design of rc building

8/11/2019 Design of RC Building

1/54


2/54


3/54


4/54


5/54


6/54


7/54


8/54


9/54


10/54


11/54


12/54


13/54


14/54


15/54


16/54


17/54


18/54


19/54


20/54


21/54


22/54


23/54


24/54


25/54


26/54


27/54


28/54


29/54


30/54


31/54


32/54


33/54


34/54


35/54


36/54


37/54


38/54


39/54


40/54

Column size = 11 12 , Effective depth of footing = d

Punching Shear area A p = 2 (11 + d + 12 + d) d = 4 (11.5 + d) d

Punching Shear strength = 4 (f c ) = 4 0.85 (3000) = 186 psi = 0.186 ksi

0.186 {4 (11.5 + d) d} = 253.2 2.751 (11 + d) (12 + d)/(12) 2

d2 + 11.5 d = 339.9 (11 + d) (12 + d)/38.99 d = 12.48

Take footing thickness, t = 16.5 d = 12.5

Flexural Shear strength = 2 (f c ) = 2 0.85 (3000) = 93.1 psi = 13.41 ksf

Maximum shear force (according to SFD) = 141.6 k

Maximum flexural shear force = 141.6 27.51d [d is in ft]

13.41 10 d = 141.6 27.51d d = 0.876 = 10.52 12.5 , OK.

Maximum bending moment (according to BMD) = 288.6 k

Depth required by M is = (M/R u b) = {288.6/(0.781 10)} = 6.08 12.5 , OK.

As( ) = (f c/f y)[1 {1 2M

( )/( f c bd2)}]bd = 8.03 in 2

and A s(+) = (f c/f y)[1 {1 2M (+)/( f c bd 2)}]bd = 3.47 in 2 [using M (+) = 127.8 k ]

As(min) = (0.2/f y)bd = (0.2/40) 10.00 12 12.5 = 7.50 in2, which is A s

( ) but A s(+)

Provide 13 #7 bars at top and bottom.

Width of the transverse beams under columns = 11 + 12.5 = 23.5Load per unit length under C 8 = 253.2/10 = 25.32 k/

Maximum bending moment = 25.32 [(10 12/12)/2] 2/2 = 256.4 k

Depth required by M is = (M/R u b) = {256.4 12/(0.781 23.5)} = 12.95 11.5 .

Provide d = 13.5 , increase the thickness to 17.5 and put transverse rods at bottom

As(+) = (f c/f y)[1 {1 2M (+)/( f c bd 2)}]bd = 7.86 in 2 [Note: d = 13.5 here]

Similarly, A s(+) under C 6 = 6.21 in

2 and A s(+) under C 7 = 2.30 in

2

As(min) = (0.2/f y)bd = (0.2/40) 23.5 13.5 = 1.59 in2 As

(+) for all columns

Provide 11#7, 4#7 and 13#7 bars at bottom within the width of the transverse beams (23.5 )

under each column (in placing 13 #7 bars in 23.5 , be careful about minimum spacing).

Elsewhere, provide (0.2/f y)bd = 0.81 in 2/ft (i.e., #7 @ 9 c/c).


41/54

Details of Individual Footing

11 # 7 bars

9.00

11 # 7 bars

9.00

11 # 7 bars

11 # 7 bars

Details of Combined Footing

11 # 7 bars 4 # 7 bars 13 # 7 bars

# 7@ 9 c/c 13 # 7 bars at top & bottom

15.5

10

20

17.5


42/54

Seismic Detailing of RC Structures

1. Materials

Specification Possible Explanation

C o n c r e

t e f c 20 Mpa ( 3 ksi) for

3-storied or taller buildings

Weak concretes have low shear and bong strengths andcannot take full advantage of subsequent design provisions

S t e e

l f y 415 Mpa ( 60 ksi), preferably 250 Mpa (

36 ksi)

Lower strength steels have (a) a long yield region, (b) greaterductility, (c) greater f ult/f y ratio

2. Flexural Members (members whose factored axial stress f c /10)


S i z e

b/d 0.3 To ensure lateral stability and improve torsional resistance b 8 To (a) decrease geometric error, (b) facilitate rod placement

d L c/4Behavior and design of deeper members are significantly

different

L o n g

i t u d i n a

l R e i n f o r c e m e n

t

N s(top) and N s(bottom) 2 Construction requirement0.1 f c /f y

(f c , f y in ksi)at both top and bottom

To avoid brittle failure upon cracking

0.025 at top or bottom

To (a) cause steel yielding before concrete crushing and(b) avoid steel congestion

As(bottom) 0.5A s(top) at joint and A s(bottom)/(top)

0.25A s(top) (max) at anysection

To ensure (a) adequate ductility, (b) minimum reinforcement

for moment reversal

Both top and bottom barsat an external joint must be anchored L d +10d b

from inner face ofcolumn with 90 bends

To ensure (a) adequate bar anchorage, (b) joint ductility

Lap splices are allowedfor 50% of bars, only

where stirrups are provided @ d/4 or 4 c/c

Closely spaced stirrups are necessary within lap lengths because of the possibility of loss of concrete cover

Lap splice lengths L d and are not allowed

within distance of 2dfrom joints or near

possible plastic hinges

Lap splices are not reliable under cyclic loading into theinelastic range


43/54

2. Flexural Members (continued)


W e b

R e i n f o r c e m e n

t

Web reinforcements mustconsist of closed verticalstirrups with 135 hooks

and 10d t ( 3 ) extensions

To provide lateral support and ensure strength developmentof longitudinal bars

Design shear force is themaximum of (a) shear

force from analysis, (b)shear force due to verticalloads plus as required forflexural yielding of joints

It is desirable that the beams should yield in flexure beforefailure in shear

Spacing of hoops within2d (beginning at 2 ) at

either end of a beam must be d/4, 8d b; elsewhere

St d/2

To (a) provide resistance to shear, (b) confine concrete toimprove ductility, (c) prevent buckling of longitudinal

compression bars

3. Axial Members (members whose factored axial stress f c /10)


S i z e

bc/hc 0.4 To ensure lateral stability and improve torsional resistance

bc 12To avoid (a) slender columns,

(b) column failure before beams

L o n g i

t u d i n a


t Lap splices are allowedonly for 50% of bars,only where stirrups are provided @ b c/4 or 4

Closely spaced stirrups are necessary within lap lengths because of the possibility of loss of concrete cover

Lap splice lengths L d and only allowed in thecenter half of columns

Lap splices are not reliable under cyclic loading into theinelastic range

0.01 g 0.06To (a) ensure effectiveness and (b) avoid congestion of

longitudinal barsMc,ult 1.2 M b,ult

at jointTo obtain strong column weak beam condition to avoid

column failure before beams

T r a n s v e r s e

R e i n

f o r c e m e n

t

Transverse reinforcementmust consist of closedspirals or rectangular/

circular hoops with 135hooks with 10d t ( 3 )

extensions


Parallel legs ofrectangular hoops must be spaced @ 12 c/c


Spacing of hoops withinL0 ( d c, h c/6, 18 ) at each

end of column must be bc/4, 4 ; else S t bc/2

To (a) provide resistance to shear, (b) confine concrete toimprove ductility, (c) prevent buckling of longitudinal

compression bars


44/54

3. Axial Members (continued)


T r a n s v e r s e


t


force from analysis, (b)

shear force required forflexural yielding of joints

It is desirable that the columns should yield in flexure beforefailure in shear

Special confiningreinforcement (i.e., S t

bc/4, 4 ) should extend atleast 12 into any footing

To provide resistance to the very high axial loads and flexuraldemands at the base

Special confiningreinforcement (i.e., S t

bc/4, 4 ) should be provided over the entire

height of columnssupporting discontinued

stiff members and extendLd into the member

Discontinued stiff members (e.g., shear walls, masonry walls, bracings, mezzanine floors) may develop significant forces

and considerable inelastic response

For special confinement,area of circular spirals

0.11 S td (f c /f y)(A g/Ac 1),of rectangular hoops

0.3 S td (f c /f y)(A g/A c 1)

To ensure load carrying capacity upto concrete spalling,taking into consideration the greater effectiveness of circular

spirals compared to rectangular hoops.It also ensures toughness and ductility of columns

4. Joints of Frames


T r a n s v e r s e

R e i n f o r c e m e n t

Special confiningreinforcement (i.e., S t bc/4, 4 ) should extend

through the joint

To provide resistance to the shear force transmitted byframing members and improve the bond between steel and

concrete within the joint

St bc/2, 6 through jointwith beams of width b

0.75b c

Some confinement is provided by the beams framing into thevertical faces of the joint


45/54

Seismic Detailing of a Typical Frame

Original Design of Frame (3)

Beams SF 1 (D) SF 2 (D) BM 1(D) BM 0(V=D) BM 2(D)B16 9.3 -9.6 -23.0 12.4 -22.9

B20 6.5 -5.6 5.5, -17.6 0.7 8.2, -13.7

The reinforcements are arranged as follows

2 #5 through 1 #7 extra

2 #5 through

C2 B16 C6 B20 C10

13

12

b f

12

4

12

10 #6 bars

#3 ties @ 12 c/c

Column Section(with reinforcements)

Beam + Slab Section(without reinforcement)

Beam Sections(with reinforcements)

Section a-a Section b-b Section c-c Section d-d Section e-e

#3 stirrups@ 6.75 c/c

e

e

c

c

a

a

b

b

d

d

12

4

All #5 barsExtra #7 bar


46/54

1. Materials

Specification Design Condition Comments

C

o n c r e t e

f c 20 Mpa ( 3 ksi) for3-storied or taller

buildingsf c = 3 ksi OK

S t e e

l f y 415 Mpa ( 60 ksi), preferably 250 Mpa (

36 ksi)f y = 40 ksi OK

2. Flexural Members (members whose factored axial stress f c /10)


S i z e

b/d 0.3 b/d = 12/13.5 = 0.89 OK b 8 b = 12 OK

d L c/4 d = 13.5 , L c/4 = 3 and 1.5 OK

L o n g i

t u d i n a


t

N s(top) and N s(bottom) 2 N s(top) 2, N s(bottom) = 2 OK

0.1 f c /f y (= 0.0043)

at both top and bottom

As(top) = 0.62, 1.22 in 2 As(bottom) = 0.62 in

2 (top) = 0.0038, 0.0075

(bottom) = 0.0038

Both (top) and (bottom) are not OK

0.025 at top or bottom Maximum = 0.0075 OKAs(bottom) 0.5A s(top) at joint

and A s(bottom)/(top)0.25A s(top) (max) at any

section

As(top) = 0.62, 1.22 in2

As(bottom) = 0.62 in2

(through)OK

Both top and bottom barsat an external joint must be anchored L d +10d b

from inner face of columnwith 90 bends

Not specified in design Needs to be specified

Lap splices are allowedfor 50% of bars, only

where stirrups are provided @ d/4 (= 3.38 )

or 4 c/c


Lap splice lengths L d and are not allowed within

distance of 2d (= 27 )from joints or near

possible plastic hinges



47/54

2. Flexural Members (continued)


W e b


t

Web reinforcements mustconsist of closed verticalstirrups with 135 hooks,

10d t 3 extensions



force from analysis, (b)shear force due to verticalloads plus as required forflexural yielding of joints

Design shear force is takenonly from analysis Needs to be checked

Spacing of hoops within2d (= 27 ), beginning at2 , at either end of a beam

must be d/4, 8d b (=3.38 , 5 ) elsewhere

St d/2 (= 6.75 )

St = 6.75 c/c (= d/2)throughout the beams

Not OK

3. Axial Members (members whose factored axial stress f c /10)


S i z e bc/hc 0.4 bc/hc = 12/13 = 0.92 OK

bc 12 b c = 12 OK

L o n g

i t u d i n a


t Lap splices are allowedonly for 50% of bars,only where stirrups are

provided @ bc/4 or 4


Lap splice lengths L d and only allowed in thecenter half of columns


0.01 g 0.06As = 4.40 in 2

g = 0.029OK

M c,ult 1.2 M b,ult at joint

Not specified in design Needs to be checked

T r a n s v e r s e


tStirrups must be closed

rectangular/ circular hoopswith 135 hooks with 10d t

( 3.75 ) extensions


Parallel legs of rectangularhoops must be spaced

@ 12 c/c

Parallel legs of rectangularhoops spaced @ 10 c/c

OK

Hoop spacing within L 0dc, H c/6, 18 (= 10 , 18 ,

18 ) at each end of column bc/4 (= 3 ), 4 ; else S t

bc/2 (= 6 )

St = 12 c/c throughout thecolumns

Not OK


48/54

3. Axial Members (continued)


T r a n s v e r s e


t


force from analysis, (b)

shear force required forflexural yielding of joints

Design shear force is takenonly from analysis Needs to be checked

Special confiningreinforcement (i.e., S t 3 )should extend at least 12

into any footing


Special confiningreinforcement (i.e., S t 3 )

should be provided overthe entire height ofcolumns supporting

discontinued stiff

members and extend L d into the member

Column supports no particular stiff member, but

soft first storey

Special confiningreinforcement (i.e., S t 3 )

over the entire height ofground floor columns

For special confinement,area of circular spirals

0.11 S td (f c /f y)(A g/Ac 1),of rectangular hoops

0.3 S td (f c /f y)(A g/A c 1)[= 0.3 3 9.5 (3/40)(156/90 1) = 0.47 in 2]

No special confinement provided and 0.22 in 2 hoop

area provided @ 12 c/c

Atleast 2-legged #4 or 4-legged #3 bars needed as

special confinement

4. Joints of Frames


T r a n s v e r s e


t Special confiningreinforcement (i.e., S t 3 )should extend through the

joint


St bc/2 (= 6 ) and 6through joint with beams

of width b 0.75b c

b = 12 , b c = 12 , but nostirrups specified within

joints

Since b 0.75b c St 6 through joint


49/54

Correct A s(min) for BeamsThe longitudinal steel ratio is (min) = 0.1 f c /f y (= 0.0043) in some cases

If all the #5 bars are changed to #6 barsAs(top) = 0.88, 1.48 in 2, A s(bottom) = 0.88 in 2

(top) = 0.0054, 0.0091, (bottom) = 0.0054, which are all (min) and (max) (= 0.025)

Check Shear Capacity of Beams

Beams SF (V) SF (D) SF (USD)B16 9.6 9.6 14.4B20 4.3 6.5 9.8

For Beam B 16, a = A sf y/0.85f c b = 1.48 40/(0.85 3 12) = 1.93Mult at joint1 = A sf y (d a/2) = 1.48 40 (13.5 1.93/2)/12 = 61.83 kSimilarly, A s = 0.88 in

2 M ult at joint2 = 37.91 kVDesign = 1.4 (61.83 + 37.91)/12 + 9.6 1.5 = 26.04 k

For Beam B 20, V Design = 1.4 (61.83 + 37.91)/6 + 4.3 1.5 = 29.72 k

Vc = 2 f c bd = 2 (3/1000) 12 13.5 = 17.75 k, V c1 = 6 f c bd = 53.24 kSmax = A sf yd/(V n Vc) = 0.22 40 13.5/(29.72/0.85 17.75) = 6.89

Stirrup spacing provided @6.75 is just adequate, but special transverse reinforcements arespaced @3 closer to the joints.

Check Shear Capacity of ColumnsSince the bending moments from analysis are small, shear forces are also assumed small;therefore the shear force is checked for flexural yielding of joints.Also the column size is 13 12 ; i.e., the dimension 12 works as h.

For Column C 2 at ground floor, P = 130.8 1.4 = 183.12 kP/( A gf c ) = 183.12/(0.7 13 12 3) = 0.56, = 7/12 0.6, = (40/2.55) 0.023 = 0.36

M/( bchc2f c ) = 0.18 M = 0.18 (0.9 13 12

2 3) = 909.79 k = 75.82 kVDesign = 1.4 (75.82 + 75.82)/9 = 23.59 k

For Column C 6, P = 140.2 1.4 = 196.28 kP/( A gf c ) = 196.28/(0.7 13 12 3) = 0.60, with = 0.6, = 0.36

M/( bchc2f c ) = 0.17 M = 0.17 (0.9 13 12

2 3) = 859.25 k = 71.60 kVDesign = 1.4 (71.60 + 71.60)/9 = 22.28 k

Vc = 2 f c bd = 2 (3/1000) 13 9.5 = 13.53 k, V c1 = 6 f c bd = 40.59 kSmax = A sf yd/(V n Vc) = 0.22 40 9.5/(23.59/0.85 13.53) = 5.88

Stirrup spacing provided @12 throughout the columns is not adequate; i.e., provide #3 ties @5c/c, moreover special transverse reinforcements are spaced @3 closer to the joints.


50/54

Check Moment Capacity of Joints (for Weak Beam Strong Column)Since the first joint (between C 2 at two floors and B 16) consists of one beam only with twocolumns, the second joint (between C 6 at two floors and B 16, B 20) is checked first for the WeakBeam Strong Column.

For ground floor Column C 6, P = 140.2 1.4 = 196.28 kP/( A gf c ) = 0.60, with = 0.6, = 0.36 M = 71.60 k

For first floor Column C 6, P/( A gf c ) 3/4 0.60 = 0.45, with = 0.6, = 0.36M/( b chc

2f c ) = 0.19 M = 0.19 (0.9 13 122 3) = 960.34 k = 80.03 k

M c,ult = 71.60 + 80.03 = 151.63 k , 1.2 M b,ult = 1.2 (37.91 + 37.91) = 90.98 kM c,ult 1.2 M b,ult ; i.e., the Weak Beam Strong Column condition is satisfied.

The following sections are chosen as columns and beams

12

10 #6 bars

4-legged #3 ties @ 3 c/c (special) or 5 c/c

Column Section(with reinforcements)

13

135 hooks with 4 extensions

Beam Sections(with reinforcements)

Section a-a Section b-b Section c-c Section d-d Section e-e

12

4

All #6 barsExtra #7 bar 135 hooks with 4 extensions


51/54

The reinforcements are arranged as follows

Sp 2-legged #3 stirrups @ 3 (length 27 )C-Sp 4-legged #3 stirrups @ 3 (through)

SpSpSp


C - S p

2 #6 through

2 #6 through


1 #7 extra

12

18

Anchorage at end joints L anch = L d + 10 d b Ld for #7 bars = 0.04 A s f y/ f c = 0.04 0.60 40/ (3/1000) 1.4 = 24.53Lanch = 24.53 + 10 7/8 = 33.29 ; i.e., 34

2.25

2.25

Lap-splices not allowed hereElsewhere, it is only allowed for 50% bars with special confinement


52/54

Design of Shear Walls

ACI Provisions for Shear Wall Design

1. The thickness of the shear wall must be atleast 8 inch; i.e., h 8

2. Sections taken at L w/2 or H w/2 (whichever is less) from base is considered as critical for shear

3. In designing the horizontal shear forces or bending moments, the depth of the section is takenas d = 0.8L w

4. The nominal shear force, i.e., the permissible shear force in the section is(i) greater the design shear force, i.e., V n = V design / [V = V design in WSD](ii) summation of the shear force capacities of concrete and steel , i.e., V n = V c + V s(iii) cannot be greater than 10 f c hd; i.e., V n 10 f c hd [V 5 f c hd in WSD]

5. If N u is taken as negative for tensile forces (lbs)Vc can be taken as 2(1 + N u/500A g) f c hd [V c 1.1(1 + N u/500A g) f c hd in WSD]Using a more detailed analysis,

Vc 3.3 f c hd + N ud/(4 L w)and [0.6 f c + {1.25 f c + 0.2 N u/(h L w)}/{M u/(V uLw) 0.5}] hd[In WSD, take V c to be about half of these]

6. If V n V c/2, then the minimum horizontal reinforcements are provided

7. If V n V c, then the horizontal reinforcements are spaced ats2 = A vhf y d/(V n Vc) [s 2 = A vhf sd/(V V c) in WSD] or ; where s 2 L w/5, 3h, 18However, h = A vh/(hs 2) 0.0025; i.e., s 2 Avh/(0.0025 h)

8. The vertical reinforcements are spaced ats1 L w/3, 3h, 18

v = A vv/(hs 1) = 0.0025 + 0.5 (2.5 H w/Lw) ( h 0.0025)However, v 0.0025 and h

9. Flexural reinforcements are provided like normal beams

hLw

Hw s1

s2

Avv

Avh


53/54

Shear Wall Design in the Long Direction of the BuildingThe long direction includes two Frame(1)s and two Frame (2)s.Here, the height of the wall H w = 40 , the length of the wall L w = 7Assuming only the rear wall to take all the loads (due to the large opening in the front wall), withsimilar seismic coefficients as taken initially for the frame analysis, the following lateral forcesare calculated in the long direction of the building.

1. Assumed thickness of the wall, h = 8

2. Sections taken at L w/2 (= 3.5 which is H w/2 = 20 ) from base can be considered as criticalThe nominal shear force, V = 28.54 + 21.40 + 14.27 + 7.13 = 71.24 k

3. The depth chosen for design is, d 0.8L w = 0.8 7 12 = 67.2

4. V = 71.24 k, while 5 f c hd = 5 (3/1000) 8 67.2 = 147.23 k V 5 f c hd OK

5. Since there is no net tensile force on the section, V c can be taken as = 1.1 f c hd = 32.39 k

6. Here V is not V c/2, then the minimum horizontal reinforcements are not sufficient

7. Using 2-legged #3 bars, the horizontal reinforcements are spaced ats2 = 0.22 20 67.2/(71.24 32.39) = 7.61Using 2-legged #4 bars, s 2 = 13.84Also s 2 (L w/5 =) 16.8 , (3h =) 24 , 18 , (A vh/0.0025 h =) 20

Use 2-legged #4 bars @ 14 c/ch = A vh/(hs 2) = 0.40/(8 14) = 0.0036

8. Using 2-legged #4 bars as vertical reinforcementsv = A vv/(hs 1) = 0.0025 + 0.5 (2.5 H w/Lw) ( h 0.0025)

0.40/(8s 1) = 0.0025 + 0.5 (2.5 40/7) (0.0036 0.0025) = 0.0008 s 1 = 64.26However, s 1 (L w/3 =) 28 , (3h =) 24 , 18 , (A vv/0.0025 h =) 20

Use 2-legged #4 bars @ 18 c/ch = A vv/(hs a) = 0.40/(8 18) = 0.0028, which is 0.0025 and h

2 (7.70 + 6.57) = 28.54 k

21.40 k

14.27 k

7.13 k

3.5

7

4@10= 40

Critical Section


54/54

9. Flexural reinforcements are provided like normal beamsMmax = 7.13 10 + 14.27 20 + 21.40 30 + 28.54 40 = 2140 k-ft

As = M/(f s jd) = 2140 12/(20 0.87 67.2) = 21.97 in 2,Use 18 #10 bars on each side, to be curtailed over the height

7

40

# 4 @ 1 4 c / c

#4 @14 c/c#4 @18 c/c

#4 @18 c/c

18 # 10 bars 18 # 10 bars

design of rc building

Documents