978-0073397948 Chapter 4 Part 1

7.5vf’c.

f’c4000psi Es29000000psi

24“

36″

typ

a)f’c=4000psi

12 41472 in4

Mcr

7.5 f’cpsi

Ig



h

2

1639 in kip

Reinforcement results in a 15% increase in cracking moment.

Mcr

1.153

7.5vf’c.

12″ f’c5000psi Ec57000 f’cpsi4.031 106

psi

Es29000000psi

n

Es

Ec

7.2 As2A

s10

2.54 in2

 nA

s

18.3 in2

 ys2.5in

bh

2



s

ys



4.3Comparethecrackingmomentbasedonthegrosssectionpropertiesandthetransformed

sectionpropertieswith4‐No.9(No.29)barsinFigureP4.1cbasedonaconcretetensilecapacityof

7.5vf’c.

f’c6000psi

28″

8″

Es29000000psi













n

Es

Ec

6.6 As4A

s9

4.00 in2

 nA

s

26.3 in2

 ys2.5in

yb

b

hf

2

bwhh

f





hh

f





2hf















 nA

s

ys



bh

f

bwhh

f





 nA

s

 12.418in

Igc

bh

f

3



12

bwhh

f





3



12

bh

f

yb

hf

2















2

 bwhh

f





hh

f



2hf

yb















2

 52269 in4



IgIgc nA

s

ybys





2

 54853 in4

 Ig

Igc

1.049Reinforcement adds 12% to Ig

Reinforcement adds about

10% to the cracking moment

7.5 f’cpsi

Ig





yb

Mcr

4.4Comparethecrackingmomentbasedonthegrosssectionpropertiesandthetransformed

sectionpropertieswith2‐No.11(No.36)barsinFigureP4.1dbasedonaconcretetensilecapacity

of7.5vf’c.

36″













4″

Es29000000psi

n

Es

Ec

9.3 As2A

s11

3.12 in2

 nA

s

29 in2

 ys2.5in

yb

bw

hh

f





2

bh

f

h

hf

2















 nA

s

ys



bwhh

f





bh

f

 nA

s

 21.13 in

Igc

bwhh

f





3



12

bh

f



3



12

bwhh

f





yb

hh

f



2















2

 bh

f

h

hf

2

yb















2

 52626 in4



IgIgc nA

s

ybys





2

 62685 in4



Reinforcement increases the cracking

7.5 f’cpsi

Ig





4.5Determinethecrackingmomentbasedonthegross sectionpropertiesinFigureP4.1difthe

sectionisprestressedsuchthatthereisa300psicompressionstressintheextremetensionzone

andtheconcretetensilecapacityis7.5vf’c.

36″

4″

Es29000000psi fc300psi

yb

Mcr0

7.5 f’cpsi

Ig



yb

948 in kip

Mcr

1.73Adding 300 psi prestressing compression to the section increases the cracking

4.6DeterminetheservicelevelmomentcapacityofthesectionsinFigureP4.1iftheallowablestress

forconcreteis0.45f’candtheallowablestressforthereinforcementis30,000psi.Usetheareasof

reinforcementfromProblem4.2.

a)f’c=4000psi

s

fc0.45f’c1800psi fs30000psi

Reinforcement details

Calculation of cracked section

Es

s11

s

4.7DeterminetheservicelevelmomentcapacityofthesectioninFigureP4.1biftheallowable

stressforconcreteis0.45f’candtheallowablestressforthereinforcementis30,000psi.Usethe

36″

ys2.5in

s

fc0.45f’c2250psi fs30000psi

n

Es

Ec

7.2 As2A

s10

2.54 in2

 nA

s

18.3 in2

 ys2.5in

ρ

As

bd0.00632 kρn()

22ρn ρn 0.259

4.8DeterminetheservicelevelmomentcapacityofthesectioninFigureP4.1ciftheallowable

stressforconcreteis0.45f’candtheallowablestressforthereinforcementis30,000psi.Usethe

8″

c)f’c=6000psi

ys2.5in

s

fc0.45f’c2700psi fs30000psi

n

Es

Ec

6.6 As4A

s9

4.00 in2



ρ

As

bwd0.01493 kρn()

22ρn ρn 0.355

4.9DeterminethenominalmomentcapacityofthesectionsininFig.4.1ausingthe

reinforcementareasfromproblem4.1.Thereinforcementyieldstressis60,000psi.

24“

typ

a)f’c=4000psi

ys2.5in

dhy

s

21.5 in

s11

a

0.85f’cb3.06 in

2









4.10DeterminethenominalmomentcapacityofthesectionsininFig.4.1busingthe

reinforcementareasfromproblem4.2.Thereinforcementyieldstressis60,000psi.

12″ f’c5000psi

fy60000psi

b12in h36in

ys2.5in

s10

a

0.85f’cb2.99 in

2









4.11DeterminethenominalmomentcapacityofthesectionsininFig.4.1cusingthe

reinforcementareasfromproblem4.3.Thereinforcementyieldstressis60,000psi.

c)f’c=6000psi

β10.75

a

0.85f’cbw

5.88 in ca

β1

7.843 in c

d0.234

2









4.12DeterminethenominalmomentcapacityofthesectionsininFig.4.1dusingthe

reinforcementareasfromproblem4.4.Thereinforcementyieldstressis60,000psi.

8″

d)f’c=3000psi

typ

f’c3000psi

ys2.5in

dhy

s

33.5 in

s11

a

0.85f’cb2.04 in a is less than hf, therefore this section can be treated as a rectangular

2





4.13Determinetherequiredareaofreinforcementandthecorrespondingreinforcementratiofor

thesectionsinFigP4.1aiftheultimatemomentisa)10,000in‐kipandb)5,000in‐kip.The

24“

typ

a)f’c=4000psi

ys2.5in

dhy

s

21.5 in

Mu1 10000in kip





s10

β1

d0.273<.375 OK





2









10 (No. 32) bars

Calculation of As for Mn2 Try: a13in

As2

fyd

a1













4.17 in2

 n

As11

2.671 Try 3 No. 11 (No. 36) bars

s11

a

0.85f’cb2.29 in Solution, use 3 No. 11 (No.

36) bars

4–No. 10 also is OK but

less efficient

ϕMnϕAs

fy

da

2















5144 in kip >Mu2 5000 in kip

4.14 . Determine the required area of reinforcement and the corresponding reinforcement

ratio for the section in Fig P4.1 b if the ultimate moment is (a)

ys2.5in

ca

0.175m

2









d0.205<0.375 OK

>Mu1 7000in kip Solution, use 3 No. 11 (No. 36) bars

Calculation of As for Mn2 Try: a13in

2





s9

2

a

0.85f’cb2.35 in

Solution, use 2 No. 9

is within 1%

2





4.15Determinetherequiredareaofreinforcementandthecorrespondingreinforcementratioforthe

sectionsinFigP4.1ciftheultimatemomentisa)10,000in‐kip,b)5,000in‐kip.Thereinforcement

8″

ys2.5in

β10.75

dhy

s

33.5 in

2





0.85f’cbw

10.62 in ca

β1

d0.423>0.375 must adjust ϕ

c

ϕ0.9 0.25

0.003









 0.825

ϕMnϕAs

fy

da











10074 in kip >Mu1 10000in kip Solution, use 2 No. 10 (No. 32) plus

2





a

0.85f’cb1.31 in









4.16. Determine the required area of reinforcement and the corresponding reinforcement ratio for the

section in Fig P4.1 d if the ultimate moment is (a) 10,000 in–kips and (b) 5000 in-kips. f y = 60,000

psi. Comment on your solutions.

8″

d)f’c=3000psi

typ

ys2.5in

dhy

s

33.5 in

Reinforcement details

As1

ϕfy

da













5.879 in2

 As2

ϕfy

da













2.939 in2



n1

As11

3.769 n2

As11

1.884

a1

0.85f’cb4.078 in a2

0.85f’cb2.039 in

β1

d0.143<0.375 OK





a2

2









ϕMn1 ϕAs1

fy

d

2















10601 in kip

>Mu2 5000in kip OK

Mu1 10000in kip

>OK

Problem 4.17 A rectangular beam made using concrete with f’c = 6000 psi and steel with fy = 60,000 psi

had a width b = 20 in., and an effective depth of d = 17.5 in and an h =20 in. The Concrete modulus of

rupture fr = 530 psi. The elastic modulus of the steel and concrete are, respectively Ec = 4,030,000 psi

and Es = 29,000,000 psi. The area of steel is four No. 11(No. 36) bars.

fr7.5 f’cpsi fr581 psin

Es

Ec

 n 6.6

(a) Find the maximum service load that can be resisted without stressing the concrete

fc0.45f’c

 fc2700 psi

fs0.60fy

 fs36000 psi

ρ

As

bd

 ρ0.018

kρn()

22ρn ρn k 0.381

k

Msmin Mss Msc





 Ms229 ft kip





(b) Determine if the beam will show cracking before reaching the service load

Ig

bh

3



 Ig13333 in4



frIg



nA

s

dh









s

Iut Igbd∆y2

 nA

s

dh

2

∆y













2

 Iut 15400 in4



frIut



0.85f’cb

MnAsfy

da

2















 Mn489 ft kip

(d) Compute the ratio of the nominal capacity of the beam to the maximum service level capacity and

4.18. A rectangular, tension–reinforced beam is to be designed for dead load of 500 lb/ft plus

self-weight and service live load of 1200 lb/ft, with a 22 ft simple span. Material strengths will be fy =

60 ksi and fc

・

= 3 ksi for steel and concrete, respectively. The total beam depth must not exceed 16

in. Calculate the required beam width and tensile steel requirement, using a reinforcement

ratio of 0.60

ρ

0.005 . Use ACI load factors and strength reduction factors. The

effective depth may be assumed to be 2.5 in. less than the total depth.

16″

b

2‐½”

wd500plf f’c3000psi β10.85

εu0.003

wl1200plf fy60000psi

L22ft

wu1.2 wdwo





1.6 wl

 3120plf

Calculate the maximum reinforcement ratio

and then 60% of that ratio. Mu

wuL2



8188.8ft kip

ρ005 0.85β1

f’c

εu









0.014 ρ0.60ρ005 0.0081

Rρfy

1 0.588 ρfy

f’c







441psi Because ρ is 0.6ρ0.005, it is reasonable to assume ϕ0.90

d231.309in

Use b = 32 in. and then check assumptions. wo

150 pcf

bd440.3plf

s10

Problem 4.19 A rectangular reinforced concrete section has dimension b=14 in., d=25 in, and h =

28 in., and is reinforced with 3 No. 10 (No. 32) bars. The material strengths are f’c = 5000 psi, fy =

60,000 psi.

(a) Find the moment that will produce first cracking at the bottom surface of the section

fr7.5 f’cpsi fr530 psin

Es

Ec

 n 7.2

(a) Find the moment that will produce first cracking at the bottom surface of the section

basing your calculations on Ig, the moment of inertial of the gross section.

bh

3



frIg



Mcr2

h

2∆y

 Mcr2 95.5 ft kip Mcr2

Mcr1

1.181

0.45 f’c or the steel stress exceeding 0.60 fy.