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1 .gE ,.vinU ainiM-lE ,.tpeD gnE liviC ,deyaS-lE .A rmA /:yB deraperP

53 \ -1

53 / bals fo naps trohs = sT . 01

revoc + bals fo thgiew nwo = daol daeD mc 21 = St emussA . 51.0 + 5.2 * St =

2/005 2/004 2/002= .2/ 0001

.2m/.gk 003 = daol eviL

.2m/t 57.0 = 03.0 + 51.0 + 21.0*5.2 = daol latoT

: ( )

y

xmLRmL

* =1*

: erehW .oitaR ytiralugnatceR : R 67.0 ro 78.0 eb dluoc m dna 1m

78.0 = )1m ro/dna m( 67.0 = )1m ro/dna m(

.snoitcerid Y dna X ni snoisnemid bals eht era yL dna xL

2> R 2< R

: : 2/ 003 -1

R = 0.050.51 2 dnA

0.53R

=

. ffohsarG 2/ 003< -2

R =+ 41 dnA R

11

R =+ 4

.sucraM -3

Prepared By:/ Amr A. El-Sayed, Civil Eng Dept., El-Minia Univ., Eg. 2

R 1.0 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.0 0.395 0.473 0.543 0.606 0.660 0.705 0.745 0.778 0.806 0.830 0.849 0.395 0.323 0.262 0.212 0.172 0.140 0.113 0.093 0.077 0.63 0.055

1 .

Design of R.C. sections:

d`

d

b

t

Fc

Fs / n

CsCc

T

Yct

c

s

z

Notations:

b Breadth of the rectangular section

t Total depth of the cross section

d Theoretical depth of cross section, from the center of tension steel to the outside fiber of the compression zone.

d` Distance from compression steel to the outside fiber of the compression zone.

z Distance from the Neutral Axis to the outside fiber of the compression zone.

YCT Arm of resisting moment of internal force = distance between the compression and tension internal forces.

AS Cross sectional area of tension steel

AS` Cross sectional area of compression steel Notations, Cont.

M External moment acting on the section

CC Compression force on concrete

CS Compression force on compression steel

C = CS + CC = Total compression force on the section


T Tensile force on tension steel

C Strain of outside fibers of concrete

S Strain of tension steel

FC Maximum compressive stress of concrete in compression

FS Tensile stress of steel in tension

FS` Compressive stress of steel in compression

n Modular ratio =

C

S

EE

, where E is the modulus of elasticity

(zeta) = dz

(beta) = dd`

r =

C

S

FF

Ratio of tension steel reinforcement in the cross section

= dbAS.

` Ratio of compression steel reinforcement in the cross

section = dbAS.

`

Assumptions:

1- .

2- . (Strain) .

zdz=S

C

.. for the elastic stage . (1)

and EconstStrainStress == .

SS EF =

S and C

C EF =

C

C S )1(


dzz

EFE

F

S

S

C

C

dz ro =z

EE

FF

C

S

S

C

.=

dzz

FF

nS

C

)2( . .=

A.N ( 2)

dzz

nF

)3( . . = CS F

(3) SA . SF = T T

zA CSznA CS) ( ro = ... TnFdz

)4( . = ... TFdz . SA.n SA

r dna d . = zFF

C

= S (2)

.

-dd.-d

rn

1 ==

)r + n( = n r. = .n n r. = )-1(n

nrn )5( . =+

1 esaC -1 :ylno leets noisnet htiw noitces ralugnatceR .SA dna ,d ,b ,M : neviG .CF dna ,SF : deriuqeR

.

:

dohtem noitceS demrofsnarT -1 dohtem secroF lanretnI -2

-1 dohtem noitces demrofsnarT -2


:

`d

d

b

t

cF

sF

z

sA.n

Ytc

: SA.n = etercnoc fo aera tnelaviuqE

= )lairetam eno fo edam saw noitces eht fi sa( noitces eht fo aera lautriv ehT .SA.n + t.b =

z :

0.0 = A.N eht tuoba saera fo M )z-d( SA.n = )2/z( z.b 0.0 = d SA.n z )SA.n( + 2z )2/b(

: )z(

axbbca

2 = 24

2

..2.... 22b

nAbnAdb =+ SSS znA

=++

S

SS

nAbd

bnA

bnA

z.

.12....2

22


S

SS

Andb

bAn

bAn

z.

..21... ++=

++=

S

S

Andb

bAnz

...211.

. . (1-1)

z = .d (AS / b.d) =

++=

.211....nb

dbnd.

++= .

211...n

dn . (1-2)

.

( )223 ..2

..12. zdAnzzbzbI SX +

+=

xC I

zMF .= and ( )X

S IzdMF = . . (1-3)

3- Internal forces method

Internal moment = external applied moment

T. C.YCT = M but the force = (stress * area)

= dAFC C . = dybFC C .. ) ( = dyFbC C .. = b . area of stress diagram

the compression force ( C ) = width of the section * area of stress diagram C = (0.5 FC.z) . b M = (0.5 FC.z) . b . YCT and YCT = d z/3

( )32zdbz

MFC = . (1-4)

substituting for z with .d


( ) 2..31

2

3...2

dbM

dddbMFC

== ..

putting 131

2 C=

.

21 ..dbMCFC = . (1-5)

where .2

1 =C taking the moment about the line of action of compression force: T . YCT = M ( ) MzdFA SS = 3.

( )3zdAMF

SS = . (1-6)

Putting AS = . b . d & z = . d

222 .

.

31.dbMC

db

MFS =

=

. . (1-7)

Where

=

31

12 .C & 31

=

then .1

2 =C

2- Case (2) Given : M, b, d, AS & AS` required : FS , FC Moment of areas about the N.A = 0.0

( ) ( )zdAndzAnzb SS =+ ..`..2 `2


C

Fc

d`

Z

Z - d

`

Cs Z/3

2/3

Z

( ) ( ) 0.0`......2 ``

2 =+++ dAdAnzAAnzb SSSS

z

( ) ( ) ( )`...2. `2

`` dAdAbn

bAAn

bAAnz SSSSSS ++

+++= . (2-1)

z = . d , AS = .b.d , AS` = `.b.d = `/ , = d`/d

( ) ( ) ( )`..`...2`... 22 ddbdbbn

bdbn

bdnbd ++

+++= `.

( ) ( ) ( ) .1...21..1... 22222 +++++= dndndnd.

( ) ( ) ( )( )

+++++= 22222 1..

.1211...1..

ndndnd.

( ) ( )( )

+++++= 21..

.1211.1.

nn . (2-2)

z.

Stresses in concrete and steel: C = CS + CC M about tension steel = 0.0

( )`3

ddCzdCM SC +

=

( )`.3

...21 `` ddFAzdbzFM SSC +

= . (2-3)

but `` dz

z

nFF

S

C

=


( )zdzFnF CS

`` .. = or ( )

dddFnF CS .....`

=

( ) = ..` CS FnF . (2-4)

FS` )2-4 ( )2-3(

( ) ( )zdzFn

ddAzdbzFM CSC`

`` ..3

...21 +

=

( )`` `..32

. ddzdzAnzdzb

MFS

C

+

= . (2-5)

z = . d , AS = .b.d , AS` = `.b.d = `/ , = d`/d

( )ddddddbndddb

MFC.

.`.....

3.

2.. ` +

=

( ) 22 ..1.....3

1.21 dbndb

MFC

+

=-

( )

+

=1...

31.

21

1.. 2

-

ndbMFC . (2-6)

21 ..dbMCFC = . (2-7)

where : ( )

+

=1...

31.

21

11

-

nC . (2-8)

FC.

FC )2-7 ( )2-4.( ( )

21`

....dbMCnFS

= . (2-9) to find FS


zdz

nFFS

C

= (z = .d)

( )CS F

nF .1.= . (2-10)

FC )2-6 ( )2-10(

( ) n

ndbMFS

)1(.1...

31.

21

1.. 2

-

+

=

( )

+

=1...

31.

21

)1(.. 2

-

2 n

ndbMFS

22 ..dbMCFS = . (2-11)

where:

( )

+

=1...

31.

21

)1(2

-

2 n

nC . (2-12)

Case 3 Design of Rectangular section with tension steel only: 3-1 Free design (the depth is unlimited) Given : M, b, FC, and FS Required : d, and AS


d`

d

bt

Fc

Fs / n

C

T

Yct

:

zdz

nFFS

C

= . (3-1)

of moment about tension steel C.YCT = M

MzdbzFC =

3

....21 . (3-2)

and from equation 3-1 d

FFn

nz

C

S.

+= and r

FF

C

S =

drnnz .+=

z (3-2)

Mrnndd

rnnbFC =

++ 31.....

21 and =+ rn

n

MdbFC =

3

1.....21 2

bM

Fd

C

.

31.

2

=

. putting 1

31.

2 KFC

=

.


bMKd .1= . (3-3)

(3-3) (Balanced Depth)

K1 FC FS.

M about C = 0.0 M = T . YCT

=3

.. zdFAM SS dz .=

=

31..dF

MAS

S = 31

dFMAS

S ..= 2. KFS =

dKMAS .2

= . (3-4) 3-2 Fixed design (the depth is given) 3-2-1 The given depth > the balanced depth

d`

d

b

t

Fc

Fs / n

C

T

Yct

Given : M, b, d, FC, and FS Required : AS

(Balanced Depth)

FS AS.


:

bM 1K wonk nac ew os 1= Kd

.CF teg nac ew sixA-X no CF dna ,sixA-Y no 1K neewteb sevruc eht morf -1 CF ngised fo esac siht ni( ton ro elbawolla si ti fi enimreted nac ew CF gniwonk morf -2 .).lla CF < eb tsum

.2K dnif nac ew ,2K dna ,CF neewteb sevruc eht morf -4

-5Kd

2. SAM =

)htpeD decnalaB(

SF CF .

:

bM 1K wonk nac ew os 1= Kd

owt ward ,sixA-Y no 1K fo eulav eht dna ,sixA-X no CF elbawolla mumixam eht morf htiw evruc a no tcesretni lliw senil owt eht dna ,ylevitcepser senil lacitrev dna latnoziroh .2K teg nac ew os , niatrec fo SF mumixam eht snoitceS-T fo ngiseD = ceS-T fo htdiw evitceffe = ro

:

CF T :

B -1

. ) T ( -2

( ) ( ) .

) ) T ( -3

(

].ceS-R fo esac ni[ ).lla( CF fo 57.0 ~ 5.0 = ].ceS-T fo esac ni[ ).lla( CF

. CF


Examples: For the plan shown in the following figure, find the loads on all beams knowing

that: 1- Slab thickness = 12.0 cm 2- Weight of cover = 150.0 kg/m2 3- Live load = 200.0 kg/m2 4- Wall height = 3.0 m 5- Depth of all beams = 60.0 cm 6- Breadth of beams ab, cde = 12.0 cm 7- Breadth of the remaining beams = 25.0 cm

And the draw the absolute shearing force and bending moment of the main beam (fbd), and design the critical sections, the draw to scale 1:20 the details of reinforcement of a longitudinal section of the beam (fbd).

4.00A

5.00

2

3

1

B2.00 2.00

5.00

3

2

1

A B

2.006.00

5.00

5.00

60 60

3.00

a

b

c

d

e

f


-8.27

16.45

-9.28

17.48

-9.63-0.3

21.41

-19.22

20.9

-1.8

-4.63

23.75

-18.38

20.25

-2.12 -0.73

20.54

-18.2

23.82

-6.47

3.0 t/m

3.4

-7.1

13.9

-2.59

-16.09

15.91

-4.09-10.09

3.912.89

-4.11

9.89

-1.83

-15.33

16.67

-4.96

-13.96

7.04

2.99

-4.01

9.99

-2.17

-15.67

16.33

-4.23-10.23

3.772.5

-8.0

13.0

-1.45-10.45

10.55

-3.74 -12.74

8.26

3.0 t/m2.0 t/m

3.0 t

Absolute bending moment of a continues beam

19.22

-23.75 -23.82

9.638.27

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