Lec.9 strength design method doubly reinforced beams

Analysis & Design of Reinforced Concrete Structures (1) Lecture.9 Strength Design Method

73

Dr. Muthanna Adil Najm

When compression steel is used, the nominal resisting moment of the steel is

assumed to be consisted of two parts.

1- The part due to the compression concrete and the balancing area of tensile

reinforcement.

211

adfAM ysn

2- The part due to compression steel and the balancing area of tensile

reinforcement.

ddfAM ysn 2 if compression steel is yield

ddfAM ssn 2 if compression steel is not yield

ddfAa

dfAMMM ysysnnn

2121

The strain in the compression steel is checked to determine whether or not it has

yielded. With the obtained strain, the compression stress is determined.

ss Ec

dcf

003.0

And the value of 2sA is computed with the following expression:

ssys fAfA 2

Also, the net tensile strain in the extreme tensile steel ( εt ) should be calculated to

check the ductility of the section.

005.0t Ø = 0.9

005.0004.0 t Ø < 0.9

To determine the strain and thus the stress in both tensile and compression steel,

the location of the N.A ( the value of 'c') should be determined using the following

quadratic equation:

Strength Design Method

Analysis & Design of Doubly Reinforced Beams.

SA'

SA

=

S1A

d + SA'

S2A

d-d'

d'


74


sscys Ec

dcAcbffA 003.085.0 1

Balanced and maximum steel ratio requirements:

y

sbb

f

f ( Balanced steel ratio)

y

s

f

f maxmax ( Maximum steel ratio)

Analysis Procedure:

1- Find the location of the N.A ( the value of 'c' ) using the quadratic equation:

sscys Ec


2- Solve the above equation for 'c' then find 'a':

ca 1

3- Compute the strains in compression steel and tensile steel.

a- Check the yielding of compression steel; 003.0

c

dcs

if s

yys

E

f Then compression steel yields and ss AA 2

if s

yys

E

f Then compression steel is not yield and ss E

c

dcf

003.0

and sy

ss A

f

fA

2

b- Calculate the strain of the extreme tensile steel t to check section ductility.

003.0

c

cdt

4- 21 sss AAA

5- Calculate the design moment strength:

ddfA

adfAM ysysn

21 if

s

yys

E

f and;

ddfA

adfAM ssysn

21 if

s

yys

E

f


75


Analysis Examples:

Ex.1) Determine the design moment capacity of the beam section shown below. Use

MPafc 21 and MPaf y 420 .

Sol.) 2407210184 mmAs

212326162 mmAs

Find the location of the N.A ;

sscys Ec


000,200003.065

123235085.02185.04204072

c

cc

4804800073920037.53101710240 2 ccc

04804800097104037.5310 2 cc

096.904786.1822 cc

mmc 4.223

2

87.26386.182

2

96.9047486.18286.1822

mmca 9.1894.22385.01

Check the yielding of compression steel;

0021.0200000

4200021.0003.0

4.223

654.223003.0

ys

c

dc

Compression steel yields ( ys ff ) and 22 1232mmAA ss

221 284012324072 mmAAA sss

Check section ductility;

005.000519.0003.04.223

4.223610003.0

c

cdt

Section is ductile and 9.0

65 mm

545 mm

680 mm

2 Ø 28

4 Ø 36

350 mm


76


ddfA

adfAM ysysn

21

mkNM n .72.806106561042012322

9.18961042028409.0 6

Ex.2) Determine the design moment capacity of the beam section shown below. Use

MPafc 28 and MPaf y 420 .

Sol.) 228287074 mmAs

27603802 mmAs

Find the location of the N.A ;

sscys Ec


000,200003.065

76035085.02885.04202828

c

cc

2964000045600057.70801187760 2 ccc

0296400007317605.7080 2 cc

015.418635.1032 cc

mmc 48.134

2

61.16535.103

2

15.4186435.10335.1032

mmca 3.11448.13485.01

Check the yielding of compression steel;

0021.0200000

42000155.0003.0

48.134

6548.134003.0

ys

c

dc

Compression steel is not yield and ;

MPaEc

dcf ss 31020000000155.0003.0

22 561760

420

310mmA

f

fA s

y

ss

221 22675612828 mmAAA sss

Check section ductility;

65 mm

545 mm

680 mm

2 Ø 22

4 Ø 30

350 mm


77


005.00106.0003.048.134

48.134610003.0

c

cdt

Section is ductile and 9.0

ddfA

adfAM ssysn

21

mkNM n .3.58910656103107602

3.11461042022679.0 6

Design of doubly reinforced concrete beams:

Sufficient tensile steel can be placed in most beams and bigger section can be used

so that compression steel is not needed. But, if section dimensions is restricted and

the required steel area exceeded the maximum steel ratio max , then compression

steel should be used.

Design Procedure:

1- Calculate the factored applied moment.

2- Calculate max and check if max then design as a doubly reinforced beam.

Or calculate Mu max and compare with Mu

3- calculate 1sA where: bdAs max1

4- Assume 9.0 and calculate 1nM :

c

yynu

f

fbdfMM max

2max1 59.01

5- 12 uuu MMM

6- Find the location of the neutral axis and check the yielding of compression steel

bf

fAa

c

ys

85.0

1 and

1

ac

003.0

c

dcs

a- If s

yys

E

f Then compression steel yields and

ddf

MAA

y

uss

22

b- If s

yys

E

f Then compression steel is not yield and ss E

c

dcf

003.0

ddf

MA

s

us

2 and sy

ss A

f

fA

2


78


7- 21 sss AAA

Design Examples:

Ex.3)

A beam is limited to the dimensions shown below. If mkNMu .1225 , determine

the required steel area. Use MPafc 21 and MPaf y 420 .

Sol.)

0135.0005.0003.0

003.085.0 1max

y

c

f

f

94.7

7003509.0

1012252

6

2

bd

MR u

u

53.23

2185.0

420

85.0

c

y

f

f

y

u

f

R

211

1

0135.00284.0420

53.2394.7211

53.23

1max

Design as a doubly reinforced concrete beam. 2

max1 33077003500135.0 mmbdAs

c

yyu

f

fbdfM max

2max1 59.01

mkNMu .7.7351021

4200135.059.017003504200135.09.0 62

1

mkNMMM uuu .3.4897.735122512

Check to see if compression steel yields;

32.2223502185.0

4203307

85.0

1

bf

fAa

c

ys

mma

c 55.26185.0

32.222

1

70 mm

700 mm

350 mm


79


0021.0200000

4200022.0003.0

55.261

7055.261003.0

s

yys

E

f

c

dc

compression steel yields.

2

62

2 2054707004209.0

103.489mm

ddf

MAA

y

uss

Use 3Ø30 221217073 mmAs 2

21 536120543307 mmAAA sss

Use 8Ø30 256567078 mmAs

Ex.4) Design a rectangular beam with maximum permissible dimensions shown in the

Figure below for mkNMD .230 and mkNM L .305 . Use MPafc 28 and

MPaf y 420 .

Sol.)

mkNMu .7643056.12302.1

0181.0005.0003.0

003.085.0 1max

y

c

f

f

mkNMu .7.5381028

4200181.059.015003754200181.09.0 62

.max)

mkNMmkNM uu .7.538.764 max)

Design as a doubly reinforced concrete beam.

2max1 33945003750181.0 mmbdAs

c

yyu

f

fbdfM max

2max1 59.01

100 mm

500 mm

375 mm

70 mm

700 mm

350 mm

2 Ø 30

8Ø 30


80


mkNMMM uuu .3.2257.53876412

Check to see if compression steel yields;

7.1593752885.0

4203394

85.0

1

bf

fAa

c

ys

mma

c 9.18785.0

7.159

1

0021.0200000

4200014.0003.0

9.187

1009.187003.0

s

yys

E

f

c

dc

compression steel is not yield.

MPaEc

dcf ss 68.2802000000014.0003.0

2

62 2230

10050068.2809.0

103.225mm

ddf

MA

s

us

Use 6Ø22 222803806 mmAs

22 14902230

420

68.280mmA

f

fA s

y

ss

221 488414903394 mmAAA sss

Use 8Ø28 249286168 mmAs

100 mm

500 mm

375 mm

6 Ø 22

8Ø 28

Education

Lec.9 strength design method doubly reinforced beams