B U I L D I N G D E S I G N - SKYSCRAPERSpriodeep.weebly.com/uploads/6/5/4/9/65495087/6_storied... · 2018-09-02 · P R I O D E E P C H O W D H U R Y C E @ 2 K 8 . 0 1 7 2 6 9 4

B U I L D I N G D E S I G N

P R I O D E E P C H O W D H U R Y

C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

2

10.1 DESIGN OF SLAB

DESIGN BY COEFFICIENT METHOD

Loads:

DL = 150 pcf

LL = 85 pcf

Material Properties:

fc = 3000 psi

fy = 60,000 psi

Design the panel (S4) as two-way slab with beam by coefficient Method.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

3

Figure 10.2: Top floor plan

Top F

loor

Pla

n

14 f

t 15 f

t 8 f

t 15 f

t 15 f

t

20 ft 19 ft

S

3

S

4

S

2

S

1


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

4

Step1: Determination of Minimum Thickness

h = 180

840

180

p = 4.67 in.

whwre, P = 2 (15+20) = 840

Select h = 5 in. as trial depth

Step 2 : Calculation of Factored Load

DL = wc * 12

h psf

DL = 150 x psf635.6212

5 ; where wc = 150 pcf

LL = 85 psf

W = 1.4D + 1.7 L = (1.4 * 63 + 1.7 * 85) = 232.7 psf

Step 3 : Determination of Moment Coefficient

Length ratio, m = 75.20

15

b

a

l

l

From the end condition case type is ‘Case 4’

From Appendix D-1

Ca, neg = 0.076; Cb, neg = 0.024

From Appendix D-2

Ca,dl,pos = 0.043; Cb,dl,pos = 0.013

From Appendix D-3

Ca,ll,pos= 0.052; Cb,ll,pos= 0.016

Step 4 : Calculation of Moment

Middle Strip Moment:

Positive Moments at Midspan

Ma,pos = Ca,dl Wla2 + Ca, ll Wla

2


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

5

Ma,pos = 0.043 * 232.7 * 152 + 0.052 * 232.7 * 15

2

= 4974 ft-lb

Mb,pos = Cb, dl Wlb2 + Ca, ll Wlb

2

Mb,pos = 0.013 * 232.7 * 202 + 0.016 * 232.7 * 20

2

= 2699 ft-lb

Negative Moments at Continuous Edge

Ma,neg = Ca,neg W la2 = 0.076 * 232.7 * 16

2 = 4527 ft – lb

Mb,neg = Cb,neg W lb2 = 0.024 * 232.7 * 20

2 = 2234 ft – lb

Negative Moment at Discontinuous Edge

Ma,neg, discontinuous = 3

1 * Ma,pos =

3

1 * 4527 = 1509 ft – lb

Mb,neg, discontinuous = 3

1 * Mb,pos =

3

1 * 2234 = 745 ft – lb

Column Strip Moment:

Column strip moments are 2/3 of corresponding middle strip’s moments in respective

direction.

Step 5 : Check the Design Thickness

d =

)59.01('c

y

y

u

f

ff

M

Here = max = 0.75ρ b = 0.75 * 0.85*β 1

y

c

f

f/

*yf87000

87000= 0.016

d =

)3

60*016.0*059.01(12*000,60*016.0*90.0

12*4974

= 2.66 in.

hrequired = (d + clear cover= 1 in ) = 3.66 in.

hrequired hdesign, design is ok


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

6

Step 6 : Calculation for Reinforcement for Middle Strip

In Short Direction:

Minimum reinforcement:

As = 0.0018*b*d

= 0.0018*12*4

= 0.0864=0.09in2/ft

Spacing, Smax= 2*h= 10”

Midspan

Mu = 4974 *12 lb- in.

By Iteration process

Trial No.

Trial, a As =

2adf

M

y

u

a =

bf

fA

c

ys

85.0

Remark

1 2 0.36 0.72 not ok

2 0.72 0.28 0.55 not ok

3 0.55 0.3 0.58 nearly ok

4 0.58 0.3 0.58 ok

Using # 3 bar required spacing:

Spacing = ccx

/44.430.0

11.012

Continuous Edge

Mu = 4527 *12 lb- in.


As = 0.27 in2/ft.


Spacing = ccx

/5.4".88.427.0

11.012

Discontinuous Edge

As = 0.084 in2/ft <As(min)


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

7


Spacing = ccx

/"10"7.15084.0

11.012

In long Direction:

As = 0.0018*b*d

= 0.0018*12*3.5

= 0.0756=0.08in2/ft

Midspan

Mu = 2699 *12 lb- in.


As = 0.15 in2/ft.


Spacing = ccx

/"5.88.815.0

11.012

Continuous Edge

Mu = 2234 *12 lb- in.


As = 0.12 in2/ft.


Spacing = ccx

/"101112.0

11.012

Discontinuous Edge

As = 0.04 in2/ft <As(min)


Spacing = ccx

/"3304.0

11.012 = 10 in.c/c


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

8

Cornar reinforcement

# 3 @ 4 in c/c upto L/5 in respective directions.

Step 7 : Calculation for Reinforcement for Column Strip

The average moments in columns being two-third of the corresponding moments in the

middle strips, adequate steel will be furnished if the spacing of this steel is 2

3 times that

in the middle strip.

Using # 3 bar spacing for column strip

Midspan = 4 x 2

3 = 6 c/c

Continuous edge = 4.5x2

3 = 7.5 c/c

Discontinuous edge = 10x2

3 = 15 c/c

But maximum allowable spacing = 2h = 10 in.

Use10 c/c

Step 8: Check for shear

Total load = 232.7*20*15

= 69810 lb

From Appendix D-4

m= .75’ case-4

Wa= .76

Wb= .24

Vu,a= ftlb /748)1520(*2

76.*69810

Vu,a= ftlb /236)1520(*2

24.*69810

Slab strength = ØVc= Ø*2* √fc’*b*d

= 0.85*2*√3000*12*4

= 4470 lb >>Vu ; Ok.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

9

Step 9: Detailing

Wh

ole

Sla

b R

ein

forc

emen

t

Figure 10.3: Detailing of slab


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

10

Figure 10.3: Detailing of slab (continued)

A

A

B

B

20 ft (Long Direction) 15 ft (Long Direction)


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

11

5in

#3 @

8.5

in c

/c

#3 @

4 i

n c

/c

#3 @

10 i

n c

/c

#3 @

10 i

n c

/c

#3 @

10 i

n c

/c

#3 @

10 i

n c

/c

Sec

tion A

-A


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

12

Figure 10.3: Detailing of slab (continued)

10.2 DESIGN OF BEAM

B2

B3

B4

B5

B1

Figure10.4: Beam for Design


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

13

Step 1: Determination of load on the Beam (we have used E-TABS data, but

conventional approach is like this)

For beam B2:

Slab area )4

*

2

*(

aaba

)4

14*14

2

19*14( +( sft25.170)

4

15*15

2

19*15

Slab load = 170.25*(5/12)*150= 10641 lb

= 10641/14= 760 lb/ft

Choosing of beam is a matter of trial and error or experience

For this beam let assume a 12 inch X 16 inch section.

Weight of beam ftlb /20015012

16

12

12

Wall load = 1000 lb/ft

ftlbDeadload /1960)1000200760(

Live load=85 psf

Total live load= 10341425.17085 lb/ft

Factored load U=1.4D+1.7L= 450210347.119604.1 lb/ft

So for this case,

Total load, uw = 4.5 k/ft


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

14

Beam Lenth

(ft)

Section

(in*in)

Area

contributi-

ng (ft2)

DL

(slab)

(lb/ft)

DL

(beam)

(lb/ft)

Wall

load

(lb/ft)

Total

DL

(lb/ft)

Total

LL

(lb/ft)

Total

factore-

d load

(lb/ft)

B1 14 12*16 170.25 760 200 1000 1960 1034 4502

B2 15 12*16 184.75 770 200 1000 1970 1047 4538

B3 8 12*16 10.75 84 200 1000 1284 114 1591

B4 15 12*16 184.75 770 200 1000 1970 1047 4538

B5 14 12*16 170.25 760 200 1000 1960 1034 4502

Step 2: Determination of approximate moment (we have used E-TABS data,

but conventional approach is like this)

Approximate Moment

208.0)( lwMu

2045.0)( lwMu

The Beam should be designed for highest negative moment, that is 81.72 ft –kips.

Step 3: Selection of the cross sectional dimension of the (we have used ETABS

data, but conventional approach is like this)

Minimum thickness

"6.821

1215

21

lin = 12 in. (Keeping the assumed size)

Width from equation

c

y

yuf

fbdfM 59.012


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

15

c

y

y

u

f

fdf

Mb

59.012

3

60016.059.01)5.212(60016.0

1272.81

2

= 13.95 in = 14 in can be taken.

Here,

uM = 980.64 in-kips

75.0max b

yy

cb

ff

f

87000

8700085.0 1

85.0

85.065.01000

4000300005.085.0

1

11

016.0021.075.0

021.06087

87

60

385.085.0

max

b

Using ETABS the analysis this step can be done. A beam section of 12in x16in

and column of 18in x 18 in is chosen and analyzed. The followed moments

were found.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

16

Figure 10.5: BMD of B2 from ETABS (top floor)

Step 4: Determination whether the Beam has to be designed as Singly or

Doubly reinforced

If we add the slab thickness fh =5 in with the Beam thickness 16 in,

Then we get total depth, h=21 in.

Effective depth=17in.

And web width, b=12 in.

Assuming singly reinforced beam,

Steel area, As= bd

As = 0.0161217

As = 3.264 in2

Here,

yy

c

b

b

ff

fwhere

87000

8700085.0,

75.0

1

max

b = 0.016

Now,

bf

fAa

c

ys

85.0

12385.0

60264.3

a

=6.4 in

And Mn = As fy (d-a/2)

Mn = 3.264*60*(17-6.4/2)

Mn = 2702.5 in-kips > Imposed moment = 185.5*12 k-ft.

So the beam is to be design as singly reinforced beam.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

17

Step 5: Determination whether the beam has to be designed as Rectangular

Beam or a T Beam

Effective flange width b

Smallest of the following three condition will determine the effective flange width b.

Span/4= 15*12/4= 45in.

921251616 wf bhb in.

c/c distance/2= 120in.

So effective flange width b=45 in.

Let a= hf= 5 in

Now,

2

adf

MA

y

us

2

517609.

12508.185sA

= 2.83 in2

So, =1745

83.2

=.0037

Now, 48.1385.0

17600037.0

85.0

c

y

f

dfa

in.

as fha

So the beam will be designed as rectangular beam which would have the following cross section.

Step 6: Rectangular Beam Design

Now

2

adf

MA

y

us


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

18

And bf

Asfa

c

y

85.0

Trial No.

Assumed a As =

2adf

M

y

u

a =

bf

fA

c

ys

85.0

Remark

1 2 2.57 1.29 not ok

2 1.29 2.52 1.26 ok

Here Mu= 185.508 k-ft (considering ETABS data)

Beam Mu(+) Reinforcement(in2) Mu(-) Reinforcement(in

2)

B1 83.46 1.11; 2 # 7 √ 166.92 2.25; 4#6 √

B2 92.75 1.13; 2 # 7 √ 185.508 2.50; 4#8 √

B3 25.4 0.17;2 # 32#7 √ 50.8 0.68; 2#5 ×

B4 92.75 1.13; 2 # 7 √ 185.508 2.50; 4#8 √

B5 83.46 1.11; 2 # 7 √ 166.92 2.25; 4#6 √

Step 7: Shear Reinforcement

dbfV wcc 2

= 35.22171230002 kips.

uV 58.27 kips.

1935.2285.0 cV kips.

Now wherever of the beam the uV exceeds cV we need to provide shear reinforcement.

For this case nowhere of the beam the shear exceeds cV but minimum reinforcement required

where shear force exceeds 94.82

87.17

2cV

kips.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

19

cu

yv

VV

dfAs

Using # 3 bar

ins 5.485.41927.58

176022.85.

up to 5 ft from support on each side. The rest with minimum

spacing.

As dbfV wcs 4

ind

S 5.82

17

2max

First stirrup at 5.225.22

5.4

2

s in.

Step 8 & 9: Design for Torsion and Serviceability

Normally in buildig design torsion and Serviceability is not considered because the tortional

effect is normally countered by shear reinforcement.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

20


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

21

Step 10: Detailing

A

C

B

A

B

C


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

22

Section A-A

Section B-B

12 in

17 in

12 in

21 in

47 in

5 in

12 in

16 in

47 in

5 in

4 no # 8 Bar

# 3 @ 4.5in c/c

# 3 @ 4.5in c/c

2 no # 7 Bar

Section B-B


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

23

Figure 10.6: Detailing of beam (continued)

Section C-C

12 in

21 in

17 in

47 in

4 no # 8 Bar

2 no # 7 Bar

# 3 @ 4.5in c/c


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

24

10.3 DESIGN OF COLUMN

Design of column C subject to axial load as a tied column with following data:

'

cf = 4 ksi (as column requires more importance in design and better quality)

yf = 60 ksi

Step 1: Determination of Factored Load

P u = 1.4 D + 1.7 L

These values were collected from E TABS.

Step 2: Steel Ratio Assumption

Since value of axial load is low, take g = 0.04

Step 3: Determination of Concrete Gross Area

Pu = 0.80 A g yggc ff )1(85.0

For tied column ,70.0 380up kips

686=.80 A g * 0.70 [0.85 * 4(1- 0.04) + 0.04 * 60]

A g = 216 in 2

Step 4: Selection of Column Size

Let us choose a square column of size = 14in. x 18 in.

A g = 252 in 2


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

25

Step 1, 2, 3 & 4 is done by following table (Data From ETABS)

Combination Pu(kips) Mxip in) My(kip in) Section

1 563 628 648 14”*16”

2 845 845 782 14”*18”

3 595 595 39 14”*14”

4 672 672 40 14”*16”

5 410 410 26 12”*14”

6 442 442 27 12”*14”

7 734 734 709 14”*18”

8 784 784 714 14”*18”

9 288 288 18 12”*14”

10 300 300 18 12”*14”

Step 5 : Check for Steel Ratio

uP = 0.80 A g yggc ff )1(85.0

= (0.80 * 252 * 0.70) 40*)1(4*85.0 gg

g = 0.026 = 2.6%; Limit of g is ok.

Step 6: Calculation of Reinforcement

A st = A g * g = 252 * 0.026 = 6.55 in 2

Let us choose # 8 bar (A b = 0.79 in 2 )

No of bar, .879.0

55.6nos

A

AN

b

st

Step 7 : Selection of Ties

Use # 3 bar for ties

Step 8 : Determination of Vertical Spacing of Ties

16 d b of longitudinal reinforcement = 16 x 1 = 16in (dia # 8 = 1 in)

48 d b of tie bar = 48 x 0.375 = 18 in

Least dimension of column section = 14 in

Choose vertical spacing of ties = 14 in


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

26

Step 9: Arrangement of Ties

Clear spacing between longitudinal bars in y direction

=[Column dimension – (2 x clear cover) – (2 x dia of ties) – (3 x dia of bar)] 2

= [18 – (2 x 1.5) – (2* 0.3750) – (3* 1)] 2

= 5.625 in < 6 in

No additional ties are required

DESIGN OF COLUMN FOR BIAXIAL BENDING USING LOAD

CONTOUR METHOD

Step 1 : Determination of Factored Load and Moment

Pu = 1.4D + 1.7L = 686kip (ACI Code-00)

Mnx = Mux=845kip-in and Mny = Muy =782kip-in

Step 2 : Selection of Chart from Appendix F for Bending About X- Axis

Find: = 75.72.18

5.22182

h

dh

Selecting the design chart of Appendix F chart corresponding to

Step 3 : Determination of Moment Parameter for Bending About X – Axis

Find: g

n

A

P= 72.2

1418

686

and g=0.04

Intersection of these two points on the chart given the value of hA

M

g

nxo=0.425

Determine Mnxo = inkip 192818252425.0

Step 4 : Selection of Chart from Appendix F for Bending About Y- Axis

Find = 6.064.014

5.22142

h

dh

Selecting chart corresponding to


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

27

Step 5 : Determination of Moment Parameter for Bending About Y Axis

Find: 72.2252

686

g

n

A

P and

g =0.04 (same as step 3)

Intersection of these two points give hA

M

g

nyo= 325.0

Determine Mnyo = inkip 6.114614252325.0

Step 6 : Checking of Design Strength

15.115.1

nyo

ny

nxo

nx

M

M

M

M

Replacing all the values in equation

=

15.115.1

1147

782

1928

845

=1.03 1.0 (Design is ok)


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

28

Step 7 : Detailing

1 of 8 # 8 bar

#3 ties @14in.

Vertical spacing

18in

.

14 in.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

29

Figure 10.7: Detailing of column

10.4 DESIGN OF FOOTING

Rectangular Footing

Data: Unfactored column dead load, DL =666.8 K

Unfactored column live load, LL = 144.4 K

Column size = 18 in x 14 in.

Allowable soil pressure, qa = 4400 psf

fc = 3000 psi

fy = 60,000 psi

Step 1 : Determination of Effective Bearing Pressure (qa)

Let, bottom of the footing is at 3 depth from grade. So, Df = 3

qe = (qa – 125 x Df) psf

= (4400 – 125 x 3)

= 3775 psf

Step 2 : Determination of Area of Footing

A= 23

2153775

10)4.1448.666(ft

q

LLDL

e

With width 14 ft,

Size of the footing = 16 ft x 14 ft

Step 3 : Determination of Bearing Pressure for Strength Design

qu = psfb

LLDL

*

7.14.1

=

psf526314*16

104.144*7.18.666*4.1 3


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

30

Step 4 : Determination of Effective Depth from Punching Shear Consideration

2

12*4

dcbqdbf uoc

0.85 * 4 *

2

12

1614165263164*3000

ddd d = 32.533 inch.

Step 5 : Check for Flexural Shear

Long Direction

)12(2 bdfV cc

= 2 * 0.85 * 3000 (12*14*33) = 516217 lb

122

/* 12 dc

bqV uu

= 5263 x 14

12

33

2

12

1816

= 331519 lb

Vu < Vc , design is ok.

Short Direction

)12(2 ldfV cc

= 2 * 0.85 * 3000* (12 * 16* 33)

= 589963 lb

122

/* 12 dcb

qV uu lb

= 5263 x 16

12

33

2

/1414 12

= 308762 lb


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

31

Vu < Vc design is ok.

Step 6 : Check for Bending Moment

max = 0.75 b * 0.85 1

yy

c

ff

f

000,87

000,87= 0.016

Long Direction Moment

bcq

M uu */1

8

2

12 lb-ft

= 14*12

1816

8

52632

= 1936455 lb-ft

= 23237 k-in

d =

1259.01

12

maxmax

c

y

y

u

f

fbf

M

inch

=

123

60016.0*59.011460*016.0*90.0

23237

xx

= 14.05 in.

d < d (provided) , design is ok.

Short Direction Moment

*/28

2

12, Cbq

M usu lb-ft

= 16*12

1414

8

52632

= 1733574 lb –ft

= 20803 k-ft

d =

123

60016.0*59.0116*60*016.0*90.0

20803

x

= 12.43 in


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

32

d < d(provided0 , design is ok.

Step 7 : Estimation of Thickness

tmin = d + 3 in + 2

1 db

= 33+ 3 + 0.5 = 27.5 28 [Using #8 bar]

Provide, t = 37 in.

Step 8 : Determination of Steel Area for Long Direction (As, l )

For Iteration Method assuming a = 4 inch

Trial Assumed, a lsA , =

)/( 2

,

adf

M

y

lu

(in

2)

[Mu= 23237k-in]

a = bf

fA

c

yls

85.0

, (in)

Remark

1 11 15.56 2.19 Not ok

2 2.19 13.488 1.89 Not ok

3 1.89 13.42 1.88 ok

As= 13.42 in2

Check for Minimum Steel

15000,60

3312*143000312*

3min,

db

f

fA

y

c

s in2

But not less than

As, min = 5.183312*14000,60

20012*

200db

f y

in2

As,l < As, min


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

33

So As= 18.5 in2

Number of Bar

Select # 9 bar (Ab = 1 in2)

N = 195.181

5.18,

b

ls

A

A

Spacing

S = 5.88.819

121412

x

N

bx in c/c

Use 19 # 9 bar @ 8.5 in c/c

Step 9 : Determination of Steel Area for Short Direction (As,s )

Required Steel Area

As,s = 2

,,

/

12*

adf

M

y

su

As,s =

122/88.13360*90.0

20803

in

2

Check for Minimum Steel

As, min = 3312*16000,60

20012*

200d

f y

= 21.12 in2

As,s < As, min

Use As,s = As,min = 21.12 in2

Steel for Band Width

Steel Area


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

34

As,bw = As,s x %1

2

= 143.114

16

933.01143.1

2

1

2

As,bw= 17.28 x 0.933

= 19.7 in2

Band width = 14 ft.

No of Bar

Select # 9 bar (Ab = 1 in2)

No of bar, N = 207.191

7.19,

b

bws

A

A

Spacing

S = 84.820

1214

N

12

xbx in c/c

Use 20 # 9 bar 8 in c/c

Steel for Outside of Band

Steel Area

As, out of band = 54.02

7.1912.21

2

,,

bwsss AA in

2

No of Bar

Select # 6 bar (Ab = 0.44 in2)

No of bar, N= 223.144.0

54.0 on each side

Spacing


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

35

Spacing, S = 2

121x6 in c/c

Use 3 # 6 bar 8 in c/c on each side.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

36

Step 10: Reinforcement Detailing

Figure 10.8: Detailing of footing

18 in

16 ft

4in

20 # 9 2 # 6 19 # 9 2 # 6

18 in

37 in

16 ft

14 ft


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

37

10.5 DESIGN OF WATER TANK

10.5.1 DESIGN OF A ROOF TOP WATER TANK

Step 1: Water requirement

Water consumption rate = 40 gpcd.

Number of persons= no of flats X 6 =12 X 6 =72 persons (considering six persons per

flat)

Total water requirement=72 X 40=288 gal/day

Daily requirement = 46224.6

2880cft cft/day

Step 2: Tank dimension

Let inside Dimension

L=15 ft =4.572 m

B =7 ft =2.1336 m

So L/B=15/7>2

Height of water level= 4.4715

462

ft

Free board = 0.5 ft.

So final height = 4.4+0.5=4.9 ft 5 ft

Step 3:

Part 1

Here, h =H/4 or 1 m (larger height to be considered)

Part 2

(H-H/4) or (H-1) m to be considered.

For L/B>2

We have, h = H/4=5/4=1.25 ft 0.381 m <1m

h =1 m =3.28 ft


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

38

Step 4: Design for long wall

Moment M =6

3wH

Here w= 9.81 kN/ 3m 10 kN/ 3m

H=5ft=1.524 m

M = 9.56

524.110 3

kN-M

= 35.43513043.0448.4

10009.5

lb ft

=4.351 kft

=52.3 kin [1 kn-m= 738.8 lb-ft]

Check for,

inchd

d

kjbdfM c

98.4

12904.0288.013505.0

10003.52

2

1

2

2

max

Overall depth=4.98+1+1.5X4/8=6.75 inch (considering # 4 bars)

Here,

904.03

1

288.022.229

9

22.2235.1

30

9300057500

1029

305.0

60

1350300045.045.0

6

'

kj

rn

nk

f

fr

E

En

ksiff

ksif

psiff

c

s

c

s

ys

y

cc

Let us take overall thickness of wall=7.0 inch

So effective depth =7.0-1-1.5 X4/8=5.25 inch


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

39

Then,

ftin

A

ftinjdf

MA

s

s

s

/252.0120.7100

3.0

%3.0(min)

/37.025.5904.030000

10003.52

2

2

(of cross sectional area)

So, ftinAs /63.0 2 (provide # 4 bar @ 4 inch c/c)

Direct tension in the wall,2

)(B

hHwTL

ftinA

ftinF

TA

ftlb

mkN

s

s

Ls

/252.0(min)

/013.030000

06.383

/06.383448.4

3048.0100059.5

/59.5

2

1336.2)1524.1(10

2

2

So # 3 bar @ 5 inch c/c to be provided

Since steel is provided on both faces therefore steel to be provided on both faces as

# 3 bar @ 10 inch c/c.

Step 5: Design for short wall

Force P=w(H-h)=10 X(1.524-1)= 5.24 kN/2m =359.07 lb/ft (per m run)

Effective span in horizontally spaced slab=7+6.5/12=7.54ft=2.3 m

Bending moment at end, M=12

)(

12

22 BhHwPl

33.212

31.224.5 2

M KN-m (per m run)

=1721.83 lb ft=20.66k in (per ft run)

Reduction in moment due to tensile steel

= Tx=383.06 x 1.5/12=47.88 lb ft=0.575 k in


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

40

Design moment

Design moment = M-Tx

= 20.66-0.575

= 20.1 k-in

Steel requirement

ftinA

ftinA

ftinjdf

TxMA

s

s

s

s

/252.0

/252.0(min)

/14.025.5904.030

1.20

2

2

2

We will use # 3 bar @ 5 inch c/c.

At mid section

(min)/07.025.5904.030

33.10

33.1012

31.224.5

2

1

24

2

22

ss AftinA

kinPL

M

So sA will be provided as # 3 bar @ 5 in c/c at mid section.

Step 6: Cantilever effect on short column

Maximum moment

mkNmkNwHh

M .54.26

1524.110.

6

22

max

=1873.50 lb ft

=22.5 k-in

Steel requirement

ftinAftinjdf

MA s

s

s /252.0(min)/16.025.5904.020

5.22 22

So use # 3 bar @ 5 inch c/c.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

41

Step 7: Design of base slab

L/B>2, so we will design for one way slab

Minimum thickness

inL

t 920

1215

20

(For 60 grade steel)

Let thickness = 9 inch

Total weight of base slab= 425.05)10005.62(15012

9 ksf

Effective width, B= 75.7212

927

ft

Moments, 19.38

75.7425.0

2

max M kft=38.29 k-in

Depth 12288.0904.035.19.0

29.38

d =3.17 inch (OK)

inchd

ftinAs

25.58

45.117

/27.025.5904.020

29.38 2

Use # 3 bar @ 4.5inch c/c. So # 3 bar @ 9 inch c/c should be used at each face.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

42

Step 8: Detailing

Figure 10.9: Detailing of over head water tank

L=15 ft

7 in

18 ft

10 ft

5 ft


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

43

Figure 10.9: Detailing of over head water tank (continued)

# 4 bar @ 4 in c/c

# 3 bar @ 9 inch c/c


A

A

Section A-A


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

44

10.5.2 DESIGN OF UNDERGROUND WATER TANK

General data

Volume to be stored= 9244622 cft

(For two days store daily requirement 462 cft)

Angle of repose 6,30 wetdry

Unit weight of soil=w=125 pcf=20 kN/2m

Most critical condition: Empty water tank and wet soil.

Step 1: Tank dimension

Let inside dimension, L=15 ft=4.512 m.

B=7 ft=2.1336 m.

So height of water level= ft8.8715

924

Free board=0.3 ft

Final height=8.8+0.5=9.3ft 2.896 m.

Step 2: Design of long walls

Pressure exerted by wet soil =

sin1

sin1

wh

=2/96.46

6sin1

6sin1896.220 mkN

2/96.46 mkNp

Tension near the water face= 76.115.33

896.296.46

5.33

22

ph

kN.m

ink

05.104

1000

12

3048.0448.4

100076.11

(Per inch run)

So tension near water face/ ft run=104.05 X0.3048=31.72 kip inch


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

45

Tension away from water face

26.2615

896.296.46

15

22

max

ph

M kN/m

=232.8 k-in (per inch run)

=70.96 kin (per ft run)

From cracking consideration the thickness of wall is determined.

22

2

18.8612411.0

80.7066

6

inbf

MD

bDFM

et

ct

D=9.28 in9.5 in

Here '86cct ff

Let, psiffcct 41130005.75.7 '

maxM = 70.84 k-in

D= Total thickness

Effective depth =9.5-1.5=8 inch

d=8 inch

Step 3: Vertical reinforcement (long walls)

Steel requirement, ftinjdf

MA

s

s /33.08904.030

84.70 2

ftinA

ftinbtA

s

s

/33.0

/342.05.912003.003.(min)

2

2

So use # 4 bar @ 7 inch c/c (inner force)

Steel requirements for M=31.72 kip in

ftinA

AftinA

s

ss

/342.0

(min)/15.08904.030

72.31

2

2

So use # 4 bar @ 7 inch c/c (Outer force)


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

46

Step 4: Horizontal reinforcement (long walls)

Minimum steel requirements

ftinbtAs /342.0003.(min) 2

Use # 4 bar @ 7 inch c/c

Step 5: Design of short wall

Earth pressure at the bottom P=46.96 kN/2m

Max moment at the center M=12

2PL

12

375.296.46 2M

=22.07 kin (per m length)

=195.35k-in (per in length)

=59.55 k-in (per ft length)

375.279.712

5.97 ftL m

Now

inchinchd

kjbdf

M c

83.51228.3874.037.035.1

36.1952

2

2

max

Step 6: Vertical reinforcement (short wall)

ftinjdf

MA

s

s /275..08904.030

55.59 2

Use # 4 bar @ 8 inch c/c.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

47

Step 7: Horizontal reinforcement (short wall)

ftinAs /342.0(min) 2

Use # 4 bar @ 7 inch c/c that is 14 inch c/c both side.

Step 8: Design of base slab

Thickness provided=9.5 inch (let)

Minimum reinforcement=0.003bt=0.342 ftin /2

Use # 4 bar @ 7 inch c/c.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

48

Step 7: Detailing

18ft

7 ft

15 ft

9.5 ft

9.5 in

10 ft


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

49

Figure 10.10: Detailing of under ground water tank (continued)



# 4 bar @ 8 inch c/c.



C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

50

10.6 DESIGN OF SHEAR WALL

Given data:

6 storied building.

Height of each floor 10 ft.

Maximum wind pressure per sft= 10 lbs

Zone 1

wR =4

Total seismic load 20 kips/ft.

I= Importance factor =1.25.

Figure 10.11: Plan of a floor

29 ft 8 ft 29 ft

6.5 ft

39 ft


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

51

Step 1: Horizontal loads on the structure

Wind loads

rewindpressubFw 2

132010662 lb/ft =1.32 kips/ft

Seismic loads

WR

ZICV

w

Where,

W= Total seismic load 20 kips/ft

4wR

Z=0.075

I=1.25

Now

3

2

25.1

T

SC

431.06002.0 4

3

4

3

T

hCT nt

3

2

431.0

125.1 C [s=1 for stiff soil]

=2.19

02.1204

19.225.1075.0

V

083.0

2

)110(1010

60)26.002.1(26.002.1

26.025.0

xx

xxe

t

hw

hwF

VF

(wx=wx)

Step 2: Calculation of moment for vertical flexural reinforcement

84060604.16

1uM ft-kips

kipM u 42604.12

1


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

52

Step 3: Design of vertical flexural reinforcement

Balanced steel ratio

016.0021.075.0

021.0147

87

60

385.085.0

max

b

Area of steel if singly reinforced

2

max 57.113)596(125.6016.01

inbdAs

Then inbf

fAa

c

ys27.34

125.6300085.0

6000057.113

85.0

1

Maximum moment that can be developed

5033302

27.34916057.113

21

adfAM ysn in-kips

=41944 ft-kips > Mu [So no compression steel.]

Using minimum

22

min 66.2360000

91)125.6(2002004.1991125.6

60000

30003in

f

dbinA

y

ws

6.5 ft

8 ft

10 in

10 in

Figure 10.12: Shear Wall Cross Section


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

53

Now using 266.231

inAs maximum moment that can be developed is

8192983082

73.5916066.23

1

inkipM n

ft- kips > Mu

As, ina 14.7125.6300085.0

6000066.23

Therefore 266.23 inAs which can be provided by 11#14 bars in back face and with

24.75 square inch of steel area.

Step 4: Shear reinforcement

Values of cV :

5.77791125.6300022 hdfV cc Kip.

Shear uV in no case exceed 66077785.0 cV kips.

Therefore minimum shear reinforcement should be provided in the flange.

Horizontal shear reinforcement:

Maximum spacing,

52

wlS , 3h or 18 in

6.152 S , 288 or 18 in

So, 182 S inch (can be provided by # 3 bars)

Vertical shear reinforcement:

Maximum spacing

31

wlS , 3 h or 18 in

1S 15.6 in, 288 in or 18 in

So, 1S =18 in (can be provided by # 3 bars)


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

54

Step 5: detailing

# 3 bar @ 18 in

11# 14 bars

8 ft

6.5 ft

10 in


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

55

Figure 10.13: Detailing of shear wall (continued)

# 3 bars spaced @ of 15.2

inches (Vertical bars)

10 ft x 6 storey


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

56

10.7 DESIGN OF DOGLEGGED STAIR CASE

Step 1: General arrangement

8ft

Figure 10.14: Dog legged stair case (general arrangement)

The figure above shows the plan of the stair hall. Let the rise be 6 inch and trade be 6.5inch. The

width of each flight is 7.5/2=3.75ft

Height of each flight =2

10= 5 ft.

No of risers required = 6

125= 10 risers in each flight.

No of tread in each flight = 10-1 = 9.

Space occupied be trades = 109 =90in= 7.5 ft.

Width of landing =6.25 ft.

Width of passage =6.25ft.

Size of stair hall = 7.5 ft 20 ft.

LANDING

PASSAGE


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

57

Step 2: Design constants

For steel yf = 60,000 psi

And for concrete cf = 3000 psi

Step 3: Determination of loading

The landing slab acts together with the going as a single slab. The bearing of the slab into the

wall may be considered 8 inch.

Then the effective span = 75.1712

825.65.7 ft.

Considering one-way slab with both end continuous minimum thickness is 28

l.

So, t = ininchesl

86.728

1275.17

28

inches.

Self weight of the slab = 115012

8 = 100 plf.

Self weight of the steps =12

15012122

1 TreadRiserTread

=12

10150

12

8

12

105.0

=50 plf.

Floor finish = 20 plf.

Total dead load =100+50+20=170plf

Live load = 100 plf.

So, Design factored load = 1007.11704.1 =408 plf.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

58

Step 4: Bending Moment Calculation

Maximum Moment

1606875.174088

1

8

22

max wl

M lb-ft =192.8 k-in.

Check for depth

016.06087

87

60

385.085.075.075.0max

b

c

y

yf

fbf

Md

59.01

max2

=

3

60016.059.011260016.09.0

8.192

d =4.8 inch

And t = 4.8+1=5.8 inch (with 1 inch clear cover)

t =5.8 inch < 8 inch (Ok)

availabled = 8-1 = 7 inch

Step 5: Reinforcement Calculation

Distribution Bar.

Minimum reinforcement is provided as temperature and shrinkage reinforcement.

Temperature and shrinkage reinforcement,

ftintbAst /173.08120018.00018.0 2

# 3 bar can be used.

The spacing will be,

S =

173.0

1211.0 7.63=7.5 inch c/c.


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

59

Longitudinal Steel.

This is selected by trial.

Trial

No

Assumed

‘a’ (inch) Steel Area,

2

adf

MA

y

s

(inch2)

bf

fAa

c

ys

85.0

(inch)

Comments

Trial-1

a=1.0

0.55 1.07

nearly Ok

Trial-2

a=1.07

0.55 1.08

Ok

So, 255.0 inchAs is provided.

It can be furnished by using # 4 bar.

Spacing = 5.48.455.0

1222.0

inch center to center


C E @ 2 K 8 . 0 1 7 2 6 9 4 4 1 8 3

60

Step 6: detailing

4 ft 7.67 ft

4.5 ft

# 4 bar @5 inch c\c

# 4 bar

@5 inch c/c

11.5 inch

6 inch

# 3 Bar @10 inch c/c

Documents

B U I L D I N G D E S I G N - SKYSCRAPERSpriodeep.weebly.com/uploads/6/5/4/9/65495087/6_storied... · 2018-09-02 · P R I O D E E P C H O W D H U R Y C E @ 2 K 8 . 0 1 7 2 6 9 4