127

Lecture Note – 24

Prediction of long time deflection

ptmtt φφφ +=

( ) imt φφφ += 1

Where, mtφ Change of curvature due to transverse load.

ptφ Change of curvature due to prestress

tφ Total curvature

φ Creep coefficient

iφ Initial curvature immediately after the application of transverse load.

According to ACI committee, ( ) ( )

EIePP

EIePP

EIPie titi

pt 2φ

φ+

−−

+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+−−=⇒ φφ

i

p

i

ppt P

LPL

EIPie

211

( )tip PPL −=

( )

Where, Initial prestress iP

tP Prestress after time t

e Eccentricity of the prestressing force at the section EI Flexural rigidity

Shrinkage and creep

Thus the total deflection after time t obtained from the above expression is

1 1 12

p pf it ip

i i

L La a a

P Pφ φ

⎡ ⎤⎛ ⎞= + − − + −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦ita

ipa

Where, Initial deflection due to transverse load

Initial deflection due to prestress

fa Final deflection after time t

Simplified method

( )φ+⎥⎦

⎤⎢⎣

⎡−= 1

i

tipitf P

Paaa

Here , +ve(positive) sign refers upward deflection. -ve(negative) sign refers downward deflection.

128

Example: A concrete beam having a rectangular section 100 mm wide and 300 mm deep is prestressed by a parabolic cable carrying an initial force of 240 kN. The cable has an eccentricity of 50 mm at the centre of the span is concentric at the supports. If the span of the beam is 10m and live load is 2 kN/m. Estimate the long time deflection after 6 months if E = 38 kN/mm2 & creep coefficient 0.2=φ , loss of prestress = 20% of the initial stress after 6 months. Estimate the long time deflection at the center of the span at the stage assuming that the DL & LL are simultaneously applied after the release of prestress. Solution:- Here, Given

iP = 240 kN.

I = 225 610×e = 50 mm

0.2=φ DL = 0.72kN/m =×× 243.01.0LL = 2 kN/m Loss of prestress = 20% of iP

Short time deflection:

⎟⎟⎠

⎞

⎝

⎛EIeLPi

485 2

⎜⎜ = ( )⎟⎟⎠

⎞⎜⎜⎝

⎛

×××××××

6

2

102253848100010502405 =14.7 mm (↑) Initial deflection due to prestress =

Deflection due to self weight and live loads =( )

⎥⎦

⎤⎢⎣

⎡ +EI

Lqg384

5 4

= ( )( )⎥⎦

⎤⎢⎣

⎡

××××+××6

4

1022538384100010002.000072.05

= 41.5 mm (↓) Therefore, net deflection = 41.5 – 14.7 = 26.8 mm (↓). Long time deflection:

The long time deflection, ( )1 1 12

p pf it ip

i i

L La a a

P Pφ φ

⎡ ⎤⎛ ⎞= + − − + −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

= 41.5 (1+2) - 14.7 0.2 0.21 12

i i

i i

P PP P

2⎡ ⎤⎛ ⎞− + − ×⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

= (124.5 - 38) = 86.5 mm(↓)

129

Using simplified formula

( )φ+⎥⎦

⎤⎢⎣

⎡−= 1

i

tipitf P

Paaa

= (41.5 -14.7 × 0.8)(1 + 2) = 89.1 mm(↓) Deflection of cracked member

Load

Camber due to prestress

Deflection

Short time Unilinear method:

rc IEMLa

2β=

cE

rI

a Maximum deflection L Effective span M Maximum moment

Modulus of elasticity of concrete Second moment of area of equivalent or transformed cracked section

β A constant depending upon the end conditions, positions of the given section and load distribution.

130

Support condition & load β 5/48

1/12

1/24 1/16 1/3 Bilinear method

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

cc

cr

cc

cr

IEMM

IEM

La85.0

2β

crI

Where,

crM Cracking moment

M Moment at which deflection is required

cI Second moment of area of uncracked equivalent concrete section

Second moment of area of cracked equivalent concrete section

β constant

131

Mom

ent

Curvature

Uncracked Section

Cracked SectionMo

M

crcr g

t

fM Iy

=

3 3

1cr cre g cr

M MI I IM M

⎡ ⎤⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

Effective moment of inertia,

Where,

crM Cracking moment

aM Moment at which deflection is required

Ig Moment of inertia of the gross concrete section neglecting the reinforcement

crI Second moment of area of cracked equivalent concrete section

yt Distance from the C.G. to the extreme fibre in tension

crf Modulus of rupture of concrete

Long time deflection of the cracked members As per ACI-318:1989

Multiplying factor, ⎟⎟⎠

⎞⎜⎜⎝

⎛+

= /501 ρξλ

Deflection = λ× immediate deflection Where,

132

⎟⎟⎠

⎞⎜⎜⎝

⎛=

bdAs

//ρ At mid-span for simple and continuous beam and at support for cantilevers

/sA Area of compression reinforcement

b Width of section d Effective depth ξ A time dependent factor having the following values Age ξ Five years or more(≥5) 2.0 12 months 1.4 6 months 1.2 3 months 1.0 Requirement as per IS1343 1. The final deflection, due to all loads including the effects of temperature, creep & shrinkage should not exceed span/250. 2. The deflection including the effects of temperature, creep & shrinkage occurring after the erection of partition and the application of the finishes should not exceed span/350 or 20 mm whichever is less. 3. If finishes are to be applied to the prestressed concrete members, the total upward deflection should not exceed span/300, unless uniformity of camber between adjacent units can be ensured.

133

Lecture Note – 25 Example:- The floor slab shown in the figure below comprises precast pretensioned beams & an in-situ concrete slab. If the span of the beam is 5 m & the imposed load is 5 kN/m2 (including finish), determine the stress distribution at the various load stages. Assume all long-term losses have occurred before the beams are erected & that the net force in each end is 19.4 kN.

600

75

250

40

Section properties of the beams:

251013.1 mmAc ×= 87.5 10cI = × mm4 ;

66 10t bZ Z= = × mm3

Solution:- Given, Eccentricity of the wires, e = 125 - 40 = 85 mm Self weight of the beams = 24113.0 × = 2.7 kN/m. Moment due to self weight, ( ) 8/57.2 2×=oM = 8.4 kN/m.

The total prestress force after all losses have occurred is given by 4.196×=oPβ =116.4

kN. Stress distribution in the beams is thus given by

6

6

6

3

5

3

106104.8

10610854.116

1013.1104.116

××

+×

××−

××

=tσ = 1.03 - 1.65 + 1.40 = 0.78 N/mm2

3 3

5 6

116.4 10 116.4 85 10 8.4 101.13 10 6 10 6 10bσ

× × ×= + −

× × ×

6

6

× =1.03 + 1.65 - 1.40 = 1.28 N/mm2

134

i) The weight of the slab is supported by the beams acting alone, so 8/5246.0075.04.8 2×××+=dM =11.8 kNm

The stress distribution within the beams is now given by

6

6

106108.1165.103.1

××

+−=tσ = -0.62 + 1.97 = 1.35 N/mm2

97.165.103.1 −+=bσ = 0.71 N/mm2

iii) The imposed load of 5 kN/m2 is supported by the composite section & the section properties of this are now required. To find the neutral axis of the composite section, taking moments about the soffit of the beams gives:

( ) ( )288600751251013.1600751013.1 55 ××+××=×+× y

mmy 171=∴

( ) ( ) ( ) ( )23258 171288/6007512/600751251711013.1105.7 −×+×+−×+×=compI

491063.1 mm×=

The imposed load bending moment, ( ) 8/556.0 2××=− ddes MM = 9.4 kNm

The stress distribution within the composite section under this extra bending moment is given by

( )1713251063.1104.9

9

6

, −×××

=slabtσ = 0.89 N/mm2

( )1712501063.1104.9

9

6

, −×××

=beamtσ = 0.46 N/mm2

6

, 9

9.4 10 1711.63 10b beamσ − ×

= ××

= -0.99 N/mm2

The total stress distributions under the three load cases are shown in the fig. below

135

1.28

0.78

0.71

1.35

-0.28

1.81

0.89

0.46

(a) (b)

(c)

Fig. Stress distribution for composite section

[(a) for beam, (b) for beam & slab (c) for beam , slab & imposed load ]

The maximum compressive stress occurs at the upper fibers of the beams, but is significantly lower than the level of stress had the beam carried the total imposed load alone. This explains the advantage of inverted T-section in composite construction, where only a small compression flange is required for bending moments and , the

compression flange for bending moment being provided by the slab. oM dM

desM

136

Lecture Note – 26Lecture Note – 26

Example: Find the profile of prestressing cable with an effective force of 500kN, which can balance the DL & LL on the beam. Sketch the profile of the cable along the length of the beam.

6m

250

150

50

50

50

550

4kN/m

Solution: Cross sectional area of the beam = 150×50 + 250×50+450×50 = 850×50 = 42500 mm2

y1

1250 50 25 150 50 525 450 50 275

42500y × × + × × + × ×

∴ =

1 245.59y∴ = mm

2 304.41y∴ = mm y2

38BR wl=

238 2X

wxM wlx= −

0XM =

If , 3 2 38 4

x l l×= =

Now, Dead load = A×25 kN/m = 1.02 kN/m Live load = 4 kN/m Therefore, total load, w = (4+1.02) kN/m = 5.02 kN/m

2

22.598s

wlM kNm= − =

137

Location 22.59 0.04518 45.18500

ss

Me m mmP

= = = =

Maximum moment

For maximum moment, 0dMdx

=

23 0

8 2d wxwlxdx

⎛ ⎞⇒ −⎜ ⎟

⎝ ⎠=

38

x l∴ =

So maximum moment, Mmax = 23 3 98 8 2 64

wwl L l× × × − ×

= 29 1164 2

wl ⎛ ⎞−⎜ ⎟⎝ ⎠

= 29128

wl

= 202 6128

9 5.× ×

=12.71 kNm

Hence, e = 12.71500

= 0.02541 m = 25.41 mm

25.41

245.6

304.4

2.25m1.5m

45.18

Fig. Cable profile

138

Lecture Note –27

Flexural strength of prestressed concrete member

Types of flexural failure 1. Fracture of steel:- A minimum longitudinal reinforcement of 0.2% of the total concrete area shall be provided in all the cases except in the case of prestressed units of small sections. This reinforcement may be further reduced to 0.15% in the case of HYSD bars. The percentage of steel provided, both tensioned & un-tensioned taken together should be sufficient so that when the concrete in precompressed tensile zone cracks, the steel is in position to take up the additional tensile stress, transferred on to it by the cracking of the adjacent fiber of concrete & a sudden failure is avoided. 2. Failure of over reinforced section:- When the effective reinforcement index , which is expressed in terms of the percentage of reinforcement, the compressive strength of the concrete and the tensile strength of the steel, exceeds a certain range of values, the section is said to be over reinforced. Generally, over-reinforced members fall by the sudden crushing of concrete, the failure being characterized by small deflection and narrow cracks. The area of steel being comparatively large, the stresses developed in steel at failure of the member may not reach the tensile strength & in many cases it may well be within the proof stress of the tendon. 3. Failure of under reinforced section:- If the cross-section is provided with a steel greater than the minimum prescribed above, the failure is characterized by an excessive elongation of steel followed by crushing of concrete. This type of behaviors is generally desirable since there is considerable warning before the impending failure. As such, it is common practice to design the under-reinforced sections, which become more important in case of statically indeterminate structure. Strain compatibility method Assumption

1. The stress distribution in the compression zone of concrete can be defined by means of coefficients applied to the characteristic compressive strength & the average compressive stress & the position of the centre of compression can be assessed.

2. The distribution of concrete strain is linear. (i.e. the plane section normal to axis remains plane after bending)

139

3. The resistance of concrete in tension is neglected. 4. The maximum compressive strain in concrete at failure reaches at a particular

level.

b k 1f ck

C u

k 2x

A psT u

d

x

εcu

Fig. Stress-St

Total compressive force, 1u ckC k f bx=

u ps pbf=Total tensile force, T A

The ultimate flexural strength of the c

Where,

ckf Characteristic st

b Effective width x Neutral axis depth

Area of prestres psA

Characteristic tepbf

1k A constant whose

2k A constant whose

The typical stress strain characteristicconcrete as recommended by IS1343

( )su seε ε−

rain Distribution at Failure

oncrete, ( )1 2u ckM k f bx d K x= −

rength of the prestressing concrete

sing tendons

nsile strength of the prestressing tendon

value varies from 0.5-0.7 for ckf = 60-20 Mpa

value varies from 0.42-0.47 for ckf = 60-20 Mpa

of different types of tendons used in prestressed is given in the figure below.

140

Fig. Characteristic of Hognestad et al’s stress block Steps

1. Compute the effective strain seε in steel due to prestress after allowing for all

losses from the stress-strain curve of steel.

141

2. Assume a trial value for neutral axis depth x & evaluate ( )su seε ε−

3. Using the stress-strain curve for steel, determine the value of stress in steel at failure corresponding to pbf suε

4. Compute total compression C & tension T. 5. If C = T the assumed value of x is OK. Otherwise if tension is less than the

compression decrease the value of x & vice versa to repeat the process 2-4 again. ( )2ps pbM A f d k x= −6. Evaluate moment of resistance (ultimate moment),

Example A pretensioned concrete beam with a rectangular section, 100 mm wide by 160 mm deep, is prestressed by 10 high–tensile wire 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 the concrete is 40 N/mm2. Given the load-strain curve of 2.5 mm diameter steel wire (fig below), estimate the ultimate flexural strength of the section using strain compatibility method.

142

Solution:- From the above figure (Hognestad stress block) we can have for fck = 40 N/mm2,

0.0033cuε = ,;k1 = 0.57 ; k2 = 0.45.

As per figure strain due to load of 6.8 kN in wire is 0.0073 So effective strain in steel after all losses is given by

( )0.0073 0.0012seε = − =0.0061

First trial Assuming x = 60 mm From the strain diagram ( )su seε ε− =0.0033

Therefore, =0.0094 ( )0.0033 0.0061suε = +

Corresponding force in the wire =8.4 kN Total tensile force = 10 × 8.4 = 84 kN.

( )1 ckk f b x⋅ ⋅ ⋅ Total compressive force =

= ( )0.57 40 100 601000

× × ×=136.8 kN

Since tension is less than compression, x is decreased for second trial Second trial Assuming x = 43 mm From the strain diagram ( )su seε ε−

( )0.0059 0.0061suε = +

=0.0059

= 0.012

Corresponding force in the wire =9.9kN Total tensile force = ( )10 9.9× =99 kN

Total compressive force = ( )0.57 40 100 431000

× × ×=98 kN

Since tension is nearly equal to compression, strain compatibility is established. ( )2u ps pbM A f d k x= −

= 99 × 103 (120 - 0.45 × 43) =9.96×106 Nmm =9.96 kNm

143

Lecture Note –28

Design of Prestressed Concrete Member

Indian code (as per IS 1343 1980) Assumption

1. The plane sections normal to the axis remain plane after bending. 2. The maximum strain in concrete at outermost compression fibre is taken as 0.35%

in bending regardless of the strength of concrete. 3. The relationship between stress & strain distribution in concrete is assumed to be

parabolic. The maximum compressive stress is equal to 0.446 ckf

Where, fck Characteristic strength of the concrete 4. The tensile strength of concrete is ignored. 5. The steel & concrete are bonded completely. 6. The stresses in bonded prestressing tendons are derived from the respective stress-

strain curve for the particular steel.

If uM Ultimate moment of resistance of the section

puf Tensile strength developed in tendons at the failure stage of the beam

pf Characteristic tensile strength of the prestressing steel

pef Effective prestress in tendon after losses

144

pA Area of prestressing tendons

d Effective depth b Effective width

ux Neutral axis depth

Then moment of resistance, ( )0.42u pu p uM f A d x= −

The value of depends upon the effective reinforcement ratio puf

p p

ck

A fbdf

⎛ ⎞⎜ ⎟⎝ ⎠

For pre-tensioned & post-tensioned members with an effective bond between concrete & tendon, the value of & puf ux are given in table 11 of IS1343. The effective prestress

after all losses should not be less than 0.45 . For post-tensioned rectangular beams with

un-bonded tendons, the value of &

pef

pf

puf ux are influenced by the effective span to depth

ratios, and their values for different span/depth ratios are given in Table-12 of IS1343.

145

Moment of resistance of flanged section 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 web & flange portion & considering the stress block is shown below.

146

d

Aps

b

bw

0.42xu

xu

0.45fck

fpuApw

Df

fpApf

0.45fck

0.45xu(b-bw)Df

(d-0.5Df)

Fig. Moment of Resistance of flanged sections (xu>0.5Df)

If Apw Area of prestressing steel for web Apf Area of prestressing steel for flange Df Thickness of flange

( )p pw pfA A A= +

( )

Then,

0.45 fpf ck w

p

DA f b b

f⎛ ⎞

= − ⎜ ⎟⎜ ⎟⎝ ⎠

But,

After evaluating Apf , the value of Apw is obtained as

( )pw p pfA A A= −

For effective reinforcement ratio ofpw p

w ck

A fb df

⎛ ⎞⎜⎝ ⎠

⎟

( / 0.87pu p

the corresponding values of

) uxd⎞⎟

⎝ ⎠& f f ⎛

⎜ are obtained from Table – 11 of IS 1343 1980. The moment of

resistance of the flange section is obtained from the expression

( ) ( ) ( )0.42 0.45 0.5u pu pw u ck w f fM f A d x f b b D d D= − + − −

147

Example A pretensioned prestressed concrete beam having a rectangular section 150 mm wide 350 mm deep has an effective cover of 50 mm If fck = 40 N/mm2 & fp = 1600 N/mm2 & the area of prestressing steel Ap = 461 mm2, calculate the ultimate flexural strength of the section using IS code method. Solution:- Given data, Characteristic strength of concrete, fck=40 N/mm2

Characteristic strength of tendon, fp = 1600 N/mm2

Area of tendon, Ap =461 mm2; Width, b = 150 mm Effective depth, d = 300 mm The effective reinforcement ratio is given by

1600 461 0.4040 150 300

p p

ck

f Af bd

⎛ ⎞ ×⎛ ⎞= =⎜ ⎟ ⎜ ⎟× ×⎝ ⎠⎝ ⎠

From Table- 11, the corresponding values of the ratios are

0.87pu

p

ff

⎛ ⎞=⎜⎜

⎝ ⎠⎟⎟ 0.9 and ux

d⎞⎛ =⎜ 0.783 ⎟

⎝ ⎠

( )0.87 0.9 1600puf∴ = × ×

)=1253 N/mm2

(0.783 300ux∴ = × = 234.9 mm

Hence, the ultimate flexural strength of the section is ( )0.42pu p uM f A d x= −

= ( )1253 461 300 0.42 234.9× − ×

= Nmm = 116 kNm 6116 10×

Example A pretensioned T-section has a flange, which is 300 mm wide 200 mm 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 & fp = 1600 N/mm2, estimate the ultimate moment capacity of the T-section using IS code method. Solution:- Given data, Characteristic strength of concrete, fck = 50 N/mm2

Characteristic strength of tendon, fp = 1600 N/mm2

148

Area of tendon, Ap =200 mm2; Width, b = 300 mm; Depth, d = 500 mm Assuming that the neutral axis falls within the flange, the value of b = 300 mm for computations of effective reinforcement ratio.

1600 200 0.0450 300 500

p p

ck

f Af bd

⎛ ⎞ ×⎛ ⎞∴ = =⎜ ⎟ ⎜ ⎟× ×⎝ ⎠⎝ ⎠

From Table- 11, the corresponding values of the ratios are

1.00.87

pu

p

ff

⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠ and 0.09ux

d⎛ ⎞ =⎜ ⎟⎝ ⎠

(0.87 1600puf )∴ = × =1392 N/mm2

(0.09 500ux )∴ = × = 45 mm

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

( )0.42pu p uM f A d x= −

( )200 500 0.42 45× − ×610×

1392 =

= 134 Nmm = 134 kNm

149

Lecture Note –29

Shear Resistance of Prestressed Concrete Failure due web-shear cracks The ultimate shear resistance of prestressed concrete section with web-shear cracking but without flexural cracks is mainly governed by the limiting value of the principal tensile stress developed in concrete. The failure is assumed to take place when the principal tension exceeds the tensile strength of the concrete. If Ultimate shear resistance of concrete in a section due to web-shear cracks. cwV

Breadth of web of the member wb

D Overall depth of the member

cpf Compressive prestress at the centroid of a section

tf Tensile strength of concrete

S Statical moment (i.e. first moment of area) I Second moment of area

vτ Maximum shear stress at failure

For the condition at failure, equating the minor principal stress with the tensile strength of concrete, we get

2 21 42 2cp

cp v t

ff fτ

⎡ ⎤⎛ ⎞− + = −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

cwv

V SIb

τ = Here,

Hence, 2

21 42 2cp cw

cp t

f V Sf fIb

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥− + = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

On simplification, 2cw w cp cp t

IV b f f fS

⎛ ⎞= +⎜ ⎟⎝ ⎠

The values of (I/S) varies from 0.67D (for rectangular section) to 0.85D (for flanged section) The British code (BS:8110:1985) & the Indian standard code specify a modified version of the above relation given by

150

20.67 0.8cw w cp cp tV b D f f= + f

In the above expression for computing , the tensile strength of concrete may be

assumed as cwV

0.24t ckf f=

Where, ckf Characteristic strength of concrete

Design of shear reinforcement (AS per IS 1343) At any given section, the ultimate shear resistance Vc of the concrete alone should be taken as the lesser value of & . When V, the shear force due to the ultimate loads,

is lesser than VcwV cfV

c the shear force which can be carried by the concrete, minimum shear reinforcement should be provided in the form of stirrups such that

0.870.4

svv

A fS

b= y [Ref. clause 22.4.3.1 of IS1343:1980]

Where,

vS Spacing of stirrups along the length of a member

svA Total cross sectional area of stirrup legs effective in shear

b Breadth of the member for T, I & L beams should be taken as the breadth of the rib bw

yf Characteristic strength of stirrups reinforcement, which shall not be

taken greater than 415 N/mm2

If the shear force V is less than 0.5Vc & in a member of minor importance shear reinforcement need not to be provided. When V exceeds Vc, [as per clause 22.4.3.2 of IS1343:1980] shear reinforcement should be provide conforming to the relation

( )0.87sv y

vc

tA f dS

V V=

−

Where, dt Depth from the extreme compression fiber either to the longitudinal bars or to the centroid of tendons, which ever is greater.

The spacing should exceed neither 0.75dt nor 4 times the web thickness for flanged members. When V exceeds 1.8Vc, the maximum spacing should be reduced to 0.5dt. The lateral spacing of the individual legs of the stirrups provided at a cross section should not exceed 0.75dt. The maximum shear stress permissible for different grades of concrete are

151

shown in Table:-1 . If the nominal shear stress (V/bd) exceeds these vales the section has to be redesigned. Table:1 (Ref. Table 7 of IS1343:1980)Maximum shear stress (N/mm2) Concrete Grade M30 M35 M40 M45 M50 M55& above Max. Shear stress 3.5 3.7 4.0 4.3 4.6 4.8 The value of minimum shear reinforcement as per clause 22.5.4.4 of IS1343:1980 is given by

0.870.4

sv ysv

A fb =

Where, bsv Centre to centre distance between corner bars in the direction of the width Example The support section pf a prestressed concrete beam, 100 mm wide & 250 mm deep is required to support an ultimate shear force of 60 kN. The compressive prestress at the centroidal axis is 40 N/mm2. The cover to the tension reinforcement is 50 mm. If the characteristic tensile strength of steel in stirrups is 250 N/mm2, design suitable reinforcements at the section using IS recommendation. Solution:- Given: Breadth of web, =100 mm; Depth, D = 250 mm; d = 200 mm wb

Compressive prestress at the centroid of a section, cpf = 5 N/mm2

Characteristic compressive strength of concrete ckf = 40 N/mm2

Characteristic tensile strength of steel, yf = 250 N/mm2

Ultimate shear force, V = 60 kN For the support section uncracked in flexure

20.67 0.8cw w t cp tD f f f= +V b

0.24 0.24 40t ckf f= = =1.517 N/mm2

So 20.67 100 250 1.517 0.8 5 1.517cwV = × × + × × = 48407 N = 48.407 kN

∴Balance shear = (V-Vc ) = (60-48.4) kN = 11.6 kN

152

Using 6 mm diameter two-legged stirrups, the spacing is obtained as

( )0.87sv y

vc

A fS

V V⎡ ⎤

= ⎢ ⎥−⎣ ⎦ = 2 28.2 0.87 250 200

11600× × × ×⎡ ⎤

⎢ ⎥⎣ ⎦=211.5 mm

Maximum permissible spacing =0.75d = 0.75×200 =150 mm ∴ Adopting 6 mm diameter two-legged stirrups at 150 mm centre to center.

153

Lecture Note – 30

Statically Indeterminate Prestressing Structure

Advantage of continuous member 1. The BMs are more evenly distributed between the centre of span & the supports

of members. 2. Reduction in the size of members results in lighter structure. 3. The ultimate load-carrying capacity is higher than in a statically determinate

structure due to the phenomenon of redistribution of moments. 4. Continuity of beams in framed structures leads to increased stability. 5. Continuous girder is formed by segmental construction using pre-cast units

connected by prestressing cables. 6. In continuous post-tensioned girders, the curved cables can be suitably positioned

to resist the span & support moments. 7. A reduction in the number of anchorage in a continuous pre-stressed beam in

comparison with a series of simply supported beams. 8. Only one pair of post-tensioning anchorages & a single stressing operation can

serve several members. 9. The deflections are comparatively small as compared to simply supported spans.

Disadvantage of continuous structure 1. Loss of prestress due to friction is appreciable in the long cables with reversed

curves & considerable curvature. 2. Secondary stresses due to prestressing, creep, shrinkage and temperature &

settlements of supports may induce very high stresses unless controlled. 3. Cables positioned to cater for secondary moments are not generally suitable to

provide the required ultimate moment under a given system of loads. 4. The computation of collapse or ultimate load is influenced by the degree of

redistribution of moments in the continuous structure. Effect of prestressing indeterminate structure When an indeterminate structure is prestressed, redundant reactions will develop due to the redundancies exercising a restraint at the supports. While a statically determinant structure is free to deform when prestressed, a continuous structure cannot deform freely. However, the deflections should conform to the law of consistent deformation. The redundant reactions, which develop as a consequence of prestressing an indeterminate

154

structure, result in secondary moments. The formation of redundant reactions & secondary moments are shown below:-

P e P

RL/2 Secendary moment

A

A

B

B

B

C

C

C

A

L L

R

Figure:- Redundant reactions secondary moments in a continuous PSC beam Under the action of the prestressing force, P, the beam will deflect if it is not restrained at the central support B. A redundant reaction R develops at the central support if the beam is restrained at B so that deflections are not possible at this support. As a consequence of this redundant reaction acting downwards, secondary moments develop in the continuous beam ABC.

Codal provision for moment redistribution

As per clause 21.1.1 of IS1343:1980

(a) Equilibrium between the internal forces & the external loads is maintained for appropriate load combination.

(b) The ultimate moment resistance provided at any section of a member is not less than 80% of the moment at that section obtained from an elastic maximum moment diagram covering all appropriate combination of loads.

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(c) The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 20% of the numerically largest moment covering all appropriate combination of load.

(d) At section where the moment capacity after redistribution is less than that from the elastic maximum moment diagram, the following relation should be satisfied:-

5.0100

≤⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ Mu

dx δ

Where,

ux Depth of the neutral axis

d Effective depth

Mδ Percentage reduction in moment (e) In multistory framed structures involving lateral stability, the reduction in moment allowed is restricted to 20% for structures up to four storeys & 10% for structures over four storeys. The British code of practice BS:8110-1985 permits the redistribution of peak moments up to a maximum of 30%. The code formally defines the moment distribution ratio bβ as

⎟⎟⎠

⎞⎜⎜⎝

⎛

tionredistribubeforetionaMomentattionredistribuaftertionaatMoment

secsec

1≤ =bβ

The neutral axis depth x should be checked to ensure the condition that

( )4.0−≤⎟⎠⎞

⎜⎝⎛

bu

dx β or 0.6 which ever is less.

Guyon’s theorem “In a continuous prestressing beam, if the tendon profile is displaced vertically at any intermediate supports by any amount, but without altering its intrinsic shape between the supports, the resultant line of thrust is unchanged.” Guyon’s theorem can be proved by considering a continuous member prestressed by a straight cable.

156

e

L

A B C

Primary moment Pe

Secondary moment MB

Resultant momentPe

MB-Pe

P Pe

MB/P-e

P P

If P Prestressing force The resultant moment at the central support = Pe/2. The shift of pressure line above centroidal axis = e/2 If the cable is linearly transformed by displacing it vertically at the central support so that the eccentricity is zero at this section, the resultant moment has the same value Pe/2 & the shift of pressure line is also be same. However, it is important to note that as a result of linear transformation of the tendons, additional reactions are induced while there is no change in the resultant moment distribution.

157

Example:- A prestressed concrete continuous beam of two equal spans AB = BC = 10 m is prestressed by a continuous cable having a parabolic profile between supports. The eccentricity of the cable is zero at all three supports & 100 mm towards the soffit at the centre of the span. The beam is of rectangular section, 100 mm wide & 300 mm deep. The effective force in the cable is 100 kN. Determine the resultant thrust line in the beam. Show there is no change in the thrust line if the cable is linearly transformed with a vertical lift of 100 mm towards the top of beam at interior support B. Solution:-

A B C

P P

L=10m L=10m

e=100mm

A B C

e=100mm 100mm

Equivalent load = 8Pe/L2

⎟⎠⎞

⎜⎝⎛

×××

=⎟⎠⎞

⎜⎝⎛=

10101.010088

2LPewe =0.8 kN/m

Resultant moment at B = 0.125 =10 kNm 22 108.0125.0 ××=Lwe

Shift of thrust line at B = ⎟⎠⎞

⎜⎝⎛ ×

=100

100010PM =100 mm from centroidal axis.

If the cable is linearly transformed with a vertical shift of 100 mm towards the top fibers at support B, the equivalent load is unchanged since the eccentricity e measured from the line joining the cable centers at A & B to the bottom of the cable at the mid-span still remains unchanged at 100 mm. Hence, there is no change in the resultant moment & position of trust line at B.

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Concordant cable profile Tendon profiles which do not induce secondary moments are referred to as concordant profile. Prestressing a statically indeterminate structure generally results in secondary moments due to the redundant reactions developed at the immediate supports. However, it is possible to arrange the cable profile in such a way that the structure does not deform at the supports or at other points of restrain. In such a case, redundant reactions & secondary moments are not induced by prestressing the cables.

x 1

A B CL A B

IA B

L B C

IB C

P

x 2

M =P e

P e

C GC G

Figure: Analysis of Indterminate beam

Depending upon the degree of inderminancy of the structure, concordant tendon profiles satisfy a set of geometrical conditions, derived from Mohr’s moment area theorems. Thus for a two-span continuous beam ABC with a degree of indetminancy one, the condition of concordant profile is given by,

( ) ( ) 02

22

1

11 =⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡EI

xAEI

xA

Where,

A1, The free BMD representing the area between the cable profile & the longitudinal axis of the beam in the adjacent span. x1 The distance of the centroid of the moment diagram from the exterior support. (EI)1 The flexural rigidity of the span AB A2, x2, (EI)2 the corresponding values for the span BC

159

Example:- A continuous concrete beam ABC (AB = BC) has a uniform cross-section throughout its length. The beam is prestressed by a straight cable carrying an effective force P. The cable has an eccentricity e towards the soffit at end supports A & C & e/2 towards the top fibre at the central support B. Show that the cable is concordant.

A B CL L

e

e /2

2 L /3

L /3

C a b le p r o f i le

Solution:- Refereeing to the figure, the geometrical condition of concordancy is expressed as

02211 =+ xAxA

( )∫=AB

exdxPxA0

11

Since EI is constant for spans AB & BC

1 2 1 2 1 82 3 3 3 2 3 2 9

L L L e LP e⎡ ⎤⎛ ⎞ ⎛ ⎞= − × × × × + × × ×⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 2 0A x =

= P.0 = 0

Similarly

Hence the given cable profile is concordant.