Lec.3 working stress 1

Analysis & Design of Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method

23

Dr. Muthanna Adil Najm

1- Working Stress Design Method or elastic method or alternative design method

or allowable stress design method : This method was the principal one used since

1900s to 1960s.

The working stress design method maybe expressed by the following:

Load = service (unfactored) load

aff

Where:

f = an elastically computed stress, such as by using flexural stress = I

cM . for

beams.

fa = a limiting allowable stress prescribed by ACI code , as a percentage of cf for

concrete and as a percentage of fy for steel.

2- Ultimate Strength Design Method:

In this method, service loads are increased by factors to obtain the load at which

the failure is considered to be " imminent". Also, the section strengths are reduced

by a safety reduction factors.

The ultimate strength design method maybe expressed by the following:

Strength provided ≥ Strength required to carry factored loads

Types of Beams:

1- Types of beams according to section reinforcement:

a- Singly Reinforced concrete beams : main steel reinforcement used at tension

zone only.

b- Doubly reinforced concrete beams: main steel reinforcement used at tension

zone and compression zone.

2- Types of beams according to section Shape:

a- Beams of rectangular section.

b- Beams of ( T, L & I ) section.

Design Methods

b

d sA'

sA

b

d

sA

h

Singly Reinforced Beam Doubly Reinforced Beam


24


c- Beams of irregular sections.

Assumptions:

1- Plain section before bending remains plain after bending.

2- Both concrete and steel obey to Hook's law.

E

3- Strain and stress are proportional to the distance from neutral axis.

4- Concrete strength in tension is negligible.

5- Perfect bond must be maintained between steel and concrete.

6- Allowable stress:

For concrete: cca ff 45.0

For steel : 140saf for fy = 300 and 350 MPa.

170saf for fy = 420 MPa

Structural Behavior of R.C. Beams:

Working Stress Design Methods

b

d

sA

cε

sε

cf

sf

tε tf

cracks

crushing


25


Three stages maybe noticed for concrete beam tested in laboratory to failure:

1- Uncracked concrete stage: Full concrete section still works.

2- Cracked concrete stage.

Where : fr = Modulus of rupture of concrete.

3- Ultimate concrete stage.

Transformed Section Method:

From Hook's law:

ccc Ef

sss Ef

The basic concept of transformed section is that the section of steel and concrete is

transformed into a homogenous section of concrete by replacing the actual steel

area to an equivalent concrete area.

Two conditions must be satisfies:

1- Compatibility:

b

d

sA sε

tε

cε

sf

cf

b

d

sA

cε

sε

ac≤ f cf

as≤ fsf tε

crt fff 7.0

N.A kd

cε

tε

ac< f cf

as< fsf

r< f tf

sε

b

d

sA

N.A


26


sc (at the same level: same distance from neutral axis)

c

cc

E

f and

s

ss

E

f

s

s

c

c

E

f

E

f c

c

ss f

E

Ef

Or cs nff

Where : n is the modular ratio and c

s

E

En

2- Equilibrium:

Force in transformed concrete section = Force in actual steel section

sscc AfAf

scsccc nAfAnfAf

sc nAA

There are two cases of transformed section:

1- Uncracked Section: where rt ff

2- Cracked Section: where rt ff

For doubly reinforced beams, the cracked section is:

b

d

sA

snA

b

kd s= b.kd + nA c)totalA

b

d

sA snA

s1)A-=(n sA-snA

b b


27


Ex.1) For the beam section shown below, if the applied moment is 35 kN.m , fr = 3.1

MPa and n = 9.

1- Calculate the maximum flexural stresses in concrete at top fiber and bottom

fiber and in steel reinforcement.

2- Calculate the cracking moment of the section.

Sol.)

22

18474

283 mmAs

21647761847195003001 mmAnbhA s

Find N.A. location by taking moment of areas about top fiber.

mmy 265

164776

420184719250500300

492

33

10513.32654201847193

235300

3

265300mmxI

1- Flexural stresses:

a- Tension stress at bottom fiber of concrete:

22

9

6

/1.3/34.210513.3

2351035mmNfmmN

I

Mcf rt

Since tension stress at bottom fiber of concrete < modulus of rupture ( rct ff ),

then section is not crack and hence assumption is true.

b- Compression stress at top fiber of concrete:

s1)A-=(n sA-snA

300 mm

420

mm 3Ø28

500

mm

265 mm

N.A

b

d s'A

snA

b

kd s1)A'-(2n= b.kd + c)totalA

snA +

sA

s1)A'-(2n


28


2

9

6

/64.210513.3

2651035mmN

I

Mcfc

c- Stress in steel:

2

9

6

/9.1310513.3

1551035mmN

I

Mcnf s

2- Cracking moment ( Mcr ):

mkNmmN

c

IfM r

cr .34.46.1034.46235

10513.31.3 69

Ex.2) Calculate the maximum flexural stresses for the beam section shown below, if the

applied moment is 95 kN.m , and n = 9. Compare with allowable stresses if fy =

420 MPa and f 'c = 25 MPa.

Sol.)

Assume cracked section. Find kd by taking moments about the N.A.

kddnAkd

kd s 2

300 kdkd

kd 420184792

300

kdkd 166236981660150 2

0465441112 kdkd

mmkd 167

2

445111

2

4654441111112

492

3

1053.1167420184793

167300mmI

a- Tension stress in concrete:

300 mm

420

mm 3Ø28

500

mm

2= 9(1847) = 16623mm snA

kd = 167 mm N.A

300 mm

420 - kd =

253 mm


29


MPaffMPa

I

Mcf crt 5.3257.07.07.20

1053.1

16750010959

6

Assumption is true, section is cracked.

b- Compression stress in concrete:

MPaffMPa

I

Mcf ccac 25.112545.045.037.10

1053.1

16710959

6

.OK c- Stress in steel:

MPafMPa

I

Mcnf sas 1704.141

1053.1

25310959

9

6

.OK

Ex.3)

Calculate the maximum flexural stresses in concrete and steel for the T beam

shown below. M = 100 kN.m , n = 10 and f 'c = 25 MPa.

Sol.)

Assume that N.A. lies within the flange.

214734913 mmAs

Find kd by taking moments about N.A.

kdkd

6001473102

9002

kdkd 147308838000450 2

0196407.322 kdkd

mmmmkd 100125

2

1964047.327.322

kd

600-kd

900 mm

nAs

N.A. 100

500

900 mm

3Ø25

250

680


30


Assumption is false and N.A. lies at the web.

Find kd by taking moments about N.A.

2

10010025050100900

kdkdkd

kd 600147310

0967046382 kdkd

mmmmkd 100126

2

9670446386382

492

33

10906.31266001473103

26250900

3

126900mmI


MPaffMPa

I

Mcf crt 5.3257.07.02.14

10906.3

126680101009

6



MPa

I

Mcfc 23.3

10906.3

126101009

6

c- Stress in steel:

MPa

I

Mcnf s 35.121

10906.3

1266001010010

9

6

Ex.4) Calculate the maximum stresses in concrete and steel for the beam section shown

below. M = 160 kN.m , n = 10 and f 'c = 25 MPa.

Sol.) 224637.6154 mmAs

kd

600-kd

900 mm

nAs

N.A.

250 mm

350 mm

360

70

70

500mm

2Ø28

4Ø28 430-kd snA

350

kd

s1)A'-(2n

N.A.


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212317.6152 mmAs

Find kd by taking moment about the N.A.

kddnAdkdAnkd

kdb ss 122

kdkdkd

kd 43024631070123111022

350

kdkdkd 1416051993551342

0698742752 kdkd

mmkd 160

2

6987442752752

4922

3

10463.21604302463107016012311203

160350mmI


MPaffMPa

I

Mcf crt 5.3257.07.022

10463.2

160500101609

6



MPa

I

Mcfc 39.10

10463.2

160101609

6

c- Stress in tension steel:

MPa

I

Mcnfs 4.175

10463.2

1604301016010

9

6

d- Stress in compression steel:

MPa

I

Mcnfs 93.116

10463.2

70160101601022

9

6

Education

Lec.3 working stress 1