1

CIVE3139 Prestressed Concrete (after MBH) PART A THE BASICS 1. Axial prestress For a member subject to moments varying between Mmin and Mmax the net stresses are: Outer fibre stresses subject to Mmax

at the top (11.1)

at the bottom (11.2)

Outer fibre stresses subject to Mmin

at the top (11.3)

at the bottom (11.4)

For FULL prestressing (i.e. no tension) The critical case is fb = 0 in equation 11.2:

Thus the minimum required prestressing force:

Hence:

Using equation 11.1 the corresponding maximum stress in the top can be calculated:

Hence:

⁄

(

)

2

2. Eccentric prestress The stress distributions are similar to axial but with the addition of a ⁄ term:

Outer fibre stresses subject to Mmax

at the top (11.5)

at the bottom (11.6)

Outer fibre stresses subject to Mmin

at the top (11.7)

at the bottom (11.8)

As the prestressing force P lies below the NA its effect is to cause hogging moments. For FULL prestressing (i.e. no tension) The critical case is fb = 0 in equation 11.6 so:

Thus the minimum required prestressing force:

( )

Hence:

Compared with an axial prestress, the moment capacity is increased by Pe.

Using equation 11.5 the corresponding maximum compressive stress can be calculated:

Hence:

(

)

The same maximum stress as axial prestress: Eccentric prestress maximises the moment capacity.

Stress blocks show: Eccentric prestress increases the range of stresses, and thus the range of moments (Mmin to

Mmax), that can be carried.

Minimum moment that can be resisted (Mmin) generally governed by the need to avoid tension at the top (equation 11.7): Condition important for prestressed beams, particularly with straight tendons.

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DESIGN FOR SLS

1. The Governing Equations Single span simply supported beams are normally governed by: Minimum moment at transfer Maximum moment in service. The governing equations are:

At transfer:

(11.9)

(11.10)

In service:

(11.11)

(11.12)

2. Minimum Section Properties Combining governing equations 11.9 and 11.11:

(11.11)

(11.9)

Factor then subtract:

(11.11)

(11.9)

( ) ( )

Where: and

Thus: ( ) (

) (11.13) Similarly, combining equations 11.10 and 11.12:

(11.12)

(11.10)

( ) ( ) (11.14) equation (11.14) is incorrect in MBH

Define the moment variation, ( )

( )

(11.15)

( )

(11.16)

4

When using equations 11.15 and 11.16 for preliminary sizing of the cross-section, it can be assumed that:

Mmax = imposed and dead (self weight) load moment

Mmin = dead (self weight) load moment Effectively Mv is independent of the self weight. Equations 11.15 and 11.16 give minimum values of section moduli, a margin above this should be allowed. As the maximum moment on the section is not directly included, it is possible that the resulting prestressing force is uneconomic or impractical (not usually the case). Span to depth ratios guidelines are less easily set than for RC but:

Span 36m h = span/25 + 0.1m

Span > 36m h = span/20 m

Generally results in conservative designs for post-tensioned members. Much greater span-depth ratios are possible for short spans but the required prestressing force may become very high. Should also consider the slenderness ratio of beams (as RC) and the possibility of web and flange splitting in flanged members.

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3. Design of Prestress force The governing equations 11.9 to 11.12 can also be rearranged to give limits on the minimum required prestress for a given eccentricity:

( )

( ⁄ ) (11.17)

(

)

( ⁄ ) (11.18)

( )

( ⁄ ) (11.19)

(

)

( ⁄ ) (11.20)

If the ( ⁄ ) term is negative the sense of the inequality in equations 11.17 and 11.18 is reversed. If the prestress lies within this defined range, the critical allowable stress conditions will be satisfied. For a simply supported beam the design prestress force will be the minimum value that satisfies these inequalities at the critical section for bending. The above make no allowance for ensuring that the eccentricity lies suitably within the beam. The minimum cover to the tendon will depend on the exposure and the structural class assumed in the design. This requirement will be most severe with the maximum moment, corresponding to equations 11.11 and 11.12. If emax is the maximum eccentricity due to cover requirements, equation 11.11 gives:

(

) (11.21)

And equation 11.12 gives:

(

) (11.22)

These two inequalities define linear relationships between Mmax and P0. For a beam subject to a sagging moment, emax will generally be positive thus equation 11.22 represents a lower

limit on P0. Since (

) is generally negative, equation

11.21 also represents a lower limit on P0. As the figure illustrates, providing a prestress force in excess of gives only limited increase in Mmax . At the intersection:

(

) (

)

Thus:

(

)

(11.23)

This value for P0 can be regarded as the maximum economic prestress beyond which consideration should be given to increasing the section size.

Having selected a value for the pre-stressing force based on the (rare) characteristic loads, the compressive stress at the top of the section should then be checked for the lower allowable service stress under quasi-permanent loads as this might prove to be the critical case.

6

4. Magnel Diagrams Magnel diagrams are a powerful tool for representing graphically the inequalities (such as equations 11.17 to 11.20) governing prestress force and eccentricity. The governing equations 11.9 to 11.12 can be re-arranged to express linear relationships between e and ⁄ :

( ⁄ ⁄ )

( ⁄ ) (11.24) from 11.11 for service @ top

( ⁄ ⁄ )

( ⁄ )

(11.25) from 11.9 for transfer @ top

( ⁄ ⁄ )

( ⁄ ) (11.26) from 11.12 for service @ bottom

( ⁄ ⁄ )

( ⁄ )

(11.27) from 11.10 for transfer @ bottom

Note that the sense of inequality 11.25 has been reversed as a tensile

makes the denominator negative. Two examples of these inequalities plotted as a Magnel Diagram follow:

In both cases the area bounded by the four inequalities will provide a satisfactory design in terms of prestress force and eccentricity.

Two examples are illustrated because inequality can have either a positive or a negative slope depending upon whether fmax is greater or less than ⁄

The additional line represents an additional physical constraint on the maximum eccentricity arising from the overall depth of the section, the cover to the tendons, possible provision of shear links etc.

A Magnel diagram provides a very useful design tool as it covers all possible solutions to the inequalities for a particular problem, thus allowing a range of prestressing force and eccentricity values to be investigated. The following can readily be identified: The minimum prestressing force The maximum economic prestressing force The maximum prestressing force The range of valid eccentricities for any prestressing force

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5. Design of Tendon Profiles Having now selected a value of prestress force that will satisfy the critical section, it is necessary to determine the eccentricity (e) throughout the length of the member. As the eccentricity is the only unknown in the governing equations 11.9 to 11.12, these will yield two upper and two lower limits on the acceptable value. These requirements must be satisfied at each point along the member, accommodating possible variations in moment, prestressing force and section properties. Rewriting the governing equations … At Transfer:

[

]

(11.28)

[

]

(11.29)

In Service:

[

]

(11.30)

[

]

(11.31)

Where the values of Mmax and Mmin relate to that particular section. For members with a constant cross section, providing the variation in prestressing force along the length can be ignored, the terms within the brackets are constant and the zone within which the centroid can lie is governed by the shape of the BM envelope:

For a uniform distributed loading the BM envelopes are parabolic, which is why a parabolic tendon profile is generally used when a straight profile will not fit within the zone. At the critical section the cable zone is generally narrow (zero if the minimum prestressing force has been used), elsewhere it gets increasingly wider.

6. Width of Cable Zone The Magnel diagram shows that for a particular prestress force, the centroid of the prestressing force must lie within a corresponding eccentricity range. At this this eccentricity the concrete section must also provide a sufficient width of cable zone to be able to practically accommodate the tendons (allowing for cover, minimum spacing and available tendon sizes).

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7. Prestress Losses • Elastic shortening of concrete • Creep of concrete under sustained compression • Relaxation of steel under sustained tension • Shrinkage of concrete Post-tensioned only: • friction in ducts • losses at anchorages

Elastic shortening is generally significantly less for post-tensioning because this is a sequential process. Creep and shrinkage are both largely dependant on the properties of the concrete at transfer (pre-tensioned concrete is likely to be less mature). Total losses are typically 20 to 30% with post-tensioned concrete incurring additional system and profile dependant losses. Elastic shortening

When the concrete shortens elastically at transfer, the strain in the steel is similarly reduced (and thus so is the steel prestress).

The corresponding stress in the concrete at the tendon level:

( )

Where is the stress due to self weight. This term is relatively small when averaged over the

length of the member, thus:

(

)

Concrete strain (and thus the reduction in steel strain) = ⁄

Reduction in steel stress = (

) where the modular ratio.

If Ap is the area of the tendons, the loss in restressing or e

(

)

Thus:

(

)

Or:

(

)

In pre-tensioned construction this full loss applies. When post-tensioning, it only applies to the previously tensioned cables – normally the loss can be assumed as 50%:

(

)

The appropriate value of is used in subsequent loss calculations.

The modular ratio should be calculated using the table for based on the transfer strength of the concrete.

time dependant losses

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Creep of concrete

Concrete is also shortened - over a longer timescale - by creep under sustained compressive stress, again with a corresponding reduction in compressive stress. Stress in the concrete at steel level:

(

)

Loss of steel stress specific creep strain

Loss of prestress force

(

) specific creep strain

Where:

Specific creep strain ( )

N/mm2

Where the concrete stress does not exceed 0.45fck at transfer, the value of the final creep coefficient φ(,t0) can be found from the table:

Final creep coefficient for normal weight concrete (class C25/30)

Relaxation of steel

Relaxation losses are typically 4-10% of the tendon load after transfer (if stressed to ~70% of characteristic strength). The value depends upon the characteristics of the prestressing steel and whether pre or post tensioned. Section 3.3.2(7) of EC2 gives guidance on applying manufacturer’s 1000hour relaxation data.

Shrinkage of concrete

Loss in steel stress

Loss in prestress force

Where εcs is the shrinkage/unit length of concrete, this is influenced by curing conditions and the transfer maturity. Typically shrinkage strains range from 230 x 10-6 to 550 x 10-6:

Final shrinkage strains for normal weight concrete [10-6] (Grade C25/30)

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Friction in ducts (post-tensioned)

Friction reduces the effective prestress force at locations remote from the jacking point.

(a) Friction losses due to unintended profile variations (wobble):

Where:

P0 = jack force

Px = cable force at distance x

e = base on napierian logs (2.718)

μ = coefficient of friction (typically 0.17 for cold drawn wire and 0.19 for strand)

k = unintentional angular displacement per unit length (typically 0.005 to 0.01 radians/m)

(b) Friction losses due to duct curvature:

Where:

θ = sum of angular displacements over distance x

Split the duct into sections if the curvature changes.

Generally curvature friction losses are greater than wobble friction losses. High friction losses can be substantially reduced by jacking the cable simultaneously from both ends. Losses at anchorages (post-tensioned)

Manufactures supply data for the wedge-set that occurs when a tendon is locked-off at the anchorage.

The manufactures of post-tensioning systems provide guidance on calculating the frictional and anchorage losses.

Time-dependant losses

EC2 also provides a combined formula for calculating the time dependant losses

(creep, shrinkage and relaxation) at location x and at time t.

8. Calculation of Deflections

See PART B.

9. End Blocks

See PART B.

11

CIVE3139 Prestressed Concrete (after MBH) PART B DESIGN FOR SLS (continued from PART A)

8. Calculation of Deflections

As EC2 provides no guidance on acceptable span to effective depth ratios, deflections must be explicitly checked. A linear elastic analysis based on the gross concrete section will give a realistic estimate of deflection for uncracked sections (see EC2 for cracked sections).

The requirement to be satisfied are similar to RC:

Deflection under quasi-permanent loads span/250 Additional deflection due to elements that are susceptible to deflection span/500 Deflection due to eccentric prestress loading by double integrating over the length of the

member:

But for uniform member with straight tendons:

Note that e positive if tendons below the NA, thus the prestress deflection is upward. And for uniform member with parabolic tendons:

but the induced bending moment is equivalent to a uniform load, thus midspan:

Mid-span deflection due to distributed load w over span L:

( )

If the prestressing force is eccentric at the ends of the member, an adjusted formula applies:

( )

( )

Deflections can be estimated for more complex cases. Deflection can be critical and may need to be checked:

At transfer (a useful physical check during construction)

Under dead load (to provide a baseline)

Long-term under full quasi-permanent actions (allow for creep)

Thus deflection due to we is …

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9. End Blocks Unlike the gradual transfer of stress into the concrete at the ends of pre-tensioned members, in post-tensioned members the force is concentrated over a small area(s) on the end face of the member.

In the resulting “end block” region the spreading high compressive stresses give rise to high tensile stresses (normal to the flow lines) requiring heavy reinforcement to resist the bursting forces. Typically empirically designed closed links and/or a spiral is provided. The bearing stress behind the anchorage plate must be checked:

( ⁄ ) Where: Ac0 = the area of the anchorage plate Ac1 = the maximum area of the same shape that can be inscribed on Ac

Applying a partial safety factor γp =1.2 to the prestressing force, the lateral tensile bursting forces can be calculated by assuming concrete struts and reinforcement links as ties as shown above, and limiting the compressive stress in the struts to no more than:

(

)

A design strength of 0.87fyk may be used for sizing the reinforcement (or limit to 300N/mm2 to avoid the need to check for cracks) and provide at least this area as closed links spread over an end-block length taken as equal to the greatest lateral dimension of the member.

13

ANALYSIS AND DESIGN AT ULS When the serviceability requirements have been satisfied, the ultimate moment of resistance and the shear resistance must be checked at ULS. The partial safety factors on loads and materials are as for RC, but with the addition of γp the partial safety factor for prestress. Details of the partial safety factor for prestress can be found in section 2.4.2.2 of EC2 where the symbols γp,fav and γp,unfav are introduced for favourable and unfavourable partial safety factors. Normally prestress is a favourable effect at ULS and a value of γp = 0.9 is specified in the UK national annex for this situation. The γp = 1.2 used on the previous page is the value used for unfavourable local effects.

1. Analysis of Section

Beyond working loads cracking of the concrete occurs (or increases) and the prestressing steel starts to behave like conventional reinforcement. At ULS the behaviour is exactly like a RC member except that the initial strain in the prestressing steel must be taken into account. The same equivalent rectangular stress block can be used to simplify hand calculation:

Parabolic-rectangular design stress-

strain for concrete in compression

Short-term design stress-strain

for reinforcement

Equivalent rectangular stress block

lever

arm

14

For a given depth of the neural axis x, the compressive force Fc developed by a uniform stress block of 0.567fck (=0.85fck /1.5) is equated to the total tensile force Fs in the prestressing steel. Fs is a consequence of the strain due to the prestress plus the additional strain due to bending:

Total strain in prestressing steel ( ) = prestress strain + bending strain ( )

When computing Fs, the prestress strain is normally reduced by applying the partial factor 0.9 which has the effect of reducing the total strain in the prestressing steel. The corresponding stress, and hence the value of Fs, is determined from the stress-strain diagram:

How to calculate the Ultimate Moment of Resistance (Mu)

If it is known that the prestressing steel is actually yielding at the ULS, the value of Fs can be computed and thus the neural axis depth x to give in an equal value of Fc. The lever arm can then be determined and Mu calculated. Unfortunately the stress in the prestressing steel at ULS is not normally known and an iterative process must be followed to determine x:

Preamble: Calculate the strain in the prestressing steel after losses (allowing for γp = 0.9) Guess an initial value for x (try 0.5d)

Evaluate for a value of x: Knowing that = 0.0035, calculate the bending strain

and thus the total strain in the prestressing steel

Using the stress-strain curve, calculate the stress in the prestressing steel Calculate and then compare the corresponding values of Fc and Fs If the two values are close, determine the lever arm and calculate Mu Otherwise select a new value for x and re-evaluate:

o Fc/Fs < 1 Fc<Fs increase x concrete stress block grows and reduces o Fc/Fs > 1 Fc>Fs reduce x concrete stress block shrinks and increases o Interpolate for x when two values of r = Fc/Fs

( ) ( )

( )

Short-term design stress-strain for prestressing steel

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2. Design of additional untensioned Reinforcement Should the ultimate moment of resistance prove to be insufficient this can usually be increased simply by adding conventional untensioned reinforcement. The how to section that follows covers the typically iterative calculation process when there is additional untensioned reinforcement. It addresses two similar cases: Calculating the quantity of reinforcement that needs to be added to achieve a given Mu.

o also follow the preamble steps in the box Calculating the value of Mu when a specified amount of reinforcement is provided.

o omit the preamble steps in the box Particular care needs to be taken because: To accommodate the prestressing steel and the reinforcement within the section they are

very likely to be located at different depths (either could be the lower one), therefor they will have different values for bending strain (

). Only the prestressing steel will have a prestress strain (and this is subject to γp = 0.9). Prestressing steel and reinforcement have different stress-strain diagrams. How to calculate when additional reinforcement

Preamble: Calculate the strain in the prestressing steel after losses (allowing for γp = 0.9) Determine the effective depth at which the reinforcement is added. Based on the value of x without reinforcement, estimate the lever arm for the additional

reinforcement. Estimate the required additional tension force Fsr by dividing the shortfall in ultimate

moment capacity by the estimated lever arm. Assume that the reinforcement is yielding, calculate the corresponding required area of

reinforcement. Convert this minimum area of reinforcement to specific reinforcement bars (round up) and

recalculate the value of Fsr. Assume that the prestressing steel is also yielding, calculate the values of Fsp. Equate Fs to the sum of Fsp and Fsr and determine the required value of x. Calculate the corresponding strains in the prestressing steel and in the reinforcement and

consult the relevant stress-strain diagrams. If they are both yielding, calculate Mu by simply taking moments about Fc for Fsp and Fsr. Otherwise select a new value for x and re-evaluate. Guess an initial value for x (try 0.5dmax).

16

Evaluate for a value of x: Prestressing steel: knowing that = 0.0035, calculate the bending strain

and thus the total strain then use the stress-strain curve to calculate the stress and compute Fsp.

Reinforcement: knowing that = 0.0035, calculate the bending strain and then use the

stress-strain curve to calculate the stress and compute Fsr. Calculate Fs = Fsp + Fsr and compute Fc Compare the values of Fc and Fs If the two values are close, calculate Mu by taking moments about Fc for Fsp and Fsr. Otherwise try a different value for x:

o Fc/Fs < 1 Fc<Fs increase x concrete stress block grows and reduces o Fc/Fs > 1 Fc>Fs reduce x concrete stress block shrinks and increases o Interpolate for x when two values of r = Fc/Fs

3. Shear Shear resistance is also checked at ULS. The approach is again similar to RC but with significant additional strength resulting (a) from the prestress induced longitudinal compression, and (b) from any vertical force reaction created by the tendon profile (or tendon deviation). In either case allowance is made for γp = 0.9. MBH considers three cases: Sections not requiring designed shear reinforcement Shear strength without shear reinforcement (special case of single span beams uncracked in

bending) Variable strut inclination method (for sections requiring shear reinforcement) Shear will not be covered to any further in CIVE3139.

CIVE3139 Prestressed Concrete (after MBH) Schedule of Numbered Equations

Page 1 of 2

Reference Correct Equation Incorrect Equation

(11.1)

(11.2)

(11.3)

(11.4)

(11.5)

(11.6)

(11.7)

(11.8)

(11.9)

(11.10)

(11.11)

(11.12)

(11.13) ( ) ( )

(11.14) ( ) (

)

Note the dash is missing in MBH ( ) (

)

CIVE3139 Prestressed Concrete (after MBH) Schedule of Numbered Equations

Page 2 of 2

(11.15)

( )

(11.16)

( )

( )

(11.17) ( )

( ⁄ )

(11.18) (

)

( ⁄ )

( )

( ⁄ )

(11.19) ( )

( ⁄ )

(11.20) (

)

( ⁄ )

( )

( ⁄ )

(11.21) (

)

(11.22) (

)

(11.23)

(

)

(11.24)

( ⁄ ⁄ )

( ⁄ )

(11.25)

( ⁄ ⁄ )

( ⁄ )

(11.26)

( ⁄ ⁄ )

( ⁄ )

( ⁄ ⁄ )

( ⁄ )

(11.27)

( ⁄ ⁄ )

( ⁄ )

( ⁄ ⁄ )

( ⁄ )

(11.28) [

]

(11.29) [

]

[

]

(11.30) [

]

[

]

(11.31) [

]

[

]