1. Estimate the thickness of the floor

Using ACI table 9.5 (a) the minimum thickness is estimated:

End bay:

h =

l18 .5 =

62 ´ 1218 .5

=40.22 in . (ACI Table 9-5a)

2. Compute the unfactored load

The dead load is load imposed by the slab and the superimposed dead load:

Slab:

w=10 .38in .12 in ./ ft .

´ 150 lb / ft 3=129 .75 psf

Superimposed dead loads = w = 45 psf

Total dead load = wD = 174.75 psf

Live load = wL = 40 psf

3. Select load and strength reduction factors

U=1.4D (ACI Eq 9-1)

U=1.2D+1.6 L (ACI Eq 9-2)

U=1.2D+1.6 L+0 .5 S (ACI Eq 9-2)

U = Total factored load

D = Dead load

L = Live load

S = Snow load

U=1.4 (174 .75)=244 .65 psf

U=1.2(174 .75)+1 .6 (40 )=273 .7 psf

U=wu=1 .2(174 .75 )+1.6 (40 )+0.5(22 .68 )=285 .04 psf

The third load combination governs and the slab is assumed to be tension controlled and so for flexure f=0 .9 and for Shear f=0 .75 .

Tributary width = 28 ft

The factored load per foot from the slab is

28 ft ´ 285 .04=7 .98k−ft

The weight of the beam stem is estimated.

The factored DL of the stem is taken as 12 to 20 percent of the other factored loads on the beam. This gives .96 k-ft and 1.60 k-ft.

The factored weight of the stem is assumed to be 1 k-ft.

The trail load per foot = 7.98 + 1 = 8.98k-ft. ~ 9 k-ft

4. The actual size of the beam is determined

1. The minimum depth based on the above calculations is

h =

l18 .5 =

28 ´ 1218 .5

=18 .16 in .

h= l21

=28 ´ 1221

=16 in . (ACI Table 9-5a)

b. The minimum depth based on the negative moment at the exterior face of the first interior support.

M u=-wu ln2

10=-9 ´ (27 )2

10=- 656 .1k−ft

A reinforcement ratio, which will result in a tension-controlled section, is determined.

ρ=β1 f

'c

4 f y

=. 825 ´ 4 .54 ´ 60

=. 0154

w=ρf y

f c'=. 0154(60

4 .5)=.205

R=wf c' (1−. 59w )=. 81ksi

bd 2>M u

fR=656 .1´ 12.9 ´ . 81

=10800 in3

d=32.5in.

b=15in.

h=35in.

This fulfills the minimum h criteria.

5. The thickness of the floor is estimated

End bay:

h=l24

=28 ´ 1224

=14 in .

Interior bay:

h=l28

=28´ 1228

=12in .

h=13 in .

Try a 15 in. wide slab by 35 in. deep extending 22 in. below the slab stem with a d = 32.5 in.

6. Calculate the shear capacity of the T beam

V u=f (V c+V u )

Maximum shear from beam loads at interior end

V u=1 .15wu ln /2

V u=1 .15( 9)(27 /2 )=139 .73kips

V c=2 λ√ f c' bwd=2´ 1 ´ √4500 ´ 15 ´ 32.5=65.40 kips

V s=8 √ f c' bwd=261 .62kips

fV n=(0.75 )(65. 4+139 .73 )=153 .85kips

The maximum factored Vu due to the applied loads and dead loads is 139.73 kips. Therefore the beam stem selected has ample capacity for shear.

7. The dead load of the stem is computed

Weight per foot of the stem below the slab =

15 ´ 22144

´ .15=.34 kip / ft

Total Dead Load = .175x28 + .34 = 5.24 kip/ft

Live Load =

Beam: wL=

401000

´ 28=1.12kip / ft

Negative Moment: wL=

401000

´ 28=1.12kip / ft

Positive Moment: wL=

401000

´ 28=1.12kip / ft

Snow Load

Beam: wS=

22.681000

´ 28=. 64 kip / ft

Negative Moment: wS=

22.681000

´ 28=. 64 kip / ft

Positive Moment: wS=

22.681000

´ 28=. 64 kip / ft

Summary of factored loads:

Beam: wu=1.2(5 .24 )+1 .6(1.12)+0.5( . 64 )=8 .4 kip / ft

Negative Moment: wu=1.2(5 .24 )+1 .6(1.12)+0.5( . 64 )=8 .4 kip / ft

Positive Moment: wu=1.2(5 .24 )+1 .6(1.12)+0.5( . 64 )=8 .4 kip / ft

8. Calculate the flange width for the positive moment regions

. 25ln=. 25(61 ´ 12)=183 in .bw+2 (8´ 13 )=15+2(8´ 15 )=255 in .

bw+27 (12)2

+27 (12)2

=15+27(12)=339 in .

be=183in.

9. Compute the beam moments

ln ft 27 27 27 27wu, kip/ft 8.4 8.4 8.4 8.4wul2

n 6123.6 6123.6 6123.6 6123.6cm -1/24 1/14 -1/10 , 1/16

Mu kip-ft -255.15 437.4 -612.36 382.73

10. Design flexural reinforcement

1.Compute the area of steel required at the point of maximum negative moment (first interior support).

As>

Mu

ff y ( jd )

A s=612.36 ´ 12.9 ´ 60 ´ . 9(32 .5 )

=4 .65in2

a=As f y. 85 f c

' b= 4 .65´ 60.85( 4 .5 )(15 )

=4 .86 in .

A s ³M u

ff y (d−a2)=612 .36 ´ 12

.9 ´ 60 ´ (32 .5−4 .862

)=4 .53 in .

a=4.50in.

c= aβ1

=4 .50.825

=5 .45 in

Confirm if it is tension controlled

c<3d /85 .45<12 .19

A s

Mu

= 4 .53612 .36

=. 0074 in2 /kip− ft

2. Compute the area of the steel required at the point of maximum positive moment

As>Mu

ff y ( jd )

As=437 .4 ´ 12.9 ´ 60 ´ . 95(40 )

=2.56 in2

a=As f y. 85 f c

' b= 2.56 ´ 60.85( 4 .5 )(183 )

=. 22in .<hf=13 in .

A s ³M u

ff y (d−a2)=437 .4 ´ 12

.9 ´ 60 ´ (32 .5−.222

)=2 .40 in .2

a=.21in.

A s ³M u

ff y (d−a2)=429.59 ´ 12

.9 ´ 60 ´ (40−. 212

)=2.39 in .2

A s

Mu

= 2.39429 .59

=. 0056 ( in2 /kip−ft )

3. Calculate the minimum reinforcement

A smin=3√ f c'f y

bw d=1.64 in2

A smin ³200bw d

f y

=. .63 in2

11.Check for under reinforcement

. 003(d−cc

)>. 002

.003(32 .5−5. 455 .45

)>. 002

.015>. 002

Fulfills under-reinforcement criteria.

Ccf=85 f c' (be−bw )h f=. 85( 4 .5 )(2184 )=8353 .8

Mu kip-ft -255.15 437.4 -612.36 382.73As coefficients .0074 .0056 .0074 .0056As required (in2)

-1.89 2.45 4.53 2.14

As>Amin in2 Yes Yes Yes YesBars Selected 4 No. 7 5 No. 7 8 No.7 4 No.7As provided in2 2.4 3 4.8 2.4Bw o.k - Yes - Yes

Ccw=. 85 f c' bwa=.85 (4 .5 )(15 )(4 .5 )=258 .19

M n=ccf (d−h f

2)+Ccw(d−a

2)=225003 .3k−ft

. 9(225003 .3 )=202502 .97

Demand>Capacity

12.Check the distribution of the reinforcement

a. Positive Moment Region

Cc=2.5in.

Maximum Bar Spacing:

s=15(40000f s

)−2 .5Cc≤12(40000f s

)

f s=. 67(60 )=40 .2ksi8 .75≤12

b. Negative Moment Region

s=15(40000f s

)−2 .5cc=8 .75in .

13.Temperature and Shrinkage Design

ACI code 7.12.2.1 requires shrinkage and temperature reinforcements perpendicular to the span of the one-way slab.

A s( s∧t )=A s

Ac

³ . 0018

Ac=Concrete−Area=12´ 13=156 in .A s=Shrinkage−Temp−AreaAs ³ . 281

MaxSpacing≤5h∧18 in ,

5h=65in .

No. 4 bars at 18in.

10. Design the shear reinforcement

Ln ft 27 27wuk/ft 8.4 8.4wlukip/ft 1.6 x 1.12 1.6 x 1.12wuln/2 113.4 113.4Cv 1.0 0.125 1.15,

1.00.125 1.0

Vu 113.4 14.175 130.41,

113.4

14.175 113.4

Vu/ 151.2 18.9 173.88,

151.2

18.9 151.2

a. V c=2 λ√ f c' bwd=2(1)√4500(15 )(32.5)=64 .4kips

V c

2=32 .2kips

Because Vu/=151.2 kips exceeds Vc/2 stirrups are required.

Try No. 3 Grade 40 double leg stirrups with a 900 hook enclosing a No.7 stirrup

support bar. Since the required value of Vs=Vu/ - Vc does not exceed 4 √ f c' bwd

ACI Code Section 11.4.5.1 limits the maximum stirrup spacing to the smaller of d/2 and 24 in.

d2=32.5

2=16 .25 in .

s=Av f yt

50bw=1320050 (15 )

=17 .6 in .

s=Av f yt

.75√ f c' bw=21 .87 in .

13in. is used as the minimum spacing.