2/2
fs0.60fy
 fs36000 psi
ρ
As
bd
 ρ0.011kρn()
22ρnρn k 0.325j1
k
3
 j 0.892
1.5 times the width, to carry a service live load of 2000 lb/ft in addition to its own weight, on a 24
ft simple span. The ACI Code load factors are to be applied as usual. With f y = 60,000 psi and fc
= 4000 psi, determine the required concrete dimensions b, d, and h, and steel reinforcing bars ( a )
d = 1.5 b
wu1.2 wo
1.6 wl
 wu3.515 kip
ft
l2
8
ρ0.6 ρmax
 ρ0.0124From table Rρfy
1 0.588
f’c

b
2.25 ϕR13.134 in d 1.5 b d 19.7 in
As4 0.79in2
 b14in d 20.5 in giving a design capacity of
a
Asfy
0.85 f’c
b3.98 in
ϕMnϕAs
fy
da
 ϕMn263 ft kip>Mu Mu253.1 ft kipOK
fy
0.003 0.005
ρρ
 ρ0.0181
1.52ϕR
Asρbd As3.78 in2
This is satisfied by 4–#9 As = 4.0 in2. Beam dimensions would be b= 12″, d= 18” and h=21”
0.85 f’c
b
ϕMnϕAs
fy
da
 ϕMn271.1 ft kip>Mu OK
4.21 Afourspancontinuousbeamofconstantrectangularcross‐sectionissupported
atA,B,C,D,andE.Thefactoredmomentsresultingfromanalysisare
AtSupports,ft‐kipAtmidspanft‐kip
Ma=138
Mab =158
β10.85 0.05
1000psi
0.75 ϕ0.90
ϕ
ρ005
0.85 β1
f’c
εu
0.024 ρ0.6ρ005 0.014
Rρfy
1 .588
ρfy
f’c
 R 788 psi
Asbdρ2.86 in2
 n
As
As8
3.6
Try the following dimensions and then check the solution
Problem 4.22 A two span continuous beam is supported on three concrete walls spaced 30 ft.
on centers. A service live load of 1.5 kip/ft is to be carried in addition to the self weight of the
beam and is to be applied in a pattern loading. The dimensions of the beam should be
approximately d=2 b, and the reinforcement is to be varied according to the demand. Determine
1000psi
wl1.5 kip
ft
 L30ft γc150pcf
Solution: Begin by estimating a beam depth and width, then compute the girder load.. Compute the
maximum negative moment due to both spans being loaded then the maximum positive moment
2
Asfy
2
Asfy
0.85 f’c
b2.94 in MnAsfy
2
ca
β1
3.92 in c
d0.182<3/8 therefore phi = 0.9 is OK
21.5″
3-#9 (#29)
Section at midspan Section at interior support
3/3
CommentsonProblem4.22
4.23. A rectangular concrete beam of width b = 24 in. is limited by architectural considerations to a
maximum total depth h = 16 in. It must carry a total factored load moment M u = 400 ft-kips. Design
the flexural reinforcement for this member, using compression steel if necessary. Allow 3 in. to the
center of the bars from the compression or tension face of the beam. Material strengths are fy =
60,000 psi and fc’
‘・
= 4000 psi. Select reinforcement to provide the needed areas, and show a
sketch of your final design, including provision for No. 4 (No. 13) stirrups.
16″
24″ 2.5″typ.topand
bottom
f’c4000psi fy60000psi
Mu400kip ft
d 16in 2.5in13.5 in
d’ 3in
b24in

εu0.003
Es29000000psi β10.85
Reinforcement details
a
0.85f’cb4.312 in ca
β1
5.07 in
MnAsfy
da
2
332.6 kip ft
The capacity is less than the applied load, therefore compression reinforcement must be provided.
s11
Problem 4.24 For the beam with a triangular cross section shown in Figure P4.24,
determine a) the balanced reinforcement ratio and b) the maximum reinforcement ratio if
t =0.005. The dimensions of the triangle are such that the width of the triangle equals
the distance from the apex. The width at the effective width b equals the effective depth
d. Draw the strain distribution, stress distribution, and define your notation.
b
c
d
A
s strain stress
C=T
0.85f’c (1/2 a ba) = Asfy
Substitution As = b d, where b = d and ba = a gives
2
)(
‘
85.0
a
f
c
tu
u
d
c
y
c
f
f‘
060.0 2
<=Solution for reinforcement ratio when t = 0.005
4.25. A precast T beam is to be used as a bridge over a small roadway. Concrete dimensions are b =
48 in., b w = 16 in., h f = 5 in., and h = 25 in. The effective depth d = 20 in. Concrete and steel
strengths are 6000 psi and 60,000 psi, respectively. Using approximately one–half the maximum
tensile reinforcement permitted by the ACI Code (select the actual size of bar and number to be used),
determine the design moment capacity of the girder. If the beam is used on a 30 ft simple span, and if
in addition to its own weight it must support railings, curbs, and suspended loads totaling 0.475 kip/ft,
what uniform service live load limit should be posted?
b=48″
25″
20″
5″
16″
f’c6000psi β10.75
fy60000psi
L30ft
wD475plf
hf5in bw16in b48in
d20in
Reinforcement details
ρ
20.014 Assume that the compression block remains in the flange, then
a
0.85f’cb3.059in less than 5 in hf, so proceed. ca
β1
4.078in
2
d0.204<0.375, therefore
ϕ0.90
Mo
1.2woL2
879 ft kip
1.6 L2
w
wL
b1242 psf
Posted load
4.26UsingEq.(4.27)andassumingthatd=0.9h,showthatAsisapproximatelyequaltoMu/4hfor
Grade60reinforcementandwhereMuisinkip‐ft.
SOLUTION
MU=Mn=Asfy(d‐a/2)
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./”.etl.ooi { = o,1 ‘o,?5( o;#)
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.:
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J:
//cw
bw
/a>
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O, E> l. oofr
6.75 l,ol?a
o.7o l,o t{7
Gorh
O,?%
t,7z
1,5?o
PROBLEM4‐28
f’c= 4000 psi
rbal 0.0289
r0.004= 0.0206
et 
Mn/bd2Mn/bd2
0.045 0.90 0.0030 176 158
0.035 0.90 0.0038 221 198
0.025 0.90 0.0052 296 266
0.015 0.90 0.0080 448 403
0.009 0.90 0.0120 646 581
0.007 0.90 0.0145 757 681
0.0059 0.90 0.0162 835 751
0.0057 0.90 0.0166 851 766
0.0055 0.90 0.0170 867 780
0.0053 0.90 0.0174 884 796
0.0051 0.90 0.0178 902 812
0.0049 0.89 0.0183 920 821
0.0047 0.87 0.0188 940 822
0.0045 0.86 0.0193 960 823
0.0043 0.84 0.0198 980 824
0.0041 0.82 0.0204 1002 825
0.0039 0.81 0.0209 1024 826 0.65%
0
200
400
600
800
1000
1200
0.000 0.005 0.010 0.015 0.020 0.025
Reinforcementratio,