Sample Reinforced concrete mechanics and design Wight 7th edition solutions manual pdf PDF

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Authors: James K. Wight
Published: Pearson 2015
Edition: 7th
Pages: 278
Type: pdf
Size: 9.7MB
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Chapter 3 https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 3-1

What is the significance of the “critical stress”? (a)

FOLFNKHUHWRGRZQORDG

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. @solutionmanual1

3-3 The concrete containing Type I cement in a structure is cured for 3 days at 70° F https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 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: C 

FOLFNKHUHWRGRZQORDG 5 F 32  , so 70° F = 21.1° C and 40° F = 4.4° C 9

From Eq. 3-6: n

M   (Ti  10)(ti ) i 1

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

3-5

Use Fig. 3-12a to estimate the compressive strength 2 for bi-axially loaded concrete subject to: (a)  (b)

1 = 0.0, 2 = fc'

(c)

1 = 0.5 fc' in compression, 2 = 1.2 fc'

1 = 0.75 ft' in tension, 2 = 0.5 fc'

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:

@solutionmanual1

3-6 What factors affect the shrinkage of concrete? https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ (a)

Relative humidity. Shrinkage increases as the relative humidity decreases, reaching a maximum at RH ≤ 40%.

FOLFNKHUHWRGRZQORDG

3-7

(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.

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.

@solutionmanual1

3-8 A structure is made from concrete containing Type 1 cement. The average ambient https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 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 crossFOLFNKHUHWRGRZQORDG 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:

,

@solutionmanual1

https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 2. Compute the creep coefficient for the time since loading, , using Eq. (3-35).

FOLFNKHUHWRGRZQORDG

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, (

)

()

(

)

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@solutionmanual1 3-5

Chapter 4

https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 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 FOLFNKHUHWRGRZQORDG 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  0.3 kips/ft 144

2. Compute the factored moment, M u : Factored load/ft: wu = 1.2(0.30 + 1.40) + 1.6(1.50) = 4.44 k/ft M u  wu

2 44  20 8  222 kip-ft

3. Compute the nominal moment capacity of the beam, M n 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)):

   1c 

A f 3.00  60000 s y   5.04 in. '  3500 12 0.85 0.85 f b c

For fc' 3500 psi, 1  0.85 . Therefore, c  

5.04  5.93 in. 1  0.85

Check whether tension steel is yielding:  d  c  21.5  5.93    Using Eq.(4-18)        0.003  0.00788 s t  c  cu  5.93  Thus,  s > 0.002 and the steel is yielding ( fs  fy ).

 

Compute the nominal moment strength, using Eq. (4-21): 5.04   3.00  60000   21.5     2   M  A f d   285 kip-ft n s y 2 12000

@solutionmanual1 Since, t  0.00788  0.005 the section is clearly tension-controlled and  =0.9. Then,  M n  0.9 285 kip-ft  256 kip-ft. Clearly, M n  M u

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

https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ concentrated service (unfactored) live load of 12 kips as shown. The concrete is psi and the steel is Grade 60. Use load and normal-weight concrete with strength-reduction factors form ACI Code Section 9.2 and 9.3. For the end section FOLFNKHUHWRGRZQORDG 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  0.563kips/ft 144

2. Compute the factored moment, M u . Factored distributed load/ft: wu = 1.2(0.563 + 1.0) = 1.88 k/ft Factored live load is a concentrated load: Pu  1.6  12  19.2 kips



M u   wu

u





1.88 10 2  19.2  9   267 kip-ft 2

3. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 6 No. 8 bars = 6  0.79 in.2 =4.74 in.2 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f s y 4.74 60000   2.79 in. 1 '  4000 30 0.85 0.85 f b c '  3.28 in. For fc 4000 psi, 1  0.85 . Therefore, c    2.79 1 0.85

  c

Check whether tension steel is yielding:  d  c  15.5  3.28    Using. Eq.(4-18)        0.003  0.011 > 0.0021 s t  c  cu  3.28  Thus,  > 0.002 and the steel is yielding ( fs  fy ). s

 

Compute the nominal moment strength, using Eq. (4-21):

  M  A f d  n s y 2

2.79   4.74  60000  15.5  2    334 kip-ft 12000

@solutionmanual1 Since, t  0.011  0.005 the section is clearly tension-controlled and  =0.9. Then,  M n 0.9 334 301 kip-ft  267 kip-ft. Clearly, M n  M u

4-3

(a)

Compare Use for singly reinforced rectangular beams the following properties. strength reduction factors from ACIhaving Code Sections 9.2 and

https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 9.3.

FOLFNKHUHWRGRZQORDG

Beam

b

d

No.

(in.)

(in.)

1 2 3 4 5

12 12 12 12 12

22 22 22 22 33

Bars 3 No. 7 2 No. 9 plus 1 No. 8 3 No. 7 3 No. 7 3 No. 7

f c'

fy

(psi) 4,000 4,000 4,000 6,000 4,000

(psi) 60,000 60,000 80,000 60,000 60,000

Beam No.1 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding.

For

1  0.85 . Therefore,

,

( ) (

)

⁄

⁄

)

(

Thus,  > 0.002 and the steel is yielding ( fs  fy ). s Since, t  0.005 the section is tension-controlled and  =0.9. (

)

( For Beam 1,

)

-

Beam No.2 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding. (

)

@solutionmanual1 For

,

1  0.85 . Therefore,

⁄

⁄

https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ ) ( ) ( > 0.002 and Thus, ) the ( steel is yielding ( f s  f y ). s Since, t  0.005 the section is clearly tension-controlled and  =0.9. FOLFNKHUHWRGRZQORDG (

(

)

For Beam 2,

)

)

(

-

Beam No.3 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding.

For

1  0.85 . Therefore,

,

( ) (

)

(

⁄

⁄

)

Thus,  > 0.002 and the steel is yielding ( fs  fy ). s Since, t  0.005 the section is clearly tension-controlled and  =0.9. (

)

( For Beam 3,

)

-

Beam No.4 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding.

For

,

⁄

. Therefore,

⁄

@solutionmanual1 ( ) (

)

(

)

Thus,  s > 0.002 and the steel is yielding ( fs  fy ). https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ Since, t  0.005 the section is tension-controlled and  =0.9.

FOLFNKHUHWRGRZQORDG )

(

(

)

For Beam 4,

-

Beam No.5 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding.

For

1  0.85 . Therefore,

,

( ) (

)

(

⁄

⁄

)

Thus,  > 0.002 and the steel is yielding ( fs  fy ). s Since, t  0.005 the section is tension-controlled and  =0.9. )

( For Beam 5, (b)

(

)

Taking beam 1 as the reference point, discuss the effects of changing and d on . (Note that each beam has the same properties as beam 1 except for the italicized quantity.)

Beam No.

M

n

(kip-ft) 1 167 2 250 3 219 @solutionmanual1 4 171 5 257

Effect of As (Beams 1 and 2) An increase of 55% in As (from 1.80 to 2.79 in. ) caused an increase of 50% in  M n . Increasing https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/ 2

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.

FOLFNKHUHWRGRZQORDG Effect of f y (Beams 1 and 3) An increase of 33% in f y caused an increase of 31% in  M n . Increasing the steel yield strength has essentially the same effect as increasing the tension steel area. Effect of f c' (Beams 1 and 4) An increase of 50% in f c' caused an increase of 2% in  M n . 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  M n . 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 effective way?

? What is the least

Disregarding any other effects of increasing d , As or f y such as changes in cost, etc., the most effective way to increase M n is to increase the effective flexural depth of the section, d , followed by increasing f y and As . Note that increasing f y and As too much may make the beam over-reinforced and thus will result in a decrease in ductility. The least effective way of increasing  M n is to increase f c' . Note that increasing fc' will cause a significant increase in curvature at failure.

@solutionmanual1

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 FOLFNKHUHWRGRZQORDG –reduction factors from ACI Code Sections 9.2 and 9.3. Also check .

https://gioumeh.com/product/reinforced-concrete-mechanics-and-design-solutions/

1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor, . Tension steel area: As = 4 No. 7 bars = 4  0.60 in.2 =2.40 in.2 Compute the depth of the equivalent rectangular stress block, , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f 2.4  60000 s y   c   2.65 in. 1 0.85 f 'b 0.85 4000 16 c For fc' 4000 psi, 1  0.85 . Therefore, c    2.65  3.1 in. 1

0.85

Check whether tension steel is yielding:  d  c  15.5  3.1   0.003  0.012   Using Eq.(4-18)      s t  c  cu  3.1  Thus,  s > 0.002 and the steel is yielding ( fs  fy ).

 

Compute the nominal moment strength, using Eq. (4-21): 2.65   2.4 60000  15.5    2   M  A f d    170 kip-ft n s y 2 12000 Since, t  0.012  0.005 the section is clearly tension-controlled and,  M n  0.9 170 kip-ft = 153 kip-ft 2. Compute Live Load Set Mu  M n  153 kip-ft 16  18  0.15  0.3kips/ft 144 Factored dead load = 1.2 0.3  0.5   0.96 kips/ft

Weight/ft of beam =

2 2 Factored dead load moment = wl 2  0.96  12 2  69.1 kip-ft

@solutionmanual1 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...