6 k 24.4 k-ft 58 in0.75 8 4,500 psi

12 in 21.5 in 12 in 21.5 in1.7 174 in

0.344 ksi 0.503 ksi




At column A-1: 0 ksV

At column A-2: 25.6 k

34.6 k 0 k0.75

us c

VV V


24.4 k-ft

32.5 k-ft0.75 0.75

un

TT


2

2

32.5 k-ft 12 in/ft in0.0330

2 cot in2 0.85 174 in 40 ksi cot 45

t n

o yt

A T

s A f


For strength: 2 2in in

2 0.0330 0.066in in

Minimum required:

' 20.75 50.3 psi 12 in in0.015

40,000 psi in

c w

yt

f b

f


7-13


max

12 in

/ 8 58 in / 8 7.25 in


hs P

d


2

2 0.2 in6.1 in

in0.066

in

s


2

2 0.31 in9.4 in

in0.066

in

s

Select closed #4 stirrups at 6.0 in spacing at the ends of the beam. While in some cases it is

possible that stirrups are not required along the full length of the beam, the torsion in this case is

constant along the length of the beam, as is the shear. Therefore, closed #4 stirrups are required

along the full length of the beam.



For strength: 2

2 2in 2cot 0.0330 58 in 1.0 1.28 in

in 3

yttl h

yl

fAA P

s f


5 'c cp yttl h

yl yl

f A fAA P

f s f

Since 2 2

,min

25in in0.0330 0.0075

in in

t w

yt

A b

s f

, we must use

2in0.0330

in

2 22

,min

5 4,500 psi 396 in in 20.0330 58 in 0.94 in

60 ksi in 3lA

Use 21.28 inlA





-At column A-1, use #5 bars in the bottom corners and halfway up the vertical face of the beam.

As top reinforcement, 2 2 2 2/ 1.08 in 2 0.21 in / 3 1.50 in / 3 0.50 insA bar , so specify 3 #7

bars along the top of the section.


bottom reinforcement, 2 2 2 2/ 1.39 in 2 0.21 in / 3 1.81 in / 3 0.60 insA bar , so specify 3 #8

bars along the bottom of the section.




7-14

7-4 The two parts of this problem refer to the floor plan shown in Fig. P7-3. Assume that

the entire floor system is constructed with sand light-weight concrete that has a

compressive strength, f’c = 4,000 psi. Also assume that the longitudinal steel has a

yield strength of fy = 60 ksi and that the transverse steel has a yield strength of

fyt = 60 ksi.

a) Design the spandrel beam between columns B1 and C1 for bending, shear,

and torsion. Check all of the appropriate ACI Code requirements for


satisfied.


Note that this problem is very similar to the previous problem. One noticeable change is the use of

lightweight concrete, which will affect the dead weight used in the calculation of the design loads.

Although no guidance is given on the density of the sand-lightweight concrete used in this

problem, any reasonable assumption would be acceptable. Here a density of 3120 lb/ft is assumed.


2 3

2

12 in 18 in lb lb120 180

in ft ft144ft


3

6 in 6.5 ft lb lb120 390

in ft ft12ft


2

lb lb6.5 ft 20 130

ft ft


2

lb lb6.5 ft 50 325

ft ft

So, using 1.2 1.6uw DL LL , the factored load is, per foot of beam:


With that, we can determine the design moments using ACI Moment Coefficients:

At columns:

( )

( )

At midspan:

( )

( )

Note that n is taken as the average of the adjoining clear spans for calculating the negative

moment at the columns.

7-15

The design shear is:

At midspan:

At d away from the column face:

⁄(

)

⁄(

)

Note that n is taken as the clear span length of the interior span when calculating the shear

acting on the face of the support. Do not reduce this value to the n used for calculating

the negative moment at column B-1.

To calculate the design torsion, we first have to determine the moment and shear that the slab is


1.2 0.060 k/ft 0.020 k/ft 1.6 0.050 k/ft 0.176 k/ftuw

12 ft -12 in 11 ftn

22 0.176 k/ft 11 ft0.887 k-ft

24 24

u nu

wM

0.176 k/ft 11 ft

0.968 k2 2

u nu

wV

Therefore, the torsion applied to our beam by the slab is:

6 in

0.887 k-ft 0.968 k 1.37 k-ft/ft12 in/ft

t

And thus our design torsion, at d away from the ends of our beam, is:

( )

( )


At the column:

Refer to Table A-3. For f’c = 4,000 psi and R = 204 psi, . So, at the column,

At midspan:

Refer to Table A-3. There are no R values below 190 psi, which corresponds to the

minimum reinforcement ratio. Therefore,

7-16



torsion.


4 24 inf

fh


Therefore,

224 in 12 in 6 in 18 in 396 incpA


222 396 in

' 0.75 0.85 4,000 psi 58,500 lb-in 4.88 k-ft108 in

cp

th c

cp

AT f

P




following:

2

22

,

396 in4 ' 0.75 4 0.85 4,000 psi 234,000 lb-in 19.5 k-ft

108 in

cp

u comp c

cp

AT f

P

Unfortunately, since ,u u compT T , we cannot reduce our design torsion. Since our design torsion is

not being reduced, no redistribution of design moments is required in the adjoining slab.





From Eq. (7-33):

√(

)

(

)

(

√

)

√(

)

(

)

( √

√ )



√ √

7-17



( )


For strength:

Minimum required:

√



max

12 in

/ 8 58 in / 8 7.25 in


hs P

d



Select closed #3 stirrups at 7 in. spacing at the ends of the beam. It is also possible that stirrups are

not required along the full length of this beam. For torsion, determine where /u thT T . By

similar triangles, this occurs at 75 in. away from the face of the column. However, torsional

reinforcement must be continued for ( ) (21.5 in 12 in) 33.5 intd b past this theoretical point.

Therefore, no torsional reinforcement is required beyond 108.5 in. away from the face of the

column. Since / 2u cV V at this point, no shear reinforcement is required beyond this point either.



One stirrup at 3.5 in from either column face, followed by stirrups spaced at 7 in. until beyond 109

in. from the face of the column.

7-18



For strength:

Minimum required:

,min

5 'c cp yttl h

yl yl

f A fAA P

f s f

Since 2 2

,min

25in in0 0.0075

in in

t w

yt

A b

s f

, we must use

2in0.0075

in

2 22

,min

5 4,000 psi 396 in in0.0050 58 in 1.0 1.80 in

60 ksi inlA

Use 21.80 inlA





-At the columns, use #5 bars in the bottom corners and halfway up the vertical face of the beam.

As top reinforcement, ⁄ ( ) ⁄ , so specify 3 #7 bars along the

top of the section.


bottom reinforcement, ⁄ ( ) ⁄ , so specify 3 #7 bars along the

bottom of the section.

7-19

b) Design the spandrel beam between columns A1 and A2 for bending, shear,

and torsion. Check that all of the appropriate ACI Code requirements for


satisfied.


Again, assume the density of the concrete is 3120 lb/ft .


2 3

2

12 in 24 in lb lb120 240

in ft ft144ft

Superimposed dead load, per foot of spandrel beam:

2lb1 ft 20 20 lb/ft

ft


2lb1 ft 50 50 lb/ft

ft

Dead weight from joist, applied as a point load at midspan:

2 3

2

12 in 18 in lb120 14.5 ft 2.61 k

in ft144ft

Dead weight from slab, applied as a point load at midspan:

3

6 in 12 ft lb120 14.5 ft 10.4 k

in ft12ft

Superimposed dead load, applied as a point load at midspan:

2

lb12 ft 14.5 ft 20 3.48 k

ft


2

lb12 ft 14.5 ft 50 8.7 k

ft

So, using 1.2 1.6uw DL LL , the factored load is:

1.2 0.240 k/ft 0.02 k/ft 1.6 50 k/ft 0.392 k/ftuw , per foot of beam

, int 1.2 2.61 k 10.4 k 3.48 k 1.6 8.7 k 33.7 ku pow , applied as a point load

24 ft -16 in 22.67 ftn

With that, we can determine the design moments, but structural analysis software must be used

since the ACI Moment Coefficients cannot be applied when not all loads are distributed.

At column A-1: 90.4 k-ftuM

At midspan: 114 k-ftuM

At column B-1: 123 k-ftuM

The design shear at d away from the supports is:

At column A-1:

, int2 0.392 k/ft 22.67 ft 2 21.5 in 33.7 k

20.6 k2 2 2 2

u pou n

u

ww dV

At midspan: 0 kuV

At column B-1: , 11.15 23.7 ku u AV V

7-20

To calculate the design torsion, we first have to determine the moment and shear that the beam is



30 ft -12 in 29 ftn

22 2.33 k/ft 29 ft81.6 k-ft

24 24

u nu

wM

, int 33.7 ku u poV w

Thus our design torsion, at d away from the ends of our beam, is:

6 in 6 in

81.6 k-ft 33.7 k 98.5 k-ft12 in/ft 12 in/ft

u u uT M V


At column A-1:

22

90.4 k-ft 12 in/ft217 psi

0.9 12 in 21.5 in

uMR

bd


column A-1,20.0037 12 in 21.5 in 0.95 insA bd .

At midspan:

22


0.9 12 in 21.5 in

uMR

bd


midspan,20.0048 12 in 21.5 in 1.24 insA bd .

At column A-2:

22


0.9 12 in 21.5 in

uMR

bd


column A-2, 20.0052 12 in 21.5 in 1.34 insA bd .



torsion.


4 24 inf

fh


Therefore,

224 in 12 in 6 in 18 in 396 incpA


222 396 in

' 0.75 0.85 4,000 psi 58,500 lb-in 4.88 k-ft108 in

cp

th c

cp

AT f

P


7-21



following:

2

22

,

396 in4 ' 0.75 4 0.85 4,000 psi 234,000 lb-in 19.5 k-ft

108 in

cp

u comp c

cp

AT f

P

Since ,u u compT T , we can reduce our design torsion to 19.5 k-ft. Since our design torsion for the

spandrel beam is being reduced, it is necessary to redistribute the design moments for the joist that

frames into the spandrel beam. See chapter 7 for further discussion of why this is required.





From Eq. (7-33):

√(

)

(

)

(

√

)

√(

)

(

)

( √

√ )



√ √

At column A-1:

At column A-2:


19.5 k-ft

26.0 k-ft0.75 0.75

un

TT


2

2

26.0 k-ft 12 in/ft in0.0176

2 cot in2 0.85 174 in 60 ksi cot 45

t n

o yt

A T

s A f

7-22


For strength:

Minimum required: 250 50 12 in in

0.01060,000 psi in

w

yt

b

f



max

12 in

/ 8 58 in / 8 7.25 in


hs P

d



Select closed #3 stirrups at 5 in. spacing for the full length of the beam. While in some cases it is

possible that stirrups are not required along the full length of the beam, the torsion in this case is

constant along the length of the beam, as is the shear. Therefore, closed #3 stirrups are required

along the full length of the beam.



One stirrup at 2.5 in from either column face, followed by stirrups spaced at 5 in.



For strength: 2

2 2incot 0.0176 58 in 1.0 1.0 1.02 in

in

yttl h

yl

fAA P

s f


5 'c cp yttl h

yl yl

f A fAA P

f s f

Since 2 2

,min

25in in0.0176 0.0050

in in

t w

yt

A b

s f

, we must use

2in0.0176

in

2 22

,min

5 4,000 psi 396 in in0.0176 58 in 1.0 1.07 in

60 ksi inlA

Use 21.07 inlA

7-23






As top reinforcement, 2 2 2 2/ 0.95 in 2 0.18 in / 3 1.31 in / 3 0.437 insA bar , so specify 3

#6 bars along the top of the section.


bottom reinforcement, 2 2 2 2/ 1.24 in 2 0.18 in / 3 1.60 in / 3 0.53 insA bar , so specify 3

#7 bars along the bottom of the section.




8-1

Chapter 8

8-1 Figure P8-1 shows a cantilever beam with containing three No. 7 bars

that are anchored in the column by standard hooks. (normal-

weight) and . If the steel is stressed to at the face of the column,

can these bars:

(a) be anchored by hooks into the column? The clear cover to the side of the

hook is 2 ¾ in. The clear cover to the bar extension beyond the bend is 2 in.

The joint is enclosed by ties at 6 in. o.c.

Except for the side cover of 2 ¾ in., a clear cover of 2 in. is assumed throughout the beam

and joint. See Fig. S8-1 below.

18 in.

18 in.

2 in.7.5 in. 2 in. (clear cover)

48 in.

6 in.

2 in. (clear cover)

Side cover = 2¾ in. for columns and 2 in. for beams;

Beam width = 12 in.

3 #7

#3, double legs

Fig. S8-1

1. Check if a 90 standard hook can be used

Length of the hook tail including the bend:

12 3 0.5 15.5 15.5 0.875 in. 13.6 in.b b b bd d d d

Available vertical room for the hook tail:

18 in. 2 2 in. 14 in. 13.6 in. , OK.

2. Required development length of 90 standard hooks:

Basic development length of 90 standard hooks:

0.02 0.02 1 600000.875 14.8 in.

1 5000

e y

hb b

c

fd

f

Note: Coating factor 1e for uncoated reinforcement (assumed)

Lightweight-aggregate-concrete factor 1 for normal-weight

concrete

8-2

Applicable factors:

ACI Code Section 12.5.3(a) applies (factor of 0.7) because:

bar size smaller than bar No. 11

side cover: 2 3/4 in. 2 1/2 in.

tail cover: 2 in. 2 in.

ACI Code Section 12.5.3(b) does not apply because:

tie spacing: 6 in. 3 2.6 in.bd

ACI Code Section 12.5.3(d) does not apply because 60000 psiyf

0.7 14.8 in. 0.7 10.4 in.dh hb

Minimum development length

,min max(8 ,6 in.) max(8 0.875 in., 6 in.) 7 in. 10.4 in.dh bd , OK.

3. Available horizontal room for 90 standard hook:

18 in. 2 in. 16 in. 10.4 in. , OK.

Conclusion: The bars can be anchored in the column by using 90˚ standard hook.

(b) be developed in the beam? The bar ends 2 in. from the end of the beam. The

beam has No. 3 double-leg stirrups at 7.5 in.

1. Available room in the beam for straight anchorage: 48 in. 2 in. 46 in.

2. Required development length of straight anchorage conforming to ACI Code Section

12.2.2.:

Determine which formulas to use in the Table 8-1

12 in. 2 2 in. 3 0.875 in.Clear spacing: 2.7 in. 2 1.75 in.

2

Clear cover: 2 in. 0.875 in.

Bar #7

b

b

d

d

1.3 1 60000

0.875 48.3 in. 46 in.20 20 1 5000

e t y

hb b

c

fd

f

, NG.

Note: Bar-location factor 1.3t is used because the bars are top

reinforcement with a depth of fresh concrete below: 18 in. 2 in. 0.875 in. 15.1 in. 12 in.

Per ACI Code Section 12.2.2, the bars cannot be developed in the beam by

using straight anchorage. This problem can be solved by reducing the bar size.

However, ACI Code Section 12.2.3 allows another method to calculate the

development length which is likely to yield a shorter development length. Let try

this.

3. Required development length of straight bars conforming to ACI Code Section

12.2.3.

Bar-size factor 1s for #7 bars

8-3

Bar-spacing factor bc

One-half center-to-center spacing of the bars:

12 in. 2 2 in. 0.875 in.0.5 1.78 in.

2

Smallest distance from beam surface to centers of bars:

0.875 in.2 in. 2.44 in.

2

min(1.78 in.,2.44 in.) 1.78 in.bc

Transverse reinforcement index

240 2 0.11 in.400.39 in.

7.5 in. 3

trtr

AK

sn

Required development length per ACI Code Section 12.2.3

3 3 60000 1.3 1 10.875 29.2 in.<46 in.

1.78 0.3940 40 1 5000

0.875

y t e sd b

b trc

b

fd

c Kf

d

(Note that 2.48 2.5,OK.b tr

b

c K

d

)

Per ACI Code 12.2.3, the bars can be developed in the beam by using straight

anchorage.

8-2 Give two reasons why the tension development length is longer than the

compression development length.

1. A bar stressed in compression transfers some of its force to the concrete by bearing

on the end of the bars.

2. A bar stressed in tension transfers its tensile stress into concrete. As a result the

concrete is cracked and “in-and-out” bond stresses exist. In such a case there are

localized bond stresses which are several times greater the average bond stress. A bar

stressed in compression transfers its stress into concrete which is compressed and

hence un-cracked. There are no “in-and-out” bond stresses in such a case.

8-3 Why do bar spacing and cover to the surface of the bar affect bond strength?

The lugs on deformed bars transfer forces to the concrete. The radial component of these

lug forces causes a tensile stress in an annulus of concrete around the bar. The thicker the

wall of this annulus the lower the tensile stresses are in it. The thickness of the wall is

governed by the minimum distance to the surface of the concrete or to the next bar. Thus,

the larger the cover and bar spacing are, the larger the bond stresses can be developed.

8-4

8-4 A simply supported rectangular beam with and and No. 3

Grade 40 stirrups at , spans 14 ft and supports a total factored uniform

load of 6.5 kips/ft, including its own dead load. It is built of 4000 psi light-weight

concrete and contains 2 No. 10 Grade 60 bars that extend 5 in. past the centers of

the supports at each end. Does this beam satisfy ACI Section 12.11.3? If not, what is

the largest size bars which can be used?

The question asks to check if nd a

u

M

V at the support.

1. Calculation of d .

Bar-location factor 1t for bottom bars

Coating factor 1e for uncoated reinforcement (assumed)

Bar-size factor 1s for #10 bars

Lightweight-aggregate-concrete factor for light-weight concrete

Bar-spacing factor min 2.5 in.,0.5 14 in. 2 2.5 in. 2.5 in.bc

Transverse reinforcement index trK .

( )

Calculate the development length

3 3 60000 1 1 11.27 50.2 in.

2.5 0.5540 40 0.75 4000

1.27

y t e sd b

b trc

b

fd

c Kf

d

(Note that 2.4 2.5,OK.b tr

b

c K

d

)

2. Calculation of na

u

M

V .

( ⁄ ) ( ⁄ )

6.5 k/ft 14 ft45.5 kips

2uV

nd a

u

M

V So, the use of 2 #10 bars satisfies ACI Code Section 12.11.3.

8-5 Why do ACI Section 12.10.3 and 12.12.3 require that bars extend past their cut-

off points?

Inclined cracking due to shear increases the tension in the flexural reinforcement at all

points except the points of maximum moments. As a result, the tensile force in the

flexural reinforcement computed from the moment at a given section actually exists at a

point about 0.75d to d from that point in the direction of decreasing moment.

8-5

8-6 Why does ACI Section 12.10.2 define “points within the span where adjacent

reinforcement terminates” as critical sections for development of reinforcement in

flexural members?

If flexural reinforcement is cut off according to the moment diagram, the flexural cut-off

point is the point in the beam where the remaining steel not cut off is just adequate for the

moment if stressed to yf . Due to the effect of shear, this point actually occurs about d

farther away and hence this point is critical.

8-7 A rectangular beam with cross section , , and supports a total factored load of 3.9 kips/ft, including its own dead load. The beam is

simply supported with a 22-ft span. It is reinforced with 6 No. 6 Grade 60 bars, two

of which are cut off between midspan and the support and four of which extend 10

in. past the centers of the supports. . The beam has No. 3 stirrups

satisfying ACI Code Sections 11.4.5 and 11.4.6.

CL

cut-off point

4 bars #6 6 bars #6

d =21.7 in.h = 24 in.

11 ft.12 in.

M =

23

6 k

-ft.

øMn = 238 k-ft.

M =

16

3 k

-ft.

øMn = 163 k-ft.

ℓd = 28.5 in. ℓd = 28.5 in.

21.7 in.37 in.

factored moment

diagram M

12 in.

cut-off point

flexural

cut-off point

Fig. S8-2

(a) Plot to scale the factored moment diagram. ( )

( ), where x is

the distance from the support and is the span.

Maximum moment at the midspan: 3.9 11 ft

/ 2 22 ft 11 ft 236 k-ft2

M x

The factored moment diagram is shown in Fig. S8-2.

10 in.

10 in. 35.6 in. 21.5 in.

236 k-ft

160 k-ft

21.5 in.

160

k-f

t

8-6

(b) Plot a resisting moment diagram and locate the cut-off point for the two cut-

off bars.

1. Moment capacity of beam sections with 6 #6 bars

26 0.44 in. 60000 psi3.3 in.

0.85 0.85 4000 psi 14 in.

s y

c

A fa

f b

( ) ( ⁄ ) ( ⁄ )

1

3.3 in.3.9 in.

0.85

ac

( 1 0.85 for 4000 psicf )

( )

2. Moment capacity of beam sections with 4 #6 bars

24 0.44 in. 60000 psi2.2 in.

0.85 0.85 4000 psi 14 in.

s y

c

A fa

f b

( ) ( ⁄ ) ( ⁄ )

1

2.2 in.2.6 in.

0.85

ac

( 1 0.85 for 4000 psicf )

( )

3. Development length for straight bars. ACI Code Section 12.2.2 will be used to

determine the development length.

Minimum clear spacing between bars:

14 in. 2 2.3 in. 5 0.75 in.1.1 in. 0.75 in.

5bd

Since the shear reinforcement is provided such that ACI Code Sections 11.5.4

and 11.5.5.3 will be satisfied, shear reinforcement is not less than the minimum

code requirement.

Bars used are #6

1 1 60000

0.75 28.5 in.25 25 1 4000

e t y

d b

c

fd

f

4. Locate the cut-off point of the 2 #6 bars (see Fig. S8-2.)

Determine the flexural cut-off point of the 2 #6 bars. This point is the intersection

of the factored moment diagram and the line of moment capacity (4#6)nM ,

i.e.:

( ) (4#6)nM x M

( )

8-7

Extend the flexural cut-off point a distance d toward the left support to take into

account the shear effect. The distance from the left support to the cut-off point is

The resisting moment diagram is shown in Fig. S8-2.

8-8 Why does ACI Code Section 12.10.5 require extra stirrups at bar cut-off points in

some cases?

A severe discontinuity in longitudinal bar stresses exists in the region of a cut-off point in

a zone of flexural tension. This causes a reduction in the inclined cracking load in that

region. To compensate, more stirrups are required.

The beam shown in Fig. P8-9 is built of 4000 psi normal-weight concrete and Grade 60 steel.

The effective depth The beam supports a total factored uniform load of 5.25

kips/ft, including its own dead load. The frame is not part of the lateral load-resisting

system for the building. Use Figs. A-1 to A-4 to select cut-off points in Problems 8-9 to 8-11.

8-9 Select cut-off points for span A-B based on the following requirements:

(a) Cut off two No. 6 positive moment bars when no longer needed at each end.

Extend the remaining bars into the columns.

1. Development length for bottom #6 and #7 bars

From the effective depth of 18.5 in, assume a concrete cover of 2.5 in. to centers of the

bars.

Clear spacing:

Shear reinforcement: 2

2 2

min

500.22 in. 50 120.028 in. /in. 0.015 in. /in.

8 in. 40000

v v w

yt

A A b

s s f

(Note that 0.75 47.4 psi 50 psi, use 50 psicf )

The development length is computed following Case 1

1 1 60000

#6, bottom 0.75 28.5 in.25 25 1 4000

e t y

d b

c

fd

f

1 1 60000

#7, bottom 0.875 41.5 in.20 20 1 4000

e t y

d b

c

fd

f

2. Locate the cutoff point for the #6 bars

After the 2 #6 bars are cutoff, the remaining #7 bars will provide an of

, which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff

when the moment demand is less than 0.577 at midspan. From Fig. A-2, this

occurs at from the face of the exterior support. The actual

cutoff has to be the larger of and beyond the theoretical cutoff point.

8-8

and , so the actual cutoff is from

the face of the exterior support.

3. Verify that the distance from midspan to the actual cutoff point is larger than the

development length.

Therefore, the cutoff for the 2 #6 bars is 34 in. from the face of the exterior support.

4. Do the same calculations for the interior support (B) for this exterior span:

After the 2 #6 bars are cutoff, the remaining #7 bars will provide ,

which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff when the

moment demand is less than 0.577 at midspan. From Fig. A-2, this occurs at

from the face of the interior support. The actual cutoff has

to be the larger of and beyond the theoretical cutoff point. and

, so the actual cutoff is from the face of the

interior support.

Verify that the distance from midspan to the actual cutoff point is larger than the

development length.

Therefore, the cutoff for the 2 #6 bars can be 36 in. from the face of the interior

support. To limit the potential for errors on the jobsite, given the very small

difference between exterior and interior support cutoff locations, we will select the

more conservative value for both supports of the exterior span. Cutoff 2 #6 bars at 34

in. from the face of the columns in the exterior span.

5. Verify that enough bars are extended into the column:

( ) ( )

This requirement is satisfied. Also, specify that the #7 bars extend 6 in. into the

columns.

6. With the #7 bars extended 6 in. into the columns, verify that the #7 bars are

sufficiently developed. To check this, ensure that the continuous #7 bars extend more

than past the theoretical cutoff point of the #6 bars.

( ) This requirement is satisfied.

7. Check if nd a

u

M

V at the positive moment point of inflection near end A.

( ⁄ ) ( ⁄ )

8-9

From Fig. A-2, the positive moment point of inflection closest to end A is

0.1 2 ft 24 in.n

5.25 k/ft 20 ft5.25k/ft 2 ft 42 kips

2uV

{ ( )

}

8. Check if nd a

u

M

V at the positive moment point of inflection near end B

Everything remains the same as for end A, except the point of inflection.

From Fig. A-2, the positive moment point of inflection closest to end B is

0.104 2.08 ft.n The change is not significant, so nd a

u

M

V is ensured.

9. Check whether ACI Code Section 12.10.5.1 is satisfied.

⁄

⁄

⁄

√ √

( ) ( )

Therefore, we must add stirrups near the point of termination for the #6 bars.

Additional stirrup area required is:

If we keep the same #3 U-shaped stirrups, . Therefore, Specify a stirrups spacing of 3 in. within d of the #6 bar termination

point.

(b) Extend all negative moment bars past the negative moment point of

inflection before cutting them off.

1. Development length for # 6 and #7 top bars

Note that the clear spacing of top #7 bars is:

The development length is computed following Case 1, with 1.3t for top bars.

1 1.3 60000

#6, top 0.75 37.0 in.25 25 1 4000

e t y

d b

c

fd

f

1 1.3 60000

#7, top 0.875 54.0 in.20 20 1 4000

e t y

d b

c

fd

f

8-10

2. Determine the cut-off point conforming to reinforcement continuity for negative

reinforcement near column A

Since all reinforcement is required to extend past the point of inflection, the

requirement of greater than (face)

3

sAto extend past the point of inflection is

automatically satisfied.

All reinforcement has to extend past the inflection point a length:

{

⁄ }

Negative moment point of inflection from face of column A is

0.164 3.28 ft = 39.4 in.n

Total length from face of column A to the reinforcement cut-off point is:

( ) Choose 58 in.


reinforcement near column B



3




{

⁄ }

Negative moment point of inflection from face of column B is

0.24 4.8 ft = 57.6 in.n

Total length from face of column B to the reinforcement cut-off point is:

( ) , so choose 78 in.

(c) Check the anchorage of the negative moment bars at A and modify the bar

size if necessary.

Reinforcement continuity and structural integrity require that all negative moment

reinforcement be fully anchored into column A. Note that the development length of

straight #6 bars is 37.1 in., which far exceeds the available column width of 18 in. A

hook must be used for anchorage.

Development length for a standard 90º hook for #6 bars:

0.02 0.02 1 600000.75 14.2 in.

1 4000

e y

hb b

c

fd

f

Applicable factors. The problem does not give sufficient information to decide

if reduction factors per ACI Code Section 12.5.3 shall apply. To be

conservative, assume that no reduction of hb will be made.

14.2 in.dh hb

Minimum development length

8-11

,min max(8 ,6 in.) max(8 0.75 in., 6 in.) 6 in. 14.2 in.dh bd , OK.

Available horizontal room for the 90º standard hook:

18 in. 2.4 in. 15.6 in. 14.2 in., OK

Extend the hook past the column face and to the other side of the column.

Fig. S8-3

8-10 Repeat Problem 8-9(a) and (b) for span B-C

(a) Cut off two No. 6 positive moment bars when no longer needed at each end.

Extend the remaining bars into the columns.



bars.

Clear spacing:


2 2

min

500.22 in. 50 120.028 in. /in. 0.015 in. /in.

8 in. 40000

v v w

yt

A A b

s s f



1 1 60000

#6, bottom 0.75 28.5 in.25 25 1 4000

e t y

d b

c

fd

f

1 1 60000

#7, bottom 0.875 41.5 in.20 20 1 4000

e t y

d b

c

fd

f

2. Locate the cutoff point for the #6 bars. Both supports are interior supports.







interior support.


development length.

21 in.

58 in. 78 in.15 in.

34 in. 6 in.34 in.6 in.

8-12

Therefore, the cutoff for the 2 #6 bars can be 52 in. from the face of the support.


( ) ( )


columns.





6. Check if nd a

u

M

V at the positive moment point of inflection.

( ⁄ ) ( ⁄ )


{ ( )

}


⁄

⁄

⁄

√ √

( ) ( )



If we keep the same #3 U-shaped stirrups, . Therefore, Specify a stirrup spacing of 4 in. within d of the #6 bar termination point.

(b) Extend all negative moment bars past the negative moment point of

inflection before cutting them off.

8-13

1. Similar to Problem 8-9(b), extend all negative reinforcement past the negative

moment point of inflection a length of 18.5 in.

2. Negative moment point of inflection from face of column B and C is

0.24 5.04 ft = 60.5 in.n

3. Total length from face of column B to the reinforcement cut-off point is

( ) Select 80 in.

See Fig. S8-4 for reinforcement details.

Fig. S8-4

8-11 Assume the beam is constructed with 4000 psi light-weight concrete. Select cut-off

points for span A-B based on the following requirements:

(a) Extend all negative moment bars at A past the negative moment point of

inflection.

1. Development length for #6 top bars

Note that the clear spacing of top #6 bars is:

The development length is computed following Case 1, with 1.3t for top bars.

( )

√

√


reinforcement near column A



3




{

⁄ }

Negative moment point of inflection from face of column A is

0.164 3.28 ft = 39.4 in.n

Total length from face of column A to the reinforcement cut-off point is:

( ) Choose 58 in.

21 in.

80 in.

52 in. 6 in.6 in.

80 in.

52 in.

8-14

(b) Cut off the two No. 6 positive moment bars when no longer needed at each

end. Extend the remaining bars into the columns.



bars.

Clear spacing:


2 2

min

500.22 in. 50 120.028 in. /in. 0.015 in. /in.

8 in. 40000

v v w

yt

A A b

s s f



( )

√

√

( )

√

√

2. Locate the cutoff point for the #6 bars

After the 2 #6 bars are cutoff, the remaining #7 bars will provide an of

, which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff

when the moment demand is less than 0.577 at midspan. From Fig. A-2, this

occurs at from the face of the exterior support. The actual

cutoff has to be the larger of and beyond the theoretical cutoff point.

and , so the actual cutoff is from

the face of the exterior support.


development length.

Therefore, the cutoff for the 2 #6 bars is 34 in. from the face of the exterior support.

4. Do the same calculations for the interior support (B) for this exterior span:







interior support.

Verify that the distance from midspan to the actual cutoff point is larger than the

development length.

8-15

Therefore, the cutoff for the 2 #6 bars can be 36 in. from the face of the interior

support. To limit the potential for errors on the jobsite, given the very small

difference between exterior and interior support cutoff locations, we will select the

more conservative value for both supports of the exterior span. Cutoff 2 #6 bars at 34

in. from the face of the columns in the exterior span.


( ) ( )


columns.





7. Check if nd a

u

M

V at the positive moment point of inflection near end A.

( ⁄ ) ( ⁄ )


0.1 2 ft 24 in.n

5.25 k/ft 20 ft5.25k/ft 2 ft 42 kips

2uV

{ ( )

}

8. Check if nd a

u

M

V at the positive moment point of inflection near end B

Everything remains the same as for end A, except the point of inflection.

From Fig. A-2, the positive moment point of inflection closest to end B is

0.104 2.08 ft.n The change is not significant, so nd a

u

M

V is ensured.


⁄

⁄

⁄

√ √

( ) ( )

8-16



If we keep the same #3 U-shaped stirrups, . Therefore, Specify a stirrup spacing of 3 in. within d of the #6 bar termination point.

(c) Cut off two of the negative moment bars at B when no longer needed.

Extend the remaining bars past the point of inflection.

1. Calculate the development length of a #7 bar:

( )

√

√

2. Cut-off points for the remaining 3 #7 bars:

These bars need to be extended past the negative point of flexure. Problem 8-9(b)

shows that the required length is 76.2 in. Use 78 in. as before.

3. Cut-off points for 2 #7 bars:

Ratio of remaining reinforcement area after cutting off the 2 #7 bars is 3/5 = 0.6

2 #7 bars can be cut off at the location x where Moment at x

0.6Maximum moment

. This

occurs at 0.08 1.6 ft.=19.2 in.nx from B

To consider the shear effect, extend 2 #7 bars a distance d past the flexural cut-

off point away from B. The distance from the face of column B to the cut-off point is

( ) , which is not OK. An option is to

specify a length of 72 in. However, to simplify construction, another option is to

extend all the bars a distance of 78 in. as in Item 1. Select a cutoff point of 78 in.

from the column face for all 5 #7 bars.

No bars are terminated within a flexural tension zone, so there is no need to check

shear at the cut off point.

2 #6

58 in. 78 in.

6 in. 6 in.

4 #6 5 #7

18 in.240 in.

18 in.

12 in.

A B2 #7

4 #3 @ 5 in. (extra)

4 #3 @ 5 in.(extra)

Cut-off points for

bars #6

58 in.

21 in.

#3 stirrups @ 3 in.

#3 stirrups @ 3 in.

8-17

Fig. S8-5

9-1

Chapter 9

9-1 Explain the differences in appearance of flexural cracks, shear cracks, and torsional

cracks.

Flexural cracks are approximately vertical cracks extending from the tensile face of the member

towards the level of the zero-strain axis (neutral axis).

Shear cracks are inclined to the axis of the element. Most often these cracks start at a flexural

crack on the tensile face of the member and extend diagonally through the member toward the

point of maximum moment.

Torsional cracks spiral around the member and, for pure torsion, are roughly 45 deg. on all faces.

For combined torsion and shear, the cracks tend to be pronounced on the face where the direct

shear stress and the shear stresses due to torsion add, and less pronounced (or even absent) on the

opposite face, where the stresses counteract.

9-2 Why is it necessary to limit the width of cracks?

Cracks are of concern for three main reasons:

appearance: wide cracks are unsightly and sometimes lead to a concern by owners and

occupants;

leakage of air or fluids through the crack: crack control is important in the design of liquid-

retaining structures;

corrosion of reinforcement: traditionally corrosion of reinforcement has been related to crack

width, but more recent studies suggest that the factors governing the eventual development of

corrosion are independent of the crack width, although the period of time required for corrosion

to start is a function of crack width.

9-3 Does the beam shown in Fig. P9-3 satisfy the ACI Code crack control provisions

(Section 10.6.4)? .

ACI Code Eq. (10-4) : 40,000 40,000

15 2.5 12c

s s

s cf f

using 1.5 in. (3 / 8) in. 1.875 in.cc

2 2 60,000 psi=40,000 psi3 3s yf f ,

(The value for sf can be also calculated as the stress in the reinforcement closest

to the tension face at service load based on the unfactored moment.)

So,

40,000 40,00015 2.5 1.875 10.3 in.< 12 12 in. OK.

40,000 40,000s

From Fig. P9-3, bar spacing

1312 2 1.5 1.128

8 22.37 in. < 10.3 in.

3

Thus, the beam satisfies ACI Code Section 10.6.4.

9-2

9-4 Compute the maximum spacing of No. 5 bars in a one-way slab with 1 in. of clear

cover that will satisfy the ACI Code crack-control provisions (Section 10.6.4).

.

ACI Code Eq. (10-4) : 40,000 40,000

15 2.5 12c

s s

s cf f

using 1.0 in..cc

2 2 60,000 psi=40,000 psi3 3s yf f

So,

40,000 40,00015 2.5 1.0 12.5 in.> 12 12 in.

40,000 40,000s

Thus, the maximum spacing for the No. 5 bars that satisfies ACI Code Section 10.6.4 is 12 in.

9-5 and 9-6 For the cross sections shown in Figs. P9-5 and P9-6, compute:

(a) the gross moment of inertia, ;

(b) the location of the neutral axis of the cracked section and ; and

(c) for .

The beams have a 1.5 in. of clear cover and No. 3 stirrups. The concrete strength is

; .

9-5 (a) the gross moment of inertia, ; 3

416 2418432 in.

12gI

(ignoring the effect of reinforcement for simplicity)

(b) the location of the neutral axis of the cracked section and ; The distance from the extreme tension edge of the section to the centroid of the lowest

layer of steel is 13 81.5 in.+ in.+ in. 2.38 in.8 82

Assuming that the spacing between the

centers of the layers is 2 in., find:

Centroid = 4 0.79 2.38 2 0.79 4.38

3.05 in. above bottom4.74

and 20.95 in.d

629 10 psi8.04

57000 4000 psi

s

c

En

E

Transformed area of steel = 28.04 6 0.79 38.11 in.

Compute location of neutral axis

Let depth of neutral axis be c and sum moments about the neutral axis to zero.

Part Area, in.2 y , in. Ay ,in.

3

Compression

zone 16c 2c 28c

Tension steel 38.11 20.95c 38.11 798.4c

28 38.11 798.4 0 and 7.89 in.Ay c c c

Thus, the neutral axis is 7.89 in. below the top of the beam

9-3

Compute Icr

Part Area, in.2 y , in.

ownI 2Ay ,in.4

Compression

zone 126.2 3.94 654.9 1959.1

Tension steel 38.11 -13.06 6500

crI 9114 in.4

(c) for

First calculate nM for the beam:

4.74 600005.23 in.

1 ' 0.85 4000 160.85

A fs y

cf bc

and 5.23 6.150.85

c in.

20.95 6.150.003 0.00722 0.002 O.K.

6.15

d c

s cuc

and 0.005 0.90 t

5.234.74 60000 20.95

25214 kips-in.

2 1000

aM A f d

n s y

and 0.9 5214 4693 kips-in.nM

Thus, -

47.5 4000 psi 18432 in.728.6 kips-in.

lbs12 in. 1000

kips

r g

cr

t

f IM

y

So, using Eq. (9-10a) find

( )

[ ( )

] (

)

[ (

)

]

9-6 (a) the gross moment of inertia, ;

Part Area, in.2

topy , in. topAy ownI , in.

4 2Ay ,in.4

Web 288 12 3456 13824 2592

Flanges 144 3 432 432 5184

432 3888 gI 22032 in.4

38889 in. 15 in.

432top btmy y

422032 in.gI

9-4

(b) the location of the neutral axis of the cracked section and ; 629 10 psi

8.0457000 4000 psi

s

c

En

E


24 in. 2.5 in. 21.5 in.d

Assume that the neutral axis is less than 6 in.

Compute location of neutral axis.


3

Compression

zone 36c 2c 218c

snA 25.4 21.5c 25.4 546c

218 25.4 546 0 and 4.85 in.Ay c c c

Thus, the neutral axis is 4.85 in. below the top of the beam (i.e. the compression zone is

rectangular).

Compute Icr


ownI 2Ay ,in.4

Compression

zone 174.6 2.42 342 1022.5

snA 25.4 -16.65 7041

crI 8405 in.4

(c) for

First calculate nM for the beam:

3.16 600001.55 in.

1 ' 0.85 4000 360.85

A fs y

a cf bc

and 1.55 1.820.85

c in.

The steel is yielding and it is a tension-controlled section 0.90 .

1.553.16 60000 21.5

23930 kips-in.

2 1000

aM A f d

n s y

and 0.9 3930 3537 kips-in.nM

Thus, -

47.5 4000 psi 22032 in.697 kips-in.

lbs15 in. 1000

kips

r g

cr

t

f IM

y

So, using Eq. (9-10a) find

( )

[ ( )

] (

)

[ (

)

]

9-5

9-7 Why are deflections limited in design?

Deflections are limited for several reasons.

1. Deflections greater than 250 of the span are visible and may be unsightly.

2. Excessive deflections may cause cracking of partitions, malfunctioning of doors and windows

and similar damage to non-structural elements.

3. Excessive deflections may interfere with the use of the structure, particularly if the structure

supports machinery that must be carefully aligned.

4. Ponding of water on deflected roofs may overload the roofs.

5. Very large deflections may damage structural members and change the load path.

9-8 A simply supported beam with the cross section shown in Fig. P9-5 has a span of 25

ft and supports an unfactored dead load of 1.5 kips/ft, including its own self-weight

plus an unfactored live load of 1.5 kips/ft. The concrete strength is 4500 psi.

Compute

(a) the immediate dead load deflection.

(b) the immediate dead-plus-live load deflection.

(c) the deflection occurring after partitions are installed. Assume that the

partitions are installed one month after shoring for the beam is removed and

assume that 20 percent of the live load is sustained.


From question 9-5 we found: 418432 in.gI , 728.6 kips-in.crM

629 10 psi7.58

57000 4500 psi

s

c

En

E


Compute location of neutral axis

Let depth of neutral axis be c and sum moments about the neutral axis to zero.


3

Compression

zone 16c 2c 28c

Tension steel 35.93 20.95c 35.93 752.7c

28 35.93 752.7 0 and 7.71 in.Ay c c c

Thus, the neutral axis is 7.71 in. below the top of the beam

Compute Icr


ownI 2Ay ,in.4

Compression

zone 123.4 3.85 611 1829

Tension steel 35.93 -13.24 6298

crI 8738 in.4

9-6

Unfactored dead load moment: 2 21.5 kip ft 25 ft

117.2 kips-ft 1406 kips-in.8

DLM

Thus, a crM M cracked section and need to calculate effI

3 3 3 3

4728.6 728.61 18432 1 8738 10087 in.

1406 1406

cr cre g cr

a a

M MI I I

M M

This is a simply supported beam with distributed loading, so using deflection Case 1 from Fig. 9-

13, the immediate dead load deflection can be calculated as:

22

pos

iD 6

1406 1000 25 125 50.342 in.

48 48 3.824 10 10087

M

EI


Unfactored dead plus live load moment:

2 23.0 kip ft 25 ft in.12 2812 kips-in.

8 ftD LM

3 3

4728.6 728.618432 1 8738 8907 in.

2812 2812eI

So using again the deflection Case 1 from Fig. 9-13, the immediate dead-plus-live-load deflection

can be calculated as:

2

iD+L 6

2812 1000 25 1250.774 in.

48 3.824 10 8907




The deflection occurring after the partitions are installed can be calculated from Eq. (9-14):

iL 0 iD iLS,t

The immediate dead load deflection, iD , was found from part (a) to be 0.342 in. However, after

the live load has been applied and the beam has cracked, the deflection due to dead load will be

increased by an amount equal to the ratio of the eI values used in part (a) and (b). Thus the

immediate dead load deflection on the structure which has been loaded to D+L will be calculated

and used in Eq. (9-14).

iD

100870.342 =0.387 in.

8907

The immediate live load deflection, iL , is found as:

iL i,L+D iD 0.774 in. 0.387 in. 0.387 in.

Twenty percent of this results from sustained live loads, so:

iLS 0.20 0.387 in. 0.077 in.

9-7

Since the beam has no compression reinforcement, and with the partitions installed 1 month after

removal of the shoring,

( )

The deflection occurring after the partitions are installed is found as:

9-9 Repeat Problem 9-8 for the beam section in Fig. P9-6.


From question 9-6 we found:

,

,

, and

The unfactored dead load moment: 2 21.5 kip ft 25 ft

117.2 kips-ft 1406 kips-in.8

DLM

Thus, a crM M cracked section and need to calculate effI

( )

[ ( )

] (

)

[ (

)

]

This is a simply supported beam with distributed loading, so using deflection Case 1 from Fig. 9-

13, the immediate dead load deflection can be calculated as:

( )


Unfactored dead plus live load moment:

2 23.0 kip ft 25 ft in.12 2812 kips-in.

8 ftD LM

(

)

[ (

)

]

So using the deflection Case 1 from Fig. 9-13, the immediate dead-plus-live-load deflection can

be calculated as:

( )





iL 0 iD iLS,t



9-8





Twenty percent of this results from sustained live loads, so:

Since the beam has no compression reinforcement, and with the partitions installed 1 month after

removal of the shoring,

( )


9-10 The beam shown in Fig. P9-10 is made of 4000-psi concrete and supports unfactored

dead and live loads of 1 kip/ft and 1.1 kips/ft. Compute:

(a) the immediate dead-load deflection.

Compute Ig for the T-section (ignore the effect of the reinforcement for simplicity):

Assume flange width = effective flange width from ACI Code Section 8.12.2. = 72 in. 629 10 psi

8.0457000 4000 psi

s

c

En

E

Part Area, in.2


4 2Ay ,in.4

Web 240 10 2400 8000 4234

Flanges 360 3 1080 1080 2822

600A 3480Ay gI 16136 in.4

34805.8 in. 14.2 in.

600top btmy y

Compute Icr at the left end:

The positive-moment reinforcement is not developed for compression at the support and will

therefore be neglected. The section is a rectangular section, with 3 No. 7 bars , 17.5 in.d ,and

the following properties: 2 21.80 in. , 14.47 in.

1.800.0086, 0.069

12 17.5

s sA nA

n

Using Eq. (9-3): 2

2 0.309 and 5.41 in.k n n n c kd

9-9


ownI 2Ay ,in.4

Compression

zone 64.9 2.71 158 477

snA 14.47 5.41 17.5 12.1 2119

crI 2754 in.4

Compute Icr at the right end:

2 23.16 in. , 25.41 in.

3.160.0150, 0.121

12 17.5

s sA nA

n

Using Eq. (9-3): 2

2 0.386 and 6.75 in.k n n n c kd


ownI 2Ay ,in.4

Compression

zone 81 3.37 307.5 920

snA 25.41 6.75 17.5 10.75 2936

crI 4163 in.4

Compute Icr at the midspan:

2 22.18 in. , 17.53 in.s sA nA

Assuming that c is less than 6 in.


3

Compression

zone 72c 2c 236c

snA 17.53 17.5c 17.53 307c

236 17.53 307 0 and 2.7 in.Ay c c c


ownI 2Ay ,in.4

Compression

zone 194 1.35 118 354

snA 17.53 2.7 17.5 14.8 3840

crI 4312 in.4

The moment at the end and the midspan can be calculated using the ACI Moment Coefficients

9-10

Unfactored dead load moments

Negative moment at left end = 2

211 22.67 32.1 kips-ft

16 16

nw

Positive moment at midspan = 2

211 22.67 36.7 kips-ft

14 14

nw

Negative moment at right end = 2

211 22.67 51.4 kips-ft

10 10

nw

Cracking moments

7.5 4000 psi 474 psirf

Ends: 474 16136

110 kips-ft5.8 12,000

crM

Midspan: 474 16136

45 kips-ft14.2 12,000

crM

Compute the average effective moment of inertia

for both ends and the midspan, a crM M and thus the sections are uncracked under dead loads

and 416136 in.eff gI I

Compute immediate dead-load deflection

This is an exterior beam span with a column as the exterior support. Therefore, the deflection can

be calculated using row 3 of Table 9-2, which requires a combination of Cases 2 and 8 from Fig.

9-13.

For deflection Case 2:

44

6

10001 24 1212

=0.0533 in.185 185 3.6 10 16136

w

EI

(down)

For deflection Case 8:

22

6

32.1 12000 24 12= 0.0343 in.

16 16 3.6 10 16136

M

EI

(up)

Thus, iD 0.0533 0.0343 0.019 in.


Unfactored dead plus live-load moments

Negative moment at left end = 2

211 1.1 22.67 67.4 kips-ft

16 16

nw

Positive moment at midspan = 2

211 1.1 22.67 77.1 kips-ft

14 14

nw

Negative moment at right end = 2

211 1.1 22.67 107.9 kips-ft

10 10

nw

Compute the average effective moment of inertia

left end: a crM M416136 in.e gI I

midspan: a crM M

3 3

445 4516136 1 4312 6663 in.

77.1 77.1eI

9-11

right end: a crM M416136 in.e gI I

Thus, the weighted average value of effI using Eq. (9-11a) is:

4( ) 0.70 6663 0.30 16136 9505 in.e avgI

Immediate plus live-load deflection

Using the same procedure as in part (a),

for deflection Case 2:

44

6

10002.1 24 1212

=0.190 in.185 185 3.6 10 9505

w

EI

(down)

for deflection Case 8:

22

6

67.4 12000 24 12= 0.122 in.

16 16 3.6 10 9505

M

EI

(up)

Thus, iD+L 0.190 0.122 0.068 in.


partitions are installed two months after the shoring is removed and

assumed that 15 percent of the live load is sustained.


iL 0 iD iLS,t






iD

161360.019 =0.032 in.

9505


iL i,L+D iD 0.068 in. 0.032 in. 0.036 in.

Fifteen percent of this results from sustained live loads, so:

Since the beam has no compression reinforcement at midspan, and with the partitions installed 2

months after removal of the shoring,

( )


10-1

Chapter 10

10-1 A five-span one-way slab is supported on 12-in.-wide beam with center-to-center

spacing of 16 ft. The slab carries a superimposed dead load of 10 psf and a live load

of 60 psf. Using and , design the slab. Draw a cross

section showing the reinforcement. Use Fig. A-5 to locate the bar cut-off points.

slab design strip

Plan

12 in.16 ft. 15 ft.

in.12

ft

n

Estimate slab thickness

Assume partitions are not sensitive to deflections. Will require recheck if sensitivity is

established later.

Table A-9:

End bay: 15 12Min 7.50 in.24 24

nh

Interior bay: 15 12Min 6.43 in.28 28

nh

Note that slab thickness was chosen on the basis of deflection control, since flexure and shear

probably won’t govern the design (this will be checked later).

Try 7.0 in.h (this may need to be checked for deflections in the end span).

Assuming a cover of 0.75 in. and No. 4 bars as the slab reinforcement,

0.57.0 0.75 6.0 in.2

d

Compute factored loads

Considering a 1-ft wide strip of slab:

Slab self weight: 7.0

150 87.5 psf12

Dsw

Superimposed dead load: 10 psfDiw

Total dead load: 87.5 10 97.5 psfDw

10-2

Live load: Factored load:

Load per foot along design strip

3L Dw w , so we can use the ACI Moment coefficients for the calculation of the positive and the

negative moments.

Thickness for flexure

The maximum value for uM is at the first interior support since 15 ft.n throughout. Using the

appropriate moment coefficient from ACI Code Section 8.3.3,

( )

For a reinforcing ratio of 0.01 , which is a reasonable upper limit for a slab, the reinforcing

index can be found from Eq. (5-21),

0.01 600000.15

4000

From Eq. (5-22) calculate the flexural resistance factor, R.

0.15 4000 1 0.59 0.15 547 psiR

Using this value of R, the required value of 2bd can be determined using Eq. (5-23a), assuming

that 0.9 (will check it later).

For , √

Therefore, the minimum d to keep 0.01 is Actual 6 in.d will be less

than 0.01 (O.K. for flexure).

Thickness for shear

The max shear uV is at the exterior face of the first interior support. Using the appropriate shear

coefficient from ACI Code Section 8.3.3,

( √ ) ( √ ) OK

So, use a 7 in. slab.

Flexural reinforcement

Max 4.8 kips-ft/ftuM

Assuming that 0.95 and 2

s y

ad d

, find the required reinforcement for a 1-ft wide strip

of slab.

( )

( )

10-3

Iterate to find the depth of the compression stress block and recompute the value of the required

reinforcement:

Since the depth to the neutral axis, c, is less than 3 8 of d , the section is tension controlled

, 0.9s y .

The minimum reinforcement required by ACI Code Section 10.5.4, is 2

,min 0.0018 0.0018 12 7 0.15 in. /ftsA bh

The maximum spacing of the bars is, by ACI Code Section 7.6.5,

max

3 21 in.

18 in

hs

Select No. 4 bars at 12 in.

⁄

⁄

Temperature and shrinkage steel as required by ACI Code Section 7.12.2,

2,min 0.0018 0.15 in. /ftsA bh and max

5 35 in.

18 in

hs

So provide No. 4 bars at 16 in. 2

20.2 in. in.12 0.15 in.ft ft16 in.

sA

The flexural reinforcement for the supports and the midspan for all the spans is calculated in the

following table.

10-4

Calculation of reinforcement required in the slab.

1. ft.n 15.0 15.0

2. 2u nw 47.9 47.9 47.9 47.9 47.9 47.9

3. Moment Coef. 1 24 1 14 1 10 1 11 1 16 1 11 1 16

4. kips-ft/ftuM 2.0 3.4 4.8 4.4 3.0 4.4 3.0

5. 2 reqd. in. /ftsA 0.08 0.13 0.19 0.12 0.17 0.12

6. 2,min in. /ftsA 0.15 0.15 0.15 0.15 0.15 0.15

7. Reinforcement #4 @

16 in.

#4 @

16 in.

#4 @

12 in.

#4 @

16 in.

#4 @

12 in.

#4 @

16 in.

8. 2 provided in. /ftsA

0.15 0.15 0.2 0.15 0.2 0.15

Fig. S10-1.1 shows a cross-section of the slab showing the reinforcement. The bar cut-off points

were located using Fig. A-5(c).

Fig. S10-1.1 Slab reinforcement detailing.

16”

12”

12”

10-5

10-2 A four-span one-way slab is supported on 12-in.-wide beams with center-to-center

spacing of 14, 16, 16, and 14 ft. The slab carries a superimposed dead load of 20 psf

and a live load of 100 psf. Design the slab, using and

. Select bar cut-off points using Fig. A-5 and draw a cross-section

showing the reinforcement.

slab design strip

Plan

Short span clear length: 12 in.14 ft. 13 ft.in.

12ft

n

Long span clear length: 12 in.16 ft. 15 ft.in.

12ft

n

Average clear span length for first interior support:

,

13 ft. 15 ft.14 ft.

2n avg

Estimate slab thickness

Assume partitions are not sensitive to deflections. Will require recheck if sensitivity is established

later.

Table A-9:

End bay: 13 12Min 6.50 in.24 24

nh

Interior bay: 15 12Min 6.43 in.28 28

nh

Note that slab thickness is chosen on basis of deflection control, since flexure and shear probably

won’t govern the design (will be checked later).

Try 6.5 in.h

Assuming a cover of 0.75 in. and No. 4 bars as the slab reinforcement,

0.56.5 0.75 5.5 in.2

d

10-6

Compute factored loads

Considering a 1-ft wide strip of slab:

Slab self weight: 6.5

150 81.25 psf12

Dsw

Superimposed dead load: 20 psfDiw

Total dead load: 81.25 20 101.25 psfDw

Live load: Factored load: Load per foot along design strip =

3L Dw w , so we can use the ACI Moment coefficients for the calculation of the positive and the

negative moments (ACI Code Section 8.3.3).

Thickness for flexure

The maximum moment will occur at either:

(1) exterior face of the first interior support, or

(2) face of the middle support

For negative moments at the face of an interior support, ACI Code Section 8.0 defines n as the

average of the clear spans of the two adjacent spans. Using the appropriate moment coefficient

from ACI Code Section 8.3.3,

( )

( )

For a reinforcing ratio of 0.01 , which is a reasonable upper limit for a slab, the reinforcing

index can be found from Eq. (5-21),

0.01 600000.171

3500


0.171 3500 1 0.59 0.171 538 psiR



For , √

i.e., min d to keep 0.01 is Actual 5.5 in.d will be less than 0.01 (O.K. for

flexure).

Thickness for shear

The max shear uV will occur in one of the two locations discussed for the maximum moments.

Using the appropriate shear coefficient from ACI Code Section 8.3.3,

10-7

'0.75 2 0.75 2 3500 12 5.5 5860 lbs/ft ok for shearc c wV f b d

Flexural reinforcement

Max -

Assuming that 0.95 and 2

s y

ad d

, find the required reinforcement for a 1-ft wide strip

of slab.

( )

( )


reinforcement:

Since the depth to the neutral axis, c, is less than 3 8 of d , the section is tension controlled

, 0.9s y .

The minimum reinforcement required by ACI Code Section 10.5.4, is 2

,min 0.0018 0.0018 12 6.5 0.14 in. /ftsA bh

The maximum spacing of the bars is, by ACI Code Section 7.6.5,

max

3 19.5 in.

18 in

hs

Select No. 4 bars at 9 in.

⁄

⁄

Temperature and shrinkage steel as required by ACI Code Section 7.12.2,

2,min 0.0018 0.14 in. /ftsA bh and max

5 32.5 in.

18 in

hs

So provide No. 4 bars at 16 in. 2

20.2 in. in.12 0.15 in.ft ft16 in.

sA

The flexural reinforcement for the supports and the midspan for all the spans is calculated in the

following table.

10-8

Calculation of reinforcement required in the slab.

1. n 13.0 13.0 14.0 15.0 15.0 15.0

2. 2 kips-ftu nw 47.7 47.7 55.3 63.5 63.5 63.5

3. Moment Coef. 1 24 1 14 1 10 1 11 1 16 1 11 1 16

4. kips-ft/ftuM 2.0 3.4 5.5 5.0 4.0 5.8 4.0

5. 2 reqd. in. /ftsA 0.09 0.14 0.23 0.17 0.25 0.17

6. 2,min in. /ftsA 0.14 0.14 0.14 0.14 0.14 0.14

7. Reinforcement #4 @

16 in.

#4 @

16 in.

#4 @

9 in.

#4 @

11 in.

#4 @

9 in.

#4 @

11 in.

8. 2 provided in. /ftsA 0.15 0.15 0.27 0.18 0.27 0.18

Fig. S10-2.1 shows a cross-section of the slab showing the reinforcement. The bar cut-off points

were located using Fig. A-5(c).

Fig. S10-2.1 Slab reinforcement detailing.

16”

9”

11”

10-9

10-3 A three span continuous beam supports 6-in.-thick one-way slabs that span 20 ft

center-to-center of beams. The beams have clear spans, face-to-face of 16-in.-square

columns, of 27, 30, and 27 ft. The floor supports ceiling, ductwork, and lighting

fixtures weighing a total of 8 psf, ceramic floor tile weighting 16 psf, partitions

equivalent to a uniform deal load of 20 psf, and a live load of 100 psf. Design the

beam, using . Use for flexural reinforcement and

for shear reinforcement. Calculate cut-off points, extending all

reinforcement past points of inflection. Draw an elevation view of the beam and

enough cross-sections to summarize the design.

1. Compute the trial factored load on the beam

(a) Dead load

slab self-weight: 6 in.

150 pcf 75 psfin.12

ft.

ceiling, tile, partitions: 8 psf 16 psf 20 psf 44 psf

The beam size is not known at this stage, so it must be estimated for preliminary design purposes.

Once the size of the beam has been established, the factored load will be corrected and then used

in subsequent calculations. The beam size will be estimated in step 2.

(b) Live Load

The ASCE/SEI 7-10 recommendations allow live-load reductions based on tributary area

multiplied by a live-load element factor, 2LL , to convert the tributary area to an influence

area.

Positive moment at span AB and negative moment at exterior support, A.

215 ft. 27 ft. 405 ft.TA

15 150.25 100 0.25 100 0.777 77.7 psf > 0.5 100 psf

2 405o

LL T

L LK A

10-10

Note that L shall not be less than 0.50 oL for members supporting one floor (O.K.)

Negative moment at interior support, B.

215 ft. 27 30 ft. 855 ft.TA

15 15

0.25 100 0.25 100 0.613 61.3 psf > 0.5 100 psf (O.K.) 2 855

o

LL T

L LK A

Positive moment at span BC. 215 ft. 30 ft. 450 ft.TA

15 150.25 100 0.25 100 0.75 75.0 psf > 0.5 100 psf (O.K.)

2 450o

LL T

L LK A

The size of the beam will be chosen on the basis of negative moment at the first interior support.

For this location, the factored load on the beam, not including the beam stem below the slab, is:

1.2 75 44 1.6 61.3 241 psfuw

The tributary width for the beam is 15 ft and the factored load from the slab per foot of beam is

kipslbs241 psf 15 ft. 3,615 3.6ft ft

Two approximate methods can be used to estimate the weight of the beam stem:

(a) the factored dead load of the stem is taken as 12 to 20 percent of the other factored loads

on the beam. This gives 0.43 to 0.72 kip/ft.

(b) the overall depth of beam h is taken to be 1/18 to 1/12 of the larger span, , and wb is

taken to be 0.5h . This gives the overall h as 20 to 30 in., with the stem extending 14 to 24 in.

below the slab, and gives wb as 10 to 15 in. The factored load of such sizes ranges from 0.17 to

0.45 kip/ft.

As a first trial, assume the factored weight of the stem to be 0.50 kip/ft. Then, total trial load per

foot 3.6 0.5 4.1 kip/ft

2. Estimate the size of the beam stem

(a) Calculate the minimum depth based on deflections.

ACI Table 9.5(a) (Table A-9) gives the minimum depths, unless deflections are checked. For

partitions flexible enough to undergo some deflection, minimum depth for beam BC is

min18.5

h , where the span center-to-center of supports16 in.

27 ft 28.3 ftin.12

ft

Thus, min

28.3 1218.4 in.

18.5h

10-11

(b) Determine the minimum depth based on the negative moment at the exterior face of the

first interior support.

The beam fits the requirements in ACI Code Section 8.3.3 and can use the moment and shear

coefficients. For the support at B,

22 4.1 28.5

333 kips-ft10 10

u nu

wM

where 27 ft 30 ft

28.5 ft2

n

(the average of the two adjacent spans).

Using the procedure developed in Chapter 5 for the design of singly reinforced beam sections, the

reinforcement ratio that will result in a tension-controlled section can be estimated from

Eq. (5-18) as,

'

1 0.825 4.5initial 0.0155

4 4 60

c

y

f

f

For this reinforcement ratio, use Eq. (5-21), to find the reinforcing index,

0.0155 600.207

4.5


0.207 4500 1 0.59 0.207 818 psiR



2 333312,000 5430 in.

0.9 818bd

Since columns are 16 in., try a 14 or 16 in. wide stem. Let’s try 14 in.b

Then, 5428

19.7 in. 21.5 in.14

d d

With one layer of steel at supports, 21.5 2.5 24 in.h (O.K. for deflections).

So, try a 14-in. wide-by-24-in. beam.

(c) Check the shear capacity of the beam

Maximum shear is at the exterior face at support B,

,max

1.15 1.15 4.1 2763.6 kips

2 2

u nu

wV

From ACI Code Section 11.2.1.1,

' 12 2 1 4500 14 21.5 40.4 kips1000c c wV f b d

ACI Code Section 11.4.7.9 sets the maximum nominal sV is

' 18 8 4500 14 21.5 162 kips1000s c wV f b d

Thus, 0.75 40.4 161 151 kipsnV (O.K. for shear)

10-12

(d) Summary

Use :

14 in.

24 in. 18 in. below slab

21.5 in.,assuming one layer of steel

b

h

d

3. Compute the dead load of the stem, and recompute the total load per foot.

Weight per foot of the stem below slab 18 14

0.15 0.26 kip/ft144

Corrected total factored load for 1st internal support moment: 3.6 1.2 0.26 3.9 kip/ft

Since this is less that the 4.1 kip/ft used earlier to estimate the beam size, the section chosen will

be adequate.

Factored Total dead load:

11.2 119 15 0.26 2.5 kips/ft1000Dw

Factored total loads:

(a) Positive moment at span AB and negative moment at exterior support, A.

12.5 1.6 77.7 15 4.4 kips/ft1000uw

(b) Negative moment at interior support, B.

12.5 1.6 61.3 15 4.0 kips/ft1000uw

(c) Positive moment at span BC.

12.5 1.6 75.0 15 4.3 kips/ft1000uw

4. Calculate the beam flange width for positive-moment regions

From ACI Code Section 8.12.2,

0.25 based on the shorter span for simplicity 0.25 27 12 in. 81 in.

2 8 6 14 2 8 6 110 in.

15 12 180 in.

n

e wb b

Therefore, the effective flange width is 81 in. and shown in Fig. S10-3.1

Fig. S10-3.1 Beam cross-section

10-13

5. Can we use the ACI Code Moment Coefficients?

Ratio of successive spans = 30 ft. / 27 ft. = 1.11<1.20 O.K. ;

Live load/ft 0.1 15

0.6 3.0 O.K.Dead load/ft 2.5

;

There are more than two spans;

The loads are uniformly distributed.

Thus, we can use the ACI Code Moment Coefficients

6. Compute the beam moments

, ft.n 27.0 27.0 28.5 30.0

, kip/ft.uw 4.4 4.4 4.0 4.3

2 ,kips-ftu nw 3210 3210 3250 3870

mC 116

114

1 1 and 10 11

116

2 ,kips-ftu m u nM C w -200 230 -325 242

7. Design the flexural reinforcement

(a) Max negative moment (first interior support)

Max _,max 325 kips-ftuM

Because the beam acts as a rectangular beam with compression in the web, we can assume that

0.902

ad d

.For and =0.9s y , the required reinforcement for that section is,

212,000 325 12,000

3.73 in.0.90 60,000 0.90 21.5

2

us

y

MA

af d


reinforcement: '

3.73 60000 4.184.18 in., 5.07 in.0.8250.85 4500 140.85

s y

c

A fa c

f b

10-14

Since the depth to the neutral axis, c, is less than 3 8 of d , clearly the section is tension

controlled , 0.9s y , and

2,

325 12,0003.72 in.

4.180.90 60,000 21.5

2

s redA

The other negative moment sections have a lower design moment, so it will be conservative to

use the ratio of s

u

AM

obtained here to quickly determine the area of tension steel required at

those other locations. That ratio is

2

23.72 in.0.0114 in. /kips-ft (eq. A)

325 kips-ft

s

u

A

M

(b) Max positive moment

Max ,max 242 kips-ftuM

Because the beam acts as a T-shape beam with compression in the top flange, assume that the

compression zone is rectangular, i.e. 6 in.fa h , use 0.952

ad d

.For and =0.9s y ,

the required reinforcement that section is,

212,000 242 12,000

2.63 in.0.90 60,000 0.95 21.5

2

us

y

MA

af d


reinforcement:

'

2.63 60000 0.510.51 in., 0.62 in.0.8250.85 4500 810.85

s y

c e

A fa c

f b

The section is tension controlled , 0.9s y , and doing one iteration for the negative

moment section results in 22.53 in.sA

The other positive moment section has a lower design moment, so it will be conservative to use

the ratio of s

u

AM

obtained here to quickly determine the area of tension steel required at those

other locations. That ratio is

2

22.53 in.0.0104 in. /kips-ft (eq. B)

242 kips-ft

s

u

A

M

10-15

(c) Calculate the minimum reinforcement


'

,min

3 200 and

c ws w

y y

f b dA b d

f f .

For 4500 psi concrete, '3 201 psicf , thus

2,min

3 450014 21.5=1.0 in.

60,000sA

(d) Calculate the area of steel and select the bars.

The remaining calculations are done in the following table.

,kips-ftuM -200 230 -325 242

sA coef. Eq. A and B 0.0114 0.0104 0.0114 0.0104 2

( .) ,in.s redA 2.28 2.39 3.70 2.52

.mins sA A Yes Yes Yes Yes

Bars selected 4 No. 7 4 No. 7 2 No. 8 and

4 No. 7

3 No. 7 and

2 No.6

sA provided, 2in. 2.40 2.40 3.98 2.68

min14 in.wb b Yes Yes

Note that in the negative moment regions some of the bars can be placed in the slab besides the

beam and it is not necessary to check whether they will fit into the web width.

8. Check the distribution of the reinforcement

(a) Positive moment region

From ACI Code Section 10.6.4, the maximum bar spacing is

40,000 40,00015 2.5 12c

s s

s cf f

, where

2 40,000 psi and 1.5 in. cover 0.375 in. stirrups 1.875 in.3s y cf f c

Thus, 15 2.5 1.875 10.3 in. 12 in.s

10-16

Bar spacing 714 2 1.5 0.375

83.1 in. <10.3 in. OK.

3

It was also clear that the bar spacing is smaller than 10.3 in., since there are four bars and

14 in.wb

(b) Negative moment region

ACI Code Section 10.6.6 says “part” of the negative moment steel shall be distributed over a

width equal to the smaller of the effective flange width (81 in.) and 10 34.2 in.n At the

interior supports, there are 6 top bars. Place the two No. 8 bars at the corners of the stirrups, two

No. 7 bars over the beam web, and the other two No. 7 bars in the slab. Within a width of 34.2

in. we must place six bars. These cannot be further apart than 10.3 in. (as calculated in part a).

We shall arbitrarily place two bars at 5 in. outside the web of the beam.

ACI Code Section 10.6.6 requires “some” longitudinal reinforcement in the slab outside this

band. We shall assume that the shrinkage and temperature steel in the slab will satisfy this

requirement.

9. Design the shear reinforcement

The shear force diagrams are calculated in the following table and shown at the bottom of the

table. The shear coefficients for the supports are from ACI Code Section 8.3.3 and the coefficient

fro the midspan of the beam is based on Eq. (6-26).

,ftn 27 30

,kips/ftuw 4.4 4.3

,kips/ftLuw 1.9 1.8

vC at support and

midpsan 1.0 0.125 1.15 1.0 0.125 1.0

2u nw 59.4 59.4 64.5 64.5

Lu nw 51.3 54

, kipsuV 59.4 6.4 68.3 64.5 6.7 64.5

n uV V 79.2 8.5 91.1 86.0 8.9 86.0

10-17

(a) Exterior end of beam AB

Because the beam is supported by a column, the critical section is located at d away from the face

of the support.

Equation for uV : kips 5.24 ft 79.2uV x

21.5 in.

@ from A 5.24 79.2 69.8 kips in.12

ft

uV d

12 1 4500 14 21.5 40.4 kips1000cV , 20.2 kips

2

cV

Because 69.8 kipsuV

exceeds 20.2 kips

2

cV , stirrups are required.

Try No. 3 Grade 40 stirrups double-leg stirrups with a 90o hook 22 0.11 0.22 in.vA

Max spacing: '4uc c w

VV f b d

and from ACI Code Section 11.4.5.1,

max

10.75 in.10.75 in.2

24 in.

d

s

To satisfy the minimum stirrup requirement in ACI Code Section 11.4.6.3, the stirrup spacing

must be,

max

0.22 40,00012.6 in.

50 50 14

v yt

w

A fs

b

Note that 0.75 4500 50.3 psi > 50 psi, so use 50 psi in ACI Code Eq. (11-13) .

Thus, use 10.5 in. as maximum stirrup spacing.

The spacing required to support the shear force at the support is,

0.22 40 21.56.45 in.-say 6 in. on centers

69.8 40.4s

We can change the stirrup spacing to 10.5 in. when 0.22 40 21.5

40.4 58.4 kips10.5

uV

This occurs at about 4 ft from face of support A.

We can stop the stirrups when 11.3 ft2

u cV Vx

from face of support A.

Place the first stirrup at 3 in. from support A, then 9 stirrups at 6 in. and 9 stirrups at 10.5 in.

10-18

(b) Interior end of beam AB


21.5 in.

@ from B 6.12 91.1 80.1 kips > ,stirrups required in. 212

ft

cu

VV d

The spacing required to at this point is,

0.22 40 21.54.8 in.-say 4.5 in. on centers

80.1 40.4s

Change the stirrup spacing to 10.5 in. when 58.4 kipsuV

.

This occurs at 5.3 ft from face of support B.

We can stop the stirrups at 11.6 ft from face of support B.

Place the first stirrup at 2 in. from support B, then 15 stirrups at 4.5 in. and 9 stirrups at 10.5 in.

(c) Ends of beam BC


21.5 in.

@ from C 5.14 86 76.8 kips > ,stirrups required in. 212

ft

cu

VV d

The spacing required to at this point is,

0.22 40 21.55.2 in.-say 5.0 in. on centers

76.8 40.4s

Change the stirrup spacing to 10.5 in. when 58.4 kipsuV

.

This occurs at 5.4 ft from face of support.

We can stop the stirrups at 12.8 ft from face of support.

Place the first stirrup at 2.5 in. from support B, then 14 stirrups at 4.5 in. and 9 stirrups at 10.5 in.

10. Bar cutoffs

(a) Detailing requirements:

21.5 in.

12 12 in. for No. 8 bar

20.25 in. for 27 ft span, 22.5 in. for 30 ft span16

b

n

d

d

Thus, d exceeds 12 and 16

nbd for AB span, while

16

n governs for span BC.

The bottom and top bars have clear spacing and cover of at least bd and are enclosed by at least

minimum stirrups. Therefore, this is Case 1 in Table 8-1 (ACI Code Section 12.2.2).

10-19

'

33.5 in. for No. 660,000 1.0 1.0

44.7 39.1 in. for No. 720 1 450020

44.7 in. for No. 8

y t e bd b b

c

f dd d

f

(b) Cutoffs for bottom steel

Span AB

4 No. 7-Extend 2 full length into each support, cut off the other two at the positive moment point

of inflection so that extra stirrups are not required.

Exterior end: From Fig. A-2, inflection point at 0.10 27 12 32.4 in. from face of column.

Rule 3-a - Extend 21.5 in. d past the flexural cutoff point, i.e. 32.4 in.-21.5 in = 10.9 in. from

face of column at A. Say 10 in.

Rule 4-a - Distance from midspan to cutoff point greater thand.

Rule 1-b - This is an interior beam with open stirrups. Since this is a discontinuous end use 90

deg. standard hooks on 2 No. 7 bars.

Rule 4-d – At the inflection point, the remaining steel is two No. 7, 21.2 in.sA Thus,

1.2 601.57 in.

0.85 4.5 12a

1.57 1.2 60 21.5 =1490 kip-in=124 kip-ft

2nM

The shear at 32.4 in. from the exterior end is,

32.45.24 79.2 65 kips

12uV

21.5 in.a

Thus, 1490

21.5 44.4 in.65

na

u

M

V

This exceeds d- therefore, OK.

Interior end: From Fig. A-2, inflection point at 0.104 27 12 33.7 in. from face of column

Rule 3-a - Extend bars to 33.7 in.- 21.5 in. = 12.2 in. from face of column. Use 10 in. to match

other end.

Rule 4-a - Satisfied

Rule 1-b - This is an interior beam with open stirrups. Rule 1-b applies. Lap splice 2#7 bars from

the exterior span with 2 No. 7 bars from the interior span with a Class A tension lap splice 1.0 39.1 in. 3 ft - 2 in.

Rule 4-d

1490 kip-in=124 kip-ftnM


33.76.12 91.1 73.9 kips

12uV

21.5 in.a

Thus, 1490

21.5 41.7 in.73.9

na

u

M

V


10-20

Span BC

2 No. 8 and 3 No. 7 at midspan – Extend 2 No. 7 into supports. Cutoff 2 No. 8 and 1 No. 7 bars at

the positive moment point of inflection so that extra stirrups are not required. Inflection point at

0.146 30 12 52.6 in. from face of column

Rule 3-a - Extend 22.5 in. 16

n past the flexural cutoff point, i.e. 52.6 in.-22.5 in. = 30.1 in.

from face of column. Say 30 in.

Rule 4-a - Satisfied.

Rule 1-b – Lap splice 2#7 bars 3 ft - 2 in at support

Rule 4-d

2650 kip-in=221 kip-ftnM


52.65.14 86 63.5 kips

12uV

21.5 in.a

Thus, 2650

21.5 63.3 in.63.5

na

u

M

V


Cutoffs for top steel

Span AB, exterior end, 4 No. 7

Use standard hook, 0.02 1.0 60,000

0.75 13.4 in.1.0 4500

dh

Set tail cover = 2 in., then 0.7 13.4 9.4 in.dh . This is OK in a 16 in. column.

Negative moment point of inflection at 0.164 27 12 53.1 in.

Rule 3-b - Extend d to 53.1+21.5 = 74.6 in. Cutoff at 75 in. = 6 ft-3 in. from face of column.

Rule 4-b - Since 75 in. > 39.1 in. Rule 2 is satisfied.

Span AB, interior end, 2 No. 8 and 4 No. 7 bars

Point of inflection at 0.24 27 12 77.8 in.

Rule 3-b - Extend d to 77.8+21.5 = 99.3 in. Say 8 ft.-4 in. from face of column

Rule 4-b – Since 100 in. > 44.7 in. Rule 2 is satisfied.

Span BC, 2 No. 8 and 4 No. 7 bars

Negative moment point of inflection at 0.24 30 12 86.4 in.

Rule 3-b – Extend to 108.9 in. Say 9 ft- 1 in, from face of column

Rule 4-b – OK

Since all cutoffs are past points of inflection, they are not in zones of flexural tension, therefore

extra stirrups are not needed.

Provide 2 No. 4 top bars as stirrup support, lab splice with negative moment steel.

10-21

Fig. S10-3.2 Beam reinforcing detailing.

10-22

10-4 Repeat Problem 10-3, but cut off up to 50 percent of the negative- and positive-

moment bars in each span where they are no longer required.

Bat cutoff for positive moment steel in AB span

Flexural reinforcement: 4#7 bars

Extend two bars full length and into each support, cutoff the other two where they no longer

required 50% of uM .

From Fig. A-2, flexural cutoff point is at 0.21 0.21 27 12 68 in.n from exterior end

(support A), and 0.22 0.22 27 12 71 in.n from interior end (support B).

Rule 3-a: Extend 21.5 in.d past the flexural cutoff point

68 in. 21.5 in. 46.5 in. from exterior end

71 in. 21.5 in. 49.5 in. from interior end

Rule 4-a: Distance between midspan to cutoff point =

46.5 115.5

13.5 12 in. in. in.49.5 112.5

d for #7 bar 115.5

39.1 in. in.112.5

Bar cutoff for negative moment steel

Flexural reinforcement: 2#8 and 4#7 bars

AB span, interior end

From Fig. A-2, flexural cutoff point is at 0.104 0.104 27 12 34 in.n

Rule 3-b: Extend 21.5 in.d past the flexural cutoff point

34 in.+ 21.5 in. 55.5 in.

Rule 4-b: d for #7 top bar 39.1 in.< 55.5 in.

Therefore, use 55.5 in. for cutoff point.

BC span

From Fig. A-1, flexural cutoff point is at 0.1 0.1 30 12 36 in.n

Rule 3-b: Extend 22.5 in.d past the flexural cutoff point

36 in.+ 22.5 in. 58.5 in.

Rule 4-b: d for #7 top bar 39.1 in.< 58.5 in.

Therefore, use 58.5 in. for cutoff point.

11-1

Chapter 11

11-1 The column shown in Fig. P11-1 is made of 4000 psi concrete and Grade-60 steel.

a) Compute the theoretical capacity of the column for pure axial load.

First compute the gross area of the section, and the area of steel in the column.

218 in 18 in 324 ingA

2 26 6 1.0 in 6 inst bA A

Now we can compute the theoretical capacity of the column.

2 2 20.85 ' 0.85 4 ksi 324 in 6 in 60 ksi 6 in 1440 ko c g st y stP f A A f A

b) Compute the maximum permissiblenP for the column.

,max 0.80 0.80 0.65 1440 k 749 kn oP P

Note that the factors 0.80 and 0.65 would change to 0.85 and 0.75 for spiral columns.

11-2 Why does a spiral improve the behavior of a column?

As any column is loaded, and thus shortened, the concrete will expand laterally. When this

expansion occurs, transverse reinforcement is engaged and will tend to react against any further

expansion of the concrete within the core. This results in a state of tri-axial compression within the

core of the column, which significantly improves both strength and ductility.

The circular spiral is much more effective than tied reinforcement at confining this

expansion for two reasons. First, spirals are often spaced more closely together than tied stirrups,

so the confinement is more uniformly applied to the core. Second, the confinement stresses are

transformed directly into hoop stresses in the spiral, which is a much more efficient mechanism for

reacting to the core’s expansion than the straight legs of tied stirrups can provide (see section 11-2).

11-3 Why are tension splices required in some columns?

Even when columns are subjected to axial loads, reinforcing bars can still often be stressed

in tension when moments are concurrently applied to the section. When this is the case, tension

splices (either Class A or Class B) are required for those bars expected to be resisting tension.

However, since it is most practical from a construction standpoint to use the same length of lap

splices on all bars within a section, all splices should be specified as tension splices when some of

the bars are expected to be in tension.

11-2

11-4 Compute the balanced axial load and moment capacity of the column shown in

Fig. P11-1. Use and .

Assume bending around an axis parallel to the two layers of steel. To calculate the

balanced point of the interaction diagram, set the extreme compression fiber strain to 0.003c and

the extreme steel tensile strain to / 60 ksi / 29,000 ksi 0.00207y yf E .

Begin by calculating the depth of the compression zone:

c

c y

c d

0.003 1.128 in18 in 1.5 in 0.375 in

0.003 0.00207 2c

9.21 inc

1 0.85 9.21 in 7.83 ina c

Now calculate the strain and stress in each layer of steel:

1 0.00207s

1 60 ksisf

Fig. S11-3

2

9.21 in 1.5 in 0.375 in 1.128 in / 2'0.003 0.00221

9.21 ins c

c d

c

2

2

60 ksismaller of

29,000 ksi 0.00221 64.1 ksis

s

fE

Calculate the force in the concrete and in each layer of steel:

0.85 ' 0.85 4 ksi 7.83 in 18 in 479 kc cC f ab

2

1 1 1 3 in 60 ksi 180 ks s sF A f

2

2 2 2 0.85 ' 3 in 60 ksi 0.85 4 ksi 170 ks s s cF A f f

Now we can calculate the nominal axial load and moment at the balanced point:

2 1 479 k 170 k 180 k 469 kn c s sP C F F

1 2 '

2 2 2 2n c s s

h a h hM C F d F d

18 in 7.83 in 18 in 18 in

479 k 180 k 15.56 in 170 k 2.44 in2 2 2 2

nM

4730 k-in 394 k-ftnM

And, since 1s y , the section is compression controlled, and 0.65 .

0.65 469 k 305 knP

0.65 394 k-ft 256 k-ftnM

11-3

11-5 For the column shown in Fig. P11-5, use a strain-compatibility solution to compute

five points on the interaction diagram corresponding to points 1 to 5 in Fig. 11-22.

Plot the interaction diagram. Use and .

Begin by calculating the depth and area of each layer of steel, and other constants.

1 18 in 1.5 in 0.375 in 1.128 in / 2 15.56 ind

2 2

1 3 3 1.0 in 3.0 inbA A

2 / 2 18 in / 2 9.0 ind h

2 2

2 2 2 1.0 in 2.0 inbA A

3 1.5 in 0.375 in 1.128 in / 2 2.44 ind

2 2

3 3 3 1.0 in 3.0 inbA A

218 in 18 in 324 ingA hb

2 28 8 1.0 in 8.0 inst bA A

1 0.80

Point 1: Pure Axial Load:

2 2 20.85 ' ( ) 0.85 5 ksi (324 in 8 in ) 60 ksi 8 in 1830 ko c g st y stP f A A f A

Note that this 0.85 is k3, not β1.

Since the column is compression controlled, 0.65 .

0.65 1830 k 1190 knP

,max 0.80 0.65 1830 k 949 knP

0.65 0 k-ft 0 k-ftnM

Point 2: Zero tension on one face:


18 inc h

1 0.80 18 in 14.4 ina c


11

18 in 15.56 in0.003 0.00041

18 ins c

c d

c

1

1

60 ksismaller of

29,000 ksi 0.00041 11.9 ksis

s

fE

22

18 in 9 in0.003 0.00150

18 ins c

c d

c

2

2

60 ksismaller of

29,000 ksi 0.00150 43.5 ksis

s

fE

33

18 in 2.44 in0.003 0.00259

18 ins c

c d

c

3

3

60 ksismaller of

29,000 ksi 0.00259 75.1 ksis

s

fE


0.85 ' 0.85 5 ksi 14.4 in 18 in 1100 kc cC f ab

2

1 1 1 0.85 ' 3 in 11.9 ksi 0.85 5 ksi 23.0 ks s s cF A f f

11-4

2

2 2 2 0.85 ' 2 in 43.5 ksi 0.85 5 ksi 78.5 ks s s cF A f f

and, since3d a ,

2

3 3 3 3 in 60 ksi 180 ks s sF A f

Now we can calculate the nominal axial load and moment:

1 2 3 1100 k 23.0 k 78.5 k 180 k 1380 kn c s s sP C F F F

1 1 2 2 1 3

2 2 2 2 2n c s s s

h a h h hM C F d F d F d

14.4 in

1100 k 9 in 23.0 k 9 in 15.56 in 78.5 k 9 in 9 in 180 k 9 in 2.44 in2

nM


And, since1s is in compression, the section is compression controlled, and 0.65 .

0.65 1380 k 897 knP

0.65 251 k-ft 163 k-ftnM

Point 3: Balanced Point:


1

0.00315.56 in 9.21 in

0.003 0.00207

c

c y

c d

1 0.80 9.21 in 7.37 ina c


11

9.21 in 15.56 in0.003 0.00207

9.21 ins c

c d

c

1

1

60 ksismaller of

29,000 ksi 0.00207 60 ksis

s

fE

22

9.21 in 9 in0.003 0.00007

9.21 ins c

c d

c

2

2

60 ksismaller of

29,000 ksi 0.00007 1.98 ksis

s

fE

33

9.21 in 2.44 in0.003 0.00221

9.21 ins c

c d

c

3

3

60 ksismaller of

29,000 ksi 0.00221 64.1 ksis

s

fE


0.85 ' 0.85 5 ksi 7.37 in 18 in 564 kc cC f ab

2

1 1 1 3 in 60 ksi 180 ks s sF A f

and, since2d a ,

2

2 2 2 2 in 1.98 ksi 3.96 ks s sF A f

2

3 3 3 0.85 ' 3 in 60 ksi 0.85 5 ksi 167 ks s s cF A f f

11-5

Now we can calculate the nominal axial load and moment at the balanced point.

1 2 3 564 k 180 k 3.96 k 167 k 555 kn c s s sP C F F F

1 1 2 2 1 3

2 2 2 2 2n c s s s


7.37 in

564 k 9 in 180 k 9 in 15.56 in 3.96 k 9 in 9 in 167 k 9 in 2.44 in2

nM


And, since 1s y , the section is compression controlled, and 0.65 .

0.65 546 k 355 knP

0.65 440 k-ft 286 k-ftnM

Point 4: Tension control limit


1

1

0.00315.56 in 5.84 in

0.003 0.005

c

c s

c d

1 0.80 5.84 in 4.67 ina c


11

5.84 in 15.56 in0.003 0.00500

5.84 ins c

c d

c

1

1

60 ksismaller of

29,000 ksi 0.00500 145 ksis

s

fE

22

5.84 in 9 in0.003 0.00162

5.84 ins c

c d

c

2

2

60 ksismaller of

29,000 ksi 0.00162 47.1 ksis

s

fE

33

5.84 in 2.44 in0.003 0.00175

5.84 ins c

c d

c

2

2

60 ksismaller of

29,000 ksi 0.00175 50.8 ksis

s

fE


0.85 ' 0.85 5 ksi 4.67 in 18 in 357 kc cC f ab

2

1 1 1 3 in 60 ksi 180 ks s sF A f

2

2 2 2 2 in 47.1 ksi 94.2 ks s sF A f

2

3 3 3 0.85 ' 3 in 50.8 ksi 0.85 5 ksi 140 ks s s cF A f f

Now we can calculate the nominal axial load and moment at the balanced point.

1 2 3 357 k 180 k 94.2 k 140 k 223 kn c s s sP C F F F

1 1 2 2 1 3

2 2 2 2 2n c s s s


11-6

4.67 in

357 k 9 in 180 k 9 in 15.56 in 94.2 k 9 in 9 in 140 k 9 in 2.44 in2

nM


And, since1 0.005s , the section is tension controlled, and 0.9 .

0.9 223 k 201 knP

0.9 373 k-ft 336 k-ftnM

Point 5: Pure Tension

28in 60 ksi 480 kn st yP A f

0 k-ftnM

And, since the section is in pure tension, it is tension controlled, and 0.9 .

0.9 480 k 432 knP

0.9 0 k-ft 0 k-ftnM

Fig. S11-4

-500

0

500

1000

1500

2000

0 100 200 300 400 500

Axia

l L

oad

(kip

)

Moment (k-ft)

11-7

11-6 Use the interaction diagrams in Appendix A to compute the maximum moment, Mu,

that can be supported by the column shown in Fig. P11-1 if (Use and

):

(a) Pu = 583 kips.

Calculate the parameters required to use the interaction diagrams:

26 1.0 in

0.018518 in 18 in

stg

g

A

A

18 in 1.5 in 0.375 in 1.128 in / 2 1.5 in 0.375 in 1.128 in / 2'

0.72918 in

d d

h

583 k

1.80 ksi18 in 18 in

n uP P

bh bh

Now go to the interaction diagrams:

From Fig. A-7a: 2

0.46 ksinM

bh

for 0.60

From Fig. A-7b: 2

0.50 ksinM

bh

for 0.75

By interpolation for 0.729 ,2

0.49 ksinM

bh

Finally, we can compute the maximum moment carried by this section:

220.49 0.49 18 in 18 in 2860 k-in 238 k-ftu nM M bh

(b) Pu = 130 kips.


26 1.0 in

0.018518 in 18 in

stg

g

A

A

18 in 1.5 in 0.375 in 1.128 in / 2 1.5 in 0.375 in 1.128 in / 2'

0.72918 in

d d

h

130 k

0.40 ksi18 in 18 in

n uP P

bh bh


From Fig. A-7a: 2

0.48 ksinM

bh

for 0.60

From Fig. A-7b: 2

0.55 ksinM

bh

for 0.75


0.54 ksinM

bh


11-8


(c) e = 4 in.


26 1.0 in

0.018518 in 18 in

stg

g

A

A

18 in 1.5 in 0.375 in 1.128 in / 2 1.5 in 0.375 in 1.128 in / 2'0.729

18 in

d d

h

4 in

0.22218 in

e

h


From Fig. A-7a: 2

0.42 ksinM

bh

for 0.60

From Fig. A-7b: 2

0.44 ksinM

bh

for 0.75


0.44 ksinM

bh



11-7 Use the interaction diagrams in Appendix A to select tied-column cross sections to

support the loads given in the accompanying list. In each case, use and

. Design the ties and calculate the required splice lengths, assuming

that the bars extending up from the column below are the same diameter as in the

column you have designed. Draw a typical cross section of the column showing the

bars and ties.

(a) Pu = 390 kips, Mu = 220 k-ft, square column with bars in two faces.

First estimate the size of the section required. For the first iteration, assume 0.02g .

2

,

390 k188 in 13.7 in 13.7 in

0.40 4 ksi 60 ksi 0.020.40 '

ug trial

c y g

PA

f f

So try a 14 in square column, assuming #3 stirrups and #8 bars.

Then calculate the parameters required to use the interaction diagrams, and reference them to select

an appropriate reinforcement ratio:

14 in 2 1.5 in 0.375 in 0.5 in

0.6614 in

390 k

1.99 ksi14 in 14 in

n uP P

bh bh

11-9

22 2

220 k-ft 12 in/ft0.962 ksi

14 in 14 in

n uM M

bh bh

From Figs. A-6a and A-6b, the section would require , so a larger section is

required.

Therefore, we will try a square column with 18 in sides. Re-calculate the parameters required to

use the interaction diagrams, and reference them to select an appropriate reinforcement ratio:

18 in 2 1.5 in 0.375 in 0.5 in

0.73618 in

390 k

1.20 ksi18 in 18 in

n uP P

bh bh

22 2

220 k-ft 12 in/ft0.453 ksi

18 in 18 in

n uM M

bh bh

From Fig. A-6a: 0.021g for 0.60

From Fig. A-6b: 0.016g for 0.75

By interpolation for 0.736 , 0.017g

Finally, we can compute the area of steel required to reinforce this section:

2

, 0.017 0.017 18 in 18 in 5.51 ins requiredA bh

Now select the bars:

28#8 bars 6.32 in fits in an 18 in section



Select #7 bars, placing 5 along the top and bottom faces.

As indicated on the interaction diagrams, 0.5s yf f in the extreme tensile layer of steel. Therefore,

Class B tension splices are required. From Table A-13, the splice length must be 54 in.

As transverse reinforcement we are permitted to use #3 bars since the longitudinal bars are not

larger than #10s. Now select the vertical spacing of the ties:

16 16 0.875 in 14 in

48 48 0.375 in 18 in

min( , ) min(18 in,18 in) 18 in

b

bt

d

s d

b h

Select #3 stirrups spaced at 14 in o.c.

11-10

Fig. S11-7a

(b) Pu = 710 kips, Mu = 50 k-ft, square column with bars in four faces.

First estimate the size of the section required. For the first iteration, set 0.02g :

2

,

710 k341 in 18.5 in 18.5 in

0.40 4 ksi 60 ksi 0.020.40 '

ug trial

c y g

PA

f f

So try a 20 in square column, assuming #3 stirrups and #8 bars.



20 in 2 1.5 in 0.375 in 0.5 in

0.7620 in

710 k

1.78 ksi20 in 20 in

n uP P

bh bh

22 2

50 k-ft 12 in/ft0.075 ksi

20 in 20 in

n uM M

bh bh

From Figs. A-9a and A-9b, the section would require 0.01g , so a smaller section

would be desirable. Try a square column with 18 in sides.

Re-calculate the parameters required to use the interaction diagrams, and reference them to select


18 in 2 1.5 in 0.375 in 0.5 in

0.73618 in

710

2.19 ksi18 in 18 in

n uP P k

bh bh

22 2


18 in 18 in

n uM M

bh bh




11-11


2

, 0.015 0.015 18 in 18 in 4.86 ins requiredA bh


212#6 bars 5.28 in fits in an 18 in. section

Select #6 bars, placing 4 along each face.

From the interaction diagrams, the extreme tensile layer of steel is under compression. Therefore,

from Table A-13, a splice of length 0.83 23 in 19.1 in 20 in is required.

As transverse reinforcement, we have selected #3 bars since the longitudinal bars are not larger

than #10 bars. Now select the vertical spacing of the ties:

16 16 0.75 in 12 in

48 48 0.375 in 18 in

min( , ) min(18 in,18 in) 18 in

b

bt

d

s d

b h

Select #3 stirrups spaced at 12 in o.c.

Fig. S11-7b

(c) Pu = 200 kips, Mu = 240 k-ft, square column with bars in four faces.


( )

( )

However, since the moment is relatively high, try a 16 in square column, assuming #3

stirrups and #8 bars.



11-12

16 in 2 1.5 in 0.375 in 0.5 in

0.7016 in

22 2


16 in 16 in

n uM M

bh bh

From Figs. A-9a and A-9b, the section would require a very high , so a larger section is

required. Try a square column with 18 in sides.

Re-calculate the parameters required to use the interaction diagrams, and reference them to select


18 in 2 1.5 in 0.375 in 0.5 in

0.73618 in

22 2


18 in 18 in

n uM M

bh bh

From A-9a: for 0.60

From A-9b: for 0.75

By interpolation for 0.736 ,




Select #8 bars, placing 3 along the each face.

From the interaction diagrams, 0.5s yf f in the extreme tensile layer of steel. Therefore, Class B

tension splices are required. From Table A-13, the splice length must be 62 in.

As transverse reinforcement, we have selected #3 bars since the longitudinal bars are not larger

than #10 bars. Now select the vertical spacing of the ties:

16 16 1.00 in 16 in

48 48 0.375 in 18 in

min( , ) min(18 in,18 in) 18 in

b

bt

d

s d

b h

So we must use #3 stirrups spaced at 16 in o.c.

11-13

Fig. S11-7c

11-8 Use the interaction diagrams in Appendix A to select spiral-column cross sections to

support the loads given in the accompanying list. In each case, use and

. Design the spirals and calculate the required splice lengths. Draw a

typical cross section of the column showing the bars and spiral.

(a) Pu = 600 kips, Mu = 65 k-ft.

First estimate the size of the section required. For the first iteration, set 0.02g .

2

,

600 k194 in 15.7 in dia.

0.50 5 ksi 60 ksi 0.020.50 '

ug trial

c y g

PA

f f

Try a 16 in diameter column, assuming #3 spiral and #8 bars.



16 in 2 1.5 in 0.375 in 0.5 in

0.70316 in

2

600 k2.99 ksi

201 in

n u

g g

P P

A A

2


16 in 201 in

n u

g g

M M

hA hA





11-14

22

2

,

16 in0.013 0.013 2.61 in

4 4s required

dA


26#6 bars 2.64 in

Select #6 bars, spaced evenly around the column.

From the interaction diagrams, the extreme tensile layer of steel is under compression. Therefore,

from Table A-13, a splice of length 0.75 23 in 17.3 in 18 in is required.

As transverse reinforcement, we have selected a #3 spiral since the longitudinal bars are not larger

than #10 bars. Now select the pitch of the spiral:

22

2 2

3 in

0.375 in 60 ksi1.81 in

0.45 ' / 1 0.45 13 in 5 ksi 201 in /134 in 1

sp yt

c c g ch

d fs

D f A A

Use a 16 in diameter column reinforced with 6 #6 bars. Use 18 in lap splices, and #3

spirals with a pitch of 1.75 in as transverse reinforcement.

(b) Pu = 200 kips, Mu = 150 k-ft.


( )

( )

Since there is also a moment applied, try an 18 in diameter column, assuming #3 spiral and

#8 bars.



18 in 2 1.5 in 0.375 in 0.5 in

0.73618 in

2

2000.79 ksi

18

2

n u

g g

P P

A A

2

150 k-ft 12 in/ft0.394 ksi

18 in 254 in

n u

g g

M M

hA hA





22

2

,

18 in0.012 0.012 3.05 in

4 4s required

dA

11-15


28#6 bars 3.52 in

26#7 bars 3.6 in

Select #6 bars, spaced evenly around the column.

From the interaction diagrams, the extreme tensile layer of steel is in tension, but only requires a

Class A splice as long as not all of the bars are spliced at the same location. Since, in reality, it is

likely that all bars will be spliced in the same plane, specify a Class B splice regardless. Therefore,

from Table A-13, a splice of length 33 in is required.

As transverse reinforcement, we have selected a #3 spiral since the longitudinal bars are not larger

than #10 bars. Now select the pitch of the spiral:

22

2 2

3 in

0.375 in 60 ksi1.81 in

0.45 ' / 1 0.45 15 in 5 ksi 254 in /177 in 1

sp yt

c c g ch

d fs

D f A A

Use an 18 in diameter column reinforced with 8 #6 bars. Use 33 in lap splices, and #3 spirals with

a pitch of 1.75 in as transverse reinforcement.

11-9 Design a cross section and reinforcement to supports Pu = 450 kips, Mux = 100 k-ft,

and Muy = 130 k-ft. Use and .

Although the strain compatibility method (shown in example 11-5) is the most

theoretically correct method for designing columns for biaxial loading, it is seldom used in design.

Here we will use two more common methods for designing a column, the equivalent eccentricity

method and the Bresler reciprocal load method. Any method outlined in section 11-7 is

appropriate for the solution to this problem.

Equivalent eccentricity method:

First select the dimensions of the trial section, assuming 0.015 :

2

,

450 k230 in 15.2 in by 15.2 in

0.40 4 ksi 60 ksi 0.0150.40 '

ug trial

c y g

PA

f f

Since biaxial moments are also applied to the section, select an 18 in square section.

Assuming #3 ties and

#9 bars:

18 in 2 1.5 in 0.375 in 0.56 in

0.7318 in

Now compute ex, ey, and eox.

130 k-ft 12 in/ft

3.47 in450 k

uy

x

u

Me

P

100 k-ft 12 in/ft

2.67 in450 k

uxy

u

Me

P

11-16

Since our trial section is square,yx

x y

ee

2

450 k0.347 0.4

' 4 ksi 18 in

u

c g

P

f A

40,000 psi

0.5' 100,000 psi

yu

c g

fP

f A

2

450 k 60,000 psi 40,000 psi0.5 0.847

100,000 psi4 ksi 18 in

0.847 2.67 in 18 in

3.47 in 5.73 in18 in

y x

ox x

y

ee e

Therefore, we can design our column for: 450 kuP , and

450 k 5.73 in 2580 k-in 215 k-ftoy u oxM P e

Now use the interaction diagrams in Appendix A to determine g .

2

450 k1.39 ksi

324 in

n u

g g

P P

A A

2

2580 k-in0.442 ksi

18 in 324 in

oyn

g g

MM

hA hA

Since 0.73 0.75 we can use Fig A-9b without interpolating for this design.

So use 0.023g

Finally, we can compute the area of steel required to reinforce this section, and select the bars:

2 20.023 324 in 7.45 inst g gA A

Select 8 #9 bars, with 3 bars along each face. Select ties and splice lengths as appropriate.

We can check this solution using the Bresler reciprocal load method. Remember that we have an

18 in square column reinforced with 8 #9 bars.

2

2

8 in0.0247

324 in

stg

g

A

A

18 in 2 1.5 in 0.375 in 0.56 in

0.7318 in

ComputenxP , the factored load capacity corresponding to and x ge .

130 k-ft 12 in/ft

3.47 in450 k

uy

x

u

Me

P

, and

3.47 in0.193

18 in

x

x

e .

From Fig. A-9b, 1.9 ksinxP

bh

Therefore 616 knxP

Compute nyP , the factored load capacity corresponding to and y ge .

11-17

100 k-ft 12 in/ft

2.67 in450 k

uxy

u

Me

P

, and

2.67 in0.148

18 in

x

x

e .

From Fig. A-9b, 2.2 ksinxP

bh

Therefore 713 knyP

From equ. 11-7 calculate noP :

2 2 2

3 ' 0.85 4 ksi 324 in 8 in 60 ksi 8 in 1560 kno c g st y stP k f A A f A

1 1 1 1 1 1 1 1

616 k 713 k 0.65 1560 ku n nx ny noP P P P P

490 kuP , so our section is sufficient for the loading defined here.

12-1

Chapter 12

12-1 A hinged end column 18-ft tall supports unfactored loads of 100 kips dead load and

60 kips live load. These loads are applied at an eccentricity of 3 in. at bottom and 5

in. at the top. Both eccentricities are on the same side of the centerline of the

column. Design a tied column with at least three bars per face using

and .

Factored loads and moments

1.2 100 kips +1.6 60 kips 216 kipsuP

1 2/ 0.6M M (Note that the column is bent in single curvature)

Estimate column size

Assume 0.015g

2uP 216 kips

(trial ) = 110 in.0.4 4 ksi + 60 ksi 0.0150.4

g

c y g

Af f

Choose a column cross section of 12 in. x 12 in.

Check whether the column is slender

1.0 (18 12) in.60

(0.3 12 in.)

uk

r

1

2

34 12 34 12 0.6 26.8M

M

60uk

r > 1

2

34 12 26.8M

M

The column is quite slender, increase column size to 16 in. x 16 in.

Are moments greater than the minimum?

2,min 2(0.6 0.03h) =216 kips 0.6 in. 0.03 16 in. 233 k-in. < uM P M

2use M

Compute EI

0.4

1

c g

d

E IEI

357,000 4000 3605 10 psicE

4

416 in.

5460 in.12

gI

factored dead load 1.2 100 kips0.56

all factored load 216 kipsd

12-2

3 4

9 20.4 3605 10 psi 5460 in.

5.05 10 lb-in.1 0.56

EI

Magnified moment

2c nsM M

1.0

1

mns

u

c

C

P

P

1

2

0.6 0.4 0.6 0.4 0.6 0.84m

MC

M

2 9 223

2 2 2

5.05 10 lb-in.1070 10 lb =1070 kips

1.0 18 12 in.c

EIP

k

0.841.15

216 kips1

0.75 1070 kips

ns

Select reinforcement

Assume # 8 bars for longitudinal reinforcement, # 3 bars for the ties, and a clear concrete

cover of 1.5 in.

16 in. 2 1.5 in. 2 0.375 in. 1 in.0.70

16 in.

2

216 kips0.84 ksi

(16 in.)

n uP P

bh bh

Assume E-type reinforcement

Fig.A-9a shows

Fig.A-9b shows

. Use . 20.01 2.56 in.st gA A

Use 6 bars # 6 22.64 in.stA for the column of 16 in. 16 in.

12-2 Repeat Problem 12-1, but with the top eccentricity to the right of the centerline and

the bottom eccentricity to the left.


216 kipsuP

1 2/ 0.6M M (Note that the column is bent in double curvature)

12-3


Assume 0.015g

2110 in.gA


Check if the column is slender

1.0 (18 12) in.51.4

(0.3 14 in.)

uk

r

1

2

34 12 34 12 ( 0.6) 41.2 40 (40 governs)M

M

51.4uk

r > 1

2

34 12 40M

M

The column is slender.


2,min 2(0.6 0.03h) =216 kips 0.6 in. 0.03 14 in. 220 k-in. < uM P M

2use M

Compute EI

0.4

1

c g

d

E IEI

33605 10 psicE

4

414 in.

3200 in.12

gI

0.56d

3 4

9 20.4 3605 10 psi 3200 in.

2.96 10 lb-in.1 0.56

EI

Magnified moment

2c nsM M

1.0

1

mns

u

c

C

P

P

1

2

0.6 0.4 0.6 0.4 ( 0.6) 0.36m

MC

M

2 9 223

2 2 2

2.96 10 lb-in.P 626 10 lb =626 kips

1.0 18 12 in.cr

EI

k

0.360.67 1, use 1

216 kips1

0.75 626 kips

ns ns

12-4


Assume # 8 bars for longitudinal reinforcement,

# 3 bars for ties, and a clear concrete

cover of 1.5 in.

14 in. 2 1.5 in. 2 0.375 in. 1 in.0.66

14 in.

2

216 kips1.10 ksi

(14 in.)

n uP P

bh bh

Assume R-type reinforcement

Fig.A-9a shows

Fig.A-9b shows

Use 8 #6 bars ( ) for the column of 14 in. 14 in.

12-3 Figure P12-3 shows an exterior column in a multistory frame. The dimensions are

center-to-center of the joints. The beams are 12 in. wide by 18 in. in over-all depth.

The floor slab is 6 in. thick. The building includes a service core which resists the

majority of the lateral loads. Use cf = 5000 psi and yf = 60,000 psi. The loads and

moments on column AB are:

Factored dead load:

Axial force = 260 kips

Moment at top = 60 kip-ft

Moment at bottom = -80 kip-ft

Factored live load:

Axial force = 200 kips

Moment at top = 50 kip-ft

Moment at bottom = -75 kip-ft

Design column A-B using a square cross section with at least three bars per face.

Since the building has a service core which resists the majority of the lateral loads, the

frame is braced, or non-sway.


| |

The column is in double curvature.

⁄ ⁄

12-5

C

A

B

D

22 ft

12 ft

24 ft

12 ft

Fig. P12-3


Assume 0.015g

( )

( )


Check if the column is slender

From Table 12-2, 0.90k

24 12 in. 18 in. 270 in.u

0.9 270 in.50.6

(0.3 16 in.)

uk

r

(

) ( )

50.6uk

r > 42.5

Column is slender.


( ) ( )

2use M

12-6

Compute k

Assume the columns CA and BD also have a cross section of 16 in. x 16 in.

4

416 in.

column 0.7 column 0.7 3820 in.12

gI I

3

412 in. 18 in.

beam 0.35 (beam) 0.35 2 0.7 4080 in.12

g webI I I

382026.5

12 12

cc

c CA

EEIE

382013.3

12 24

cc

c AB

EEIE

408015.5

12 22b

beams

cc

EEIE

26.5 13.3

top joint 2.57 bottom joint 15.5

c c

c

E E

E

From nomograph, read 0.87k

Compute EI

2.5(1 )

c g

d

E IEI

357,000 5000 4030 10 psicE 45460 in.gI

2600.57

460d

3 49 24030 10 psi 5460 in.

5.61 10 lb-in.2.5 (1 0.57)

EI

Magnified moment

2c nsM M

1.0

1

mns

u

c

C

P

P

1

2

0.6 0.4 0.6 0.4 ( 0.71) 0.32m

MC

M

2 9 223

2 2

5.61 10 lb-in.1000 10 lb =1000 kips

0.87 270 in.c

u

EIP

k

0.320.83 1, use 1

460 kips1

0.75 1000 kips

ns ns

12-7


Assume # 8 bars for longitudinal reinforcement, # 3 bars for ties, and a clear concrete

cover of 1.5 in.

16 in. 2 1.5 in. 2 0.375 in. 1 in.0.70

16 in.

Assume E-type reinforcement

Fig. A-10a yields

Fig. A-10b yields

Use 8 #8 bars ( ) for the column of 16 in. 16 in.

12-4 Use the ACI moment-magnifier method to redesign the columns in the main floor of

Example 12-3 assuming that the floor-to-floor height of the first story is 16 ft 0 in.

rather than 18 ft 0 in. Also, assume the lateral wind forces are 15 percent larger

than those used in Example 12-3.

The floor plan and elevation is shown in Fig. S12-1

The following design shall be performed for a typical interior frame, for instance the fram

along line 2. The design of the columns will follow the following steps:

A. Calculate loads

B. Calculate the beam and column properties and modulus of elasticity

C. Select preliminary column size

D. Check with gravity load case

E. Check with gravity plus wind load case

F. Finalize the design of column reinforcement

A. Calculate loads for a typical interior frame

Dead load from roof

Distributed loads on beams

3 2 3

2 2

18 in. 24 in.6 in.0.150 k/ft 0.025 k/ft 10 0.150 k/ft

12 in/ft 144 in /ft

1.45 k/ft

DL

Distributed load on girders (beams supporting other beams)

2 2

18 in. 18 in. 30 in0.025 0.150 0.6 k/ft

12 in./ft 144 in /ftDL

Concentrated on the interior columns

(

)

12-8

A B C D

1

2

3

4

5

6

32' 30' 32'

20'

20'

20'

20'

20'

11' 6"

11' 6"

11' 6"

11' 6"

11' 6"

16' 0"

Roof

5 th floor

4 th floor

3 th floor

2 th floor

Ground

floor

Slab thickness: 6"; Column size: 18" x 18"; Beam size: 18" x 30"

a) Plan

b) Section 1-1

11

N

Fig. S12-1

12-9

Concentrated on the exterior columns

(

)

Dead load from other floors

Distributed loads on beams

3 2 3

2 2

18 in. 24 in.6 in.0.150 k/ft 0.020 k/ft 10 0.150 k/ft

12 in/ft 144 in /ft

1.40 k/ft

DL

Distributed load on girders (beams supporting other beams)

2 2

18 in. 18 in. 30 in0.020 0.150 0.593 k/ft

12 in./ft 144 in /ftDL

Concentrated on the interior columns

(

)

Concentrated on the exterior columns

(

)

For each column of 11.5 ft, a weight of 3.88 kips is added for the column self-weight.

For each column of 16.0 ft, a weight of 5.40 kips is added for the column self-weight.

Live load from roof

Concentrated live load on the interior columns

(

)

Concentrated live load on the exterior columns

(

)

Live load from other floors

Concentrated live load on the interior columns

(

)

Concentrated live load on the exterior columns

(

)

B. Calculate the beam and column properties and modulus of elasticity

Column: 218 in. 18 in. 324 in.A

4

418 in.

0.7 0.7 6120 in12

c gI I

12-10

Beam:

Effective flange width 18 in. 8 6 in. 66 in.

26 66 18 24 828 inA

3

418 in. 30 in.

0.7 beam-web 0.7 28350 in.12

b gI I

Note that the selection of rigid end zones follows Example 12-3.

Modulus of elasticity 357000 4000 3600 10 psiE

C. Select preliminary column size

Columns are sized based on the gravity load 1.2 1.6 0.5 rD L L . In this load combination, live

load can be reduced. From separate analyses of dead load, live load from the roof floor, and live

load from the other floors, the axial loads of the columns in the ground floor are shown in the

Table P12-1.

The axial load in the exterior column from the live load from the other floors then can be reduced

with a reduction factor as follows:

150.25 0.46

4 4 16.75 20

The axial load in the interior column from the live load from the other floors then can be reduced

with a reduction factor as follows:

150.25 0.40

4 4 31 20

The calculation of reduced axial live load and factored load is shown in Table P12-1.

Table P12-1

All unit are in kips

Exterior

column

Interior

column

Dead load 285 487

Live load from the roof floor 10.0 18.5

Live load from the other floors 107 198

Reduced live load from the other floors 48.8 79.2

Total factored load 425 720

Assume 0.015g

Exterior column

( )

( )

( ) ( )

12-11

Interior column

( )

( )

( ) ( )

Based on the result of Example 12-3, select a column cross section of 18 in. 18 in. ( 2324 in.gA ) for both exterior and interior columns.

D. Check with gravity load case 1.2 1.6 0.5 rD L L

1. Is the story being designed sway or non-sway? In order to answer this question, we need

to calculate the stability index u oh

us c

PQ

V

.In order to have the terms from the same

analysis, we need to analyze a frame with an arbitrary lateral load of 20 kips applied at

the 2nd

floor level in conjunction with the factored dead load and live load as shown in

Fig. S12-2. In order to take into account the live load reduction, an average of live load

reduction factor for exterior and interior columns (0.43) is multiplied with the live load

factor 1.6, yielding 0.69. Therefore, the load combination used is:

1.2 0.69 0.5 20 k lateral loadrD L L .

A structural analysis gives the following results:

0.127 in.oh 414 722 723 418 2277 kipsuP

16 ft 192 in.c

2277 kips 0.127 in.0.075 0.05

20 kips 192 in.Q

The first story is a sway story.

Note that the 2277 kipsuP does not differ significantly from

( )

2. Are the columns slender?

1.2 (192 30) in.36 22

(0.3 18 in.)

uk

r

The columns are slender.

3. Compute the factored axial loads and moments from a first-order frame analysis. As

explained in Example 12-3, the unfactored moments for exterior columns can be

determined based on the live load pattern shown in Fig. S12-3 while those for interior

columns based on the live load pattern shown Fig. S12-4. After a structural analysis is

made, live load reduction factors will be applied. All results and calculation are shown in

Table P12-2.

12-12

Factored dead load and live load plus arbitrary lateral load to evaluate stability index, Q

20 kips

Fig. S12-2

All span loaded with live load Fig. S12-3

12-13

Staggered live load pattern Fig. S12-4

Table P12-2

(Forces in kips Moments in k-ft) Exterior Column Interior Column

DP 285 487

reducedLP 48.8 79.9

LrP 10 18.5

topDM 37.0 -4.9

bottomDM -36.2 5.5

top, reducedLM 20.9 0.67 14 18.2 0.46 8.4

bottom, reducedLM 20.5 0.67 13.7 2.8 0.46 1.3

The factored load on exterior and interior columns are as follows:

Exterior column

2top 1.2 37 1.6 14 66.8 k-ftM M

1bottom 1.2 36.2 1.6 13.7 65.4 k-ftM M

12-14

Interior column

2top 1.2 4.9 1.6 8.4 19.3 k-ftM M

1bottom 1.2 5.5 1.6 1.3 8.68 k-ftM M

4. Find ns for the exterior and interior column?

1.0

1

mns

u

c

C

P

P

Exterior column

1

2

65.40.6 0.4 0.6 0.4 ( ) 0.21

66.8m

MC

M

2

2Pc

EI

k

0.2

1

c g s se

dns

E I E IEI

48750 in.gI

3600 ksicE

4150 in.seI

( )

6 25.88 10 kip-inEI

2 6 22

2 2

5.88 10 kip-in2210 kips

1.0 162 in.c

EIP

k

66.8 k-ftcM

Interior column

1

2

8.680.6 0.4 0.6 0.4 ( ) 0.42

19.3m

MC

M

( )

Since (int)dns does not change significantly, EI and Pcwill

remain essentially the same.

67.8 k-ftcM

12-15

5. Check initial column sections for gravity load case

Exterior column

1.9

0.1118

e

h

0.015g

Fig. A-9b yields

Interior column

0.015g

Because reading from the graph may not be accurate given the two

values are so close, we need to select reinforcement for the column and

check its capacity against the demand. Select 8 bars #8, Example 12-3

shows , OK.

E. Check with gravity plus wind load case 1.2 1.6 0.5 rD W L L

Wind loads are given in Fig. S12-5

1. Calculation of the stability index u oh

us c

PQ

V

Similar to the gravity load case, we need to do one single analysis with the wind load plus

gravity load case. To take into account of the live load reduction, an average live load

reduction factor of 0.43 will be multiplied with the live load factor 0.5, yielding 0.22.

Therefore, the load combination becomes 1.2 1.6 0.215 0.5D W Lr L . A structural

analysis yields the following results:

0.357 in.oh

352 635 633 385 2005 kipsuP

16 ft 192 in.c

4.46 3 6.61 8.49 32.9usV kips

2005 kips 0.357 in.

0.113 0.0532.9 kips 192 in.

Q

1 1

1.131 1 0.113

sQ

The first story is a sway story.

12-16

Wind load

4.46 kips

6.61 kips

6.61 kips

6.61 kips

8.49 kips

Fig. S12-5

2. Factored axial loads and moments

A structural analysis of the frame subjected to the wind load (without the load factor) yields

the following results, as shown in Table P12-3.

Table P12-3

Forces in kips

Moments in k-ft Exterior Column Interior Column

WP 10.4 0.5

topWM 46.2 65

bottomWM -46 -64.6

Exterior column

( )

2 2 2ns s sM M M

2 1.2 0.5 1.2 37 0.5 14 51.4 k-ftns D LM M M

2 1.6 1.6 46.2 73.9 k-fts WM M

2 51.4 1.13 73.9 135M k-ft

Interior column

12-17

( )

2 2 2ns s sM M M

2 1.2 0.5 1.2 4.9 0.5 8.4 10.1ns D LM M M k-ft

2 1.6 1.6 65 104 k-fts WM M

2 10.1 1.13 104 128M k-ft

Note that the 2005 kipsuP does not differ significantly from

2 387 632 2038 kipsuP

3. Check column sections for axial loads and moments

Exterior column

0.01g

Fig. A-9b yields

At this point, we can select 8 bars #6 ( 23.52 in. , 0.011st gA ) for the exterior columns.

Interior column

0.02g

Conclusion: The cross section of exterior and interior columns is 18 in. x 18 in. Use 8 bars #6 for

exterior columns, and 8 bars #8 for interior columns.

13-1

Chapter 13

13-1 Compute for the edge beam shown in Fig. P13-1. The concrete for the slab and

beam was placed in one pour.

Because the slab and the beam have the same elastic modulus, Eq. (13-9) reduces to bf

s

II

1. ComputebI . The cross section of the beam is shown in Fig. S13-1.1 and

bI is computed for the

shaded area.

Fig. S13-1.1 Section through edge of slab.

Part Area, in.2


4 2Ay ,in.4

Web 320 10 3200 10,670 663

Flanges 91 3.5 319 372 2331

411 3519 bI 14,030 in.4

35198.56 in.

411topy

414,030 in.gI

2. ComputesI . The cross section of the slab is shown in Fig. S13-1.2 and

sI is computed for the

shaded portion of the slab.

3

4(108 8) 73316 in.

12sI

Fig. S13-1.2 Edge beam.

16 in.

7 in.

20 in.

13 in.

45

≤ 28 in.

13-2

3. Compute f .

14,030 4.233316

bf

s

II

13-2 Compute the column-strip and middle-strip moments in the long-span direction for

an interior panel of the flat-slab shown in Fig. 13-25. Assume the slab is 6 in. thick,

the design live load is 40 psf and the superimposed dead load is 5 psf for ceiling,

flooring, and so on, plus 25 psf for the partitions. The columns are 10 in. 12 in., as

shown in Fig. 13-25.

1. Compute the factored load.

6

1.2 150 5 25 1.6 40 190 psf12

uq

Note: if the local building code allows a live-load reduction, the 40-psf live load could be

multiplied by the appropriate reduction factor.

2. Compute the static moment in the long span of the slab.

12

14.5 13.5 ft12

n

2 13.2 ft

The column strip extends the smaller of 2 14 or 4 on each side of the column centerline

(ACI Code Section 13.2.1). Thus, the column strip extends 13.2 4 3.3 ft on each side of column

centerlines. The total width of the column strip is 6.6 ft. Each half-middle strip extends from the

edge of the column strip to the centerline of the panel. The total width of two half-middle strips is

13.2 6.6 6.6 ft

The static moment oM can be calculated from Eq. (13-5),

2 2

2 190 13.2 13.5 1 57 kip-ft10008 8

u no

qM

3. Divide oM into negative and positive moments.

From ACI Code Section 13.6.3.2, for an interior span, the total moment is divided as follows:

Negative moment = 0.65 0.65 57 37 kip-ftoM

Positive moment = 0.35 0.35 57 20 kip-ftoM

4. Divide the moments between the column and middle strips.

Negative moments

From Table 13-3 for 1 2 1 0f (since there are no beam between the columns).

Column-strip negative moment = 0.75 37 27.8 kip-ft

Middle-strip negative moment = 0.25 37 9.3 kip-ft

13-3

Half of the middle-strip negative moment, -4.7 kip-ft, goes to each of the adjacent half-middle

strip. Because the adjacent bays have the same width, 2

, a similar moment will be assigned to

the other half of each middle strip so that the total middle-strip negative moment is 9.3 kip-ft.

Positive moments

From Table 13-4 for 1 2 1 0f ,

Column-strip positive moment = 0.60 20 12 kip-ft

Middle-strip positive moment = 0.40 20 8 kip-ft

13-3 Use the direct-design method to compute the moments for the column-strip and

middle-strip spanning perpendicular to the edge of the exterior bay of the flat-plate

shown in Fig. P13-3. Assume the slab is 7.5 in. thick and supports a superimposed

dead load of 25 psf and a live load of 50 psf. There is no edge beam. The columns are

all 18 in. square.

1. Compute the factored load.

7.5

1.2 150 25 1.6 50 223 psf12

uq


multiplied by the appropriate factor.

2. Compute the static moment for the span perpendicular to the edge of the exterior bay

18

20 18.5 ft12

n

2 19 ft

The column strip extends the smaller of 2 14 or 4 on each side of the column centerline

(ACI Code Section 13.2.1). Thus, the column strip extends 19 4 4.75 ft on each side of column

centerlines. The total width of the column strip is 9.5ft. Each half-middle strip extends from the

edge of the column strip to the centerline of the panel. The total width of two half-middle strips is

19 9.5 9.5 ft

The static moment oM can be calculated from Eq. (13-5),

2 2

2 223 19.0 18.5 1 181 kip-ft10008 8

u no

qM


From ACI Code Section 13.6.3.3, for a “slab without beams between interior supports and

without edge beam”, the total moment is divided as follows:

Interior negative moment = 0.70 0.70 181 127 kip-ftoM

Positive moment = 0.52 0.52 181 94 kip-ftoM

Exterior negative moment = 0.26 0.26 181 47 kip-ftoM

13-4

4. Divide the moments between the column and middle strips.

Interior negative moments

From Table 13-3 for 1 2 1 0f (since there are no beam between the columns),

Interior column-strip negative moment = 0.75 127 95 kip-ft

Interior middle-strip negative moment = 0.25 127 32 kip-ft

Half of the middle-strip negative moment, -16 kip-ft, goes to each of the adjacent half-middle

strip. Because the adjacent bays have the same width, 2

,a similar moment will be assigned to

the other half of each middle strip so that the total middle-strip negative moment is 32 kip-ft.

Positive moments

From Table 13-4 for 2 1 0f ,

Column-strip positive moment = 0.60 94 56 kip-ft

Middle-strip positive moment = 0.40 94 38 kip-ft

Exterior negative moment

From Table 13-5, for 1 2 1 0f (since there is no beam parallel to 1) and for 0t (since

there is no edge beam),

Exterior column-strip negative moment = 1.0 47 47 kip-ft

Exterior middle-strip negative moment = 0 47 0 kip-ft

13-4 For the slab configuration and loading conditions in P13-3, use the direct-design

method to compute moments for the edge-column strip and the middle strip

spanning parallel to the edge of the slab.

1. Compute the factored loads.

7.5

1.2 150 25 1.6 50 223 psf12

uq


multiplied by the appropriate factor.

2. Compute the static moment for the span parallel to the edge of the slab.

18

19 17.5 ft12

n

For the definition of 2 refer to Fig. 13-22 in the textbook.

Edge frame: 2,

20 910.8 ft

2 12e

13-5

Interior frame: 2, 20 fti

Generally, the column strip extends the smaller of 2 14 or 4 on each side of the column

centerline (ACI Code Section 13.2.1).Thus; the width of the edge-column strip is 19 9

5.5 ft4 12

The half-middle strip extends from the edge of the column strip to the centerline of the panel. The

total width of two half-middle strips is 20 9.5 10.5 ft .

The static moment oM can be calculated from Eq. (13-5).

Edge frame: 2 2

2, 223 10.8 17.5 1 92.2 kip-ft10008 8

u e n

o

qM

Interior frame: 2 2

2, 223 20 17.5 1 171 kip-ft10008 8

u i n

o

qM


From ACI Code Section 13.6.3.2, for the edge frame, the total moment is divided as follows:

Negative moment = ,0.65 0.65 92.2 60 kip-fto eM

Positive moment = ,0.35 0.35 92.2 32 kip-fto eM

For the interior frame, the total moment is divided as follows:

Negative moment = ,0.65 0.65 171 111 kip-fto iM

Positive moment = ,0.35 0.35 171 59.9 kip-fto iM

4. Divide the moments between the edge-column and middle strips.

Exterior negative moment

From Table 13-3, for 1 2 1 0f (since there is no beam between the columns),

Edge column-strip negative moment = 0.875 60 52.5 kip-ft

Middle-strip negative moment = 1

0.25 60 0.25 111 =21.4 kip-ft2

Note that 1

0.25 0.875 1.02

Positive moments

From Table 13-4 for 2 1 0f ,

Edge column-strip positive moment = 0.80 32 25.6 kip-ft

Middle-strip positive moment = 1

0.40 32 0.40 59.9 =18.4 kip-ft2

13-6

Note that 1

0.40 0.80 1.02

13-7

13-5 A 7-in. thick flat-plate slab with spans of 20 ft in each direction is supported on 16

in. 16 in. columns. The average effective depth is 5.6 in. Assume the slab supports

its own dead load, plus 25 psf superimposed dead load and 40 psf live load. The

concrete strength is 4000 psf. Check two-way shear at a typical interior support.

Assume unbalanced moments are negligible.

1. Compute the factored uniform load.

(

)

2. Check one-way shear.

One-way shear is critical at a distance from the face of the column. Thus, the critical

sections for one-way shear are A-A and B-B in Fig. S13-5.1. The loaded areas causing shear on

these sections are cross hatched. Their outer boundaries are lines of symmetry on which 0uV .

We will only check the shear for section A-A, since the check for section B-B is the same.

Fig. S13-5.1 Critical section for one-way shear at interior column.

(a) Compute uV at section A-A.

6

6 6

6

13-8

(

)

(b) Compute cV for one-way shear.

Because there is no shear reinforcement, we have n cV V and from Eq. (13-27),

( √ ) ( √ ) ⁄

Thus, the slab is OK for one-way shear.

3. Check two-way shear

Punching shear is critical on a rectangular section located at 2d away from the column face, as

shown in Fig. S13-5.2. The critical perimeter is 21.6 in. by 21.6 in. The average d value for

determining the shear strength of the slab is 5.6 in.d

Fig. S13-5.2 Critical section for two-way shear at interior column.

(a) Compute uV on the critical perimeter for two-way shear.

(

)

(b) Compute cV for the critical section.

13-9

The length of the critical perimeter is 2 21.6 21.6 86.4 in.ob

Now, cV is to be taken as the smallest of the following. From Eq. (13-24),

' 14 4 1 4000 86.4 5.6 122 kips 1000c c oV f b d

For Eq. (13-25), 1.0 (since column is square). Therefore,

'4 12 2 4 1 4000 86.4 5.6 184 kips1000c c oV f b d

For Eq. (13-26), 40s for this interior column. Therefore,

' 40 5.6 12 2 1 4000 86.4 5.6 140 kips100086.4

sc c o

o

dV f b d

b

Therefore, the smallest values is 122 kipscV , so 0.75 122 91.5 kips > Vc uV and the slab

is OK in two-way shear.

13-6 Assume the slab described in Problem 13-5 is supported on 10 in. 24 in. columns.

Check two-way shear at a typical interior support. Assume unbalanced moments

are negligible.

1. Compute the factored uniform load.

(

)

2. See the solution to problem 13-5 for one-way shear calculations.

3. Check two-way shear

Punching shear is critical on a rectangular section located at 2d away from the column face, as

shown in Fig. S13-6.2. The critical perimeter is 29.6 in. by 15.6 in. The average d value for

determining the shear strength of the slab is 5.6 in.d

(a) Compute uV on the critical perimeter for two-way shear.

(

)

(b) Compute cV for the critical section.

The length of the critical perimeter is 2 29.6 15.6 90.4 in.ob

Now, cV is to be taken as the smallest of the following. From Eq. (13-24),

' 14 4 1 4000 90.4 5.6 128 kips 1000c c oV f b d

13-10

Fig. S13-6.2 Critical section for two-way shear at interior column.

For Eq. (13-25), 24

2.410

(since column is 10 in. by 24 in.). Therefore,

'4 4 12 2 1 4000 90.4 5.6 117 kips10002.4

c c oV f b d

For Eq. (13-26), 40s for this interior column. Therefore,

' 40 5.6 12 2 1 4000 90.4 5.6 143 kips100090.4

sc c o

o

dV f b d

b

Therefore, the smallest values is 117 kipscV , so 0.75 117 88 kips > Vc uV and the slab is

OK in two-way shear.

13-7 The slab shown in Fig. P13-7 supports a superimposed dead load of 25 psf and a live

load of 60 psf. The slab extends 4 in. past the exterior face of the column to support

an exterior wall that weighs 400 lbs/ft of length of wall. The story-to-story height is

9.5 ft. Use 4500-psi concrete and Grade-60 reinforcement.

(a) Select slab thickness.

Determine the thickness to limit deflections.

From Table 13-1, the minimum thicknesses of the four typical slab panels are as follows:

Panel 1-2-A-B (corner; treat as exterior), and panels 2-3-A-B and 1-2-B-C (exterior)

13-11

Maximum 20 12 16 224 in.n

Minimum 224

7.47 in.30 30

nh

Panel 2-3-B-C (interior)

Maximum 224 in.n

Minimum 224

6.78 in.33 33

nh

Try 8.0 in.h

Check the thickness for shear. We should check the shear at columns A2 and B2

The tributary area for column A2 is cross-hatched in Fig. S 13-7.1 The factored uniform load can

be calculated as:

81.2 150 25 1.6 60 246 psf

12uq

Note that if the area of any of the panels exceeded 2400 ft , it would be possible to reduce the live

load before factoring it.

Fig. S 13-7.1 Initial critical shear perimeters and tributary areas for column A2.

The critical shear perimeter is located at 2d away from the interior column face and 4 in. from

the exterior column face, as shown in Fig. S 13-7.1. In the following calculation for the factored

shear force transmitted to column A2, the shear force multiplier of 1.15 required for the first

interior support will be applied directly to the appropriate tributary lengths. Then,

8 0.75 0.5 6.75 in.avgd (assuming 34

in. clear cover and No. 4 bars as slab

reinforcement).

22.75 2 23.38 69.5 in.ob

12 22.75 23.38

246 9 1.15 9 10 1.2 400 9 1.15 9 60700 lbs 61 kips12 144

uV

13-12

From Eq.(13-25),

16

116

4

2 6.0 4

(does not govern)

From Eq. (13-26),

30s , for an exterior slab-column connection

30 6.75

2 2 4.91 4.069.5

s

o

d

b

(does not govern)

Thus, using Eq. (13-24):

' 14 0.75 4 1 4500 69.5 6.75 94.4 kips > V1000c c o uV f b d

Note: 0.65 0.75u cV V

For this ratio, ACI Code Section permits modification of f for moment transfer about an axis

parallel to the edge of the slab. With that information and because this ratio is below 0.8, the slab

thickness at this connection should be sufficient for checking shear and moment transfer about an

axis perpendicular to the edge of the column. Shear check for column B2 follows the same

procedure as for column A2. Thus, use an 8-in. slab. Final shear checks will be made in part (c)

after completing the flexural design of the slab.

(b) Use the direct design method to compute moments, and then design the

reinforcement for the column and middle strips associated with column line

2.

Because there is no edge beams, 0f

Compute moments in the slab strip along column line 2

A2 B2 C2

1 (ft) 20.0 20.0

(ft)n 18.67 18.67

2 (ft) 18.0 18.0

(ksf)uq 0.25 0.25 2

2 (kip-ft)

8

u no

qM 196 196

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -51 +102 -137 -127 +69 -127

13-13

Compute moments in the slab strip along column line 1

A1 B1 C1

1 (ft) 20.0 20.0

(ft)n 18.67 18.67

2 (ft) 10.0 10.0

(ksf)uq 0.25 0.25 2

2 (kip-ft)

8

u no

qM 109 109

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -28 +57 -76 -71 +38 -71

Wall load (kip/ft) 0.48 0.48

Wall 2

8

wall no

qM

21 21

Moments from wall

(kip-ft) -5.5 11 -15 -14 +7 -14

Distribute the negative and positive moments to the column and middle strips and design the

reinforcement.

In each panel, the column strip extends 1 20.25 min , 0.25 18 12 54 in. on each side

of the column lines. The total width of the column strip is 2 54 in.= 108 in. 9 ft . The width of

the middle strip is 9 ft. The edge strip has a width of 54 in. 12 in. 66 in. 5.5 ft .

Place the steel in the long direction close to the surface of the slab. Try No. 4 bars. Thus,

8 0.75 0.25 7.0 in.d

Compute trial sA required at the section of maximum moment (column strip at B2). The largest

uM is 102.3 kip-ft. Assuming that 0.95jd ,

2102.3 12,000(trial) 3.42 in.

0.9 60,000 0.95 7.0sA

Compute a and check whether the section is tension controlled:

3.42 60,0000.82 in.

0.85 4500 5.5 12a

0.82

1.00 in.0.825

c

Clearly, the section is tension-controlled; therefore, 0.9 .

Compute the value of jd :0.82

7.0 6.59 in.2

jd

Assuming that a is constant for all sections (conservative assumption), compute a constant for

computing sA :

2 12,000(in. ) 0.0337 (kip-ft)

0.9 60,000 6.59

us u

MA M

(Eq. A)

13-14

The values of sA required in the following table are computed from Eq. (A).


,min 0.0018sA bh for Grade-60 reinforcement. Maximum bar spacing is 2h (ACI Code

Section 13.3.2), but not more than 18 in. (ACI Code Section 7.12.2.2). Therefore maximum is

16 in.

Edge column strip:

2,min 0.0018 5.5 12 8 0.95 in.sA

Minimum number of bar spaces5.5 12

4.116

Therefore, the minimum number of bars is 5.

Other strips:

2,min 0.0018 9 12 8 1.56 in.sA

The minimum number of bars is 8.

Division of moment to column and middle strip: north-south strips

Edge

Column Strip

Middle

Strip

Column

Strip

Middle

Strip

Column

Strip

Strip Width, ft 9.0 9.0 9.0 9.0 5.5

Exterior Negative Moments A1 A2 A3

Slab moment (kip-ft) -28 -51 -51 Moment Coef. 1.0 0.0 0.0 1.0 0.0 0.0 1.0

Distributed moments to

strips -28 0.0 0.0 -51 0.0 0.0 -51

Wall moment (kip-ft) -5.5 Total strip moment (kip-ft) -33.5 0.0 -51 0.0 -51

Required 2 (in. )sA 1.13 0.0 1.72 0.0 1.56

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 6 #4 9 #4 9 #4 9 #4 9 #4

2 (in. )sA provided 1.20 1.80 1.80 1.80 1.80

End Span Positive

Moments

Slab moment (kip-ft) 57 102 102 Moment Coef. 0.6 0.4 0.2 0.6 0.2 0.2 0.6


strips 34.2 22.8 20.4 61.2 20.4 20.4 61.2

Wall moment (kip-ft) 11 Total strip moment

(kip-ft) 45.2 43.2 61.2 40.8 61.2

Required 2 (in. )sA 1.52 1.45 2.06 1.37 2.06

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 8 #4 10 #4 10 #4 10 #4 10 #4 2 (in. )sA provided 1.60 2.00 2.00* 2.00 2.00*

13-15

First Interior Negative

Moments B1 B2 B3

Slab moment (kip-ft) -76 -137 -137 Moment Coef. 0.75 0.25 0.125 0.75 0.125 0.125 0.75


strips -57 -19 -17.1 -102.3 -17.1 -17.1 -102.3

Wall moment (kip-ft) -15 Total strip moment

(kip-ft) -72 -36.1 -102.3 -34.2 -102.3

Required 2 (in. )sA 2.43 1.22 3.45 1.15 3.45

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 8 #5 9 #4 11 #5 9 #4 11 #5 2 (in. )sA provided 2.48 1.80 3.41* 1.80 3.41*

Interior Positive Moments Slab moment (kip-ft) 38 69 69

Moment Coef. 0.6 0.4 0.2 0.6 0.2 0.2 0.6 Distributed moments to

strips 22.8 15.2 13.8 41.1 13.8 13.8 41.1

Wall moment (kip-ft) 7 Total strip moment (kip-

ft) 29.8 29 41.1 27.6 41.1

Required 2 (in. )sA 1.00 0.98 1.36 1.04 1.36

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 6 #4 9 #4 9 #4 9 #4 9 #4 2 (in. )sA provided 1.20 1.80 1.80 1.80 1.80

Interior Negative Moments C1 C2 C3

Slab moment (kip-ft) -71 -127 -127 Moment Coef. 0.75 0.25 0.125 0.75 0.125 0.125 -0.75


strips -53.2 -17.8 -15.9 -95.2 -15.9 -15.9 -95.2

Wall moment (kip-ft) -14 Total strip moment (kip-

ft) 67.2 -33.7 -95.2 -31.8 -95.2

Required 2 (in. )sA 2.26 1.14 3.21 1.07 3.21

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 8 #5 9 #4 11 #5 9 #4 11 #5 2 (in. )sA provided 2.48 1.80 3.41 1.80 3.41

*As(provided) < As(required) is o.k. because adjacent positive moment regions are over-designed

and some moment redistribution can occur in ductile slabs.

(c) Check two-way shear and moment transfer at columns A2 and B2. Neglect

unbalanced moments about column line 2.

Column A2

13-16

The critical perimeter is at 2d from the face of the column, where d is the average depth. At all

exterior ends, the reinforcement is No. 4 bars and 6.75 in.avgd The shortest perimeter results

from the section shown in Fig. S13-7.2 and the perimeter dimensions are,

1

2

20 in. 2 23.38 in.

16 in. 22.75 in.

b d

b d

For moments about the z z axis,

2 23.38 5.69 23.38 27.86 in.

2 23.38 5.69 22.75 5.69ABy

Therefore, 7.86 in. and 15.52 in.AB CDc c

For moments about the w w axis,

22.7511.38 in.

2CB ADc c

As calculated in part (a), 61 kipsuV .

For slabs designed by the direct-design method, the moment transferred from the slab to the

column axis z-z is 0.3 oM , and using the moments calculated from part (b),

0.3 0.3 196 58.8 kip-ftoM (acting about the centroid of the shear perimeter).

Fig. S 13-7.2 Critical section- ColumnA2

From part (b), we found that the unbalanced moment due to the wall moments is 7 kip-ft and

assuming that the loads acts at 2 in. from the edge of the slab,

23.38 2.0 7.86 13.52 in. from centroid

The total moment to be transferred is,

13.52

58.8 7 50.9 kip-ft12

z zM

13-17

Note that the unbalanced moment about column line 2 w wM is neglected as stated in the

problem.

From Eq. (13-32), calculate the fraction of moment transferred by flexure,

1 2

1 10.60

2 21 1 23.38 22.75

3 3

f

b b

ACI Code Section 13.5.3.3 allows f to be increased to 1.0 if u cV V and the resulting is less

than 0.375 b within a width of 2 3c h centered in the column. From part (a), 0.65u cV V .

Therefore, take 1.0f and check the reinforcement required.

Width effective for flexure2 3 16 3 8 40 in.c h

Effective width 2 2 12 2tc c c c (second expression governs)

Effective width 16 2 16 48 in.

Assume that 0.95jd d , with 7.0 in.d

250.9 12,0001.70 in.

0.9 60,000 0.95 7.0sA

The steel provided is 9 No. 4 in a column strip width of 9 ft 108 in. , or roughly 13.5 on centers.

The bars within the 40 in. effective width can be used for the moment transfer. Place four

column-strip bars into this region and add 5 No. 4 bars, giving 21.8 in.sA in the effective width.

Recompute sA ,

1.80 60,000

0.71 in.0.85 4500 40

a

2 50.9 12,000(in. ) 1.70 kip-ft

0.710.9 60,000 7

2

sA

(steel chosen OK)

The reinforcement ratio is,

1.8

0.006440 7

sA

bd

and from Eq. (4-24),

0.85 0.825 4500 0.003

0.031160,000 0.003 0.00207

b

,

and thus, 0.375 0.0117 0.0064b and we can use 1.0f . As a result, it is not necessary to

transfer any of the moment about z-z axis by eccentric shear stresses.

Column B2

The critical perimeter is shown in Fig. S 13-7.3 and the centroidal axes pass through the centers

of the sides.

2 22.75 22.75 91 in.ob

13-18

The factored shear force transmitted to column B2 is,

22.75 22.75

246 9 1.15 9 10 1.15 10 101500 lbs 102 kips144

uV

Fig. S 13-7.3 Critical section- Column B2

From Eq.(13-25),

4

2 6.0 4

(does not govern)

From Eq. (13-26),

40s , for an interior slab-column connection

40 6.75

2 2 4.97 4.091

s

o

d

b

(does not govern)

Thus, using Eq. (13-24):

' 14 0.75 4 1 4500 91 6.75 124 kips > V1000c c o uV f b d

Since, 0.82 0.4u cV V no adjustment will be permitted in the ratio of unbalanced moment

resisted by eccentric shear.

The moment about x-x axis to be transferred comes from part (b) and is the difference between

the negative moments on the two sides of column B2, i.e. , 137 127 10 kip-ftu x xM .

From Eq. (13-32), calculate the fraction of moment transferred by flexure (x-x axis),

1 2

1 10.6

2 21 1 1

3 3

f

b b

The torsional moment of inertia can be calculated from Eq. (13-34),

23 3

1 1 122 2

12 12 2c

b d db bJ b d

Where 1 26.75 in. and 22.75 in.d b b Thus, 454150 in.cJ

By inspection, the reinforcement that is already in the slab is adequate for moment transfer.

13-19

From Eq. (13-30) and neglecting unbalanced moment about column line 2 (i.e. about axis y-y),

(shear transfer) 1 0.6 10 4 kip-ft 48,000 lb-in.u uM M

Then,

48,000 11.38

12.1 psi45,150

u

c

M c

J

So,

102,000 124,000

(max) 12.1 166 psi 12.1 psi=178 psi 202 psi91 6.75 91 6.75

u c

Thus, the shear is OK at this column.

13-8 Refer to the slab shown in Fig. P13-7 and the loadings and material strengths given

in Problem 13-7.

(a) Select slab thickness.

Problems 13-7 and 13-8 refer to the same flat-slab. As a result, the thickness of the slab was

chosen in part (a) of problem 13-7. Thus, use an 8 in. thick slab.

(b) Use the direct design method to compute moments, and then design the

reinforcement for the column and middle strips associated with column line

A.

Because there is no edge beams, 0f

Compute moments in the slab strip along column line A

A1 A2 A3

1 (ft) 18.0 18.0

(ft)n 16.67 16.67

2 (ft) 11.0 11.0

(ksf)uq 0.25 0.25 2

2 (kip-ft)

8

u no

qM 96 96

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -25 +50 -67 -62 +34 -62

Wall load (kip/ft) 0.48 0.48

Wall 2

8

wall no

qM

16.7 16.7

Moments from wall

(kip-ft)

-4.3 8.7 -11.7 -10.9 +5.8 -10.9

Compute moments in the slab strip along column line B

B1 B2 B3

1 (ft) 18.0 18.0

(ft)n 16.67 16.67

2 (ft) 20.0 20.0

(ksf)uq 0.25 0.25

13-20

22

(kip-ft)8

u no

qM 174 174

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -45 +90 -122 -113 +61 -113

Distribute the negative and positive moments to the column and middle strips and design the

reinforcement.

In each panel, the column strip extends 1 20.25 min , 0.25 18 12 54 in. on each side

of the column lines. The total width of the column strip is 2 54 in.= 108 in. 9 ft . The width of

the middle strip is 9 ft. The edge strip has a width of 54 in. 12 in. 66 in. 5.5 ft .

Place the steel in the long direction close to the surface of the slab. Try No. 4 bars. Thus,

8 0.75 0.25 7.0 in.d

Compute trial sA required at the section of maximum moment (column strip at B2). The largest

uM is 91.5 kip-ft. Assuming that 0.95jd ,

Compute a and check whether the section is tension controlled:

The section is tension-controlled; therefore, 0.9 .

Compute the value of jd : ⁄

Assuming that a is constant for all sections (conservative assumption), compute a constant for

computing sA (Eq. A):

The values of sA required in the following table are computed from Eq. (A).


,min 0.0018sA bh for Grade-60 reinforcement. Maximum bar spacing is 2h (ACI Code

Section 13.3.2), but not more than 18 in. (ACI Code Section 7.12.2.2). Therefore maximum is

16 in.

Edge column strip:

2,min 0.0018 5.5 12 8 0.95 in.sA

Minimum number of bar spaces5.5 12

4.116

Therefore, the minimum number of bars is 5.

Other strips:

2,min 0.0018 9 12 8 1.56 in.sA

The minimum number of bars is 8.

13-21

Division of moment to column and middle strip: east-west strips

Edge

Column Strip

Middle

Strip

Column

Strip

Strip Width, ft 9.0 9.0 9.0

Exterior Negative Moments A1 B1

Slab moment (kip-ft) -25 -45 Moment Coef. 1.0 0.0 0.0 1.0


strips -25 0.0 0.0 -45

Wall moment (kip-ft) -4.3 Total strip moment (kip-

ft) -29.3 0.0 -45

Required 2 (in. )sA 0.99 0.0 1.52

Minimum 2 (in. )sA 0.95 1.56 1.56

Selected Steel 5 #4 8 #4 8 #4

2 (in. )sA provided 1.00 1.60 1.60

End Span Positive Moments Slab moment (kip-ft) 50 90

Moment Coef. 0.6 0.4 0.2 0.6 Distributed moments to

strips 30 20 18 54

Wall moment (kip-ft) 8.7 Total strip moment (kip-

ft) 38.7 38 54

Required 2 (in. )sA 1.31 1.28 1.82

Minimum 2 (in. )sA 0.95 1.56 1.56

Selected Steel 7 #4 8 #4 10 #4 2 (in. )sA provided 1.40 1.60 2.00

First Interior Negative

Moments A2 B2

Slab moment (kip-ft) -67 -122 Moment Coef. 0.75 0.25 0.125 0.75


strips -50 -17 -15 -91.5

Wall moment (kip-ft) -11.7 Total strip moment (kip-

ft) -51.7 -32 -91.5

Required 2 (in. )sA 1.75 1.08 3.09

Minimum 2 (in. )sA 0.95 1.56 1.56


13-22

Interior Positive Moments

Slab moment (kip-ft) 34 61 Moment Coef. 0.6 0.4 0.2 0.6


strips 20.4 13.6 12.2 36.6

Wall moment (kip-ft) 5.8 Total strip moment (kip-ft) 26.2 25.8 36.6

Required 2 (in. )sA 0.89 0.87 1.24

Minimum 2 (in. )sA 0.95 1.56 1.56


13-9 For the corner column (A1) in Fig. P13-7 and the loadings and material strengths

given in Problem 13-7, select a slab thickness to satisfy ACI Code strength

requirements for two-way shear and moment transfer, and deflection control.

(a) Make the check for moment transfer in only one principal direction (use the

more critical direction).

First determine the thickness to limit deflections:

From Table 13-1, the minimum thicknesses of the corner panel is (treat it as an exterior

column):

Maximum in

20 ft 12 16 in 224 inft

n

Minimum 224 in

7.47 in30 30

nh

Like Problem 13-8, try 8.0 inh

Now check whether this thickness is OK for shear at column A1, considering moment transfer in

the more critical direction:

Note that if the area of any of the panels exceeded 2400 ft , it would be possible to

reduce the live load before factoring it, but this is not the case for column A1. Therefore the

factored uniform load acting on the corner panel is:

3

8 in lb1.2 150 25 psf 1.6 60 psf 246 psf

in ft12ft

uq

Although the shear force acting on the first interior column face is amplified by 1.15, the ACI

Code does not permit reducing the tributary shear area for an exterior column to below 0.5 in

either principal direction. Also note that the critical shear perimeter is located at 2d away from

the interior column faces.

Therefore,

8 in 0.75 in 0.5 in 6.75 inavgd

13-23

2 16 in 4 in 6.75 in / 2 46.8 inob

2

2

2

23.4 in246 psf 10 ft 1 ft 9 ft 1 ft 26100 lb 26.1 k

in144

ft

uV

Now calculate the ultimate shear stress given this Vu and moment transfer in the more critical

direction.

u v uu

o c

V M cv

b d J

where:

1 2

1 11 1 0.4

2 21 / 1 1

3 3

v

b b

1 2 23.4 inb b

16 inc

2 2

1

1 2

6.75 in 23.4 in5.85 in

2 4 23.4 in 6.75 inAB

d bc

b d b d

23 321 1 1

1 2

3 3 2

12 12 2

23.4 in 6.75 in 6.75 in 23.4 in 23.4 in23.4 in 6.75 in 5.85 in

12 12 2

c AB AB

c

b d db bJ b d c b dc

J

2

4

23.4 in 6.75 in 5.85 in

18,600 incJ

And, the more critical moment transfer axis is along column line 1, which is presented in the

solution to Problem 13-7.

28 k-ftuM

Finally, we can calculate the maximum combined shear stress considering moment transfer in the

more critical direction as follows:

4

in0.4 28 k-ft 12 16 in

26.1 k ft 0.198 ksi 198 psi46.8 in 6.75 in 18,600 in

u v uu

o c

V M cv

b d J

Now compare compute vn to check whether this level of shear stress is acceptable.

'

'

'

4 0.75 4 1 4500 psi 201 psi

4 42 0.75 2 4500 psi 302 psi

1

20 6.75 in2 0.75 2 4500 psi 246 psi

46.8 in

c

n c

sc

o

f

v f

df

b

13-24

Therefore,

201 psi 198 psin uv v OK

(b) Make the check for moment transfer in both principal directions, but permit

a 20 percent increase in the maximum permissible shear stress calculated at

the corner of the critical shear perimeter.

From part (a),

2

2

2


in144

ft

uV

1 28 k-ftuM

Following the same procedure shown in Problem 13-7, the moment in the E-W direction is:

2 25 k-ftuM

Now we can calculate the maximum combined shear stress considering moment transfer in both

principal directions as follows:

1 2

4 4

in in0.4 28 k-ft 12 16 in 0.4 25 k-ft 12 16 in

26.1 k ft ft

46.8 in 6.75 in 18,600 in 18,600 in

0.301 ksi 301 psi

u v u v uu

o c c

u

u

V M c M cv

b d J J

v

v

Now compare compute vn to check whether this level of shear stress is acceptable.

'

'

'

4 0.75 4 1 4500 psi 201 psi

4 42 0.75 2 4500 psi 302 psi

1

20 6.75 in2 0.75 2 4500 psi 246 psi

46.8 in

c

n c

sc

o

f

v f

df

b

Even allowing for a 20% increase in the maximum permissible shear stress, a slab thickness of 8

in does not seem to be deep enough to satisfy this shear check. Repeat with h = 9.5 in.

9.5 in 0.75 in 0.5 in 8.25 inavgd

2 16 in 4 in 8.25 in / 2 48.3 inob

2

2

2


in144

ft

uV

1 2 24.1 inb b

2 2

1

1 2

8.25 in 24.1 in6.03 in

2 4 24.1 in 8.25 inAB

d bc

b d b d

13-25

23 321 1 1

1 2

3 3 2

12 12 2

24.1 in 8.25 in 8.25 in 24.1 in 24.1 in24.1 in 8.25 in 6.03 in

12 12 2

c AB AB

c

b d db bJ b d c b dc

J

2

4

24.1 in 8.25 in 6.03 in

25,200 incJ

So, now the shear stress demand is:

1 2

4 4

in in0.4 28 k-ft 12 16 in 0.4 25 k-ft 12 16 in

26.1 k ft ft

48.3 in 8.25 in 25,200 in 25,200 in

0.227 ksi 227 psi

u v u v uu

o c c

u

u

V M c M cv

b d J J

v

v

If a 9.5 in deep slab is used, then 227 psi 1.2 241 psiu nv v .

(c) Check one-way shear for a critical diagonal section across the corner near

the corner column.

Assume that the critical section for one-way shear at this corner column occurs at a 45 degree

angle from either of the exterior edges, at a distance, d, from the most interior corner. If we

assume that d = 6.75 in, as in part (a), the length of the critical section is as follows:

2 4 in 16 in 2 6.75 in 70 in

Therefore,

2

2

2


in144

ft

uV

The capacity of the section is:

6.75 in 70.0 in 0.75 1.0 2 4,500 psi 6.75 in 70.0 in 47,500 lb 47.5 kn cV V

Therefore, a slab with h = 8 in will not fail in one-way shear along this diagonal failure plane.

13-26

13-10 For the slab system shown in Fig. P13-7, assume the slab has four equal spans in the

north-south direction and three equal spans in the east-west direction. Use the

loading and material strengths given in Problem 13-7 and assume a slab thickness of

7.5 in.

(a) Use an equivalent-frame method to analyze the factored design moments

along column line 2 and compare the results with the moments used in part

(b) of Problem 13-7.

For this solution, structural analysis software is used to model an equivalent frame used

to represent the column line in question. Before this can be done, an appropriate equivalent frame

must be defined, which is done here following recommendations from the text.

Column properties:

216 in 16 in 256 ingA

3

416 in 16 in

5460 in12

gI

45460 ine gI I

Note that a full 9.5 ft of column will be modeled above and below the slab. These

elements will be fixed at their ends.

Beam properties:

In the positive bending regions and the negative bending regions near interior columns:

0.5 , 0.5

33342

0.5 18 ft 12 in/ft 7.5 in3800 in

12 12 12g

hbhI

40.5 1900 ine g gI I I

In the negative bending regions near exterior columns (assumed to be 10.2 ft long):

0.2 , 0.33

33342

0.2 18 ft 12 in/ft 7.5 in1520 in

12 12 12g

hbhI

40.33 500 ine g gI I I

Loads to be applied:

3 2

7.5 in lb lb lb k18 ft 150 25 2140 2.14

in ft ft ft ft12

ft

Dq

2

lb lb k18 ft 60 1080 1.08

ft ft ftLq

use 1.2 1.6U D Lq q q with appropriate pattern loading schemes for each location

13-27

Location Equivalent Frame Analysis with

Software-Based Model

Direct Design Method

(From Problem 13-7)

Column A-2 -62.8 k-ft -51 k-ft

Midspan A-B 80.3 k-ft 102 k-ft

Exterior of col. B-2 -156 k-ft -137 k-ft

Interior of col. B-2 -130 k-ft -127 k-ft

Midspan B-C 64.5 k-ft 69 k-ft

Column C-2 -126 k-ft -127 k-ft

Note that the remaining locations along column line 2 are symmetrical across column C-2.

Some discussion should be presented in the solution, in addition to this table, commenting on the

relative precision of the two approaches, noting that neither is necessarily accurate nor “correct”,

although they are hopefully close to reality. It might be noted that similar discrepancies show up

when the results of a software analysis of a frame system are compared with the results from

using an ACI moment coefficient approach.

(b) Use an equivalent-frame method to analyze the factored design moments

along column line A and compare the results with the moments used in part

(b) of Problem 13-8.

For this solution, structural analysis software is used to model an equivalent frame used

to represent the column line in question. Before this can be done, an appropriate equivalent frame

must be defined, which is done here following recommendations from the text.

Column properties:

216 in 16 in 256 ingA

3

416 in 16 in

5460 in12

gI

45460 ine gI I

Note that a full 9.5 ft of column will be modeled above and below the slab. These

elements will be fixed at their ends.

Beam properties:

In the positive bending regions and the negative bending regions near interior columns:

0.5 , 0.5

( )

In the negative bending regions near exterior columns (assumed to be 10.2 ft long):

0.2 , 0.33

( )

13-28

Loads to be applied:

( ) (

)

( ) (

)

use 1.2 1.6U D Lq q q with appropriate pattern loading schemes for each location

Location Equivalent Frame Analysis with

Software-Based Model

Direct Design Method

(From Problem 13-7)

Column A-1 -52.5 k-ft -29.3 k-ft

Midspan A1-A2 46.9 k-ft 58.7 k-ft

Exterior of col. A-2 -89.6 k-ft -78.7 k-ft

Interior of col. A-2 -78.8 k-ft -72.9 k-ft

Midspan A2-A3 39.4 k-ft 39.8 k-ft

Note that the remaining locations along column line A are symmetrical across Midspan A2-A3.

Some discussion should be presented in the solution, in addition to this table, commenting on the

relative precision of the two approaches. It should be mentioned that neither is necessarily

accurate nor “correct”, although they are hopefully close to reality. It might be noted that similar

discrepancies show up when the results of a software analysis of a frame system are compared

with the results from using an ACI moment coefficient approach.

13-11 For the same floor system described in Problem 13-10 and the loading and material

strengths given in Problem 13-7, assume the slab thickness has been selected to be

6.5 in.

(a) For a typical interior floor panel, calculate the immediate deflection due to

live load and compare to the limit given in ACI Code Table 9.5 (b).

This solution will follow the approach shown in Example 13-16.

Step 1: Compute the immediate deflection of an interior column strip, which should be taken in

the N-S direction, as these are the longer span column strips. Take the span between columns B-

2 and C-2 as the interior span in question.

First compute Ma. The loads we must consider for deflection calculations are:

Dead load: 3

6.5 in lb1.0 Dead Load 150 25 psf 106 psf

in ft12

ft

Service load: 1.0 Dead Live 106 psf 60 psf 166 psf

Construction load: 2.0 Slab Dead Load 2 81.3 psf 163 psf

Therefore, cracking will be governed by the service load. Since we know that the moments

calculated in Problem 13-7b are based on an area load of 246 psf, we can take Ma to be 166/246 =

0.675 times the column strip moments calculated in Problem 13-7b, as follows:

Negative moment at B-2 and C-2: 0.675 95.2 k-ft 64.3 k-ft

Positive moment at midspan: 0.675 41.1 k-ft 27.7 k-ft

13-29

Now compute Mcr:

3in 17.5 4500 psi 9 ft 12 6.5 in12ft 383,000 lb-in 31.9 k-ft

3.25 in

r g

cr

t

f IM

y

It is a safe bet that the slab will be cracked in both the positive and negative moment

regions regardless, due to temperature and shrinkage effects.

Next, we have to compute Icr and Ie. For simplicity, we will assume that the reinforcement

selected in Problem 13-7 is used in this slab as well.

Negative moment region:

211 0.31 in

0.00486in

9 ft 12 6.5 inft

29,000,000 psi

7.5857,000 4500 psi

n

0.0368n

2 2

2 2 0.0368 0.0368 0.0368 0.237k n n n

3 2

3 22 4

1

3

1 in 9 ft 12 0.237 5.4 in 3.41 in 7.58 5.4 in 0.237 5.4 in 514 in

3 ft

cr stI b kd A n d kd

3 3

4 4 4 431.9 k-ft514 in 2470 in 514 in 753 in

64.3 k-ft

cre cr g cr

a

MI I I I

M

Positive moment region:

42470 ine gI I

So, the weighted average value of Ie is:

4 4 4 4

( ) 1 20.7 0.15 0.7 2470 in 0.15 753 in 753 in 1950 ine average em e eI I I I

Now we can calculate the immediate deflection due to live load. Using the same logic described

in Example 13-16, assume that 67.5% of the loads will be carried by the column strip.

60 psf 18 ft 0.675 0.729 k/ft 60.8 lb/inLw

106 psf 18 ft 0.675 1.29 k/ft 107 lb/inDw

4

4

(column strip max) 4

in60.8 lb/in 20 ft 12

ft0.0048 0.0048 0.130 in

57,000 4500 psi 1950in

LL

w

EI

4

4

(column strip max) 4

in107 lb/in 20 ft 12

ft0.0026 0.0026 0.124 in

57,000 4500 psi 1950in

DD

w

EI

Step 2: Compute the immediate deflection of an interior middle strip, which should be taken in

the E-W direction, as these are the shorter span middle strips. Take an interior span between

column lines 2 and 3.

13-30

First compute Ma. The loads we must consider for deflection calculations are:

Dead load: 3

6.5 in lb1.0 Dead Load 150 25 psf 106 psf

in ft12

ft

Service load: 1.0 Dead Live 106 psf 60 psf 166 psf

Construction load: 2.0 Slab Dead Load 2 81.3 psf 163 psf

Therefore, cracking will be governed by the service load. Since we know that the moments

calculated in Problem 13-7b are based on an area load of 246 psf, we can take Ma to be 166/246 =

0.675 times the slab strip moments calculated using the loading described in Problem 13-7b.

Negative moment at B-2 and C-2: 0.675 28.3 k-ft 19.1 k-ft

Positive moment at midspan: 0.675 24.4 k-ft 16.5 k-ft

Now compute Mcr:

3in 17.5 4500 psi 11 ft 12 6.5 in12ft 468,000 lb-in 39.0 k-ft

3.25 in

r g

cr

t

f IM

y

It is unlikely that the middle strip will be significantly cracked given that the moments

which control the cracking state are so far below the cracking moment for the strip section.

Therefore, we can assume that:

Negative moment region:

43020 ine gI I

Positive moment region:

43020 ine gI I

And therefore the weighted average value of Ie is obviously:

4

( ) 3020 ine averageI

Now we can calculate the immediate deflection due to live load. Using the same logic described

in Example 13-16, assume that 32.5% of the loads will be carried by the middle strip.

60 psf 20 ft 0.325 0.390 k/ft 32.5 lb/inLw

106 psf 20 ft 0.325 0.689 k/ft 57.4 lb/inDw

4

4

(middle strip max) 4

in32.5 lb/in 18 ft 12

ft0.0048 0.0048 0.029 in

57,000 4500 psi 3020in

LL

w

EI

4

4

(middle strip max) 4

in57.4 lb/in 18 ft 12

ft0.0026 0.0026 0.028 in

57,000 4500 psi 3020in

DD

w

EI

Step 3: Compute the maximum immediate total deflection in the panel due to the live load, and

compare against allowable deflections according to ACI Code limits.

(max) 0.130 in 0.029 in 0.159 inL

13-31

in20 ft 12

ftACI Code Limit / 360 0.667 in360

We are OK.

(b) For the same floor panel, calculate the total deflection after the attachment

of partitions and compare to the limit given in ACI Code Table 9.5(b) for

partitions that are not likely to be damaged by long term deflections.

Assume that 85 percent of the dead load is acting when the partitions are

attached to the structure and assume that 25 percent of the live load will be

sustained for a period of one year.

We will assume that the total expected deflection after the attachment of partitions is a

sum of the following:

1. Dead load

Assuming that 85% of the dead load is already acting when the partitions are

attached, consider 15% of the immediate deflection due to dead load affects the

partitions.

Dead immediate 0.1 (0.124 in 0.028 in) 0.015 in

2. Live load

Assume that the total instantaneous deflection due to live load could occur at any

time once the partitions are attached, so include the full instantaneous deflection

due to live load.

Live immediate 0.159 in

3. Long term deflections

Assume that the full dead load acts plus 25% of the live load over the

course of a year. Use the suggested deflection coefficient of 3.0 to adjust

for long-term amplification effects.

Long term 3.0 0.124 in 0.028 in 0.25 0.159 in 0.575 in

Therefore, the total expected deflection felt by the partitions is:

partitions Dead immediate Live immediate Long term

partitions 0.015 in 0.159 in 0.575 in 0.749 in

Compare this to the ACI limit for slab systems supporting partitions which are not

likely to be damaged by large deflections:

in20 ft 12

ftACI Code Limit / 240 1.0 in240

We are OK.

13-12 Repeat the questions in Problem 13-11 for the following panels.

(a) An exterior panel along the west side of the floor system.

13-32

This solution will follow the same lines as the solution for Problem 13-11. The column

strip in question will still be taken along the N-S direction. The major change will lie in the

definition of moments acting in the column and middle strips.

(b) An exterior panel along the north side of the floor system.

This solution will follow the same lines as the solution for Problem 13-11. The column

strip in question will still be taken along the N-S direction. The major change will again lie in the

definition of moments acting in the column and middle strips.

(c) A corner panel.

This solution will again follow the same lines as the solution for Problem 13-11. The

column strip in question will still be taken along the N-S direction. The major change will still lie

in the definition of moments acting in the column and middle strips.

14-1

Chapter 14

14-1 For the slab panel shown in Fig. P14-1, use the yield-line method to determine the

minimum value of the area load, , at the formation of the critical yield-line

mechanism.

We may need to investigate two yield-line mechanisms to find the minimum qf.

1. Select first plastic mechanism.

2. Assume a virtual displacement equal to along the positive yield line from A to B.

3. Compute the internal work. For slab segment I the internal work is:

( ) ( )

Similarly, the internal work for slab segment II is:

( ) ( )

Therefore, the sum of the internal work is:

∑ (

) (

)

4. Compute the external work. The two end regions outside of the points A and B are essentially two

half-pyramids that can be combined into a single pyramid with a base area 18 ft x 2 x 20 ft. The

central region between points A and B has a triangular cross-section and extends over a length of

20 ft x (1 – 2 ). Thus, the external work is:

[

( )

] ( )

18 ft.

L1 = 22 ft.

Mechanism No. 1

9 ft.

L1

I

II

A B

L1 = 20 ft

14-2

5. Equate the external and internal work. Set EW = ∑ and solve for qf.

(

)

( )

6. Solve for the minimum value of qf. The solution table below starts with = 0.5 and slowly

decreases .

Numerator, kips Denominator, ft2 qf, ksf

0.50 57.4 120 0.478

0.48 59.1 122 0.482

0.46 60.8 125 0.488

From this table the minimum value of qf is 0.478 ksf and it occurs for a value of 0.50. Because of

this result, we should investigate a second possible yield-line mechanism.

7. Select a second possible plastic yield-line mechanism.

8. Assume a virtual displacement of along positive yield line C – D.


( ) ( )


( ) ( )


∑ (

) (

)

10. Compute the external work. The two end regions outside of the points C and D are essentially

two half-pyramids that can be combined into a single pyramid with a base area 20 ft x 2 x 18 ft.

L2 = 18 ft.

22 ft.

Mechanism No. 2

11 ft

I

II

C

D

L2

20 ft

10 ft

14-3

The central region between points C and D has a triangular cross-section and extends over a length of

18 ft x (1 – 2 ). Thus, the external work is:

[

( )

] ( )


(

)

( )

12. Solve for the minimum value of qf. The solution table below starts with = 0.5 and slowly

decreases .


0.50 57.2 120 0.477

0.48 57.9 122 0.475

0.46 58.7 125 0.47

0.44 59.6 127 0.469

0.42 60.6 130 0.468

0.4 61.6 132 0.467

0.38 62.8 134 0.467

From this table the minimum value of qf is 0.467 ksf and it occurs for an value of approximately

0.47. In this case it was important for us to investigate this second possible yield-line mechanism.

14-4



mechanism.

1. Select plastic yield-line mechanism.

2. Assume a virtual displacement equal to along the positive yield line from A to B.


( ) ( )


( ) ( )


∑

( ) (

)

4. Compute the external work. The end region below the point A is essentially a half-pyramid with a

base area of 20 ft x x 20 ft. The region between points A and B has a triangular cross-section and

extends over a length of 20 ft x (1 – ). Thus, the external work is:

[

( )

] ( )


(

)

( )

L1 = 20 ft.

20 ft.

Assumed yield-line mechanism

free edge

I

II

10 ft.

L1

A

B

14-5

6. Solve for the minimum value of qf. The solution table below starts with = 0.75 and increase in

increments of 0.05.


0.75 34.7 150 0.231

0.80 33.6 147 0.229

0.85 32.7 143 0.228

0.90 31.9 140 0.228

0.95 31.2 137 0.228

1.00 30.5 133 0.229

From this table the minimum value of qf is 0.228 ksf and it occurs for a value of approximately

0.90.

14-6



mechanism.

1. Select plastic yield-line mechanism.

This is a complex mechanism with many potential variables. We will select values for some of the

distances shown in this figure as an initial trial. A few points should be noted regarding this

mechanism. As stated in the text, a positive yield line between two adjacent plate segments will

extend to or toward the intersection of the axes of rotation for those two plate segments. Thus, the

positive yield line A-D starts at the intersection between the axes of rotation for segments I and II (at

point A). Similarly, the other two positive yield lines branching out from point D aim toward the

points E (intersection of axes of rotation for segments I and III) and F (intersection of axes of rotation

for segments II and III). Experience with this particular mechanism indicates that the minimum value

of qf will occur when the point D is relatively close to the corner column support.

2. Select trial values for the assumed mechanism. The following values were selected for the

mechanism shown above.

B-E = 24 ft C-F = 18 ft Y = 15 ft X = 20 ft

With these values selected, geometry can be used to determine the additional values shown on the

following figure. It should be noted that the three lines extending from the slab to line E-F (the axes

of rotation for segment III) are all perpendicular to that line.

3. Assume a virtual displacement equal to occurs at the point D.

; , and 15 ft 20 ft 4.80 ft

I II III

18 ft

24 ft


A B

C

Y

XLIII

D

E

F

I

II

III

14-7

4. Compute the internal work. For the three slab segments the internal work is:

( ) ( ) 24 ft

k-ft (6 8) 24 ft (22.4 k)

ft 15 ft

( ) ( ) 18 ft

k-ft (6 8) 18 ft (12.6 k)

k 20 ft

( )

k-ft 6 8.60 ft (10

k 4.80 ft

p n I

p n II

p III III

IW I m m

IW II m m

IW III m L

.8 k)


45.8 kIW

5. Compute the external work. Rather than calculate a displaced volume, we will determine the

magnitude of load acting on each plate segment and then multiply that load times the displacement at

the centroid of the segment. If we did this directly for the defined segments, it would be difficult to

find the centroids of each segment. Thus, for plate segment I we will calculate the work done by the

distributed load acting on the triangular segment A-D-E, and then subtract to work for the triangular

piece outside the actual slab panel. So, for segment I:

(12.9 15)

( ) (total) (outside)3 3

fEW I A A q

2

1 1 1 12.9 15( ) 48 15 24 12.9

2 3 2 3

(75.6 ft )

f

f

EW I q

q

18 ft

24 ft


A B

C

12.9'

17.1'

4.5'

D

E

F

I

II

III4.80

'

4.11

'

4.09

'

26.9'

24.5'

4.1'

14-8

Similarly, for segment II:

2

1 1 1 17.1 20( ) 36 20 18 17.1

2 3 2 3

(76.1 ft )

f

f

EW II q

q

And for segment III, where some additional small triangles are used outside the slab panel:

2

1 1 1 4.11 4.80 1 4.09 4.80( ) [ 60 4.80 26.9 4.11 24.5 4.09

2 3 2 3 2 3

1 4.11 4.80 1 4.09 4.80 3.10 4.11 5.5 4.09 ]

2 3 2 3

( ) (13.0 ft )

f

f

EW III

q

EW III q

Thus, the sum of external work is:

2(165 ft ) fEW q

6. Setting the sum of the internal work equal to the sum of the external work gives:

2

45.8 kips(trial) 0.278 ksf

165 ftfq

7. Estimate for the minimum value of qf. Rather than doing several trials, we will assume that the

selected trial mechanism is reasonable and should give an answer within ten percent of the true

minimum value for qf. Thus, we can say:

(min) 0.9 0.278 0.250 ksffq

14-9



mechanism.

1. Select a plastic yield-line mechanism. Try the mechanism shown below. Note that there is a single

variable for this mechanism, the value of .

2. Assume a virtual displacement equal to occurs at the point A.

( )


( ( )) ( ( ))

[

( )

]


( ) ( )

( ) [

]


∑ [ ( )

]

4. Compute the external work. The deflected shape is essentially a half-pyramid, with a combined

base area (18 ft x 24 ft)/2. Thus, the external work is:

[

]


( ( )

)


L1 = 18 ft.

L2

A

I

II

L2 = 24 ft.

L1

free edge

14-10

6. Solve for the minimum value of qf. Because the denominator is constant, we need to minimize the

numerator. It is expected that ≈ 0.5.


0.48 29.0 72.0 0.403

0.50 28.5 72.0 0.396

0.52 28.1 72.0 0.390

0.54 27.7 72.0 0.385

0.56 27.5 72.0 0.382

0.58 27.4 72.0 0.381

0.60 27.5 72.0 0.382

From this table the minimum value of qf is 0.381 ksf and it occurs for a value of 0.58.

14-11

14-5 The slab panel shown in Fig. P14-5 is 7.5 in. thick and is made of normal weight

concrete. The slab will need to support a superimposed dead load of 25 psf and a live

load of 60 psf. Assume that the slab will be designed with isotropic top and bottom

reinforcement.

(a) Find the critical yield-line mechanism and then use the appropriate load factors

and strength-reduction factors to determine the required sum of the nominal

negative- and positive-moment capacities ( ) in units of - ⁄ .

1. Determine the total factored load to be resisted.

Slab weight is (7.5/12) x 150 = 94 psf; Thus, the total dead load = 94 + 25 = 119 psf.

The live load can be reduced based on the panel area (20 x 26 = 520 ft2). Note: KLL = 1.0. For

this area, the reduced live load is equal to 54.5 psf.

So, the total factored load, qu = 1.2 x 119 + 1.6 x 54.5 = 230 psf = 0.230 ksf

2. Select a possible plastic yield-line mechanism. A couple items should be noted. First, in text

Example 14-4, the value of was close to 0.5, so having a virtual displacement of only at a single

point (point A) will give a relatively good answer. Also, because of the fixed supports at the support

edges for plate segments I and II, we would like to make the rotations of those plate segments smaller

than for plate segments III and IV. Thus, will be greater than 0.5.

3. Assume a virtual displacement equal to at point A.

, and 20 ft 26 ft

, and (1 ) 20 ft (1 ) 26 ft

I II

III IV

20 ft

26 ft


A

x 20 ft

x 26 ft

II

III

IV

I

14-12

3. Compute the internal work. For the four slab segments the internal work is:

1.3( )( ) ( ) 26 ft

0.77( )( ) ( ) 20 ft

n p

n p I

n p

n p II

m mIW I m m

m mIW II m m

1.3( ) 26 ft

1

0.77( ) 20 ft

1

p

p III

p

p IV

mIW III m

mIW IV m

From these, the sum of the internal work is:

∑ [ ( )

]

At some point in the design process need to make a decision on the ratio between the flexural strength

of the bottom reinforcement, mp, vs. the strength of the top reinforcement, mn. Reading ahead to

problem 14.5(b), we are asked to assume that . Thus, the internal work can be expressed

as a function of mp.

∑ [ ( )

]

4. Compute the external work. In previous problems we have solved for the distributed load, qf, at the

development of a failure mechanism. In this case, qf is equivalent to qu divided by the strength

reduction factor, , which will be assume to be equal to 0.9. Thus,

20.230 ksf

20 ft 26 ft (173 ft ) (44.2 k)3 0.9

uqEW

5. Equate the external and internal work.

[ ( )

]

[

]

6. Solve for the minimum required value of . The solution table below starts with = 0.5 and

increase in increments of 0.05.

Required , kip-ft/ft

0.50 2.97

0.55 3.07

0.60 3.13

0.65 3.11

From this table the minimum required value of is 3.13 kip-ft/ft. Assuming that ,

kip-ft/ft.

14-13

(b) Start with approximately equal to , and then select top and bottom

isotropic reinforcement to provide the required combined flexural strength

calculated in part (a). Be sure to check reinforcement spacing requirements

given in ACI Code Section 13.3.2.

1. From part (a), and kip-ft/ft. Select top and bottom bars to provide the

required moment strength. Assume and . Also, assuming that #4 bars will

suffice, use

For negative bending,

( )

( )

(

)

Using #4 bars,

Using #3 bars,

For positive bending,

( )

( )

(

)

Using #4 bars,

Using #3 bars,

2. Check reinforcement detailing requirements and select reinforcement:

Minimum reinforcement area:

. This is easily satisfied by the top and

bottom reinforcement required for strength purposes.

Maximum spacing: {

} To satisfy the strength requirements from step 1

and this maximum spacing requirement, select #3 bars for both top and bottom reinforcement. Bars

shall be spaced at 7.5 in. and 11 in. along the top and bottom faces of the slab, respectively, in both

orthogonal directions. Clear cover requirements of 0.75 in. shall be enforced. Proper anchorage shall

be provided for all slab reinforcement.

15-1

Chapter 15

15-1 Design wall footings to be supported 3 ft below grade for the following conditions.

Assume a soil density of ⁄ and normal weight concrete for both problems.

Service dead load is 6 kips/ft, service live load is 8 kips/ft. Wall is 12 in. thick.

Allowable soil pressure, , is 4000 psf. and .

The following design is done for a 1 ft length of footing. Although not included for simplicity, the

“per foot” notation applies to all units in the following solution.

1) Estimate size of footing and factored net pressure

The bottom face of the footing is 3 ft below the ground surface.

Assume the thickness of the footing is 12 in.

The soil density is 120 3lb/ft

Allowable net soil pressure: 3 34 ksf (1.0 ft 0.15 k/ft 2.0 ft 0.12 k/ft ) 3.61 ksfnq

Required footing area:

2

.

6 k 8 k3.88 ft

3.61 ksfreqA

Try 4 ft wide strip footing.

Factored net pressure:

1.2 6 k 1.6 8 k5 ksf

4 ft 1 ftnuq

2) Check one-way shear

Assume a concrete cover of 3 in., and #8 bars being used.

12 in. 3 in. 0.5 in. 8.5 in.d

The distance the footing extends to each side from the column face:

48 in. 12 in. / 2 18 in.

18 in. 8.5 in.5 ksf 1 ft 3.96 k

12 in/ftuV

2 0.75 2 3500 12 8.5 9050 lb 9.05 kc c wV f b d

Since cV is significantly greater than uV , we reduce the footing thickness to the

minimum allowable by ACI Section 15.7, i.e.:

6 in. + bar dia. + 3 in. cover. = 10 in.

10 in. 3 in. 0.5 in. 6.5 in.d

18 in. 6.5 in.5 ksf 1 ft 4.79 k

12 in/ftuV

√ √

Use 10 in. thick footing

15-2

3) Design reinforcement

21.5 ft

5 ksf 1ft 5.63 k-ft2

uM

Assume 0.90j

25.63 k-ft 12 in/ft0.21 in.

0.9 60 ksi (0.9 6.5 in)sA

2

,min 0.0018bh 0.0018 12 in. 10 in. 0.22 in.sA

Use #4 bars at with a spacing of 10 in. (20.24 in.sA )

10 in. 3 in. 0.25 in. 6.75 in.d 20.24 in. 60 ksi

0.4 in.0.85 3.5 psi 12 in.

a

6.75 0.4/ 0.850.003 0.040 0.005, 0.9

0.4/ 0.85t cm

d c

c

20.9 0.24 in. 60 ksi 6.75 in. 0.4 in./2

84.9 k-in. 7.07 k-ft

n

u

M

M

4) Check development

Bar spacing exceeds b2 and cover exceeds dbd This is Case 2 development.

1 1 60000 psi0.5 in. 20 in. 18 in. 3 in., NG.

25 25 1 3500 psi

e t y

d b

c

fd

f

We shall use 90º standard hooks to anchor the bars. The required development length for

a 90º standard hook is:

0.02 0.02 1 60000 psi0.5 in. 10 in. < 18 in. 3 in., OK.

1 3500 psi

e y

d b

c

fd

f

5) Temperature requirement for longitudinal reinforcement: 2

, . 0.0018bh 0.0018 48 in. 10 in. 0.86 in.s reqA

Use 5 #4 bars for longitudinal reinforcement.

Use 10 in. thick by 4 ft. wide footing with #4 bars at 10 in. o.c. with 90º hooks at both ends in the

transverse direction, and 5 #4 bars in the longitudinal direction.

15-2 Design wall footings to be supported 3 ft below grade for the following conditions.

Assume a soil density of ⁄ and normal weight concrete for both problems.

Service dead load is 15 kips/ft, service live load is 8 kips/ft. Wall is 16 in. thick.

Allowable soil pressure, , is 6000 psf. and .


The bottom face of the footing is 3 ft below the ground surface.

Assume the thickness of the footing is16 in.

The soil density is 120 3lb/ft

15-3

Allowable net soil pressure:

( ⁄ ⁄ ) Required footing area:

Try 4.5 ft wide strip footing.

Factored net pressure:


Assume a concrete cover of 3 in., and #8 bars being used.

16 in. 3 in. 0.5 in. 12.5 in.d


54 in. 16 in. / 2 19 in.

19 in. 12.5 in.6.84 ksf 1 ft 3.71 k

12 in/ftuV

2 0.75 2 3000 12 12.5 12.3 kc c wV f b d

Since cV is significantly greater than uV , we reduce the footing thickness to 12 in.:

12 in. 3 in. 0.5 in. 8.5 in.d

19 in. 8.5 in.6.84 ksf 1 ft 5.99 k

12 in/ftuV

√ √

Use 12 in. thick footing


21.583 ft

6.84 ksf 1ft 8.58 k-ft2

uM

Assume 0.90j

28.58 k-ft 12 in/ft0.25 in.

0.9 60 ksi (0.9 8.5 in)sA

2

,min 0.0018bh 0.0018 12 in. 12 in. 0.26 in.sA

Use #5 bars at with a spacing of 12 in. (20.31 in.sA )

12 in. 3 in. 0.31 in. 8.7 in.d 20.31 in. 60 ksi

0.61 in.0.85 3.0 psi 12 in.

a

8.7 0.61/ 0.850.003 0.033 0.005, 0.9

0.61/ 0.85t cm

d c

c

20.9 0.31 in. 60 ksi 8.7 in. 0.61/ 2 in.

141 k-in. 11.7 k-ft

n

u

M

M

15-4


Bar spacing exceeds b2 and cover exceeds dbd This is Case 2 development.

1 1 60000 psi0.625 in. 27.4 in. 19 in. 3 in., NG.

25 25 1 3000 psi

e t y

d b

c

fd

f

We shall use 90º standard hooks to anchor the bars. The required development length for

a 90º standard hook is:

0.02 0.02 1 60000 psi0.625 in. 14 in. < 19 in. 3 in., OK.

1 3000 psi

e y

hb b

c

fd

f

5) Temperature requirement for longitudinal reinforcement: 2

, . 0.0018bh 0.0018 54 in. 12 in. 1.17 in.s reqA

Use 6 #4 bars for longitudinal reinforcement.

Use 12 in. thick by 4 ft – 6 in. wide footing with #5 bars at 12 in. o.c. with 90º hooks at both ends

in the transverse direction, and 6 #4 bars in the longitudinal direction.

15-3 Design a square spread footing for the following conditions:

Service dead load is 350 kips, service live load is 275 kips. Soil density is ⁄ .

Allowable soil pressure is 4500 psf. Column is 18 in. square.

(normal

weight) and . Place bottom of footing at 5 ft below floor level.


Estimate thickness between 18 and 36 in., say 30 in.

Assume the floor thickness is 6 in.

Allowable net soil pressure: 3 34.5 ksf (2.5 ft 0.5 ft) 0.15 k/ft 2.0 ft 0.13 k/ft 3.79 ksfnq

Required footing area:

2

.

350 k 275 k165 ft

3.79 ksfreqA

Try 13 ft square footing.

Factored net soil pressure:

2

1.2 350 k 1.6 275 k5.09 ksf

169 ftnuq

2) Check thickness for two-way shear

Assume a concrete cover of 3 in., and #8 bars being used. Average effective depth:

30 in. 3 in. 1 in. 26 in.d 2

2 18 in. 26 in.5.09 ksf 169 ft 792 k

12 in./ftuV

15-5

cV is taken as the smaller of the following:

4 42 2 6

1c c w c w c w

c

V f b d f b d f b d

(Note c is the ratio of the column dimensions)

40 26 in.

2 2 7.94 44 in.

sc c w c w c w

o

dV f b d f b d f b d

b

(Note 40s for a column placed at the center of its footing)

4c c wV f b d

√ √

Use 30 in. thick footing.



156 in. 18 in. / 2 69 in.

[

]

√ √


25.75 ft

5.09 ksf 13 ft 1095 k-ft2

uM

Assume 0.90j

Use 18 #7 (210.8 in.sA )

( )


Bar spacing exceeds b2 and cover exceeds dbd This is a Case 2 development.

1 1 60000 psi0.875 in. 35.5 in.<69 in. 3 in., OK.

25 25 1 3500 psi

e t y

d b

c

fd

f

Use 13 ft square footing, 30 in. thick with 18 #7 bars each way.

6) Design footing-column joint

Factored load at base of column:

15-6

1.2 350 1.6 275 860 kips

Allowable bearing on footing:

0.65 0.85 3.5 18 18 2 1250 k

Allowable bearing on column (assume column 4000 psicf )

0.65 0.85 4 18 18 716 k

2860 716Area of dowels required 3.7 in.

0.65 60

Depending on how the column reinforcement is selected, the dowel bars will be selected

accordingly for construction simplicity.

15-4 Design a rectangular spread footing for the following conditions.

Service dead loads are: axial = 350 k, moment = 80 k-ft; service live loads are:

axial = 250 k, moment = 100 k-ft. The moments are acting about the strong axis of

the column section. Soil density is ⁄ . Allowable soil pressure is 5500 psf.

Place the bottom of the foundation at 4.5 ft below the basement floor. Assume the

basement floor is 6 in. thick and supports a total service load of 80 psf. Assume the

column section is 24 in. x 16 in., is constructed with 5000 psi normal-weight concrete

and contains 6 No. eight bars (Grade 60) placed to give maximum bending

resistance about the strong axis of the column section. Assume the footing will use

and .

1) Select footing dimensions based on allowable soil pressure. Assume a footing thickness of 3

ft. The net permissible bearing pressure is:

( ) Consider only the axial loads to establish the width of the footing.

⁄

Use axial loads and moments to establish the length of the footing. Assume a linear

distribution of soil pressure as shown in Fig. 15-5(b). Thus,

Set and .

( )

Solving for the positive root,

Set

2) Calculate factored soil pressure:

Factored loads are:

With these values, the factored soil pressures are:

15-7

3) Check footing thickness for two-way shear:

Dimension of the critical perimeter are:

Therefore,

( ) (

)

Calculate the two-way shear capacity, with cV taken as the smaller of:

(

) √

(

) √

√

(Note c is the ratio of the column dimensions)

(

) √

(

) √

√

(Note 40s for a column placed at the center of its footing)

4c c wV f b d

√ √

Use 36 in. thick footing.

4) Check for combined transfer of shear and moment:

√ ⁄

√ ⁄

(

) (

)

(

) (

)

( )

√ √

5) Check for one-way shear:

(

)

√ √

6) Design flexural reinforcement for long direction:

15-8

( )

Assume jd = 0.95d, and use

Use 15 #7 ( )

( )

7) Check development:

The available length is 72 in., so this is ok.

17-1

Chapter 17

17-1 The deep beam shown in Fig. P17-1 supports a factored load of 1450 kips. The beam

and columns are 24 in. wide. Draw a truss model neglecting the effects of stirrups

and the dead load of the wall. Check the strength of the nodes and struts, and design

the tension tie. Use (normal weight concrete) and .

1290 k

644 k

1450 k

1290 k

644 k

21.1 10.5

55.3°35.8º

30 24

243

69 36

Note: Length units are in inches.

A D

B C

Node 1

Node 2 Node 3

Fig. S17-1

1. Compute /uP and reactions

To include in calculations use / 1450 kips/0.75 1930 kipsuP

Sum moments about the center of the support at A, assuming the member acts as a simple

beam and all loads and reactions act along the axes of the columns.

17-2

69 18 151930 kips 644 kips

243 12 15DR

1930 kips 644 kips 1286 kips 1290 kipsAR

2. Select size of beam

Assume AR is the maximum value of /uV , and set this equal to ( 6 to 8 ) .cf bd For

24 in.b , solve to find d values between 142 and 106 in. Try an overall beam height h

of 10 ft. and assume that d is approximately 9 ft. = 108 in.

3. Isolate D-regions

Because the distance between the load and the reactions is less than 2h at both ends, the

beam consists of two D-regions, one of each side of the load.

4. Draw a strut-and-tie model and establish base dimensions of the nodes

Use a truss with two inclined struts and a tie, as shown below.

Compute cuf

Node 1 (CCC node):

0.85 0.85 1.0 4000 psi 3400 psicu n cf f

Node 2 and 3 (CCT nodes):

0.85 0.85 0.8 4000 psi 2720 psicu n cf f

Strut 1-2 (assume sufficient web reinforcement is used to satisfy Eq. 17-10):

0.85 0.85 0.75 4000 psi 2550 psicu s cf f

Select the sizes of the nodes assuming they are hydrostatic nodes with 2550 psicuf

Node 1

Divide the load into two components equal to the two reactions.

Width of left part of node 1290 k /(2.55 ksi 24 in.) = 21.1 in.

Similarly, the width of the right part of the node = 10.5 in.

Total node width 21.1 10.5 31.6 in. < 36 in. , OK.

Node 2

Width of node = 21.1 in. < 30 in. (assume reaction is at center of column)

Node 3

Width of node = 10.5 in. < 24 in. (assume reaction is at center of column)

5. Establish geometry of truss and forces in struts and tie:

Assume that 2 ft 10 ft. 2 ft. = 8 ft. = 96 in.vd h

Then, tan ( left part of span ) = 96 in./(69 18 15) in.; so, ( left )=53.1

Total span length = 243 12 15 216 in.

Then, tan ( right part of span ) 96 in./(216 72)in.; so, ( right) 33.7

17-3

Strut 1-2

Axial force = 1290 kips/sin53.1 1610 kips

Horizontal component = 1610 kips cos53.1 967 kips force in Tie 2-3

Node 2 (Note that was included in force calculations)

Node height = force in the Tie 2-3/( ) 967 k /(2.55 ksi 24 in.)cuf b

Node height = 15.8 in.; this is also the height of Node 1.

At this point we must check the assumed value for vd

Revised: 2 (15.8 in./2)=120 in. 15.8 in. = 104.2 in.vd h (should recycle)

Assume: 104 in.vd

With this value: (left)= 55.3 and (right) = 35.8

Axial forces: Strut 1-2 = 1570 kips and Tie 2-3 = 893 kip

Heights of nodes 1 and 2 = 14.6 in., and 120 in. 2 (14.6 in./2) 105.4 in.vd , OK.

Strut 1-3

Axial force: 644 kips/sin35.8 1100 kips

Horizontal component: 1100 kips cos35.8 892 kips ( approx. 893 kips, OK.)

6. Select reinforcement for Tie 2-3 (Note that was included in force calculations) 2893 k /60 ksi = 14.9 in.sA

Use 16 #9 bars ( 216.0 in.sA )

Provide 4 layers of 4 No. 9 hooked bars. These must be anchored into the column at each

end with anchorage starting where the centroid of the tie first meets the inclined struts at

each end of the beam. The length required for a 90° standard hook is 21.4 in. for a No. 9

bar, which is less than the dimension of each support of column. Thus, this should be OK.

The centroid of the bars should be at about the mid-height of the nodal zones at each end,

i.e. about 7.5 in. above the bottom of the beam.

7. Minimum web reinforcement to control cracking inclined struts

Because 0.75s was used for the inclined struts in both the left and right spans,

minimum web reinforcement that satisfies Eq. 17-10 is required throughout the length of

the beam. For a deep beam, it is a good practice to also satisfy the reinforcement

requirements in ACI sections 11.8.4 and 11.8.5

Part of the beam to the left of the concentrated load:

For horizontal reinforcement, try #5 bars on the front and back faces at a vertical spacing

of 12 in. 22 0.31 in.

0.0022 0.0015 (OK.)12 in. 24 in.

h

90 55.3 34.7h

For vertical reinforcement, try #5 two-legged stirrups at a spacing of 10 in.

17-4

22 0.31 in.0.0026 0.0025 (OK.)

10 in. 24 in.v

55.3v

Checking Eq. 17-10:

sin sin

(0.0022) (sin34.7 ) (0.0026) (sin55.3 ) 0.0034 0.003,OK.

sii i i

si

A

b

Part of the beam to the right of the concentrated load:

Use the same reinforcement. Here h35.8 and 90 35.8 54.2v

sin 0.00174 0.00151 0.00325 0.003 (OK.)i i

Therefore, throughout the span use 2 #5 horizontal bars at a vertical spacing of 12 in. and use #5

two-legged stirrups as a spacing of 10 in.

17-2 Repeat Problem 17-1, but include the dead load of the wall. Assume that stirrups

crossing the lines AB and CD have a capacity ∑ equal to one-third or more

of the shear due to the column load.

This problem is solved similarly to Problem 17-1. The factored dead load of the wall can

be added to the factored load applied on top of the beam. In this case, the vertical

reinforcement is designed such that the dead load of the wall portion below the lines AB

and CD can be transferred to the level at least above those lines.

17-3 Design a corbel to support a factored vertical load of 120 kips acting at 5 in. from

the face of a column. You should include a horizontal load equal to 20 percent of the

factored vertical load. The column and corbel are 14 in. wide. The concrete in the

column and corbel was cast monolithically. Use 5000-psi normal-weight concrete

and .

Interface transfer shear stress is limited to the smallest of

0.2 1000 psicf

480 0.08 880 psicf and

1600 psi

The transfer shear stress is taken as 880 psi.

Compute minimum effective depth: 120,000

13 in.0.75 880 14

d

Try 16 in. and 18 in.d h

Factored shear = 120 kips = 120,000 lbs.

Normal force 0.2 120 kips 24 kips 24000 lbs.

Moment 120 5 24 2 648 k-in. (Note: 2-in. moment arm assumed between horizontal load and horizontal reinforcement)

17-5

Shear Friction Steel

2120

1.90 in.0.75 1.4 1.0 60

uvf

y

VA

f

(Note that we have used 1.4 for monolithic concrete and 1 for normal-weight

concrete)

Flexural Steel

26481.0 in.

0.75 60 (0.9 16)fA

1.0 601 in.

0.85 5 14a

26480.93 in.

0.75 60(16 1/ 2)fA

Direct Tension Steel

2240.53 in.

0.75 60nA

Area of Tension Tie will be taken as the largest of: 2(0.93 0.53) 1.5 in.f nA A

22 / 2 2 1.90/3 0.53 1.8 in.vf nA A

2

,min

0.04 0.04 514 16 0.75 in.

60

csc w

y

fA b d

f

Use 3 #7 bars (or 2 #9), 2 21.80 in. (or 2.00 in. )sA

Horizontal Stirrups

Area required: 2/3 0.63 in.vfA

Use 2 #4 closed stirrups, 20.80 in.sA

Anchor tension tie into column

Use 90° standard hooks to anchor the tension tie. Use #7 bars.

Also assume that the ties are inside the column cage; therefore a modification factor of

0.7 is multiplied to the basic development length per ACI Code Section 12.5.3(a):

0.02 0.02 1 60000

0.7 0.7 0.7 0.875 10.4 in.1 5000

e y

dh b

c

fd

f

Assume a column size of 14 in. 14 in. , #8 bars used for column longitudinal

reinforcement, and a concrete cover of 2.5 in. to the longitudinal bars. The available

space for anchorage is:

14 in. 2.5 in. 0.5 in. 12 in. 10.4 in. , OK.

Fig. S17-2 shows the detailed reinforcement of the corbel.

17-6

3 #7 welded

to the angle

plate

2 #4

closed ties

2 #4

bars

14 in. 8 in.

9 i

n.

9 i

n.

Fig. S17-2

17-4 Repeat Problem 17-3, but with a factored vertical load of 100 kips and a factored

horizontal load of 40 kips.

The problem is solved similar to Problem 17-3.

17-5 Figure P17-5 shows the dapped support region of a simple beam. The factored

vertical reaction is 100 kips, and you should include a horizontal reaction of 20 kips.

Use normal-weight concrete with and reinforcement with

(weldable). The beam is 16 in. wide.

(a) Isolate the D-region.

(b) Draw a truss to support the reaction.

(c) Detail the reinforcement.

The factored load on the dap is 100 kips. To include in the design we shall design the dap for

100/0.75 133 kipsnV

In addition, design the dap for a horizontal force of

0.2 133 26.6 kipsnH

(a) Isolate the D-region.

Fig S17-3(a) shows the D-region.

17-7

3.5 16 10.564

10 30

15

15

10.5

14.5

2.5

2.5

120

147

147 203

13

3

13

3

13

3

179

198

D-region

A

B

C

F

D

E G

47.9°

42.2°

67.2°

B F

D

C E G

A

22

48

17

top reinforcement

bottom reinforcement

2 #5 U bars @ 3.8 in.

2 #4 U bars

@ 3.5 in.

4 #5 U-stirrups @ 3 in.

4 #5 U-stirrups @ 2 in.

4 #5 closed

double-leg stirrups

5 # 7 with left ends

welded to the angle plate

2 #8 U bar

(a) Truss model

(b) Reinforcement details

Note: Force units are in kips, length units are in inches.

144

17-8

(b) Draw a truss to support the reaction.

1. Select a strut-and-tie model

There are two basic types of truss which can be drawn:

- Vertical stirrups at end of deeper part

- Inclined stirrups in the end

We shall use the former detail. Fig. S17-3(a) shows the assumed truss model.

2. Locate nodes at A, B, and C

0 133 kipsBCBV N

2

, 133/60 2.22 in.s requiredA

Use 4 #5 closed double leg stirrups 2( 2.48 in. ).sA Place them as shown in Fig. S17-3(b)

because these must be as close to the end of the deeper part of the beam as possible.

The centroid of this group is at 1.5 (2 0.625) 0.75 3.5 in. from the end of the

deeper section.

Assume 2.5 in. from the bottom of the beam to the centroid of the steel at C, 2 in. at A,

and assume the centroid of the compression force B-F is 2.5 in. below the top of the

beam.

4. Compute the strut and tie forces

Solve Join at A

10.5 in.tan 1.11 47.9

9.5 in.

133/sin47.9 179 kips ( )ABN

9.5

133 26.6 147 kips 10.5

ADN

Solve Joint at B

0 133 kips BCV N

9.5

0 133 120 kips 10.5

BFH N

Locate node at D

Strut CD is axially compressed by the horizontal tensile force in AD and the

vertical tensile force in DE. The horizontal distance from C to D is

14714.5 16.0 in.

133

The axial force in CD is 198 kips.

The horizontal distance EG is 30 in. 3.5 in. 16 in. 10.5 in.

Solve Joint at E

10.5Horizontal component in EF 133 55.9 kips

25

Axial force in EF = 144 kips

Force in EG = 147 + 55.9 = 203

17-9

Joint F

Horizontal force to the right of F = 176 kips

As a check on the calculations, cut a section parallel to EF and sum moments

about F to calculate the force in EG EF133 36 in. +26.6 10.5 = T 25

203 kips OKEFT

5. Compute strut widths and see if they will fit

0.85 0.85 0.75 5000 3190 psicu s cf f

(Note that the struts are assumed to have confinement steel, hence 0.75s )

Strut AB

Assume the concrete outside the ties spalls at B. Remaining thickness of strut

16 2 1.5 13 in.

179Width 4.32 in.

3.19 13

Strut BF

120Width 2.89 in.

3.19 13

Strut CD

Use full thickness.

198Width 3.88 in.

3.19 16

Strut BE

144Width 2.82 in.

3.19 16

Struts having these approximate widths are shown in the drawing.

6. Select reinforcement for the ties

Tie BC

We have already chosen 4 #5 double leg closed stirrups in two groups of two

stirrups.

Ties DE and FG

Use 4 #5 double leg stirrups with 135 hooks spaced at 3 in. o.c. for tie DE and 4

#5 double-leg stirrups with 135 hooks spaced at 2-in. for tie FG.

Ties AD 2147/60 2.45 in.sA

Use 5 #7 bars. Weld these to the angle at A.

We need to extend these bars past D a distance equal to the development length.

These bars are considered top bars, from Table A-6, 55.2 48.3 in.d So,

extend the bars 50 in past D.

17-10

Tie CE 22.45 in.sA

Provide 2 #8 U bars spaced 1 in. clear above the bottom steel. Lap splice them

with the tension reinforcement in the bottom of the beam using a splice length of

7. Design confinement reinforcement:

Strut AB. Use only horizontal steel. Try 2 #4 U bars spaced equally at 3.5 in. o.c. 22 0.20 in.

0.007116 in. 3.5 in.

shh

v

A

bs

47.9h

sin 0.0053 0.0030h h , OK.

Strut CD. Use only horizontal steel. Try 2 #5 U bars spaced equally at 3.83 in. o.c.

22 0.31 in.

0.010116 in. 3.83 in.

shh

v

A

bs

42.2h

sin 0.0068 0.0030h h , OK.

18-1

Chapter 18

18-1 Check moment and shear strength at the base of the structural wall shown in Fig. P18-

1. Also, show that the given horizontal and vertical reinforcement satisfies all of the ACI

Code requirements regarding minimum reinforcement percentage and maximum

spacing. The given lateral loads are equivalent wind forces, considering both direct

lateral forces and the effects of any torsion. Use a load factor of 1.6 for the wind load

effects. The given vertical loads represent dead loads, and you can assume that the

vertical live loads are equal to 60 percent of the dead loads. Assume the wall is

constructed with normal-weight concrete that has a compressive strength of 4500 psi.

Assume all of the steel is Grade-60.

1. Calculate factored loads at base of wall.

,min

(base) 1.6 (7 k 15 ft 12 k 26 ft 18 k 37 ft 22 k 48 ft 20 k 59 ft)

5310 kip-ft

(base) 1.6 7 12 18 22 20 kips 126 kips

(base) 0.9 230 k 207 kips

u

u

u

M

V

N

2. Calculate flexural strength. The total area of vertical wall reinforcement is:

2 2 2240 in.

2 0.20 in. 0.40 in. 8.00 in.12 in.

wstA

s

From Eq. (18-26a), the percentage of vertical (longitudinal) reinforcement is:

28.00 in.0.00333

10 in. 240 in.

st

w

A

h

From Eq. (18-26b), the vertical reinforcement index is:

60 ksi

0.00333 0.04444.5 ksi

y

c

f

f

From Eq. (18-27), the axial load index is:

207 kips

0.019210 in. 240 in. 4.5 ksi

u

w c

N

h f

For 4500 psi concrete, 1 = 0.825. Then, from Eq. (18-28) the depth to the neutral axis is:

1

0.0192 0.0444240 in. 19.3 in.

0.85 2 0.85 0.825 2 0.0444wc

This is very small compared to d (taken as 0.8 ℓw), so this is clearly a tension-controlled section

and = 0.9. From Eq. (18-25a), the tension force in the vertical reinforcement is:

2 240 19.3

8.00 in. 60 ksi 441 kips240

wst y

w

cT A f

18-2

And, from Eq. (18-29) the nominal moment strength at the base of the wall is:

240 in. 240 in. 19.3 in.441 k 207 k

2 2 2 2

53,000 k-in. 22,800 k-in. 75,800 k-in. 6320 k-ft

w wn u

cM T N

Using the strength reduction factor, , the design strength is:

0.9 6320 5690 k-ft (o.k.)n uM M

3. Check shear strength. Because the wall is subjected to compression, we are permitted to use

Eq. (18-41) to determine the concrete contribution to shear strength. For this calculation we will

assume d = 0.8 x ℓw = 0.8 x 240 = 192 in., as permitted in ACI Code Section 11.9.4.

2 2 1 4500 psi 10 in. 192 in.

258,000 lbs 258 kips

c cV f hd

Using = 0.75 for shear, the design strength contribution from the concrete is:

0.75 258 193 kipsc uV V

Thus, no horizontal reinforcement is required for shear strength. However, because Vu exceeds

one-half of Vc, the reinforcement requirements in ACI Code Section 11.9.9 must be satisfied.

Horizontal reinforcement:

From Eq. (18-45a), the percentage of horizontal reinforcement is:

2,

2

2 0.20 in.0.0025 0.0025 (o.k.)

10 in. 16 in.

v horiz

t

A

h s

The maximum center-to-center spacing for the horizontal reinforcement is the smallest of ℓw/5 (48

in.), 3h (30 in.) and 18 in. Thus, the spacing of 16 in. for the horizontal reinforcement is ok.

Vertical reinforcement:

Because the wall aspect ratio, hw/ℓw = 59ft/20ft = 2.95, exceeds 2.5, the minimum required

percentage of vertical reinforcement is 0.0025. From step 2, ℓ = 0.00333, which exceeds the

minimum value. The maximum center-to-center spacing for the vertical reinforcement is the

smallest of ℓw/3 (80 in.), 3h (30 in.) and 18 in. Thus, the provided spacing of 12 in. for the

vertical reinforcement is ok.

18-3

18-2 Design a uniform distribution of vertical and horizontal reinforcement for the

structural wall shown in Fig. P18-2. Your design must satisfy all of the ACI Code

requirements, as well as the requirements for minimum reinforcement percentage and

minimum spacing. The given lateral loads are equivalent wind forces, considering both

direct lateral forces and the effects of any torsion. Use a load factor of 1.6 for the wind

load effects. Assume the wall is constructed with normal-weight concrete that has a

compressive strength of 4000 psi. Assume all of the steel is Grade-60.

1. Calculate factored loads at base of wall.

,min

(base) 1.6 (75 k 12 ft 100 k 24 ft) 5280 kip-ft

(base) 1.6 75 100 kips 280 kips

(base) 0.9 (70 40) kips 99.0 kips

u

u

u

M

V

N

2. Flexural design. Based on the results from Problem 18-1 (a wall with similar design base

moment and similar dimensions), select a trial value for ℓ = 0.0035. Then, from Eq. (18-26b),

60 ksi

0.0035 0.05254 ksi

y

c

f

f

From Eq. (18-27), the axial load index is:

99.0 kips

0.010310 in. 240 in. 4 ksi

u

w c

N

h f

For 4000 psi concrete, 1 = 0.85. Then, from Eq. (18-28) the depth to the neutral axis is:

1

0.0103 0.0525240 in. 18.2 in.

0.85 2 0.85 0.85 2 0.0525wc

This is very small compared to d (taken as 0.8 ℓw), so this is clearly a tension-controlled section

and = 0.9. The total area of vertical wall reinforcement is: 20.0035 10 in. 240 in. 8.40 in.st wA h

From Eq. (18-25a), the tension force in the vertical reinforcement is:

2 240 18.2

8.40 in. 60 ksi 466 kips240

wst y

w

cT A f

And, from Eq. (18-29) the nominal moment strength at the base of the wall is:

240 in. 240 in. 18.2 in.466 k 99.0 k

2 2 2 2

55,900 k-in. 11,000 k-in. 66,900 k-in. 5570 k-ft

w wn u

cM T N

Using the strength reduction factor, , the design strength is:

0.9 5570 5020 k-ft (not o.k.)n uM M

18-4

For a wall with relatively low axial load, the nominal moment strength should increase approximately

linearly with increases in the percentage of vertical reinforcement. Thus,

5280

(req'd.) (trial) 0.0035 0.00368(trial) 5020

u

n

M

M

Round this up a little and try ℓ = 0.0038. Then, redo the calculations to find:

20.0570, 19.3 in., 9.12 in. , 503 kips, and

5350 k-ft (o.k.)

st

n u

c A T

M M

Select vertical reinforcement as:

No. 4 bars at 10 in. spacing in each face (EF), results in ℓ = 0.00400

No. 5 bars at 16 in. spacing in each face (EF), results in ℓ = 0.00388

Either selection will work. To use less total bars, select No. 5 bars at 16 in. spacing, EF. It is good

practice to replace the pair of No. 5 bars at the edges of the wall with a pair of No. 6 bars.

3. Shear Design. The aspect ratio for this wall is, hw/ℓw = 24/20 = 1.20. Thus, this is a short wall and

the shear strength contribution from the concrete is probably given by Eq. (18-43). Using d = 0.8 x ℓw

= 192 in., results in:

3.3 4

99,000 lbs 192 in. 3.3 1 4000 psi 10 in. 192 in.

4 240 in.

(401,000 19,800) lbs 421,000 lbs 421 kips

uc c

w

N dV f hd

Before accepting this value, we will check the value of Vc from Eq. (18-44). For this flexural-shear

strength equation, we need to evaluate the ratio of Mu/Vu at the critical section above the base of the

wall, as defined in Fig. 18-19. For this wall, ℓw/2 = 10 ft, governs. At that section the factored

moment is,

(crit. sect.) (base) (base)

2

5280 kip-ft 280 kip 10 ft 2480 kip-ft

wu u uM M V

Thus, the ratio of Mu/Vu = 2480/280 = 8.86 ft. Using this value, the denominator in the second term

of Eq. (18-43) is,

8.86 ft 10 ft 1.14 ft2

u w

u

M

V

Because this is a negative number, Eq. (18-44) is not valid for this wall. So, using = 0.75 and the

value for Vc from Eq. (18-43):

0.75 421 kips 316 kipsc uV V

18-5

Thus, no horizontal reinforcement is required for shear strength. However, because Vu exceeds

one-half of Vc, the reinforcement requirements in ACI Code Section 11.9.9 must be satisfied.

Horizontal reinforcement:

Use 2 No. 4 bars at 16 in. spacing in each face. Then, from Eq. (18-45a), the percentage of

horizontal reinforcement is:

2,

2

2 0.20 in.0.0025 0.0025 (o.k.)

10 in. 16 in.

v horiz

t

A

h s

The maximum center-to-center spacing for the horizontal reinforcement is the smallest of ℓw/5 (48

in.), 3h (30 in.) and 18 in. Thus, the provided spacing of 16 in. for the horizontal reinforcement is

ok.

Vertical reinforcement:

Because t = 0.0025, the minimum value for ℓ in Eq. (18-46) is 0.0025. From the flexural

design, the provided value for ℓ = 0.00388, so it is o.k. Also, the selected spacing of 16 in. is less

than the smallest of ℓw/3 (80 in.), 3h (30 in.) and 18 in. (o.k.)

18-3 The structural wall shown in Fig. P18-3 is subjected to gravity loads (

and ) and an equivalent, static, lateral earthquake load, )

(torsion effects included). Check the moment strength at the base of the structural

wall assuming that the distance, , from the base of the structure to the lateral force,

, is equal to two-thirds of the wall height. Use a capacity-design approach to check

the shear strength of the wall and assume the distance, , is equal to one-half of the

wall height for this check. Also, show that the given horizontal and vertical

reinforcement in the web of the structural wall satisfies all of the ACI Code

requirements regarding minimum reinforcement percentage and maximum spacing.

Assume the wall is constructed with normal-weight concrete that has a compressive

strength of 5000 psi. Assume all of the steel is Grade-60.

Use:

for the flexural strength check

for the capacity-based shear strength check

1. Calculate factored loads for flexural strength check at base of wall.

,min

(base) 1.0 150 k 2 3 60 ft 6000 kip-ft

(base) 0.9 150 k 135 kips

u

u

M

N

2. Flexural strength. For a boundary element in tension, use Eq. (18-30) to find:

212 0.79 in. 60 ksi 569 kipss yT A f

Assuming that the depth of the compression stress block does not exceed the size of the boundary

element, use Eq. (18-32) to find:

569 k 135 k

8.28 in. ( 20 in., o.k.)0.85 5 ksi 20 in.0.85

u

c

T Na

f b

18-6

Assume d = ℓw – 20in./2 = 180 – 10 = 170 in. Then, from Eq. (18-33), the nominal moment

strength is:

2 2

8.28 in. 180 in. 8.28 in. 569 k 170 in. 135 k

2 2

(94,400 11,600)k-in. 8830 kip-ft

wn u

aaM T d N

With a = 8.28 in., it is clear that this is a tension-controlled section, and thus = 0.9. So,

0.9 8830 7950 kip-ftn uM M

The wall is substantially over-designed in flexure and we should reduce the steel in the boundary

element to reduce the shear required to develop the flexural strength. One possible redesign is to

use eight No. 9 bars in each boundary element. This leads to the final result of Mn = 6860 kip-ft.

3. Design shear (determine capacity-based design shear using original flexural design). Assume

that the probable axial load is:

150 k 100 k 250 kipspr D LN N N

With this axial load the wall moment strength will be reevaluated and referred to as the probable

moment strength, Mpr.

First, the depth of the compression stress block is,

569 k 250 k

9.64 in.0.85 0.85 5 ksi 20 in.

u

c

T Na

f b

This is larger than calculated previously, but it is still clear that the tension steel in the boundary

element will be yielding. With this value of a, the moment strength is:

2 2

94,000 k-in. 21,300 k-in.

115,000 kip-in. 9610 kip-ft

wpr pr

aaM T d N

With this moment and assuming that x = 0.5 x 60 ft = 30 ft, the capacity-based design shear is,

9610 k-ft

(cap-based) 320 kips0.5 0.5 60 ft

pr

u

w

MV

h

4. Check shear strength. For this wall, the value of Acv in Eq. (18-48) is:

210 in. 180 in. 1800 in.cv wA h

Because this is a slender wall, c = 2.0. Eq. (18-45b) will be used to determine t for the distributed

horizontal reinforcement.

18-7

2,

2

2 0.20 in.0.0025

10 in. 16 in.

v horiz

t

A

h s

Using the values calculated here, the nominal shear strength of the wall from Eq. (11-48) is,

2

2

1800 in. 2 1 5000 psi 0.0025 60,000 psi

1800 in. 141 psi 150 psi 524,000 lbs 524 kips

n cv c c t yV A f f

Using = 0.75 for shear,

0.75 524 k 393 kips (cap-based) (o.k.)n uV V

The vertical and horizontal steel percentages in the web of the wall (both 0.0025) and the bar spacing

(16 in. both horizontal and vertical) satisfy the requirements of ACI Code Section 11.9.9, which are

applicable for this wall.

18-4 The structural wall shown in Fig. P18-4 is subjected to gravity loads ( and

) and a single, equivalent, static, lateral earthquake load ( ) at the

top of the wall (at roof level). For the given uniform distribution of vertical and

horizontal reinforcement, check the moment strength at the base of the wall and use a

capacity-deisgn approach to check the shear strength of the wall. Also, show that the

given horizontal and vertical reinforcement in the structural wall satisfies all of the ACI

Code requirements regarding minimum reinforcement percentage and maximum

spacing. Assume the wall is constructed with normal-weight concrete that has a

compressive strength of 4000 psi. Assume all of the steel is Grade-60.

1. Calculate factored loads for flexural strength check at base of wall.

,min

(base) 1.0 220 k 15 ft 3300 kip-ft

(base) 0.9 80 k 72 kips

u

u

M

N

2. Flexural strength. From Eq. (18-454a) the vertical reinforcement percentage is:

2,

1

2 0.31 in.0.00388

10 in. 16 in.

v vertA

h s

From Eq. (18-26b) the reinforcement ratio for the vertical reinforcement is:

60 ksi

0.00388 0.05824 ksi

y

c

f

f

From Eq. (18-27) the axial load ratio is:

72 kips

0.0075010 in. 240 in. 4 ksi

u

w c

N

h f

18-8

For 4000 psi concrete, 1 = 0.85. Then, from Eq. (18-28), the depth to the neutral axis is:

1

0.0075 0.0582240 in. 18.8 in.

0.85 2 0.85 0.85 2 0.0582wc

This is very small compared to d (0.8ℓw = 192 in.), so this is clearly a tension-controlled section

and = 0.9. For the longitudinal steel,

2 2240 in.

2 0.31 in. 9.30 in.16 in.

stA

Then, from Eq. (18-25a), the flexural tension force is:

2 240 in. 18.8 in.

9.30 in. 60 ksi 514 kips240 in.

wst y

w

cT A f

Finally, the nominal moment capacity is found using Eq. (18-29):

2 2

240 in. 240 in. 18.8 in. 514 k 72 k

2 2

(61,700 7960)k-in. 5810 kip-ft

w wn u

cM T N

Applying the strength factor, the design flexural strength is,

0.9 5810 5230 kip-ftn uM M

The wall is substantially over-designed in flexure and we should reduce the vertical

reinforcement in the wall to reduce the shear required to develop the flexural strength. One

possible redesign is to use two No. 4 bars at a 16 in. spacing in each face. For this reinforcement

ℓ = 0.0025 (the minimum value) and the design flexural strength, Mn = 3670 kip-ft.

3. Design shear (determine capacity-based design shear using original flexural design). Assume

that the probable axial load is:

80 k 40 k 120 kipspr D LN N N

With this axial load, the axial load ratio is:

120 kips

0.012510 in. 240 in. 4 ksi

And, the depth to the neutral axis is:

0.0125 0.0581

240 in. 20.2 in.0.85 0.85 2 0.0581

c

18-9

Then, the flexural tension force is:

2 240 in. 20.2 in.

9.30 in. 60 ksi 511 kips240 in.

wst y

w

cT A f

Finally, the probably moment capacity is:

240 in. 240 in. 20.2 in.511 k 120 k

2 2

(61,300 13,200)k-in. 6210 kip-ft

prM

With this probable moment strength, the capacity-based design shear is,

6210 k-ft

(cap-based) 414 kips15 ft

pr

u

w

MV

h

4. Check shear strength. For this wall, Eq. (18-45b) will be used to determine t for the distributed

horizontal reinforcement.

2,

2

2 0.20 in.0.0025

10 in. 16 in.

v horiz

t

A

h s

The effective wall shear area from Eq. (18-48) is:

210 in. 240 in. 2400 in.cv wA h

Because this is a squat wall (hw/ℓw = 0.75), c = 3.0. Thus, the nominal shear strength of the wall from

Eq. (11-48) is,

2

2

2400 in. 3 1 4000 psi 0.0025 60,000 psi

2400 in. 190 psi 150 psi 815,000 lbs 815 kips

n cv c c t yV A f f

Using = 0.75 for shear,

0.75 815 k 612 kips (cap-based) (o.k.)n uV V

The vertical and horizontal steel percentages in the web of the wall (both ≥ 0.0025) and the bar

spacing (16 in. both horizontal and vertical) satisfy the requirements of ACI Code Section 11.9.9,

which are applicable for this wall.

18-5 Design a uniform distribution of vertical and horizontal reinforcement for the first story

of the structural wall shown in Fig. P18-5. The given horizontal loads are strength-level

static equivalent earthquake forces, so use a load factor of 1.0. The vertical dead and

live loads are from the tributary area adjacent to the wall. Assume the given lateral

loads include the direct shear force and any torsional effects that need to be considered.

18-10

Use a compressive strength of 4000 psi and Grade 60 reinforcement. Your final design

should satisfy all ACI Code requirements for minimum reinforcement percentages and

maximum spacing.

Try using ( ) ( ) . Use 2#4 bars at 16 in. both ways.

( )

{

⁄

}

Start with flexural design. Calculate :

( ) ( )

⁄

(

) (

)

Calculate :

( ) ( )

Therefore, - - , and the two curtains of #4 vertical bars spaced at 16

in. on-center are OK.

Continue with the shear design. Calculate ( ) based on the probable moment capacity of

the wall and assuming that the lateral force resultant is located at .

( ) ( ) ( )

⁄

(

) (

)

18-11

( )

⁄

( )⁄

Calculate :

( ⁄ )

( √ ) ( √ )

Therefore,

( )

The capacity and demand are approximately equal, so the horizontal reinforcement is adequate.

Final design:

Vertical Reinforcement: 2 curtains of #4 bars spaced at 16 in. on-center.

Horizontal Reinforcement: 2 curtains of #4 bars spaced at 16 in. on-center.

19-1

Chapter 19

19-1 Use Eq. (19-28) to check the need for specially confined boundary elements in the

structural wall described in Problem 18-3. Assume the design displacement, , at the

top of the wall is 0.6 ft. Use all of the dimensions, loading information, and material

properties given in Problem 18-3. If a specially confined boundary element is required,

define the required vertical and horizontal dimensions for the boundary element.

1. Building drift ratio. The ratio, u/hw, represents the building drift ratio due to the design

earthquake for the building in question. We are given a value for the design displacement, u, at the

top of the building, so the building drift ratio is:

0.6 ft

Building drift ratio 0.010 (1%)60 ft

u

wh

2. Limit for neutral axis depth. Eq. (19-27) gives a limiting value for the neutral axis depth. For

larger values, the wall boundary elements must be confined.

15 ft 12 in./ft(limit) 30.0 in.

600 600 0.010

w

u w

ch

3. Calculated neutral axis depth. The depth of the neutral axis for this check should come from

the calculation of the probable moment strength of the wall, which was calculated in step 3 of

Problem 18-3. In that calculation the depth of the equivalent stress block, a, was found to be 9.64

in. Then, using 1 = 0.80 for 5000 psi concrete, the corresponding neutral axis depth is:

1

9.64 in.12.1 in. (limit)

0.80

ac c

Therefore, special confinement reinforcement is not required in the wall boundary elements.

19-2

19-2 Use Eq. (19-29) to check the need for specially confined boundary elements in the

structural wall described in Problem 18-3. Use all of the dimensions, loading

information, and material properties given in Problem 18-3. If a specially confined

boundary element is required, define the required vertical and horizontal dimensions

for the boundary element.

1. Factored loads at base of wall. From step 1 of Problem 18-3:

,min

(base) 6000 kip-ft 72,000 kip-in.

(base) 135 kips

u

u

M

N

2. Wall section properties. To determine the gross area and moment of inertia of the wall, divide

the wall into rectangular pieces representing the web and the two boundary elements. Then,

2 2

(web) 2 (boundary)

10 in. 140 in. 2 20 in. 2200 in.

A A A

And,

3 32 2

6 4

10 140 20 200 2 20 20 80

12 12

7.43 10 in.

i i i

i

I I A y

Also,

180 in.

90 in.2 2

wy

3. Combined stress in edge of wall at base of wall. Eq. (19-28) will be used to find the combined

stress at the edge of the wall at its base. If that stress exceeds the limit of 0.2fc', then we will need use

special confinement reinforcement in the wall boundary and continue that reinforcement up the wall

until we reach a section where the combined stress in the edge of the wall is less than 0.15fc'. Using

the values from steps 1 and 2:

2 6 4

135 k 72,000 k-in. 90 in.

2200 in. 7.43 10 in.

0.061 ksi 0.872 ksi 0.934 ksi 0.2 5 ksi 1.0 ksi

u uc

N M yf

A I

Therefore, special confinement reinforcement is not required in the wall boundary elements.

Documents

Reinforced Concrete Mechanics Design Solutions