Prestress Lecture Notesfsfsfs

7/21/2019 Prestress Lecture Notesfsfsfs

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WHAT IS A PRESTRESSED CONCRETE?

Prestressed concrete is a concrete in which internal stresses of such magnitudeand distribution have been introduced such that the stresses resulting from the givenapplied loading are counteracted to a desired degree.

METHODS OF PRESTRESSING:

1. Pre-tensioning: The tendons are arranged to pass through the mould ormoulds for a number of similar members arranged end to end, and aretensioned between external end anchorages, by which the tension ismaintained while the concrete is placed. When the concrete has hardenedsuciently, the ends of the tendons are slowly released from the anchorages,thus causing compression to the concrete.

Pre-tensioning is more suitable for mass production of standard membersin a factory.

2. Post-tensioning: The concrete member is cast incorporating ducts for thetendons. When the concrete has hardened suciently, the tendons aretensioned by jacing against one or both ends of the member, and areanchored by means of end anchorages which bear against the member at theends or are embedded in it.

Post-tensioning is generally used on site for members cast in their !nalplace.

HISTORICA DE!EOPMENT OF PRESTRESSING

The basic principle of prestressing was applied to construction perhapscenturies ago, when ropes or metal bands were wound around wooden staves toform barrels. When the bands were tightened, they were under tensile prestresswhich in turn created compressive prestress between the staves and thusenabled them to resist hoop tension produce by the internal li"uid pressure.

#$$% - &ngr. P. '. (acson of )an *rancisco, +alifornia obtained patents fortightening steel tie rods in arti!cial stones and concrete arches to serve as oorslabs.

#$$$ - +. &. W. oehring of ermany independently secured a patent forconcrete reinforced with metal that had tensile applied to it before the slab wasloaded. These applications were based on the concept that concrete, thoughstrong in compression, was "uite wea in tension, and prestressing the steelagainst the concrete would put the concrete under compressive stressed whichcould be utili/ed to counter balance any tensile stress produced by dead load orlive load.

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#12$ - +. 3. )teiner of 4. ). suggested the possibility of retightening thereinforcing rods after some shrinage and creep of concrete had taen place, inorder to recover some of the losses.

#156 - 3. &. ill of 7ebrasa tried high strength steel bars coated toprevent bond with concrete. 0fter the concrete had cured, the steel rods were

tensioned and anchored to the concrete by means of nuts.

8odern development of prestressed concrete was credited to &. *reyssinetof *rance, who in #159 started using high-tensile wires for prestressing. )uchwires possessed ultimate strength as high as 562,222 psi.

Practical application of pre-tensioning method was !rst made by &. 'oyerof ermany.

Wide application of prestressed concrete was not possible until reliableand economical methods of tensioning and end anchorages were devised.

#191 - *reyssinet developed conical wedges for end anchorages anddesigned double-acting jacs which tensioned the wires and then thrust themale cones to the female cones for anchoring them. The system is called the*reyssinet system.

#1:2 - Prof. . 8agnel of ;elgium developed the 8agnel system, whereintwo wires were stretched at a time and anchored with a simple metal at eachend. 0bout this time prestressed concrete begin to ac"uire wide acceptance inconcrete construction.

AD!ATAGES OF PRESTRESSING

#. 'igh strength concrete with high concrete with high strength steel can beused to yield economical proportions. The entire concrete section is fullyutili/ed to e<ectively resist the applied moment, whereas only the portion of the section above the neutral axis is fully e<ective in ordinary reinforcedconcrete.

5. The use of curve tendon pro!les enables part of the shear force to be carriedby tendons and the precompression in the concrete tends to reduce diagonaltension.

9. =t can connect relatively longer spans and can accommodate larger load

capacity than ordinary reinforced concrete member.:. Prestressed structures are more slender and hence more adaptable to artistic

treatment. They yield more clearance where needed.6. Prestressed structures do not crac under woring loads, and whatever cracs

may be developed under overloads will closed up as soon as the load isremoved.

%. 4nder dead load, the deection is reduced owing to the cambering e<ect of prestress. 4nder live load, the deection is smaller becauset of thee<ectiveness of the entire uncraced concrete section.

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>. Prestressed elements are more adaptable to precasting because of the lighterweight.

PRESTRESSING TERMS

#. Tr"ns#er? Transfer is the operation in pretensioning in which the force in the

tendons is transferred to the concrete by bond stress.5. An$%or"ges: 0nchorages are mechanical devices use to transmit theprestress to the concrete at each end of the member. 0nchorages usuallybear against the member or are embedded in it.

9. Con$or&"nt $"'(e: +oncordant cable or tendon is a cable in a beam thatproduces a line of pressure coincident with the tendon itself. That is themoment due to loading is annulled due to the cable tension. 0 concordantcable induces no support reactions and no secondary moments.

:. )on&e& ten&ons: ;onded tendons are tendons which are bonded to theconcrete either directly or through grouting.

6. *n'on&e& ten&ons: 4nbonded tendons are tendons which are free to

move relative to the surrounding concrete.%. ine"r +restressing: This a method of prestressing where the structure islinearly prestressed. The prestressing tendons in a linearly prestressedstructure are necessarily straight@ they can be either bent or curved but theydo not go round and round in circles.

>. Cir$,("r +restressing: This a term applied to prestressed circularstructures where restressing wires are wound around in circles.

$. P"rti"( +restressing: 0 term given to structural members whereprestressing is done partially. =n practice there is partial prestressing if thereare some tensile stress produced in the member under woring loadcondition.

1. F,(( +restressing: This is a term which denote that full prestress is appliedto the member so that no tensile stress are produced under woring loadcondition.

PRESTRESS OSSES

Initial prestress in steel minus the losses is nown as the efective prestress or the design prestress. The total amount of losses to be assumed indesign will depend on the basis on which the initial prestress is measured. *irst,there is the temporary jacing stress to which tendon may be subject for thepurpose of minimi/ing creep in steel or for which balancing frictional losses.

Then there is a slight release from that maximum stress bac to the normal jacing stress.

0s soon as the prestress is transferred to the concrete, anchorage loss willtae place. The jacing stress minus the anchorage loss will be the stress atanchorage after release and is fre"uently called the initial prestress.

*or post-tensioning, losses due to elastic shortening will gradually taeplace. *or pre-tensioning, the entire amount of loss due to elastic shortening willoccur at the transfer of prestress. Ather losses will tae place and these willinclude creep and shrinage of concrete, creep in steel and for post-tensioned

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members, friction losses will also be considered. *or average steel and concreteproperties, cured under average air condition, tabulated percentages may betaen as representative of the average losses.

0verage PercentageBosses after transfer Pre-

tensioning

Post-

tensioning&lastic shortening and bending of concrete

9C #C

+reep of concrete % 6)hrinage of concrete > %+reep in steel D stress relaxation E 5 9

Total #$C #6C

0nchorage tae-up and friction losses will apply for post-tension members only.

STRESSES AT !ARIO*S PRESTRESSING STAGES

ii

N A

sf

c

si

e

As

C T

C )

)e" Se$tion+

+

+

++ +

+

_

_ _ _

D,e to Prestress D,e to D D,e to Res,(t"nt Stress

S T R E S S ! A R I A T I O N

=nitial +ondition ;efore Transfer?

Fi G initial prestressing force∆si G initial elongation of steel

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∆si G

+ondition 0fter Transfer?

F G e<ective prestressing force right after transfer∆si G ∆sf H ∆c

∆sf G

∆c G

Bet f c G stress in concrete at the level of steel after transfer?

F DFeEe 0c =

Hf c G

f s0s f s0se5

0c =

HG

∆si G ∆sfH ∆c

f siB&s

f sB&s

f cB&c

G H

f si G f s H nf c

f si G f s H nf s0sD E# e5

0c =H

f si GF0s

H nFD E# e5

0c =H

f si G F0se

5

r5F0c

n D # H EH

1ST PRESTRESSING CONCEPT: To transform concrete into an elastic material.

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FiB0s&s

f siB&s

G

FB0s&s

f sB&s

G

f cB

&c

6



N A

e

As

C T

C )

)e". Se$tion+

+

+

+ +

+

_

_ _

D,e to Prestress D,e to o"&

Res,(t"nt Stress


e

load w

irect

Boad &<ect

ue to Prestress

&ccentricity

ue to &xternal

8oment 8

FeyI= 8yI=

FI0 FecI= 8cI=

3esultant )tress ? f G F DFeEy 0 =

H H8y=

y

2ND PRESTRESSING CONCEPT: Prestressing for combination of high-strength steelwith concrete.

N A

e

As

C T

C )

)e". Se$tion+

+ + +

+ _ _

Res,(t"nt Stress


load w

ue to irectPrestress

ue to +ouple+ - F

3esultant )tress ? f G F Fβ 0 =

H

α

β

e-

C

&xternal 8oment G =nternal 8oment

8 G +α or Fαα G 8I+

β G α - e

y

y

/RD PRESTRESSING CONCEPT: Prestressing to achieve load balancing.

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N

e

As

C T

C )

)e". Se$tion+

+ + +

+ _ _

Res,(t"nt Stress


load w

D,e to Dire$tPrestress

D,e toNet o"& 0

3esultant )tress ? f G F ∆8 0 =H

&"uivalent uniform load from tendon?

wJ G y

A- -

load wJ

h

$FhB5

7et load ? ∆w G w - wJ7et moment? ∆8 G D#I$E∆wB5

y

THE FO*R )ASIC RE*IREMENTS IN WORING STRESS DESIGN OFPRESTRESSED CONCRETE

1. *or loading condition, prestress plus beam weight, the tensile stress at top!ber must be less than or e"ual to the allowable tensile at transfer.

2. *or loading condition, prestress plus beam weight, the compressive stress atthe bottom !ber must be less than or e"ual to the allowable compressivestress at transfer.

/. 4nder the action of prestress, superimposed load including impact, thecompressive stress at the top !ber must be less than or e"ual to theallowable !nal compressive stress.

. 4nder the action of prestress, superimposed load including impact, thetensile stress at the bottom !ber must be less than or e"ual to the allowable!nal tensile stress.

e

As

3 T

3 )

)e". Se$tion+

+ + +

+ _ _

Fin"(Res,(t"nt Stress

D,e to Prestress Initi"( Res,(t"ntStress

D,e to)e".Weig%t

D,e toS,+eri.+se&

o"&

+ +

+ +

_ _

_

3e"uirements # and 5

3e"uirements 9 and :

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3e"uirement K #? f t GF0

FeLt

=8g Lt

= M H N f t allowable at transfer

F0

8g Lt

=H N f t allowable at transfer Top !ber stressf t G D # -eLt

r5 E

3e"uirement K 5? f c GF

0

FeLb

=

8g Lb

=

O f c allowable at transferH M

F0

8g Lb

=O f c allowable at transfer ;ottom !ber stressf c G D #H

eLb

r5 E M

3e"uirement K 9? f c G DF 0

FeLt

=8g Lt

=O f c allowable at !nalH

M Eη H

8s Lt

=

F 0

FeLt

=8t Lt

=O f c allowable at !nal Top !ber stressH M Eηf c G D

3e"uirement K :? f t G DF 0

FeLb

=8g Lb

=N f t allowable at !nal

H

M EηH

8s Lb

=

F 0

FeLb

=8t Lb

=N f t allowable at !nal ;ottom !ber stress

M

Eηf t G D M

r5 G =0

8t G 8g H 8s G total moment due to dead load of beam plus superimposed load

η G prestressing e<ectiveness or prestressing eciency, η G $2C for pre-tensioned

members, η G $:C for post-tensioned members.

AOWA)E STRESSES FOR CONCRETE:

Co&e

A((o0"'(e Stress "tTr"ns#er4 MP"

A((o0"'(e Stress "t Fin"(St"ge4 MP"

+ompressive

Tensile+ompressive

Tensile

00)'TA Pre-tension 2.%2 f ciJ4nbonded

reinforcements

G .56 'ci f

;ondedreinforcements

G .%5 'ci f

2.:2 f cJ 4nbonded G 2.22)evere exposure G .56

'c f

;onded G.62 'c f 00)'TA

Post-tension 2.66 f ciJ 2.:2 f cJ

0+=Pre-tension,Post-tension

2.%2 f ciJ

0t any point bet.

supports G .56 'ci f

0t ends of simple

supports G .62 'ci f

2.:6 f cJ

0t precompressed tensile

/one G .62 'c f

0nalysis base ontransformed craced

section G #.22 'c f

P+=Post-tension 2.%2 f cJ .56

'ci f 2.:6 f cJ 2.22

AASHTO Cr"$5ing Stresses 6Mo&,(,s o# R,+t,re7.P.0ncog$



*or normal-weight concrete2.%5

'c f

*or sand-lightweight concrete2.65

'c f

*or all other lightweight concrete2.:%

'c f

AASHTO An$%or"ge-'e"ring stressesPost-tensioned anchorage at service load Dbut not to exceed 2.1f ciJE 52.%2 8Pa

AOWA)E STRESS FOR STEE:

Co&eTe+or"r3 8"$5ing

For$eIe&i"te(3 A#ter

Tr"ns#erA#ter A((osses

AASHTO 2.1:f py ≤ 2.$2f pu 2.$5f py ≤ 2.>:f pu 2.>2f pu

ACI 2.1:f py ≤ 2.$2f pu 2.$5f py ≤ 2.>:f pu 2.>2f pu

PCI 2.$2f pu 2.>2f pu 2.%2f pu

Post-tensioning tendons at anchorage, immediately after tendon anchorage 2.>2f pu.f py @ 2.$6f pu

*or low relaxation prestressing steel, f py G 2.12f pu.

Ter De9nitions f py G speci!ed yield strength of prestressing tendons

f y G speci!ed yield strength on non-prestressed reinforcementsf pu G speci!ed tensile strength of prestressing tendonsf cJ G speci!ed compressive strength of concretef ci G compressive strength of concrete at time of initial prestress

Stres-Re(iee& Wire #or Prestresse& Con$rete 6ASTM A 217

7ominal ia. DinE8in. tensile strength DpsiE 8in. stress at #C extension DpsiE

Type ;0 Type W0 Type ;0 Type W02.#15 562,222 5#5,6222.#1% 5:2,222 562,222 52:,222 5#5,6222.562 5:2,222 5:2,222 52:,222 52:,2222.5>% 596,222 596,222 #11,>62 #11,>62

)ource? Post-Tensioning =nstitute

Stress-Re(iee& Seen-Wire St"n&"r& Str"n& #or Prestresse& Con$rete 6ASTMA 1;7

7ominaliameter

DinE

;reaing )trength8in. lbE

7ominal 0reaDs". inE

7ominal WeightDlbI#222 ftE

8in. load at#C &xtension DlbE

30& 562 D2.562E 1,222 2.29% #55 >,%626I#% D2.9#9E #:,622 2.26$ #1> #5,9229I$ D2.9>6E 52,222 2.2$2 5>5 #>,222

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>I#% D2.:9$E 5>,222 2.#26 9%> 59,222Q D2.622E 9%,222 2.#:: :12 92,%229I6 D2.%22E 6:,222 2.5#% >9> :6,122

30& 5>29I$ D2.9>6E 59,222 2.2$6 512 #1,662>I#% D2.:9$E 9#,222 2.##6 912 5%,962Q D2.622E :#,922 2.#69 652 96,#22

9I6 D2.%22E 6$,%22 2.5#> >:2 :1,$22

Stress-Re(iee& Seen-Wire Co+"$te& Str"n& #or Prestresse& Con$rete6ASTM A <<=7

7ominal iameterDinE

7ominal ;reaing)trength

DibE

7ominal )teel 0reaDs". inE

7ominal WeightDlbI#222ftE

Q :>,222 2.#>: %222.%2 %>,::2 2.56% $>92.>2 $6,:92 2.9:% ##>%

Rein#or$eent Gr"&es "n& Strengt%s

#1$5 )tandard Type 8inimum Lield )trength, f y DpsiE 4ltimate )trength, f u DpsiE;illet )teel D0%#6E

rade :2 :2,222 >2,222rade %2 %2,222 12,222

0xle )teel D0%#>Erade :2 :2,222 >2,222rade %2 %2,222 12,222

Bow 0lloy )teel D0>2%Erade %2 %2,222 $2,222eformed Wire 3einforced >6,222 $6,222

eformed Wire *abric >2,222 $2,222)mooth Wire 3einforced >2,222 $2,222

DESIGN OF PRESTRESSED CONCRETE:

PREIMINAR> DESIGN. Intern"( Co,+(e C- Met%o&

#. =n practice the depth h of the section is either given, nown or assumed, as is thetotal moment 8t of the section.5. 4nder woring load, the lever arm a for the internal couple could vary between 92C

to $2C of the overall height h and averages about 2.6%h.

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As

)e" Se$tionResisting Moent "n&

Stress Distri',tion

h a

C

3e"uired e<ective prestress F?

R

R

2.%6h

2 . 6

h

2.6f c

f c

F G8t 8t

a 2.%6hG

0rea of steel?

0s GF 8t

f s 2.%6hf sG

Total prestress F G 0sf s is also the force F on the section. This force will produce anaverage unit stress on the concrete f c ave.

f c ave G+ F 0sf s0c 0c 0c

G G G 2.62f c Dpreliminary designE

3e"uired area of concrete section? 0c G 0sf s2.62f c

+hec results in the preliminary design and revised as needed.

EASTIC DESIGN 6No tension in $on$rete7 s"(( r"tios o# MG @ Mt.

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Se$tionPro+erties

h

cgs

cgc

d e

t

b

ct

cb

Fo

0c

F0c

a G eHt

Co

C

a G e - b

Fo h0c ct

F h0c cb

8,st "#ter tr"ns#er C "t'otto 5ern +oint

At 9n"( st"ge C"t to+ 5ern +oint

S T R E S S D I S T R I ) * T I O N

Ste+s:

#. *or the section obtained from preliminary design, the values of 8, t, b and 0c arecomputed.

5. *rom the preliminary design section, locate c.g.s D + at the bottom ern pointE

e - b G

8

Fo

)tresses at top and bottom !bers will be?

f t G 2@ f b G

Fo h0c ct

@ 0c GFo hf b ct

9. With the location of the c.g.s., compute the e<ective F D + at the top ern pointE

F G

8t

eHtand then Fo G FD

f so f s

E@

Top and bottom !ber stresses will be?

f b G 2@ f t G

F h0c cb

@ 0c GF h f t cb

:. +ompute the re"uired 0c by?

0c G

Fo h f b ct

@ 0c GF h f t cb

6. 3evised the preliminary section to meet the above re"uirements for F and 0c.

#. 3epeat steps 5 to 6 if necessary.

EASTIC DESIGN 6No tension in $on$rete74 ("rge r"tios #or MG @ Mt.

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Se$tionPro+erties

h

cgs

cgc d

e

t

b

ct

cb

a G eHt

Co

C

e - a

8,st "#ter tr"ns#er4C "'oe 'otto

5ern +oint

At 9n"( st"ge o# (o"&ingC "t to+ 5ern +oint


0s

a G8

Fof b

f t

When the ratio of 8 I 8t is large, the value of e - b computed from e - b G 8IFo

may place c.g.s. outside of the practical limit Di.e. outside the beam sectionE. Then it isnecessary to place the c.g.s. only as low as practicable and design accordingly. (ust after transfer?

f b G

Fo

0cH

DFoe - 8 Ecb

= GFo

0cD# H

e - 8IFo

tE

3e"uired area?

0c GFo

f bD# H

e - 8IFo

tE

Ste+s:#. *rom the preliminary section, compute the theoritical location of c.g.s. by

a G e - b G 8IFo =f it is feasible to locate c.g.s. as indicated by this e"uation, follow the !rst

procedure. =f not, locate c.g.s. at the practical lower limit and proceed asfollows.

5. +ompute F by?

and then Fo G FDf so f t

E@F G8t

eHt

9. +ompute the re"uired area by the e"uations?

0c

0c GF h f t cb

GFo

f bD# H

e - 8IFo

tE

:. 4se the greater of the two 0cJs and the new value of F and revise the preliminarysection.

#. 3epeat steps # through : if necessary.

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EASTIC DESIGN4 "((o0ing "n& $onsi&ering tension s"(( r"tios o# MG@Mt.

Se$tionPro+erties

h

cgs

cgc

e

t

b

ct

cb

a G eHt

Co

C

Stress Distri',tion"t Tr"ns#er

Stress Distri',tion"t Wor5ing o"&


0s

f b

f t

e#e5

f tJ

f bJ

=f tensile stress f tJ is permitted in the top !bers, the center of compression + can be

located below the bottom ern by an amount?

e# G

f tJ=Foct

G f tJ0b

Fo

*or a given moment 8, the c.g.s. can be further located below + by the amount?

e5 G8

Fo.. . e# H e5 G

8 H f tJ0b

Fo

Bever arm acting up to the top ern point ? a G e H t

;ottom !ber stress at transfer ?

Prestress re"uired ? F G8t - f bJ0t

a

f b GFoh0cct

Hf tJcb

ctthen 0c G

Fohf bct - f tJcb

Top !ber stress under woring load? f t GFh0ccb

Hf bJct

cbthen 0c G

Fhf tcb- f bJct

Ste+s:

#. *rom the properties of the preliminary section, obtain e# H e5 and locate c.g.s.5. +ompute the net moment to be carried by the prestress? 8t - f bJ0t and obtain? F G D8t - f bJ0tEIa and Fo G FDf soIf sE9. +ompute the re"uired area by the e"uations?

0c GFoh

f bct - f tJcb

0c GFh

f tcb- f bJct

:. 4se the greater of the two 0cJs and revise the preliminary section.#. 3epeat steps # through : if necessary.

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*or large ratios of 8I8t, + will be within the ern at transfer, and the allowing of tension on top !ber will have no e<ect on the design. The c.g.s. has to be locatedwithin practical limits. Atherwise, the design is made as for the !rst case. The area of concrete re"uired is given by?

−+=

=

t

oG

b

o

c

bt

c

k

Q M e

f

Q A

c f Qh A

(1

EASTIC DESIGN4 Co+osite Se$tions

"7 Pre$"st Portion4 StressDistri',tion "t Tr"ns#er

'7 Co+osite Se$tion4 Stress Distri',tion*n&er Wor5ing o"&

cgs

cgct

b

ct

cb e

cgs

cgcJctJ

cbJ

in-place portion

f b

f tJ

f bJ

f t

The procedure of design follows closely the basic approach previously adoptedfor noncomposite sections. =t is essentially a trial-and-error process, simpli!ed by asystematic and fast converging procedure and assisted by the use of some simplerelations and formulas. Ane additional concept introduced for composite action is thereduction of moments on the composite section to e"uivalent moments on the precastportion. This is accomplished by the ratio of the section moduli of the two sections.

Ste+s:

#. Bocate the c.g.s. *or a given trial section, the c.g.s. must be so located that theprecast portion will not be overstressed and yet will possess the optimum capacityin resisting the applied external moments. The c.g.s. must be situated as low aspossible but not lower than given by the following value of eccentricity? e G e# H e5

H b , where?

e#Gf tJ=ctFo

e5 G8

Fo

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Where? f tJ G allowable tension stress on top !ber of precast portion at transfer = G moment of inertia of precast portion

ct G distance to top !ber from c.g.c. of precast portion5. +ompute the e"uivalent moment of the precast portion. *or any moment 8+ acting

on the composite section, it will produce stresses on the precast portion as follows?

f tG

8+ctJ

=J

f b G8+cbJ =J

where =J G = of composite section, ctJ and cbJ G distance to extreme !bers of theprecast portion measured from c.g.c.J of the composite section.

mt G=Ict

=JctJ

mb G

Bet

and =Icb

=JcbJ

Then we have f t Gmt8+ ct

=

f b G

Gmt8+ 0cb

mb8+ cb

=G

mb8+ 0ct

where 0c G area of the precast section b G bottom ern distance of precast section t G top ern distance of precast section.

The above indicates that 8+ can be modi!ed by the coecients mt and mb sothat it can be reduced to e"uivalent moments for computation based on the precastportion properties.

9. +ompute the amount of prestress re"uired for the moments as follows. =f 8P G thetotal moment acting on the precast portion, and f bJ G allowable tensile stress at the

bottom !ber, we have?

F0c

D- # -et

EH 8P 0ct

Hmb8+ 0ct

G f bJ

F GH mb8+ - f bJt0c

e H t

8P

or F G H mb8+ 8P

e H t

if f bJ G 2

from which compute the re"uiredinitial prestress Fo. 3evised the location of c.g.s. by this new Fo if necessary.

:. =n order to limit the bottom !ber stress to the allowable value at transfer, we have?

Fo

0cH D Foe - 8 E

0ct

f b G from which 0c G #f b

Fo HD Foe - 8 E

t

=n order to limit the top !bers of the precast portion to within allowablecompressive stress f t under woring load, we have?

F0c

Hf t G

0c G#f t

F H

H mt8+ - Fe8P

0cb

H mt8+ - Fe8P

b

.P.0ncog#%



compressiontension

123

4

fti fcsi

t1f1r

Mg

St

f2r t2

fc1fts

Mg

Sb

h

ct

cb

Concrete stress distribution at various loading stages

Sectionariable eccentricit! profile

Q Q

fcent

e

c

Q

A

i

c

Q

A

"t support "t midspan

1

3

1 #i

#i + Mg2

3 #e + Mg

4 #e + Mg + Mdl + Mll

The greater of the two formulas will control the 0c re"uired for the precast portion. The top !ber stress of the cast- in-place top ange can be computed by the formulaf G 8cI=, using the applicable values.

A(tern"tie F(e,r"( Design

A - Beam with variable Eccentricities:

Let:

f ci = permissible concrete compressive stress at initial stagef ti = permissible concrete tensile stress at initial stagef cs = permissible concrete compressive stress at service load stagef ts = permissible concrete tensile stress at service load stage

Required section modulus:

( )

( )

1

1

g dl ll

t

cs ti

g dl ll

b

ci ts

t t b b

t b

t b

M M M S

f f

M M M S

f f

I S c S c

c S

h S S

η

η

η η

− + +≥

−

− + +≥−

= =

=+

.P.0ncog

where:St = section modulus at topSb = section modulus at bottomct = top fiber distance

cb = bottom fiber distanceη = prestressing efficiency

#>



Q Q

compressiontension

1

2

h

ct

cb

Concrete stress distribution at various loadingSectionConstant eccentricit! profile

fcent

e $ em

ft

fb1 #i Stresses

#e Stresses2

The required initial prestressing force:

Qi = Ac !f cent" # where: Ac = concrete area$ f cent = aial compressive stress due to prestress%

( )t cent ci ti ti

c f f f f

h= − +

unsymmetrical section

1( %b

cent cs ts ts

c f f f f

hη

= − +

symmetrical section

That prestressing initial prestressing force applied with eccentricity e$ must produce a bending moment Q ie$ at the topfiber:

( )

( )

g icent ti

t t

g icent ci

b b

M Q e f f

S S

M Q e f f

S S

= − +

= + +

The required eccentricity:

( )

( )

ma&

g t m cent ti

i i

g bm cent ci

i i

m

M S e e f f Q Q

M S e e f f

Q Q

e e

= = − +

= = + +

=

B - Beams with constant eccentricity:

.P.0ncog#$



Required section modulus:

g dl ll

t cs ti

g dl ll

b

ci ts

M M M

S f f

M M M S

f f

η

η

+ +

≥ −+ +

≥−

Required constant eccentricity:

( )

( %

icent ti

t

t cent ti

i

bcent ci

i

Q e f f

S

S e f f

Q

S e f f

Q

= −

= −

= +

C - S3etri$"( Cross Se$tion

&ften for practical reasons$ a symmetrical cross section is chosen$ even though the limit stresses indicate otherwise% 'ncuch case$ St = Sb = S$ and the stress change at the top and bottom fiber will be identical as the transverse load isapplied% As a result$ only three of the four limit stresses will be achieved$ in general% 'n such circumstanaces$ there isno unique combination of Q i and e that will serve$ but rather a range of values% The best choice will normally be thatwhich minimi(es the prestress force required%

'f the bottom fiber stress controls: Sb ) St:

( )1

g dl ll

b t b

ci ts

M M M S S S S

f f

η

η

− + +≥ > =

−

'f the top fiber stress controls: St ) Sb:

( )1

g dl ll

t b t

cs ti

M M M S S S S

f f

η

η

− + +≥ > =

−

The concrete centroidal stress is computede based on service stress load condition *:

1( %b

cent cs ts ts

c f f f f

hη

= − +

The initial prestress Qi can be found by:

( ) g t

m cent ti

i i

M S e e f f

Q Q= = − +

.P.0ncog#1



SHEAR IN PRESTRESSED CONCRETE:

A C

-

!$

!s

Portion o# )e".

Af c f c

ν

−ν

St"te o# stress o# E(e.ent A

O

C

f c

)c )t

ν ν

m a x

Mo%rBs Cir$(e o# Stress #or E(e.ent A

Conention"( Met%o&:#. *rom the total external shear S across the section, deduct the shear Ss carried by

the tendon to obtain the shear Sc carried by the concrete? Sc G S - Ss

5. +ompute the distribution of Sc across the concrete section?

ν GScFv

=b

9. +ompute the !ber stress distribution for that section due to external moment 8,the prestress F and its eccentricity by the formula?

f c GF Fec 8c0 = =

H H

:. the maximum principal tensile stress in )t corresponding to the above ν and f c iscomputed by?

)t G D E5f c

5 ν2 H

f c5

Bimiting values for principal tensile stress in prestressed concrete?

)t N 2.2#9f cJ to 2.299f cJ for beams wIo web reinforcements

)t N 2.2:2f cJ to 2.#22f cJ for beams with web reinforcements

=n general, shear design under woring alone is not sucient.

3e"uired stirrups spacing under woring load?

.P.0ncog52



s G

0vf sv )tbw

Where? 0v G area of web reinforcement at spacing sf sv G allowable stress of web reinforcement)t G principal tensile stressbw G width of beam web

*TIMATE SHEAR DESIGN

0t loads near failure, a prestressed beam is usually extensively craced andbehaves much lie an ordinary reinforced concrete beam. 0ccordingly, many of theprocedures and e"uations developed for non-prestressed beams can be applied to thedesign of web reinforcement for prestressed beams also.

)hear design id based on the relation? Su O φSn

Where? Su G total shear force applied to the section at factored loadsSn G the sum of the contributions of the concrete and web

reinforcement.Sn G Sc H Ss and strength reduction factor φ G .$6=n computing the factored load shear Su , the !rst critical section is assumed to

be at a distance hI5 from the face of the support, and sections located a distance lessthan hI5 are designed for the shear computed at hI5.

The shear force Sc resisted by the concrete after cracing has occurred is taene"ual to the shear that caused the !rst diagonal crac. Two types of diagonal cracshave been observed in tests of prestressed concrete beams?

#. *lexural-shear cracs, occurring at nominal shear Sci, start as nearly verticalexural cracs at the tension face of the beam, then spread diagonallyupward Dunder the inuence of diagonal tensionE toward the compression

face. These are common in beams with low value of prestress force.5. Web-shear cracs, occurring at nominal shear Scw, start in the web due tohigh diagonal tension, then spread diagonally both upward and downward. These are often found in beams with thin webs with high prestress force.

Sci G 2.%2 f cJ bwd H So H Scr

Where?So G shear caused by beam self weight D without load factorEScr G additional shear due to superimposed dead load and live loads

Scr GS dHl

8 dHl

8cr

8crG =cc5

D % f cJ H f 5pe - f 5o E

Where?SdHlI8dHl , the ratio of superimposed dead load and live load shear tomoment remain constant as the load increases to cracing load.c5 G distance from concrete centroid to tension facef 5pe G compressive stress at tension face resulting from e<ective prestress force alonef 5o G bottom-!ber stress due to beam self-weight

.P.0ncog5#



S dHl

8 dHl

8crSci G 2.% f cJ bwd H So H O #.>2 f cJ bwd

the shear force causing web-shear cracing can be found from an exact principalstress calculation, in which the principal tensile stress is set e"ual to the direct tensilecapacity of concrete Dconservatively taen e"ual to :vf cJ according to 0+= code.0lternatively, the 0+= code permits the use of the approximate expression?

ScwG D9.6 f cJ H 2.9f pcEbwd H Sp

in which f pc is the compressive stress in the concrete, after losses, at the centroid of the concrete section Dor at the junction of the web and the ange when the centroidlies in the angeE and Sp is the vertical component of the e<ective prestress force.

0fter Sci and Scw have been calculated, then Sc , the shear resistance provided bythe concrete, is taen e"ual to the smaller of the two values.

+alculating 8cr, Sci, and Scw for a prestress beam is tedious matter because manyof the parameters vary along the member axis. *or hand calculations, the re"uired"uantities may be found at discrete intervals along the span, such as BI5,BI9,BI% andat hI5 from the support face, and stirrups spaced accordingly, or computer

spreadsheets may be used. To shorten calculation re"uired, the 0+= code includes, as a conservative

alternative to the above procedure, an e"uation for !nding the concrete shearresistance Sc directly?

ScG D2.% f cJ H >22 EbwdS ud8 u

O

N

5 f cJ bwd

6 f cJ bwd

in which 8u is the bending moment occurring simultaneously with shear force Su , butSudI8u is not to be taen greater than #.2

+ontribution of shear reinforcement to the shear strength of the prestressedbeam?

Ss G0vf yd s

The total nominal shear strength Sn is found by summing the contributions of thesteel and concrete, as indicated by?

Su G φSn G φDSs H ScE

G φD0vf yd s

H Sc E

7ormally, in practical design, the engineer will select a trial stirrup si/e, for which there"uired stirrup spacing is found.

N 2.>6h or 5: in.φ0vf ydSu - φSc

s G

0t least a certain minimum area of shear reinforcement is to be provided in allprestressed concrete members, where the total factored shear force is greater thanQDφScE provided by concrete. &xceptions are made for slabs, footings, concrete-joistoor construction, and certain very shallow beams. The minimum area of shearreinforcement to be provided in all other cases is to be taen e"ual to the smaller of?

.P.0ncog55



0pf pus$2f yd

bws f y

0v G 62

0v Gdbw

in which 0p is the cross-sectional area of the prestressing steel, f pu is the ultimatetensile strength of the prestressing steel, and all other terms are as de!ned before.

*TIMATE FE*RA STRENGTH OF PRESTRESSED CONCRETE

Pre(iin"r3 Design.

The amount of mathematics involved in the design of prestressed concretesection is less in ultimate design than in the elastic design, since the ultimate exuralstrength of sections can be expressed by simple semi-empirical formulas. *or

preliminary design, it can be assumed that the ultimate resisting moment of bondedprestressed sections is given by the ultimate strength of steel acting with a lever arm. This lever arm varies with the shape of section and generally ranges between 2.%2hand 2.12h, with a common value of 2.$2h. 'ence the area of steel re"uired isapproximated by?

0s G8 TDB*E2.$hf ps

G 8u

φD.$hf psE

where?B* G factor of safety or load factor

0ssuming that concrete on the compressive side is stressed to 2.$6f cJ, then there"uired ultimate concrete area under compression is ?

0cJ G8 TDB*E

2.$hf psD 2.$6f cJ E 8u

φ.$hf psD.$6f cJ EG

which is supplied by the compression ange D occasionally with the help of part of thewebE. The web area and the concrete area on the tension side are designed to providethe shear resistance and the encasement of steel, respectively. =n addition, concreteon the precompressed tension side has to stand the prestress at transfer.

Fin"( Design.

.P.0ncog59



0s

d

bεcuG2.229:

εs GfsuI&s

f cJ

c a Gβ#c

2.$6f cJ

C

T As# s,fsu

jd G d - aI5

aI5

Se$tion Con&itions "t *(ti"te Strengt% E,i"(ent Stress )(o$5

)teel stress?

f ps G εc &sβ#d - a a

N f py

a G β#c

4ltimate resisting moment? 0+=

8u G 0sf psDd - aI5 E G 2.$6f cJabD d - aI5 E

=n more convenient form?

ρG

8u G 0sf ps d D # - 2.61ω E in terms of steel area

8u Gωbd5 f cJ D # - 2.61ω E in terms of concrete section

ω Gρf ps

f cJ@

0s

bd

8u G 0sf psD d - aI5 E

a G 0sf ps2.$6f cJb

G ρf psd2.$6f cJ

8u G D 0s - 0sf Ef psD d - aI5 E H 0sf f psD d - tI5 E

0sf G 2.$6f cJf ps

D b - bJ Et a GD 0s - 0sf Ef ps

2.$6f cJbJ

3ecttangularsections and for

T and -sectionswith 7.0. fallswithin the ange.

T and -sectionswith 7.0. fallswithin the web.

f ps G f pu # -γ p ρf pu

β# f cJU members with bonded tendons

f ps G f pe H #2,222 Hf cJ

#22ρmembers with unbonded tendons with BId 96O

f ps G f pe H #2,222 H members with unbonded tendons with BId N 96 f cJ

922ρf ps N f py , f peH92,222 psi U min. for all cases

γ p G .66 for f pyIf pu not less than 2.$2, typical high strength barsγ p G .66 for f pyIf pu not less than 2.$2, typical high strength barsγ p G .66 for f pyIf pu not less than 2.$2, typical high strength bars

γ p G .:2 for f pyIf pu not less than 2.$6, typical ordinary strand

γ p G .5$ for f pyIf pu not less than 2.12, typical low-relaxation strand

PCI Co&e:

.P.0ncog5:



#. 3ectangular sections or anged sections in which the neutral axis lies within theange.

8u G 0sf ps d D # - 2.%2ω E

5. *lange sections in which 7.0. falls outside the ange.

ρ1G

H 2.$6f cJD b-bJ EtD d-.62t E

ω G ρ1f ps f cJ

@ 0s- 0sf bJd

@ 0sf G 2.$6f cJD b-bJ Et f ps

8u G D 0s-0sf Ef psdD # - 2.%2ωE

9. Where information for the determination of f ps is not available and provided that f pe

is not less than 2.62f pu the following values shall be used? ;onded members ? f ps G f puD # - 2.62ρf puIf cJ E

4nbonded members ? f ps G f pe H #6,222 psi or f ps G f pe H #29 8Pa

=n 0+= +ode, .61ω is used instead of 2.%2ω.Bimitations on steel percentages?

ω Gρf ps

f cJN 2.92, and whenω is greater than 2.92?

2.$6a

d

O 2.9%β#

#. 3ectangular or anged sections in which 7.0. lies within the ange 8u G 2.56f cJbd5

5. *langed sections in which 7.0. falls outside the ange 8u G 2.56f cJbJd5 H 2.$f cJD b - bJ EtD d - 2.6t E

PCI ACI&c G #.$2x#2% H 622f cJ psi

&c G 99w#.6 'c f psi

&c G #5.:x#29

H 622f cJ 8Pa &c G 2.2:9w#.6 'c f 8PaBoad *actors?4 G #.$D H B E 4 G #.: H #.>B4 G #.5 H 5.:B 4 G .>6D #.: H #.>B H

#.>W E4 G #.:D H B H W E+apacity 3eduction *actors?φ G 2.12 for exure φ G 2.12 for exure

φ G 2.$6 for shear φ G 2.$6 for shear

*TIMATE !S. EASTIC DESIGN

0t the present time, both the elastic and the ultimate designs are used forprestressed concrete, the majority of designers still following the elastic theory. =t isdicult to state exact preference from or the other. &ach has its advantages and shortcomings. ;ut, whichever method is used in design, the other one must often beapplied for checing. *or example, when the elastic theory is used in design, it is thepractice to chec for the ultimate strength of the section in order to !nd out whether

.P.0ncog56



the section is overstressed under certain conditions of loading and whether thedeections are excessive. Averstressing is objectionable because it may result inundesirable cracs and creep and fatigue e<ects. When the design is of conventionaltypes and proportions, such checing becomes unnecessary, because it is thengenerally nown that designing by one method will yields safe results when checedby the other. This is, in fact, the reason why such checing is not re"uired of reinforced

concrete structures designed by the usual codes. When we delve into new types andproportions, it is possible that elastic design alone will not yield a suciently safestructure under overloads, while the ultimate design by itself might give no guaranteeagainst excessive overstress under woring conditions. =t is therefore deemeddesirable to apply both the elastic and the ultimate methods, especially for structuresof unusual proportions.

0n understanding of both theories of design is also essential informing judgmentwhen designing structures. )ometimes, design based on one method will yielddi<erent proportions from those based on the other.

DESIGN EAMPES

Inestig"tion Pro'(e :

#. 0 prestressed concrete rectangular beam 52Vx92V has a simple span of 5:ft and isloaded by a uniform load of 9Ift including its own weight. The prestressing tendonis located as shown and produces an e<ective prestress of 9%2. +ompute !berstresses in concrete at the midspan.

5:J

w G 9Ift

Prestress F G 9%2

52V

9 2 V

1 V

)e". E(e:"tion

)e". Se$tion"t Mi&s+"n

So(,tion 1: Transforming concrete into an elastic material.

iven?

&<ective prestress, F 9%2 Tendon eccentricity at midspan, e G #6-1G

% in

8oment, 8 G #I$wB5 G #I$D9ED5:E5

G5#% -ft

.P.0ncog5%



)ection properties D no hole deductionE 0 G52x92G

%22 in5

= G#I#5bh9 G#I#5x52x929G :6,222in:

&xtreme !ber

yG92I5G

#6in

)tresses?

F Fey 8y0 = =

H Hf G

9%2,222 9%2,222x%x#6 5#%x#,222x#5x#6 %22 :6,222 :6,222G H H

G %22 >52 $%:G %22 - >52 H $%: G >:: psi for top !berG %22H >52 - $%: G :6% osi for bottom !ber

H H

)tress diagram?

%22psi >52psi

->52psi $%:psi

-$%:psi :6%psi

>::psi

F

0

Fey

=

8y

=

F

0

Fey

=

8y

=

H H

So(,tion 2: 0pply the second concept of prestressing using the pinciple of internalcouple

method.

Tae one-half of the beam as free body, thus exposing the internal couple. Theexternal moment at the section is?

.P.0ncog5>



w G 9 Ift

C

αβ

e1V

H"(# E(e"tion o# )e" Stress Distri',tion "tMi&s+"n

:6%psi

>::psi

8 GwB5

$G

9x5:5

$G 5#% -ft

The internal couple is furnished by the forces + G F G 9%2 which must act at a leverarm of?

8 G+α G Fα

α G 8IF G5#%x#5 9%2

G >.5 in

)ince F acts at 1in from the bottom, + must be acting at #%.5 in from it. Thus thecenter of the compressive force + is located.

β G α - e G >.5 - % G #.5 in.

)tress?

F Fβy0 =

Hf G

9%2,222 9%2,222x#.5x#6%22 :6,222

G H

G %22 #::G >:: psi for top !berG :6% psi for bottom !ber

H

So(,tion /: 0pplying the concept of load balancing.

.P.0ncog5$



N

w G 9Ift

AwJ G 5.6 Ift

hG %in

1V 1V

9 2 V

52V

5:ft

# 6 V

# 6 V

)e" E(e"tion S%o0ing S,+eri+ose& o"&"n& E,i"(ent Ten&on *+0"r& For$e

Se$tion "tMi&s+"n

The upward uniform force wJ from the tendon on the concrete is obtained from?

Fh G wJB5#$

wJ G $Fh B5

G $x9%2xD%I#5E 5:5

G 5.6 Ift

'ence the net downward load on the beam is ∆w G 9.2 - 5.6 G 2.6 Ift, and theresulting moment due to this net load is?

∆8 G∆wB5

$G

2.6x5:5

$G 9% -ft

)tresses?

F ∆8y0 =

Hf G

9%2,222 D9%x#,222x#5E#6%22 :6,222

G H

G %22 #::G >:: psi for top !berG :6% psi for bottom !ber

H

)tress iagram?52V

9 2 V

%22psi -#::psi

#::psi%22psi

:6%psi

>::psi

H G

Se$tion "tMI&s+"n

D,e to .o.ento# net (o"& 0

D,e to "i"(+restress -

Res,(t"ntstress

Design Pro'(e:

.P.0ncog51



5. esign a prestressed concrete -beam with the following data?

8 T G 952 -ft f t G #.%2 si8 G :2 -ft f b G #.$2 si

f tJ G -.92 si f so G #62 sif bJ G -2.5: si f s G #56 si

So(,tion:

Preliminary esign? h G T M @ D G #.6 - 5.2E , for G 5 , h G 96.>>, try hG 9%V

+ompute the re"uired prestress? F G 8 TID.%6hE F G 952x#5ID.%6x9%E G #%:.#2 Fo G FDf soIf sE G #%:.#D#62I#56E G #1%.15

0rea of tendon? 0s G FIf s G 8 TID.%6hf sE G #%:.#2I#56 G #.9#5$ in5

3e"uired concrete area 0c G 0sf sID.6f cE G FID.6f cE G #%:ID.6x#.%2E G 526 in5

Trial section?#%V

#%V

:V

: V

: V

9 % V

# $ V

# $ V

cgc

cgs

Tri"( Se$tion

0#

05

09

0# G :x#% G %:

05 G :x#% G %:09 G :x5$ G ##50c G 5:2 in5

t

b

=# G D#I#5Ebt9 H 0#x#%5 G #%x:9I#5H%:x#%5G #%:%1.99=5 G D#I#5Ebt9 H 05x#%5 G #%x:9I#5H%:x#%5G #%:%1.99=9 G D#I#5EbJDh-5tE9 G :x5$9I#5 G >9#>.99= G :2566.11 in:

tG b G r5Ic G D =I0EIc GD:2566.11I5:2EI#$ G 1.95V

# % V

# % V

Se$tion Pro+erties

3atio of 8I8 T G :2I952 G 2.#56, small ratio

Bocate c.g.s?

e#He5 G8 H f tJ0b

Fo

:2x#5H2.9x5:2x1.95 #1%.15

G G 6.$: in

'ence c.g.s. can be located 6.$:V below the bottom ern or 5.$: above the bottom!ber.+ompute the net moment to be carried by the prestress?

.P.0ncog92



8 T - f bJ0t G 952x#5 - .5:x5:2x1.95 G 9929.#> -in

The lever arm a G t H e G 1.95 H D 1.95H6.$: E G 5:.:$ in. and the re"uiredprestress?

F G8 T - f bJ0t

aG G #9:.19

952x#5 - .5:x5:2x1.95 5:.5$

Fo G FDf soIf sE G #9:.19D#62I#56E G #%#.15 To limit the bottom !ber stress to #.$2 si, we need an 0c of?

0cGFoh

f bct - f tJcb

#%#.15x9%#.$x#$ - 2.9x#$

G G 5#6.$1 in5

To eep the top !bers to #.%2 si, we need an 0c of?

0cGFh

f tcb- f bJct

#9:.19x9%#.%x#$ - 2.5:x#$

G G #1$.:9 in5

0c G 5#6.$1 in5 controls and indicates that our trial section of 0c G 5:2 in5 N 5#6.$1in5 is satisfactory. Aur trial section may be reduced and a new section tried over again.

9. esign a prestress concrete T- beam with the following data?8 T G 952 -ft f t G #.%2 si8 G 5#2 -ft f b G #.$2 sif tJ G -.92 si f so G #62 sif bJ G -2.5: si f s G #56 si

So(,tion:

Preliminary esign? h G T M @ D G #.6 - 5.2E , for G 5 , h G 96.>>, try hG 9%V

+ompute the re"uired prestress? F G 8 TID.%6hE F G 952x#5ID.%6x9%E G #%:.#2

Fo G FDf soIf sE G #%:.#D#62I#56E G #1%.15 0rea of tendon? 0s G FIf s G 8 TID.%6hf sE G #%:.#2I#56 G #.9#5$ in5 3e"uired concrete area 0c G 0sf sID.6f cE G FID.6f cE G #%:ID.6x#.%2E G 526 in5 Trial section?

#$V

: V

:V

9 % V

cgc

cgs

Tri"( Se$tion

0#

05

0# G :x#$ G >5 y# G 5.2 0#y# G #::.2205 G :x95 G #5$ y5 G52.2 05y5 G 56%2.220c G 522 in5 Σ0iyi G5>2:.22yc G ctG DΣ0iyi EI0c G #9.65 in, cb G 55.:$ in.d# G ##.65in, d5 G %.:$in

t

b

=# G D#I#5Ebt9 H 0#xd#5 G #$x:9I#5H>5x##.655G 1%6#.#6

=5 G D#I#5Ebt9 H 05xd55 G :x959I#5H#5$x%.:$5G #%51>.:9= G 561:$.6$ in:

tG r5Icb G D =I0EIcb GD561:$.6$I52EI55.:$ G 6.>>VbG r5Ict G D =I0EIct GD561:$.6$I52EI#9.65 G 1.61V

Se$tion Pro+erties

d #

d 5

y#G 5

y 5 G 5

2

c b

c t

y c

3atio of 8I8 T G 5#2I952 G 2.%6%56, large ratio

.P.0ncog9#



Bocate c.g.s?

e - b G8

Fo

5#2x#5 #1%.15

G G #5.>1 in.

e G #5.>1 H b G #5.>1 H 1.61 G 55.9$ in

The eccentricity e G 55.9$ in. is almost e"ual to cb which is not feasible. We !x thelocation of c.g.s. 9 in. above bottom !ber@ then no tension will be in top !ber. The newvalue of e? e G cb - 9 G #1.:$ in.and the lever arm a? a G e H t G #1.:$H 6.>> G 56.56 in.+ompute the amount of prestress re"uired?

F G8 T - f bJ0t

aG G #:#.##

952x#5 - .5:x522x6.>> 56.56

Fo G FDf soIf sE G #:#.##D#62I#56E G #%1.99

To eep the bottom !ber stress within limits, we compute the re"uired concrete areafrom?

Fo

f b

e - D8IFoE t

# HD E0c G

G#%1.99 #.$2

#1.:$ - 5#2x#5I#%1.99 6.>>

D# H E G #%1.29 in5

To eep the top !ber stress within limits, we compute the re"uired concrete area from?

0cGFh

f tcb- f bJct

#:#.##x9%#.%x55.:$ - 2.5:x#9.65

G G #66.5: in5

0c G #%1.29 in5 controls and the area furnished from our trial section is 522 in 5 which

may be reduced if desired.

o"& )"("n$ing Pro'(e:

:. 0 double cantilever beam is to be designed so that its prestress will exactly balancethe total uniform load of 59.6 7Im on the beam. esign the beam using the leastamount of prestress assuming that the c.g.s. must have a concrete protection of >6mm. =f a concentrated load P G %6 7 is applied at the midspan, compute themaximum top and bottom !ber stresses.

.P.0ncog95



922

> 6 2

%m %m#6m

w G 59.6 7Im

P G %6 7

;eam)ection

So(,tion:

=n order to balance the load in the cantilever, the c.g.s at tip must coincide withthe c.g.c. with a hori/ontal tangent.

To use the least amount of prestress, the eccentricity over the support should bea maximum. h G >62I5->6 G 922 mm or 2.92 m

The prestress re"uired?

Fh GwB5

5

F GwB5

5hG

59.6x%5

5x.92G #:#2 7

=n order to balance the load at midspan, using the same prestress F G #:#2 7, thesag of the parabola h# must be?

Fh# GwB5

$

h#GwB5

$FG

59.6x#65

$x.#:#2G 2.:%$ m

4nder the combined action of the uniform load and prestress, the beam will have nodeection anywhere and the uniform compressive stress?

F0c

f c G G#:#2x#29

D922x>62EG %.5%> 8Pa

ue to P G %6 7?

8 GPB :

G%6x#6 :

G 5:9.>6 7-m

The extreme !ber stresses?

.P.0ncog99



Hf G

#:#2x#29 %x5:9.>6x#2% D922x>62E 922x>625

G H

G %.5%> $.%%>G #:.19: 8Pa for top !berG -5.:2 8Pa for bottom !ber

F0c

%8bh5

H

%m %m#6m

hG922

>6

h# G :%$

52>

hG922

C"'(e Pro9(e

S%e"r Design Pro'(e 6Wor5ing (o"& $on&ition7

6. 0 prestressed concrete beam has a rectangular section shown and is subjected to ashear of #62 under woring loads. The e<ective prestress in the tendons totals922 and is inclined at an angle of arc sin α G #I%. The !ber stress distributionunder woring load is 626 psi throughout. 'alf-inch 4-stirrups are to be used D 0v G2.:2 in5E, f sv G 52,222 psi.

α

ScG#22

SsG62

922

Portion o# )e"

#2V

% 2 V

9 2 V

Se$tion Fi'er stress S%e"r stress*n&er Wor5ing o"& Con&ition

626 562

So(,tion:

4nder woring load, the shear carried by the tendon is Ss G 922xsinα G 922D#I%E G 62 Sc G S - Ss G #62 - 62- G #22

+onsider the state of stress at centroidal axis?

.P.0ncog9:



ν G 9S 509D#22x#222E 5D#2x%2E

G 562 psiG

fI5 G 626I5 G 565.6 psi

+omputing the principal tensile stress?

)t G ν2 + D fI5 E5 - fI5

G D562E5 H D565.6E5 - 565.6

G #25.$5 psi

3e"uired stirrup spacing?

s G0vf sv )tbw

2.:2x52,222

#25.$5x#2GG >.>$ in c.c.

S%e"r Design Pro'(e 6*(ti"te (o"& $on&ition7

%. The unsymmetrical beam shown carries an e<ective prestress force of 55$ andsupports a superimposed dead load of 9:6 plf and service live load of 122 plf, inaddition to its own weight of 556 plf, on a 62 ft simple span. 0t the maximum-moment section, the e<ective depth to the main steel is 5:.6 in. D eccentricity ##.:

in. E. The wires are deected upward #6 ft from the support, and eccentricity isreduced linearly to /ero at the support. =f concrete having strength fcJ G 6,222 psiand stirrups with fy G :2,222 psi are used, and if prestressed wires have strengthfpu G 5>6 si, what is the re"uired stirrup spacing at a point #2 ft from the support

.P.0ncog96



#$V

6 V

6V 5

: 9 . 6 V

cgc

Se$tion

0c G 5:6 in5

0s G #.>6 in5

=c G 5:,522 in: ct G #9.# in, cb G #6.1 in.

d# G ##.65in, d5 G %.:$inr5 G =cI0c G 11 in5

Se$tion Pro+erties

c b G # 6 . 1

V

c t G # 9 . #

V

6 V

#5V

# 1 V

5 1 V

e G #

# . :

eG##.:e#2

#2J#6J

62J#6J

C"'(e Pro9(e

So(,tion:

0t a distance from the support, the tendon eccentricity is?

e#2 G ##.: x#2#6

G >.% in.

The e<ective depth d at #2 ft from the support, d G #9.#H>.% G 52.> in.0ccording to the 0+= +ode, the larger value of d G .$2h G .$x51 G 59.5 in will be

used.

+alculate Sci at #2 ft from support? The bottom !ber stress due to e<ective prestress acting alone is?

c5 G cb G #6.1 in

f 5pe GF0c

# He#2c5

r5D EG5$$,222 5:6

D# H>.%x#6.1 11

EG 5,%#2.9: psi

.P.0ncog9%



The moment and shear at the section due to beam self weight alone are,respectively?

8o,#2Gwox 5

D B - x E G2.566x#2 5

D 62 - #2 E G 6# -ft

So,#2G woD BI5 - x E G 2.566D 62I5 - #2 E G 9.$56

and the bottom !ber stress due to this load is?

f 5o G8o c5

=G

D6#,222x#5E#6.1 5:,522

G :25.#2 psi tension

+ompute the cracing moment, 8cr?

8cr G=cc5

D% f cJ H f 5pe - f 5o E G5:,522 #6.1

D % 6,222 H 5,%22 - :25.#2 E

G 91121%%.%1 in-lbs I #5 G 9956$2.6% ft-lbs

The ratio of the superimposed load shear to moment at the section is?

8dHl wxI5D B - x E

SdHl wD BI5 - x EG G

B - 5x

xD B - x EG 62 - 5x#2

#2D62 - #2EG 2.2>6

Then Sci can be computed from?

Sci G 2.% f cJ bwd H So HSdHl

8dHlO #.> f cJ bwd

G 2.% 6222 D6x59.5E H 9$56 H 2.2>6D9956$2.6% U I #222G 92.95

#.> f cJ bwd G #.> 6222 D6x59.5EI#222 G #9.1: O 92.95 , Sci o

+alculate Scw for?

θ G tan-#D##.:ID#6x#5EE G 9.%59$o Sp G Fsinθ G 5$$xsin 9.%59$o G #$.52

+oncrete compressive stress at the centroid? f pc G FI0c G 5$$x#222I5:6 G ##>6.6# psi

Thus Scw is?

Scw G D9.6 f cJ H 2.9f pcEbwd H Sp

G D 9.6 6,222 H 2.9x##>6.6# ED6x59.5E H #$522 U I #222$>.$5 G

The shear force Sc is? Sc G Sci , ScwUmin G 92.95 , $>.$5 Umin G 92.95

The factored shear force Su at the section?

Su G #.:wdD BI5 - x E H #.>w lD BI5 - x E G #.:D.566H.9:6ED62I5 -#2 E H #.>D .122ED62I5 - #2E G 96.66 4sing K9 4-stirrups, 0v G 5x.##G.55 in5, and the re"uired spacing s?

.P.0ncog9>



s Gφ0vf vydSu - φSc

G .$6x.55x:2,222x59.596.66x#29 - .$6x92.95x#29

G #>.>6 in

+hec for maximum spacing code re"uirements?

0v G 62 bws

f vy

2.55 G 62 6D s E:2,222

s G 96.5 in

0v G0p f pu s d

$2 f vy d bw

2.55 G#.>6 5>6 s 59.5 $2 :2 59.5 6

s G #6.>6 in s G #6.>6 in governs.

+omparison of stirrups spacing based on conservative approach.

8u wuxI5D B - x E

Su wuD BI5 - x EG G

B - 5xxD B - x E

G 92I:22 G2.2>6

Sc G D2.% f cJ H >22 xSud 8u

O 5.2 f cJ bwd

G 2.% 6222 H >22 x 2.2>6D 59.5I#5E U D6x59.5E I #222G #%.%1

5.2 f cJ bwd G 5.2 6222 D6x59.5EI#222 G #%.:2 O #%.%1 o

E bwdN 6.2 f cJ bwd

6.2 f cJ bwd G 6.2 6222 D6x59.5EI#222 G :#.2# N #%.%1 o

3e"uired spacing s?

s Gφ0vf vydSu - φSc

G .$6x.55x:2,222x59.596.66x#29 - .$6x#%.%1x#29

G $.#5 in

The simpler but conservative approach yielded more stirrupsX

*(ti"te Design Pro'(e

>. esign a prestress unsymmetrical -beam to carry a total dead load and live loadmoment of 952 -ft having an e"uivalent ultimate moment of %:2 -ft. 4se rade562 with QV dia. strand, f ps G 552 psi, carrying an e<ective prestress of f pe G #:9

si. +oncrete strength is f cJ G :,222 psi. and consider a bonded beam.

So(,tion:

Pre(iin"r3 &esign

Trial h G T M G 5.2v D952E G 96.>> in say 9% in.

3e"uired area of tendon?.P.0ncog9$



0s G8u

φD.$2hf psEG

%:2x#52.1D.$2x9% E552

G #.9:% in5

The concrete compression area?

0cJ G8u

φD.$2hx.$6f cJEG

%:2x#52.1D.$2x9% ED.$6x:.2E

G $>.#:% in5

Tri"( Se$tion #or Fin"( Design:#$V

: V

6V

Se$tion

6 V

#5V

5 > V

9 % V

: V

d G 9

5 V

7 0

c

b G #$ inbJ G 6 int G : ind G 95 in0s G #.9:% in5 f cJ G :.2 sif ps G 552 siβ# G 2.$6ρ G 0sIDbdE G 2.22599%$

ω G ρ f psIf cJ G 2.#5$65c G #.#$ωdIβ#G 6.>21 in N t G : in7.0. is within the web0sf G D.$6f cJDb-bJEtEI f ps G 2.$29% in5

0s - 0sf G 2.6:5: in5 7ew value of ρ ?ρ G D0s - 0sf EIDbJdE G 2.2299$>6

0rea provided by compression side? 0cJ G #$x: H 6x#.>21 G $2.6:6 in5 8oment capacity of the section?

H 2.$6f cJD b-bJ EtD d-.62t E8u G φD 0s-0sf Ef psdD # - 2.61ωE

G 2.122.6:5:x552x95D # - .61x.2299$>6E H.$6x:.2D #$-6ED:ED95 - :I5EUG $65#.%# -in I #5 G >#2.#9 -ft N %:2 -ft , o

8ore exact solution?

*rom? + G T

.$6f cJabJ H 0sf f ps G D 0s - 0sf Ef ps

a Gf psD 0s - 0sf E - 0sf U .$6f cJbJ

Bet D 0s - 0sf E G 0s#then

a Gf ps 0s# - 0sf U .$6f cJbJ

G552D 0s# - .$29%E 2.$6x:.2x6

G 5#.6%$0s# - #>.9956

*rom?

.P.0ncog91



8u G φ[D 0s#f psD d - aI5 E H 0sf f psD d - tI5 EU

%:2x#5 G 2.1 0s#x552D 95 - D5#.6%$0s# - #>.9956EI5E H 2.$29%x552D 95 - :I5EU$699.99 G >2:2 0s#- 59>5.:$ 0s#

5 H #12%.6>60s# H 6929.>%9551.6> G -59>5.:$ 0s#

5H $1:%.6>60s#

0s#G 9.>>21$ D-9.>>21$E5 - :D#.9%#5%E 5

H

G D9.>>21$ 5.1%59E I 5G 2.:2:99 in5

H

0s#5 - 9.>>21$0s# H #.9%#5% G 2

The re"uired area of prestress 0s?

0s G 0s# H 0sf

G 2.:2:99 H 2.$29% G #.52>19 in5

Which is very near to the preliminary value of 0s G #.9:% in5 The trial section will beadopted with the amount of 0s G #.52>195.

The web is of course to be designed by shear. ANCHORAGE ONE DESIGN

)palling /ones ;ursting /ones

Tensi(e stresses "t t%e s+"((ing "n&',rsting ones

b

h

I&e"(i"tion o# t%e "n$%or"ge one #or"n"(3sis

3ational design for the reinforcement for end /ones must recogni/e thathori/ontal cracing is liely. =f ade"uate reinforcement is provided, so that the cracsare restricted to a few inches in length and to 2.2# in. or less in width, these cracswill not be detrimental to the performance of the beam either at service load or at theoverload stage. =t should be noted that end-/one stresses in prestressed concrete

beams do not increase in proportion to loads. The failure stress fps in the tendon atbeam failure is attained only at the maximum moment section.

For +retensione& e'ers , a very simple e"uation has been proposed forend-/one reinforcement?

0t G 2.25#Fih f slt

where 0t G total cross-sectional area of stirrups necessary, in5 Fi G initial prestress force, lb.

.P.0ncog:2



h G total member depth, in. f s G allowable stress in stirrups, psi

lt G transfer length, in.0n allowable stress of 52,222 psi has been found in tests to produce acceptably smallcrac widths.. The re"uired reinforcement having total area 0t should be distributedover a length e"ual to hI6 measured from the end face of the beam, and for most

ecient crac control the !rst stirrup should be placed as close to the end face aspractical.

For +ost-tensione& e'ers, end-/one reinforcement is often designed onthe basis of an e"uilibrium analysis of the craced anchorage /one. 0s shown in the!gure, the end region of a post-tensioned beam with an initial prestress of Fi appliedat an eccentricity e. 0t some distance l from the end, the compressive stressdistribution is linear and the forces and stresses acting on the free body are ine"uilibrium.

Fi g

y

b

h

l2

#

5

9 Fi

bhD #H %

eh

E

Fi

bhD #- %

eh

E

Fi g

e #

2Fi

bhD #H %

eh

E

y

5

9l

For$es on #ree 'o&3

T +

S

En& o# 'e" s%o0ing #ree 'o&3 (o$"tion

0ny longitudinal section through the anchorage /one at a distance y from thebottom !ber is subjected to a bending moment, which can be determined from theforces that are acting on the ends of the bloc.

The bending moment may be determined for any longitudinal section as follows.*or the end bloc shown, two cases can be considered?

#. *or y O g?

.P.0ncog:#



8 G Fiey hD E5

9

- 9H h5e

D EyhD E

5

8Fie

Gyh

D E5

h5e

yh

5 - D 9H E in dimensionless form and considersclocwise moment as positive.

5. *or y N g?

8Fie

Gyh

D E9 h

5eyh

5

D 9H E in dimensionless form and considersclocwise moment as positive.

5 - D E h e

yh

H h5e

D# - E

=n practical cases, moments can be calculated at increments of height, startingfrom at the bottom of the beam, and plotted as a function of distance from thebottom. *rom this analysis, the total re"uired area of steel reinforcement can be found?

0t G 8max

f sD h - x E

where? f s G the allowable stress in stirrups D usually taen as 52,222 psiE x G the distance in inches from the end face to the centroid of the steel

that are within hI5 from end0ll other terms are as already de!ned previously.

En& An$%or"ge Pro'(e

$. The !gure shows the end portion of a post-tensioned beam. The beam has aninterior -shaped section and solid rectangular end blocs. The arrangement of thepost-tensioning steel is as shown in the !gure. There are six cables anchored at theends, exerting %2 each on the beam. =t is re"uired to design the verticalreinforcement for the end bloc.

%2

%2

5:2

# 5 V

# 5 V

1 V

9 V

1 V

9 V

9 . 6

V

En& )(o$5 An$%or"ge one

eG#5 6V h G 9 % V

So(,tion:

8oment at various hori/ontal planes can be calculated as follows? Fi G 9%2, e G #5.6in., h G 9% in.

*or 2 O y O 9 in. ?.P.0ncog:5



8 G 9%2x#5.6y9%D E5

9

- 9H 9%5x#5.6

D E y9%D E

5

from which? 8 G - 2.#151y9 H #6.:#>y5 When 9 O y O 1 ? 8 G - 2.#151y9 H #6.:#>y5 - 5:2y H >52

When 1 O y O #5 ? 8 G - 2.#151y9 H #6.:#>y5 -922y H#5%2ad when y N #5 ?

8 G - 2.#151y9 H #6.:#>y5 - 9%2y H #1$2

The table below shows the moment at various longitudinal sections?

y,in.

8,-in

y,in.

8,-in

#.2 H#6.5 #1.2

-%#>.>

5.2 H%2.# 52.2

-61%.6

9.2 H#99.6 5#.2

-6%>.>

:.2 -6%.$ 552.

-695.9

6.2 -##$.> 59.2 -:1#.%

%.2 -52%.> 5:.2

-::%.%

>.2 -5>2.> 56.2

-91$.%

$.2 -9#5.# 5%.2

-9:$.>

1.2 -999.# 5>.2

-51$.#

#2.

2

-91#.5 5$.

2

-5:>.1

##.2

-:9#.9 51.2

-#11.5

#5.2

-:69.9 92.2

-#69.9

#9.2

-6#$.: 9#.2

-###.9

#:.2

-6%>.> 95.2

->:.9

.P.0ncog:9



#6.2

-%25.9 99.2

-:9.6

#%.2

-%59.6 9:.2

-52.#

#>.2

-%95.5 96.2

-6.5

#$.2 -%92.2 9%.2 2.2

We can assume that x G $ in., approximately hI:, and calculate the area of end-/onereinforcement?

0t G 8max

f sD h - x E

G %95.5x#22252,222D 9% - $ E

G #.#5$1 in5

4sing K: closed stirrups, 0v G 5x2.52 G 2.:2 in5 , total number needed?n G 0tI0v G #.#5$1I2.:2G 5.$5, use 9 closed stirrups.

etails of end /one reinforcement is shown below?

%2

%2

5:2

# 5 V

# 5 V

1 V

9 V

1 V

9 V

9 . 6

V

An$%or"ge one

eG#5 6V h G 9 % V

lGhG9%V

%Y6.6V G 99V

5V

xG$V

E"+(e Design o# Prestresse& Co+osite Se$tion:

The top ange of a composite section is given as a slab #22 mm thic and #.62m wide cast in place. esign a precast section with a total depth of 122 mm Dincludingthe slab thicnessE to carry the following moments? 8 T G :96 7-m, 8 G 66 7-m, 8P

G #96 7-m, 8+ G 922 7-m. 0llowable stresses are? f t G -##.22 8Pa, f b G -#5.:2 8Pa,

.P.0ncog::



1))

1))

1*))

))1))

31*,4

4-4,.

cgs

cgc

cgc/10),4

.2,.

3**

Q

inplace portion

Beam Trial Section

f tJ G H5.22 8Pa, f bJ G H #.#2 8Pa. =nitial prestress f so G #292 8Pa, e<ective prestressf se G $%2 8Pa.

So(,tion:

To assume the section, mae a preliminary design, assuming a lever arm of

2.%6h for the prestressing force in resisting the total moment.

kN x

h

M Q T *-+,043

%+))(.*,)

1)43*

.*,)

3

===

+ompute Fo?

kN

f

f QQ

se

so

o *00,-+)

-.)

1)3)*-+,043 ===

*or the inverted T-section, the concrete area re"uired can be approximate by?

23

)-,1)00314),12

1)*00,-+)*,1*,1 mm

x

f

Q A

b

c ===

*rom this preliminary section, setch a trial section and proceed to !nal design.

&rror? 3eference source not found

For t%e +re$"st +ortion4 t%e se$tion +ro+erties:

#22 x 966 G 96622 x 62 G #>>6222 >22 x #22 G >2222 x :62 G 9#622222

0c G #26622 995>6222 I #26622 G 9#6.:2 mm G cb 96622D#225I#5 H 5%6.:5E G 5.692x#21

.P.0ncog:6



>2222D>225I#5 H #9:.%5E G :.#5%x#21 = G %.%6%x#21 I #26622 G %9212.2:> G r5

t G r5Icb G %9212.2:> I 9#6.:2 G 522.29# mm b G r5Ict G %9212.2:> I :$:.%2 G #92.#$1 mm

For t%e $o+osite se$tion4 t%e se$tion +ro+erties:

#22 x #622 G #62222 x 62.2 G >622222 #26622 x 6$:.% G %#%>6922

0c G 566622 %1#>6922 I 566622 G 5>2.>: mm Z #22 G#>2.>: G ctJ

#62222D#225I#5 H 552.>:5E G >.:991x#21 #26622D2.22 H 9#9.$%5E G #2.915%x#21 = of precast G M%.%6%2x#21

=J G 5:.:$56x#21

*424,)2.,.2+1)4-2*,24

4,31*1).*.,.

''

)+*0-,)04,10)1)4-2*,24

.,4-41).*.,.

''

+

+

+

+

===

===

x

x

c I

c I m

x

x

c I

c I m

b

b

b

t

t

t

Ste+ 1: o$"tion o# $gs.

mm x

x

Qc

I f e

ot

t -4*,3)%1)*00,-+)(.,4-4

%1).*.,.(),2'3

+

1 ===

mmk eee

mm x

xQ

M e

b

o

G

0+1,2221-+,13)0*0,.1-4*,3)

0*0,.11)*00,-+)

1)**

21

3

.

2

=++=++=

===

The cgs can be located at?

fiber bottom frommmeccgs b .)+,+20+1,2224),31* =−=−=

Ste+ 2: Co+,te t4 '.

*424,)2.,.2+1)4-2*,24

4,31*1).*.,.

''

)+*0-,)04,10)1)4-2*,24

.,4-41).*.,.

''

+

+

+

+

===

===

x

x

c I

c I m

x

x

c I

c I m

b

b

b

t

t

t

Ste+ /: Co+,te t%e re,ire& .

.P.0ncog:%



kN f

f QQ

kN x x x

k e

Ak f M m M Q

se

so

o

t

ct bC b P

*.,000-.)

1)3)224,.4+

224,.4+1))31,2))0+1,222

%1)**))%()31,2))(1),1%1)3))(*424,)1)13*

'

3..

===

=+

−+=

+−+

=

−

*or Fo G >>>.6% 7 instead of $12.6>> 7, revise e# and e5 as follows?

mmk eee

mmQ

Qee

mmQ

Qee

b

newo

previo uso

previo us

newo

previo uso

previo us

2*,23.1-+,13)033,0)32-,3*

033,0)*.,000

*00,-+)0*0,.1

32-,3**.,000

*00,-+)-4*,3)

21

22

11

=++=++=

===

===

Which indicates that cgs can be located at?

fiber bottombovemmeccgs b 1*,0+2*,23.4,31* =−=−=

3evise F as?

kN f

f QQ

kN x x x

k e

Ak f M m M Q

se

so

o

t

ct bC b P

*02,0*3-.)

1)3)1+.,.2+

1+.,.2+1))31,2))2*,23.

%1)**))%()31,2))(1),1%1)3))(*424,)1)13*

'

3..

===

=+

−+=

+−+

=

−

Ste+ : To 5ee+ 'otto 9'er 0it%in "((o0"'(e stress # '.

2

.3

3 .0,11)303)31,2))

1)**%2*,23.(1)*02,0*31)*02,0*34),12

1

1

mm x x x

k

M eQQ

f A

t

Go

o

b

c

=

−+=

−+=

To eep top !ber within allowable f t.

.P.0ncog:>



2

3..3

2-,.0034

1-+,13)

%2*,23.(1)1+.,.2+%1)3))()+*0-,)1)13*1)1+.,.2+

)),11

1

1

mm

x x x x

k

Qe M m M Q

f A

b

C t P

t

c

=

−++=

−++=

0c G ##29>9.%> mm5 controls and compare to 0c of trial section G #26622.22 mm5, alittle di<erence but considered acceptable and therefore there is no need to revise thetrial section.

S%"+es o# Con$rete Se$tions

The simplest form is the rectangular shape possessed by all solid slabs and useof some short span beam. 0s far as formwor is concerned, the rectangular beam isthe most economical. ;ut the ern distances are small, and the available lever arm forthe steel is limited. +oncrete near the centroidal axis and on the tension side is note<ective in resisting moment, especially at the ultimate stage of loading.

Ather shapes are fre"uently used for prestressed concrete?

#. The symmetrical =-section5. The unsymmetrical =-section

9. The T-section:. The inverted T-section6. The box section

The suitability of these section will depend on the particular re"uirements. The=-section is has its concrete concentrated near the extreme !bers where it can moste<ectively furnish the compressive force, both at transfer of prestress and underworing and ultimate loads. The more the concrete is concentrated near the extreme!bers, the greater the will be the ern distances and the greater will be the lever armfurnished for the internal resisting couple. 'owever, this principle of concentrating theconcrete in the extreme !bers cannot be carried too far, because the width and

thicness of the anges are governed by practical considerations, and for web musthave a minimum thicness to carry the shear, to avoid bucling and to permit properplacement of concrete.

=f the 8I8 T ratio is suciently large, there is the danger of over-stressing theanges at transfer, and the concrete in the bottom ange can be accordinglydiminished. This will result in an unsymmetrical =-section, which when carried to thefullest extent becomes a T-section. 0 T-section is similar to that for a reinforced beams,is often most economical, since the concrete is concentrated at the top ange where it

.P.0ncog:$



ectangular SectionS!mmetrical Section ns!mmetrical Sections

and nverted Sections 5o& Sections

is most e<ective in supplying the compressive force. =t may not be economically used,however, where the 8I8 T ratio is small, because the center of pressure at transfermay lie below the ern point. Then tensile stresses may result in the top ange andhigh compressive stress in the bottom section.

The unsymmetrical =-section with a bigger bottom ange, lie a rail section is not

an economical one in carrying ultimate moment, since there is relatively little concreteon the compression ange. 'owever there is a great deal of material to resist initialprestress. =t can be economically used for certain composite sections, where thetension ange is precast and the compression ange is poured in place. This sectionre"uires very little girder moment to bring the center of pressure within the ern andhence is suitable when the 8I8 T ratio is small. When carried to the extreme, thissection becomes an inverted T-beam.

The box section has the same properties of =-section in resisting moments. =7fact, their section properties are identical.

=n summary?

#. *or economy in steel and concrete, it is best to put the concrete near theextreme !bers of the compression ange.

5. When 8I8 T ratio is small, more concrete near the tension ange may benecessary.

9. When 8I8 T ratio is large, there is little danger of overstressing at transfer,and concrete in the tension ange is re"uired only to house the tendonsproperly.

:. =f formwor is used only once, it may constitute a major cost of the beam,so that irregular shapes for the purpose of saving concrete or steel maynot be in the interest of overall economy.

6. When the forms can be reused repeatedly, more complicated shapes maybe justi!ed.

%. *or plants producing precast elements, it is often economical to constructforms that can be easily modi!ed to suit di<erent spans and depths

.P.0ncog:1



lo6er 7ern

ension

8igh

compression

C

M9

Section Stress Distribution

lo6er 7ern

C1

1

M9

1

Section

C

M9 + MS

:artial :restress

1 for M9 ;nl!

otal :restress

for M9 + MS

:retensioning in t6o stages to 7eep C 6ithin 7ern,

&rror? 3eference source not foundArr"ngeent o# Stee( The arrangement of steel is governed by a basic principle? in order to obtain the

maximum lever arm for the internal resisting moment, it must be placed as near the

tensile edge as possible. =f the cgs is very near the tensile edge, and if there is nosucient girder moment to bring the center of pressure near or within the ern, the

tension ange may be overcompressed at transfer while the compression ange may

be under high tensile stress.


0 heavy moment is desirable at transfer so that the steel can be placed as near the edge as possible. 'owever, noeconomy is achieved by adding unnecessary dead load weight to the structure in order to enable a bigger lever arm for the steel,

because whatever additional moment capacity was thus obtained would be used in carrying the additional dead load, althoughsome additional reserve capacity is obtained at ultimate range. Bad that can be eventually carried by the beam can be moreeconomically put on the structure before transfer rather than after, because moments produced by such loads will permit the

placement of steel nearer the tensile edge.

0nother method sometimes used in order to permit placement of steel near theedge is to prestress the structure in two or more stages@ this is nown as retensioning.0t the !rst stage, when the moment on the beam is small, only a portion of theprestress will be applied@ the total prestress will be applied only when additional deadload is placed on the beam producing heavier moment on the section. thus the centerof pressure can be ept within the ern at all times, and the excessive tension in thecompression ange, as well as high compression in the tension ange can be avoided.

.P.0ncog62



top 7ern C

Steel in 5oth <langes

C

:restressing steel in both flanges reduces lever arm for resisting moment,

Small

armcgs

cgs

top 7ern

5igarm

Steel in ension <lange ;nl!

2)=

2,*= 2,*=

1,)=

1,2*=

2= fillet,.)=

1),4)=

cgc

>!nacore Section

-,))/


*or certain sections, the tendons are placed in the compression ange as well asin the tension ange. enerally speaing, this is not an economical arrangement,because it will move the cgs nearer to the cgc and thereby decrease the resisting leverarm. 0t the ultimate range, tendons in the compressive ange will neutrali/e some of its compressive capacity, whereas only those in the tension ange are e<ective in

resisting moment. 'owever, under certain circumstances it may be necessary to puttendons in both anges in spite of the resulting disadvantages. These conditions are?

#. When the member is to be subject to loads producing both H8 and Z8 inthe section.

5. When the member might be subject to unexpected moments of oppositesign, during handling process.

9. When the 8I8T ratio is small and the tendons cannot be suitably groupnear the ern point. Then the tendons will be placed in both the tensionand the compression anges with the resulting cgs lying near the ern.

Design E"+(e4 Pre-tensione& Core& Se$tion

0 pre-tensioned cored section is to be used for roof construction carrying anadded dead load of #2 psf and a design live load of 92 psf on a simple span of >2 ft. =tis made of lightweight concrete at ##2 pcf, precast in factory with a transfer strengthof f ciJ G :222 psi and a minimum 5$-day strength of f cJ G 6222 psi, &c G 5.6x#2% psi.esign the tendons, using >I#% in. >-wire strands with 0s G 2.#2$1 in5 per strand,f sJGf su G 562,222 psi, f i G #>6,222 psi, f t G #%6,222 psi, and f e G #:6,222 psi, &s G5>x#2% psi. +hec the loss of prestress, both immediately at transfer and eventuallysay at the end of 9 years. *ollow P+= ;uilding +ode re"uirements when applicable.

.P.0ncog6#




Co+,te #or 'en&ing oents:

lb ft M M M

lb ft M M M

lb ft x x !w

M

lb ft x x !w

M

lb ft x !w

M

! "T

S G "

!

!

S

S

G

G

−−+=+=−=+=+=

−===

−===

−===

)))?41*)))?140)))?2.-

)))?2.-)))?4+)))?21+

)))?140-

0)%3)-(

-

)))?4+-

0)%1)-(

-

)))?21+-

0)3*-

-

22

22

22

E("sti$ Design Met%o&:

Co+,te to+ "n& 'otto 5erns.

inc

r k

inc

r k

in A

I r

t

b

b

t

c

c

++,*.,+

*,*0

*3,*4,1)

*,*0

*,*040)

201))

2

2

22

===

===

===

rtioel M

M

T

G arg*200,))))?41*

)))?21+==

)ince the girder load is relatively heavy, the cgs can be polaced as low as possible.

0ssuming a clear concrete protection of #.6 in for !re resistance and a protection ofcgs of 5.: in, the total available from cgs to top ern?

h G cb Z cover H t G #2.: Z 5.: H 6.69 G #9.69 in.

This will give a resisting moment 83 G *eh up to /ero tension in bottom !ber.

0llowing a maximum tension of psi f c 423*))).'. ==

P+= +ode, we have theresisting moment from /ero tension to :59 psi tension?

.P.0ncog65



lb ft x x

c

fI M −=== .))?+1

12

1

4,1)

201))423

This value in moment of 1#,%22 ft-lb may be found to be too high since its constitutes55C of the total moment of :#6,222 ft-lb. =f fully utili/ed in design, it will appreciablyreduce the re"uired amount of prestressing steel and hence the ultimate moment.

*urthermore, we will be using a great deal of tension in the bottom ange concrete,which once craced, may not be able to resist that tension. 'ence it would be well touse perhaps only half of that value in our !rst attempt to determine the re"uiredprestress@ thus?

22,2)))?14*

)))?31-

)))?31-

2

.))?+1)))?41*

2

1

12

in f

# A

lbs #

M M # h

e

e

s

e

T e

===

=

−=−=

7o. of 1.

0

in strands re"uired G 5.52I0o G 5.52I2.#2$1 G 52.5 pcs. 4sing 52 strands,we can chec the stresses at transfer and under total load as follows.

*t G 52D2.#2$1ED#%6,222E G 961,222 lb*e G 52D2.#2$1ED#:6,222E G 9#%,222 lbe G cb G cover G #2.: Z 5.: G $.22 in.

)tresses at transfer?

201))

%4,1).,+(1221+)))

201))

%4,1).,+%(-(3*+)))

40)

3*+))) or xor

I

c M

I

ec #

A

# f

c

G

c

t

c

t

±±=

±±=

Top !ber?

f T G >%9.$51 Z #2#>.9$> H 192.165 G %>>.91: psi compression

;ottom !ber?

f ; G >%9.$51 H ##25.#%1 Z #22$.69# G $6>.:%> psi compression

These stresses indicate a near rectangular stress bloc, and a small amount of camberunder the girderJs own weight, which is usually desirable.

4nder total design load?

.P.0ncog69



Strand and cgs location

. *@-=

cgs

2 3@-=1 1@2= clear

4 A0@1.= $1 B=

4=

3 A1 3@4= $* 1@4=

cgs ocation Strand ocation

201))

%4,1).,+(1241*)))

201))

%4,1).,+%(-(31.)))

40)

31.))) or xor

I

c M

I

ec #

A

# f

c

T

c

e

c

e

±±=

±±=

Top !ber?

f T G %>5.9:2 Z $16.65> H #>%:.#95 G #6:2.$:6 psi compression

;ottom !ber?

f ; G %>5.9:2 H 1>2.#6: Z #1##.#:9 G -5%$.%:1 psi tension

The top !ber compression is well within the allowable 2.:6fcJ G 5562 psi, and thebottom !ber tension is also less than the permissible :59 psi@ this is expected whenchoosing 52 Z >I#% in. strands.

To locate the strands at the end of the section, we assume that the strands arespaced at a minimum of # [ in c.c. epending on the hardware available and thepractice at the plant, this spacing may vary considerably. *urther we will assume thatone point harping is preferred for simplicity of production Dactually it fre"uentlyhappens that 5-point or 9-poinrt harping may be used, then the pro!le may be easilyarranged both for camber control and for stress control along the entire length of thespanE, then we can mae a tentative arrangement. =7 order to produce no tension inthe top, the cgs at the ends should be within the ern. =n order to produce arectangular stress bloc at the ends, the cgs should be at the cgc. =n order not to havethe intermediate points control the design cgs should be located within a limiting /one.

The cgs at the end should also be located so as to produce the best camber e<ect. Taing all this into consideration and using : strands for each web, a tentativearrangement is shown below.


.P.0ncog6:



The stress at the end can only be critical at transfer and will be computed asfollows, assuming no external moment.

fiber bottomncompressio psi

fiber topncompressio psi

or

I

ec #

A

# f

c

t

c

t

?22.,12-33+0,*1+-2+,0.3

?3-.,2-4443,40+-2+,0.3

201))

%4,1).,+%(.3,.4,1)(3*+)))

40)

3*+)))

=+==−=

−±=

±=

These stresses are well within the allowable range. 'ence the design isconsidered satisfactory so far as the exural elastic stresses are considered.

*(ti"te Strengt%:

+ompute ultimate moment, P+= +ode?

8u# G #.$D H BE G #.$D:#6,222E G >:>,222 ft-lb8u5 G #.5 H 5.:B G #.5D5%$,222E H 5.:D#:>,222E G %>6,222 ft-lb8u G 8u#, 8u5Umax G >:>222, %>6222Umax G >:>,222 ft-lb The expected actual ultimate resisting moment strength of the section may be

estimated as follows?

212-)))?*-*,

)))?*4*

'-*,'

)))?*4*%)))?2*)(1)-+,)2)'

in x f

M A

lb x f NAT

c

u

c

suo

===

===

*or width b G $ft or 1% in and thicness of #.56in, ange area G #52 in 5 with only$ in5 to be additionally furnished by web !llets. 'ence the center of compression canbe assumed at tI5 G #.56I5 G 2.%9 in. the lever arm at ultimate is approximatelycomputed as?

/ G h Z cover Z tI5 G 52 Z 5.: Z 2.%9 G #%.1> in.

lb ft lb ft $ T M −>−=== )))?040)))?00)12

+0,1.)))?*4*''

The 52->I#% in. strands are just about right, while #1 strands would not have beensucient.

C"'er "n& Dee$tion&rror? 3eference source not found

+amber at transfer due to prestress only.

.P.0ncog66



&c G 5,622,222 psi*t G D52x2.#2$1ED#%6,222E G 961,222 lb8# G *ty# G 961,222D%.%9 Z 5.:E G #,652,222 in-lb85 G *ty5 G 961,222D#2.: Z %.%9E G #,966,222 in-lb

upwrd in x x x

x M M

%I

!)+,31*2))))

3

213**)))

201))1)*,2-

%120)(

3

2

- .

2

12

2

=

+=

+=∆

eection due to beam own weight.

downwrd in x x x

x M

%I

!

lb ft w!

M

g

g

-.,2%1221+)))(.

*

201))1)*,2-

%120)(

.

*

-

)))?21+-

.

22

2

=

=

=∆

−==

'ence the net camber at transfer, #-day old is? 9.21 Z 5.$% G 2.59 in, upward.

epending on the age of concrete at which the additional dead load is placed,the deection will vary. =f the dead load is added early, then after one year, theamount of deection or camber can be approximated by using an average value of &G 9,622,222 psi, for 9,222 psi O f cJ O %,222 psi and a creep factor of #.$2 for thee<ects of prestress and dead load. 4sing the e<ective prestress of f c G #:6,222 psi,the deection can be calculated as follows?

8 G 5%$,222 ft-lb*e G n0of se G D52x2.#2$1ED#:6,222E G 9#%,222 lb8# G *ey# G 9#%,222D%.%9 Z 5.:E G #,992,222 in-lb85 G *ey5 G 9#%,222D#2.: Z %.%9E G #,#12,222 in-lb

( )

( )

downwrd in

x x x x

x

creep M M M %I

! "

)1,1

-,1123.-))).

*133))))

3

211+))))

201))1)*,2-

%120)(

-,1.

*

3

2

-

.

2

12

2

−=

−+=

=

−+=∆

which indicates that, in the course of # year, the beam will have a maximumdownward deection of #.2# in., which probably will not be objectionable on a span of

>2 ft. ;ut it does indicate that either a more careful study of the camber and deectionhistory is desirable, or that some means to reduce the deection is needed, such asplacing the cgs at ends further below the present location.

The instantaneous deection due to live load may be estimated using a highervalue of &c G, say 9,$22,222 psi. Dimmediate deectionE.

.P.0ncog6%



excessiveconsidered not length spntheof x

iswhich

downwrd in x x x

x M

%I

!

lb ft M

!

!

?..0

1

120)

2.,1

2.,1%12140)))(.

*

201))1)-,3-

%120)(

.

*

-

)))?140

.

22

=

=

=

=∆

−=

Design #or S%e"r: )hear is critical only under total load, hence,

lb xw!

& T 0))?232

0)%24)-)3*-(

2=

++==

Sertical component from the tendons, for *e G 9#%,222 lb, is?

lb x !

' # # & ee s 32))

123*

4,2.3,.)))?31.

2tan 1 =

−

=

== α

7et shear, assuming the prestress is already transmitted to the concrete, is?

Sc G S T Z Ss G 59,222 Z 9,522 G 52,622 lb

+ompute the F-value for shear, use the top portion of section with respect to the cgc.

Top ange? #52 x $.1> G #2>6*illet? #% x >.%$ G #56Web? #2: x :.#$ G :96

F G #%99 in9

psi x Ib

Q&

v

c

++%*,2*(201))

%1.33(2)*))

===

*or average prestress?

psi A

#

c

e .0240)

31.)))==

Principal tension is?

psi f f

vS cc

t 142

.02

2

.02++

22

2

2

2

2 =−

+=−

+=

which is very low. 'owever, the end bloc stresses should be considered and a certainamount of web reinforcement used near the ends.

oss o# Prestress

&lastic shortening at transfer? &s G 5>x#2% psi, &c G 5.6x#2% psi.

.P.0ncog6>



psi x

xnf f

psi A

# f

ct s

c

t

ct

-2*)0.*1)*,2

1)20

0.*40)

)))?3*+

.

.

===∆

===

The loss at the level of tendons at midspan may be slightly higher since the !berstress in concrete is somewhat higher. This also indicates that the estimated loss of #2,222 psi from the initial prestress of #>6,222 psi to #%6,222 psi at transfer is fairlygood estimation.

Boss of prestress due to creep may be estimated at 5 times the elastic shortening,thus?

psi x f s *))?1.-2*)2 ==∆Boss of prestress due to shrinage may be estimated assuming a coecient of 2.2229,at

psi x % f s s s 1))?-%1)20()))3,) . ===∆ ε

Total loss of prestress, excluding loss or gain due to bending of members, is,

psi f st 32-*)-1))1.*))-2*) =++=∆

which is fairly close to the assumed loss of 92,222 psi in the problem statement.

Design E"+(e4 Post-tensione& )ri&ge Gir&er

Precast girder of a highway bridge are to be post-tensioned, grouted, then liftedto the bridge site to be connected together by concrete poured in placed, as shown inthe !gure below. The two-lane bridge is to carry '52-)#%-:: loading, and the girdersare spaced % ft on centers. Averall length of the girder is 1% ft, with 16 ft betweencenters of supports. 8aximum live load moment for one lane, 8max G #:99 -ft,maximum live load shear, Smax G %#.9 Ilane. The dead load due to 5V bituminouspaving is #62 plf and the in-place slab and diaphragm is #99 plf. The weight of thegirder is taen e"ual to 1:2 plf. *ollowing the 0))'A specs for 'ighway ;ridges, whenapplicable, design the interior girder as follows?

DaE esign the midspan section, indicating the re"uired amount of

prestressing steel.DbE esign the end section, showing the mild steel stirrups.DcE esign the longitudinal layout of girder showing the pro!le for the cgs and

the intermediate and end diaphragms.DdE =nvestigate the factor of safety of the girder for cracing and ultimate

strengths.DeE +ompute the deection of the girder at transfer and under the woring

load.DfE etail the midspan and the end sections using the *reyssinet system.

.P.0ncog6$



.= 1-,*/ 24,)/

4-,)/

8alf Dlevation

:recast diaphragm >iaphragm:recast diaphragm >iaphragm

2= bitumen surface

4,*/

.,)/ cc of girders

Midspan Section

*,)/

C

+ompute the loss of prestress due to friction and the initial prestress re"uired atthe jac. )trength of concrete is to be :,622 psi at 5$ days and :,222 psi at transfer. The high tensile steel used is to have a minimum ultimate tensile strength of f sJG562,222 psi and a minimum yield strength of f yJ G 522,222 psi at 2.5C plastic set. Thesteel stress at transfer will be #%6,222 psi and the e<ective prestress at 56,222 psiloss is #96,222 psi. &s G 5$,222,222 psi, &c G :,222,222 psi. 4se intermediate-grade

reinforcing bars for the mild-steel reinforcement, f yJ G :2,222 psi.

&rror? 3eference source not foundSo(,tion"7

8aximum live load for # lane? 8max G #:99 -ft

*or the interior girder spacing of % ft and a lane width of #2 ft? 1)ma&

S M M !! =

ft k M !! −== -.)1)

),.1433

=mpact factor?ok

! I ?3),)220,)

12*+*

*)

12*

*)<=

+=

+=

=mpact moment? ft k I M M !! I!

−=== 1+*%220,)(-.)%(

Total live load plus impact? ft k M M I! !! −=+=+ 1)**1+*-.)


.P.0ncog61



+ompute dead load moments?

0dded dead load?

;ituminous paving G #62 plf =n-place slab G #99 plf Total G 5$9 plf

Weight of girder G 1:2 plf

0dded dead load moment? ft k w! M " −=== 31++*%2-3(,

-

1

-

1 22

irder moment? ft k !w M GG −=== 1).)+*%+4)(,

-

1

-

1 22

Total moment? ft k M M M M M I! !! "GT −=+++= 2434

)ection properties of the trial section which resulted from preliminary design andtrials.

% x 6: G 95: x 9 G 1>5 9 x 9 G 1 x > G %99$ x $ G 92: x 56 G >%22 > x > G :$ x :#.%>G 52:255 x $ G #>% x :$ G $:62

$%5 #1#56 I $%5 G 55.6 in G ct

65 Z 55.6 G 51.6 in G cb The moment of inertia of the concrete section about cgc.

95:D%5I#5 H #1.65E G #5:2221D95I#$ H #6.65E G 552292:D9$5I#5 H 5.65E G 9$%22:1D>5I#$ H #1.#>5E G #$222#>%D$5I#5 H 56.65E G ##6222

51>$22 I $%5 G 9:6 G r5

.P.0ncog%2



t G r5Icb G 9:6 I 51.6 G ##.> in.b G r5Ict G 9:6 I 55.6 G #6.: in.

el reltivel' M

M

T

G arg?43*,)2434

1).)==

, the cgs can be located as low as

practicable without producing any tension in the top !ber.

0ssume cgs is located :V above the bottom !ber, compute the total arm for theinternal resisting moment.

iner ck btt 2,304*,2+0,11cov =−+=−+=

Total e<ective prestress re"uired?

k x

M # T 0-.

2,30

122434 ===

*or a loss of prestress ofE1*1*1*,

1.*)))

2*)))≈=

, the initial prestress re"uired?

k

# # o +2*

-*,

0-.

-*,===

00)'TA specs for post tensioned members?

0llowable stress at transfer? psi f f cb 22))%4)))(**,'**, ===

0llowable stress at !nal stage? psi f f d'st ct 1-))%4*))(4),'4), 2- ===

3e"uired concrete area, at !nal stage of loading?

200)*,2+-,1

*20-.in

x

x

c f

#h A

bt

c ===

3e"uired concrete area at transfer?

( ) ( ) 2

-410,11

+2*@121).)*,2*12,2

+2*1 in

x

k

# M e

f

# A

t

oG

b

o

c =

−

+=

−+=

0ctual gross area provided is 0g G $%5 in5, which seem s to be just aboutsucient for the re"uired area of 0c G $:# in5.

=t generally taes two or three trials to arrive at this adopted section rather than just one trial as illustrated here.

.P.0ncog%#



To supply the e<ective prestress of >$% at an allowable stress of #96 si, there"uired steel area is?

2-3,*13*

0-.in

f

# A

s

s ===

'7)hearing stresses can be checed for two sections, one at the support and

another at 6J from the support where the web is $V thic. 0t the support the web is 55Vthic@ shear is evidently not controlling. 0t 6J from the support?

SBB G %#.9 Ilane

SBB for girder G girder k

S & lne @-,3.

1)

.3,.1

1)==

=mpact ? k I & & !! I! 4,-%220,)(-,3.%( ===

;ituminous paving H in-place concrete? k w! ),12%*,42(2-3,2@ ==

irder weight?k !wG ),4)%*,42(+4),2@ ==

Total shear? S T G 9%.$ H $.: H #5.2 H :2.2 G 1>.5

etermine the shear carried by tendons assuming a parabolic pro!le with rise hcomputed as follows?

h G e G cb Z cover G 51.6 Z : G 56.6 in.

Provide some eccentricity at support such that h ≈ 5: in.D 5 ftE

k !

#h !

!

#h !w& s *,.*

+.

%2%(0-.(44

2

-

2 2 =====

0t 6.6 ft from end of girder?

),*-*,.*4-

*,42

2 ma& === &

!

x& s

'ence Sc of concrete?

k & c 2,3+),*-2,+0 =−=

+ompute the value of F?

95: x #1.6 G %9#$1 x #6.6 G #91.6$ x #%.65I5G #2$1

F G >6:%.6 in9

8aximum shearing stress occurs at the cgc?

.P.0ncog%5



psi x

Ib

Q& v c 124

%-(2+0-))

%0*4.(1)))2,3+===

+ompressive !ber stress at cgc?

psi

x

A

#

f c

c +12-.2

1)))0-.

===

The principal tensile stress is?

( )

psi

f f v s cct

2)4*.4*.124

22

22

22

=−+=

−+=

0llowable value of principal tension?

psi s psi f s t c 2)13*%4*))()3,)')3,) =>===

7o stirrups are needed under woring load.

=nvestigate the ultimate strength for shear, 00)'TA specs?

k

I ! "&

1+1%4,--,3.(*,2%),*2(*,1

%(*,2*,1'

=++=++=

)hear carried by the tendons?

SsJ G 6$.2

'ence?

k & & & su 133*-1+1'' =−=−=

)hear carried by concrete?

k xb(d f & cc 0.,321)%*2%(-0%(-%(4*))()2,')2,) 3 === −

)tirrups re"uired?

( ) ( ) 23

)20*3-,)%*2%(-0(4))))

1)0.,32133

2

1

'2

1in s

s x

(d f

s& & A

'

cu

v =−

=−

=

4sing Q in 4-stirrups, 0v G 2.:2 in5?

.P.0ncog%9



in s' A

s v *),14*2*,14)20*3-,

4),)

)20*3-,)===

)imilar computations can be made for other points along the girder. )o far asshear is concerned, more stirrups are re"uired near the ends than along the middleportion of the girder, but the reverse is true when considering the e<ect of combined

moment and shear. 'ence judgment should be exercised in the actual spacing of thestirrups.

8inimum web reinforcements, 00)'TA specs?

sb Av '))2*,)=

*or bJ G $ in, 0v G 2.:2 in5?

in A

s v 2)

%-())2*,)

4),)

%-())2*,)

===

*or the end section, stirrups are re"uired to distribute the anchorage stresses.)ince the anchorage are fairly uniformly distributed, the computed tensile stresses inthe anchorage /one will be low and analysis is not re"uired. 7ormal stirrups, however,are provided.

$70 half elevation of the girder is shown in the setch layout. The midspan section

is adopted for the entire girder, except 6 ft near the ends where a uniform webthicness e"ual to the bottom ange width of 55 in. is used in order to accommodatethe end anchorages, to permit the curving up of some tendons, and to distribute the

prestress. Three intermediate diaphragms are placed along the length of the span.)ometimes transverse prestressing is employed to bind the girders together. ;ut forthis design, transverse dowels are provided in these diaphragms to be joined togetherby in-placed concrete. The amount of steel is not excessive for these diaphragms@some nominal reinforcements are employed as shown on the setch.

The most common location of cgs for a simple span is a parabola with cgs nearthe cgc at the ends. )uch a pro!le will give ample moment resistance along the entirebeam. =f cgs is above the cgc at the ends, the tendons will carry greater shear but losesome of the reserve moment resistance. =f cgs is below the cgc at the ends, thetendons will carry less shear, but the positive prestressing moment at the ends will

tend to decrease the principal tension. 0lso note that the cgcJs for the midspan andthe end sections actually di<er slightly. *or this design, the cgs will be placed a littlebelow the cgc of the end section.

&7

Co+,t"tion o# $r"$5ing oent.

3esisting moment up to /ero stress in the bottom !ber?

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:recast >iaphragm:recast >iaphragm

4,*/

Section

1F1

3= fillet

1/1)=

-=

.=

3/2= 4/4=-=

!

ft k # M r −=== 243412

2,300-.

8odulus of rupture taen e"ual to psi f f c .3)%4*))(14,)'14,) ===

ft k xc

fI M

b

−=== *3)1)))12

1

*,2+

%2+0-))(.3)

Total resisting moment at cracing?

ft M M M )cr 2+.4*3)2434 =+=+=

Averall factor of safety against cracing?

22,12434

2+.4

,, === T

cr

M

M

S #

*actor of safety for live load and impact?

( )*),1

1+*-.)

%31+1).)(2+.4,, =

++−

=+

+−=

I! !!

"Gcr

M M

M M M S #

This indicates that the girder will begin to crac only when the live load plusimpact is increased by as much as 62C.

T%e ,(ti"te resisting oent $o+,t"tion:

0ssuming that the ultimate strength of steel to be developed?

k x A f T s s 14.)1)%-3,*(2*))))' 3 === −

0verage compressive stress in concrete 2.$6f cJ G 2.$6D:622E G 9.$5 si

.P.0ncog%6



2

3-2-2,3

14.)

'-*,

'-*,

in f

T

A

T A f

T C

c

c

===

==

% x 6: G 95:9 x 9 G 1$ x c G $c

Total G 999H$c G 9$5@ c G %.#56 in.y G % H c G #5.#56 in. from top D7. 0. locationE

The cg of compression /one?

95: x 9 G 1>5

1 x > G %9:1 x 1.2%56 G :::.2%56 Total G #:>1.2%56 I 9$5 G 9.$> in. cg below top !ber.

The ultimate lever arm?

,13,44-0,34*2cov incg er h $ =−−=−−=

The ultimate moment 8u?

ft k M M

ft k T$ M

uu

u

−===−===

24,4-32%1*,*3.+(+),

1*,*3.+12

13,44

14.)

φ

*actored moment?

ft k I! !! " M u −=+=++= 40).%1)**(*,2%130+(*,1%(*,2*,1

'ence the ultimate strength is considered sucient.e7Dee$tion Co+,t"tion

ft k M

ft k #' M

ft k #e M

" −=

−=======

1).)

1-*)%2(+2*

.3,11*12

*,1

+2*

2

11

=nstantaneous deection due to prestress, upward deection?

.P.0ncog%%



cgc

cgs

parabolic! $ 2/

e1 $1,*=

Dccentricit! of :restress

< $ 2* 7

1-*) 7/

Moment >ue to :restress

11*,.3 7/

parabolic

+1).) 7/

parabolic

Moment >ue M9

Diagram for Deflection Computation

in x

x

%I

! M

%I

! M p

0-,2%2+0-))(1)4

1)%12(+.

4-

%1-*)(*

-

.3,11*

4-

*

-

.

332

2

2

2

1

=

+=

+=δ

&rror? 3eference source not foundAwing to loss of prestress of ≈#6C?

,34*,20-,21.))))

13*)))in

f

f p

so

se

pe === δ δ

ownward deection due to 8?

,40,1%2+0-))(1)4

1)%12(+.

4-

%1).)(*

4-

*.

3322

in x

x

%I

! M GG =

==δ

The immediate deection at transfer?

,31,140,10-,2 inG pi =−=−= δ δ δ

0dded dead load deection due to in-place concrete?

.P.0ncog%>



1 2

34

* .

0-

1) 11

-=

1 2 34

* . 0-

1) 11

.=

.=

-=

-=

-=

-=

-=

24,2=

20,-=

4/ 4=

cgs

4,2=

cgs

2=3=

3=

End Section End Elevation Midspan Section

Cable Location For Freyssinet System

,44,)40,11).)

31+in

M

M G

G

"

" === δ δ

Bong term deection, assume creep coecient +c G 5.2, and loss of prestress?

[ ] upwrd inC c "G pe f ,?-0,))),2%44,)40,1(34*,2%( =+−=+−= δ δ δ δ

=nstantaneous downward deection due to design live load and impact,assuming a parabolic moment diagram?

,4.,140,11).)

1)**in

M

M G

G

I !

! === + δ δ

#7 The above design will be applicable to most prestressing systems now in use

today, although minor modi!cations may be desirable for certain cases. *or purpose of illustration, detailed arrangement of the tendons is shown for the *reyssinet system.4sing cables of #$ - 2.#1% in. wires, ## tendons are re"uired, 0o G 2.6:9 in5?

)teel area as re"uired? 0st G 6.$9 in5

)teel area provided? 0st G ##D2.6:9E G 6.1>9 in5

The midspan and end sections are drawn showing arrangement of tendons to

give the re"uired locations of cgs. +urving the tendons in both hori/ontal and verticalplanes is necessary to conform with the re"uired location of the cgs. =t is noted thatthat some deviation from the re"uired parabola is permissible, because it will nota<ect the strength of the girder. 0 recommended order for tensioning the cables isindicated as shown.

Boss for anchorage slip in *reyssinet cones may be assumed to average 2.5 in.,which if average throughout the entire length of 1% ft, indicates a loss of prestresse"ual to?

.P.0ncog%$



.= 1-,*/ 24,)/

4-,)/

8alf Dlevation

:recast diaphragm

2= bitumen surface

4,*/

.,)/ cc of girders

Midspan Section

C

2/ G 4=

2/ G )=

1/ G .=

1F1

cgs

:recast girder

Continuous 7e!

:recast diaphragm

1/ G 1)=

-=

3/2=

.=

4/4=

s!mmetrical-A.= -A12= r A 2)=

irder Layout*,)/

psi x x

% !

f s 4+))1)2-12+.

2,) . ===∆ δ

To estimate the frictional loss, assume a coecient of friction µ G 2.96 for*reyssinet cables in metal sheathing and a \ G 2.22#2 per ft for wobble e<ect.. the

average rate of change in direction for the parabolic cable is given by?

rdin !

h1.0,)

+.

%2(--===θ

The maximum frictional loss at the far end, if tensioned only from one end is?

E4,1*1*4,))+.,))*-,)%+.())1),)%1.0(,3*,E1))%(E

,%(

2

%(

12

or *! *!loss

*!

e # # Ave

e # #

*!

*!

=+=+=++=

+=

=+−

+−

µθ µθ

µθ

µϑ

µϑ

The controlling point is the midspan which has a loss e"ual to Q of total #6.:C G>.>C.

,3))?12%1.))))E(0,0%(E psi f loss f so f ===∆

=f the tendons are overstressed by #5,922 psi at the anchorages, the anchorageloss of :,122 psi be balanced. 4sing the *reyssinet jac, there is an additional loss at

the jac of about ( f ∆ G $,222 psi. Therefore the maximum initial stress at the jacshould be?

psi f f f f ( f soi 3))?1-)-)))123))1.)))) =++=∆+∆+=0ccording to 00)'TA specs, the temporary jacing stress?

psi f f s ( )))?2))%2*))))(-),)'-),) === The e<ective prestress should not exceed 2.%2f sJ G 2.%2D562222EG#62,222 psi,

or 2.$2f yJ G 2.$2D522222E G #%2,222 psi. 'ence the stress in the steel is considered

safe in all conditions.

Gir&er "3o,t

.P.0ncog%1



6/

Simple 5eam

6c/

Continuous 5eam

@ 2 $ )

a/

6/

C!

T!

<5> of 8alf Span

@ 2 $ )

a/C!

T!

<5> of 8alf Span

6c/

a/C!

T!

''-'

2

T !w =

''2-

' 2

T !wc =

Moment >iagramSimple Span

Moment >iagramContinuous Span

Load"Carrying Capacity #f

$ Simple Beam

Load"Carrying Capacity #f

" Continuous 5eam

Contin,o,s )e"sContin,it34 Pros "n& Cons

Bet?

TJ G ultimate tension developed in the tendonsaJ G lever armwJ G ultimate load produced by tendons

)imple beams?

.P.0ncog>2



2

2

''-'

''-

'

!

T w

T !w

=

=

+ontinuous beams?

2

2

''1.'

''2-

'

!

T w

T !w

c

c

=

=

The ultimate load wJ carried by the simple beam is controlled by the capacity of

the midspan section and cannot be increased by any change in the end eccentricities

of the cgs.

*or the continuous beam, there are two resisting moments, one at midspan andanother over the support. 'ence the load-carrying-capacity is de!nitely a<ected bythe position of the cgs over the intermediate support. wcJ G 5wJ, this means that twicethe load on the simple beam span can be carried by the continuous span for the sameamount of concrete and steel. This represents a very signi!cant basic economy thatshould be reali/ed by engineers designing prestressed concrete structures. ;ecause of this strength inherent in continuous construction, it is possible to employ smallersections for the same load and span, thus reducing the dead weight of the structureand attaining all the resulting economies.

0lthough it is generally conceded that continuity is economical in reinforced

concrete, it is seldom nown that, from certain point of view, even greater economycan be attained in prestressed construction.

&conomical design of continuous prestressed beams can be achieved in severalways. Awing to the variation of moment along the beam, the concrete section and theamount of steel are often varied accordingly. The pea of the negative moments canbe reinforced with non-prestressed steel, thus reducing the amount of prestressingsteel. 0dvantage can be taen of the redundant reactions to obtain favorable lines of pressure in the concrete. esign can be based on the ultimate strength of such beams,applying the principles of limit design.

Bie any other type of construction, there are advantages and alsoshortcomings, which, under certain conditions, could outweigh the advantages. Thechoice of a particular type of design must be made after considering all the factorsinvolved in the job.

isadvantages inherent in continuous prestressed concrete beams can beenumerated as follows?

.P.0ncog>#



#. Fri$tion"( (oss in $ontin,o,s ten&ons. This can be serious if there are many reversed curves, if the curves possess largedeection angles, or if the tendons are excessively long. )uch losscan be minimi/ed by using relatively straight cables in undulating orhaunched beams. The usual method of overtensioning, of stressingfrom both ends, can also be used to reduce frictional losses.

5. S%ortening o# (ong $ontin,o,s 'e"s ,n&er +restress. Thismay produce excessive lateral force and moments in thesupportingcolumns, if they are rigidly connected to the beamsduring prestressing. Provisions are usually made to permitmovement at the beam bearings or rocing of the columns.

9. Se$on&"r3 stresses. )econdary stresses due to prestressing,creep and shrinage e<ects, temperature changes, and settlementsof supports could be serious for continuous structures unless theyare controlled or allowed for in the design. Ane interesting point incontinuous prestressed structures is that these secondary stressescan often be utili/ed to good advantage so that they will add to the

economy of the structure.:. Con$,rren$e o# "i, oent "n& s%e"r oer s,++orts .=t is believed that the concurrence of maximum moment and shearat the same section may decrease the ultimate capacity of thebeam. This happens over the support of most continuous beams.'ence care must be taen to reinforce such points properly for bothshear and moment if high ultimate strength is desired. The elasticstrength, however, is not a<ected by such concurrence.

6. Reers"( o# oents. =f live loads are much heavier than deadload, and if partial loadings on the spans are considered, continuousbeams can be subjected to serious reversal of moments. This can be

sometimes be overcome by proper design, such extensive used of nonprestressed steel in combination with prestressed concrete.

%. Moent +e"5s. Peas of maximum negative moments maysometimes control the number of tendons re"uired for the entirelength of the beam. These peas, however, can be strengthened byemploying deeper sections or by adding prestressed and prestressedreinforcements over the portions where they are needed.

>. Di$,(t3 in "$%ieing $ontin,it3 #or +re$"st e(eents. =t iseasy and natural to obtain continuity for cast-in-place construction,but continuity of precast elements cannot always be achievedwithout special e<ort. An account of diculties in handling precast

continuous beams, they are often precast as simple elements, to bemade continuous after they are erected in place.

$. Di$,(t3 in &esigning. =t is more dicult to design continuousrather than simple structures. ;ut with the development of simplermethods, the design of continuous prestressed concrete beams canbe made into a more or less routine procedure applying basicprinciples for continuous structures familiar to most engineers

.P.0ncog>5



ve tendons 6ith straight beam? often used for slabs or short span beams? 6here simple form6or7is more important than saving of steel and concrete, he m

nched or curve beams? often used for longer spans or heavier loads? this 6ill not onl! save concrete and steel but permit the use of straight tendons? li7e6

mpromise of the above t6o arrangements? using curve beams and slightl! curve tendons, his 6ould permit optimum depth of beam as 6ell as ideal positi

rlapping tendons, Cables protruding at intermediate points offer a possibilit! of var!ing prestressing force along the beam, he arrangement has no revers

Ten&on "3o,ts #or Contin,o,s )e"s

+ontinuous beams may be divided into two classes? fully continuous beams and partilly continuous beams. *or fullcontinuity, all the tendons are prestressed in place and are generally continuous from one end to the other., although some can be

anchored at intermediate points if found desirable. The concrete can be either poured in place or made of blocs assembled onfalsewor The tendons may be encased in the concrete during pouring, threaded through preformed holes, or placed outside the

webs. They may be either bonded or unbonded, depending on the re"uirements of the structures.

Precast elements can also be made fully continuous by coupling the tendonstogether with a high tensile rod and then stressing one or both ends of the tendons.

F,((3 $ontin,o,s 'e"s ten&on ("3o,ts:

.P.0ncog>9



ntinuous prestressed cables placed in conduits or grooves left in the structure, "fter erection? concrete is poured bet6een the beams over th

ntinuous tendons placed over the supports onl!, his saves steel but reJuires more anchorages than first la!out, More of the anchorages are

endons stressed after erection

endons stressed before erection

:ouredinplace concrete

pl!ing continuit! using cap cables, hese tendons are usuall! made of 6ires or small strands? can be convenientl! stressed from the soffit

Cap cables placed and stressed after erection

endons stressed before erection

:ouredinplace concrete

ntinuous elements over the supports transversel! prestressed, ensile elements are are inserted over the supports and attach them to the pre

P"rti"((3 $ontin,o,s 'e" ten&on ("3o,ts:

.P.0ncog>:



upler over supports, his is speciall! applicable to hightensile bars? but also to other forms of tendons to obtain continuit!, his permits th

emporar! anchoragecoupler

5eam 6ith tendons prestrressed:ouredinpl

.P.0ncog>6



nprestressed steel over the supports, Continuit! can be achieved b! emplo!ing nonprestressed over the supports, his is speciall! true for c

nplace concrete nonprestresse

Beam Elevation

T C

$ )

e

M $ e $ Ce

FBD of Section

5eam Dlevation

5ending of beam under prestress if not held b! supports,

eactions e&erted to hold the beam in place,

M

Moment diagram due to reactions,

>eviation of Cline from cgs G line due to moment produce b! secondar! reactions,

a $M

Cline

cgs G lineor line

Moent in $on$rete &,e to +restressing in " si+(e 'e":

*or simple beam, no matter how the beam is prestressed, only the internal stresses

will be a<ected by prestressing. The external reactions, being determined by statics,

will depend on the dead load and live load Dincluding the weight of the beamE. Without

load on the beam, no matter how the beam is prestress internally, the external

reactions will be /ero, hence the external moment will be /ero. With no external

moment on the beam, the internal resisting moment must be /ero, hence the +-line

Dline of pressure in the concreteE must coincide with the T-line in the steel Dwhich is the

cgs lineE. The +-line in the concrete being nown, the moment in the concrete at any

section can be determined by?

CeTe M ==

Moent o# $on$rete &,e to +restressing in " $ontin,o,s 'e":

.P.0ncog>%



When the continuous beam is prestressed, it bends and deects. The bending of the beam can be such that the beam will tend to deect itself away from the suuports.

=f the beam is refrained from deecting at these supports, reactions must be exertedon the beam to hold the it there. Thus reactions are induced when a continuous beamis prestressed. These induced reactions produce moments in the beam. To resist thesemoments, the +-line must be at a distance a from the T-line, such that the internalresisting moment e"uals the external moment 8 caused by the induced reactions.

T

M =

=n simple beam, the +-line coincides with the T-line. The stress distribution in theconcrete at any section is given by the location of the T-line. 7o secondary moment is

created. =n continuous beams, the +-line usually deviates from the T-line. The stressdistribution at any section is given by the +-line which does not coincide with the t-line. The di<erence between the two beams lies in the presence of external reactionsand moments in the continuous beam, produced as a result of prestressing. )ince theexternal moment is soley produced of external reactions, and since the reactions areonly applied at the supports, the variation of moment between the supports is a linearone.. =f T remains constant between supports, then the deviation a, being directlyproportional to 8, also has to vary linearly.

.P.0ncog>>



e1 e2 $M2< cgsline

Cline

a

%a& Beam Elevation

%b& 'rimary Moment Diagram Due To 'restress

%c& S(ear Diagram For %b&

%d& Load Diagram For %c&

%e& )esultiing Moment Diagram Due To 'restress* From Loading +n %c&

M2

M1 $ <e1

The di<erence between the simple and a continuous beam under prestress canbe represented by the existence of ]secondary momentsV. Ance these moments overthe supports are determined, they can be interpolated for any point along the beam.

8oment in the concrete given by the eccentricity of the prestressed isdesignated as the primary moment , such as would exist if the beam were simple. An

account of the primary moment acting on a continuous beam, the secondary momentscaused by the induced reactions can be computed. The resulting moment due toprestress, then is the algebraic sum of the primary moment and secondary moment.

)econdary moment H Primary moment G 3esulting moment

Ass,+tions #or t%e Met%o& o# An"(3sis "n& Design A&o+te&:

#. The eccentricities of the prestressing cables are small compared to thelength of the members.

5. *rictional loss of prestress is negligible Dwhere frictional loss is appreciable,

it should be taen into accountE.9. The same tendons run through the entire length of the member, Dvarying

steel areas can be included with some modi!cationsE.:. The axial component of the prestress is constant for the member and is

e"ual to the prestressing force *.6. The primary moment 8# at any section in the concrete is given by?

11 #e M =

Pro$e&,re o# An"(3sis:

#. *irst treat the entire beam as if it had no supports. Plot the momentdiagram for the concrete produced by the eccentricity of the prestress.

5. +ompute the loading on the beam ccorresponding to that momentdiagram. This is the loading produced by the steel on the concrete.

9. With this loading acting on the continuous beam as it is actuallysupported, compute the resulting moment by moment distribution or othersimilar method.

.P.0ncog>$



)ince the +-line deviates linearly from the cgs-line, it will have the same intrinsicshape as the cgs-line, and can be plotted. =t is usually not necessary to compute thesecondary moment, which is represented by the deviation from the +-line and the cgs-line. =f desired, it can be computed by the simple relation?

)econdary moment G 85 Z 8#

#

M M 12 −=

Hints #or +(otting t%e (o"&ing &i"gr"s &ire$t(3 #ro t%e oent &i"gr"s:

#. 0t the end of the tendons, the force F from the tendons on the concretecan be resolved into 9 components?

0. 0n axial force, Fcosφ 1 = F Dsince cos φ# G#E, acting at the end of theanchorage. This is usually has no e<ect on the bending moment butmay produce moment in a rigid frame, owing to the axial shorteninge<ect.

;. 0 transverse force Fsinφ 1 = F φ 1 = Ftanφ 1, applied at the support andbalanced by the vertical reaction from the support directly beneath. This again produces no moment in a continuous beam, unless it isapplied away from the support. =ts e<ect in a rigid frame is small.

.P.0ncog>1



< 2

< 3

< 4

< *

<orces <rom endons ;n Concrete

+. 0 moment, Fcosφ 1e = Fe, acting at the end of the beam. This willproduce moment along the entire length of the continuous beamand it must be included when following the moment distributionprocedure.

5. 0long the span of the member, where the cgs or the cgc-line of the

member curves, transverse loads are applied to the concrete. Twocommon cases can be considered?0. When the moment diagram taes the shape of a parabolic or a

circular curve Dnote? owing to the assumption of a at curvature,parabolic and circular curves are considered to have the same e<ectin producing transverse loadsE, a uniformly distributed load isapplied to the concrete along the length of the curve. The total forceon each curve is given by the change in slope between the two endtangents@ thus the total force at φ5 is given by?

22sin φ φ # # + ==

*or practical purposes, the load W can be considered as uniformlydistributed along the length of the curve.

;. When the moment diagram changes direction sharply, the force can

be considered as concentrated at one point@ the moment at f:, for

example is?

44sin φ φ # # =9. Aver the interior supports, where the moment diagram changes direction,

a load is applied directly over these supports. 0gain two cases can be

considered?

0. =f the moment diagram curves gradually over the support, again auniformly distributed load is applied as shown for φ9. This will a<ectmoments in the beam, and the load must be considered inperforming the moment distribution.

;. =f the moment diagram is bent abruptly over the supports, aconcentrated load is applied thereon. )uch a concentrated load isdirectly reacted by the support underneath and produces nomoments on the beam. =t can be neglected in performing themoment distribution.

O't"ining o"&ing Di"gr"s D,e To Prestress:

.P.0ncog$2



a% 5eam Dlevation

*)7/

2))7/

+1))7/

22*7/

),))

b% :rimar! Moment >iagram >ue o :restress

2*)(,2)% $ *),)) 7/

2*)(,-)% $2))7/

2*)(+,4)% $ +1))7/

2*)(,)% $ 22*7/

*7

+1*7

247

+2)7

c% Shear >iagram <rom b%

Mb $ Ma + "v

"v $ a $ Mb Ma

a $ 2)) G (*)%N@3) $ *

a $ +1)) G (2))%N@2) $ +1*

) G 1))N@2* $ 4

22* G )N2@2* $ 1- 22

+) G 1))N@*) $ 2

24) G (22*%N2@2* $ 1-

) G 1))N@*) $ 2

2)

b $ a + "6

:1 $ * G )N $ *:2 $ 1* G(*%N $ 2)

:3 $ 24 G(1*%N $ 3

6 $ 2)(24%N@*) $ ),-- 7@ft

:4 $ )(2)%N $ 2)

*7

+2)7

37

d% oading >iagram <rom c%

2)7

<DMF

Concentrated loadF

M"5 $ (:ab2%@2

M"5 $ 2)(3)%(2)2%@*)2 $ +.

M5" $ (:a2b%@2

M5" $ 2)(3)2%(2)%@*)2 $ 144

<DMF

niform loadF

M"5 $ M5" $ 62@12

M"5 $ M5" $ ),--(*)2%@12 $ 1-3

<DM

>ist

C;

>ist

5al

rel, 7 $ 1 rel, 7 $ 1

*) +.

.

*)

144 +1-3

03 +2

2 2

24. +24.

1-3 )

+1-3

),))

),2/),4*/

),-/

),.1/

"5

C>

cgcline

Cline

3)/ 2)/ 2*/ 2*/

),2/),-/

),4/

),/),)-

),10. rad" 5 C>

cgcline

cgsline

< $ 2*)7

d% Cline >ue o :restress <rom a% and e%

e% Moment >istribution <or oading n d%

E"+(e 1:

0 continuous prestressed concrete beam with the bonded tendons is shown inthe !gure. The cgs has an eccentricity at 0, is bent sharply at and ;, and has aparabolic curve for the span ;+. Bocate the line of pressure Dthe +-lineE in the concrete

due to prestress alone, not considering the dead load of the beam. +onsider aprestress of 562 .

.P.0ncog$#



The line of pressure for the entire beam can be computed by plotting its momentdiagram and dividing the ordinates by the value of the prestress.

.P.0ncog$5



*)/ *)/" 5 C>

cgcline

cgsline

6 $ 1,2 7@ft12=

.=

. =

. =

2 4 =

Concrete Section<DM 2*)

>ist +2*)

C; ____

),))

+2*) 2*)

+12* 12*

+30* +30*

2*)

+2*)

____

),))

+211 +211

30*

),*4/

1,*)/

),04/

Mdiag,>ue to >

Cline,>ue to M>

Cline

),)/,*2/

),13/esulting Cline,

due to combined M:+M>

Cline

+-,)2*)

24.

'2),)2*)

*)

==

==

,

A

e

e

=f desired the secondary moment over the center support can be computed as?

ft k M

M M M pri mr't resul ondr'

−=−=

−=

14.1))24.sec

tansec

E"+(e 2:

*or the prestressed beam in &x. #, a uniform load of #.5 Ift is applied to the entirelength of the two spans Dincluding weight of the beam itselfE. Bocate the line of pressure in the concrete due to the combined action of the prestress and externalload. +ompute the stress in concrete section at ;, = G 91,>22 in:, 0c G 5$$ in5.

;y dividing the moments by the prestress *, the shifting of the +-line can beobtained due to 8B.

.P.0ncog$9



Co+,ting t%e stresses:

0t ;? 8; G *e; G 562D-.65E G -#92 J

Top !ber?

ksi

x I

Mc A # f T

1.),)0)0,)-.0,)

3+0))%1-(1213)

2--2*)

−=+−=

+−=+−=

;ottom !ber?

ksi

x

I

Mc

A

# f ,

*04,10)0,)-.0,)

3+0))

%1-(1213)

2--

2*)

−=−−=

−−=−−=

Note: The prestress moments and the dead load moments can be combined together

and then mae the moment distribution. The results would obviously be the same.

ine"r Tr"ns#or"tion "n& Con$or&"n$3 o# C"'(es

The design of continuous prestress beams is more complicated problem. =nanalysis, the concrete section, the steel, and the location of the steel are alreadynown or assumed. =t is only necessary to compute the stresses for the given loadingconditions. This is not true in design, which is essentially a trial-and-error process in ane<ort to reach the best proportions. The designer must be well ac"uainted with the

method of analysis before he can perform eciently in design. =7 order to design well,we must be conversant with some of the mechanics of continuous prestressed beams.

ine"r Tr"ns#or"tion

The designer should be able to perform linear transformation with ease and silland to obtain either concordant or nonconcordant cables to satisfy the most desirableconditions.

When the position o cgs-line or o a C-line is moved over the interior supports o a continuous beam without changing the intrinsic shape (ie the curvature and bends!o the line within each individual span" the line is said to be linearly transormed

E+("n"tion:

#. *irst, the position of the line is moved only over the interior supportswhenever desired, but not at the ends of beam.

5. )trictly speaing, a line can still be termed linearly transformed if it ismoved at the ends.

9. *or purpose of design, linear transformation without involving movementst the ends is much more useful.

.P.0ncog$:



),3/

),*/

),4/),2/

),-/

1,)/ 1,)/

),./

),2/

2*/

*)/ *)/ *)/

1st cgs line

2nd cgs line

a% inear ransformation of cgs ine

-)7/

.)7/

1.)7/

1.)7/

4)7/

2)7/

14)7/

2 ) )

2 ) )

M for 1st cgs line

M for 2nd cgs line

b% :rimar! Moment due to :restress

-) 7ft

14,4 7

),.4 7@ft

c% oading diagram <or 5oth Moment >iagrams

< $ 2)) 7

:. ;L linear transformation, the intrinsic shape of the line within each spanremains unchanged@ only the amount of bending of the line over theinterior supports changed.

6. The +-line resulting from perstressing a continuous beam is linearlytransformed line from the cgs line.

T%eore:

In a continuous beam" any cgs line can be linearly transormed without changingthe position o the resulting C-line

This means that the linear transformation of cgs line does not a<ect the stresses

in the concrete, since the +-line remains unchanged. Thus the two cgs line in theabove !gure will produce the same stresses in the concrete, despite their apparentlydivergent locations.

0ny bending of cgs over the support will produce transverse forces acting on thebeam which are directly counteracted by reactions from the supports. 'ence theloading will not a<ect the moment along the beam. )ince the moment is not a<ected,the +-line is not a<ected. Thus, linear transformation involving bending of the cgs lineover the interior supports will not change the location of the +-line. An the other hand,

.P.0ncog$6



any movement of the cgs line at the ends of the beam changes the magnitude of theapplied end moment which do a<ect the moments on all spans. 'ence, lineartransformation cannot involve the movement of the cgs line over the ends of thebeam or over the exterior support of a cantilever, but it can involve movement of thecgs line over the interior supports.

The above theorem, permitting the linear transformation of the cgs line withoutchanging the +-line, o<ers many possible adjustments in the location of the cgs linewhich cannot be easily accomplished without that nowledge.

Con$or&"nt C"'(e In Contin,o,s )e"

Concordant cable in continuous beam is a cgs line which produces a C-linecoincident with the cgs line

=n other words, a concordant cable produces no secondary moment. Thus every

cable in a statically determinate structure is concordant, because no external reactionis induced, and there is no secondary moment in the structure. *or a continuous beam,on the other hand, external reactions will usually be induced by prestressing. Thesereactions will produce secondary moments in the beam, and the +-line will shift awayfrom the cgs line. When this happens, the cable is termed nonconcordant. When bychance or by purpose, no reactions are induced in a continuous beam by prestressing,then there will be no secondary moments and the cable is a concordant one.

When a concordant cable is prestressed, it will tend to produce no deection of the beam over the supports, and hence no reactions will be induced Dnot consideringthe weight of the beamE.

;esides the fact that concordant cable line is easier for analysis, there is seldoma necessity for using a concordant one.

While no signi!cant reason can be given for preferring a concordant cable, thereis even less justi!cation for locating a nonconcordant cable for the sae of nonconcordancy. The real choice of a good cgs location depends on the production of adesirable +-line and the satisfaction of other practical re"uirements, but not on theconcordancy or nonconcordancy of the cable.. 0 concordant cable, being easier tocompute, is slightly preferred, other things being e"ual.

.P.0ncog$%



$ ) due to

prestress $ )

$ )

Concordant cable? Cline

coincides 6ith cgs line

a% Continuous 5eam Iith Concordant Cable

O ) due to

prestress O )

cgs line? a non$

coincides cable

b% Continuous 5eam Iith Lonconcordant Cable

Cline due to prestress

O )Secondar! moment

0 convenient procedure in design is to obtain a concordant cable that gives goodposition of the cgs in resisting the external moment. =f that location falls outside thebeam, it can be linearly transformed to give a practical location without changing its+-line. 0ccording to this procedure, the !nding of locations for concordant cables

become a useful means to an end.

T%eore #or o't"ining " $on$or&"nt $"'(e.

. #very real moment diagram or a continuous beam on nonsetting supports" produced by any combination o e$ternal loadings" whether transverse loads or moments" plotted on a scale" is one location or a concordant cable in that beam

E"+(e /:

*or a continuous beam loaded as shown, obtain some desirable locations for

concordant cables to support that loading.

So(,tion:

7ote? &very moment diagram plotted to any scale is a concordant cable.

.P.0ncog$>



1,.) 7@ft4- 7

*)/ *)/

21

40* 7ft

3.3

a% Continuous 5eam Iith "n! oading

1/

b% Moment >iagram <or a%

*,-2=

,*)= 0,2.=

c% ;ne Concordant Cable <rom b%

2,1=

4,0*= 3,.3=

d% "nother Concordant Cable <rom b%

e% Continuous 5eam Iith niform oads "nd Dnd Moments

f% Concordant Cable <rom oading n e%

2*/

E"+(e )e":

#. =f we plot the continuous beam moment diagram for a given loading, weobtain bE

5. Two concordant cable locations are shown in cE and dE, both areproportional to the moment diagram in bE and hence both are concordant.

.P.0ncog$$



9. E gives another location of a concordant cable, which is proportional to themoment diagram for loading in eE.

:. 8any similar concordant cables can be found by drawing all inds of moment diagrams.

6. The most desirable concordant cable will be governed by practicalre"uirements of the particular problem as well as by the ability of the

cable to resist the applied loads. *or example, the location in cE giveslarger resisting arms for the steel but may overstress the concrete if theweight of the beam is light, in which case dE may be a better location.

%. E does not suit this particular loading as well but gives a symmetricallayout and may carry other loadings, such as beamJs own weight, moreeciently.

Coro(("ries o# t%e Gener"( T%eore

#. The reverse of the theorem is true? The eccentricity of any concordant

cable measured from the cgc is a moment diagram for some system of

loading on the continuous beam plotted to some scale.

5. 0ny +-line is a concordant cable, since it is obtained by computing themoments due to a system of loads on a continuous beam.

9. )uperposing two or more concordant cables will result in anotherconcordant cable. )uperposing a nonconcordant cable and a concordantcable will result in a noncocordant cable.

:. When a sudden change in direction is desired, a concentrated load isapplied. When a gradual change is desired, a uniform load is applied. Anemoment diagram can thus be modi!ed into another by the addition of loads. 'ence one concordant cable can be easily modi!ed into another.

6. =n order to obtain a concordant cable from another by lineartransformation involving moving of eccentricities over the ends of acontinuous beam, the following procedure can be used?a. 0pply an end moment on the continuous beam@ compute the

moment diagram due to that moment.

b. When one end is moved by a given moment, the entire cable mustbe transformed linearly in proportion to that moment diagram.c. =f the movement of eccentricities at both ends is desired, apply end

moments at both ends proportional to the respective amount of movements, and shift the entire cable in proportion to the momentdiagram so obtained. This will yield another concordant cable.

8uch ingenuity can be exercised in the location of concordant cables, but itshould be left to the sill of the designer after he understands the basic theoremand its main corollaries. When applied to rigid frames, the e<ect of sidesway and

.P.0ncog$1



1/

*,-2=

,*)= 0,2.=

E,- ."c&

1/

1 , ) )

),)*= ),2*=),30*=

.,2)=

-,*)= *,0.=

b% Concordant Cable <rom D&, 3c% Iith 4= >isplacement at ight Dnd

1/

a% Moment >iagram for Continuous 5eam >ue to nit Moment at ight Dnd

4=

rib shortening should be additionally considered. *or varying prestress along thebeam, the moment diagram should be divided by the corresponding prestress ateach point in order to obtain the location of a concordant cable. Ar the tendonsmay be treated separately. =f each individual tendons or group of tendons formsa concordant cable, then, when acting together, they also form a concordantcable.

E"+(e :

Abtain a new concordant cable, with its intrinsic shape the same as that of &xample 9-cE, but with the right end of the cable :V above the cgc.

So(,tion:

#. 0pply a unit moment at the right end of the beam@ by moment distributionmethod, plot the moment diagram as in aE

5. The concordant cable can now be linearly transformed in proportion to themoment diagram in aE, giving a new concordant cable as in bE.

7ote?

The same moment diagram in aE can be used to shift the end eccentricity anyother amount, not only for :V as illustrated. 0lso, owing to the symmetry of the beam,aE can be similarly used for moving the end eccentricity at the left. 0 combination of

.P.0ncog12



Mmin

M9

Mma&

Moment >iagram for ;ne Span of aContinuous 5eam

op <iber

5ottom <iber

op Pern

5ottom Pern

a9

a9

ama&

amin

imiting Qone for Cline >ue to :restress

two moment diagrams due to a moment at each end will permit the simultaneousshifting of both end eccentricities.

0n in!nite number of concordant cables can be obtained by rotating oneconcordant cable about the points of inection, because such rotation simplyrepresents the addition of one concordant cable to another, and should result in a

concordant one. The points of iniction in these moment diagram are sometimescalled ]nodal pointsV.

C"'(e o$"tion

+able location means the location of the centroid of the tendons, that is the cgs

line. 0fter the cgs line is determined, the location of the individual position of thevarious tendons is an easier problem.

esigning a continuous prestressed-concrete beam, lie that of any othercontinuous structure, is essentially a procedure of trial and error. \nowledge regardingthe analysis of such structures, together with a systematic approach to the solution,will aid greatly in arriving at desired results.

Re$oen&e& Ste+s #or Designing " Contin,o,s Prestresse& )e"

#. 0ssume section of members for dead load computation.

5. +ompute maximum and minimum moments at critical points for variouscombinations of dead, live, and other external loads. +ompute the amountof prestress re"uired for these moments and the corresponding depth of concrete. 8odify section of members and repeat steps # and 5 if necessary.

9. Plot the top and bottom ern lines for the members. *rom the bottom ernline, plot?

o

GG

#

M lso

#

M == min

min

where 8min G the algebraic smallest moment. The distance amin and a

should be plotted upward for Z8 and downward for H8.

.P.0ncog1#



*or the top ern, plot

o

GG

#

M lso

#

M == ma&

ma&

again, upward for Z8 and downward for H8.

The shaded area, between the limits of these four lines, obtained byamin, amax and a, represents the /one in which the line of pressure must lieif no tension is permitted. When the /one is too wide, an excess of prestress or of concrete section or of girder load is generally indicated. =f the limiting line from one ern crosses the a limiting line from anotherern, an inade"uacy is evident. 0n ideal layout is obtained when thereexist a narrow limiting /one within the beam where the centroid of thecables can be conveniently located.

:. )elect a trial cable location within the above /one. 7ote that, if the cablefollows the shape of some moment diagram, it will be a concordant cable.=f this location is a concordant cable, it is a satisfactory solution. =f it is anonconcordant cable, the +-line can be determined by momentdistribution method. =f the +-line still lies within the limiting /one, then thetwo locations are possible? either the trial location giving a nonconcordantcable, or a new location following the +-line. =f this +-line lies outside the/ones, a new cable location can be tried. 0n attempt should be made toget a concordant cable within the /one. =t is generally best to try aconcordant cable, because they coincide with their +-lines and give a noredirect solution.

6. The concordant cable within the limiting /one obtained in )tep : is a goodlocation for resisting external moment, but it may or may not be a goodpractical location

E"+(e J:.P.0ncog15



0 pedestrian bridge of prestressed-concrete slab, has a three-span symmetricalcontinuous layout as shown. The bridge is 1J-:V wide with a uniform thicness of #9VDneglecting curbE. The total e<ective prestressing force is #,592,222 lbs afterdeducting a loss of #6C. esign live load is 62 psf. +hoose a suitable location for thecable, allowing no tension in the concrete, f cJ G 6,222 psi.

So(,tion:

*ollowing the procedures described above and considering #-ft unit width of slab?

)tep #? The section is already chosen, and the dead load is #%5 plf or #%5 plf for a#-ft width.

)tep5? The amount of prestress is already chosen@ it is?

k #

# k # o 1*.1*,)1

132?132

33,+

123)))) =−

====η

)tep 9? \ern lines for a rectangular section are located h%&. the maximum and

minimum moment diagrams together with the girder moment diagram are shown for

one-half of the structure. These diagrams are divided by the respective prestress, F for

those with live loads, and F o for dead load only. The a values thus obtained are plotted

from the ern lines as shown, giving the limits for the /one within which the +-line due

to prestressing must lie.

)tep :? 4sing the moment diagrams as guides, select a trial cgs location withinthe /one as shown. *or purposes of illustration, assume the cgs line to posses the

following characteristics.

#. Passing through the cgc Dmid-depth of slabE at end supports.

5. Ane sharp bend for each side span.

9. Ane sharp bend over each intermediate support.

:. 0 parabolic curve for the center span.

*or the cgs location, the corresponding loading on the concrete is shown indE,moment distribution for which gives a moment diagram as in eE. ividing the moment

diagram by the prestress yields a +-line as shown in E which is very close to the trial

location and is still within the limiting /one. 'ence this +-line is a location for a

satisfactory concordant cable.

)tep 6? 0 more practical location for the cgs is shown in gE, a<ording a better

location for the steel. This is obtained by linearly transforming the concordant cable

.P.0ncog19



4)/ 4)/.)/

13=

/4=

a% 5ridge Dlevation

b% 5ridge Section

$ B C D

4)/ 3)/

a% 8alf 5ridge DlevationC

$ 5

on "5? C>

on 5C

> onl!

> onl!

$ 5

on "5? 5C

22,2 7ft

14,2 7ft

41,* 7ft

2,3 7ft

43,.

4.,0*3,

*-,4

C

C1*/

3=

4,*=

4=

7t

7b

cgc

on "5? C>

> onl! on "5? 5C

> onl! on 5C

cgs

b% Ma&imum R Minimum Moment Curves

c% imiting Qone R rial cgs ocation

*,* 7

),2)- 7@ft

>< ),. ),31

<DM+32,2

>ist32,2

C;

>ist

otal),))

1,3+.2,4

1.,11-,. -,4*4,)l *4,)

31,3

*4,)

3,.

2,-*=

4,)=

3,.)=

3,)=

4,2*=

4,2*=

d% oading R Moment >istribution for :restress in c%

e% Moment >iagram from d%

f% Cline from e%

g% Lonconcordant Cable

into a nonconcordant one. This nonconcordant cable will yield the same +-line as the

concordant one and hence will serve the same purpose as far as the stresses are

concerned.

.P.0ncog1:



$ 5 C D1 2 3

:e

D

An"(3sis o# Contin,o,s Prestresse& Con$rete )e"

1. Are"-Moent Met%o& o# An"(3sis

.P.0ncog

∆B#

"

DC)A

x

16



Pro$e&,re:

#. +onsidere the spans cut at interior supports and each span is considered as asimple beam on which moment, Pe, Dprimary momentE are imposed due toprestress..

5. The slope of the end span at the !rst interior support due to prestressing maythen be calculated as?

1

1 %(

!

x

%I

! Pe A,

∆=θ

9. 0pply a moment on the end span at ; D8);0, the !xed-end secondary momentat ;E which will rotate the beam bac to /ero slope. *or two span symmetricalstructure, this moment would be the secondary moment due to prestressing.*or three or more spans or unsymmetrical structures, it is necessary todistribute the !xed-end secondary moments. 0pplication of 8);0 will resultin a triangular moment diagram over the end span and the rotation of thesupport due to this moment diagram will be?

%I

! M S,A

A,

3

1=θ

&"uating rotations due to prestressing and due to secondary moment 8);0 can beobtained?

( ) x ! Pe !

M

%I

! M

!

x

%I

! Pe

S,A

S,A

∑

∑

∆=

=∆

12

1

1

1

1

(3

3

%(

6. =n similar manner, the rotations due to prestress moments are calculatedfor the other end span and for the interior spans.%. *or the interior spans, moments are imposed at each end of the span to

rotate the beam bac to /ero slope. =t is necessary then to solve 5simultaneous e"uations to determine values of 8);+ and 8)+;.

>. 0fter obtaining 8);+, 8)+; and 8)+ Dthe latter in a manner similar to 8);0E,the moment distribution techni"ue may be used to calculate the !nalsecondary moments at the two interior supports.

.P.0ncog

B9B5B#

8)+8)+;

8);+8);0

'DC)A

1%



:

:

-:(eb G ec%2

2

$. 0fter determining the !nal secondary moments at supports, the secondarymoment e<ect in any interior span point is readily obtained by linearinterpolation between the secondary moments at supports.

1. The !nal moments due to prestressing at any point are then calculated asthe algebraic sum of the primary moments, Pe, and the secondarymoment 8).

2. E,i"(ent o"& Met%o&

The e<ect of the prestressing force on a beam can be determined byconsidering the prestressing force to be replaced by e"uivalent external loads. 0fterobtaining the e"uivalent loads, !xed end moments can be determined and the !nal!xed end moments can be determined by applying moment distribution techni"ue orany other method of structural analysis.

Si+(e S+"n:

The uniform load w?

.P.0ncog1>



eb1eb

eb

1(1 G 1 $

5

End Span

&

e

$ 5End Span

: :

: t a n

eb

M5"

2:( 1 %eb ( %2

2:(1 1%eb

(1 %(1 1%

2:(1 1%eb

1 (1 %2

2

32

2

2

2

3

2

2

2

1

1

+

=

+

=

===

dx

d'

dx

d P

w

dx

'd

dx

d'

)

)

P w

d

ds )

ds

d P w

θ

θ

The eccentricity can be expressed as?

ccb e x

!

ee xe +

−= 2

2

%(4%(

The moment due to prestressing force P at a distance x from the origin?

+

−== c

cb e x !

ee P x Pe M 2

2

%(4%(

E,i"(ent o"& #or Contin,o,s )e":

For$es eerte& on $on$rete in t3+i$"( en& "n& interior s+"ns 0%ene$$entri$it3 "ries +"r"'o(i$"((3.

En& S+"n:

.P.0ncog1$



2eb

eb

2 2$

5

+nterior Span

eb

( 2% ( 2%

&

e

4:(1 G 2%eb

22

4:(1 G 2%eb

(1@2 G 2%2

Fie& En& Moent:

( ) ( )[ ] ( ) ( )[ ]( )α α α α α α α α β βα −−−++−++−−−= 112334

1

2

11

2

111

2b ,A

Pe M

E$$entri$it3:

x G 2 to x G αB?

bbb

e x !

e x

!

ee β

α

β β

α

β β +

−+

−−=

%(

%(2

%(

%(12

2

1

x G αB to x G D# Z α#EB

bbbb e

e x

!

e x

!

ee 1

1

1

2

1

12

2

1

1

%1%(1(

%1(

%1%(1(

%1(2

%1%(1(

%1(β

α α α

β α

α α α

β α

α α α

β +

−−−−

+−−−

−−

−−−−

=

x G D# Z α#EB to x G B

b

bbb ee

x !

e x

!

ee +

−−

+−

−+

−−

−=%1(

%1(

%1(

%1(2

%1(

%1(

1

1

1

12

2

1

1

α α

β

α α

β

α α

β

Interior S+"n:

.P.0ncog11



eb1eb

eb

(1 G %$

5

End Span

&

e

$ 5End Span

: eb

:

M5"

2:( 1 G %eb

2

12(1 1%eb

(1 %44

&2 +

24 (1 1%eb

(1 %43

&

4(2 2 + 2 1%(1 G 1%eb

(1 %42

Fie& En& Moents:

%1%(1(3

222 α β −−=−= b

,A A,

Pe M M

E$$entri$it3:

x G - BI5 to x G - D#I5 Z α5EB

bbbb e

e x

!

e x

!

ee +

−−

−−

−−=

2

2

2

22

2

2

2

2

%1(%1(2%1(2

α

β

α

β

α

β

x G -D#I5 Z α5EB to D#I5 Z α5EB

bb e x

!

ee 2

2

2

2

2

%21(

%1(2β

α

β +

−−

=

x G D#I5 Zα5EB to x G BI5

bbbb e

e x

!

e x

!

ee +

−−

−+

−=

2

2

2

22

2

2

2

2

%1(%1(2%1(2

α

β

α

β

α

β

For$es eerte& on $on$rete in t3+i$"( en& "n& interior s+"ns 0%ene$$entri$it3 "ries "s " +"r"'o(" "n& " #o,rt%-&egree $,re.

.P.0ncog#22



2eb

eb

$ 5

+nterior Span

eb

&

e

@2 @2

$ 5

+nterior Span

12(1 G 2%eb

4 1.(1 G 2%eb

2&2 +

::

M5"

M"5

Fie& En& Moent:

[ ]%32(4%1)123(*2)

2

1

2 ++++−−= α α α α β βα b ,A

Pe M

E$$entri$it3:

x G 2 to x G αB

b

bb e x !

e x

!

ee β

α

β β

α

β β +

−+

−−=

%(

%(2

%(

%( 12

2

1

x G αB to x G B

bb

bbbb

ee

x !

e

x !

e

x !

e

x !

e

e

14

1

22

4

12

24

1

23

34

14

44

1

%1(

%1%(24(

%1(

%1%(12(4

%1(

%1%(122(2

%1(

%1(4

%1(

%1(

β α

β α α α

α

β α α

α

β α α

α

β α

α

β

+−

−+−+

−

−−

+−

−−+

−−

−

+−

−

−=

.P.0ncog#2#



e"e5

e"

e

&" 5

(e5 G e"%&

(e5 G e"%

M"5

M

&" 5

(M5" G M"5%&

(M5" M"5%

M5"M"5

Fie& En& Moents:

%1(1*

-2β −=−= b

,A A,

Pe M M

E$$entri$it3:

b

bb e x !

e x

!

ee 2

2

2

24

4

2 %1(-%1(1.β

β β +

−+

−−=

0fter the prestressing moments have been determined at the supports,prestressing moments between supports may be obtained by superimposing thesimple beam moment diagram due to prestressing loads on the base line provided byconnecting the total prestressing moments at supports. 'owever, it is generally

simpler to obtain interior prestressing moments using the tendon pro!le. 4sing thisapproach, the interior moments may be expressed as?

−+−−++=

!

xeee P

!

x M M M Pe M A , A A, ,A A, %(%(

where?80; and 8;0 are the total prestressing moments at two adjacent supportsP is the tendon forcee0, e; ^ e are the eccentricities at 0, ; ^ x, respectively

&ccentricities are taen as negative when it falls below the cgc, and positive above

cgc.

/. o"& )"("n$ing Met%o& o# An"(3sis

The basic concept of load balancing also represents the inuence of the tendonsby e"uivalent loads. The tendon is selected to directly counteract the imposed loadingat the indicated eccentricity, e. )ince the moment induced by the tendon and the loado<set each other, the net stress in the beam will be the axial compression due toprestressing, = '%. =f it is desired to design the beam for /ero stress at the bottom!ber at center span Dor for any value of stress less than the modulus of ruptureE, it isonly necessary to reduce the amount of prestressing provided. The net stress on thesection may be calculated from?

.P.0ncog#25



: t a n

: t a n

: :

e

6p

oad 6

cgc

cgs

Simple 5eam

1 2 3

Continuous 5eam

I

c M

A

P f net +=

where 8net is the net DunbalancedE bending moment on the section.

*or continuous beam designs, the tendon geometry would be assumed to have aparabolic shape, as shown.

While load balancing is an extremely powerful concept for visuali/ing the e<ectof prestressing on the structure, there are some "uali!cations that should be ept inmind when using the techni"ue in !nal detailed designs.

#. =t is not practical to install tendons with a brea incurvature oversupports. The sti<ness of the tendons re"uires a reverse curvature in

the tendon pro!le with a point of contraexure some distance fromthe supports. This reverse curvature modi!es the loads imposed byprestressing from those assumed using load balancing.

5. The load balancing techni"ue should be adjusted to counter frictionlosses and the resulting variation of force in the tendon along itslength.

9. 0s in e"uivalent load method, load balancing does not accountdirectly for moment induced by tendons anchored eccentrically atends free to rotate.

Pre(i"r3 Siing o# Prestresse& Me'ers

There are no set span-depth limits for prestressed members, but some values inthe table below are provided as a guide to the preliminary si/ing of members.

T3+i$"( s+"n-&e+t% r"tios

&lement Type+ontinuous span )imple span

3oof *loor 3oof *loor#-way solid slab 62 :6 :6 :25-way solid slabs, supported on columns only :6-:$ :2-:6

.P.0ncog#29



1 / .2)=&24= cols,? 1)/ long

2)=

*0/))= *0/))=

b $ 1-2=

. =

3 ) =

14=

e & t e r i or c ol

i n t e r i or c ol

3-*,*

-)4,3

+4)2,1

12.,0

2.4,4

+132,2

*0/)=

5-way wa_e slabs D9%V pansE :2 96 96 92;eams 96 92 92 5%Ane-way joists :5 9$ 9$ 96

These ratios maybe increased if calculations verify that deection, camber, andvibration fre"uency and amplitude are not objectionable.

Design E"+(e 1: 2-S+"n C"st-In-P("$e T-)e"

esign a 5-span cast-in-place T-beam of normal weight concrete. 4se f cJ G 6222psi and f pu G 5:2222 psi D#I:V dia. wire, post-tensionedE. ;eams are 5%.1 ft on centers.

Se$tion +ro+erties:

0g G #6#5 in5

yt G $.22 inyb G 5$.22 in=g G #99222 in:

`t G #%%92 in9

`b G :>65 in9

De"& (o"& G 5:66 lbIft Dincluding full slabEie (o"& G 62 psf reduced to 92 psf on columns and beamsMoents

*rom the above information, the dead load moments were calculated Dbymoment distributionE to be Z9$6.6 -ft at the exterior columns and Z$2:.9 -ft at theinterior column and H:25.# -ft at midspan. The corresponding live load moments Dnotconsidering alternate panel loadingE are Z#5%.>, -5%:.: and H#95.5 -ft, respectively.

8oment at B BB Total&xterior columns -9$6.6 -#5%.> -6#5.5=nterior columns -$2:.9 -5%:.: -#2%$.>8id-span H:25.# H#95.5 H69:.9

.P.0ncog#2:



@12

$ 4,0*/(11@12%(*0%$*2,2*/

3=

4= - =

2 - =

1 /

a

b

4,0*/*2,2*/

Mp $ :e : $ 1 7ipa $ (2- G 3% & 1 $ 2* 7in

b $ (- G 4% & 1 $ 4 7in

Post-tensioning

The e<ects of post-tensioning are treated separately from the e<ects of loadsand are dependent on the magnitude and position of the tendon force and on frameaction. 0ssume two-end stressing. 0 tendon pro!le is selected?

Se$on&"r3 Moents

)econdary moments induced by post-tensioning are accounted for byconsidering the e<ects of restrained rotations. *or simplicity, a post-tensioning force of # ip is used.

4sing conjugate beam method?

3otation Y #?

3,401*0

1%0*,4(

-

340*,4

3

20*,4

2

2*,*22*,*22*

3

21 −=

−

+= x x x x x x %I θ

3otation Y 5?.P.0ncog#26



+,3-.%0*,4(-

*2*,*240*,4

3

22*

2

2*,*2

3

2 2

2 −=

+−−= x x x x x %I θ

*ixed end moments?

*&8 Y #? G

ink !

−+==−

*),1+*,2-

0,***

2

2 12 θ θ

*&8 Y 5? G

ink !

−+==−

.1,1)*,2-

*,3)2

2

2 21 θ θ

8oments Dper ip of forceE

;y moment distribution, moments at exterior columns are H##.9# -in, atinterior column, H #:.># -in, and at mid-span, H #9.2# -in. These secondarymoments can be considered to be?

%( e P M S ∆=

)o the resulting ∆e values, in this example, are ##.9#V, at columns # ^ 9, #:.>#V atcolumn 5, and #9.2#V at mid-span.

Re,ire& Post-Tensioning For$e:

0llow a tensile stress of? ) G psi f c 424*))).'. ==

8id-span ∆8 due to allowable tensile stress?

ft k - f c

I f M b

b

−====∆ +,1.0%1)))(12

40*2424''

)upport ∆8 due to allowable tensile stress?

ft k - f c

I f M t

t

−====∆ .,*-0%1)))(12

1..3)424''

*inal 3e"uired Prestress Y &xterior )upport?

.P.0ncog#2%



1 /

2 * =

4 =

2-,*2-,*

1 2

k

ee A

-

M M P

g

t

port T .,4)

31,11)1*12

1..3)

12%.,*-02,*12(sup −=++

−=

∆++

∆−=

*inal 3e"uired Prestress Y =nterior )upport?

k

ee A

-

M M P

g

t

port T 3,1+4

01,1441*12

1..3)

12%.,*-00,1).-(sup +=++

−=

∆++

∆−=

*inal 3e"uired Prestress Y 8id-span?

k ee

A

-

M M

P

g

b

midspnT

*,2+))1,132*

1*12

40*2

12%+,1.03,*34(

=−+

−=∆−+

∆−=

The !nal prestressing re"uired is the maximum computed?

[ ] *,2+)*,2+)?3,1+4?.,4)?? ma&ma&int =++−==

midspnerior exterior P P P P

8id-span prestressing force governs.

A++roi"te Prestress oss !"(,es

Post-tensioning Tendon 8aterial

Prestress Boss, psi)labs ;eams ̂ (oists

)tress relieved 5>2\ strand, stress relieved 5:2\ wire

92,222 96,222

;ar 52,222 56,222

This Table of approximate pretsress losses were developed to provide common Post-Tensioning =ndustrybasis for determining tendon re"uirements on projects in which the magnitude of prestress losses is notspeci!ed by the designer. These loss values were based on the use of normal weight concrete and onaverage values of concrete strength, prestress lebvel, and exposure conditions. 0ctual values of lossesmay vary signi!cantly above or below the Table values in cases where the concrete is stressed at low

strengths, where the concrete is highly prestressed, or in a very dry or very weight exposure conditions. The table values do not include losses due to friction.

Fri$tion osses

µ G 2.52, \ G 2.22#

.P.0ncog#2>



)tressing simultaneously at supports # ^ 9?

%( µα += *!

xo e P P

*or circular curve?

= −

2

2tan 1

!

eα

rdins14*,*,2-

%12@2*(2tan 1

1 =

= −α

rdins10),*,2-

%12@2+(2tan 1

2 =

= −α

*or mid-span?

µα# G 2.52D.#:6E G 2.2512

\B G 2.22#D5$.6E G 2.25$62.26>6

x x

*!

xo P e P e P P )*+,1%)*0*,)(%( 1 === + µα

pu pu

pu

xo

pu f f

f

P P

f 0,)0**,)

)*+,1

-,)

@

-,)>==

, so use 2.>f pu.

=nitial steel stress G 2.>f pu G 2.>D5:2E G #%$.22 siBosses taen G 56.22 si*inal steel stress G #:9.22 si

*or interior support?

µΣα G 2.52D.#:6 H 5D2.#%$EE G 2.21%5\B G 2.22#D6>E G 2.26>2

2.#695

.P.0ncog#2$



x x

*!

xo P e P e P P 1..,1%1*32,)(%( 1 === + µα

ksi f f

P P

f pu

pu

xo

pu0,1.4%24)(.-.,).-.,)

1..,1

-,)

@

-,)====

=nitial steel stress G 2.%$%>f pu G 2.%$%D5:2E G #%:.>2 siBosses taen G 56.22 si*inal steel stress G #91.>2 si

Deterine N,'er o# Wires:

4se V dia. wires, 0o G 2.2:1# in5, mid-span control the number of wiresre"uirements.

?30,41%)4+1,)(143

*,2+)

%(===

oe A f

P N

use :5 Z V dia. wires.

+hec total prestress at interior support?

k k A Nf P oe 3,1+41,2--%)4+1,)%(0,13+(42 >===, oX

*(ti"te Moent An"(3sis:

4ltimate load wu? #.:B H #.>BB G #.:D5:66E H #.>D$2>E G :$2$ plf Total service load? w G B H BB G 5:66 H $2> G 95%5 plf

0t supports # ^ 9?

ft k M w

w M T

uu −=== 0**%2,*12(

32.2

4-)-

0t support 5?

ft k M w

w M T

u

u −=== 1*0*%0,1).-(32.2

4-)-

.P.0ncog#21



0t mid-span?

ft k M w

w M T

uu −=== 0--%3,*34(

32.2

4-)-

8oment provided?D;onded reinforcementE0t supports # ^ 9?

( ) ft k x xd f A M

f

f

ksi f

f f f

bd

NA

bd

A

s su

c

s

c

pu

pu s

o s

−=−=−=

<===

=

−=

−=

====

0,0-+12@%N221,)*,1%(2-%(0,2)+%()4+1,)42(+M,*,1

,ma&3,)221,)*

0,2)+%))*2.,)(

'

0,2)+*

24)%))*2.,)(*,124)%

'(

2

11

))*2.,)%2-(14

%)4+1,)(42

ρ ρ

ρ

ρ

ρ

ω φ

ρ ω

ρ

ρ

8u provided G >$1.> -ft N 8u re"uired G >66 -ft, o

0t mid-span?

( ) ft k x xd f A M

f

f

ksi

f

f f f

bd

NA

bd

A

s su

c

s

c

pu

pu s

o s

−=−=−=

<===

=

−=

−=

====

12)312@%N)1.,)*,1%(33%(),23-%()4+1,)42(+M,*,1

,ma&3,))1.,)*

),23-%)))343,)(

'

),23-

*

24)%)))343,)(*,124)%

'

(

2

11

)))343,)%33(1-2

%)4+1,)(42

ρ ρ

ρ

ρ

ρ

ω φ

ρ ω

ρ

ρ

8u provided G #529 -ft N 8u re"uired G >$$.2 -ft, o

0t support 5?

( ) ft k x xd f A M

f

f

ksi f

f f f

bd

NA

bd

A

s su

c

s

c

pu

pu s

o s

−=−=−=

<===

=

−=

−=

====

),+3412@%N1+.,)*,1%(32%(*,213%()4+1,)42(+M,*,1

,ma&3,)1+.,)*

*,213%))4.),)(

'

*,213*

24)%))4.),)(*,124)%

'(

2

11

))4.),)

%32(14

%)4+1,)(42

ρ ρ

ρ

ρ

ρ

ω φ

ρ ω

ρ

ρ

8u provided G 19:.2 -ft O 8u re"uired G #6>6 -ft, so add rade %2 rebars.

.P.0ncog##2



+alculate moment at face of column?

k !

M M !w&

umidspnu

uu 4,1*14,14),130*0

%0**1*0*(%*,2-(-)-,4

2

2A =+=−

+=−

+=

0pprox. moment reduction? ∆8 G 3

2

SuDaE@ a G QDcol. si/eE G Q D52E G #2 in.

ft k M −==∆ ),-412

1)%4,1*1(

3

2

8oment at face of column? 8u G #6>6 Z $: G #:1# -ft

3e"uired additional rebars?

Tma&?3),)2++,)1)3,)1+.,)

1)3,*

.)

%32(14

-0,3

''

-0,3%32%(.)(+,)

%12%(+3414+1( 2

ok

f

f

bd

A

f

f

ind f

M M A

c

' sb

c

'

b

'

uprovided u

sb

<=+=+

==

==

=−=−

=

ω ω

ρ ω

φ

ρ

C%e$5 stresses "t seri$e (o"&:

0t exterior supports?

f se G #:9.22 si

P G 70of se G :5D2.2:1#ED#:9E G 516

*iber )tress? -

M

-

ee P

A

P f T ±

∆+±= %(

Top !ber stress?

ncompressioksi

-

M

-

ee P

A

P f

t

T

t

t

?)20,)3.+,)2)1,)1+*,)

1..3)

%12(*12

1..3)

%31,11)(2+*

1*12

2+*

%(

+=−+=

−+

+=

−∆+

+=

;ottom !ber stress?

.P.0ncog###



ncompressioksi

-

M

-

ee P

A

P f

b

T

b

b

?0-.,)2+3,10)2,)1+*,)

40*2

%12(*12

40*2

%31,11)(2+*

1*12

2+*

%(

+=+−=

++

−=

+∆+

−=

0t mid-span?

*iber stress? -

M

-

ee P

A

P f T ±

∆−±= %(

Top !ber stress?

ncompressioksi

-

M

-

ee P

A

P f

t

T

t

t

?3.0,)3-*,)213,)1+*,)

1..3)%12(*34

1..3)%)1,13)),2*(2+*

1*122+*

%(

+=+−=

+−−=

+∆−

−=

;ottom !ber stress?

T?424*))).'.'4)+

?4)+,)34-,1044,)1+*,)

40*2

%12(*34

40*2

%)1,13)),2*(2+*

1*12

2+*

%(

ok psi f f psi

tensionksi

-

M

-

ee P

A

P f

c

b

T

b

b

===<

−=−+=

−−

+=

−∆−

+=

0t interior support

f se G #91.>.22 siP G 70of se G :5D2.2:1#ED#91.>E G 5$$

*iber )tress? -

M

-

ee P

A

P f T ±

∆+±= %(

Top !ber stress?

.P.0ncog##5



22/))= transversedesign strip

3 ba!s A 22/))=

1 ) =

1 ) =

cols, 1.= & 1.=

tensionksi

-

M

-

ee P

A

P f

t

T

t

t

?2*0,)001,)324,)1+),)

1..3)

%12(1).+

1..3)

%01,144(2--

1*12

2--

%(

−=−+=

−+

+=

−∆+

+=

;ottom !ber stress?

ok psi f f psi

ncompressioksi

-

M

-

ee P

A

P f

cc

b

T

b

b

?22*)%*)))(4*,'4*,)10**

?0**,1.++,2134,11+),)

40*2

%12(1).+

40*2

%01,144(2--

1*12

2--

%(

===<+=+−=

++

−=

+∆+

−=

These stresses computed are all within allowable limits. 3epeat stress chec forinitial prestress force of 2.$2f pu, allowable stress for compression f c G 2.%f cJ, and

tension fJ G 9 'c f

.

To complete the design, chec mid-span deection and design the shearreinforcements.

Design E"+(e: F("t P("te

esign a two-way post-tensioned at plate with 55J-22V s"uare bays as shown.

o"&ing:

)uperimposed dead load DpartitionE 52 psf Bive load :2 psf

.P.0ncog##9



3 ,2 * =

3 ,2 * =

3 =

3 =

3 =

1,2*=

1,2*=

),)= ),)=2,)= 2,)=1,)=+2,)= +2,)=

endon :rofile

M"teri"(s:

+oncrete Dnormal weightE #62 pcf +ompressive strength, f cJ :222 psi

Prestressing stee(:

4se QV dia. strand unbonded, f pu 5>2 si

Non-+restresse& rein#.4 0)T8 0%#6-%2, f y %2 si

Se(e$t s("' t%i$5ness:

4se span-depth ratio of :2

,U*,.

.,.4)

%12(22

4) slbuse

in !

t ===

o"&ing #or 22 #t stri+:

)lab dead load?klf 0+,1

1)))

22%1*)(

12

*,.=

)uperimposed dead load DpartitionE?klf 44,)

1)))

222) =

Bive load?klf --,)

1)))

224) =

Se(e$t ten&on +ro9(e:

*or [V min. cover and QV dia. tendons?

8in. distance face of slab to cgs of tendons G 2.>6 H 2.6I5 G #.22, use #.56V

.P.0ncog##:



With e"ual length spans, simple beam moments are identical for all spans,therefore set total drape Dnet vertical pro!le distanceE of the tendons the same foreach span as shown in the tendon pro!le.

)ince the tendon is anchored at slab mid-depth, the end span sets the max. mid-span drape as the sum of the negative eccentricity and the average positive

eccentricity?

8id-span drape G 5 H D5 H 2EI5 G 9V

8id-span eccentricity of Z#V is then calculated for the interior span to give e"ualtotal drape in all spans.

*or une"ual span frames, the tendon eccentricity is usually set at the maximumallowed over each support to get maximum ultimate strength capacity.

Se(e$t '"("n$ing +restress #or$e:

;alance total dead load? #.>1 H 2.:: G 5.59 lf

oment spn simple !w

drpetotl elossesll fter force prestress P where !w

Pe

b

b

=

===

-

'??-

'

-

2

stripdesign for k e

!w P b '22*4)

%3(-

%12(%22(23,2

'-

22

===

0t this stage, the balanced load portion of the design is complete. 0 prestressforce of 6:2 following the pro!le shown in the !gure for tendon pro!le, will produce auniform upward load of 5.559 lf, which is e"ual and opposite to the uniform totaldead load. =7 this ideali/ed state, bending moment stresses are /ero, and the slab is indirect compression caused by the prestress.

f c G PI0 at balanced condition

)imilarly, bending stresses at time of prestressing are calculated by summationof initial compressive stress DPoI0E and bending stress due to the distributedoverbalanced moment D8obE caused by the di<erence between balanced uniform loadand the uniform load at time of prestressing Dusually slab dead loadE.

-

M

A

P f ubo

c ±=

I&e"(ie& stress "t '"("n$e& $on&ition

.P.0ncog##6



2,2,

42,) 42,)30,1 30,1

2), 2),1*,

nbalanced Moments

P G 6:2 0 G %.6D55x#5E G #>#% in5

ncompressio psi x

A

P f c ?31*

101.

1)))*4)===

I&e"(ie& stress "t seri$e (o"&ing

Perform moment distribution for the unbalanced uniform live load of 2.$$ lf andresults are plotted as follows?

+hec stresses at maximum moment?

ncompressio psi

x -

M

A

P f ub

c

?4420131*

.@%*,.(1222

%1)))%(12(),42

101.

%1)))(*4)2

=−=

−=−=

therefore, the slab has no tensile tress under full service load.

I&e"(ie& stress "t initi"( (o"&ing:

0ssume initial tendon stress G 2.>fpu, and e<ective tendon stress G 2.%fpu.

Then initial PoI0G psi3.0%31*(

.,

0,=

, and initial wb Gupwrd klf .),2%23,2(

.,

0,=

4niform slab dead load G #.>1 lf downward7et loading G 5.%2 Z #.>1 G 2.$# lf upward DoverbalancedE

)ince the overbalanced initial loading D2.$# lfE is less than the unbalanced service

loading D2.$$ lfE, and A

P o

is greater than A

P

, it is apparent that the slab has no tensile

stresses at initial loading.

Co+"rison o# i&e"(ie& &esign 0it% ore rigoro,s "n"(3sis.

)ince tendons cannot be placed with sharp peas of tendon pro!le, the actualtendon pro!le produces load reversals as it changes from concave upward to concavedownward, and resultant moments change accordingly. 4sing e"uivalent loading andtendon iniction points at #I#6 of span lengths and from structural analysis Dmomentdistribution, computer program, etc.E, the following results are?

.P.0ncog##%



&xt. +ol. 8id-span =nt. +ol. 8id-span-5>.: -#$.$ -6#.6 -1.1

+hec stresses at max. moment?

tension psi -

M

A

P f ub

c

?10%201(),42

*,*131* −=−=−=

C%e$5 *(ti"te F(e,re

*rom moment distribution Dmoments in -ftE8oment &xt. col 8id-span =nt. col 8id-span

8B -%#.5 H69.> -#22.> H:5.>8BB -5$.$ H#1.> -9$.: H#>.:8u actual -#9:.9 H#2$.> -52%.9 H$1.:8u

capacity-#99.6 H55%.6 -59#.5 H#$2.%

C"($,("tion o# ,(ti"te oent $"+"$it3:

0ssume e<ective prestress of prestressing steel, f se G 2.%f ρs G 2.%D5>2E G #%5 si, then

233,31.2

*4)in

f

P A

se

ps ===

0t interior column?

d G %.6 Z #.56 G 6.56 in

))24,)%2*,*%(1222(

33,3 === xbd

A ps ρ

ksi f

f f c se s 1-+

%))24,)(1))

41)1.2

1))

'1) =++=++=

ρ ρ

, DspecsE

ok f

f

c

s?3),)113,)

4

1-+))24,)

'<=== ρ

ρ ρ ω

in xb f

f A

c

s ps0),)

%1222%(4(-*,

%1-+(33,3

'-*,=== ρ

ft k

d f A M s psu −=−=

−= 2,23112@%N2@0,2*,*%(1-+(33,3M+,

2 ρ φ

C%e$5 +,n$%ing s%e"r stress:

.P.0ncog##>



) , * - =

. , * =

),).

+),.

*or 55J x 55J panel?

SB G #.:D5.59 lfED55E G %$.> SBB G #.>D2.$$ lfED55E G 95.1

Su G #2#.%

*or a #%V x #%V cols., bo G :Db H dE G :D#% H 6.56E G $6 in.

+oncrete shear capacity? psi f v cuc .3,2.*4)))2,4'2,4 ===

0ctual shear stress?

.3,2.*-*,2.0%2*,*%(-*(-*,

1))).,1)1≈===

x

d b

& v

o

u

uφ taen o.

0lthough the value obtained is somewhat higher, empirical e"uations developedfrom the following tests indicate ade"uate punching shear capacity.

in-S$or&e(is-M"3 Tests:

'%@))))2,)'))))242,)10*,)( ccu bdf S P f & +−=P G 6:2 , ) G 55J x #5, b G :D#%E G %: in., d G 6.56 in.

k x x x x& u ),1.1%),4%(2*,*%(.4%%N(1222@(1)*4)(1)),2%4)))(1)42,210*,)M 3** =+−= −−

N #2#.%

Gro0-!"n&er'i(t Tests:

Su G D9%2 H2.92fcEbd G 9%2 H 2.9D9#6EUD%:ED6.56EI#222 G #69.22 N #2#.%

Mini, 'on&e& rein#or$eents:

*rom 0+= +ode, when tension exists in the precompressed tensile /one at a loadof B H #.5BB, a minimum amount of bonded reinforcement is re"uired. 0t interiorcolumn, the moment at B H #.5BB G #22.> H #.5D9$.:E G #:>.2 -ft?

-

M M

-

M

A

P f !! "! s

c

%2,1( +±±= ρ

P G 6:2 , 0 G #>#% in5, ` G #$61 in9 , 8ρs G $>.% -ft D from momentdistributionE

Top !ber?ksi

x x f c ).+,)

1-*+

%12(140

1-*+

12.,-0

101.

1)*4) 3

−=−+=

;ottom !ber?ksi

x x f c .++,)

1-*+

%12(140

1-*+

12.,-0

101.

1)*4) 3

+=+−=

.P.0ncog##$



' '

$

$ Section at "" ransformed

Section

:restressed Concrete ension Member,

(n1%"s

Tension in concrete?

21-,)%.)(*,)

2-,*

*,)

2-,*%1222%(*-,)%().+(,2

1'

2

1

in f

T A

T f A

k xcb f T

'

c

s

c s s

c

===

=

===

Provide? 5 - K9 top bars at int. columns.

Tension Me'ers

Prestressed tension members combine the strength of a high-tensile steel withthe rigidity of concrete and provide a uni"ue resistance to tension consistent withsmall deformations that cannot be obtained by either steel or concrete acting alone. The rigidity of prestressed concrete serves well, especially for long tension memberssuch as tie rods for arches or staybacs for wharves and retaining walls. Whenprestressed, concrete is given strength to resist any local bending and at the sametime steel is sti<ened and protected.

T%e '"si$ 'e%"ior o# +restresse& tension e'ers:

#. The member can be considered as essentially made of concrete which is putunder uniform compression so that it can carry tension produced by externalloads. =f the concrete has not craced, it is able to carry a total tensile forcee"ual to the total e<ective precompression plus the tensile capacity of theconcrete itself.

5. The member can be considered as essentially made of high-tensile steelwhich is pre-elongated to reduce its deection under load. *rom this viewpoint, the ultimate strength is often limited by excessive elongation of thesteel which usually taes place at the cracing of the concrete.

9. The member can be considered as a combined steel and concrete member

whose strains and stresses before cracing can be evaluated, assumingelastic behavior and taing into account the e<ect of plastic ow.

.P.0ncog##1



=nitial prestress?

c

o

co

A

# f =

&<ective prestress?

c

c A

# f =

Transformed section?

c st AnA A += unbonded section

s g t An A A %1( −+= grouted section

)tress produce by P?

t

c A

P f =

for concrete

t

c s

A

nP nf f ==

for steel

c

s

%

% n =

consider the e<ect of creep ^ shrinage3esultant stress Ddue to e<ective prestress plus external loadE?

t c

c A

P

A

# f +=

for concrete

t

c s A

nP f f +=

for steel

To !nd load P to produce resultant stress to /ero?

%1() ρ n # A

A # P

A

P

A

#

c

t

t c

+−=−==+

.P.0ncog#52



=nstantaneous unit strain?

c

oo

%A

# =δ

4nder the action of external load P?

t %A

P =δ

+omparison of strains in a prestressed concrete and in ordinary steel member?iven? f s G 52,222 psi, &s G 92x#2% psi

))).0,)1)3)

2)))). ===

x %

f

s

sδ

iven? f c from Z#,222 psi to 2, &c G :x#2% psi

)))2*,)1)4

1))). ===

x %

f

c

cδ

iven? f se G #56,222 psi, &s G 92x#2% psi

))410,)1)3)

12*))). ===

x %

f

s

seδ

, which % times that of structural steel and #%

times that of prestress concrete.

E"+(e 1: Tension Me'er

0 straight concrete member #62 ft long is prestressed with a high tensile steelstrand through the centroid of the section. The strand is anchored to the concrete withend anchorages but separated from it by bond-breaing agents along the length. 0 c G$2 in5, 0s G 2.$2 in5, f cJ G :,222 psi, f sJ G 562,222 psi, f o G #62,222 psi, f se G #5>,622psi, &c G :x#2% psi, &s G 92x#2% psi.

aE +ompute the allowable external load on the member, allowing no tension inthe concrete.

bE +ompute the shortening of concrete due to prestress, assuming a creepcoecient of #.6.cE +ompute the lengthening of the member due to the external load obtained

in aE, neglecting creep.dE =f the member were designed of structural steel with an allowable stress of

52,222 psi, compute the lengthening under the load.eE +ompute the lengthening if the strand is used alone by itself with an

allowable stress of #5>,622 psi.

.P.0ncog#5#



So(,tion:

aE 0llowable external load, allowing no tension?

lbs x

xn A f n # P s se )))?11)%

-)

-),)

1)4

1)3)1%(-)(,120*))%1(%1(

.

.

=+=+=+= ρ ρ

bE 4nder initial prestress, compute the shortening?

,.0*,)%-)(1)4

%121*)%(-),)(1*)))).

in x

x

A %

! #

cc

oo ===δ

=f e<ective prestress is considered?

,*03,)1*))))

120*)).0*,) in

f

f

o

se

o === δ δ

=f creep is considered?

,-.),)%*,1(*03,) inC cc === δ δ

cE 4nder the external load of ##2,222 lbs?

2

.

.

-.%-),)(1)4

1)3)-) in

x

xnA A A sct =+=+=

?

,*0*,)%-.(1)4

%121*)(11)))).

in x

x

A %

P!

t c

===δ

, a closely chec the shortening of concrete in bE.dE *or a structural steel stressed to 52,622 psi, the elongation is?

in. x

x

%

! f

s

s

s 2),11)3)

%121*)(2)))).

===δ

eE *or high tensile steel stressed to #5>,622 psi?

,.*,0

1)3)

%121*)(120*)).

in

x

x

%

! f

s

se s ===δ

enerally speaing, pretsressed-concrete tension members have a very lowreserve strength above the point of /ero stress. =n order to get a sucient factor of safety, it may be necessary to design the member so that, under woring loads, therewill be some residual compression in the concrete.

.P.0ncog#55



<o<o C $ <o C $ <o

>ue to prestress? <o

Cline $ cgc linecgs line

J

E"+(e 2:

*or the tension member in &xample #, what woring load can it carry using afactor of safety of 5.2 against the cracing of concrete, assuming the direct tensilestrength of concrete to be 2.2$f cJ G 952 psi +ompute the residual compression inconcrete under the woring load.

So(,tion:

*or f c G 952 psi? * G f se0s G #5>622D2.$2E G #25222 lbs, 0c G $2 in5, 0t G $% in5?

,?)))?130

-.-)

1)2)))32)

lod crcking theiswhichlbs P

P

A

P

A

#

t c

=

+−

=+

*or a factor safety of 5.2?

lbs #S

P P

crcking

working *))?.-),2

130)))===

The residual compression computed?

psi A

P

A

# f

t c

c 4-)0+*120*-.

.-*))

-)

1)2)))−=+−=+

−=+=

Cir$,("r Prestressing

+ircular prestressing is employed to denote the prestressing of circularstructures such as pipes and tans where the prestressing wires are wound in circles.=n most prestressed circular structures, prestress is applied both circumferentially andlongitudinally, the circumferential prestressing being circular and the longitudinalprestress actually linear.

ue to circular prestress, the stress in concrete is always axial an e"ual to?

c

o

co A

# f

−=

, at transfer

.P.0ncog#59



c

c A

# f

−=

, at !nal stage of loading after all losses.

With application of internal li"uid pressure?

sct

t

c An A A A

/) f %1( −+==

The resultant prestress in the concrete under the e<ective prestress * and the internalpressure "?

t c

c A

/)

A

# f +

−=

=n order to be exact, the value of n has to be chosen correctly, considering the level of

stress and the e<ect of creep.

The criteria for designing prestressed tans vary. The usual practice is to providea slight residual compression in the concrete under the woring pressure. This isaccomplished as follows?

#. 0ssume that hoop tension produced by internal pressure is entirely carried by

the e<ective prestress in steel@ then /) f A # s s == and the total re"uired steelis?

s

s f /) A =

, the total initial prestress is then, so so f A # =

The concrete area to resist the initial prestress *o?

c

o

c f

# A

−=

, from the value of the re"uired 0c, the thicness t of tan can beobtained.

5. +orresponding to the adopted value of 0c, the stress in the concrete and steel

under internal pressure " can be obtained as?

t c

c A

/)

A

# f +−=

, the stress in concrete

c s f s nf f f +=, the stress in steel

)ince the serviceability of the tan is impaired as soon as the concrete begins tocrac, it is of utmost importance that ade"uate factor of safety be provided against

.P.0ncog#5:



cracing. =f a factor of safety m against cracing is re"uired, the design procedureadopted will be as follows?

#. 0ssuming f t G tensile strength in concrete at cracing Dwhich averages 2.2$f cJbut may be /ero if the concrete has previously craced or if precast blocs areusedE?

t c

t A

m/)

A

# f +−=

5. 0t the same time in order to limit the maximum compression in concrete to f c,then?

c

o

c f

# A −=

, and

sc so so

c s se

t nA f A f

m/)

A f

f A f

f ++−= %(

9. etermine the re"uired steel area?

%@1N(%@(M ococt se

s f nf f f f f

m/) A

−−=

:. 0fter 0s is obtained, *o and 0c can be computed by?

c

o s

c

o

c f

f A

f

#

A −=−=6. )tresses can be evaluated using the following e"uations?

t c

c A

/)

A

# f +−=

, the stress in concrete

c s f s nf f f +=, the stress in steel

Prestress (osses #or t"n5s:

0verage loss of prestress G 56,222 psi4sual allowance of loss G 96,222 psi, considered conservative0t extreme adverse condition G :2,222 psi+reep and shrinage allowance?

elastic and creep strain in concrete G 2.2226shrinage G 2.2226

Total loss G 2.22#2which amounts to about 5$,222 psi, taing &s G 5$x#2% psi.

.P.0ncog#56



E"+(e /: T"n5 Prestressing

etermine the area of steel wire re"uired per foot of height of a prestressedconcrete water tan %2 ft inside diameter to resist 52 ft of water pressure. +omputethicness of concrete re"uired, f cJ G 9,222 psi, f c G >62 psi, n G #2, f o G #62,222 psi, f se

G #52,222 psi. 7eglect mortar coating in the calculations. esign both steel and

concrete on the following two bases?

#. 0ssuming all hoop tension carried by e<ective prestress.5. for a load factor of #.56, producing /ero stress in the concrete.

So(,tion:

Pressure of 52 ft of water?

psf h/ 124-%2)(4,.2 ===ω

aE 0ssuming all hoop tension carried by the e<ective prestress?

3e"uired steel area?

2312,)12))))

%3)(124-in

f

/) A

se

s ===

3e"uired concrete area?

2) *,.2

0*)

%1*))))312,)(in

x

f

# A

c

c =

−

−=

−=

*or a height of #5 ], the thicness re"uired is?

,*,*,?2,*12

*,.2indopt in

b

At c ===

and under the action of the internal pressure?

22 12,.+%312,)(1)....%12(*,* innA A Ain A A

/)

A

# f sct c

t c

c =+=+===+−=

psi x

f c

2.12,.+

%3)(124-

..

12))))312,)

−=+−=

bE *or a load factor of #.56, producing /ero stress?

3e"uired steel area?

.P.0ncog#5%



load carried b!

ring tension

load carried b!

vertical element

8

ertical Sectionof an7 Iall

and >eflectionunder oad

oad >istribution bet6een 8oriKontal

R ertical Dlements

Moment inertical Dlement

Moment and deflection in vertical element of tan7 6all

),)

),1

),2

),3

),4

),*

),.

),0

),-

),

1,)

,) ,1 ,2 ,3 ,4 ,* ,. ,0 ,- , 1,)

8 e i gh t ?8

Coefficient? C

<i&ed 5ase? <ree op

8P$ 82

2t

ing ension $ C 8

8

t

2 P$ ) ,4

P$ ) , -

P$1 , .

P$ 3 , )

P$ . , )

P$1 2 , )

2302,)%1*))))@%0*)(1)1%N(1*))))%%(0*)@()(12))))M

%3)%(124-(2*,1

%@1N(%@(M

in

f nf f f f f

m/) A

ococt se

s

=−−−−

=

−−=

3e"uired concrete area?

24,04

0*)

%1*))))(302,)in

f

# A

c

o

c =−

−=−=

Thicness re"uired?

,2,.12

4,04in

b

At c ===

, adopt t G %.6 in.

3esultant stress in concrete under full water pressure? 0c G %.6D#5E G >$ in5

0t G 0c H n0s G >$ H #2D2.9>5E G $#.>5 in5

psi A

/)

A

# f

t c

c 11*02,-1

%3)(124-

0-

%12))))(302,−=+−=+−=

Note: ;y designing by the second method, gives heavier sections for both concrete and steel. The design can be economi/ed if some tension in the concrete is allowed at 56C overload.

!erti$"( Prestressing in T"n5s

.P.0ncog#5>



),)

),1

),2

),3

),4

),*

),.

),0

),-

),

1,),) ,1 ,2 ,3 ,4 ,* ,. ,0 ,- , 1,)

8 e i gh t ?8

Coefficient? C

8inged 5ase? <ree op

8P$ 82

2t

ing ension $ C 8

8

t

2

P$1 2 , )

P$ . , )

P$ 3 , )

P$1 , .

P$ ) ,4

P$ ) , -

'ori/ontal elements of the tan wall are subjected to hoop tension and thevertical elements are subjected to bending. The variation of bending in the verticalelements will depend on several factors.

#. The condition of support at the bottom of the wall, whether !xed, hinged, freeto slide, or restrained by friction.

5. The condition of support at the top of the wall, whether fully or partiallyrestrained or free to move.

9. The variation of concrete thicness along the height of the wall.

:. The variation of pressure along the depth, whether triangular or trape/oidal.6. The ratio of the height of tan to its diameter.

Sertical prestressing should be designed to stand the stresses produced by variouspossible combinations of the following forces?

#. The vertical weight of the roof and the walls themselves.5. The vertical moments produced by internal li"uid pressure.9. The vertical moments produced by the applied circumferential prestress.

.P.0ncog#5$



" "

2 ) /

2,0*=

-=

1 2 =

1*)) lbs

ertical Dlement of Iall Section ""

inside of tan7

cgs

=n addition to the above, stresses may be produced as a result of di<erentialtemperature between the inner and outer faces of the wall, and by shrinage of concrete walls unless they are entirely free to slide on the foundation. These factorscannot be easily evaluated and hence are often neglected or provided for indirectly inan overall factor of safety..

=t must be noted that the maximum stresses in the concrete usually exist whenthe tan is empty, because then the circumferential prestress would have its fulle<ect. When the tan is !lled, the li"uid pressure tends to counterbalance the e<ect of circumferential prestress and the vertical moments are smaller. )ince it is moreconvenient to use the same amount of vertical prestress throughout the entire heightof the wall, the amount will be controlled by the point of maximum moment. ;yproperly locating the vertical tendons to resist such moment, a most economicaldesign can be obtained. 'owever, e<orts are seldom made to do so, and the amountof prestress as well as the location of tendons is generally determined empiricallyrather than by any logical method of design.

E"+(e : Design o# T"n5 W"((

0 #-ft vertical element of a water tan is shown in the !gure below. =t carries#622 lbs of weight from the roof. 0t a point 52 ft below the top, the vertical momentsare? for initial circumferential prestress 8 G 9522 ft-lb Dtension on the inside !bersEwhich reduces to 5622 ft-lb eventually. *or full uid pressure, 8 G 5:22 ft-lb Dtensionon the outside faceE. The vertical prestressing wire is located 5.>6 in from the insideface and exerts an initial prestress of ##,222 lbIft, which reduces to $,222 lbIfteventually. +ompute the stresses in the extreme vertical !bers of the concrete underthe initial and !nal conditions, considering both an empty and a full tan.

So(,tion:

.P.0ncog#51



> om e pr e s t r e s s i n g 6i r e s

a n d pr e s t r e s s t e m p or a r !

e r e c t i on b a r s

C i r c umf e r e n t i a l pr e s t r e s s i n g 6i r e s

>ome shell

:neumatic mortar

ertical prestressing 6ires

eservoir 6all

!pical Section of >ome ing for an7s

+omputation for stresses in concrete?

)tages of loading condition =nitial *inal

*iber =nsideAutsid

e=nsid

eAutsid

e

0. Weight of roof? f G PI0 G #622ID$x#5E -#% -#% -#% -#%

;. Weight of wall? f G γ hI0 G #62D52EID$x#5E -5# -5# -5# -5#+. 0xial component of vertical prestress? f G

PI0

D##,222EID$x#5E G D$222EID$x#5E G

-##6 -##6-$9 -$9

. &ccentricity of vertical prestress? f G %8IDbd5E G %PeIDbd5E

%D##222x#.56EID#5x$5E %D$222x#.56EID#5x$5E

-#2> H#2>->$ H>$

&. Sertical moments due to circumferentialprestress

f G %8IDbd5E G %D9522x#5EID#5x$5E %D5622x#5EID#5x$5E

Tot"( #or t"n5 e+t3

H922

K1

-922

-/J

H59:

K/;

-59:-2<;

*. Sertical moment due to li"uid pressure f G %8IDbd5E G %D5:22x#5EID#5x$5E

Tot"( #or t"n5 #,((-556-1L

H556-12

-556-

1=L

H556-J1

)light tension of 9% to :# psi exists on the inside vertical !bers when the tan isempty. Atherwise, compressive stresses are obtained throughout.

Doe Ring Prestressing

.P.0ncog#92



otal dome load I

>esign for :restress in Ddge ing of >ome

rical dome

The dome roof itself is made of concrete or pneumatic mortar with thicnessvarying from 5V to %V. *or domes of large diameter, variable thicness may beemployed and thicness greater than %V are used for the lower portion. ;eforeconcreting the dome, some erection bars are prestressed around the base of thedome. 0fter the hardening of the shell concrete, wires are prestressed around it.uring this operation, the dome shell rises from its forms as it is compressed, thussimplifying the careful procedure for decentering re"uired for non-prestressed domes.

8ethods and formulas, though available for the analysis of dome stresses underuniform loads, are applicable only to points on the domes removed from the

discontinuous edge. The computation of stresses in the edge ring becomes verycomplicated problem if the edge ring is prestressed. 'owever, for purposes of design,a conventional method is available. =t consists of prestressing the ring to inducesucient compressive stresses to counteract the tensile stresses set up in the ringunder the maximum live and dead loads. With this prestress, it is usually possible toraise the dome from its falsewor, since only the dead load is actually acting on thedome.

For " s+%eri$"( &oe:

Sertical reaction per unit length along the edge member?

φ π sin2 )

+ & =

The hori/ontal reaction per unit length?.P.0ncog#9#



4*o

12) ft

otal I $ ))7

spherical dome

φ π

φ φ

sin2

cotcot

)

+ & 0 ==

3e"uired prestressing force * to resist the hori/ontal reaction?

φ π

φ cot2

sin + 0) # ==

The e<ective prestressing * having been determined, the cross-sectional area of the ring concrete can be obtained by?

c

o

c f

# A =

3 G radius of domeW G total vertical load on the dome

*o G initial prestressing force, f c G allowable compressive stress in concrete.

=t is desirable to eep f c at a relatively low value, say 2.5f cJ and not greater than$22 psi. This is necessary in order to minimi/e excessive strain in the edge ring whichmight in turn produce high stresses in the shell. =t must be further observed that thisprocedure of design is satisfactory only when there is no possibility of heavyoverloads, because the prestressed edge ring does not posses a high factor of safetyagainst overloads, although the factor of safety is sucient for ordinary roof loading.

E"+(e J: Prestress Doe Ring

0 spherical dome as shown the !gure, carries a total live and dead load of 122.esign the prestress in the edge ring and the cross-sectional area of the concretere"uired for the edge ring. Boss of prestress G 52C, f c G %22 psi.

3e"uired prestress?

.P.0ncog#95



: :

"

"

: :

Column under load :

Moment at "" due to deflection

5

5

<C

Column prestressed 6ith force <

Lo moment at 55 due to deflection

Column $ction Due To Load Column $ction Due To 'restress

"

"

"

"

" prestressed concrete member

<

<

<

<

Concrete as freebod!

Steel as freebod!

Balancing $ction of Concrete and Steel

k +

# o 1434*cot2

+))cot

2===

π φ

π

=nitial prestressing force?

k #

# o

10+-),

143

%2,1( ==−=

3e"uired area of concrete?

22+-.))

1)))143in

x

f

# A

c

o

c ===

Prestressing Co+ression Me'ers "n& Pi(es

.P.0ncog#99





Bent Members 0nder

Concentric 'restress

Eccentric 'restress

$nd Column $ction

into contact with the concrete and the two will begin to deect together. 'ence thecolumn action is limited to the di<erential deection of the two materials.

=f the steel is in contact with the concrete at several points, say at & and *, butnot along the entire length, the column action is limited to the length between thepoints of contact. =f such length is short, column action will not be serious.

C,re Me'ers

=f a curve or bent member is subjected to internal prestress, and if the prestressis concentric at all sections Dthe cgs line coinciding with the cgc lineE, then theconcrete is behaving lie an arch subject to axial force with the correspondingexception that the applied force from the steel will move with the deection of theconcrete and will always remain concentric. 'ence there is no tendency to bucle as inordinary arch under external loads, whose line of pressure is determined by the loadsand may not shift together with the deection of the arch.

=f the prestress is eccentric, the compression in the concrete is still e"ual andopposite to the tension in the steel. 0ny deection of the member will still displaceboth of them together, and there will be no column action due to prestress. The e<ectof an eccentric prestress on the concrete, however, will produce deection of themember should be used in computing column e<ects due to external loads.

0s far as column action is concerned, it is immaterial whether there is anyfrictional loss along the length of the prestressing tendon, because the tension in thestell is always balanced by the compression in the concrete at any section, whateverfrictional loss may occur. 'ence, whether there is frictional loss or not, there will be nocolumn action due to prestress.

Co+ression Me'ers

0 prestressed-concrete compression member is one that carries externalcompressive load. 0 member that is simply compressed by its prestress is not acompression member. 0 prestressed member is not under column action due to itsown prestress, but it is subject to column action under the external compressive load just lie a column of any other material.

.P.0ncog#96



Co(,n ,n&er +restress F 0it% "n e$$entri$it3 e4 (o"&e& 0it% $on$entri$ (o"&P +(,s etern"( oent M.)hort column condition?

t t c

c I

Mc

A

P

I

#ec

A

# f ±+±=

Ee$t o# "i"( +restressing on t%e ,(ti"te strengt% o# $o(,ns.

4nder the action of an external compressive load, the column will shorten andthe prestress in the steel will be decreased.

εuc G 2.229 for concretef e G e<ective prestressf s G remaining prestress&s G modulus of elasticity of steel

suce s % f f ε −=

7umerical values? εuc G 2.229, f e G #52,222 psi, &s G 92x#2%

psi3emaining prestress, f s G #52222 Z 2.229D92x#2%E G 92,222 psi

Re"r5:

The major part of prestress may be lost at the ultimate compressive strength of the concrete. This means that the ultimate load carrying capacity of the column is notmuch decreased by prestressing. =f the column fails on the tensile side as the result of bending or bucling, the steel on that side can be stressed to near its ultimatestrength.

The ultimate strength of concentrically prestressed slender columns under axialloads has been investigated both theoretically and experimentally at variousuniversities. The general conclusion is that the axial prestressing of a slender columnhas no e<ect on the superimposed axial load which will cause that column to bucle. =f the prestressing exceeds the di<erence between the bucling stress and the ultimatestrength of the concrete, the column will fail in compression before it will bucle. Whenthe superimposed load is not axial, prestressing could increase both the cracing andthe ultimate strength.

The bucling of the compressive ange of prestressed beam is subject to thesame reasoning. There is no danger of ange bucling produced by internal prestress

in a beam. *or external loads, the tendency to bucle in the ange is governed by theusual theory of elasticity, so long as there is no cracs in the concrete. 0fter cracingor near ultimate load, little is nown about the bucling of the compressive ange inprestressed beams.

E"+(e 1

0 concrete column #%V by #%V in cross section and #$ ft high is pre-tensionedwith $ Z 9I$ in. wires, which are end anchored to the concrete. The e<ective prestress

.P.0ncog#9%



+ /

+ /

-) 7 - 7

Column

1.=

1 . =

2= to center of steel

2 = t o c e n t e r of s t e e l umn Sectionis #22,222 psi in the steel. *or a concentric compressive load of $2 and a hori/ontal

load of $ at the midheight of the column, compute the maximum and minimumstresses in the column, assuming it to be hinged at the ends. =nvestigate thesecondary moments in the column due to deection. iscuss the safety of the columnunder such loads and also during handling. 0ssume n G >, f cJ G :,222 psi, f sJ G522,222 psi, &c G :x#2% psi.

So(,tion:

)tress in concrete due to prestress?

psi x x

x

A

#

c

344%11,)-1.1.(

%)))?1))%(11,)-(−=

−−

=

)tress due to axial load P G $2 , disregarding deection of column?

psi x x An A

P

A

P

s g t

3).2.1

)))?-)

11,)-%10(1.1.

)))?-)

%1(−=−=

−+−

=−+

=

The maximum bending moment occurs at midheight?

ft k P!

M −=== 3.4

%1-(-

4

=t of the transformed section?

424

23

*.)2142*4.)%.%(11,)3%(10(212

1.%1(2

12in x x An

bh I

i st =+=−+=−+=

&xtreme !ber stresses?

psi x

I

Mc

t

.1.*.)2

%-%(12%(1)))3.(±=

.P.0ncog#9>



8aximum and minimum stresses?

ncompressio psi f

ncompressio psi f

34.1.3).344

12...1.3).344

min

ma&

−=+−−=−=−−−=

8aximum deection of the column due to the hori/ontal load?

,)0*,)%*.)2%(1)4(4-

%12(%1-%(1)))-(

4- .

333

in x

x

I %

P!

t c

==

=ncrease in moment due to axial load?

%*),)(.)))%)0*,)%(1)))-)( ft k lbin x P M −−===∆ δ

Sery small to materially a<ect the column stresses, and may be neglected. The

maximum compressive stress of #5%% psi would appear to be high for a reinforced-concrete column but is not excessive for a prestressed member which more a beamthan a column in this example.

The safety of the column can be determined only if we now the ultimatestrength of the column under such combined axial and transverse loads and also if wenow the possibilities of overloading, that is, to what extent the axial or the hori/ontalloads may be increased, and whether eccentricity of the applied axial load may bepossible.

*or the purpose of investigation, let us assume that both the hori/ontal and the

axial load are increased by 62C while, in addition, there will be an eccentricity of 5 in.for the axial load. The stresses can be computed?

)tress due to axial load? #.6D-92%E G - :61 psi

8oment due to eccentricity? 8 G #.6D$2ED5E G 5:2 -in D52 -ftE

)tress due to eccentricity? psi342%

3.

2)(.1. ±=±

)tress due to hori/ontal load? #.6D psi+24%.1. ±=±

3esulting stresses?

psi f

psi f

4.3+243424*+344

2).++243424*+344

min

ma&

+=++−−=−=−−−−=

Re"r5:

The maximum compressive stress of 52%1 psi is only about 2.65f cJ, while the

tensile stress is below the modulus of rupture of about 2.#5f cJ, hence the column

.P.0ncog#9$



e

:

:

Column undereccentric load

would not have craced, and the midheight deection can still be computed by the

elastic theory not to be more than 2.5 in., which is not a signi!cant value. Thus it can

be concluded that the column is safe.

=nvestigating the handling stresses?

0ssume the column is piced up at the midheight?

3esulting moment?

ft k x ! A !w

M c g c −=== -,1)

2

%+N(144@%1*)(1.1.M

22

222 γ

8aximum tensile stress?

psi psi x

I

Mc

t

3441-**.)2

%-%(12%(1)))-,1)(−<+==

)ince the maximum tensile stress during handling, H #$6 psi is much less than

the precompression stress of Z 9:: psi, the column is safe during handling.

Co(,n *n&er E$$entri$ o"&

Precast bearing walls and columns can be prestressed to improve their elasticbehavior and handling characteristics, and to increase their resistance to lateral forces

both in the elastic and the ultimate ranges. They cannot be designed following rules of

thumb applied to reinforced concrete walls and columns. ;ut they can be designed on

the basis of principles of mechanics and properties of materials. The behavior and

strength of prestressed columns under eccentric loading can be predicted with fair

precision. The degree of accuracy will depend on the choice of values of modulus of

elasticity, the modulus of rupture, and the compressive strength of the concrete.

;efore cracing, the stresses and deections can be calculated assuming the column

to behave elastically. The stress at any section is the sum of the stresses due toprestress, direct axial load, moment due to the eccentricity, and the moment due to

the deection.

.P.0ncog#91



Co(,n stresses:

t t t t I

c P

I

Pec

A

P

A

# f

∆±±−−=

minma&

where?

* G e<ective total pretress including all losses except elastic shortening of concrete due to superimposed load.

P G superimposed load.e G eccentricity of load from the centroid of section.c G distance to the extreme !ber from the centroid of the section0t G area of transformed section=t G moment of inertia of the transformed section∆ G deection of column at the section.

+ritical stress occur at the midheight of the column, where the deection of thecolumn is given by the well-nown secant formula?

−=∆ 1

4sec

2

t c I % P!e

;y the elastic theory, cracing can be assumed to occur when the !ber stressreaches the modulus of rupture. ;eyond cracing, the elastic theory is no longeraccurate. 0n estimation of the ultimate load can be made by the elastic theory,assuming it to be the load at which the extreme !ber stress reaches the compressivestrength of concrete. )ince such an approximate analysis can be way o< Dby perhapssome #2C even for ordinary casesE, it is desirable to apply plastic analysis, taing into

.P.0ncog#:2



C

C

1,*=2 =

"sF . 3@-= strands

vation

ection ""

account the cracing of concrete under tension, the plasticity of concrete undercompression, and the plasticity of steel. *urthermore, while the elastic analysismentioned above will generally err on the conservative side, it is conceivable that,under unusual conditions, erratic conclusions could be reached unless plastic analysisis applied.

E"+(e 2: Co(,n 0it% E$$entri$ o"&

0 pre-tensioned concrete pin-ended column has elevation and section as shown. The e<ective prestress in the six 9I$ in. >-wire strands Das G 2.2$ in5 eachE is #62,222psi or #5,222 lbs per strand. &s G 92x#2% psi. concrete has cylinder strength of 6,>22psi, modulus of rupture of %22 psi, and &c G :x#2% psi. =t is loaded by load P with aneccentricity of #.6 in. along the wea direction. +ompute the cracing and the ultimatevalue of P using the elastic theory, assuming noncraced section.

So(,tion:

aE +ompute properties of the section?

223

2

2

.

.

*311+*12%*,2%()-,3%(1*,0(212

-12%(%1(2

++3+.%)-,).%(1*,0(%12-(%1(

*,01)4

1)3)

in x x

x An I I

in x x An A A

x

x

%

% n

x g t

s g t

c

s

=+=−+=−+=

=+=−+=−+=

===

bE +ompute deection of the column at midheight by the secant formula for severalvalues of P, say P G 62, #22, #52, #:2, and #%2 . 7umerical computation is givenfor P G #52 .

.P.0ncog#:#



0))) 1))) 3)))2)))1))))2)))3)))4)))*))).))) ) 1 2 3)

2)

4)

.)

-)

1))

12)

14)

1.)

1-)

2))

oad :

7ips

Ma&imum compressive stress? psiMa&, tensile stress? psi >eflection? inches

f c / $ * ? 0 )

) p s i

S t r e s s d u e t o< $ 0 2 - p s i

M o d ul u s of r u p t ur e $ . ) ) p s i p s i

Crac7ing load $ 13) 7

ltimate load $ 144 7

esulting tensile stress

esulting compressive stress

>eflection at midheight

of column

a& Stresses by Elastic T(eory b& Deflection by Elastic T(eory

>ue to : >ue to : + :e >ue to :

>ue to : :e

,)3,1

%1.--,1(*,1%10,*3(sec*,1%1+3*,)(sec*,1

1%*31%(1)4(4

%24+(1)12)sec*,11

4sec

.

232

in

x

x

I %

P!e

t c

=−=−=−=

−=

−=∆

cE +ompute stresses in concrete at midspan?

t t t t I

e P

I

Pec

A

P

A

# f

∆±±−−=

minma&

* G 0sf se G D%x.2$ED#62,222Ex#2-9 G >5 , assuming it is not reduced by thepresence of P, and using P G #52 , G #.29 in?

psi f

psi f

x x x x f

340

4223

+3)13**121)02-

*31

%4%()3,1(1)12)

*31

%4%(*,1(1)12)

++

1)12)

++

1)02

min

ma&

3333

minma&

+=−=

±±−−=

±±−−=

dE Plot elastic stresses and deections for various loads P, assuming noncracedsection. *rom the graph, determine cracing load and ultimate load.

.P.0ncog#:5



:

M

t2

t2

12

C

fc/

s1

s2c

Column

Strain at

Section

Stress bloc7 in concrete

C of column

Steel

!c

0ltimate strengt( under combined a,ial load plus bending moment

*(ti"te Strengt% o# Prestresse& Co(,n *n&er Co'ine& Ai"( P(,s)en&ing

0pplying e"uilibrium e"uation?

ct

v

C' '

T T M

T T C P #

+−=Σ

−−==Σ

2%(F)

F)

21

21

;y assuming a location for the neutral axis at ultimate load, setting εc as theultimate strain in concrete and f cJ as the ultimate stress of concrete, and by assigningultimate stress distribution curves for concrete, it is possible to compute thecombination of P and 8 that results in this ultimate failure.

*or slender columns, the value of 8 just computed should include the e<ect of deection, which can be computed by numerical procedure provided the load-momentcurvature relationship of the column section is nown.

E"+(e /: *(ti"te $o(,n (o"&

*or the column section shown in example 5, if the ultimate neutral axis werelocated at 9 in. from one edge, compute the combined P and 8 producing that failure.

.P.0ncog#:9



:

M

t

2

t

2

12

C

,-*fc/

s1

s2c

Column

Strain at

Section

Stress bloc7 in concrete

C of column

Steel

!c

3=*=

2,*= 2,*=

a

So(,tion:

0ssume εc G 2.229, by proportion, changes strains for the steel are?

&s G 92x#2% psi, ec G 2.229?

)))+,)%))3,)(*

*,1

*

*,1

))21,)%))3,)(*

*,3

*

*,3

2

1

+===

−===

c s

c s

ε ε

ε ε

0s G 9D2.2$E G 2.5: in5

k x A % f T

k x A % f T

s s s se

s s s se

-,2)%24,)%%(1)3)%())21,)(1*))))(%(

*,42%24,)%%(1)3)()))+,)(1*))))(%(

.

22

.

11

=−=−=

=+=+=

ε

ε

4sing trape/oidal stress distribution with average ultimate stress at 2.$6f cJ and a G 9V?

.P.0ncog#::



V

d /

d t

:

M

b

:

M

se

ucse G s/

se + s

s/s

s/

s

C Cs/

s

),-*fc/ ),-*fc/

Wensen

dealiKation8ognestad

dealiKation

Cross Section Dlevation Strain ariation Stress 5loc7 dealiKation

Conditions at ultimate in a prestressed concrete column- b

t

:uMu

1

2

cu

Cross section

Dlevation

Strain distribution

in P

M e

ink C'T T M

k T T C P

in '

k xc

b f C

c

t

c

c

)11.,31.+

+0,*)-

+0,*)-%+.,1(232%*,2%(-,2)*,42(

2

%(

1.+-,2)*,42232

+.,1

2321)%2

*3%(12%(*0))(-*,%

2('-*,

21

21

3

===

−=+−=+−=

=−−=−−==

=+

=+

= −

γ

which indicate that a load of #%1 with a total eccentricity of 9.2##% in. Dincludingcolumn deection if anyE will produce failure in the column, when the ultimate neutralaxis is assumed located at 9 in. from the edge.

Sarious combination of P and 8 can be plotted and a column interaction diagramcan be drawn for the column with given amount of prestress.

S3ste"ti$ &eterin"tion o# t%e inter"$tion &i"gr" ,sing "rio,s str"in&istri',tion.

.P.0ncog#:6



Typical ultimate column interaction diagram-

Mu

:u

1

2

3

0 systematic way to determine points of the interaction diagram is illustrated inthe above picture. Ane may start with a uniform strain distribution over all the section,at the concrete strain εo and maximum concrete stress, and determine the maximumaxial force that can be supported by the section. This is given by position # of thevarious strain-distribution planes. The failure criterion that re"uires that the middepthstrain be e"ual to εo is satis!ed by rotating the strain-distribution plane about the

middepth section. 0ny strain-distribution contained between planes # and 5 satis!esthe failure criterion.

The top branch of the interaction diagram shown above between points # and 5 was

obtained from the strain-distribution planes bound by planes # and 5. The top branch

is characteri/ed by failures due to large axial loads and small bending moments. The

rest of the interaction diagram, between points 5 and 9, may be obtained by satisfying

the criterion that failure of the section occurs when the crushing strain εuc is attained

at the extreme !ber, between limits set by planes 5 and 9. The later corresponds to

failure of the section due to bending moment alone, such as in the conventional case

of a beam section, and cannot be predicted a priori. =t can be obtained by considering

the column as pure beam action, no axial load.

E"+(e :

.P.0ncog#:%



1 . =

1)=

2= 2=

o $ ),))2

:u

s s/C

:lan

Strain

Stresses

- G / strand"s $ "s/ $ ),*0*se $ ),))*fc/ $ . 7si

73fc/ $ ,-*(.% $ *,1 7si

Strains in steelFes/ $ es $ eo $ ),))2

s/ $ s $ se $ s $ ),))* G ),))2 $ ),))3

Stresses in steelFfs/ $ fs $ s Ds $ ),))3(3)&1)3% $ ) 7si

<orcesF

C $ 73fc/bt $ *,1(1.&1)% $ -1. 7

s $ s/ $ "sfs $ ),*0*()% $ *1,- 7

ertical eJuilibriumF

:u $ C G s G s/ $ 012 7 Mu $ )

0 prestressed-conrete column with a #%V x #2V rectangular cross section is

reinforced with $ Z Q in. strands, four in each wide face. The strength of the concrete

is f cJ G %222 psi. the strain at which maximum stress occurs, εo, and the crushing strain

εuc are assumed e"ual to 2.225 and 2.229, respectively. The stress-strain diagram for

concrete is described by #, 5, and 9. *or this problem assume # G 2.>6, 5 G 2.:5,

and 9 G 2.$6. *or the steel, εse G 2.226, f su G 52$ si.

So(,tion:

Two levels of initial prestrain in the prestressing steel, ese are considered. =n the !rst

case, full prestressing to εse G 2.226 is applied while the second case the section has

no prestressing at all, εse G 2.

P("ne 1:

.P.0ncog#:>



oc $ ),))3

:u

s s/C

:lan

Strain

Stresses

1 . =

1)=

2= 2=

- G / strand

"s $ "s/ $ ),*0*se $ ),))*fc/ $ . 7si

7173fc/ $ (,0*%(,-*%(.% $ 3,- 7si

Strains in steelFes/ $ uc(d@c% $ ),))3(-@1)% $ ),))24

s/ $ se G s/ $ ),))* G ),))24 $ ),))2.

es $ euc(d/@c% $ ),))3(2@1)% $ ),))).

s $ se s $ ),))* G ),))). $ ),))44

Stresses in steelF

fs/ $ s/ Ds $ ),))2.(3)&1)3% $ 0- 7sifs $ sDs $ ),))44(3)&1)3% $ 132 7si

<orcesFC $ 7173fc/bt $ 3,-(1.&1)% $ .)- 7

s $ "s fs $ ),*0*(132% $ 0. 7 s/ $ "sfs/ $ ),*0*(0-% $ 4* 7

ertical eJuilibriumF:u $ C G s G s/ $ 4-0 7

Mu $ C(t@2 G 72c% + s(t@2 G d/% G s/(t@2 G d/%Mu $ .)-(* ,42&1)% + 0.(* G 2% G 4*(* G 2%Mu $ *-) 7in

ss/

c $ 1)=

e

1 . =

1)=

2= 2=

oc $ ),))3

s s/

C

:lan

Strain

Stresses

- G / strand"s $ "s/ $ ),*0*

se $ ),))*fc/ $ . 7si

7173fc/ $ (,0*%(,-*%(.% $ 3,- 7si

Strains in steelFes/ $ uc(cd/%@cN $ ),))3(c G 2%@c

s/ $ se G s/ $ ),))* G ),))3(c G 2%@cN

es $ euc(d G c%@cN $ ),))3(- G c%@cNs $ se + s $ ),))* G ),))3(- G c@cN

Stresses in steelF

fs/ $ s/ Ds $ (3)&1)3%X),))* G ),))3(c G 2%@cNY

fs/ $ 1*) G )(c G 2%@cN

fs $ sDs $ (3)&1)3%X,))* + ),))3(- G c%@cNY $ fsu

<orcesF

C $ 7173fc/bc $ 3,-(1.c%s $ "s fs $ ),*0*fss/ $ "sfs/ $ ),*0*fs/>etermination of cF

<v $ )F $ C G s G s/ $ 4-0 7 ) $ 3,-(1.c% G ),*0*(1*) G )(c G 2%@cN ),*0*fs

"ssume c $ 3 in,? then fs $ fsu $ 2)- 7si and substitutingF

s

s/

c

Mu

3,-(1.&3% $ ),*0*(1*) G )(3 G 2%@3N + ),*0*(2)-%1-2 Z . + 11 $ 1--? therefore c $ 3 in,? tr! c $ 3(1-*@1-2% $ 3,)* in,

then fsu $ 2)- 7si?3,-(1.&3,)*% $ ,*0*(1*) G )(3,)* G 2%@3,)*N + ),*0*(2)-%

1-. [ .- + 11? c $ 3,)* in,:u $ )Mu $ 7173fc/bc(h@2 G 72c% G "s/fs/(h@2 G d/% + "sfs(h@2 G d/%Mu $ ,0*&,-*&.&1.&3,)*(* ,42&3,)*% ,*0*(1*) G )(3,)* G 2%@3,)*N(* G 2% + ),*0*(2)-%(* G 2%

Mu $ -44 7in

P("ne 2:

P("ne /:

.P.0ncog#:$





:

M

1

23

4ouc

Dlevation

Straindistribution

Systematic determination of t(e ero tensile"strain interaction diagram using various strain distribution-

:lan

Strain

Stresses

1 . =

1)=

2=2=

- G /strand"s $ "s/

$ ),*0*se $ ),))*

fc/ $ . 7si

Strains in steelFes/ $ 4(d@c% $ ),))1*(-@1)% $ ),))12

s/ $ se G s/ $ ),))* G ),))12 $ ),))3-

es $ e4(d/@c% $ ),))1*(2@1)% $ ),)))3 s $ se s $ ),))* G ),)))3 $ ),))40

Stresses in steelF

fs/ $ s/ Ds $ ),))3-(3)&1)3% $ 114 7si

fs $ sDs $ ),))40(3)&1)3% $ 141 7si

<orcesF

C $ ,-*fc/bt(e4@eo%(1 G 1@3(e4@eo% $ ,-*(.%(1.%(1)%(,))1*@,))2%(1 G 1@3(,))1*@,))2% $ 4* 7 s $ "s fs $ ),*0*(114% $ .*,. 7 s/ $ "sfs/ $ ),*0*(141% $ -1,1 7

ertical eJuilibriumF: $ C G s G s/ $ 312 7

M $ (1@.%Ct(1 G ( 4@ o%%@(1 G 1@3( 4@ o%% + s(h@2 G d/%

G s/(h@2 G d/%

M $ 4*(1)@.%(1 G (,))1*@,))2%%@(1 G 1@3(,))1*@,))2% + .*,.(* G 2% G -1,1(* G 2% $ .3- + 243 G 10 $ .-4 7in

4 $ ),))1*

:u

s s/C

ss/

c $ 1)=

e

−=

−=

oo

cc

oo

c

bt f M

bt f C

ε

ε

ε

ε

ε

ε

ε

ε

442

44

2

11'-*,)(

.

1

3

11%'-*,)(

where all symbols have been de!ned as before with the exception of ε: which is thestrain in the concrete at the near end !ber. The preceding expressions are valid for therange 2 ε: εo when the strain in the concrete at the far end is /ero.

P("ne :

.P.0ncog#62



Prestresse& Con$rete Pi(es

)ince piles are subjected to tensile stresses during transportation, driving and undercertain service conditions, the desirability of prestressing is evident.

Design o# +restresse& +i(es

&xperience seems to indicate that a prestress of about >22 psi in the piles willensure safety during handling and driving under normal conditions. While the amountof prestress re"uired will vary with the si/e and shape of the pile, the hammer blow,and the cushioning e<ects, as well as the soil conditions, it is obviously impractical tovary the prestress in each pile. Af course, higher or lower values than >22 psi may bedesirable for special cases.

The bearing capacity of concrete piles is seldom if ever considered by theirstrength under direct compression, but it is convenient to express the bearing capacity

in terms of the compressive strength or stresses. )trictly speaing, if the bearingcapacity were limited by the compressive stress, there would be no need forprestressing. Therefore current formulas are empirical in nature.

The design load on such piles is often based on ultimate strength using arbitraryfactor of safety of about :. )uch a higher factor of safety is hardly necessary so far asthe service load is concerned, but it is believed that for piles so designed, thecompressive stress during driving will seldom be critical, and it should be possible toattain the desired bearing value without damaging the pile.

=f the cylinder strength of concrete is f cJ, the ultimate strength of the concrete in

a pile can be safely assumed 2.$6f cJ. 0t ultimate load, the amount of prestressremaining in the tendons is approximately %2C of the e<ective prestress. Thus if a%,222 psi concrete is prestressed to an e<ective prestress of >22 psi, the ultimatestrength can be computed by the formula?

7J G D2.$6f cJ Z 2.%2f cE0c G D2.$6f cJ Z 2.%2D>22EE0c

where 0c G cross-sectional of the concrete pile. 4sing a factor of safety of :.2, thedesign load 7 G 7J.

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(oint standard set up by 00)'TA and the Prestressed +oncrete =nstitute statethat the maximum compressive stress f pc on prstressed concrete piles Din addition tothe e<ective prestressE for f cJ G 6,222 psi shall not exceed the following?

f pc G #222 psi for 2 O BI O #2

f pc G ##%2 Z #%BI for #2 O BI O 56

where B G e<ective length of pile, taen as the actual length of pile when bothends are hinged

G diameter or width of pile

*or 6,222 psi O f cJ O 1222 psi, these stresses may be increased indirectproportion to the design concrete strength. The bucling load of a pre-tensionedconcrete pile can be computed using the &ulerJs column formula?

2

2

%(k!

%I N

cr

π =

where 7cr G critical bucling load& G modulus of elasticity of concrete= G moment of inertia of the concrete pile sectionB G length of pile G e<ective column length factor

The value of & should be chosen to !t the duration of loading Z that is, a highervalue should be used for dynamic load and a reduced value for sustained load. )incepossibility of an increase in actual load is remote, a factor of safety of 5.2 isconsidered sucient. Thus the allowable load is often set at?

2

2

%(22 k!

%I N N cr π ==

=f no tensile stress is allowed, a high factor of safety is obtained forconcentrically prestressed members subjected to bending. =t is therefore oftenpermissible to allow tensile stresses in the concrete under design moments.

The joint 00)'TA-P+= )tandards allow 562 psi tension for normal design loads,

while values up to %22 psi have been permitted for earth"uae and other infre"uentloads. Thus if f e is the e<ective prestress in the pile concrete, the allowable momentunder earth"uae loading is?

c

I f f M t e %( +=

where = G moment of inertia of concrete sectionc G distance from neutral axis to extreme !ber

.P.0ncog#65



The ultimate moment is computed by?

d f CA M s su '=

where + G a coecient depending on shape of pile section, etc. varying between

D2.95 Z 2.9$E0s G total area of prestressing steelf sJ G ultimate strength of steeld G diameter or si/e of pile

The design moment, based on the ultimate load, should have a factor of safety of 5.2,while a factor of safety of #.6 to #.> will be sucient for earth"uae and wind loads.

0ccording to the elastic theory, the existence of direct external loads delayscracing of the concrete pile and thereby increase the moment-carrying capacity. A7the other hand, the ultimate moment capacity is reduced by the presence of directexternal loads. 'ence, when the design is for combined moment and direct loads, the

moment capacity of the pile should be checed by both the elastic and the ultimate-load theories, as described in columns under eccentric loads.

REFERENCES

#. 7arbey \hachaturian, and erman urfunel, ]Prestressed +oncreteV, 8craw-'ill,#1%1.

5 T L Bin ]esign of Prestressed +oncrete )tructuresV (ohn Wiley ^ )ons 5e #1%9

Documents

Prestress Lecture Notesfsfsfs