Ce2404 Notes Qn Answers Rejinpaul

www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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www.rejinpaul.com UNIT 1 INTRODUCTION THEORY AND BEHAVIOUR

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UNIT-II FLEXURAL STRENGTH OF PRESTRESSED CONCERTE SECTIONS

TYPES OF FLEXURAL FAILURE When prestressed concrete members to bending loads, different types of flexural failures are possible at critical sections, depending upon the principle controlling parameters, Such as the percentage of reinforcement in the section, degree of bond between tendons and concrete, compressive strength of concrete and the ultimate tensile strength of the tendons. In the post-cracking stage, the behavior of a prestressed concrete member is more akin to that of a reinforced concrete section may as well be used for prestressed concrete sections. The Various types of flexural failures encountered in prestressed concrete Members are examined in the light of recommendations of various codes of practise.

Fracture of steel in tension The sudden failure of a prestressed member without any warning is generally due to the fracture of steel in the tension zone. This type of failure is imminent when the percentage of steel provided in the section is so low that when the concrete in the tension zone cracks, the steel is not in a position to bear up the additional tensile stress transferred to it by the cracked concrete. This type of failure can be prevented by providing a certain minimum percentage of steel in the cross section The Indian standard code IS: 1343 prescribes a minimum longitudinal reinforcement of 0.2 per cent of the cross-sectional area in all cases except in the case of pretensioned units of small sections. When a high-yield strength deformed reinforcement

is used, the minimum

steel percentage is reduced to 0.15 per cent. The percentage of steel provided , both tensioned and untensioned

taken together , should be sufficient so that when the concrete in the

precompressed tensile zone cracks, the steel is in a position to bear the additional tensile stress transferred to it by the cracking of the adjacent fibres of the concrete , thereby preventing a sudden failure of the beam due to fracture of steel in tension In contrast, the British code BSEN :1992-1-1 prescribes that the number of prestressing tendons should be such that cracking of the concrete precedes the PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

failure of the

Mr.M.KALIRAJ

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www.rejinpaul.com beam. This requirement will be satisfied if the ultimate moment of resistance of the section exceeds the moment necessary to produce a flexural tensile stress in the concrete at the extreme tension fibres of magnitude equal to 0.6√fcu . In these computations, the effective prestress in concrete should be considered after allowing for the various losses. The American Concrete Institute code ACI: 318M-2011 specifies that the minimum area of bonded reinforcement should be not less than 0.004 times the area of that part of crosssection which is between flexural tension face and the centre of gravity of the gross concrete section.

Failure of under-reinforced sections If the cross section is provided with an amount of steel greater than the minimum prescribed in case 1, the failure is characterised by an excessive elongation of steel followed by the crushing of concrete. As bending loads are increased, excessive elongation of the steel raises the neutral axis closer to the compression face at the critical section. The member approaches failure due to the gradual reduction of the compression zone, exhibiting large deflections and cracks. Which develop at the soffit and progress towards the compression face. When the area of concrete in the compression zone is insufficient to resist the resultant internal compressive force, the ultimate flexural failure of the member takes place through the crushing of concrete. Large deflections and wide cracks are the characteristic features of under-reinforced section at failure (Fig.7.1). This type of behaviour is generally desirable since there is considerable warning before the impending failure. As such, it is a common practise to design under-reinforced sections which become more important in the case of statically indeterminate structures. An upper limit on the maximum area of steel is generally prescribed in various codes for under-reinforced sections.

Failure of over-reinforced sections When the effective reinforcement index, which is expressed in terms of the percentage of reinforcement, the compressive strength of concrete and the tensile strength of steel, exceeds a certain range of values, the section is said to be over-reinforced. Generally, over-reinforced members fail by the sudden crushing of concrete, the failure being characterised by small deflections and narrow cracks (Fig. 7.1). The area of steel being comparatively large. The stresses developed in steel at failure of the member may not reach the tensile strength and in many cases it may well be within the proof, stress of the tendons. PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com In structural concrete members, it is undesirable to have sudden failures without any warning in the form of excessive deflections and widespread cracks, and consequently the use of over-reinforced sections are discouraged. The amount of reinforcement used in practise should, preferably, not exceed that required for a balanced section. In this connections, most of the codes follow a conservative approach in formulating the evaluation procedures for flexural strength calculations of over-reinforced sections. The redistribution of moments in an indeterminate structure depends upon the rotation capacities of the critical sections of the member under a gives system of loads. The use of overreinforced sections in such structures curtails the rotation capacity of the sections, consequently affecting the ultimate load on the structure. Other modes of failure Prestressed concrete members subjected to transverse loads may fail in shear before their full flexural strength is attained, if they are not adequately designed for shear. Web shear cracks may develop occur due to web crushing. In the case of pretensioned members, the failure of the bond between the steel and the surrounding concrete is likely due to the inadequate transmission lengths at the ends of members. In post-tensioned members, anchorage failures may take place if the end block is not properly designed to resist the transverse tensile forces. 7.2 STRAIN COMPATIBILITY METHOD The rigorous method of estimating the flexural strength of prestressed concrete section is based on the compatibility of strains and equilibrium of forces acting on the section at the stage of failure. The basic theory is applicable to all structural concrete sections, whether reinforced or prestressed, and generally the following assumptions are made: 1. The stress distribution in the compression zone of concrete can be defined by means of coefficients applied to the characteristic compressive strength and the average compressive stress and the position of the centre of compression can be assessed. 2. The distribution of concrete strain is linear (plane sections normal to axis remain plane after bending) 3. The resistance of concrete in tension is neglected. 4. The maximum compressive strain in concrete at failure reaches a particular value.

PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

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www.rejinpaul.com PROBLEM-1 A pretensioned concrete beam with a rectangular section, 100mm wide by 160 mm deep, is prestressed by 10 hightensile wires of 2.5 mm-diameter located at an eccentricity of 40 mm. The initial force in each wire is 6.8 kN. The strain loss in wires due to elastic shortening. Creep and shrinkage of concrete is estimated to be 0.0012 units. The characteristic cube strength of concrete is 40 N/mm2. Given the load-strain curve of 2.5 mm-diameter steel wire (Fig. 7.6), estimate the ultimate flexural strength of the section using the strain compatibility method. Solution. For fck = 40 N/mm2, read out ɛcu = 0.0033, k1 = 0.57 and k2 = 0.45 from Fig. 7.3. Strain due to load of 6.8 kN in wire is 0.0073. Effective strain in steel after all losses is given by ɛse = (0.0073 – (0.0012) = 0.0061 First trial Assume x = 60 mm From the strain diagram (Fig. 7.2), (ɛsu - ɛse ) /

ɛsu = (0.0033 + 0.0061) = 0.0094

Corresponding force in the wire = 8.4 kN Total tensile force = (10 x 8.4) =84 kN Total compressive force = (k1 . fck . b . x)

(0.57 x 40 x 100 x 60) =

= 136.8 kN

1000

Since tension is less than compression, decrease x for the second trial.

Second trial Assume x= 43 mm

(ɛ su - ɛse) = 0.0059 ɛsu = (0.059 + 0.0061) = 0.012 Corresponding force in the wire = 9.9 kN Total tensile force = (10x9.9) =99 kN Total compressive force =

(0.57 x 40 x 100 x 43)

= 98 kN

1000 PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com Since tension is nearly equal to compression, strain compatinbility is established. Mu = Apsfpb (d – k2x) = (99 x 103) x (120 – 0.45 x 43) = (9.96 x 106) N mm = 9.96 kN m

PROBLEM-2 If the number of wires in Example 7.1 is increased to 16, estimate the flexural strength of the section.

Solution. First trial Assume x = 60 mm (εsu – εse) = 0.0033 εsu = (0.0033 + 0.0061) = 0.0094 Corresponding force in the wire = 8.4 kN Total tensile force = (16 x 8.4) = 134.4 kN

Total compressive force =

(0.57 x 40 x 100 x 60)

= 136.8 kN

1000 Since tension is nearly equal to compression, strain compatibility is established.

(120 – 0.45 x 60) Mu = Apsfpb(d – k2x) = 134.4

1000

= 12. 5 kN m

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 = fpu Ap (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 PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com Ap = area of prestressing tendons d = effective depth xu = neutral-axis depth The value of fpu depends upon the effective reinforcement ration

Apfp bdfck For pretensioned and post-tensioned members with an effective bond between concrete and tendons, the values of fpu and xu are 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 fpu and xu are 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 Ap = 461 mm

d = 300 mm

2

The effective reinforcement ratio is given by

fpAp

=

fckbd

1600 x 461

= 0.40

40 x 150 x 300

From Table 7.1, the corresponding values of

fpu 0.87 fp

= 0.9 and xu

= 0.783

d

fpu = (0.87 x 0.9 x 1600) = 1253 N/mm2 PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com xu = (0.783 x 300) = 234.9 mm Mu = fpuAp (d – 0.42 xu) = 1253 x 461 (300 – 0.42 x 234.9) = 116 x 106 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.

fpAp

=

fckbd

1600 x 200

= 0.04

50 x 300 x500

From Table 7.1, the corresponding values of the ratios are

fpu

= 1.0 and

0.87fp

xu

= 0.09

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. PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com 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 4700mm2 and 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 Fck=40 N/mm B=1200 mm

d=1600 mm 2

Dt=150 mm bw=300 mm

Ap=(Apw+Apf) Apf =0.45 fck(b- bw)

Df fp

=(0.45 x 40) (1200-300)

=1518 mm2

150 1600

Apw =(4700-1518)=3182 mm2 Also

Apwfp

3182

Bwdfck

300 x1600 x40

fpu

=1.00

x

1600

=0.265

/ fpu = (0.87 x1600) =1392 N/mm2

0.87 fp Xu

=

0.56

/ Xu=(0.56 x 1600) =896 mm

d Mu = fpu Apw (d-0.42 xu ) + 0.45 fck ( b-bw) D t(d – 0.5 Dt) = (1392 x3182) (1600-0.42 x 896) + 0.45 x40 x 900 x150 (1600-75) = [(5420 x106) + (3705 x 106)] = (9125 x 106) Nmm = 9125 K Nm PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com 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 2000mm 2 of high – tensile steel located at an effective depth of 1600 mm. if fck = 40 N / mm2 and fp = 1600 assuming span / depth radio as 20 and fpe = 1000 N /mm2

Solution. b

= 1200 mm

bw = 300mm,

Df

= 200mm ,

( L / d ) = 20

fck

= 40 N / mm2 , fp = 1600 N / mm2

Ap

= 2000 mm2

d = 1600mm

fpe

= 1000 N/ mm2

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

Ap fpe bd fck

=

2000 x 1000

= 0.26

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

And [ xu / d ] = 0.10

/ fpu = [ 1.34 X 1000 ] = 134o N / mm2 / 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

PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

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www.rejinpaul.com Solution : Given data : = 1700 mm2

Fpu = 1600 N/ mm2

B

= 400mm

Fpe = 960 N / mm2

D

= 500mm

Fpu

Aps

The radio ,

fpe fpu

Radio

960

=

= 50 N / mm2

= 0.6

1600

fpe Aps

1600 x 1700

fcu bd

=

50 x 400 x 500 = 0.272

From table fpb

= 0.865

0.87 fpu fpu

= ( 0.865 X 0.87 X 1600 ) 2

= 1204 N / mm X

= 0.515

D x

= (0.1515 X 500) = 257.5 mm

dn

=0.45 x = 115.87mm

Mu

= Fpb Aps ( d – d n ) = ( 1204 X 1700 ) ( 500 – 115 .87) = ( 786 X 106 ) N mm = 786 kNm.

PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

Mr.M.KALIRAJ

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www.rejinpaul.com 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 mm2 of stands of 2

characteristics strength 1700 n / mm2 with an effective prestress of 910 N / mm the 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 : Given data :

fpu

= 1700 mm2

L = 16m

fcu

= 60 N / mm2

b = 400 m

2

Aps

= 3300mm

fpe

= 910 N / mm2

( L / d ) = 18.39

For bonded tendons, the radio fpu Aps

=

1700 X 3300

fcu bd

= 0.27

60 x 400 X 870

And

fpe

910

=

fpu

= 0.54

1700

From table 7.3. by interpolation

fpb

= 0.86 and

0.87 fpu

x

= 0.50

d 2

Hence ,

fpb = ( 0.86 X 0.87 X 1700 ) = 1272 N / mm

And

x

/

= ( 0.50 X 870 ) = 435 mm

Mu = Fpb Aps ( d – 0.45x) = ( 1272 X 3300 )(870 – 045 X 435 ) Nmm = 2830 kNm

For unbonded tendons, PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

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www.rejinpaul.com fpb= fbe + 7000 1- 1.7

fpuAps fcubd

(L/d)

= 910 + 7000 [1 -1.7 (0.27)] =1116 N/mm2 18.39 x= 2.47

=2.47

fpuAps

fpb

fcubd

fpu

0.27 1116

d

870

1700

=380.8 mm

Hence, Mu =fpb Aps (d – 0.45 x ) =(1116 X 3300) (870-0.45 X380.8) Nmm =2573 KNm

PROBLEM-10 A post-tensioned ,prestressed concrete girder is of T-section with an effective flange width and depth of 1500 mm and 250 mm , respectively .Thickness of the web is 200 mm. The area of prestressing steel is 5000 mm2 , located at an effective depth of 1600 mm. Given fpu =1600 N / mm2

,

fcu =

40 N/mm2 and fpe = 960 N/mm2 ,estimate the ultimate moment or resistance of the T-Section .Assume the effective span of the girder as 32m Solution Given date : Aps= 5000 mm2

b=1500 mm

Fcu = 40 N/ mm2

bw=200 mm

Fpu=1600 N/mm2

Dt=250 mm

Fpe=960 N/mm2

d =1600 mm

L=32 m

L d

PRESTRESSED CONCRETE

= 32

= 20

1.6

DEPARTMENT OF CIVIL ENGINEERING

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www.rejinpaul.com fpe

=

0.6

fpu Thus Aps

=Apw

+ Apt

Apf

=0.45 fck

(b-bw)

Df fpu

=(0.45 X 40 ) (1500 -200)

250

=3656 mm2

1600 = (5000 - 3656 )=1344 mm2

Apw

and

fpuApw

= 1600 X 1344

fcubwd

=0.168

40 X200 X1600

From Table 7.3, by interpolation fpb

= 0.95

0.87 fpu fpb =(0.95 X0.87 X1600) =1322 N/mm2 X

= 0.35

d x=(0.35 X1600) =560 mm Mu=fpbApw (d -0.45x) + 0.45 fcu (b – bw) Dt (d-0.45 Dt) =[ (1322 X1344) (1600 -0.45 X560) + 0.45 X40 (1500-200) X 250 (1600-0.5 X250) ] =(11,023 X106) Nmm =11,023 kNm

PRESTRESSED CONCRETE

DEPARTMENT OF CIVIL ENGINEERING

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www.rejinpaul.com SRI VIDYA COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING SUBJ. CODDE AND NAME: CE 1402 – PRESTRESSED CONCRETE STRUCTURES FACULTY NAME: C.D.M.Kaliraj

CLASS &SEC :IV yr CIVIL

Academic year: 2014-’15

Semester :VI UNIT 3

Part A 1. Sketch the loop reinforcement, hair-pin bars in end blocks.(NOV-DEC 2009)

2. Sketch the correct arrangement of sheet cage in anchorage zone.(NOV-DEC 2009)

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www.rejinpaul.com 3. Define two stage constructions.(NOV-DEC 2012)  One-stage construction: Construct and initialize the object in one stage, all with 

the constructor.   Two-stage construction: Construct and initialize the object in two separate stages. The constructor creates the object and an initialization function initializes it.  4. Write any two general failures of prestressed concrete tanks.(NOV-DEC 2012)



 deformation of the pre-cast concrete units during construction   Manufacturing inaccuracies led to out of tolerance units being delivered to the site under investigation and may have affected the ability to achieve a good seal.  5. Mention the importance of shrinkage in composite construction?(NOV-DEC 2010) The time dependent behavior of composite prestressed concrete beams depends upon the presence of differential shrinkage and creep of the concretes of web and deck, in addition to other parameters, such as relaxation of steel, presence of untensioned steel, and compression steel etc.

Part B 1. The end block of a post-tensioned PSC beam, 300 x 300 mm is subjected to a concentric anchorage force of 832.8 kN by a Freyssinet anchorage of area 11720 2

mm . Design and detail the anchorage reinforcement for the end block.(NOVDEC 2009) 2

2ypo = (π/4xd )

(1/2)

= (11720)

(1/2)

=

108.25mm 2yo = 300/2 = 150mm Ypo/yo = 0.72 Fc = P/A = 832.8/(300x300) = 9.25N/mm

2 2

Fv(max) =fc(0.98 – 0.825 ypo/yo) = 9.25(0.98 – 0.825x0.72) = 3.57N/mm Fbst = p(0.48 – 0.4 ypo/yo) = 832.8(0.48-0.4x0.72) = 159.89kN 3

Ast = Fbst/0.87fy = (159.89x10 )/(0.87x260) = 706.85mm

2

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www.rejinpaul.com 2. Explain the different types of joints between the walls and floor slab of prestressed concrete tanks.(NOV-D EC 2009) 3. Explain the effect of va rying the ratio of depth anchorage to the de pth of end block on the distribution of b ursting tension. (8) (NOV-DEC 2012) Bursting tensile forces a) The bursting tensile forces in the end blocks, or regions of bonded posttensioned members, sh ould be assessed on the basis of the tendo n jacking load. For unbonded member s, the bursting tensile forces should be assessed on the basis of the tendon jacking load or the load in the tendon at the limit st ate of collapse, whichever is greater ( see Appendix B ). The bursting tensile force, Fbst existing in an individual square e nd block loaded by a symmetrically pla ced square anchorage or bearing plate, may be derived from the equation below:

b) The force Fbst will be distributed in a region extending from 0.1 yo to yo from the loaded face of the end block. Reinforcement provided to sust ain the bursting tensile force may be assumed to be acting at its design stren gth (0.87 times characteristic strength of reinforcement) except that the stress shou ld be limited to a value corresponding to a strain of 0.001 when the concrete cover to the reinforcement is less than 50 mm. c) In rectangular end blocks, the bursting tensile forces in th e two principal directions should be as sessed on the basis of 18.6.2.2. When circular anchorage or bearing plates are use d, the side of the equivalent square area s hould be used.

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www.rejinpaul.com Where groups of anchorages or bearing plates occur, the end blocks should be divided into a series of symmetrically loaded prisms and each prism treated in the above manner. For designing end blocks having a cross-section different in shape from that of the general cross-section of the beam, reference should be made to specialist literature. d) Compliance with the requirements of (a), (b) and (c) will generally ensure that bursting tensile forces along the load axis are provided for. Alternative methods of design which make allowance for the tensile strength of the concrete may be used, in which case reference should be made to specialist literature. e) Consideration should also be given to the spalling tensile stresses that occur in end blocks where the anchorage or bearing plates are highly eccentric; these reach a maximum at the loaded face. 4. (i) Explain the general features of prestressed concrete tanks. (8) (ii) Explain the junctions of tank wall and base slab with neat sketch. (8) (NOVDEC 2012) Joint in the concrete introduced for convenience in construction at which special measures are taken to achieve subsequent continuity without provision for further relative movement, is called a construction joint. A typical application is between successive lifts in a reservoir.

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The position and arrangement of all construction joints should be predetermined by the engineer. Consideration should be given to limiting the number of such joints and to keeping them free from possibility of percolations in a similar manner to contraction joints. A gap temporarily left between the concrete of adjoining parts of a structure which after a suitable interval and before the structure is put into use, is filled with mortar or concrete either completely ( Fig. 5A) or as provided below, with the inclusion of suitable jointing materials ( Fig. 5B and SC). In the former case the width of the gap should be sufficient to allow the sides to be prepared before filling. Where measures are taken for example, by the inclusion of suitable jointing materials to maintain the water tightness of the concrete subsequent to the filling of the joint, this type of joint may be regarded as being equivalent to a contraction joint ( partial or complete ) as defined above. 5. (a) What are the different types of joints used between the slab of prestressed

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www.rejinpaul.com concrete tanks. Joints shall be categorized as below: a) Movetnent Joints - There are three categories of movement joints: contraction joint - A movement joint with a deliberate discontinuity but no initial gap between the concrete on either side of the joint, the joint being intended to accommodate contraction of the concrete ( see Fig. 1 ). A distinction should be made between a complete contraction joint (see Fig. 1A ) in which both concrete and reinforcing steel are interrupted, and a partial contraction joint (. see Fig. 1B ) in which only the concrete is interrupted, the reinforcing steel running through. Expansion joint - A movement joint with complete discontinuity in both reinforcement and concrete and intended to accommodate either expansion or contraction of the structure (see Pig. 2). In general, such a joint requires the provision of an initial gap between the adjoining parts of a structure which by closing or opening accommodates the expansion or contraction of the structure. Design of the joint so as to incorporate sliding surfaces, is not, however, precluded and may sometimes be advantageous.

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b) Construction Joint-A joint in the concrete introduced for convenience in construction at which special measures are taken to achieve subsequent continuity without provision for further relative movement, is called a construction joint. A typical application is between successive lifts in a reservoir.

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The position and arrangement of all construction joints should be predetermined by the engineer. Consideration should be given to limiting the number of such joints and to keeping them free from possibility of percolations in a similar manner to contraction joints. c) Temporary Open Joints - A gap temporarily left between the concrete of adjoining parts of a structure which after a suitable interval and before the structure is put into use, is filled with mortar or concrete either completely ( Fig. 5A) or as provided below, with the inclusion of suitable jointing materials ( Fig. 5B and SC). In the former case the width of the gap should be sufficient to allow the sides to be prepared before filling. Where measures are taken for example, by the inclusion of suitable jointing materials to maintain the water tightness of the concrete subsequent to the filling of the joint, this type of joint may be regarded as being equivalent to a contraction joint ( partial or complete ) as defined above.

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www.rejinpaul.com (b) Design the circular tank (only procedure).(NOV-DEC 2010) .(NOV-DEC 2010) in the construction of concrete structures for the storage of liquids, the imperviousness of concrete is an important basic requirement. Hence, the design of such construction is based on avoidance of cracking in the concrete. The structures are prestressed to avoid tension in the concrete. In addition, prestressed concrete tanks require low maintenance. The resistance to seismic forces is also satisfactory. Prestressed concrete tanks are used in water treatment and distribution systems, waste water collection and treatment system and storm water management. Other applications are liquefied natural gas (LNG) containment structures, large industrial process tanks and bulk storage tanks. The construction of the tanks is in the following sequence. First, the concrete core is cast and cured. The surface is prepared by sand or hydro blasting. Next, the circumferential prestressing is applied by strand wrapping machine. Shotcrete is applied to provide a coat of concrete over the prestressing strands. Analysis The analysis of liquid storage tanks can be done by IS:3370 - 1967, Part 4, or by the finite element method. The Code provides coefficients for bending moment, shear and hoop tension (for cylindrical tanks), which were developed from the theory of plates and shells. In Part 4, both rectangular and cylindrical tanks are covered. Since circular prestressing is applicable to cylindrical tanks, only this type of tank is covered in this module. The following types of boundary conditions are considered in the analysis of the cylindrical wall. a) For base: fixed or hinged b) For top: free or hinged or framed. For base

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www.rejinpaul.com Fixed: When the wall is built continuous with its footing, then the base can be considered to be fixed as the first approximation. Hinged: If the sub grade is susceptible to settlement, then a hinged base is a conservative assumption. Since the actual rotational restraint from the footing is somewhere in between fixed and hinged, a hinged base can be assumed. The base can be made sliding with appropriate polyvinyl chloride (PVC) waterstops for liquid tightness. For top Free: The top of the wall is considered free when there is no restraint in expansion. Hinged: When the top is connected to the roof slab by dowels for shear transfer, the boundary condition can be considered to be hinged. Framed: When the top of the wall and the roof slab are made continuous with moment transfer, the top is considered to be framed. The hydrostatic pressure on the wall increases linearly from the top to the bottom of the liquid of maximum possible depth. If the vapour pressure in the free board is negligible, then the pressure at the top is zero. Else, it is added to the pressure of the liquid throughout the depth. The forces generated in the tank due to circumferential prestress are opposite in nature to that due to hydrostatic pressure. If the tank is built underground, then the earth pressure needs to be considered. The hoop tension in the wall, generated due to a triangular hydrostatic pressure is given as follows. The hoop tension in the wall, generated due to a triangular hydrostatic pressure is given as follows. T = CT w H Ri (9-6.15) The bending moment in the vertical direction is given as follows. M = CM w H3 (9-6.16) The shear at the base is given by the following expression. V = CV w H2 (9-6.17) In the previous equations, the notations used are as follows. CT = coefficient for hoop tension

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www.rejinpaul.com CM = coefficient for bending moment CV = coefficient for shear w = unit weight of liquid H = height of the liquid Ri = inner radius of the wall. The values of the coefficients are tabulated in IS:3370 - 1967, Part 4, for various values of H2/Dt, at different depths of the liquid. D and t represent the inner diameter and the thickness of the wall, respectively. The typical variations of CT and CM with depth, for two sets of boundary conditions are illustrated. The roof can be made of a dome supported at the edges on the cylindrical wall. Else, the roof can be a flat slab supported on columns along with the edges. IS:3370 - 1967, Part 4, provides coefficients for the analysis of the floor and roof slabs. Design IS:3370 - 1967, Part 3, provides design requirements for prestressed tanks. A few of them are mentioned. 1) The computed stress in the concrete and steel, during transfer, handling and construction, and under working loads, should be within the permissible values as specified in IS:1343 - 1980. 2) The liquid retaining face should be checked against cracking with a load factor of 1.2. σCL/σWL ≥ 1.2 (9-6.18) Here, σCL = stress under cracking load σWL = stress under working load. Values of limiting tensile strength of concrete for estimating the cracking load are Specified in the Code. 3) The ultimate load at failure should not be less than twice the working load. 4) When the tank is full, there should be compression in the concrete at all points of at least 0.7 N/mm2. When the tank is empty, there should not be tensile stress

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www.rejinpaul.com greater than 1.0 N/mm2. Thus, the tank should be analysed both for the full and empty conditions. 5) There should be provisions to allow for elastic distortion of the structure during prestressing. Any restraint that may lead to the reduction of the prestressing force, should be considered. 6. (a) What are the design considerations of prestressed concrete poles? (4) The pre stressed concrete pole for power transmission line are generally designed as member with uniform prestress since they are subjected to bending moment of equal magnitude in opposite directions. The poles are generally designed for following critical load conditions, 1. Bending due to wind load on the cable and on the exposed face. 2. Combined bending and torsion due to eccentric snapping of wire. 3. Maximum torsion due to skew snapping of wires. 4. Bending due to failure of all the wires on one side of the pole. 5. Handling and erection stresses. (b) What are the advantages of partially prestressed concrete poles?  Resistance to corrosion in humid and temperature climate and to erosion in      

desert areas.   Freeze thaw resistance in cold region.   Easy handling due to less weight than other poles   Fire resisting, particularly grassing and pushing fire near ground line.   Easily installed in drilled holes in ground with or without concrete fill.   Lighter because of reduced cross section when compared with reinforced concrete poles.   Clean and neat in appearance and requiring negligible maintenance for a number of years, thus ideal suited for urban installation. 

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www.rejinpaul.com SRI VIDYA COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING SUBJ. CODDE AND NAME: CE 1402 – PRESTRESSED CONCRETE STRUCTURES FACULTY NAME: C.D.M.Kaliraj

CLASS &SEC :IV yr CIVIL

Academic year: 2014-’15

Semester :VI

UNIT 4 Part A 1. Sketch the arrangement of Tendons & anchorages in circular prestressing of concrete pipe.(NOV-DEC 2009)

2. Give the advantages of precast prestressed units.(NOV-DEC 2009)  



The C/S is more efficiently utilized when compared with a RC section 



Effective saving in use of materials. 



Improves the ability of material for energy absorption under impact load. 

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www.rejinpaul.com   



The economy of PSC is well established for long span structures. 



There is considerable saving on the quantity of materials used in it. 

3. How do you compute the shrinkage and resultant stresses in composite member? (NOV-DEC 2012) 4. Distinguish between propped and unpropped construction methods. (NOV-DEC 2012)

Here, A = area of the precast web c = distance of edge from CGC of precast web /

c = distance of edge from CGC of composite section e = eccentricity of CGS I = moment of inertia of the precast web /

I = moment of inertia of the composite section. 5. What is circular prestressing?(NOV-DEC 2010) The term refers to prestressing in round members such as tanks and pipes. Liquid retaining structures such as circular pipes,tanks and pressure vessels are admirably suited for circular prestressing. 6. What are the advantages of prestressed concrete sleepers?(NOV-DEC 2010) 

 It is economical.   Full cross-section of member is utilized. 

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www.rejinpaul.com  Increases durability. 



 Reduces corrosion of steel. 



 Increase in shear capacity. 



 Larger spans can be constructed. 



 It allows rapid construction. 

Part B 1. A precast pre-tensioned beam of rectangular section has a breadth of 100 mm and a depth of 200 mm. The beam with an effective span of 5 m is prestressed by tendons with their centroid coinciding with the bottom kern. The initial force in the tendons is 150 kN. The loss of prestress may be assumed to be 157%. The beam is incorporated in a composite T-beam by casting a top flange of breath 400 mm and t = 400 mm. If a composite beam supports a live load of 1 kN/m. Calculate the resultant stresses developed in precast & in-situ cast concrete. (NOV-DEC 2009) Stress in pre tensioned beam: A = 20000mm

2

2

3

2

Z = (100x200 )/6 = 666.67x10 mm

Self weight of pre tensioned beam = 0.1x0.2x24 = 2

0.48kN/m Self weight moment = (0.48x5 )/8 = 1.5kNm 6

3

Stress at top & bottom = ±(1.5x10 /666.67x10 ) = ±2.25N/mm

2

Stress in cast insitu slab: A = 16000mm 2

2 3

2

Z = (400x40 )/6 = 10.6x10 mm

Self weight of pre tensioned beam = 0.4x0.04x24 = 2

3.84kN/m Self weight moment = (0.348x5 )/8 = 1.2kNm 6

3

Stress at top & bottom = ±(1.2x10 /10.6x10 ) = ±1.13N/mm

2

Stress in composite member

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www.rejinpaul.com 3

3

9

4

Ixx = (400x40 )/12 - (300x200 )/12 = 1.9x10 mm 9

6

2

Zt= (1.9x10 )/233.3 = 8.14x10 mm 9

6

2

Zb = (1.9x10 )/366.7 = 5.18x10 mm 2

Live load moment = (1x5 )/8 = 2.08kNm 6

6

Stress at top = ±(2.08x10 /8.14x10 ) = ±0.25N/mm 6

6

2 2

Stress at bottom = ±(2.08x10 /5.18x10 ) = ± 0.45N/mm

2. A composite T-girder of span 5 m is made up of a pre-tensioned rib, 100 mm wide by 200 mm depth, with an in situ cast slab, 400 mm wide and 40 mm thick. The rib is prestressed by a straight cable having an eccentricity of 33.33 mm and carrying initial force of, 150 kN. The loss of prestress is 15%. Check the composite T-beam for the limit state of deflection if its supports an imposed load of 3.2 kN/m for (i) unpropped(ii) propped. Assume modulus of Elasticity of 35 2

kN/mm for both precast & in situ cast elements.(NOV-DEC 2009) Properties

of

precast

beam A = 20000 mm

prestressed

2

g = 0.1x0.2x24 = 0.48 kN/m 3

6

4

I = (100x200 )/12 = 66.66x10 mm Properties of composite section

A = 16000 mm

2

g = 0.04x0.4x24 = 0.384 kN/m yb = 146mm yt = 94mm 6

4

I = 226x10 mm

Deflection of prestressed beam 2

ap = (pel )/(8EI) = -6.69mm(upward) Deflection of composite beam due to self 4

weight ag+q = (5wl )/(384EI) q = 8kN/m g = (0.48 + 0.384) = 0.864kN/m

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www.rejinpaul.com w = (g+q) = 0.864+8 = 8.864kN/m ag+q = 9.12mm Long term deflection aRL = (1+Ø)(ƞ ap + ag+q) = (1+1.6)(0.85(-6.69)+9.12) = 8.99mm Permissible limit according to IS 1343, Span/250 = 5000/250 = 20mm. Hence the actual deflection is within the Permissible limit 3. i) Explain the types of composite construction with neat sketch. (8) (ii) Explain the precast prestressed concrete stresses at serviceability limit state. (8)(NOV-DEC 2012) Limit State of Serviceability : Deflection — The deflection of a structure or part thereof shall not adversely affect the appearance or efficiency of the structure or finishes or partitions. The deflection shall generally be limited to the following: a) The final deflection, due to all loads including the effects of temperature, creep and shrinkage and measured from the as-cast level of the supports of floors, roofs and all other horizontal members, should not normally exceed span/250. b) The deflection including the effects of temperature, creep and shrinkage occurring after erection of partitions and the application of finishes should not normally exceed span/350 or 20 mm whichever is less. c) If finishes are to be applied to prestressed concrete members, the total upward deflection should not exceed span/300, unless uniformity of camber between adjacent units can be ensured. Limit State of Serviceability : Cracking — Cracking of concrete shall not affect the appearance or durability of the structure. The criteria of limit state of cracking for the three types of prestressed concrete members shall be as follows: a) For type 1, no tensile stresses. b) For type 2, tensile stresses are allowed but no visible cracking. c) For type 3, cracking is allowed, but should not affect the

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www.rejinpaul.com appearance or durability of the structure; the acceptable limits of cracking would vary with the type of structure and environment and will vary between wide limits and the prediction of absolute maximum width is not possible. 4. (a) What are shear connectors? Explain the different types of shear connectors in detail. (b) A PSC beam of cross section 150 mm x 300 mm is SS over a 6pan of 8m and is prestressed by means of symmetric parabolic cables @ a distance of 76 mm from the soffit @ mid span and 125 mm @ top @ support section. If the force in the cable ie 350 KN. Calculate deflection @ midspan the beam is supporting its own weight The point load which must be applied at midspan to restore the beam to the level of its support. (8)(NOV-DEC 2010) 5. A composite T-section is made up of a pretension rib 100 mm wide and 200 mm deep and a cast in situ slab 400 mm wide and 40 mm thick baring a modulus of 2

4

elasticity of 28 kN/mm . lf the differential shrinkage is 100x10 units. Determine the shrinkage stresses developed in the precast and cast in situ units. (16)(NOVDEC 2010) Properties of composite section A = 36000 mm Yt

=

2

((400x40x20)+(200x100x140))/36000

=

86.66mm Yb = 153.33mm Ixx

= 6

3

(400x240 )/12 4

3

-

(300x200 )/12

6

6

= 2

260x10 mm Zt= (260x10 )/86.67 = 3x10 mm 6

6

2

Zb = (260x10 )/153.33 = 1.7x10 mm 6

6

2

Zj = (260x10 )/40 = 6.5x10 mm

-6

3

Uniform tensile stress in cast insitu slab = ecs.Ec = 100x10 x28x10 = 2.8 2 N/mm Pre stressing force = ecs.Ec.A = 2.8x16000 = 44.8kN Eccentricity of force =86.67-20 = 66.67mm 3

6

Moment = 44.8x10 x66.67 = 2.99x10 Nmm 2

Direct compressive stress = 44.8/36000 = 1.24 N/mm

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6

Stress at top = ±(2.99x10 /3x10 ) = ±0.9N/mm 6

2

6

Stress at bottom = ±(2.99x10 /1.7x10 ) = ±1.75N/mm 6

6

Stress at junction = ±(2.99x10 / 6.5x10 ) = ±0.46N/mm

2 2

Shrinkage stresses: a. In PSC beam 2

At top = 1.24+0.46 = 1.7 N/mm

2

At bottom = 1.24 – 1.75 = -0.55 N/mm b. In insitu slab At top = 1.24+0.99-2.8 = -0.57 N/mm

2 2

At bottom = 1.24 +0.36 -2.8 = -1.2 N/mm

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