Lecture 5 - Delta UnivDr. A.Helba CIV 416 E 12 Arrangement of Radial Steel A sr 2- calculate and check A sr required at mesh outer perimeter (at circle , r 1 = R/4 = 1 m, = 0.25 and

Nov-15

Dr. A.Helba CIV 416 E 1

A Guide to Designing Reinforced Concrete Water Tanks

Helba AlaaDr.

Analysis of Circular R.C. slabs (Roofs – Floors)

Design of simply supported circular roof slabs

Arrangement of st. rft. in circular s.s. slab

Illustrative Example

Lecture 5

Nov-15


R R s.s. slab

Fixed slab

Nov-15


Moments and St. Rft. in Circular R.C Plates

Ast

Asr

Asmax

Steel required:

As max , Asr radial steel and Ast tangential

x

Y

r

t

Moments:

M max

Mr radial moment Mt tangential

R = radius

R

Analysis of circular Simply Supported Slab

• in a circular plate Radial Moments

(Mr) and Tangential Moments (Mt) could be calculated from the following equations:

Mr max = M+ve

max

B.Mr .D

at center:

2

(3 )16

wR

Mredge=0

Nov-15


Analysis of circular Simply Supported Slab

• Tangential Moments (Mt)

Mt max = M+ve

max

B.Mt.D

at center: 2

(3 )16

wR

At edge: Mt edge

2

(2 2 )16

wR

Moments at any point in S. S. circular Slab

22

22

[(3 )(1 )]16

[(3 ) (1 3 ) ]16

= Poisson's ratio

r and =

R

r

t

wRM

wRM

( = 0 at center , = 1 at edge)

Nov-15


• For concrete assume n = 0.2 then :

2

2

2

2

(1

(1 )5 2

)5

t

r

and

wRM

wRM

+ ve tension downside

- ve tension upperside

Table XII

coeffs. for Moment in Circular Slab / No Center

Support / Uniform Load / Fixed Edge

0.0

R

0.1

R

0.2

R

0.3

R

0.4

R

0.5

R

0.6

R

0.7

R

0.8

R

0.9

R

1.0

R

Fixed

Fixed

Nov-15


Asr radial steel to resist Mr

PLAN SEC. ELEV.

Asr

Ast r R

Asr Ast

Mr Mr

Ast tangential steel to resist Mt

Nov-15


Calculation and arrangement of Asr in circular S.S. slab

The radial steel is distributed on the perimeter which varies along the slab radius , the radial steel should be calculated by its total amount.

Calculate the radius (r) at which Asr (total) is maximum

by putting

0r totaldM

dr

(r) for Maximum Asr (total)

(

22 22

2 3

2(1 )(2 )

5 5

2

5

r total

wR wR rM r r r

R

wR r r

( 2 22

3 05

10.577

3 3

r totaldM w R rdr

Rr

Nov-15


Maximum Mr total For simply supported circular slab the

total radial steel (the maximum area) is only calculated at = 1/√3 = 0.58

2 22

2 3 3

1

3

2(1 )(2 ) ( )(2 )

5 5 3

4 4

15 16.515 3

r total

rat

R

wR wRM r r

rwR wR wD

Example: Design of Roof Slab

Data: Dtank(inner)=7.5 m , H=5 m, fcu=25 N/mm2 ,mild steel

Loads on roof slab:

Assume troof = D / 50 = 7500 / 50 = 150 mm

own wt. of slab = 0.15 X 25 = 3.75 kN/m2

covering = 1 kN/m2 (may be ass. 0.75 to 1 kN/m2)

L . L = 1 kN/m2 (inaccessible roof)

w roof = 5.75 kN/m2

wu = 1.5 w = 1.5 X 5.75 = 8.625 kN/m2

assume twall (for practical purposes) = H/20 = 5/20 = 0.25 m

radius of roof slab Rroof = 7.5/2 + 0.25 = 4 m

Nov-15


Moments in Cicular Roof Slab • Maximum Moments:

at middle of slab

• Radial Moment Mr :

Mr max +ve = wuR2/5 = 8.63 X 42 / 5 = 27.6 kN.m

• Tangential Moment Mt :

Mt max +ve = Mr max +ve = 27.6 kN.m

mm

bf

R

Md

c

u 88

)1000(5.1

25)214.0(

106.27 6

cumax

min

As max

Asr radial steel Ast tangential st. Asmax square mesh at center in 0.25Dx0.25D

Steel Rft. in circular roof slab

Ast

Asr

As max

Nov-15


Calculation of st. rft. in circular roof slab 1- calculate As mesh at center As max(at middle)=Mumax/ [bcrd(1-a/2)fy/s] = 27.6 X 106 / [ 0.9 X 110 X (1 - 0.23/2)(240/1.15)]

= 1511.4 mm2 chosen 8 f 16 / m' as a square mesh in middle area of 2m X 2m (area of 0.5 R X 0.5 R)

(Check As min= 0.25 % Ac = 2.5 t = 2.5 x 150 = 375 mm2)

mesh 16 f 16 in 2m

2- calculate and arrange the radial steel Asr - Calculate Maximum Total Radial Steel Asr tot max

Mur total max. = 8.625 X 83 / 16.5 = 267.7 kN.m

3

.16.5

r total MAX

wDM

62

, . . : 23

12

267.7 1012607

240 0.1421 110

1.15 2

rusr total

y

cr

s

cr

M RA N B to get use b

fd

mm

a a

b

b

Trial 1 : chosen 112 f 12 (with bcr =1)

Nov-15


Arrangement of Radial Steel Asr At circle ❷ 112 f 12 (Asr tot at r2 =

0.58 x 4 = 2.32 m) are distributed along a perimeter = 2 r2 = 14.6m at spacing s2 = 14600/112 = 130 mm < 200 mm

At the support (outer edge of roof

slab) R = 4m, and perimeter = 2 R = 25.13 m.

Spacing of 112 bars at the edge circle ❸s3 = 25130/112 = 224.4 mm > 200 mm, in this case the solution is to increase the required number of bars so as to keep sedge to 200 mm, so,

nreq = 25130/200 = 125.7 use 126 bars spaced with: sat (r=0. 58R) =2 (0.58R)/126=

14577/126 = 115.7 mm.

❶

❷

❸

❹

Trial 2: chosen 126 f 12

Arrangement of Radial Steel Asr In order to cut off half the number of radial bars (63 out of 126) try the

following: 1- calculate the perimeter required to limit spacing of bars to 80 mm, = 126

x 80/1000 = 10.08 m at r = 10.08/2 = 1.6 m (at circle ❹ you can use 126/2 = 63 bars at = 0.4)

Then As 4 total available = 63 x 113 = 7119 mm2

Check with Asr required at circle ❹ as follows: - Calculate Mr4 at r4 = 1.6 m, = 0.4 Mr4 = wR

2(1-2)/5 = 8.63(4)2[1-(.4)2]/5 = 23.2 kN.m Mr4 total = (2 r4) x 23.2 = 233.2 kN.m Then As 4 total required This trial cannot be done as As 4 total required >> As 4 total available .. Cancel it ..

624 233.2 10 10807

240 0.121 1101

1.15 22

ru

ycrcr

s

Mmm

fd

a bb

Nov-15


Arrangement of Radial Steel Asr 2- calculate and check Asr required at mesh outer perimeter

(at circle ❶ , r1 = R/4 = 1 m, = 0.25 and perimeter = 6.28 m):

- calculate Mr1 at r1 = 1.0 m , = 0.25: Mr1 = wR

2(1-2)/5 = 8.63(4)2[1-(.25)2]/5 = 25.9 kN.m Mr1 total = (2 r1) x 25.9 = 162.7 kN.m Then As 1 total required Then Asr1 total required (7540 mm2) > As cut off avilable (7119 mm2) It needs 67 f 12 at s1 = 6280/67= 94 mm > 80 mm o.K That means using 67 x 2 = 134 bars Ld = 40 f = 40 x 12 = 480 mm use extension 0.5 m for overlap.

621 162.7 10 7540

240 0.121 1101

1.15 22

ru

ycrcr

s

Mmm

fd

a bb

Finally: chosen 134 f 12

Arrangement of Radial Steel Asr Finally: Then Asr total required = 134 f 12 to be arranged as follows: 67 f 12 with length L1 and another 67 f 12

with length L2 : L1 = R – r1 – cover + v

al part + 2 hook = 4 – 1 - .030 + (0.15 -2 x .03) + 2 X 6f = 3.204 m L1 = 3.25 m L2 = R – r1 – cover + ld + 2 hook = 4 – 1 - .030 + 0.5 + 2 x 6 x 0.012 = 3.614 m

L2 = 3.75 m

❶

❷

❸

Nov-15


3- calculate and arrange the tangential steel Ast - Calculate Maximum tangential Steel Ast max/m’ (at r = R/4 , r = 0.25)

Mtu1 = wuR2(1-1

2/2)/5

=8.625 X 42 (1 – 0.252/2)/ 5 = 26.74 kN.m 6

11

27.7 10

240 0.121 1101

1.15 22

tust

ycrcr

s

MA

fd

a bb

chosen 7 f 16 / m’ = 1377 mm

2 (bcr = 0.9)

And use 6 f 16 / m’ at Mt = 1206 x 0.9 x (240/1.15) x 0.95 x110 = 23.67 kN.m (corresponding to 2 = 2(1-5M/wR2) = 0.29 , = 0.535) at r5 = .535 x 4 = 2.14 m and so on .) starting from r = 2.2 m .

150 mm

PLAN Roof Slab rft.

3.75 m

3.75 m

0.25 m

0.25 m

Nov-15


SEC. ELEV.

Roof Slab rft.

5.0 m

0.15 m

square mesh 16 f 16 in 2m

C.L.

Alternative choice and Arrangement of Radial Steel Asr

(Note: if f 16 mm is chosen bcr = 0.9 and As = 13954 mm2 )

At circle ❷ 70 f 16 (Asr tot at r2 = 0.58 x 4 = 2.32 m) are distributed along a

perimeter = 2 r2 = 14.6m at spacing s2 = 14600/70 = 209 mm > 200 mm then the number of f 16 mm bars should be increased from 70 to 14600/200 = 73 bars to keep spacing of bars =< 200 mm)

Check At the support (outer edge of roof slab) R = 4m, and perimeter =

2R = 25.13 m. Spacing of 73 bars at the edge sedge = 25130/73 = 344 mm >> 200 mm, in

this case the solution is to use 73 f 12 amid the other number of bars so as to get sedge = 25130/146 = 172 mm < 200 mm, so, these additional 73 f 12 are ended at s = 200mm at r = r2 = 2.32 m

L = R – r2 = 4 – 2.32 = 1.68 m use L = 1.75

Trial 1 : chosen 70 f 16 (with bcr = 0.9)

Trial 2 : chosen 73 f 16 & add 73 f 12

Nov-15


Finally: Then Asr total required = 73 f 16 with length L1 and 73 f 12 with length L2 : L1 = R – r1 – cover + ld + 2 hook = 4 – 1 - .030 + 0.5 + 2 X 6f = 3.614 m L1 = 3.75 m L2 = R – r2 – cover + v

al part + 2 hook = 4 – 2.32 - .030 + (0.15 -2 x .03) + 2 x 6 x 0.012

= 1.884 m

L2 = 2 m r2 = 2.3 m

❶

❷

❸


150 mm

PLAN

Roof Slab rft.

3.75 m

3.75 m

0.25 m

0.25 m


Nov-15


Plan (Sec. under floor level) An elevated Tank (Floor rests directly on 4 Columns)

D t 0.71 D

Case 1 n = number of Columns = 4 or 6 or 8

D H

(+…. m) Hw

SEC. PLAN

SEC. ELEV.

Case 1

V (Tank Capacity) = (πD2/4)x Hw

Nov-15


Loads on Wall and Floor of an elevated Tank

(Floor rests directly on columns)

Sec. Elev.

D

H

t

Water Pressure

Dead Load (D.L.)

Final B.M.D. on Walls and Floor

1- An elevated Tank rests directly on Columns

MFinal at connection >> MF wall Calculate MFinal at connection by

Moment Distribution

MFinal

Floor structural system

Nov-15


Options of wall thickness

Uniform t Tapered t1/t2 Haunch t1/t2 2Haunches in wall &

floor

Case 1

Sec. ELEV. An elevated Tank (Floor rests directly on Columns)

t1 = twall at top

t2 = twall at base

t

t t2

t1

tfloor tfloor

Plan (Sec. under floor level) An elevated Tank (Floor rests on Beams on Columns)

t 0.71 D D

0.15 D Main Beam

Paneled Beams

Main Beam

Case 2

Wall act as a Deep Beam between columns

Nov-15


Final B.M.D. on Walls and Floor

2-An elevated Tank rests on Beams on Columns

C.L

MFwall

M1

M2

MFinal at connection = MF wall

Floor is divided to slabs with Mx & My

Wall acts as D

ee

p B

eam


M1 & M2 From 3-mom. Eq.

Plan (Sec. under floor level) An elevated Tank rests on Circular Beam on Columns

D t 0.71 Db

Case 3

Db

Db diameter of C.L. of Circular beam

b=bbeam

Nov-15


Plan (Sec. under floor level) An elevated Tank rests on Circular Shaft

D t

Case 4

Dsh

Dsh diameter of C.L. of Circular shaft

tsh=thickness of shaft wall

Final B.M.D. on Wall and

Floor

3- Elevated Tank rests on Circular Beam on Columns

C.L

MFwall

Mr1

Mr2 = Mt2 MFinal at connection = MF wall

Mr1 = MF wall + M due to P wall + M due to floor cant. part

Floor is circular slab with Mr & Mt (Use superposition)


4- Elevated Tank rests on Circular Shaft

Nov-15


Final B.M.D. on Wall and Floor

An elevated Tank Floor rests directly on Columns

Study of Case 1

Tension and Shear in Wall and Floor

Nov-15


Tension and Shear in Wall and Floor

Final Results of B.M & N.F

Mr

Mt

1

2

3 4

Tring acts horizontally in Wall

Nov-15


Distribution of Mr & Mt in Floor slab (Values from table XII)

-0.15

-0.1

-0.05

0

0.05

0.1

0 0.2 0.4 0.6 0.8 1

Mr

Mt

Ring Tension in wall for Different Base Conditions

Nov-15


Sections 1 & 4 Tension due to B.M is on Air-Side (design as Cracked sec.)

Open Tank 1

4

2 3

Sections 2 & 3 Tension due to B.M is on Water-Side (design as uncracked sec.)

1

4

2

3

An elevated Tank (Floor rests directly on Columns)

B . M . D

Case 1

General Procedure for Design( 7 steps)

• Assign the overall Dimensions for tank (D & H)

• Consider the Connection (distribute DM– get Mf)

• Define sections for design & design values (M,Q,N)

• Calculate t for different sections

& Choose proper thicknesses.

• Calculate As required for different sections.

• Arrange steel at sections and connections.

• Draw steel arrangements in plan and elevation

Documents

Lecture 5 - Delta UnivDr. A.Helba CIV 416 E 12 Arrangement of Radial Steel A sr 2- calculate and check A sr required at mesh outer perimeter (at circle , r 1 = R/4 = 1 m, = 0.25 and