TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103)
Soal 1. a A. Diketahui data dan soal seperti pada gambar dibawah ini:
Pertanyaan :
Hitung Reaksi Hitung dan gambar bidang momen dari kiri dan kanan Hitung dan gambar bidang D (gaya lintang) dari kiri dan kanan Penyelesaian :
Reaksi ∑MA = 0 - P1(2) + P2(4) + q.(9)(4,5) - RB(9) + P3.(1,5) = 0 (-1)(2) + 2.4 + 2.9(4,5) – 9RB – 5.12 = 0 - 2 + 8 + 81 – 9RB + 60 = 0 – 9RB = - 147 RB =
−147 −9
RB = 16,33 ton ∑MB = 0 P3(3) + RA(9) - q.(9)(4,5) – P2(5) – P1.11 = 0 5.3 + 9RA + 2.9(4,5) – 2.5 + (–1)11 = 0 15 + 9RA + 81 – 10 - 11 = 0 9RA = 87 RA =
87 9
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103) RA = 9,67 ton
Kontrol ∑V = 0 = (RA + RB) – (P1 +P2 +P3 + q.9) = (9,67 + 16,33) – (1 + 2 + 5 + 2.9) = 26 – (1 + 2 + 5 + 18) = 26 – 26 = 0 ton ............ (ok)
Momen (M) Momen dari kiri. MC = 0 tm MA = - P1 . 2 = -1.2 = - 2 tm MD = -P1.6 + RA.4 – q.4.2 = (-1).6 + 9,67.4 – 2.4.2 = -6 + 38,68 – 16 = 16,7 tm MB = -P1.11 + RA.9 – q.9 (4,5) – P2.5 = -1.11 + 9,67.9 – 2.9(4,5) – 2.5 = - 11 + 87,03 – 81 - 10 = -15 tm ME = - P1.14 + RA.12 – q.4.(2 + 8) - q.5.(2,5 + 3) + RB.3 - P2.8 = - 1.14 + 9,67.12 – 2.4.10 – 2.5.5,5 + 16,33.3 – 2.8 = - 14 + 116,04 – 80 – 55 – 48,99 - 16 = 0 tm
Momen dari kanan. ME = 0 tm MB = - P3 . 3 = -5.3 = - 15 tm MD = - P3.8 + RB.5 – q.5.2,5
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103)
= (-5).8 + 16,33.5 – 2.5.2,5 = - 40 + 81,65 – 16 = 16,7 tm MA = - P3.12 + RB.9 – q.5. (2,5 + 4) – P2.4 – q.4.2 = - 5.12 + 16,33.9 – 2.5.6,5 – 2.4 – 2.4.2 = - 60 + 146,97 – 65 – 8 - 16 = - 2 tm MC = - P3.14 + RB.11 – q.5.(2,5 + 6) - P2.6 - q.4.(2 + 2) + RA.2 = - 5.14 + 16,33.11 – 2.5.8,5 – 2.6 - 2.4.2 + 9,67.2 = - 70 + 179,63 – 85 – 12 – 32 + 19,34 = 0 tm
Gaya Lintang (D) Gaya Lintang Dari Kiri DC1 DC2 DA1 DA2 DD1
DD2
= = = = = = = = = = = = = = =
DD2 – q.5 0=ton DC1 – P1 0–1 - 1 ton DC2 = - 1 ton DA1 + RA - 1 + 9,67 8,67 ton DA2 – q.4 8,67 – 2.4 8,67 – 8 0,67 DD1 – P2 0,67 – 2 - 1,33 ton
= = = = = = = = = = =
- 1,33 – 2.5 - 1.33 – 10 - 11,33 ton DB1 + RB - 11,33 + 16,33 5 ton DB2 5 ton DE1 – P3 5–5 0 ton
DB1
DB2 DE1 DE2
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103)
Gaya Lintang Dari Kanan DE2 DE1 DB2 DB1 DD2
DD1
= = = = = = = = = = = = = = =
DD1 + q.4 0=ton DE2 + P3 0+5 5 ton DE1 = 5 ton DB2 – RB 5 - 16,33 - 11,33 ton DB1 + q.5 - 11,33 + 2.5 - 11,33 + 10 - 1,33 ton DD2 + P2 - 1,33 + 2 0,67 ton
= = = = = = = = = = =
0,67 + 2.4 0,67 + 8 8,67 ton DA2 - RA 8,67 - 9,67 - 1 ton DA1 - 1 ton DC2 + P1 -1 +1 0 ton
DA2
DA1 DC2 DC1
Momen Maksimum (Mx) Momen maksimum dari kiri ( syarat x ≤ 4 dari titik A ). 1 P1 (2 x ) RA..x q.x. x 2 Mx = =
- 1 ( 2 + x ) + 9,67.x – 2 . x. ½. x
=
- 2 – x + 9,67x – x2
TUGAS BESAR STATIKA =
Mx 0 x
Sertin N. M. Mooy (1221103)
- 2 + 8,67x – x2
(- 2 + 8,67x - x 2 ) 0 x
- 2x + 8,67 = 0 2x = 8,67 8,67 2 x = x = 4,335 m Tidak memenuhi syarat, karena nilai x Momen maksimum dari kanan ( syarat x ≤ 5 dari titik B ). = 4,335 m lebih besar dari 4 m. 1 P3 (3 x ) RB ..x q.x. x 2 Mx = =
5 ( 3 + x ) - 16,33. x + 2 . x. ½. x
= =
5x + 5 x - 16,33x + x2 x2 – 11,33x + 15
Mx 0 x ( x 2 - 11,33x + 15) 0 x
2x – 11,33 = 2x =
Tidak memenuhi syarat, karena nilai x
x = x =
0 11,33 11,33 2 5,665 m
= 4,335 m lebih besar dari 4 m. Dengan demikian disimpulkan bahwa momen maksimum terletak di titik D. Disebabkan selain karena nilai x dari kiri dan kanan tidak memenuhi syarat, pada gaya lintang di titik D, daya lintang (D) = 0 sehingga momen berada pada titik ekstrim.
TUGAS BESAR STATIKA
Gambar Gaya Lintang (D) dan Momen (M). Skala 1 ton = 2 mm.
Sertin N. M. Mooy (1221103)
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103)
Soal 1. b A. Diketahui data dan soal seperti pada gambar dibawah ini:
Pertanyaan :
Hitung Reaksi Hitung dan gambar bidang momen dari kiri dan kanan Hitung dan gambar bidang D (gaya lintang) dari kiri dan kanan Penyelesaian :
Reaksi ∑MS2 = 0
∑MS3 = 0 q . 4 . 2 – RS3 . 4 = 0 2 . 4 . 2 – 4RS3 = 0 16 – 4RS3 = 0 - 4RS3 = -16 16 RS3 = 4
- q . 4 . 2 + RS2 . 4 - 2 . 4 . 2 + 4RS2 - 16 + 4RS2 4RS2
= = = =
RS2 =
RS3 = 4 ton
0 0 0 16 16 4
RS2 = 4 ton
∑MB = 0 PS2 . 3 + q . 3 . 1,5 – P1 . 1 + RS1. 3 = 0 4 . 3 + 2 . 3 . 1,5 – 2,5 . 1 + 3 RS1 = 0 12 + 9 – 2,5 + 3 RS1 = 0 3 RS1 = - 18,5 RS1 =
−18,5 3
RS1 = - 6,17 ton
∑MS1 = 0
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103) P1 . 2 - RB . 3 + q . 3 . 4,5 + PS2 . 6 = 0 2,5 . 2 - 3 RB + 2 . 3 . 4,5 + 4 . 6 = 0 5 - 3 RB + 27 + 24 = 0 - 3 RB = - 56 RB =
−56 −3
RB = 18,67 ton ∑MA = PS1 . 2 = - 6,17 . 2 = - 12,24 ton RA
= PS1 = - 6,17 ton
∑MC = 0 - PS3 . 4 - q . 4 . 2 - RD . 2 + P2 . 2 + P3 . 5 = 0 - 4 . 4 - 2 . 4 . 2 - 2 RD + 4 . 2 + 2,5 . 5 = 0 - 16 - 16 – 2 RD + 8 + 12,5 = 0 - 2 RD = 11,5 RD =
11,5 −2
RD = - 5,75 ton ∑MD = 0 - q . 4 . 4 + RC . 2 - PS3 . 6 + P3 . 3 = 0 - 2 . 4 . 4 + 2 RC - 4 . 6 + 2,5 . 5 = 0 - 32 + 2 RC - 24 + 7,5 = 0 2 RC = RC =
48,5
48,5 2
RC = 24,25 ton
Kontrol ∑V = 0 ( RA + RB + RC + RD ) – ( P1 + P2 + P3 + q . 11 )
= 0
( - 6,17 + 18,67 + 24,25 – 5,75 ) – ( 2,5 + 4 + 2,5 + 2 . 11 )
= 0
31 – 31
= 0
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103) 0
= 0 .... (ok)
Gaya Lintang (D) Gaya Lintang dari kiri DA1 DA2 DS1 DE1 DE2 DB1 DB2 DS2
= = = = = = = = = = = = = = = = =
0=ton 4 – 2.4 DA1 + RA 0 + ( - 6,17 ) - 6,17 ton DA2 = - 6,17 ton DS1 = - 6,17 ton DE1 – P1 - 6,17 – 2,5 - 8,67 ton DE2 = - 8,67 ton DB1 + RB - 8,67 + 18,67 10 ton DB2 – q . 3 10 - 2 . 3 10 – 6 4 ton DS2 – q . 4
= = = = = = = = = = = = = = = = =
4 - 8 - 4 ton DS3 – q . 4 -4 - 2.4 -4 – 8 - 12 ton DC1 + RC - 12 + 24,25 12,25 ton DC2 = 12,25 ton DD1 + RD – P2 12,25 + (-5,75) - 4 2,5 ton DD2 = 2,5 ton DF1 – P3 2,5 - 2,5 0 ton
DS3
DC1
DC2 DD1 DD2 DF1 DF2
Gaya Lintang dari kanan
TUGAS BESAR STATIKA DF2 DF1 DD2 DD1 DC2 DC1
DS3
DS2
= = = = = = = = = =
12,25 0 =ton DF2 + P3 24,25 0 + 2,5 2,5 ton DF1 = 2,5 ton DD2 - RD + P2 2,5 + 4 - (-5,75) 12,25 ton DD1 = 12,25 ton DC2 - RC
= = = = = = = = =
- 12 ton DC1 + q . 4 - 12 + 2 . 4 - 12 + 8 - 4 ton DS3 + q . 4 -4 + 2.4 -4 + 8 4 ton
DB2
DB1
DE1
Sertin N. M. Mooy (1221103)
=
DS DE22 ++ Pq1 .
= = = = = = = =
3 4 + 2.3 4 + 6 10 ton DB2 - RB 10 - 18,67 - 8,67 ton 12,25 ton DB1 = - 8,67 ton
= = = = = = = =
- 8,67 + 2,5 - 6,17 ton DE1 = - 6,17 ton DS1 = - 6,17 ton DA2 - RA - 6,17 - ( - 6,17 ) - 6,17 + 6,17 0 ton
DE1 DS1 DA2 DA1
Momen (M) Batang A - S1 MA = - P1 . 2 = - ( -6,17 ) . 2 = 6,17 . 2 = 12,3 tm
–
TUGAS BESAR STATIKA
Sertin N. M. Mooy (1221103)
Batang S1 - S2 Momen dari kiri MS1 = 0 tm ME = PS1 . 2 = - 6,17 . 2 = - 12,3 tm MB = PS1 . 3 – P1 . 1 = - 6,17 . 3 – 2,5 . 1 = - 18,51 – 2,5 = - 21 tm MS2 = PS1 .6 – P1 . 4 + RB . 3 – q . 3 . 1,5 = (-6,17) 6 – 2,5 . 4 + 18,67 . 3 – 2 . 3. 1,5 = - 37,02 – 10 + 56, 02 – 9 = 0 tm
Momen dari Kanan MS2 = 0 tm MB = - q . 3 . 1,5 – P2 . 3 = - 2 . 3 . 1,5 – 4 . 3 = - 9 – 12 = - 21 tm ME = - q.3.2,5 – PS2 . 4 + RB . 1 = - 2.3.2,5 – 4 . 4 + 18,67 . 1 = - 15 – 16 + 18,67 = - 12,3 tm MS1 = - q.3.4,5 – PS2 . 6 + RB . 3 – P1 . 2 = - 2.3.4,5 – 4 . 6 + 18,67 . 3 – 2,5 . 2 = - 27 – 24 + 56 – 5 = 0 tm
Batang S2 - S3 Momen dari kiri MS2 = 0 tm MS3 = – q . 2 . 4 + RS2 . 4 = –2.2.4 + 4.4 = - 16 + 16 = 0 tm
Momen dari kanan MS3 = 0 tm MS2 = – q . 4 . 2 + RS3 . 4 = –2.4.2 + 4.4 = - 16 + 16 = 0 tm
Batang S3 - F Momen Dari kiri MS3 = 0 tm MC = - PS3 . 4 – q . 4 . 2 = -4.4–2.4.2 = - 16 – 16 = - 32 tm MD = - PS3. 6 – q . 4 . 4 + RC . 2 = - 4. 6 – 2 . 4 . 4 + 24,25 . 2 = - 24 – 32 + 48,5 = - 7,5 tm MF = - PS3. 9 – q .4. 7 + RC . 5 – P2 . 3 + RD . 3 = - 4. 9 – 2 .4. 7 + 24,25 . 5 – 4 . 3 + (-5,75) . 3 = - 36 – 56 + 121,25 – 12 – 17, 25 = 0 tm
Momen Dari Kanan MF = 0 tm MD = - P3 . 3 = - 2,5 . 3 = - 7,5 MC = - P3. 5 – P2 . 2 + RD . 2 = - 2,5. 5 – 4 . 2 + (-5,75) . 2 = - 12,5 – 8 - 11,5 = - 32 tm MS3 = - P3. 9 – q .4. 2 + RD . 6 – P2 . 6 + RC . 4 = - 2,5 . 9 – 2 .4. 2 + (-5,75). 6 – 4 . 6 + 24,25 . 4 = - 22,5 – 16 – 34,5 – 24 + 97 = 0 tm
TUGAS BESAR STATIKA
Momen Maksimum (Mx)
Sertin N. M. Mooy (1221103)
Momen maksimum dari kiri ( syarat x ≤ 4 dari titik S2 ) 1 Mx = - q . 2 . x2 + RS2 . x = - 2.
1 2
. x2 + 4 . x
= – x2 + 4x
Mx 0 x
( x 2 4 x) 0 x
Substitusi nilai x ke Mx
– 2x + 4 = 0 – 2x = - 4 −4 x = −2
Mx = - x2 + 4x = - (22) + 4 . 2 = - 4+8
x = 2 m
= 4 tm
Momen maksimum dari kanan ( syarat x ≤ 4 dari titik S3 ) 1 Mx = RS3 . x - q 2 . . x2 = 4.x - 2.
1 2
. x2
= 4x – x2
Mx 0 x
(4 x x 2 ) 0 x
4 – 2x = 0 – 2x = - 4 −4 x = −2 x = 2 m
Substitusi nilai x ke Mx Mx = 4x - x2 = 4 . 2 - (22) = 8 - 4 = 4 tm
Gambar Gaya Lintang (D) dan Momen (M). Skala 1 ton = 1 mm.
Soal 2 A. Diketahui data dan soal seperti pada gambar dibawah ini:
Pertanyaan :
Hitung Reaksi Hitung dan gambar bidang momen Hitung dan gambar bidang D (gaya lintang) Penyelesaian :
Reaksi ∑MA = 0 q . 6 . 3 – VB . 11 + HB . 2
∑MB = 0 - q . 6 . 8 + VA . 11 + HA .
= 0 2 . 6 . 3 – 11 VB + 2 HB =
2 = 0 - 2 . 6 . 8 + 11 VA + 2 HA
0 36 – 11 VB + 2 HB = 0 11 VB - 2 HB = 36 .... (1) ∑MS ( Kiri ) = 0 - q . 3 . 1,5 + VA . 3 - HA. 4 = 0 - 2 . 3 . 1,5 + 3 VA - 4 HA = 0 - 9 + 3 VA – 4 HA = 0 3 VA - 4 HA = 9 .... (3) Eliminasi Persamaan (1) dan (4)
= 0 - 96 + 11 VA + 2 HA = 0 11 VA + 2 HA = 96 .... (2) ∑MS ( Kanan ) = 0 q . 3 . 1,5 – VB . 8 + HB . 6 = 0 2 . 3 . 1,5 – 8 VB + 6 HB = 0 9 – 8 VB + 6 HB = 0 8 VB - 6 HB = 9 .... (4)
11 VB - 2 HB = 36 8 VB - 6 HB = 9
x 6 x 2
66 VB - 12 HB = 216 16 VB - 12 HB = 18 50 VB = 198 198 VB = 50 VB = 3,96 ton
Eliminasi Persamaan (1) dan (4) 11 VB - 2 HB = 36 x 8 8 VB - 6 HB = 9 x 11
88 VB - 16 HB = 288 88 VB - 66 HB = 99 50 HB = 189 189 HB = 50 HB = 3,78 ton
Eliminasi Persamaan (2) dan (3) 11 VA + 2 HA = 36 x 4 3 VA - 4 HA = 9 x 2
44 VA + 8 HA = 384 6 VA - 8 HA = 18 + 50 VA = 402 402 VA = 50 VA = 8,04 ton
Eliminasi Persamaan (2) dan (3) 11 VA + 2 HA = 36 x 3 3 VA - 4 HA = 9 x 11
33 VA + 6 HA = 288 33 VA - 44 HA = 99 50 HA = 189 189 HA = 50 HA = 3,78 ton
Kontrol ∑MH = 0
∑MV = 0
HA - HB = 0 3,78 - 3,78 = 0 0 = 0 ... OK) 6 Tan 1 = 5 =
1,2
1 =
arc tan ( 1,2 )
1 =
50,190
2 =
90o - 50,19o
( VA + VB ) - q . 6 = 0 ( 8,04 + 3,96 ) - 2 . 6 = 0 12 - 12 = 0 ... (OK)
=
39,81o
VB diuraikan menurut garis tegak lurus dan sejajar terhadap batang = VB . cos 2 D1 = VB . sin 2 = 3,96 . cos 39,81 = 3,96 . sin 39,81 = 3,96 . 0,77 = 3,96 . 0,64 = 3,05 ton = 2,53 ton HA diuraikan menurut garis tegak lurus dan sejajar terhadap batang N1
N2
= = = =
HB . cos 1 3,78 . cos 50,19 3,78 . 0,64 2,42 ton
D2
Ntotal = N1 + N2 (searah) = 3,05 + 2,42 = 5,47 ton Batang C - D
DC DC DA1 2 DA2
= = = = = = = = = = = = = = = = = =
0 ton DC + VA 0 ton 1 0 + 8,04 DA1 - HA 8,04 ton 0 - 3,78 DC - q . 3 - 3,782 ton 8,04 - 2 . 3 DA2 = - 3,78 ton 8,04 - 6 2,04 ton DS - q . 3 2,04 - 2 . 3 2,04 - 6 - 3,96 ton DD1 + VB - 3,96 + 3,96 0 ton
1 Batang A-C
DS DC DD1
DD2
= = = =
HB . sin 1 3,78 . sin 50,19 3,78 . 0,77 2,91 ton
Dtotal = D1 - D2 (Lawan arah) = 2,53 - 2,91 = - 0,38 ton
Gaya Lintang (D)
Batang D – B DD1 = DD2 = DB1 = DB2 = =
0 ton - D = 0,38 ton DD2 = 0,38 ton DB1 + D 0,38 – 0,38 = 0 ton
Momen (M)
Batang A - C MA = 0 tm MC1 = - HA . 4 = - 3,78 . 4 = - 15,12 tm
Batang C - D MC2 = MC1 = - 15,12 tm MS = - q . 3 . 1,5 – HA . 4 + 3 . VA = - 2 . 3 . 1,5 – 3,78 . 4 + 8,04 . 3 = - 9 – 15,12 + 24,12 = 0 tm MD1 = - q . 6 . 3 – HA . 4 + VA . 6 = - 2 . 6 . 3 – 3,78 . 4 + 8,04 . 6 = - 36 – 15,12 + 48,24 = - 2,88 tm Batang D - B MB = 0 tm MD2 = HB . 6 - VB . 5
= 3,78 . 6 - 3,96 . 5 = 22,68 -19,8 = 2,88 tm Gaya Normal (N) Batang A - C NA1 = 0 ton NA2 = - VA = - 8,04 ton NC1 = NA2 = - 8,04 ton NC2 = NC1 + VC = - 8,04 + 8,04 = 0 tm Batang C - D NC1 NC2 ND1 ND2
= = = = = =
0 ton - HC = - 3,78 ton NC2 = - 3,78 ton ND1 + HD - 3,78 + 3,78 0 tm
Kontrol Momen Kontrol Momen MD ( Momen di D ) MD2 – MD1 = 0 28 – 28 = 0..... (OK)
Batang B - D NB1
= 0 ton
NB2
= - N = - 5,47 ton
ND1
= NB2 = - 5,47 ton
ND2
= ND1 + N = - 5,47 + 5,47 = 0 tm
Momen Maksimum (Mx) Momen Maksimum dari kiri ( Syarat x ≤ 3 Dari titik S3 ) 1 Mx = - q . ( 3 + x ) . 2 . ( 3 + x ) + ( 3 + x ) . VA - 4 . HA
= - q. = = = = = =
1 2
. ( 3 + x )2 + ( 3 + x ) . VA - 4 . HA
– ( 3 + x )2 + ( 3 + x ) VA - 4 . HA – ( 9 + 6x + x2 ) + 3 . VA + VA . x - 4 . HA - x2 – VA . x – 6x + 3 . VA - 4 HA - 9 - x2 + 8,04x – 6x + 3 . 8,04 - 4 . 3,78 - 9 - x2 + 2,04x + 24,12 – 15,12 - 9 - x2 + 2,04x
Mx 0 x
( x 2 2,04 x) 0 x
– 2x + 2,04 = 0 – 2x = - 2,04 −2,04 x = −2 x = 1,02 m Substitusi nilai x ke Mx Mx = - x2 + 2,04x = - (1,022) + 2,04 . 1,02 = - 1,0404 + 2,0808 Momen Maksimum dari kanan ( Syarat x ≤ 3 Dari titik D ) = 1,0404 tm 1 Mx = - q . x . 2 . x + ( 5 + x ) . VB - 6 . HB = -q.
1 2
. x2 + VB ( 5 + x ) - 6 HB
= -2.
1 2
. x2 + 3,96 . ( 5 + x ) - 6 . 3,78
= - x2 + 3,96 . 5 + 3,96x - 22,68 = - x2 + 3,96x - 2,88
Mx 0 x
( x 2 3,96 x 2,88) 0 x
- 2x + 3,96 = 0 - 2x = - 3,96 −3,96 x = −2
x = 1,98 m Substitusi nilai x ke Mx Mx = - x2 + 3,96x - 22,68 = - (1,982) + 3,96 . 1,98 – 2,88 = - 3,9204 + 7,8408 – 2,88 = 1,0404 tm
Gambar Free Body
Gambar gaya Normal ( N ) dan gaya Lintang (D) Skala 1 ton = 2 mm.
Gambar Gaya Lintang (D) dan Momen (M) Skala 1 ton = 2 mm.
Soal 3 Diketahui data dan soal seperti pada gambar dibawah ini:
Pertanyaan :
Hitung Reaksi secara grafis dan analisis. Hitung dan gambar bidang momen. Hitung bidang gaya lintang dan gaya normal di titik D secara grafis Penyelesaian :
Reaksi Perletakan 4+x=
√ 52−12
4+x=
√ 25−1
4 + x = 4,89 x = 4,89 – 4 x = 0,89 m ΣMA = 0 P1 . 3 + P2 . 4 + q . 4,445 . 7,2225 – HB . 1 – VB . 9,89 2 . 3 + 3 . 4 + 2 . 4,445 . 7,2225 – HB – 9,89 VB 6 + 12 + 64,208 – HB – 9,89 VB 82,208 – HB – 9,89 VB HB + 9,89 VB
= = = = =
0 0 0 0 82,208 ....... (1)
ΣMS (kanan) = 0 HB . 4 – VB . 4,89 + q . 4,445 . 2,2225 4 HB – 4,89 VB + 2 . 4,445 . 2,2225 4 HB – 4,89 VB + 19,758 4,89 VB - 4 HB
= = = =
0 0 0 19,758 ....... (2)
ΣMB = 0 VA . 9,89 - HA . 1 – P1 . 6,89 – P2 . 5,89 – q . 4,445 . 2,6675 = 0 9,89 VA - HA – 2 . 6,89 – 3 . 5,89 – 2 . 4,445 . 2,6675 = 0 9,89 VA - HA – 13,78 – 17,67 – 23,714 = 0 9,89 VA - HA – 55,164 = 0 9,89 VA - HA = 55,164 ....... (3) ΣMS (kiri) = 0 VA . 5 - HA . 5 – P1 . 2 – P2 . 1 = 0 5 VA - 5 HA – 2 . 2 – 3 . 1 = 0 5 VA - 5 HA – 4 – 3 = 0 5 VA - 5 HA – 7 = 0 5 VA - 5 HA = 7 ....... (4) Eliminasi Persamaan (1) dan (2) 9,89 VB + HB = 82,208 x 4,89 48,36 VB + 4,89 HB = 401,99 4,89 VB - 4 HB = 19,758 x 9,89 48,36 VB - 39,56 HB = 195,41 44,45 HB = 206,58 206,58 HB = 44,45 HB = 4,648 ton Eliminasi Persamaan (1) dan (2) 9,89 VB + HB = 82,208 x 4 4,89 VB - 4 HB = 19,758 x 1
39,56 VB + 4 HB = 401,99 4,89 VB - 4 HB = 19,76 + 44,45 VB = 348,59 348,59 VB = 44,45 VB = 7,842 ton
Eliminasi Persamaan (3) dan (4) 9,89 VA - HA = 55,164 x 5 5 VA - 5 HA = 7 x 9,89
49,45 VA _ 5 HA = 275,82 49,45 VA - 49,45 HA = 69,23 44,45 HA = 206,58 206,58 HA = 44,45 HA = 4,648 ton
Eliminasi Persamaan (3) dan (4) 9,89 VA - HA = 55,164 x 5 5 VA - 5 HA = 7 x 1
49,45 VA - 5 HA = 275,82 5 VA - 5 HA = 7 44,45 VA = 268,82 268,82 VA = 44,45 VA = 6,048 ton
Kontrol
∑MH = 0 HA - HB = 0 4,648 - 4,648
∑MV = 0 ( VA + VB ) - ( P1 + P2 + q . 6 ) = 0 ( 6,048 + 7,842 ) - ( 2 + 3 + 2. 4,445 ) = 0
= 0 0 = 0 ... OK) Pesamaan Parabola Yx =
= =
x
Yd
4 hx ( L−x ) L2
x
4 ( 5 ) x (2.5−x )
Yc
2
=
20 x (10−x) 102 200 x−20 x 100
=
4m
=
=
600−180 100
=
420 100
2
x
=
4 20 (¿¿ 2) 200.4− 100 ¿
=
800−320 100 4,8 m
3m 3 20(¿ ¿2) 200.3− 100 ¿
(2.5)
=
=
13,89 - 13,89 = 0 ... (OK)
=
Ys
480 100
=
4,2 m
=
5m
=
5 20(¿ ¿2) 200.5− 100 ¿
=
1000−500 100
=
5m
=
500 100
x
Ye
=
9m
=
9 20(¿ ¿2) 200.9− 100 ¿
=
1800−1620 100
Momen Momen dari kiri 180 MA == 0 100 tm MC = VA . 3 – HA . Yc = 6,048 . 3 – 4,648 . 4,2 = 18,144 – 19,5216 = - 1,38 tm MD = VA . 4 – HA . Yd – P1 . 1 = 6,048 . 4 – 4,648 . 4,8 – 2 . 1 = 24,192 – 22,3104 – 2 = - 0,12 tm MS = VA . 5 – HA . 5 – P1 . 2 – P2 . 1 = 6,048 . 5 – 4,648 . 5 – 2 . 2 – 3 . 1 = 30,24 – 23,24 – 4 – 3 = 0 tm ME = VA . 9 – HA . Ye – P1 .6 – P2 . 5 – q . 4 . 2 = 6,048 . 9 – 4,648 . 1,8 – 2 .6 – 3 . 5 – 2 . 4 . 2 = 54,432 – 8,3664 – 12 – 15 – 16 = 3,06 tm MB = VA . 9,89 – HA . 1 – P1 . 6,89 – P2 . 5,89 – q . 4,445 . 2,6675 = 6,048 . 9,89 – 4,648 . 1 – 2 . 6,89 – 3 . 5,89 – 2 . 4,445 . 2,6675 = 59,815 – 4,648 – 13,78 – 17,67 – 23,715 = 0 tm
Momen dari Kanan MB ME
MS
MD
MC
MA
= = = = = = = = = = = = = = = = = = = = = = = = =
0 tm VB . 0,89 - HB . ( Ye – 1 ) - q . 0,89 . 0,445 7,842 . 0,89 – 4,648 . ( 1,8 – 1 ) - 2 . 0,89 . 0,445 6,979 - 4,648 ( 0,8 ) - 0.198 6,979 - 3,719 - 0,198 3, 06 tm VB . 4,89 - HB . ( Ys – 1 ) - q . 4,445 . 2,2225 7,842 . 4,89 – 4,648 . ( 5 – 1 ) - 2 . 4,445 . 2,2225 38,34 - 4,648 ( 4 ) - 19,75 38,34 - 18,59 - 19,75 0 tm VB . 5,89 - HB . ( Yd – 1 ) - q . 4,445 . 3,2225 7,842 . 5,89 – 4,648 . ( 4,8 – 1 ) - 2 . 4,445 . 3,2225 46,189 - 4,648 ( 3,8 ) - 28,648 46,189 - 17,662 - 28,648 - 0,12 tm VB . 6,89 - HB . ( Yc – 1 ) - q . 4,445 . 4,2225 - P2 . 1 7,842 . 6,89 – 4,648 . ( 4,2 – 1 ) - 2 . 4,445 . 4,2225 - 3 . 1 54,031 - 4,648 ( 3,2 ) - 37,538 - 3 54,031 - 14,874 - 37,538 - 3 - 1,38 tm VB . 9,89 + HB . 1 - q . 4,445 . 7,2225 - P2 . 4 - P1 . 3 7,842 . 9,89 + 4,648 . 1 - 2 . 4,445 . 7,2225 - 3 . 4 - 2 . 3 77,557 + 4,648 - 64,208 - 12 - 6 0 tm
Momen Maksimum ( dari kiri ) dengan syarat 5 < x < 9 Mx = VA . x - HA . Yx - P1 . ( x – 3 ) – P2 ( x – 4 ) – q . ( x – 5 ) . ½ . ( x – 5 ) 2 200 x−20 x = 6,048 . x – 4,648 . - 2 . ( x – 3 ) – 3 ( x – 4 ) – 2 . ( x – 5 )2 . 100
(
)
½ = 6,048 x – 9,296 x + 0,9296 x2 - 2 x + 6 – 3 x + 12 – x2 + 10x - 25 = 0,9296 x2 – x2 + 6,048 x – 9,296 x - 2 x – 3 x + 10x + 12 + 6 - 25 = - 0,0704 x2 + 1,752 x - 7 Mx 0 x ( 0,0704 x 2 1,752 x 7) 0 x
-
0,1408 x + 1,752 = 0 0,1408 x = 1,752 x = 12,443 m ( tidak memenuhi syarat )
Momen Maksimum ( dari kanan ) dengan syarat 0 < x < 4,89 Mx = - VB . x + HB . ( Yx – 1 ) + q . x . ½ . x 2 200 x−20 x −1 = - 7,842 . x + 4,648 . + 2 . x2 . ½ 100
(
)
= - 7,842 x + 9,296 x - 0,9296 x2 - 4,648 + x2 = x2 - 0,9296 x2 - 7,842 x + 9,296 x - 4,648 = 0,0704 x2 + 1,454 x - 4,648 Mx 0 x (0,0704 x 2 1,454 x 4,648) 0 x
0,1408 x + 1,454 = 0 0,1408 x = - 1,454 x = - 10,326 m ( tidak memenuhi syarat )
Grafik Momen Max
Gaya Lintang Dan Gaya Normal bidang D secara grafis
Skala : 1 : 100 1 ton = 1 cm