Prestressed Concrete Practice Problems
1. A Prescon cable, 18.00 m long is to be tensioned from one end to an initial prestressed of 1040 MPa immediately after transfer. Assume that there is no slack in the cable, that the shrink shrinkage age of concre concrete te is 0.000 0.0002 2 at the time time of transf transfer er,, and that that the aerage aerage compression in concrete is !.!0 MPa along the length of tendon. " c # 2$.2 %Pa& " s # 200 %Pa. 'ompute the length of shims re(uired, neglecting any elastic shortening of the shims and any friction along the tendon. Ans) 100.*8mm Fig. 1
length of shims
shortening of concrete
end of beam after transfer
elongation of steel
end of beam before transfer
18.00 m
Solution: Elastic elongation of steel:
∆ s =
f s L E s
=
100(18 x10 3 ) 200 x10
= 93.6mm
3
Shortening of concrete due to t o shrinkage:
∆ c shri nkage= 0.0002(18 x10 3 ) = 3.6mm Elastic shortening of concrete:
∆ c elastic=
f c L E c
3
=
"."(18 x10 ) 26.2 x10 3
= 3.!8mm
Length of shims required:
∆ T = ∆ s elastic+ ∆ c shri nkage+ ∆ c elastic= 93.6 + 3.6 + 3.!8 = 100.98mm 2. A pret pretens ensio ione ned d memb member er has has a sect sectio ion n sho+n sho+n 200m 200mm m-0 -00m 0mm. m. t is conc concen entri trica call lly y 2 prestressed +ith !1$mm of high tensile steel +ire +hich is anchored to the bulkheads of a unit stress of 1040 MPa. Assuming n # $, compute the stresses in the concrete and steel immediately after transfer. Ans) f c # 8.!/! MPa& f y # *88.!! MPa Fig. 2
200
Qi
Qi
1
3 0 0
#. P. $ncog
Prestressed Concrete Practice Problems
Solution: Exact Method f c
nf c
=
Qo Ac
+ ( n − 1) A s
=
"16 x100
( 200 x300 ) + ( 6 − 1)"16
= 8."!"MPa
= (6)8."!" = "1." MPa
tress in steel after transfer f s
= f so − nf c = 100 − "1." = 988."" MPa
Approximate Method he loss of prestress in steel due to elastic shortening of concrete is approimated by) Q "16 x100 = "3.66 MPa f s = n o = (6) A g 200 x300 tress in steel after loss f s
= f so − nf c = 100 − "3.66 = 986.33" MPa
tress in concrete is) f c
=
net stress of steel x A s A g
=
Qnet A g
=
986.33" x"16 (200 x300)
= 8.82 MPa
Approximations introduced: 1. using using gross gross area area inste instead ad of net net area 2. using initial initial stress in steel steel instead instead of the the reduced reduced stress stress -. A pretensio pretensioned ned member member has a sectio section n 200mm 200mm-0 -00mm 0mm.. t is eccentr eccentrica ically lly prestress prestressed ed 2 +ith !1$mm of high tensile steel +ire +hich is anchored to the bulkheads at a unit stress of 1040 MPa. he c.g.s. is 100mm aboe the bottom fiber. Assuming n # $, compute the stresses in the concrete immediately after transfer tr ansfer.. Ans) f # 0.00 MPa& f # 31$.*18 MPa 200
Fig. 3
3 0 0
initial cgc
cgc
e
c final cgc
's
c%
cgs 100
%eam &ection
2
ransformed section #. P. $ncog
Prestressed Concrete Practice Problems
Solution Exact Method ( n − 1) A s
= (6 − 1)("16) = 2"80 mm 2 A g = 200 x300 = 60000 mm 2 e = (300 2) − 100 = "0mm umming up moment at initial cgc) AT y o = A1 y1 + A2 y 2 y o
=
A1 y1
+ A2 y 2 AT
=
(200 x300)(0) + 2"80 ("0) 60000 + 2"80
= 2.06 mm
= (300 2 − y o ) = 1!.9 mm cT = (300 2 + y o ) = 1"2.06 mm e = c B − 100 = !.9 mm c B
'ompute transformed section moment of inertia) I T
= =
1 12
bh
3
+ A g ( yo ) 2 + ( n − 1) A s y s
200 x300
3
12
iber stresses) Q Q ey f = i ± i AT I T
+ 60000(2.06) 2 + 2"80(!.9) 2 = ."62 x108 =
"16 x100 60000 + 2"80
±
mm
("16 x100)(!.9) y ."62 x10
8
= 8."!" ± 0.0"6393 y op fiber stress) f T = 8."!" − 0.0"6393(1"2.06) = 0.00 MPa ottom fiber stress) f B = 8."!" + 0.0"6393(1!.9) = 16.918 MPa
Approximate Method 5oss of prestress) nQi 6("16 x100) = f sL = A g 60000
= "3.66 MPa
6et prestress) f sn = f si
− f sL = 100 − "3.66 = 986.336 MPa Qnet = f sn A s = 986.336("16 x10 −3 = "08.99 kN
iber stresses)
3
#. P. $ncog
Prestressed Concrete Practice Problems
f =
=
Qnet A g
±
Qnet ey I c
"08.99 x10 3 60000
±
"08.99 x10 3 ("0) y 200(300) 3 12
= 8.828 ± 0.0"6"98 y op fiber stress) f T = 8.828 − 0.0"6"98 (1"0) = 0.00 MPa ottom fiber stress) f B = 8.828 + 0.0"6"98 (1"0) = 16.96 MPa Approimation introduce) 1. using approimat approimate e alues alues of reduced reduced prestressed prestressed 2. using using the the gross gross area area of concre concrete te 4. A post7tensioned beam has a mid span cross7section +ith a duct of !0mm /!mm to house the +ires. t is pretensioned +ith !1$mm2 of steel to an initial stress of 1040 MPa. mmediately after transfer, the stress is reduced by ! o+ing to anchorage loss and elastic shortening of concrete. 'ompute the stresses in the concrete at transfer. Ans) f # 4.82* MPa, f # 32-.*1- MPa 200
Fig. 0 0 3
'o "0*!" cgs
cgs
%eam &ection
Solution Method 1: Using net section of concrete Ac
= A g − Aduct = 200 x300 − "0 x!" = "62"0 mm 2
5ocate the cg of net section)
#. P. $ncog
Prestressed Concrete Practice Problems
=
yo
Aduct (!") Anet
=
("0 x!")(!") "62"0
= ".00 mm
= !" + yo = !" + " = 80 mm cT = 1"0 − y o = 1"0 − " = 1" mm c B = 1"0 + y o = 1"0 + " = 1"" mm
y s
'ompute the moment of inertia of net section) bh 3
I =
+ bh( y o ) − 2
12
200 x300
=
b+ h + 3 12
− b+ h+ (80) 2
3
+ 60000(") − 2
"0 x !"
12 otal prestress in steel)
3
− 3!"0(80) 2 = ."2! x108
12
Q = η ( A s f s ) = 9",("16 x100) x10
−3
mm
= "09.808 kN
iber stresses) f =
Q Ac
±
(Qe) y I
=
"09.808 x10 3 "62"0
±
"09.808 x10 3 (80) .2"! x10 8
y
= 9.063 ± 0.09"806 y op fiber stress) f T
= 9.063 − 0.09"806 (1") = −.828 MPa
ottom fiber stress)
= 9.063 + 0.09"806 (1"") = 23.913 MPa
f B
Method 2: Using gross section of concrete f =
Q A g
±
Qec I
=
"09.808 x10 3 200 x300
"09.808 x10 3 (!")(1"0) ± 1 (200 x300 3 ) 12
= 8.968 ± 12.!"2 op fiber stress) f T
= − .28 MPa
ottom fiber stress) f B
= 21.22 MPa
"
#. P. $ncog
Prestressed Concrete Practice Problems
f eccentricity does not occur along one of the principal aes of the section, it is necessary to further resoled the moment into t+o components along the t+o principal aes. Q Qe x y Qe y x ± f = ± A I x I y
!. A post post7t 7ten ensi sion oned ed bond bonded ed conc concre rete te beam beam has has a pres prestr tres esss of 1!$0 1!$0 k6 in the the stee steell immediately after prestressing +hich eentually reduces to 1--0 k6. he beam carries t+o lie loads of 4! k6 each in addition to its o+n +eight of 4.40 k69m. 'ompute the etreme fiber stresses at mid7span) a: under the initial initial conditi condition on +ith full full prestress prestress and no lie load load b: under final final condition condition after all all the losses losses hae taken taken place and and +ith full lie lie load. Ans)) nitial condition) f # 2.2-4 MPa, f # 1!.10 MPa& inal condition) f # 1-.80- MPa, f Ans # 0.*/! MPa "- Fig. "
."0m
"- 3.00m
."0m 300 6 0 0
1!"
/id san section
Solution o be theoretically eact, net concrete section should be used up to the time of grouting, after +hich the transformed section should be considered. ection Properties) A g
= bh = 300 x600 = 180000 mm 2
I g
=
1 12
bh 3
=
1 12
(300 x600 ) = ". x10 mm 3
9
nitial condition 2
M =
wL 8
=
. x12 8
2
= !9.2 kN − m 6
#. P. $ncog
Prestressed Concrete Practice Problems
f =
Qo A g
±
Qo ey I g
±
My I g
=
1"60 x10 3 180000
±
1"60 x10 3 (12")(300) ". x10 9
±
!9.2 x10 6 (300) ". x10 9
= 8.66! ± 10.833 ± .0 op fiber stress) f T
= 8.66! − 10.833 + . = 2.23 MPa
ottom fiber stress) f B
= 8.66! + 10.833 − . = 1".10 MPa
Final condition 5ie load moment at mid7span) M L
= Pa = "(.") = 202." kN − m
;ead load moment at mid7span)
M
=
wL2 8
=
.(12 2 ) 8
= !9.2 kN − m
otal moment) M # /*.20 3 202.! # 281./ k67m tresses) f =
f =
Q A g
±
Qey I g
1330 x10 3
±
±
M T y I g
1330 x10 3 (12")(300)
180 x10 3 ". x10 9 = !.389 ± 9.236 ± 1".6"
±
281.! x10 6 (300) ". x10 9
op fiber stress) !
#. P. $ncog
Prestressed Concrete Practice Problems
f T
= !.389 − 9.236 + 1".6" = 13.803 MPa
ottom fiber stress) f B
= !.298 + 9.236 − 1".6" = 0.9!" MPa
ote: or pre7tensioned beam, steel is al+ays bonded to the concrete before any eternal moment is applied.
" - 1."0 m
."0 m
13.8202 /Pa
e
' # , M # ' α # α α =
M
281.! x10
T 0.9!6 /Pa
6
= = 211.8 mm 3 T 1330 x10 β = α − e = 211.8 − 12" = 86.8 mm !
=
M
C
tresses)
8
#. P. $ncog
Prestressed Concrete Practice Problems
f =
=
! A
±
! β y I
1330 x10 3
±
1330 x10 3 (86.8)(300)
180 x10 3 = !.389 ± 6.13
". x10 9
op fiber stress) f T
= !.389 + 6.13 = 13.802 MPa
ottom fiber stress)
= !.389 − 6.13 = 0.9!6 MPa
f B
!omputation of a"erage strain for un#onded #eams: ε =
f E
=
My E c I
My
∆ = ∫
dx E c I
ε a"e
=
1 My
∫
dx L E c I
Aerage stress in steel is) f s Fig. 6
MyE s
= E s ε a"e = ∫
dx =
n My
∫
LE c I 01".0 L-m I 11.00.01".0 11.00.
dx
300 $. A post7ten post7tensioned sioned simple simple beam on a span of 12 m carries carries a superimposed superim posed load load of 11.00 k69m in addition to its o+n +eight of 4.40 k69m. he initial prestress in the steel is *!0 MPa, reducing to 8-0 MPa after deducting all loses and assuming no bending of the 1!" beam. he parabolic cable has an area of 1$12.*mm2, n # $. 'ompute the stresses in /id san section the steel at mid7span assuming) 12.00 m &ection roerties4 a: the stee steell is bond bonded ed by grouti grouting ng $ bh 300*600 180000.00 mm2 b: the steel steel is is unbonded unbonded and and entirely entirely free free to slip. 5 bh312 300(600)312 ".*109 mm c h2 6002 300 mm Ans) onded condition) f s # 84!.2!8 MPa, =nbonded condition) f s # 8-8.1-/ MPa /o 6 0 0
/
* Parabolic moment diagram
* '
'o
Parabolic ' diagram
9
#. P. $ncog
Prestressed Concrete Practice Problems
Solution 1: Moment at mid7span) 2
M o
=
wL 8
=
1".(12) 8
2
= 2!!.2k N − m
Moment at mid7span due to prestress) M s
= Qe = (1612.9 x830) (12") x10 −6 = 16!.3 kN − m
6et moment at mid7span) M6 # 2//.2 > 1$/.-4 # 10*.8$ k67m tress in concrete at the leel of steel due to bending) =sing for gross section
10
#. P. $ncog
Prestressed Concrete Practice Problems
f c
=
My I
=
109.86 x10 6 (12") ". x10 9
= 2."3 MPa
he stress in steel is increased by) f s
= nf c = 6(2."3) = 1".2"8 MPa
?esultant stress in steel) f sf # 8-0 3 1!.2!8 # 84!.2!8 MPa sf #
Solution 2: $f the ca#le is un#onded and free to slip%
f s
=
n My L
∫ I dx
x 2 M = M o 1 − L 2 x 2 y = y o 1 − L 2 2
x 2 n dx f s = M o y o 1 − − L 2 LI ∫ L 2 L 2 nM o y o 2 x 3 x " = + x − 3 LI ( L 2) 2 "( L 2 ) 2 − L 2 L 2
=
8 nM o y o 1"
I
=
8 1"
(1".2"8) = 8.13! MPa
?esultant stress in steel) f sf # 8-0 3 8.1-/ # 8-8.1-/ MPa sf #
=ltimate strength analysis) .8"f c
11
#. P. $ncog
Prestressed Concrete Practice Problems
εc 0.003 a2
a β1c
C
c d
7 T
εs
! = .8" f c + ba = T = A s f s + a
=
! .8" f c + b
=
A s f s + .8" f c + b
# = d − a 2 M = A s f s + ( d − a 2 )
/. A rectang rectangula ularr sectio section n -00mm -00mm $00mm $00mm deep deep is prestres prestressed sed +ith *-/.! *-/.! mm 2 of steel +ires for an initial stress of 1040 MPa. he cgs of the +ires is 100mm aboe the bottom fiber. or the tendons, f s@ # 1$!0 MPa, f c@ # -4.4 MPa. ;etermine the ultimate resisting moment. Ans) Mu # $-/.0! k67m
Solution
otal tension of steel at rupture T = 93!."(16"0) x10 −3
= 1"6.8!" kN
! = T .8" f c + ba = T a=
T .8" f c + b
=
1"6.8!" x10 3 .8"(3.)(300)
= 1!6.3 mm
=ltimate moment M u
= A s f su ( d − a 2) = 1"6.8!"( "00 − 1!6.3 2) x10 −3 = 63!.0" kN − m 12
#. P. $ncog
Prestressed Concrete Practice Problems
8. A post7tensioned bonded beam +ith a transfer prestress of t # 1!$0 k6 is being +rongly picked up at its mid7span point. 'ompute the critical fiber stresses. f the top fiber cracks cracks and the concrete concrete is assume assume to take take no tensio tension, n, compute compute the bottom bottom fiber stres stresse ses. s. f the the beam beam is picke picked d up sudd suddenl enly y so that that an impa impact ct fact factor or of 100 100 is considered compute the maimum stresses. Ans) 'ase 1) f # 7$.!$$ MPa, f # 2-.*0 MPa& 'ase 2) f # 2/.*0! MPa& 'ase -) f # -!.-* MPa 6.00m
Fig. !
6.00m 300 6 0 0
1!"
Solution ection properties) A = 300 x600 = 180 x10 3 mm 2 I = c
=
1
= ". x109
(300)(600) 3
12 600 2
mm
= 300 mm
"ternal moment at pick7up point
M = −
wL2 2
=−
.(6) 2 2
= −!9.2 kN − m
a& Fi#er stress at mid'span f =
=
Q A
±
Qey I
1"60 x10 180 x10 8.66!
±
My
3
3
±
I 1"60 x10 3 (12")300 ". x10 10.833
9
±
!9.2 x10 6 (300)
= ± ± f T = 8.66! − 10.833 − . = −6."66 MPa f B = 8.66! + 10.833 + . = 23.9 MPa 13
". x10 .
9
#. P. $ncog
Prestressed Concrete Practice Problems
#& $f the fi#er cracks and concrete is assume to take no tension%
e "0.!! Q
3!2.69
C
12.23
rianglar &tress %loc-
5ocate center of pressure, ') M = Qe e=
M Q
=
!9.2 x10
6
1"60 x10
3
= "0.!! mm
rom bottom) 1/!7!0.// # 124.2- mm Assuming a triangular stress block, height y) y
= 3(12.23) = 3!2.69 mm
T = ! = f c
=
2T by
1 2
=
f c by 2(1"60 x10 3 ) 300(3!2.69)
= 2!.90" MPa
c& 1(() impact factor
1
#. P. $ncog
Prestressed Concrete Practice Problems
M T e=
= M + 100, M = 2 M = 2(!9.2) = 1"8. MPa M T Q
=
1"8. x10 6 1"60 x10 3
= 101."38 mm
$rom bottom 4 1!" − 101." = !3.6 mm Assu min g a triangular stress block 4 y f c
= 3(!3.6) = 220.38 mm =
2T by
3
=
2(1"60 x10 ) 300(220.38)
= !.19 MPa
Assu min g a rec tan gular gular stress block 4
= 2(!3.6) = 16.92 mm T = ! = f c by
y
f c
=
T by
=
1"60 x10 3 300(16.92)
= 3".39 MPa
*. ;etermine the total dead and lie uniform load moment that can be carried by the beam +ith a simple span span of 12m)1. for for ero tensile tensile stress in the bottom fibers. fibers. 2. for crac cracki king ng in the the botto bottom m fiber fiberss at a modul modulus us of ruptu rupture re of 4.14.1- MPa MPa and and assum assumin ing g concrete to take up tension up to that alue. Ans) 'ase 1) + # 1$.21 k69m& 'ase 2) + # 20.-4 13.8"3 18."3 .13 Fig. 8 k69m 300 C 1296.8 -
2 2 "
-t 100
6 0 0
12"
1296.8 - %eam &ection
0 291.!8 -m
.13 !.3 -m
.13 366.12 -m
Solution ection properties) A = bh = 300(600) = 180 x10 3 mm 2 I =
1
bh
3
=
1
(300)(600)
12 12 h 600 c= = = 300 mm 2 2
3
= ". x109
1"
mm
#. P. $ncog
Prestressed Concrete Practice Problems
Prestress B)
= 1"62."(830) x10 −3 = 1296.8 kN
Q = A s f s
1. Moment for ero tensile stress at the bottom)
=
f B
Q A
+
Qey I
−
1296.8 x10 3
0=
My I
+
=0
1296.8 x10 3 (12")(300)
180 x10 3 M = 291.!8 kN − m w=
8 M 2
L
=
8(291.!8) 12
2
". x10 9
−
Mx10 6 (300) ". x10 9
= 16.21 k N m
op fiber stress)
f T
= =
Q A
−
Qey I
+
1296.8 x10
My I
3
180 x10 3 = 13.8"3 MPa
−
1296.8 x10 3 (12")(300) ". x10 9
+
281.!8 x10 6 (300) ". x10 9
2. or cracking in the bottom fibers. Additional moment carried by the section up to the beginning of crack.
∆ M =
f + I c
=
.13(". x10 9 ) 300
x10 − 6
= !.3 kN − m
otal moment capacity) M T
= M 1 + ∆ M = 291.!8 + !.3 = 366.12 kN − m
w=
8 M L2
= 8(3662.12) = 20.3 kM m 12
10. A concrete beam of 10m simple span is post7tensioned +ith a /!0mm 2 of high tensile steel to an initial prestress of *$! MPa immediately after prestressing. 'ompute the initial deflection at the mid7span due to prestress and the beam@s o+n +eight assuming "c # 2/.! %Pa. "stimate "stimate the deflection deflection after - mos. Assumin Assuming g creep coefficient coefficient of c c # 16
#. P. $ncog
Prestressed Concrete Practice Problems
1.8 and an effectie prestress of 8-0 MPa at that time. f the beam carry a 4! k6 concentrated load applied at mid7span +hen the beam is - mos. old after prestressing, +hat is the deflection at mis7spanC Ans) After - mos. δ # 0.!!/*mm up+ard& Dhen 4! k6 is added after - mos. δ # 14.40/ mm do+n+ard. P
Fig. 9
arabola 300 200
" 0
100
10.00 m Q 96"(!"0)*103 !23.!" - !23.!"*2"
/oment de to restress
!23.!"*1"0
:28
/oment de to beam eight
P:
/oment de to load P
Solution ection properties) A = bh = 300 x "0 = 13" x10 3 mm 2 I =
1 12
bh 3
=
1 12
(300)("0) 3
= 2.2!8 x109
mm
he parabolic tendon +ith 1!0mm mid7ordinate is replaced by a uniform load acting along the beam.
1!
#. P. $ncog
Prestressed Concrete Practice Problems
Qh =
=
w P
w P L2 8 8Qh L2
=
8(!23.!")(1"0) x10 −3 10 2
= 8.68" kN m
Moment due to eccentric load at the end of the beam M = Qe+ = !23.!"(2") x10 −3
= 18.093 kN − m
;ead load uniform load
= γ A = 23."(300 x "0) x10 −6 = 3.1! kN m
w
6et uniform load)
∆w = wQ − w = 8.68" − 3.1! = "."1" kN m =p+ard deflection at mid7span due to net uniform load)
=
δ %
"(∆w) L 38 EI
=
"("."1")(10 ) x1012 38(2!." x10 3 )(2.2!8 x10 9 )
= 11.62 mm
;o+n+ard deflection at mid7span due to end moment)
=
δ
ML2 8 EI
=
(18.093)(10 2 ) x1012 3
9
8(2!." x10 )(2.2!8 x10 )
= 3.61 mm
nitial deflection due to pretsress and beam +eight)
δ net
= δ % − δ = 11.62 − 3.61 = !.8"2 mm; u&ward
;eflection due to prestress alone)
δ P
=
"( w P ) L 38 EI
−
ML2 8 EI
"(8.68")(10 ) 18.093(10 2 ) 1012 = − (2!." x10 3 )(2.2!8 x10 9 ) = 1. mm 38 8 18
#. P. $ncog
Prestressed Concrete Practice Problems
;eflection due to dead load alone)
δ L
=
"w L L 38 EI
=
"(3.1!)(10 ) x1012 38(2!." x10 3 )(2.2!8 x10 9 )
= 6."9 mm
he initial deflection is modified by t+o factors) 1. loss of prestress 2. creep effect +hich tend to increase deflection ;eflection after - months)
δ f
f 830 = δ P s + = − 6."9 (1.8) = 0.""!9 mm; u&ward c 1 . δ ( ) L c f 96" so
;eflection due to applied concentrated load of 4!k6)
δ LL
=
PL3 8 EI
=
"(10 3 ) x1012 8(2!." x10 3 )(2.2!8 x10 9 )
= 1.96" mm; downward
he resultant deflection) δ '
= δ LL + δ f = 1.96" − .""!9 = 1.0! mm; downward
11.A double cantileer beam is to be designed so that its prestress +ill eactly balance the total uniform load of 2-.! k69m on the beam. ;esign the beam using the least amount of prestressed assuming that the cgs must hae a concrete protection of /! mm. f a concentrated load P # $! k6 is applied at the mid7span, compute the maimum top and bottom fiber stresses. stresses. Ans) # 1410 k6& f # 14.*-4 MPa, f # 72.40 MPa
19
#. P. $ncog
Prestressed Concrete Practice Problems
Fig. 10
6" - 23." -m 300 ! " 0
1".00 m
6.00 m
6.00 m
Solution ection properties) A = bh = 300 x!"0 = 22" x10 3 mm 2 I =
1 12
bh 3
=
1 12
(300)(!"0) 3
= 1.0"68!" x1010
mm
n order to balance the load on the cantileer, the cgs at the tip must coincide +ith the cgc cgc +ith +ith a hori horio onta ntall tang tangen ent. t. o use use the the least least amou amount nt of prets pretsre ress ss,, the eccentricity oer the support should be a maimum. Assume a gross coer of /!mm, ema # /!0927/! # -00 mm. 6" - 23." -m 300
e
1".00 m
6.00 m
! " 0
e
h
6.00 m
he prestress re(uired)
Qe =
wL2 2
=
wL2 2e
=
23."(6 2 ) 2(300 x10 −3 )
= 110 kN
20
#. P. $ncog
Prestressed Concrete Practice Problems
n order to balance the load at the mid7span, using the same prestress B, the sag of the parabola must be)
Qh =
wL2 8
wL2 8Q
23."(1" 2 )
=
8(110)
x10
−3
= 68.!" mm
he result +ill be a concordant concordant cable and under the action of the uniform load load and prestress, prestress, the beam +ill hae no deflection deflection any +here and +ill only hae a uniform compressie stress.
f c
30
=
Q A
=
110 x10 3 22" x10 3
= 6.26! MPa
;ue to concentrated load P)
30
2 0 0
M =
PL
=
6"(1")
= 23.!" kN − m
he etreme fiber stresses) 200
f =
Mc
= 8.66! MPa 1.0"68!" x1010 Mc f T = + = 6.26! + 8.66! = 1.93 MPa A I Q Mc = 6.26! − 8.66! = −2. MPa f B = − A I I Q
=
23.!" x10 6 (3!")
1 0 0
12. A 12. A hollo+ member is reinforced +ith 4 +ires of $2.! mm 2 each pretensioned f sisi # 10-0 100 MPa. f f c@ # f cici # -4.4 MPa, n # /, determine the stresses +hen the +ires are cut bet+een members. ;etermine the moment that can be carried at a maimum tension of 0.!EFf c@: and a maimum of f c # 0.4!f c@. f 240 MPa of the prestressed is lost Fin addition to the elastic deformation: determine this limiting moment. Ans) +hen the +ires are oen cut, f s # *-$.*8 MPa& 5imiting moment, M # *.$$! k67m Fig. 11
200 2 0 0
21
100 1
(n1)$s (!1)(62.") 3!" mm2
0 0
oen
ransformed &ection
#. P. $ncog
Prestressed Concrete Practice Problems
Solution ransformed section) AT
= 200 x 200 − 100 x100 + (n − 1) A s = 31." x10 3 mm 2
I T
=
Qi
= A st f si = ( x62.")(1030) x10 −3 = 2"!." kN
1
[200 − 100 ] + (n − 1) A (!0 ) = 1.323" x10
2
s
12 nitial prestressing force, B i before transfer)
f c
=
Qi AT
=
2"!." x10 3 31." x10 3
8
mm
= 8.1!" MPa
∆ f s = nf c = !(8.1!") = "!.22" MPa
6et stresses right after transfer Floss due to elastic shortening:)
= 8.1!" MPa f so = f si − nf c = 1030 − "!.22" = 9!2.!!" MPa f c
Allo+able concrete stresses)
= 0." f c + = 0."(3.) = 1".8 MPa f t = 0." f c + = 0." 3. = 2.93 MPa f c
otal moment ) M # M;3M5 Additional concrete stress on top)
∆ f t = 1".8 − 8.1!" = !.30" MPa c
compression
22
#. P. $ncog
Prestressed Concrete Practice Problems
Additional concrete stress on botton)
∆ f b = 2.93 + 8.1!" = 11.10" MPa tension otal moment that can be carried)
f =
M T c I
< M T
=
∆ f t I c
=
!.30"(1.323 x10 8 ) 100
x10 − 6
= 9.66" kN − m
'oncrete stress on top reach full allo+able limit) f T
= f c = 1".8 MPa compression
'oncrete stress at the bottom) f B
= 8.1!" − ∆ f t = 8.1!" − !.30" = 0.8! MPa compression 8.1!" /Pa
1".8 /Pa
!.30" /Pa
2.93 /Pa
$llo $lloabl ablee =ale =ale of stre stress ss
8.1!" /Pa
11.10" /Pa
5niti 5nitial al conc concret retee stres stresss
concrete stress at the le=el of steel
6et stress in steel)
f sn
$dditional concrete stress
!0 = f so ± nf cs = 9!2.!!" ± ! !.30" = 9!2.!!" ± 3".!9".113" 100
op steel) f snT
= 9!2.!!" − 3".!9" = 936.98 MPa
ottom steel) f snB
= 9!2.!!" + 3".!9" = 1008."69" MPa 23
#. P. $ncog
Prestressed Concrete Practice Problems
After 240 MPa of prestress is lost Fin addition to elastic deformation:
= f snet A st = (1030 − 20)( x62.") x10 −3 = 19!." kN Q 19!." = 6.2! MPa f c = i = 8.1!" AT 2"!." f se = f snet − nf c = (1030 − 20) − !(6.2!) = !6.11 MPa
Qi
Additional concrete stress on top)
∆ f t = 1".8 − 6.2! = 9.21 MPa c
compression
Additional concrete stress on botton)
∆ f b = 2.93 + 6.2! = 9.2 MPa tension otal moment that can be carried) f =
M T c I
< M T
=
∆ f t I c
=
9.2(1.323 x10 8 ) 100
= 12.1! kN − m
6.2! /Pa
1".8 /Pa
9.21 /Pa
2.93 /Pa
$llo $lloabl ablee =ale =ale of str stress ess
x10 −6
6.2! /Pa
9.20 /Pa
5nitia 5nitiall concr concrete ete stress stress
$dditional concrete stress
herefore the limiting moment) M T
= [ 9.66"; 12.1!] min = 9.66" kN − m
2
#. P. $ncog
