AE401- Tee Spandrel and Slabs

7/31/2019 AE401- Tee Spandrel and Slabs

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AE401 Tee Beams, One Way Slabs and Spandrel Beams

Floor systems with slabs and beams are placed in a monolithic pour.

The slab acts as the top of the beam below it;

Creating flanged T-beams, and Spandrel Beams.

Pictures are from MacGregor text, Chpt 4

Tee Beam and One Way Slab example

Given: Live Load = 100 psfDead Load = concrete plus 15 psffc=5 ksi fy=60 ksi

N-S Bay size 30 Assume beams run N-SE-W Bay size 28

Assume all spandrels and girders are 18 wide.Assume there will be 4 floor beams per bay.

Minimum depth of beams (table 9.5(a)) 19.5 on the outsidebays, and 17.1 on the inside bay. Assume 18 and checkoutside bay deflection. N


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N

With 4 floor beams, the slab span c-c of beams

is 5.6 = 67.2. Using table 9.5(a) the minimumslab thickness is L/24 for the end spans and

L/28 for the interior spans. This gives minimum

slab thicknesses of 2.8 and 2.4. For ease andpracticality, use 3 for the slab thickness.

Since the beams are 18 deep, assume they

are 10 wide (approx. h/2) and also that the18 is the total depth to the top of the slab.

Slab thickness


N

In this example we are spacing the c-c of

beams 5.6 = 67.2 with 10 beams so the clearopening is 57.2. This is not typical or good

construction practice. It would be better to use

a standard pan of 5 (53 inside) and size thebeams appropriately.

A Note about pans.


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ACI Code Provisions for Calculating bf

For this beam, rule (a) controls,8*3 = 24 and bf = 24+10+24 = 58

Once bf has been set, the normal As

calculations can be made assuming

a is within the slab.


Tee Beam and One Way Slabexample

Total concrete area per beam= 351.6 sq. In.

Concrete weight = 366.25# / ft of beamExtra DL = 15 psf * 5.6 = 84 # / ft of beamLive Load = 100 psf * 5.6 = 560 # / ft of beam

Factored wu = 1.44 K / ft of beam.

Use ACI Moment Coefficients (Section 8.3.3) toget maximum negative and positive moments.

in-K1,404ft-K11710

528441

C

LwMu-Max

2

m

nu====

.*.2

in-K1,000ft-K83.314

528441

C

LwMuMax

2

m

nu====+

.*.2

Beam Load and moment calculation

in-K585ft-K48.824

528441

C

LwMu-Ext

2

m

nu====

.*.2


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Moment Analysis of T-Beam - rebar selection

For the positive Mu of 83.3 K-ft, use d=18-2.5=15.5 and bf= 58

solve for As= 1.21 2 # 7 or 3 # 6 would be possible choices.

For the negative Mu of 117 K-ft, use d=15.5 and bw= 10

solve for As= 1.83 3 # 7 or 2 # 9 would be possible choices.

For the positive Mu in the middle bay where the moment coeff is 1/16,

Mu=73 K-ft, using d=15.5 and bf= 58 solve for As= 1.1

Use the same rebars as the other positive moment.

For the negative Mu at the end, where the moment coeff is 1/24, Mu=48.8 K-ft.

Solve for As= 0.72 2 # 6 or 3 # 5 would be possible choices.


Limitations on Reinforcement for Flanged Beams

Lower Limits

Minimum As is ALWAYS based on bw, not bf

fy and fc in psi. If fc >4,444 the second term controls

The minimum As (0.55) is less than the calculated As for all cases.

d*b*fy

d*b*fy

wwcf'3

..or..200

:oflargesttheA mins =

Lower Limit consideration

If As(provided) > 4/3 As(reqd) based on analysis then

As(min) is not required. See ACI 10.5.3



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1: Calculate the Vu values at the critical sections (d away from the faces of

supports). Since the worst midspan shear occurs when the LL is on only halfthe beam, the text book suggests using a midspan shear value of :


Shear in the Tee Beam:

K238

.L*LLw nu

=

This is not required by the ACI code but it does follow the spirit of the code

in using pattern loads to find the worst shear and moment values.

Vu

Vu at outside criticalsection = 18.9K

3.2

23.5

3.2

20.5

Vu at first inside criticalsection = 21.7K

2: Calculate Vn values at the critical sections, and Vc:


Shear in the Tee Beam

Vu

Vu = 18.9K . Vn = 25.2

3.2

23.5

3.2

20.5

Vu = 21.7K . Vn =28.9

3: 21.9K(10)(15.5)50002 ==Vc


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4: Calculate shear stirrup requirements:There are three possible results at this stage:

i) If Vn < Vc/2 then no stirrups are needed. Calculationwould stop here if that was the case.

ii) If Vc/2 < Vn < Vc then stirrups can be placed at maximum

spacing (see step 5 and then go to step 6b and finally step 7)

iii) If Vc < Vn then Vs (shear strength of stirrups, where Vs =

Vn Vc) and stirrup spacing (s) must be calculated:This is the case for this example.


Outside Vs= 25.2 - 21.9= 3.3K

First inside Vs = 28.9 21.9 =7K


5: Pick a bar size to use for the stirrup (normally a #3 or a #4 bar ) and

calculate the required spacing for the stirrup. s = (Av* fy *d) / Vs. In these

calculations Av is normally twice the area of the stirrup bar but it could belarger depending on stirrup placement.

The maximum spacing is d/2. There are two possible results:

i) If calculated s > d/2 then the stirrups must be placed at s = d/2.

ii) If calculated s < d/2 then the stirrups are placed at calculated s.


Outside s = 62.4 > s max of 7.75

First inside s = 29.2 > s max of 7.75

Both ends of this beam require stirrups at the maximum spacing of d/2 = 7.75

Note: There is a Minimum Av/s

calculated by (ACI 11.4.6.3)


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6: If the s had calculated to be less than the maximum, we wouldRepeat steps 1 to 5 until:

a) the calculated s reaches maximum spacing (d/2) then,

b) find the point where no stirrups are required, (when Vn = Vc/2)(step 4i).

For this example, all that is needed is to solve for where stirrups areno longer needed (when Vn=Vc/2= 11K, therefore when Vu=8.2K).

For the outside, that is 10.1 from the face of the column. For the

inside, it is 10.7 from the face of the column.

7: Start the first stirrup within a distance of d/2 from the face of the

support and calculate the total number of stirrups required.For the outside, 16 stirrups, starting the first 5 from the face

For the inside, 17 stirrups, starting the first 5 from the face.

Slab reinforcement is calculated based on the moments in the slab.

Designing for a 12 wide strip, the 3 slab weight is 37.5 #/ft plus the 15 #/ftmisc DL and the 100 #/ft LL. The factored load is 223 #/ft = 18.6 #/in.

The clear slab span is 57.2 so the largest Mu = 1/10(18.6)(57.2)^2 = 6.1 in-KThe As required for Mu, assuming d=2, is 0.058 sq. inches.

However, the minimum As =0.0018bh = 0.0018*12*3= 0.065 sq. inches.

Use minimum As: Ratio to find spacing of #3 bar: 0.065 / 12 = 0.11 / s

s = 20.4> 3*h=9 Create a grid of #3 bars at 9 spacing perpendicular to

the beams (ACI 10.5.4) and 15 parallel to the beams (ACI7.12.2.2).

Slab Moment Reinforcement



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Shear in the Slab, clear end span=4.43, clear inside spans = 4.47This Vu diagram is for the end span. wu = 0.223 K/ft, wuLL = 0.16 K/ft


Max Vu=1.15 (wu ln)/2 = 0.568 K Vn = Vu/ = 0.76 K

K4321250002dbcf'2Vc .** ===

No slab shear reinforcement needed.

Vu

0.09

.568

0.09

.494

Loads: Use same floor loads as before plus add 400#/ft curtain wall. The curtainwall is a dead load. Assume the curtain wall acts along the center line of the beam.

Total Factored Spandrel load = 1.57 K/ft, Factored LL=.45 K/ft

Compute flange width for moment

Spandrel Beam example: The N-S Spandrel that carries the slab.

"..bb

"*bhb

"bL

b

wf

wff

wf

5442

5318beamnexttodistclear5

3618366

481812

360

12

=+=+

=+=+

=+=+

bf is the smallest of these values. Therefore bf = 36

Use the 36 for the positive moment As calculations for negative moment use bw = 18

Moment and shear calculations are done the same as for the interior Tee beams.Stirrup spacing is not calculated until torsion reinforcement is calculated.

Flange width = 36

Contrib loadwidth = 33.6


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Shear and Torsion in the Spandrel Beam

Shear in the Spandrel Beam, clear span =28.5

Total wu = 1.6 K/ft, wu LL = .47 K/ft

Vu

Vu at exterior critical section = 20.4K

. Vn = 27.2

1.67

26.2

1.67

22.8

Vu at interior critical section = 23.4K

. Vn = 31.2 K

Calculate Vc = 49.6 K

Vc/2 < Vn < Vc so minimum stirrups arerequired.

However, since torsion is also present, do notassume minimum applies to shear alone.


The moments from the end of the floor slab, plus any effects of offset

shear, create a torsion in the spandrel that acts along with the shearforce.

Reinforcement must be designed to carry both the shear and the torsion

combined. The reinforcement is composed of closed hoop stirrupsthat are put inside the web, so all shear and torsion steel calculations

are based on bw.

Design for torsion is covered in section 11.5 of the 2008 ACI 318 Code.Design for torsion shall be in accordance with 11.5.1. through 11.5.6, or

11.5.7.


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Spandrel flange torsion width

The calculation of the flange width used

for torsion is not the same as used formoment. The overhang is defined by

section 13.2.4 of the code.

hb= 19 > 4*3 = 12

So flange width = 18+12 = 30


30


8.6 Lightweight concrete8.6.1 To account for the use of lightweight concrete,unless specifically noted otherwise, a modification factor appears as a multiplier of in all applicable equations andsections of this Code, where = 0.85 for sand-lightweightconcrete and 0.75 for all-light-weight concrete. Linear

interpolation between 0.75 and 0.85 shall be permitted, onthe basis of volumetric fractions, when a portion of thelightweight fine aggregate is replaced with normal weightfine aggregate. Linear interpolation between 0.85 and 1.0shall be permitted, on the basis of volumetric fractions, forconcrete containing normal weight fine aggregate and ablend of lightweight and normal weight coarse aggregates.

For normal weight concrete, = 1.0.

The change bar is because the 2008code added the lightweight concretefactor . The entire section 8.6 is new.

pcp = outside perimeter of concrete crosssection. 104 for this example.

Acp = area enclosed by outside perimeter ofconcrete cross section. 432 sq in. for thisexample.

30

Threshold Tu = 7.93 K-ft. for this example


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Calculate Tu values at face of supports and at the critical sections:

Torsion in the Spandrel Beam is caused by the slab moment (0.182 K-ft per foot of beam), shear offset(slab shear is 0.494 K per foot of beam) , and any other eccentric loads (for example if the curtain wallwas not centered on the beam).

TuK-ft

At critical section,

Tu = 7.2 K-ft < Threshold valueTorsion calculation NOT required

So minimum stirrups are based on

shear alone.

1.4

7.9


1.4

9.1 (based on 1.15 increasefor first interior end)

At critical section,Tu = 8.2 K-ft > Threshold valueTorsion calculation required, check

minimum based on torsion.

.494 K / ft

.182 K-ft / ft

Note: The step in the middle is based on the torsion fromthe offset factored live load shear of 0.35 K/ft and factoredlive load moment of 0.13 K-ft/ft from the floor slab.


a) For non prestressed members, Tu max=

This is just 4 times the threshold

value. 31.7 K-ft for this example.

Our maximum torsion (8.2) is less

than 31.7, so no adjustment is

required.


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11.5.2.3 Unless determined by a more exactanalysis, it shall be permitted to take the torsionalloading from a slab as uniformly distributed along themember.

11.5.2.4 In nonprestressed members, sectionslocated less than a distance d from the face of asupport shall be designed for not less than Tucomputed at a distance d. If a concentrated torqueoccurs within this distance, the critical section fordesign shall be at the face of the support.

ph = perimeter of centerline ofoutermost closed transverse torsionalreinforcement. Assuming 1.5 coverand #3 sti rrup rebar, ph = 66.5

Aoh = area enclosed by centerlineof the outermost closed transversetorsional reinforcement. Aoh = 272sq in. for this example.

30

85 < 530 so eq 11-18 is satisfied.


11.4.6.3 Where shear reinforcement isrequired by 11.4.6.1 or for strength andwhere 11.5.1 allows torsion to be neglected,Av, min shall be computed by equation 11-13.

11.5.5.2 Where torsional reinforcement isrequired by 11.5.5.1, the minimum area oftransverse closed stirrups shall be computed byequation 11-23

Therefore, only calculate minimum shearstirrups if torsion is not required.

Therefore, the minimum shear stirrups canbe left as zero as long as the total torsionplus shear is greater then the minimum.

Shear stirrup maximum spacing is d/2 or24. Shear plus torsion closed hoop maximum

spacing is ph/8 or 12


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8.2"8

pofspacing

maximumtheathoopsclosed#3space

130160s

2A

0.016ofminimumthethanlessisthis

010000052s

A

s

2A

005602182

6131

in-K6131ft-K1175

28

h

vt

vt

2

=

>==

=+=+

===

====

maxs".

.)(.

.**

.T

s

A

..

.T

T

ytofA

nt

u

n


Longitudinal reinforcement is also required for torsion. It is added at each section to thelongitudinal reinforcement required for bending moment that acts at the same time as the torsion. Thelongitudinal reinforcement is then chosen for this sum, but should not be less than the amount required for the

maximum bending moment at that section if this exceeds the moment acting at the same time as the torsion.

If the maximum bending moment occurs at one section, such as the midspan, while the maximum torsional

moment occurs at another, such as the support, the total longitudinal steel required may be less than thatobtained by adding the maximum flexural steel plus the maximum torsional steel. In such a case, the requiredlongitudinal steel is evaluated at several locations.

AE401- Tee Spandrel and Slabs

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