Tugas Besar.docx

TUGAS POLIGON DAN WATERPAS ILMU UKUR TANAH

OLEH : ALGAIL GREZELDA GUMINTANG 1541320027 / 03 1 MRK 3

JURUSAN TEKNIK SIPIL POLITEKNIK NEGERI MALANG 2016

BAB I SOAL 1. Data hasil pembacaan rambu pada pengukuran waterpas jalur tertutup sebagai berikut : RAMBU BELAKANG

RAMBU MUKA DAN

(cm)

PROFIL (cm)

NO TITIK

BA

BT

BB

BA

BT

BB

P

153

148

143

A

232.5

225.7

219

A

122.4

116.4

110.5

B

109.3

106

102.5

B

57.3

48.1

38.8

C

101.9

95.8

89.5

C

88.9

82.5

76.1

D

25.5

16.2

7.3

D

124.4

118.9

113.3

E

82.9

76.7

70.4

E

140

136.8

132.9

P

132.2

128.3

124.4

Hitung Elevasi (H) titik A,B,C,D,E sesuai prosedur, pada jaringan tersebut, jika Hp = 100,325 meter. (Hp + No. urut absen = 100,325 + 3= 103,325 m)

2. Data hasil pengukuran “sudut dalam” dan jarak mendatar suatu polion

A

A

B

C

D

E

A

TINGGI

PEMBACAAN ARAH HORIZONTAL

(meter)

TARGET

T.PSWT

tertutup sbb :

1.500

ALAT

1.450

1.420

1.474

1.410

1.502

1.602

JARAK (meter) ͦ

‘

"

U

0

0

0

B

179

53

0

E

50

30

45

B

179

53

0

A

216

15

0

C

306

17

40

B

287

56

30

D

18

28

20

C

56

39

20

E

148

22

15

D

321

58

45

A

100

20

30

E

50

34

20

B

179

1

15

Jika diketahui koordinat titik a : XA = -50,000 meter ; YA=15,000 meter A. Gambarlah (sket) poligon tersebut sesuai data B. Hitunglah koordinat poligon sesuai prosedur catatan : koordinat awal (XA,YA) + no. urut absen (meter) (XA = -47,000 meter ; YA = 18,000 meter)

42.550

37.020

56.350

19.730

23.130

BAB II PENYELESAIAN

SOAL 1 Menghitung Jarak

D = 100 x (BA − BB) 

DAP = 100 x ((BAP + BAA ) − (BBP + BBA )) DAP = 100 x ((153,0 + 232,5) − (143,0 + 219,0)) DAP = 2350 cm



DAB = 100 x ((BAA + BAB ) − (BBA + BBB )) DAB = 100 x ((122,4 + 109,3) − (110,5 + 102,5)) DAB = 1.870 cm



DBC = 100 x ((BAB + BAC ) − (BBB + BBC )) DBC = 100 x (( `157,3 + 101,9) − (38,3 + 89,5)) DBC = 3.090 cm



DCD = 100 x ((BAC + BAD ) − (BBC + BBD )) DCD = 100 x ((88,9 + 25,5) − (76,1 + 7,3)) DCD = 3.100 cm



DDE = 100 x ((BAD + BAE ) − (BBD + BBE )) DDE = 100 x ((124,4 + 82,9) − (133,3 + 70,4)) DDE = 2.360 cm



DEF = 100 x ((BAE + BAF ) − (BBE + BBF )) DEF = 100 x ((140,0 + 132,2) − (132,9 + 124,4))

DEF = 1.490 cm

∑ D = DAB + DBC + DCD + DDE + DEF ∑ D = 1.870 + 3.090 + 3.100 + 2.360 + 1.490 = 14.260 cm

Menghitung Beda Tinggi ∆h = BTblk − BTmk 

∆hPA = BTP − BTA ∆hPA = 148 − 225,7 = −77,7 cm ≈ −0,777 m



∆hAB = BTA − BTB ∆hAB = 116,4 − 10,6 = 10,4 cm ≈ 0,104 m



∆hBC = BTB − BTC ∆hBC = 48,1 − 95,8 = −47,7 cm ≈ −0,477 m



∆hCD = BTC − BTD ∆hCD = 82,5 − 16,2 = 66,3 cm ≈ 0,633 m



∆hDE = BTD − BTE ∆hDE = 118,9 − 76,7 = 42,2 cm ≈ 0,422 m



∆hEP = BTE − BTP ∆hEP = 136,8 − 128,3 = 8,5 cm ≈ 0,085 𝑚

∑ ∆h = ∆hPA + ∆hAB + ∆hBC + ∆hCD + ∆hDE + ∆hEP ∑ ∆h = −77,7 + 10,4 + −47,7 + 66,3 + 42,2 + 8,5 = 2 cm ≈ 0,002 m

Menghitung Koreksi Beda Tinggi δh = 

δhPA =

dPA ∑d

d x(−fh) ∑d

x(−fh)

2.350

δhPA = 14.260 x(−2) = −0,3296 cm ≈ -0,0033 m 

δhAB =

dAB ∑d

x(−fh)

1.870

δhAB = 14.260 x(−2) = −0,2623 cm ≈ -0,0026 m 

δhBC =

dBC ∑d

x(−fh)

3.090

δhBC = 14.260 x(−2) = −0,4334 cm ≈ -0,0043 m

 δhCD = δhCD = 



dCD ∑d

x(−fh)

3.100 14.260

δhDE =

dDE

δhDE =

2.360

δhEP =

∑d

x(−fh)

14.260 dEP ∑d

1.490

x(−2) = -0,4348 cm ≈ -0,0043 m

x(−2) = -0,3310 cm ≈ -0,0033 m

x(−fh)

δhEP = 14.260 x(−2) = −0,2090 cm ≈ −0,0021m

Menghitung Elevasi Titik

H(titik) = Hawal + ∆h + δh 

HA = HP + ∆hAP + δhAP HA = 103,325 + (−0,77) + (−0,0033) = 102,545 m



HB = HA + ∆hAB + δhAB HB = 102,545 + 0,104 + (−0,0026) = 102,646 m



HC = HB + ∆hCB + δhCB HC = 102,646 + (−0,477) + (−0,0043) = 102,165 m



HD = HC + ∆hCD + δhCD HD = 102,165 + 0,663 + (−0,0043) = 102,823 m



HE = HD + ∆hDE + δhDE HE = 102,823 + 0,422 + (−0,0033) = 103,242 m



HP = HE + ∆hEP + δhEP HP = 103,242 + 0,0085 + (−0,0021) = 103,325 m (sebagai kontrol hitungan)

SOAL 2 A.(dibaliknya) B. MENGHITUNG BACAAN SUDUT HORIZONTAL (β) 

βA = JAB − JAE βA = 179°53′ 0′′ − 50°30′ 45′′ = 129°22′15′′



βB = JBC − JBA βB = 306°17′ 40′′ − 216°15′ 0′′ = 90°2′40′′



βC = JCD − JCB βC = 18°28′ 20′′ − 287°56′ 30′′ = −270°31′ 50′′ + 360° = 90°31′50′′



βD = JDE − JDC βD = 148°22′ 15′′ − 56°39′ 20′′ = 91°42′55′′



βE = JEA − JED βE = 100°20′ 30′′ − 321°58′ 45′′ = −222°21′ 45′′ + 360° = 138°21′45′′

MENGHITUNG KESALAHAN TOTAL SUDUT UKURAN ATAU CLOSSING ERROR POLYGON (fβ) fβ={(∑ β)-n.180°}={(∑ β)-(N-2).180°} (memakai sudut dalam) fβ={(βA +βB +βC +βD +βE )-(5-2).180°} fβ={(129°22'15''+90°2'40''+90°31'50''+91°42'55''+138°21'45'')-(5-2).180°} fβ={(540°1'25'')-540°} fβ=0°1'25''

Menghitung nilai koreksi sudut dan nilai sudut terkoreksi. 

Nilai koreksi total = -fβ = -0°1′25′′



Besarnya koreksi setiap sudut ukuran (Δβ) Δβ = -fβ / N Δβ = -0°1′25′′ / 5 = -0°0′17′′



Nilai sudut terkoreksi : β = βu + ∆β o βA = βA u + ∆β βA = 129°22′ 15′′ + (−0°0′ 17′′ ) = 129°21′58′′ o βB = βB u + ∆β βB = 90°2′40′′ + (−0°0′ 17′′ ) = 90°2′23′′ o βC = βC u + ∆β βC = 90°31′50′′ + (−0°0′ 17′′ ) = 90°31′33′′ o βD = βD u + ∆β βD = 91°42′55′′ + (−0°0′ 17′′ ) = 91°42′38′′ o βE = βE u + ∆β βE = 138°21′45′′ + (−0°0′ 17′′ ) = 138°21′28′′

MENGHITUNG AZIMUTH/SUDUT JURUSAN (lihat sket) 

αAE = JAE − JAU αAE = 50°50′ 45′′ − 0°0′ 0′′ = 50°50′ 45′′



αEA = αAE + 180° αEA = 50°50′ 4 5′′ + 180° = 230°50′45′′



αAB = JAB − JAU αAB = 179°53′ 0′′ − 0°0′ 0′′ = 179°53′ 0′′



αBC = αAB + βB − 180° αBC = 179°53′ 0′′ + 90°2′ 23′ − 180° = 89°55′40′′



αCD = αBC + βC − 180° αCD = 89°55′ 40′′ + 90°31′ 33′′ − 180° = 0°27′30′′



αDE = αDC + βD αDE = (αCD + 180°) + βD αDE = (0°27′ 30′′ + 180°) + 91°42′ 38′′ = 272°10′25′′

MENGHITUNG KESALAHAN TOTAL JARAK UKURAN DALAM ARAH X : (fx) dan ARAH Y : (fy)

1. Menghitung kesalahan total jarak ukuran arah Absis (fx) : fx = {∑(d . sin α)} 

fx = {dAB . sin αAB + dBc . sin αBC + dCD . sin αCD + dDE . sin αDE + dEA . sin αEA }



fx = {42,550 . sin 179°53′ 0′′ + 37,020 . sin 89°55′40′′ + 56,350 . sin 0°27′ 30′′ + 19,730 . sin 272°10′25′′ + 23,130 . sin 230°50′45′′}



fx ={42,550 . 0,0020 + 37,020 . 0,9999 + 56,350 . 0,0080 + 19,730 . (-0,9993) + 23,130 . (-0,7718)}



fx = 0,0293

2. Menghitung kesalahan total jarak ukuran arah Ordinat (fy) : fy = {∑(d . cos α)}  fy = {dAB . cos αAB + dBc . cos αBC + dCD . cos αCD + dDE . cos αDE + dEA . cos αEA }  fy = {42,550 . cos 179°53′ 0′′ + 37,020 . cos 89°55′40′′ + 56,350 . cos 0°27′ 30′′ + 19,730 . cos 272°10′25′′ + 23,130 . cos 230°50′45′′}  fy ={42,550 . (-0,9999) + 37,020 . 0,0013 + 56,350 . 0,9999 + 19,730 . 0,0379 + 23,130 .0,6360}  fy = -0,0157

MENGHITUNG NILAI KOREKSI SETIAP JARAK UKURAN

Besarnya koreksi setiap jarak ukuran dalam arah X : δx = (d / ∑d) . (-fx) ∑d = dAB + dBc + dCD + dDE + dEA ∑d = 42,550 + 37,020 + 56,350 + 19,730 + 23,130 = 178,78 m dAB

δx1 = (

) . (−fx)

∑d

42,550

δx1 = (

) . (−0,0293) = -0,0070 m

178,78

dBC

δx2 = (

∑d

) . (−fx)

37,020

δx2 = (

) . (−0,0293) = -0,0061 m

178,78

dCD

δx3 = (

∑d

) . (−fx)

56,350

δx3 = (

) . (−0,0293) = -0,0092 m

178,78

dDE

δx4 = ( δx4 = (

∑d

) . (−fx)

19,730

) . (−0,0293) = -0,0032 m

178,78

dEA

δx5 = (

∑d

23,130

) . (−fx)

δx5 = (178,78) . (−0,0293) = -0,0038 m

Besarnya koreksi setiap jarak ukuran dalam arah Y : δy = (d / ∑d) . (-fy) dAB

δy1 = (

) . (−fy)

∑d

42,550

δy1 = (

) . (0,0157) = 0,2647 m

178,78

dBC

δy2 = (

∑d

) . (−fy)

37,020

δy2 = (

) . (0,0157) = 0,2303 m

178,78

dCD

δy3 = (

∑d

) . (−fy)

56,350

δy3 = (

) . (0,0157) = 0,3506 m

178,78

dDE

δy4 = ( δy4 = (

∑d

) . (−fy)

19,730

) . (0,0157) = 0,1228 m

178,78

d

δy5 = ( EA) . (−fy) ∑d 23,130

δy5 = (

) . (0,0157) = 0,1439 m

178,78

MENGHITUNG KOORDINAT POLIGON Xx = Xp + dPX sin αPX + δx Yx = Yp + dPX cos αPX + δy

XB = XA + dAB . sin αAB + δx1 = (-47,000) + 42,550 . sin 179°53′ 0′′ + (-0,0070) = -46,922 m XC = XB + dBC . sin αBC + δx2 = (-46,922) + 37,020 . sin 89°55′40′′ + (-0,0061) = -9,867 m XD = XC + dCD . sin αCD + δx3 = (-9,867) + 56,350 . sin 0°27′ 30′′ + (-0,0092) = -9,425 m XE = XD + dDE . sin αDE + δx4 = (-9,425) + 19,730 . sin 272°10′25′′ + (-0,0032) = -29,145 m XA = XE + dEA . sin αEA + δx5 = (-29,145) + 23,130 . sin 230°50′45′′ + (-0,0038) = -47,000 m (sebagai kontrol hitungan)

YB = YA + dAB . cos αAB + δy1 = (-47,000) + 42,550 . cos 179°53′ 0′′ + 0,2647 = -25,281 m YC = YB + dBC . cos αBC + δy2 = (-46,922) + 37,020 . cos 89°55′40′′ + 0,2303 = -25,003 m YD = YC + dCD . cos αCD + δy3 = (-9,867) + 56,350 . cos 0°27′ 30′′ + 0,3506 = 31,696 m YE = YD + dDE . cos αDE + δy4 = (-9,425) + 19,730 . cos 272°10′25′′ + 0,1228 = 32,567 m YA = YE + dEA . cos αEA + δy5 = (-29,145) + 23,130 . cos 230°50′45′′ + 0,1439 = 18,000 m (sebagai kontrol hitungan)