Prestressed Prestresse d 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$.2 %Pa& " s # 200 %Pa. 'ompute the length of shims re(uired, neglecting any elastic shortening of the shims and any friction along the tendon. Ans) 100.*8mm
#i. 1
s!ortenin o" concrete
end o" beam a"ter trans"er
lent! o" s!ims elonation o" steel
end o" beam be"ore trans"er
18.00 m
Solution: Elasttic elongation of steel:
∆ s =
f s L E s
=
1040(18 x10 3 ) 200 x10 3
= 93.6mm
Shortening of concrete due to t o shrinkage: ∆c
shrinkage =
0.0002(18 x10 3 ) = 3.6mm
Elastic shortening of concrete: δ D =
ML2 8 EI
=
(18.093)(102 ) x1012 3
9
8(27.5 x10 )(2.278 x10 )
= 3.61 mm
Length of shims required: ∆ T = ∆ s
elastic +
∆c
shrinkage shrinkage +
∆c
elastic =
93.6 + 3.6 + 3.78 = 100.98mm
2. A pretens pretension ioned ed member member has a sectio section n sho+n sho+n 200mm 200mm-0 -00m 0mm. m. t is concen concentric tricall ally y 2 prestressed +ith !1$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 #i. 2
200
$i
$i
0 0 3
Solution: Exact Method f c
nf c
=
Qo Ac
+ ( n − 1) A s
=
516 x1040
( 200 x300) + ( 6 − 1) 516
= 8.575MPa MPa
= (6)8.575 = 51.45 MPa
1
G. P. Anco
Prestressed Concrete Practice Problems
tress in steel after transfer f s
= f so − nf c = 1040 − 51.45 = 988.55 MPa
Approximate Method he loss of prestress in steel due to elastic shortening of concrete is approimated by) f s
=n
Qo A g
= (6)
516 x1040 200 x300
= 53.664 MPa
tress in steel after loss f s
= f so − nf c = 1040 − 53.664 = 986.335 MPa
tress in concrete is) f c
=
net stress of steel x A s A g
=
Qnet A g
=
986.335 x516 (200 x300)
= 8.482 MPa
Approximations introduced: 1. using gross area instead of net area 2. using initial stress in steel instead of the reduced stress -. A pretensioned member has a section 200mm-00mm. t is eccentrically prestressed +ith !1$mm2 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 # $, compute the stresses in the concrete immediately after transfer. Ans) f # 0.00 MPa& f # 31$.*18 MPa 200
#i. 3
initial cc
cc e
0 0 3
c "inal cc
s
c&
cs 100
&eam 'ection
rans"ormed section
Solution Exact Method ( n − 1) A s
= (6 − 1)(516) = 2580 mm 2 A g = 200 x300 = 60000 mm 2 e = (300 % 2) − 100 = 50mm umming up moment at initial cgc)
= A1 y1 + A2 y 2 A y + A2 y 2 ( 200 x300)(0) + 2580(50) = = 2.06 mm y o = 1 1 60000 + 2580 AT c B = (300 % 2 − y o ) = 147.94 mm cT = (300 % 2 + y o ) = 152.06 mm e = c B − 100 = 47.94 mm
AT y o
'ompute transformed section moment of inertia)
2
G. P. Anco
Prestressed Concrete Practice Problems
=
I T
1 12
+ A g ( yo ) 2 + (n − 1) A s y s
bh 3
200 x300
=
3
12
+ 60000(2.06) 2 + 2580(47.94) 2 = 4.562 x108
mm 4
iber stresses) Qi
f =
AT
Qi ey
±
I T
=
516 x1040 60000 + 2580
±
(516 x1040)(47.94) y 4.562 x10 8
= 8.575 ± 0.056393 y op fiber stress) f T
= 8.575 − 0.056393(152.06) = 0.00 MPa
ottom fiber stress)
= 8.575 + 0.056393(147.94) = 16.918 MPa
f B
Approximate Method 5oss of prestress) f sL
=
nQi
=
A g
6(516 x1040)
= 53.664 MPa
60000
6et prestress)
= f si − f sL = 1040 − 53.664 = 986.336 MPa −3 Qnet = f sn A s = 986.336(516 x10 = 508.949 kN
f sn
iber stresses) f =
=
Qnet A g
±
Qnet ey I c
508.949 x10 60000
3
3
±
508.949 x10 (50) y 200(300) 3 12
= 8.48248 ± 0.0565498 y op fiber stress) f T
= 8.48248 − 0.0565498(150) = 0.00 MPa
ottom fiber stress) f B
= 8.48248 + 0.0565498(150) = 16.964 MPa
Approimation introduce) 1. using approimate alues of reduced prestressed 2. using the gross area of concrete 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
#i. 4 o
0 0 3
50*75 cs
75
cs
75
&eam 'ection
Solution Method 1: Using net section of concrete Ac
= A g − Aduct = 200 x300 − 50 x75 = 56250 mm 2 3
G. P. Anco
Prestressed Concrete Practice Problems
5ocate the cg of net section) Aduct (75)
(50 x75)(75)
y o
=
y s
= 75 + y o = 75 + 5 = 80 mm = 150 − yo = 150 − 5 = 145 mm = 150 + y o = 150 + 5 = 155 mm
cT c B
Anet
=
56250
= 5.00 mm
'ompute the moment of inertia of net section) bh 3
I =
+ bh( y o ) 2
12
200 x300
=
3
−
b+ h+ 3 12
− b+ h+ (80) 2
+ 60000(5) − 2
12
50 x753 12
otal prestress in steel) Q = η ( A s f s ) = 95,(516 x1040) x10 −3 iber stresses) f =
Q Ac
±
(Qe) y
=
I
509.808 x10 3 56250
±
− 3750(80) 2 = 4.527 x10 8
mm 4
= 509.808 kN
509.808 x10 3 (80) 4.257 x10 8
y
= 9.063 ± 0.095806 y op fiber stress) f T
= 9.063 − 0.095806(145) = −4.828 MPa
ottom fiber stress)
= 9.063 + 0.095806(155) = 23.913 MPa
f B
Method 2: Using gross section of concrete f =
Q A g
±
Qec I
=
509.808 x10 200 x300
3
±
3
509.808 x10 (75)(150) 1 (200 x300 3 ) 12
= 8.4968 ± 12.7452 op fiber stress) f T
= − 4.2484 MPa
ottom fiber stress) f B
= 21.242 MPa
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. f =
Q A
±
Qe x y I x
±
Qe y x I y
!. A post7tensioned bonded concrete beam has a prestress of 1!$0 k6 in the steel 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 condition +ith full prestress and no lie load b: under final condition after all the losses hae taken place and +ith full lie load. Ans) nitial condition) f # 2.2-4 MPa, f # 1!.10 MPa& inal condition) f # 1-.80- MPa, f # 0.*/! MPa 45- #i. 5
4.50m
45- 3.00m
4.50m 300
4
0 0 6
G. P. Anco 175
/id san section
Prestressed Concrete Practice Problems
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
bh
12
3
=
1 12
3
(300 x600 )
= 5.4 x10 9
mm
4
nitial condition M =
f =
L2
=
8 Qo
A g
4.4 x122 8
Qo ey
±
I g
= 79.2 kN − m 3
± My = 1560 x10 ± 1560 x10 I g
180000
3
(125)(300)
5.4 x10
9
± 79.2 x10
6
(300 )
5.4 x10 9
= 8.667 ± 10.833 ± 4.40 op fiber stress) f T
= 8.667 − 10.833 + 4.4 = 2.234 MPa
ottom fiber stress)
= 8.667 + 10.833 − 4.4 = 15.10 MPa
f B
Final condition 5ie load moment at mid7span) M L
= Pa = 45(4.5) = 202.5 kN − m
;ead load moment at mid7span) M D
=
L2 8
=
4.4(12 2 ) 8
= 79.2 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 (125)(300)
180 x10 3 5.4 x10 9 7.389 ± 9.236 ± 15.65
5
±
281.7 x10 6 (300) 5.4 x10 9
G. P. Anco
Prestressed Concrete Practice Problems
op fiber stress) f T
= 7.389 − 9.236 + 15.65 = 13.803 MPa
ottom fiber stress) f B
= 7.298 + 9.236 − 15.65 = 0.975 MPa
ote: or pre7tensioned beam, steel is al+ays bonded to the concrete before any eternal moment is applied.
45 - 1.50 m
4.50 m
13.8202 /Pa
β e
C α
T 0.976 /Pa
' # , M # ' α # α α =
M
=
M
±
! β y
281.7 x10
6
= = 211.8 mm 3 T 1330 x10 β = α − e = 211.8 − 125 = 86.8 mm !
tresses) f =
= =
! A
I
1330 x10 3
±
1330 x10 3 (86.8)(300)
180 x10 3 7.389 ± 6.413
5.4 x10 9
op fiber stress) f T
= 7.389 + 6.413 = 13.802 MPa
ottom fiber stress) f B
= 7.389 − 6.413 = 0.976 MPa
!omputation of a"erage strain for un#onded #eams:
6
G. P. Anco
Prestressed Concrete Practice Problems
ε =
f E
=
My E c I
My dx E c I
∆ = ∫ ε a"e
=
1
My
dx L ∫ E I c
Aerage stress in steel is) f s
MyE s
= E s ε a"e = ∫
LE c I
dx
=
n My L
∫ I dx
$. A post7tensioned simple beam on a span of 12 m carries a superimposed 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 beam. he parabolic cable has an area of 1$12.*mm2, n # $. 'ompute the stresses in the steel at mid7span assuming) a: the steel is bonded by grouting b: the steel is unbonded and entirely free to slip. Ans) onded condition) f s # 84!.2!8 MPa, =nbonded condition) f s # 8-8.1-/ MPa 11.004.4015.40 -%m
#i. 6
300 0 0 6
175
/id san section 12.00 m
'ection roerties A b! 300*600 180000.00 mm 2 b!3%12 300(600) 3%12 5.4*109 mm 4 c !%2 600%2 300 mm
/o
/
* Parabolic moment diaram
*
o
Parabolic diaram
Solution 1: Moment at mid7span) M o
=
L2 8
=
15.4(12) 2 8
= 277.2kN − m
Moment at mid7span due to prestress) M s
= Qe = (1612.9 x830)(125) x10 −6 = 167.34 kN − m
6et moment at mid7span) M6 # 2//.2 > 1$/.-4 # 10*.8$ k67m tress in concrete at the leel of steel due to bending) 7
G. P. Anco
Prestressed Concrete Practice Problems
=sing for gross section
=
f c
My I
6
=
109.86 x10 (125) 5.4 x10
9
= 2.543 MPa
he stress in steel is increased by)
= nf c = 6(2.543) = 15.258 MPa
f s
?esultant stress in steel) f sf # 8-0 3 1!.2!8 # 84!.2!8 MPa
Solution 2: $f the ca#le is un#onded and free to slip%
=
f s
n
My
dx L ∫ I
x 2 M = M o 1 − L 2 x 2 y = y o 1 − L 2 2
x 2 dx f s = M o y o 1 − − L 2 LI ∫ L 2 L % 2 nM o y o 2 x 3 x 5 = + x − 3 LI ( L 2 ) 2 5( L 2) 2 − L 2 n
=
L 2
8 nM o y o 15
I
=
8 15
(15.258) = 8.137 MPa
?esultant stress in steel) f sf # 8-0 3 8.1-/ # 8-8.1-/ MPa =ltimate strength analysis) .85" c
εc 0.0034 a%2
a β1c
C
d
T
! = .85 f c + ba a
=
! .85 f c + b
c
εs
= T = A s f s +
=
A s f s + .85 f c + b
# = d − a 2 M
= A s f s + ( d − a 2)
/. A rectangular section -00mm $00mm deep is prestressed +ith *-/.! mm 2 of steel +ires for an initial stress of 1040 MPa. he cgs of the +ires is 100mm aboe the bottom fiber. or the tendons, f s@ # 1$!0 MPa, f c@ # -4.4 MPa. ;etermine the ultimate resisting moment. Ans) Mu # $-/.0! k67m
8
G. P. Anco
Prestressed Concrete Practice Problems
Solution
otal tension of steel at rupture
= 937 .5(1650) x10 −3 = 1546.875 kN ! = T .85 f c + ba = T T
a
=
T .85 f c + b
=
1546 .875 x10 3 .85(34 .4)(300 )
= 176 .34 mm
=ltimate moment M u
= A s f su ( d − a 2 ) = 1546.875( 500 − 176.34 2 ) x10 −3 = 637.05 kN − m
8. A post7tensioned bonded beam +ith a transfer prestress of t # 1!$0 k6 is being +rongly picked up at its mid7span point. 'ompute the critical fiber stresses. f the top fiber cracks and the concrete is assume to take no tension, compute the bottom fiber stresses. f the beam is picked up suddenly so that an impact factor of 100 is considered compute the maimum stresses. Ans) 'ase 1) f # 7$.!$$ MPa, f # 2-.*0 MPa& 'ase 2) f # 2/.*0! MPa& 'ase -) f # -!.-* MPa 6.00m
#i. 7
6.00m 300
0 0 6
175
Solution
ection properties) A = 300 x 600 I = c
=
1
mm 2
= 5.4 x10 9
(300)(600) 3
12 600 2
= 180 x10 3
mm 4
= 300 mm
"ternal moment at pick7up point 2
M
=−
L 2
=−
4.4(6)
2
2
= −79.2 kN − m
a& Fi#er stress at mid'span f =
=
Q A
±
Qey I
1560 x10 3 180 x10 3 8.667
±
My
±
I 1560 x10 3 (125)300 5.4 x10 9 10.833
±
79.2 x10 6 (300)
= ± ± f T = 8.667 − 10.833 − 4.4 = −6.566 MPa f B = 8.667 + 10.833 + 4.4 = 23.9 MPa
9
5.4 x10 9 4.4
G. P. Anco
Prestressed Concrete Practice Problems
#& $f the fi#er cracks and concrete is assume to take no tension% e 50.77 Q
372.69
C
124.23
rianlar 'tress &loc-
5ocate center of pressure, ')
= Qe
M e
=
M
79.2 x10 6
=
Q
1560 x10 3
= 50.77 mm
rom bottom) 1/!7!0.// # 124.2- mm Assuming a triangular stress block, height y)
= 3(124.23) = 372.69 mm
y
T = ! =
1
2T
=
=
f c
by
2
f c by 3
2(1560 x10 ) 300(372.69)
= 27.905 MPa
c& 1(() impact factor M T e
=
= M + 100, M = 2 M = 2(79.2) = 158.4 MPa M T Q
=
158.4 x10 6 1560 x10 3
= 101.538 mm
$rom bottom 175 − 101.54
= 73.46 mm
Assu min g a triangular stress block y
= 3(73.46) = 220.38 mm
f c
=
2T by
=
2(1560 x10 3 ) 300(220.38)
= 47.19 MPa
Assu min g a rec tan gular stress block
= 2(73.46) = 146.92 mm T = ! = f c by y
f c
=
T by
=
1560 x10 3 300(146.92)
= 35.39 MPa
*. ;etermine the total dead and lie uniform load moment that can be carried by the beam +ith a simple span of 12m)1. for ero tensile stress in the bottom fibers. 2. for cracking in the bottom fibers at a modulus of rupture of 4.1- MPa and assuming concrete to take up tension up to that alue. Ans) 'ase 1) + # 1$.21 k69m& 'ase 2) + # 20.-4 k69m #i. 8
13.853
300
4.13
18.534
C 1296.8 - - t 100
0 0 6
125
5 2 2
1296.8 - &eam 'ection
0 291.78 -m
10
4.13 74.34 -m
4.13 366.12 -m
G. P. Anco
Prestressed Concrete Practice Problems
Solution ection properties) A = bh 1
I = c
12 h
=
2
= 300(600) = 180 x10 3
=
1
=
bh 3
600 2
(300)(600) 3
12
mm 2
= 5.4 x10 9
mm 4
= 300 mm
Prestress B)
= A s f s = 1562.5(830) x10 −3 = 1296.8 kN
Q
1. Moment for ero tensile stress at the bottom)
=
f B 0
=
A
+
Qey I
−
1296.8 x10 3
=
M
Q
I
+
=0
1296.8 x10 3 (125)(300)
180 x10 3 291.78 kN − m
8 M
=
My
=
2
L
8( 291.78) 12 2
5.4 x10 9
−
Mx10 6 (300) 5.4 x10 9
= 16.21 kN % m
op fiber stress)
=
f T
=
Q A
−
Qey I
+
1296.8 x10
My
3
180 x10 3 = 13.853 MPa
I
−
1296.8 x10 3 (125)(300) 5.4 x10 9
+
281.78 x10 6 (300) 5.4 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
=
4.13(5.4 x10 9 ) 300
x10 −6
= 74.34 kN − m
otal moment capacity)
P
#i. 9
M T
= M 1 + ∆ M = 291.78 + 74.34 = 366.12 kN − m
=
8 M
L2
=
8(366.12) 12 2
arabola 300
= 20.34 kM % m 200
0 5 4
100
10. A concrete beam of 10m simple span is post7tensioned +ith a /!0mm 2 of high tensile 10.00 m steel to an initial prestress of *$! MPa immediately after prestressing. 'ompute the $ 965(750)*10 723.75 -and the beam@s o+n +eight assuming initial deflection at the mid7span due to3prestress "c # 2/.! %Pa. "stimate the deflection after - mos. Assuming creep coefficient of c c # 723.75*25 /oment de 1.8 and an effectie prestress of 8-0 MPa at that time. f the beam carry a 4! k6 to restress concentrated load applied at mid7span +hen723.75*150 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. :2%8
/oment de to beam ei!t G. P. Anco
11 /oment de to load P
P:%4
Prestressed Concrete Practice Problems
Solution ection properties) A
= bh = 300 x 450 = 135 x10 3
I =
1 12
bh 3
=
1 12
(300)( 450) 3
mm 2
= 2.278 x10 9
mm 4
he parabolic tendon +ith 1!0mm mid7ordinate is replaced by a uniform load acting along the beam. Qh P
= =
P L2 8 8Qh L2
=
8(723.75)(150) x10 −3 10 2
= 8.685 kN % m
Moment due to eccentric load at the end of the beam M
= Qe+ = 723.75(25) x10 −3 = 18.093 kN − m
;ead load uniform load D
= γ A = 23.5(300 x 450) x10 −6 = 3.17 kN % m
6et uniform load)
∆ = Q − D = 8.685 − 3.17 = 5.515 kN % m =p+ard deflection at mid7span due to net uniform load) δ %
=
5(∆) L4 384 EI
=
5(5.515)(10 4 ) x1012 384(27.5 x10 3 )(2.278 x10 9 )
12
= 11.462 mm
G. P. Anco
Prestressed Concrete Practice Problems
;o+n+ard deflection at mid7span due to end moment)
=
δ D
ML2 8 EI
(18.093)(10 2 ) x1012
=
= 3.61 mm
8(27.5 x10 3 )(2.278 x10 9 )
nitial deflection due to pretsress and beam +eight) δ net
= δ % − δ D = 11.462 − 3.61 = 7.852 mm; u&ard
;eflection due to prestress alone)
=
δ P
5( P ) L4 384 EI
−
ML2 8 EI
5(8.685)(10 4 ) 18.093(10 2 ) 1012 = − (27.5 x10 3 )(2.278 x10 9 ) = 14.44 mm 384 8
;eflection due to dead load alone) 4
δ DL
=
4
5 DL L
=
384 EI
12
5(3.17)(10 ) x10 3
9
384(27.5 x10 )(2.278 x10 )
= 6.59 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 + − 6.59 (1.8) = 0.5579 mm; u&ard δ DL ( cc ) = 14.44 965 f so
;eflection due to applied concentrated load of 4!k6) δ LL
=
PL3 48 EI
=
45(10 3 ) x1012 3
9
48(27.5 x10 )(2.278 x10 )
= 14.965 mm; donard
he resultant deflection) δ '
= δ LL + δ f = 14.965 − .5579 = 14.407 mm; donard
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. Ans) # 1410 k6& f # 14.*-4 MPa, f # 72.40 MPa #i. 10
65 - 23.5 -%m 300 0 5 7
6.00 m
15.00 m
13
6.00 m G. P. Anco
Prestressed Concrete Practice Problems
Solution ection properties) A
= bh = 300 x750 = 225 x10 3
I =
1 12
=
bh 3
1 12
(300)(750) 3
mm 2
= 1.0546875 x1010
mm 4
n order to balance the load on the cantileer, the cgs at the tip must coincide +ith the cgc +ith a horiontal tangent. o use the least amount of pretsress, the eccentricity oer the support should be a maimum. Assume a gross coer of /!mm, ema # /!0927/! # -00 mm. 65 - 23.5 -%m 300
e
!
e
0 5 7
he prestress re(uired) 15.00 m
6.00 m
Qe =
L2
2
=
L2 2e
=
23.5(6 2 ) 2(300 x10 −3 )
6.00 m
= 1410 kN
n order to balance the load at the mid7span, using the same prestress B, the sag of the parabola must be) Qh
=
L2 8
=
L2 8Q
23.5(15 2 )
=
8(1410)
x10 −3
= 468.75 mm
he result +ill be a concordant cable and under the action of the uniform load and prestress, the beam +ill hae no deflection any +here and +ill only hae a uniform compressie stress. f c
=
Q A
=
1410 x103 225 x10
3
= 6.267 MPa
;ue to concentrated load P) M =
PL 4
=
65(15) 4
= 243.75 kN − m
he etreme fiber stresses) f =
Mc
= 8.667 MPa 1.0546875 x1010 Mc + = 6.267 + 8.667 = 14.934 MPa f T = A I Q Mc − = 6.267 − 8.667 = −2.4 MPa f B = A I I Q
=
243.75 x10 6 (375)
12. A hollo+ member is reinforced +ith 4 +ires of $2.! mm 2 each pretensioned f si # 10-0 MPa. f f c@ # f ci # -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 14
G. P. Anco
Prestressed Concrete Practice Problems
to the elastic deformation: determine this limiting moment. Ans) +hen the +ires are cut, f s # *-$.*8 MPa& 5imiting moment, M # *.$$! k67m #i. 11
200
200
100
100
(n1)A s (71)(62.5) 0 0 2
0 0 2
0 0 1
oen
375 mm2
0 0 1
oen
30
rans"ormed 'ection
30
Solution ransformed section) AT
= 200 x 200 − 100 x100 + 4(n − 1) A s = 31.5 x10 3
I T
=
1 12
mm 2
[ 200 − 100 ] + 4(n − 1) A (70 ) = 1.3235 x10 4
4
2
s
8
mm 4
nitial prestressing force, B i before transfer) Qi
= A st f si = (4 x62.5)(1030) x10 −3 = 257.5 kN
f c
=
Qi AT
=
257.5 x10 3 31.5 x10 3
= 8.175 MPa
∆ f s = nf c = 7(8.175) = 57.225 MPa 6et stresses right after transfer Floss due to elastic shortening:)
= 8.175 MPa f so = f si − nf c = 1030 − 57.225 = 972.775 MPa f c
Allo+able concrete stresses) f c f t
=
=
0.45 f c + 0.5 f c +
=
0.45(34.4)
=
0.5 34.4
=
=
15.48 MPa 2.93 MPa
otal moment ) M # M;3M5 Additional concrete stress on top) ∆ f t = 15.48 − 8.175 = c
7.305 MPa compression
Additional concrete stress on botton)
∆ f b = 2.93 + 8.175 = 11.105 MPa tension otal moment that can be carried) f =
M T c I
< M T
=
∆ f t I c
=
7.305(1.323 x10 8 ) 100
x10
−6
= 9.665 kN − m
'oncrete stress on top reach full allo+able limit) 15
G. P. Anco
Prestressed Concrete Practice Problems
= f c = 15.48 MPa compression
f T
'oncrete stress at the bottom)
= 8.175 − ∆ f t = 8.175 − 7.305 = 0.87 MPa compression
f B
8.175 /Pa
15.48 /Pa
7.305 /Pa
2.93 /Pa
Alloable =ale o" stress
8.175 /Pa
11.105 /Pa
nitial concrete stress
concrete stress at t!e le=el o" steel
6et stress in steel) f sn
Additional concrete stress
70 = f so ± nf cs = 972.775 ± 7 7.305 = 972.775 ± 35.7945.1135 100
op steel) f snT
= 972.775 − 35.7945 = 936.98 MPa
ottom steel) f snB
= 972.775 + 35 .7945 = 1008.5695 MPa
After 240 MPa of prestress is lost Fin addition to elastic deformation:
= f snet A st = (1030 − 240)(4 x62.5) x10 −3 = 197.5 kN Q 197.5 = 6.27 MPa f c = i = 8.175 AT 257.5 f se = f snet − nf c = (1030 − 240) − 7(6.27) = 746.11 MPa Qi
Additional concrete stress on top) ∆ f t = 15.48 − 6.27 = 9.21 MPa compression c
Additional concrete stress on botton)
∆ f b = 2.93 + 6.27 = 9.2 MPa tension otal moment that can be carried)
16
G. P. Anco
Prestressed Concrete Practice Problems
f =
M T c I
< M T
=
∆ f t I c
9.2(1.323 x10 8 )
=
100
x10 −6
= 12.17 kN − m
6.27 /Pa
15.48 /Pa
2.93 /Pa
Alloable =ale o" stress
9.21 /Pa
6.27 /Pa
9.20 /Pa
nitial concrete stress
Additional concrete stress
herefore the limiting moment) M T
= [ 9.665; 12.17] min = 9.665 kN − m
17
G. P. Anco
