SIMPLIFIED CODE PROCEDURES 1 Indian Code Provisions

The Indian standard code method (IS:1343) for computing the flexural strength of rectangular sections or T-sections in which neutral axis lies within the flange is based on the rectangular and parabolic stress block as shown in Fig. 7.7.

The moment of resistance is obtained from the equation, Mu= fpuAp(d – 0.42 xu)

Where Mu= ultimate moment of resistance of the section

fpu= tensile stress developed in tendons at the failure stage of the beam

fp= characteristic tensile strength of the prestressing streel

fpe= effective prestress in tendons after losses

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Ap= area of prestressing tendons

d = effective depth xu= neutral-axis depth

The value of fpudepends upon the effective reinforcement ration

A

p

f

p

bdf

ck

For pretensioned and post-tensioned members with an effective bond between concrete and tendons, the values of fpuand xuare given in Table 7.1. The effective prestress fpe after all

losses should be not less than 0.451 fp. For post-tensioned rectangular beams with unbonded

tendons, the values of fpuand xuare influenced by the effective span to depth rations, and their

values for different span/depth rations are shown in Table 7.2.

The ultimate moment of resistance of flanged sections in which the neutral axis falls outside the flange is computed by combining the moment of resistance of the web and flange portions and considering the stress blocks shown in Fig.7.8.

PROBLEM-3

A pretensioned prestressed concrete beam having a rectangular section. 150 mm wide and 350 mm deep, has an effective cover of 50 mm. If fck= 40 N/mm2, and the areaof

prestressing steel Ap= 461 mm2, fp= 1600 N/mm2, and the area of prestressing steel Ap= 461

mm2, calculate the ultimate flexural strength of the section using IS: 1343 code provisions.

Solutions.

Given date : fck= 40 N/mm2 b = 150 mm

fp= 1600 N/mm2 d = 300 mm

Ap= 461 mm2

The effective reinforcement ratio is given by

f

p

A

p

= 1600 x 461

= 0.40

f

ck

bd

40 x 150 x 300

From Table 7.1, the corresponding values of

f

pu

=

0.9 and

x

u

= 0.783

0.87 f

p

d

f

pu

= (0.87 x 0.9 x 1600) = 1253 N/mm

2

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x

u

= (0.783 x 300) = 234.9 mm

M

u

= f

pu

A

p

(d – 0.42 x

u

)

= 1253 x 461 (300 – 0.42 x 234.9)

= 116 x 10

6

N mm = 116 kN m.

PROBLEM-4

A pretension, T-section has a flange which is 300 mm wide 200mm thick. The rib is 150 mm wide by 350 mm deep. The effective depth of the cross section is 500 mm. Given Ap= 200

mm2, fck= 50 N/mm2, estimate the ultimate moment capacity of the T-section using the Indian

standard code regulations.

Solution.

Given data : fck= 50 N/mm2 b = 300 mm

fp= 1600 N/mm2 d = 500 mm

Ap= 200 mm2

Assuming that the neutral axis falls within the flange, the value of b = 300 mm for computations of effective reinforcement ratio.

f

p

A

p

=

1600 x 200

= 0.04

f

ck

bd

50 x 300 x500

From Table 7.1, the corresponding values of the ratios are

f

pu

= 1.0 and

x

u

= 0.09

0.87f

p

d

fpu= (0.87 x1600) = 1392 N/mm2

Xu = (0.09 x500) = 45mm

The assumption that the neutral axis falls within the flange is correct, Hence the ultimate flexural strength of the section is

Mu= fpuAp(d-0.42xu)

= (1392 x200) (500-0.42 x 45) = (134 x106) Nmm =134 kN m.

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PROBLEM-5

A pretensioned, T-section has a flange 1200mm wide and 1500 mm thick. The width and depth of the rib are 300 and 1500 mm respectively. The high-tensile steel has an area of 4700mm2and is located at an effective depth of 1600mm. If the characteristic cube strength of the concrete and the tensile strength of steel are 40 and 1600 N/mm2, repectively, calculate the flexural strength of the T-section

Solution. Given data: Ap=4700mm2 d=1600 mm Fck=40 N/mm2 Dt=150 mm B=1200 mm bw=300 mm

A

p

=(A

pw

+A

pf

)

A

pf

=0.45 f

ck

(b- b

w

)

D

f

f

p

=(0.45 x 40) (1200-300) 150

=1518 mm

2

1600

A

pw

=(4700-1518)=3182 mm

2

Also

A

pw

f

p

3182 x 1600

=0.265

B

w

df

ck

300 x1600 x40

f

pu

=1.00

/ f

pu

= (0.87 x1600) =1392 N/mm

2

0.87 f

p

X

u =

0.56

/ X

u

=(0.56 x 1600) =896 mm

d

M

u

= f

pu

A

pw

(d-0.42 x

u

) + 0.45 f

ck

( b-b

w

) D

t

(d – 0.5 D

t

)

= (1392 x3182) (1600-0.42 x 896) + 0.45 x40 x 900 x150 (1600-75)

= [(5420 x10

6

) + (3705 x 10

6

)]

= (9125 x 10

6

) Nmm = 9125 K Nm

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PROBLEM-7

A Post – tensioned prestressed concrete tee beam having a flange width of 1200 mm flange thickness of 200 mm. thickness of being 00 mm is prestressed by 2000mm2 of high – tensile steel located at an effective depth of 1600 mm. if fck = 40 N / mm2and fp =1600

assuming span / depth radio as 20 and fpe =1000 N /mm2

Solution.

b = 1200 mm bw = 300mm, d = 1600mm

Df = 200mm , ( L / d ) = 20

fck = 40 N / mm2, fp = 1600 N / mm2 fpe = 1000 N/ mm2

A

_p

= 2000 mm

2

Assuming the neutral axis to fall within the flange , we have the radio of

A

p

f

pe

=

2000 x 1000

= 0.26

bd f

ck

1200 x 1600 x 40

From table 7.2, corresponding to radio [ L / d ] = 20 , by interpolation we have the radio. [ fpu/ fpe] = 1.34 / fpu= [ 1.34 X 1000 ] = 134o N / mm2

And [ xu/ d ] = 0.10 / xu= ( 0.1 X 1600 ) = 1600mm2< Dr = 200mm

Hence the neutral axis falls within the flange.

The ultimate flexural strength of the unbounded beam is computed as Mu = Apfpu[ d – 0.42 xu] = [ 2000 X 1340 ] [1600 – 0.42 X 160 ]

= ( 4107 X 106) Nmm = 4107 kNm.

PROBLEM-8

A pretensioned beam of rectangular section 400 mm wide and 600 mm overall depth is stressed by 1700 mm2 of high – tensile steel wires located 100 mm from the sofift of the section. If the characteristics cube strength of concrete is 50 N / mm2 and tensile strength of prestressing steel is 1600 N / mm2 , estimate the flexural strength of the section using the British cord recommendations. Assume the effective prestress after all losses as 960 N / mm2

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Solution : -

Given data : Aps = 1700 mm2 Fpu = 1600 N/ mm2 B = 400mm Fpe = 960 N / mm2 D = 500mm Fpu = 50 N / mm2

The radio ,

f

pe = 960 = 0.6

f

pu

1600

Radio

fpe

A

ps

1600 x 1700

f

cu

bd

=

50 x 400 x 500

= 0.272

From table

f

pb = 0.865 0.87 fpu fpu = ( 0.865 X 0.87 X 1600 ) = 1204 N /

mm

2 X

= 0.515

D

x

= (0.1515 X 500) = 257.5 mm

d

n

=0.45 x = 115.87mm

M

u

= F

pb

A

ps

( d – d

n

)

= ( 1204 X 1700 ) ( 500 – 115 .87)

= ( 786 X 10

6

) N mm = 786 kNm.

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PROBLEM-9

A prestressed concrete beam of effective span 16m is of rectangular section 400mm wide by 1200 mm deep. The tendons consist of 3300 mm2of stands of

characteristics strength 1700 n / mm2 with an effective

prestress of 910 N / mm

2the stands are located 870 mm from the top face of the beam . if Fcu=60 N / mm2 , estimate

the flexural strengths of the section as per British code provisions for the following cases.

a) bonded tendons b) unbonded tendons

solution

: -

In document Ce2404 Notes Qn Answers Rejinpaul (Page 38-44)