catatan dan tugas prestress

8/16/2019 Catatan Dan Tugas Prestress

1/88

1 SONDRA RAHARJA, ST

CONTOH SOAL BETON PRESTRESS

Balok prestress diatas dua tumpuan :

Panjang balok = 15

Initial gaya prestress = 1100

Beban hidup = 12Eksentrisitas cgs cgc = !25

"itung tegangan di tengah bentang#

$ehilangan jangka pendek = 10%

$ehilangan jangka panjang = 10%

&imana'

(cu = 50

(ci = !0

Jawaban

Luas penampang balok

b1 = )00 mmh1 = 200 mm

b2 = 150 mm

h2 = !50 mm

b! = )00 mm

h! = 200 mm


2/88


*1 = b1#h1 = +0000mm2

*2 = b2#h2 = 52500mm2

*! = b!#h! = +0000mm2

* = *1 , *2 , *! = 212500mm2

* = 2#1!E,05mm2


3/88


y1 = -50 1.2#h1 = /50 mmy2 = 1.2#h2 , h! = !-5 mm

y! = 1.2#h! = 100 mm

+0000 /50 , 52500 !-5 , +0000 100

2#1!E,05

!-5 mm dari tepi baah bottom3

' dimana " = -50 mm

!-5 mm dari tepi atas top3

Menghitung ne!tia

dimana 4

= 2-5 mm= 0 mm

= 2-5 mm

= 1.12 )00 200 ! , +0000 2-5 2

= 1.12 150 !50 ! , 52500 0 2

Menghitung "a!ak titik be!at #$ent!oi%& %a!i tepi bawah #bottom& ' (b

yb = Σ *i#yi.*

yb =

yb =

ya

= " yb

ya =

I = ΣIi, *

i#d

i23

d1

= ya

1.2#h1

d2 = y

a 1.2#"

d! = y

b 1.2#h!

I1 = 1.12#b1#h1!3 , *1#d12

I2 = 1.12#b2#h2!3 , *2#d22


4/88


= 1.12 )00 200 ! , +0000 2-5 2

= /!1//////- mm)

= 5!56!-500 mm)

= /!1//////- mm)

Inertia total = 62-0+!! mm)

I! = 1.12#b!#h!!3 , *!#d!2

I1

I2

I!


5/885 SONDRA RAHARJA, ST

Momen lawan #se$tion mo%ulus ) mo%ulus penampang&

7 pada bagian atas

7 pada bagian baah

= 1!1/62-0+!! = 3.512E+07mm! = mm!

!-5

= 1!1/62-0+!! = 3.512E+07mm! = mm!

!-5

Ja!ak ke!n #ke!n %istan$e&

= 1/5#2/ mm

= 1/5#2/ mm

81 = I . y

a

82 = I . y

b

81 !5#12 10/

82 !5#12 10/

kt = 82 . *

kb = 8

1 . *


6/88


Menghitung Beban

5#1k9.m1

2)k9.m!

Beban hidup = 12 k9.m

otal ser;ice load = Berat sendiri balok , Beban hidup

= 1-#1 k9.m


7/88


0#+0

dimana' Pi = 1100 k9

0#6 1100 = 660 k9

0#+ 1100 = ++0 k9

dan β =

αPi =

βPi =



*+ Pa%a kon%isi t!ans,e! #at t!ans,e!&

= )#// - 6#1/ . )#0+= 0#)2 9.mm2

= )#// . 6#1/ - )#0+

= 6#-) 9.mm2

/+ Pa%a kon%isi se!0is #at se!0i$e&

= )#1) - +#1) . 1!#/6

= 6#/6 9.mm2

= )#1) . +#1) - 1!#/6

= 1#)1 9.mm2

, *t ' P

i ) A - P

i+e)1

* . M

i ) 1

*

( 1t = 66010!.2#1!1053 - 66010!!25.!5#1210/3 . 1)!#)10/ . !5#1210/3

, /t ' P

i ) A . P

i+e)1

/ - M

i ) 1

/

( 2t = 66010!.2#1!1053 . 66010!!25.!5#1210/3 - 1)!#)10/ . !5#1210/3

, *s

' Pi ) A - P

i+e)1

* . M

s ) 1

*

( 1s

= ++010!.2#1!1053 - ++010!!25.!5#1210/3 . )+0#610/ . !5#1210/3

, /s

' Pi ) A - P

i+e)1

/ . M

s ) 1

/

( 2s

= ++010!.2#1!1053 . ++010!!25.!5#1210/3 - )+0#610/ . !5#1210/3


9/88


Tegangan (ang %ii1inkan #st!ess limits&


@lass 2' posttensioned3

= 0#!/ !0 1.2

= 1#6- 9.mm2

( tt = 0#!/(

ci31.2 9.mm2



= 0#5 !0

= 15#00 9.mm2

( ct = 0#5#(

ci 9.mm2 (leural members3


11/88




= 0#!/ 50 1.2

= 2#55 9.mm2

= 0#!! 50

= 1/#50 9.mm2

2esimpulan

1# Pada kondisi trans(er at trans(er3

egangan yang terjadi

0#)2 9.mm2 1#6- 9.mm2

6#-) 9.mm2 15#00 9.mm2

2# Pada kondisi ser;is at ser;ice3


6#/6 9.mm2 2#55 9.mm2

1#)1 9.mm2 1/#50 9.mm2

( ts = 0#!/(

cu31.2 9.mm2

( cs

= 0#!!#( cu

9.mm2 (leural members3

( 1t =

( 2t =

( 1s

=

( 2s =

Bandingkan dengan te"ada "e#it$ngan %e&


12/88


m

k9

k9.mmm

9.mm2

9.mm2


13/88




y&

ya


15/88



16/88



17/88



18/88



19/88



20/88



21/88



22/88


angan yang te!adi'$(nya


23/88


T34AS * BETON PRESTRESS

Auatu jembatan beton prategang dengan penampang sebagai berikut :

"itung besaran penampang untuk kondisi trans(er dan ser;ice berupa :

1# ?uas penampang

2# itik berat penampang

!# Inertia penampang)#


24/88


Initial gaya prestress = 1100 k9


Eksentrisitas cgs cgc = !25 mm

"itung tegangan di tengah bentang#




25/88


&imana'

(cu = 50 9.mm2

(ci = !0 9.mm2

Jawaban

*+ 2ON6S TRANS7ER


b1 = )00 mm "1 = -50 mm

h1 = 200 mm

b2 = 150 mm

h2 = !50 mm

b! = )00 mm

h! = 200 mm

*1 = b1#h1 = +0000 mm2

*2 = b2#h2 = 52500 mm2 *! = b!#h! = +0000 mm2

* = *1 , *2 , *! = 212500mm2

* = 2#1!E,05mm2

y1 = "1 1.2#h1 = /50 mmy2 = 1.2#h2 , h! = !-5 mm

y! = 1.2#h! = 100 mm

+0000 /50 , 52500 !-5 , +0000 100

2#1!E,05


yb = Σ *i#yi.*

yb =

650

375

100

150

y1 =

y2 =

y3 =

375y2 =


26/88


!-5 mm dari tepi baah bottom3

' dimana "1 = -50 mm

!-5 mm dari tepi atas top3

yb =

ya = "1 y

b

ya =

400

375

100

y2 =

y3 =


27/88


Menghitung ne!tia

dimana 4

= 2-5 mm

= 0 mm karena berhimpitan dg garis netral penampang

= 2-5 mm

= 1.12 )00 200 ! , +0000 2-5 2

= 1.12 150 !50 ! , 52500 0 2

= 1.12 )00 200 ! , +0000 2-5 2

= /!1//////- mm)

= 5!56!-500 mm)

= /!1//////- mm)

Inertia total = 62-0+!! mm)


7 pada bagian atas

7 pada bagian baah

= 1!1/62-0+!! = 3.512E+07mm! = mm!

!-5

= 1!1/62-0+!! = 3.512E+07mm! = mm!

I = ΣIi, *

i#d

i23

d1 = y

a 1.2#h

1

d2 = y

a 1.2#"1

d! = y

b 1.2#h!

I1 = 1.12#b1#h1!3 , *1#d12

I2 = 1.12#b2#h2!3 , *2#d22

I! = 1.12#b!#h!!3 , *!#d!2

I1

I2

I!

81 = I . y

a

82 = I . y

b

81 !5#12 10/

82 !5#12 10/


28/88

650y1 =


29/88


y! = 1.2#h! = 100 mm

y) = "2 1.2#h) = +50 mm

+0000 /50 , 52500 !-5 , +0000 100 , 200000 +50

)#1!E,05

/05#!0 mm dari tepi baah bottom3

yb =

yb =

400

375

100

y2 =

y3 =


30/88


' dimana "2 = 650 mm

!))#-0 mm dari tepi atas top3

Menghitung ne!tia

dimana 4

/50 /05#!0 = ))#-0 mm

/05#!0 !-5 = 2!0#!0 mm

/05#!0 100 = 505#!0 mm

+50 /05#!0 = 2))#-0 mm

ya = " y

b

ya =

I = ΣIi, *

i#d

i23

d1 = y

1 y

b=

d2 = yb y2 =d

! = y

b y

! =

d) = y

) y

b =

2on%isi se!0i$e

400

200

200

750

1000

150

850

650

375

100

y4 =

y1 =

y2 =

y3 =

44

230

d1 =

d2 =


31/88


= 1.12 )00 200 ! , +0000 ))#/6/6/6/6/6/6- 2

= 1.12 150 !50 ! , 52500 2!0#!0!0!0!0!0! 2

= 1.12 )00 200 ! , +0000 505#!0!0!0!0!0! 2

= 1.12 1000 200 ! , 200000 2))#/6/6/6/6/6- 2

= )#2/E,0+ mm)

= !#!2E,06 mm)

= 2#0-E,10 mm)

= 1#2/E,10 mm)

Inertia total = !#-1E,10 mm)


7 pada bagian atas

7 pada bagian baah

= !-0+21)6/21 = 1.076E+08mm!

!))#-0

= !-0+21)6/21 = 6.126E+07mm!

/05#!0

I1 = 1.12#b1#h1!3 , *1#d12

I2 = 1.12#b2#h2!3 , *2#d22

I! = 1.12#b!#h!!3 , *!#d!2

I) = 1.12#b)#h)!3 , *)#d)2

I1

I2

I!

I)

81 = I . y

a

82 = I . y

b

81

82


32/88


Menghitung Beban

5#1k9.m1

2)k9.m!


otal ser;ice load = Berat sendiri balok , Beban hidup

= 1-#1 k9.m


33/88


0#+0


0#6 1100 = 660 k9

0#+ 1100 = ++0 k9

dan β =

αPi =

βPi =


34/88

34 SONDRA RAHARJA ST


&ari perhitungan besaran penampang dapat diketahui :

* = 2#1!E,05mm2 e =

!#512E,0-mm! e = !-5 1.2 200

!#512E,0-mm! e = 2-5 mm

= )#// - -#-5 . )#0+

= 0#66 9.mm2

= )#// . -#-5 - )#0+

= +#!! 9.mm2

yb h!

81 =

82 =

, *t ' P

i ) A - P

i+e)1

* . M

i ) 1

*

( 1t = 66010!.2#1!1053 - 66010!!25.!5#1210/3 . 1)!#)10/ . !5#1210/3

, /t ' P

i ) A . P

i+e)1

/ - M

i ) 1

/

( 2t = 66010!

.2#1!105

3 . 66010!

!25.!5#1210/

3 - 1)!#)10/

. !5#1210/

3

cgc = cen

cgs = centre gravity of steel

+4.66 N!!2

400

200

200

750

150

275e =

α

α#i

+4.66 N!!2

$7.75 N!!2

+7.75 N!!2

cgs = centre gravity of steel200

+4 66 N!!2 +7 75 N!!2


35/88

35 SONDRA RAHARJA ST

400

+4.66 N!!2 +7.75 N!!2


36/88



* = )#1!E,05 mm2 e =

1#0-/E,0+ mm! e = /05#!0!0!0!0!0! 1.2 200

/#12/E,0- mm! e = 505.30 mm

= 2#1! - )#1! . )#)-

= 2#)- 9.mm2

= 2#1! . -#2/ - -#+5

= 1#5) 9.mm2

yb h!

81 =

82 =

, *s

' Pi ) A - P

i+e)1

* . M

s ) 1

*

( 1s

= ++010!.)#1!1053 - ++010!!25.1#0-/10+3 . )+0#610/ . 1#0-/10+3

, /s

' Pi ) A - P

i+e)1

/ . M

s ) 1

/

( 2s

= ++010!.)#1!1053 . ++010!!25./#12/10-3 - )+0#610/ . /#12/10-3

cgc = centre gravity of concrete


+2.13 N!!2

+2 13 N!!2

$ 4.13 N!!2

+ 7 26 N!!2

1000

150 505e =

200

200

750

200

β#i

cgs = centre gravity of steel200


37/88

+2.13 N!!2 + 7.26 N!

400


38/88





= 0#!/ !0 1.2

= 1#6- 9.mm2

= 0#5 !0

= 15#00 9.mm2


@lass 2' posttensioned3= 0#!/ 50 1.2

= 2#55 9.mm2

= 0#!! 50

= 1/#50 9.mm2

2esimpulan



0#66 9.mm2 1#6- 9.mm2

( tt = 0#!/(

ci31.2 9.mm2

( ct = 0#5#(


( ts = 0#!/(

cu31.2 9.mm2

( cs

= 0#!!#( cu


( 1t =



39/88


+#!! 9.mm2 15#00 9.mm2



2#)- 9.mm2 2#55 9.mm2

1#5) 9.mm2 1/#50 9.mm2

( 2t =

( 1s =( 2s

=


40/88



41/88



42/88


on%isi t!ans,e!

200

200

750

375

275

275

50

yb =

375ya =

d1 =

d3 =

375

275

yb =

d3 =


43/88


00

200


44/88



45/88


200

200

750

200

1000

150

605.30

344.70

44.70

230.30

244.70

505.30

yb =

ya =

d1 =

d2 =

d3 =

d4 =

2on%isi se!0i$e

605.30

.

yb =

=


46/88

400

200


47/88


200

200

750

200

605.30

344.70

44.70

230.30

244.70

505.30

yb =

ya =

d1 =

d2 =

d3 =

d4 =


48/88



49/88



50/88



51/88

cgc = centre gravity of concrete

275

"i

α#i

N!!2 +4.08 N!!2

$4.08 N!!2

α#i

$4.08 N!!2


52/88


53/88

+ 4.47 N!!2

"s

$ 7.85 N!!2

β#i


54/88

7.85 N!!2


55/88


56/88


T34AS / BETON PRESTRESS


57/88


Auatu jembatan beton prategang dengan penampang sebagai berikut :

6engan %ata sebagai be!ikut 5

Panjang jembatan = !0 m

Initial prestress = 2000 k9


Eksentrisitas cgs cgc e3 diketahui setelah masingmasing ttik berat cgc dan cgs diketahui

catatan : cgc = center gra;ity o( concrete titik berat penampang beton3

cgs = center gra;ity o( steel titik berat penampang baja.besi3

"itung tegangan :


2on%isi t!ans,e!

1000

400



58/88



59/88


60/88

Menghitung ne!tia

2


61/88


dimana 4

= )00 mm

= 0 mm karena berhimpitan dg garis netral penampang= )00 mm

= 1.12 )00 200 ! , +0000 )00 2

= 1.12 150 /00 ! , 60000 0 2

= 1.12 )00 200 ! , +0000 )00 2

= //////- mm)

= 2-00000000 mm)

= //////- mm)

Inertia total = !!!!!!! mm)


7 pada bagian atas

7 pada bagian baah

= 2++!!!!!!!! = 5.767E+07mm!

500

= 2++!!!!!!!! = 5.767E+07mm!

I = ΣIi, *

i#d

i23

d1 = y

a 1.2#h

1

d2

= ya

1.2#"1

d! = y

b 1.2#h!

I1 = 1.12#b1#h1!3 , *1#d12

I2 = 1.12#b2#h2!3 , *2#d22

I! = 1.12#b!#h!!3 , *!#d!2

I1

I2

I!

81 = I . y

a

82 = I . y

b

81

82

150%00

500

y1 =

y2 =

500


500y2 =


62/88



= 2!0#/- mm

= 2!0#/- mm

/+ 2ON6S SER8CE


b1 = )00 mm b! = )00 mm "2 = 1200 mm

h1 = 200 mm h! = 200 mm

b2 = 150 mm b) = 1000 mm

h2 = /00 mm h) = 200 mm

*1 = b1#h1 = )00 200 = +0000 mm2

*2 = b2#h2 = 150 /00 = 60000 mm2

*! = b!#h! = )00 200 = +0000 mm2

*) = b)#h) = 1000 200 = 200000mm2

* = *1 , *2 , *! , *) = )50000mm2

* = )#50E,05 mm2

y1 = "1 1.2#h1 = 600 mm

y2 = 1.2#h2 , h! = 500 mm

kt = 8

2 . *

kb = 8

1 . *


yb = Σ *i#yi.*

400

100y3 =

100

150%00

500

y1 =

y2 =

1100y4 =


63/88

Menghitung ne!tia

I = ΣI , * d 23


64/88


dimana 4

600 -//#/- = 1!!#!! mm

-//#/- 500 = 2//#/- mm -//#/- 100 = ///#/- mm

1100 -//#/- = !!!#!! mm

= 1.12 )00 200 ! , +0000 1!!#!!!!!!!!!!!! 2

= 1.12 150 /00 ! , 60000 2//#///////////- 2

= 1.12 )00 200 ! , +0000 ///#///////////- 2

= 1.12 1000 200 ! , 200000 !!!#!!!!!!!!!!!! 2

= 1#/6E,06 mm)

= 6#10E,06 mm)

= !#5+E,10 mm)

= 2#26E,10 mm)

Inertia total = /#65E,10 mm)

I = ΣIi, *

i#d

i23

d1 = y

1 y

b=

d2 = yb y2 =d

! = y

b y

! =

d) = y

) y

b =

I1 = 1.12#b1#h1!3 , *1#d12

I2 = 1.12#b2#h2!3 , *2#d22

I! = 1.12#b!#h!!3 , *!#d!2

I) = 1.12#b)#h)!3 , *)#d)2

I1

I2

I!

I)


7 pada bagian atas8 = I . y


65/88


7 pada bagian atas

7 pada bagian baah

= /6500000000 = 1.604E+08mm!

)!!#!!

= /6500000000 = %.065E+07mm!

-//#/-

Menghitung Beban

/ k9.m1

2)k9.m!

12k9.m

)#+ k9.m 7 luas penampang plat 1 m B> Beton = 1000 200

otal ser;ice load = Berat sendiri balok , beban hidup , Berat plat di atas balok

= 22#+ k9.m

Beton γ con

=

Beban hidup' ??

=

Beban plat' &?

=

?2 . +


66/88


22#+ !0 2

+

25/5k9#m

<s =

<s =

Tegangan (ang Te!"a%i

$ehilangan tegang jangka pendek = 10%


67/88

$ehilangan tegang jangka pendek = 10%

$ehilangan tegang jangka panjang = 5%

0#60

0#+5


0#6 2000 = 1+00 k9

0#+5 2000 = 1-00 k9



* = 2#50E,05 mm2 e = 0 mm

5#-/-E,0- mm! e = 500 1.2 200

5#-/-E,0- mm! e = 400mm

1+00 10! . 2500003 1+00 10! )00 .5-//////#//////-3 , /-5 . 5-//////#//////-3

= -#20 - 12#)6 . 11#-1

= /#)2 9.mm2

1+00 10! . 2500003 ,1+00 10! )00 .5-//////#//////-3 /-5 . 5-//////#//////-3

= -#20 . 12#)6 - 11#-1

= -#6+ 9.mm2

jadi α =

dan β =

αPi =

βPi =

yb h!

81 =

82 =

, *t ' P

i ) A - P

i+e)1

* . M

i ) 1

*

( 1t =

, /t ' P

i ) A . P

i+e)1

/ - M

i ) 1

/

( 2t =

+7.20 N!!2 $12.4% N!!2

200

500

400d1 =

ya =

+7.20 N!!2

200

500

400d1 =

ya =


68/88


cgc = centre gravity of concrete α#i

+7.20 N!!2 +12.4% N!400

200

1000

150%00

500

100

y1 =

y2 =

y3 =

500

400

400

d1 =

d3 =

yb =


+7.20 N!!2 +12.4% N!!2400

200

500

100

y2 =

y3 =

500

yb =

α#i


69/88



* = )#50E,05 mm2 e =

1#/0)E,0+ mm! e = -//#///////////- 1.2 200

6#0/5E,0- mm! e = 666.67 mm

1-00 10! . )500003 1-00 10! ///#///////////- .1/0!+)/15#!+)/153 , 25/5 . 1/0!+)/15#!+)/153= !#-+ - -#0- . 15#66

= 12#-0 9.mm2

1-00 10! . )500003 ,1-00 10! ///#///////////- .60/521-!#61!0)!53 25/5 . 60/521-!#61!0)!53

= !#-+ . 12#50 - 2+#26

= 12#02 9.mm2

yb h!

81 =

82 =

, *s

' Pi ) A - P

i+e)1

* . M

s ) 1

*

( 1s

=

, /s

' Pi ) A . P

i+e)1

/ - M

s ) 1

/

( 2s

=

cgc = centre gravity of concrete α#i

+3.78 N!!2 $ 7.07 N!!2

200

1000

200

1100y4 =

433.33

133.33

266.67

333.33

ya =

d1 =

d2 =

d4 =

1000150%00

500

y1 =

y2 =

1100y4 =

766.67

266.67

666.67

yb =

d2 =

d3 =


70/88


+3.78 N!!2 + 12.50 N!!2

200100y3 =




71/88



= 0#!/ !0 1.2

= 1#6- 9.mm2

= 0#5 !0

= 15#00 9.mm2



= 0#!/ 50 1.2= 2#55 9.mm2

= 0#!! 50

= 1/#50 9.mm2

2esimpulan



/#)2 9.mm2 1#6- 9.mm2

( tt = 0#!/(

ci31.2 9.mm2

( ct = 0#5#(


( ts = 0#!/(

cu31.2 9.mm2

( cs

= 0#!!#( cu


( 1t =


-#6+ 9.mm2 15#00 9.mm2


( 2t =


72/88



12#-0 9.mm2 2#55 9.mm2

12#02 9.mm2 1/#50 9.mm2

( 1s

=

( 2s =


73/88



74/88



75/88


400

200

200

1000

150

500

500

400

400

d1 =

d3 =

yb =

ya =

2on%isi t!ans,e!


76/88



77/88


200

200

1000

150

500

500

400

400

d1 =

d3 =

yb =

ya =

2on%isi t!ans,e!

200

500

400d3 =

yb =


78/88


400

200

1000

200

1000

150

766.67

433.33

133.33

266.67

666.67

333.33

yb =

ya =

d1 =

d2 =

d3 =

d4 =

2on%isi se!0i$e

1000

150

766.67

666.67

yb =

d3 =


79/88

400

200

100


80/88



81/88


82/88



83/88

!!2 +11.71 N!!2 +6.42 N!!2


84/88

$11.71 N!!2

"i

+7.%8 N!!2

i


85/88


86/88

$ 28.2% N!!2 $ 12.02 N!!2


87/88


angan yang te!adi'$(nya


88/88

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