1Ex 5.3

Integrated Design of T-beamNilson, Darwin and Dolan

2004

T-Beam Design Floor system consists of single span T-beams 8

ft on centers supported by 12 in. masonry walls spaced at 25 ft between inside faces.

A 5 in. monolithic slab carries a uniformly distributed service load of 165 psf.

The T-beams in addition to its own weight must carry two concentrated 16,000 lb equipment loads over the stem of the T-beam 3 ft from the span centerline as shown.

Complete design of the T beams using fc = 4000 psi and bars with 60,000 psi yield stress

Description Steps in Design

Size beam for shear Calculate loads Find effective width Find As and select steel Bar cut-off Design shear steel Add cut-off shear steel

Span

Effective Span Clear span + depth not

exceeding c/c distance Beam web estimated as

12 x 24 Smaller is 26 ft

Dead Load The beam supports its own weight and that of

the slab spanning 8 ft between the beams It is easier to determine this load from:

Weight of 8ft slab section Weight of beam section below the slab. This has the

dimensions 12 x (24 5) = 12 x 19 in

2Calculation of Dead Load Weight of 8 ft section of

slab: (5/12) x 150 x 8 = 500 lb/ft

Weight of beam section (12 x 19)/144 x 150 =

237.5 lb/ft

Total dead load = 737.5 lb/ft

The book answer is 740 lb/ft since they rounded up 437.5 lb/ft to 440 lb/ft

Live Load

Live load consists of a 165 psf distributed load and two symmetric 16,000 lb equipment loads 3 ft from the center line

Pu = 1.6 x 16,000 = 25,600 lb

Calculation of wu Distributed Live Load

Intensity x tributary width 165 x 8 = 1320 lb/ft

Distributed Dead Load 737.5 lb/ft We will round up to 740

lb/ft to be consistent with book

wu = 1.2wd + 1.6wl= 1.2(740)+1.6(1320)= 3 k/ft

Shear Diagram

Sizing beam for shear

The web dimensions are sized based on shear.

Vumax = Pu + wuL/2 = 25.6 + 13(3)= 64.6k Vu@d is at distance d from the face of the

support. Book assumes d = 20 in (2 layers of steel for 24 in depth). The wall is 12 in wide, so the distance is 6 in + 20 in = 26 in

Vu@d = 64.6 3 (26/12) = 58.1 kips

Sizing beam for shear - 2

Assume Vs = 2Vc, so Vn = (Vc + 2Vc) Solve for bwd

3Vc = 6fcbwd = 58.1 kips bwd = 58.1/(6 x 0.75)4000 = 204 in2

Select bw = 12 in. and d = 18 in. Overall depth is 18 + 4 = 22 in. < 24 in. Assumed dead load is not revised

3Design for Flexure

Calculate effective width, bE Calculate Mu Calculate As Select steel

Calculate Effective Width

bE is least of: L/4 26/4 = 6.5 ft = 78 in 16hf + bw 16 x 5 + 12 = 92 in centerline spacing 8 ft = 96 in

Least is 78 in

Calculate Mu Simply supported beam with uniformly

distributed load (3 k/ft) and two symmetric point loads (25.6k) located 10 ft from the supports

Mu = wuL2/8 + Pu x 10 ft= 3(262)/8 + 25.6(10)= 253.5 + 256 = 509.5 ft k say 510 k-ft

Calculate As Since the flange is 78 in. wide and 5 in.

deep, it is unlikely to be a true T For T-beams, the approximate lever arm is

taken as 0.95d = 0.95(18) = 17.1 in (see MacGregor & Wight 2005)

As = Mu/fy x lever arm = 510 x 12/(0.9)(60)(17.1) = 6.62 in2

This is close to book value of 6.60 in2

Select Steel

Four # 9 (As = 4 in2) and 4 # 8 (3.16 in2) to provide a total area of 7.16 in2 (the book value is 7.14 in2)

Steel will be arranged in two rows, each row containing 2 #9s and 2 #8s.

Bar Cut-off

At the simple support, the moment is zero so the steel can be reduced

ACI requires at least 1/3rd of the steel to continue over the supports.

In this example, 50% (the entire top row) is cut off

4Detailing Location of Cut-Off

Find capacity of beam with the bottom steel layer (2 # 9s and 2 # 8s)

As = 7.16/2 = 3.58 in2 (book uses 3.57 in2) d = h - 1.5 in cover - 3/8 in cg of #9 & 8 bars

= 22 1.5 0.375 (1.128+1)/2= 19.06 in.

Residual Moment Capacity

inbf

fAa

ec

ys 81.078485.0

6058.385.0 '

===

kftadfAM ysn === 300121)

281.006.19(6058.390.0)

2(

Mn = 300/ = 333 ft-k

Map Location of Cut-off

Let x = distance from support where Mu = 300 ft-k (residual capacity)

Equate residual capacity = Rx wux2/2 64.6x 3x2/2 = 300

x = 5.3 ft from the support

Cut-off

Extend by larger of d or 12db d is 18 in. or 1.5 ft 12db = 12 x (1.128) = 13.54 in Here we are using the larger #9 bar since the

entire row is stopped. Otherwise use the diameter of the bar that is cutoff

Larger is 1.5 ft

Development Length

Because of the large concentrated load, check development length from that location

Check if simplified form can be used Clear cover > db, (1.5 + 0.375)/1.128 = 1.66db Clear spacing > db

[12 2 x (1.5+0.38+1.13(#9 bar) +1.0 (#8 bar)]3 = 1.33 in

minimum stirrups (will be provided) OK

5Calculate ld

ftorindf

fb

c

etyd 42.453400020

13.1000,6020

==

= l

Comment

Bars must be continued 3 + 4.42 = 7.42 ft past the mid-span point

The theoretical cutoff point is 5.3 ft. The bars must be extended by 1.5 ft so that the actual cutoff point is 5.3 -1.5 = 3.8 ft from the support.

Bottom layer

Extend 3 in. from the end of the beam

Positive Moment Anchorage

Since the positive moment rises more rapidly than the linear variation in development length an additional check regarding bar size is needed

Positive Moment Check

OKinVMl au

nd 8336.64

123333.13.1 =+=+ l 8.503.2775.0 6.58 ==sV 7.48.50 )18)(60)(22.0( === syvV dfAs Shear values from loadsBook

calculates first spacing at the maximum shear not at d

9 in

Least of

d/2

24 in

Avfy/50bw8.8726.840.68 ft

No more calculations needed since required spacing exceeds ACI maximum of 9 in

12.6418.834.610 ft

6.8334.846.66 ft

.5.5542.852.64 ft

58.6@ d

d = 18 + 6 = 24 in

CommentsMaximum spacing,

in.

Spacing, #3 U in

Vs = Vu/-VcVuSection

Shear Design

6Assigned Spacing

45 in5 spaces @ 9 in

9 ft 7 in from face of support

115 in Total

40 in8 spaces @ 5 in

This is the book spacing. We could have used a different spacing

28 in7 spaces @ 4 in

At s/2 rounded down2 in1 space @ 2 in

CommentsDistance from left support

Spacing

Extra Shear at Cut-Off

Since the shear at the cut-off location 51.7k (4.3 ft from support center is 64.6 3 x 4.3 = 51.7) > 3/4Vn (0.75 x 56.1 = 42) [ for s=5, Vs = 47.5; Vc = 27.3], additional stirrups are needed over a distance d (13.5 in. from the end of the discontinued steel)

Extra Steel Details

Spacing = d/8b = 18/(8 x ) = 4.5 in. Required Av = 60bws/fy

= 60 x 12 x 4.5/60000 = 0.054 in2

Provide 4#3 U, Av =0.22 in2 OK

Reinforcement Detail

Summary

The top layer is cutoff 3.8 ft or 45.6 in. from the center line of the support. The drawing shows 3 ft 4 in or 40 in from the face of the support

Additional 4 No 3 U stirrups are provided over the 3/4d or 13.5 in from the ends of the discontinued bars

Comment

It would have been easier to continue the bars. However, if there were a number of repetitions, it would be worthwhile to cut the bars off

There are design aids that simplify the cut-off calculations, e.g. MacGregor and Wight