Spring 2016 Dr. Alaa Helba - Delta...

Feb-16

Lec. # 3 - CIV 223 - Dr. A. Helba spring

2016 1

CIV 223

Reinforced Concrete Structures - I

Spring 2016

Office hours: Sunday 11 - 1:00 pm

بسم هللا الرحمن الرحيم

Dr. Alaa Helba

For units: Use SI units (N, kN, mm, m)

For a rectangular R.C. Beam section let: b – width of rectangular R.C. section (Beam sec.)

t – total height of R.C. section

d – effective depth (distance to c.g of steel)

As – Area of steel

(Ex.: for 4f18 As=4*p(18)2/4 = 1018 mm2)

c - cover [50 mm/(1 row)–70/(2rows)–100/(3rows)]

Es - Modulus of Elasticity of steel = 200 kN/mm2

Ec - Modulus of Elasticity of concrete = 4400 fcu

n – modular ratio (Code suggests: n =10 in deflection & Mcr calculations

n = 15 in working stage analysis Mworking)

fcu – characteristic strength of conc. (20 or 25 or 30 N/mm2)

fy - yield stress of steel (240 or 360 or 400 N/mm2)

s

c

E

En

Basic units / Dimensions / Properties Review (lecture 2)

t

Feb-16


2016 2

Calculation of Mcracking

Procedure: Transform the area of steel to

equivalent concrete, nAs

using the modular ratio:

1- Calculate the location of

the N.A. (e = ,yt = t/2 - e)

2- Calculate the total moment

of Inertia @ N.A.

3- Calculate the cracking

moment Mcr based on

fbottom = fctr = 0.6 fcu t

tctrcr

y

IfM

223

)()(12

)(

cynAebtbt

I

centrefromshifttyeccentricie

tst

10c

s

E

En

))((2

)0(

.sec...(@

enAbtct

nAbt

conofLCAreasofMoments

ss

Review (lecture 2)

Example # 1 Calculate the cracking moment Mcr

+ve and Mcr -ve for the

singly reinforced rectangular section shown with dimensions

b x t = 250 x 800 mm, if As = 5 f 16 mm. materials used are

concrete fcu=25 N/mm2 and st. grade 360/520.

Solution

a- case of Mcr +ve

1- For - section : Calculation of Mcr

+ve and Mcr-ve for beam sections

Feb-16


2016 3

410

223

223

22

1019.1

)503.383)(1005(10)7.16)(800(25012

)800(250

)()(12

3.3837.164002

7.16

)]1005(10)800)(250[(50400)1005(10

))((2

)0(

)sec..(@@

1005)4

16(5165

mm

cynAebtbt

I

mmet

y

mme

e

enAbtct

nAbt

tionconcreteofLClinemiddleAreasofMoments

mmAfor

ts

t

ss

s

pf

mkN

mmN

y

IfMMomentCracking

mmI

mmy

mmN

ff

y

IfM

MomentCracking

t

ctrcr

t

cuctr

t

ctrcr

.1.93

.1031.9

3.383

)1019.1(3

1019.1

3.383

/3256.0

6.0

7

10

410

2

Feb-16


2016 4

Example # 1 - b

- case of Mcr -ve

for the rectangular sections of the same dimensions with the

same AS (in tension side) the positive cracking moment Mcr +ve = the negative cracking moment Mcr

-ve

2- For T - section : Calculation of Mcr

+ve and Mcr-ve for beam sections

for the T - sections of the same dimensions with the same AS

(in tension side) the positive cracking moment Mcr +ve ≠ the

negative cracking moment Mcr -ve

Feb-16


2016 5

186mm70

mm510

mm120

mmb 200

mmB 800

Calculate the cracking moment Mcr +ve and Mcr

-ve for the

singly reinforced T - section shown with dimensions b x t =

200 x 700 mm and B x ts = 800 x 120 mm, if As = 5 f 16

mm. materials used are concrete fcu=30 N/mm2 and st. grade

400/600.

Example # 2

Solution

a- case of Mcr +ve

Example # 2 - b

- case of Mcr -ve

186

mm580

mm120

mmb 200

mmB 800

- Solve it yourself.

- (without calculations), For T – sections of same dimensions

and steel area, Whichever is bigger Mcr+ve or Mcr

-ve , and why?

Feb-16


2016 6

11

Lecture # 3

R.C. Fundamentals

Analysis of RC Sections under Flexure

Stage I: Uncracked Section

Stage II: Cracked Section in the working stage

Stage III: Cracked Section at ultimate stage

Analysis of cracked Section:

1- Determine the position of

the neutral axis, N.A , (z)

2- calculate the inertia of

the cracked section (Icr).

3-Calculate the working moment

Mw using the maximum allowable stresses

fc(all) in conc. and fs(all) for steel as follows:

Stage 2: Cracked section in Working stage

a rectangular section reinforced with steel As.

Steps

Feb-16


2016 7

Calculation of working moment Mw

1- for a singly reinforced section

n As

(n=15)

fc

fs/n

Calculation of working moment Mw for a cracked Doubly Reinforced section

1- Determine the position of N.A) (z): use (n = 15)

Taking moments of areas @ the N.A.

2- Calculate the inertia:

0dnAznAz2

b

0zdnA)2

z(bz

ss

2

s

2s

3

cr zdnA3

bzI

Feb-16


2016 8

z

IfM

crallc

w

)(

1

3- Calculate the working moment Mw

Mw is the lesser of Mw1 , Mw2

)(

)(

2zd

In

f

Mcr

alls

w

Calculation of working moment Mw

2- for a doubly reinforced section

n As’

n As

(n=15)

fc

fs/n

Feb-16


2016 9

Calculation of working moment Mw for a cracked Doubly Reinforced section

1- Determine the position of N.A) (z):



0)()(2

2

'''2

''

dAdAnzAAnzb

zdnAdznAz

bz

ssss

ss

2''23

3dznAzdnA

bzI sscr

z

IfM

crallc

w

)(

1



)(

)(

2zd

In

f

Mcr

alls

w

Feb-16


2016 10

Working stresses used in Calculation of working

moment (stage II - cracked section)

Allowable stresses in concrete and steel according to ECP

Code 203 – Version 2007 From Table 5 – 1:

- Concrete in Case of - sections > 200 mm under flexure:

30 25 20 fcu

10.5 9.5 8 fc all - flexure

400 360 240 fy

220 200 140 fs all

- Steel reinforcement

Table 5-1

Working

stresses in

concrete

and steel

according

to ECP 203

Feb-16


2016 11

Mw is the least of Mw1 , Mw2

z

IfM

cravc

w

)(

1

M+vew

)(avcf

)(allcf

For T - section use :

fc(av) = 2/3 fc(all)

For -section use:

fc = fc(all) = allowable compression

(bending) stress in concrete

working stage (from table)

█-section

working stresses in conc. in Case of

T- sections under flexure:

Calculation of working moment M+vew

for T-sec. 1- Determine the position of N.A) (z):


02

2

2

dnAznAzB

zdnAz

Bz

ss

s

mmz

assume z < ts

Feb-16


2016 12

in case of z < ts

the assumption is O.k.

Complete the solution, calculate Icr then Mw is the

lesser of Mw1 & Mw2

in case of z > ts then the assumption is wrong

Calculate another z Taking moments of areas @ the

new N.A. then Complete the solution, calculate Icr

then Mw is the lesser of Mw1 & Mw2

- Calculation of working moment M+vew for T-sec.

Case of z < ts


23

3zdnA

BzI scr

z

If

Mcrallc

w

)(

13

2


)(

)(

2zd

In

f

Mcr

alls

w


Feb-16


2016 13

- Calculation of working moment M+vew for T-sec.

Case of z > ts

1- Taking moments of areas @ the N.A.

0]2

)([])([2

)2

()(2

2

ssss

ss

s

tbBdnAztbBnAz

b

zdnAt

ztbBz

bz

mmz


2233

)2

()(12

)(3

zdnAt

ztbBt

bBbz

I ss

ss

cr

z

If

Mcrallc

w

)(

13

2

)(

)(

2zd

In

f

Mcr

alls

w


Mw is the lesser of

Mw1 , Mw2

3- Calculation of working moment M-vew for T-sec.

Solve as rectangular

section as before

M-vew

Feb-16


2016 14

Example - 3

‘ 2 For the section shown

Calculate Mw

Solution

‘ 2

‘ 2

1- Determine the position of N.A) (z): use (n = 15)



0)()(2

2

'''2

''

dAdAnzAAnzb

zdnAdznAz

bz

ssss

ss

As = 5 * 201 = 1005 mm2 ,

A’s = 3 * 113 = 339 mm2

mmz

zz

zz

234

0924843.161

0)]50(339)750(1005[15)3391005(15125

2

2

2''23

3dznAzdnA

bzI sscr

Feb-16


2016 15

‘ 2

‘ 2 3- Calculate the working moment Mw:

2''23

3dznAzdnA

bzI sscr

49

223

1025.5

50234)339(15234750)1005(153

)234(250

mm

mkNmmN

z

IfM

crallc

w

.1.213.101.213

234

)1025.5(5.9

6

9)(

1


Mw = 135.7 kN.m

mkN

zd

In

f

Mcr

alls

w

.7.135

10)234750(

)1025.5(15

200

)( 6

9)(

2

Example - 4

Calculation of working moment Mw – T - section

186mm70

mm510

mm120

mmb 200

mmh 700

mmB 800For the section shown

Calculate Mw

Feb-16


2016 16

Solution

1- Determine the position of N.A) (z):


02

2

2

dnAznAzB

zdnAz

Bz

ss

s

186

mmB 800

mmb 200

As = 6 * 254 = 1524 mm2

mmz

zz

zz

3.163

05.3600415.57

0)630)(1524(15)1524(15400

2

2

mmts 120

the assumption is wrong

z > ts

Taking moments of areas @ the new N.A.

> ts = 120 mm

mmz

zz

zz

7.167

01872186.948

0]2

120)200800()630)(1524(15[

)]120)(200800[()1524(15[100

2

2

2

186

mmB 800

mmb 200

mmb 200

mmts 120

0]2

)([])([2

)2

()(2

2

ssss

ss

s

tbBdnAztbBnAz

b

zdnAt

ztbBz

bz

z > ts

Taking moments of areas @ the new N.A.

Feb-16


2016 17

z > ts

49

22

33

101.6

)7.167630)(1524(15)607.167(120)200800(

12

)120)(200800(

3

)7.167(200

mm

Icr

186

mm800

mm200

mmb 200

mmts 120

2- Calculate the inertia Icr :

22

33

)2

()(

12)(

3

zdnAt

ztbB

tbB

bzI

ss

s

scr

z

IfM

cravc

w

)(

1

3- Calculate the working moment Mw :

mkN

z

IfM

cravc

w

.4.230

107.167

)101.6)(5.9(3

2

6

9

)(

1


Mw = 193.5 kN.m

mkN

zd

In

f

Mcr

alls

w

.5.193

10)7.167630(

)101.6(15

220

)( 6

9)(

2

Spring 2016 Dr. Alaa Helba - Delta...

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