Kumpulan Soal Mekrek
April 19, 2017 | Author: Widya Apriani | Category: N/A
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KUMPULAN SOAL DAN PENYELESAIAN MEKANIKA REKAYASA 1 DAN 2
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 1
GERBER
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 2
SOAL 1 P = 8 kN q = 4 kN/m
A
B 8m
D
C
S 2m
6m
P = 5 kN
4m
-33,32 kN
-7,99 kN
-0,02 kNm BMD
A
+
B
C
S
D
2,64 kNm 33,32 kNm
16,33kN
0,33 kN
2,677 kN 5,33 kN SFD
-13,327 kN
- 8 kN
JAWABAN : Ms Kiri Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 3
Rav.10+Rbv.2-4(8)(6)-4(2)(1) = 0 Rav.10+Rbv.2-192-8 = 0 Rav.10+Rbv.2 = 200……………………….(1) Ms Kanan -8(10)+Rcv.6 = 0 80= Rcv.6 Rcv = 13.333 kN ∑H = 0 Rah-5 =0 Rah = 5 ∑V = 0 Rav+Rbv+Rcv-4(10)-8 = 0 Rav+Rbv+13,33-40-8 =0 Rav+Rbv = 34,67 kN………………..(2)
Eliminasi (1) dan (2) 10.Rav +2.Rbv = 200 ..……………………..(1) Rav + Rbv = 34,67 ………………..……..(2) 10.Rav +2.Rbv 2.Rav +2.Rbv
= 200 ..……………………..(1) = 69.34 ………………..……..(2)
8.Rav
= 130.66
Rav
= 16.333 kN
Menentukan Rbv Rav + Rbv = 34,67 ………………..……..(2) 16.333 + Rbv = 34,67 Rbv = 18.34 kN
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 4
GAYA-GAYA DALAM
INTERVAL A–B 0-8
MOMEN Mx = Rav.x – qx
2
= 16.333 x - (4) x2 = 16.333 x - 2 x2
B–S 8 - 10
LINTANG Dx = = 16,333 – 4x x=0
D0 = 16.333 kN
x=0
M0 = 0
x=4
D4 = 0,33 kN
x=4
M4 = 33,332 kNm
x=8
D8 = -15,677 kN
x=8
M8 = 2,64 kNm
Mx = Rav.x – qx2 + Rbv.(x-8) – q(x-8)2
= Dx =
= 16.333 x - (4) x2 + 18.34(x-8) - (4) (x-8)2
= -8x + 66.673
= 16.333x - 2x2 + 18.34(x-8) – 2(x-8)2
x=8
D8 = 2,673 kNm
= 16.333x- 2x2 + 18.34x – 146,72 – 2(X2 - 16x +64)
x = 10
D10 = -13.327 kNm
= 16.333x - 2x2 + 18.34x – 146,72 – 2X2 + 32x – 128 = - 4x2 + 66.673 x -274.72 x=8
M8 = 2,64 kNm
x = 10
M10 = -7,99 kNm
D–C
Mx = -8x
Dx = -8
0-4
x=0
M0 = 0 kNm
x=0
M0 = -8 kNm
x=4
M4 = -32 kNm
x=4
M4 = -8 kNm
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 5
C–S 4 - 10
Mx = -8x + Rcv (x-4)
Dx =
= -8x + Rcv.x -4Rcv
Dx = 5,333
= -8x + 13,333x – 4(13,333)
x=4
M4 = 5,333 kNm
x = 10
M10 = 5,333 kNm
= 5,333x – 53,332 x=4
M4 = -32 kNm
x = 10
M10 = -0.02 kNm
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 6
PORTAL
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 7
SOAL 1 Contoh 2: gaya
Hitung dan gambarkan diagram gaya-
dalam yang terjadi pada Struktur Portal berikut ini.
Q=2 KN/m P=5 KN C
D
A
3m RAH
B
RAV
1m 3m
RBV
PENYELESAIAN : a. Menghitung Reaksi Perletakan:
H 0 M
A
P RAH 0
0
RAH 5 KN
R .3 R .1 2.3.1,5 5.3 0 BV
BH
R BV
29 kN 3
M 0 RAV .3 5.4 2.3.1,5 0 B
R AV
V 0
R R Q.3 AV
R AV
Kumpulan Soal Mekanika Rekayasa 1 dan 2
11 kN () 3
BV
11 29 6 KN 3 3
Page 8
b. Free Body Diagram dan Perhitungan GGD: Q2 = 2
15
P=5 5
5 29/3
20 11/3 11/3
29/3
0
20
5
FBD 5
11/3 29/3
Tabel Gaya-Gaya Dalam:
Bagian/
Momen
Geser
Interval A-C
(M)
(V)
M 0
dM 0 dx X
X
0-3
X 0 MA 0
(dari kiri) C-D
X 3M 0
Normal (N) N X
11 3
C
11 / 3 X 0,5 Q X
2
X 0M 0
dM 11 / 3 QX dx X
Nx = -5
C
0-3 (dari
X 3 M 20 D
X 0 V 0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
C
Page 9
kiri)
X 3 V 29 / 3
B-D
M R .4
Tdk Ada Mmak VX 5
0–4
X 0M 0
D
X
BH
N 29 / 3 X
B
X 4 M 20 D
D
2 C
c. Diagram Gaya-Gaya Dalam:
8
5 A
15
15
C
SFD
D
B
A
BMD
B
C
D 2
8
A
NFD
Kumpulan Soal Mekanika Rekayasa 1 dan 2
B
Page 10
SOAL 2
Q1 = 5 t/m
C
P = 10 t
E
D
F Q2 = 2t/m
3m
B
RBH 1m A 2m
2m
2m
RBV
RAV PENYELESAIAN ∑H=0 RBH – 2.3 = 0 RBH = 6 t (→) ∑ MB = 0 RAV.6 – 5.2.5 -10.2 – 2.3.1,5 = 0 RAV.6 – 50 – 20 – 9 = 0 RAV = 79/6 = 13,11 t ∑ MA = 0 RBV.6 - RBH.1 + 2.3.2,5 + 10.4 – 5.2.1 = 0 RBV.6 – 6.1 – 15 – 40 – 10 = 0 RBV = 41/6 = 6,8 t
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 11
DIAGRAM BENDA BEBAS Q1 = 5 t/m
P = 10 t
9
0
0 C
D
E
F 6,8
13,11 13,11
6,8
6 3m 6
B
1m A
6,8
13,11
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 12
Tabel Gaya-Gaya Dalam: Bagian/
Momen
Geser
Interval
(M)
(V)
A-C
M 0
dM 0 dx X
X
0-4
Normal (N)
NA-C = 13,11
X 0 MA 0
X 3M 0 C
B-F
Mx =6x – ½ Qx2
0-3
X 0MB 0
C-D
1 M X 13.11x qx 2 2
X 3 M F 9kNm
dM X (6 QX ) dx
NB-F = -6,8
dM X (6 2 X ) dx
X 0 VB 6
X 3 VF 0
1 V X 13.11 (5)(2) x 2
NC D 0
0–2 X 0 MC 0 1 X 2 M D 13.11(2) (5)(2) 2 2
D–E
Mx = 13.11(x+2)-10(x+1)
0-2
X 0 M D (13 .11)2 10 (2) 16 .34
X 0 VC 13 .11
X 2 V D 3.11 V X 13 .11 10
N D-E = 0
VC-D = 3.17
X 2 M E (13 .11)( 4) 10 (3) 22 .68
F–E
Mx = 6.83 x + 9
V X 6.83
0-2
X 0 MF 9
V F – E = -6.83
N F-E = -0
X 2 M E 22 .68
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 13
DIAGRAM BIDANG MOMEN (BMD) C
D
E
F
B
A DIAGRAM BIDANG GESER (SFD)
C
D
E
F
B
A
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 14
DIAGRA BIDANG NORMAL (NFD) C
D
E
F
B
A
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 15
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 16
PORTAL TIGA SENDI
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 17
SOAL 1
Gambar 5.4 Diagram gaya-gaya dalam portal sendi-rol akibat beban merata dan beban lateral terpusat
P = 5kN Q1 = 5 kN/m
E
C
S
D Q2 = 3 kN/m
4m
B 2m A 2m
2m
4m
Struktur portal tiga sendi dengan ukuran dan pembebanan seperti Gambar di bawah, hitunglah : 1. Reaksi Perletakan 2. Persamaan Gaya-Gaya Dalam dan Diagram Benda Bebas (FBD) 3. Diagram Gaya-Gaya Dalam (Momen, Geser, dan Normal)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 18
SOLUSI DIAGRAM BENDA LEPAS (DBL) P = 5 kN
R1 = 10 kN
C
R2 = 20 kN
D
S
2m R3 = 12 kN 2m
B
Rbh
2m
Rbv Rah
A
Rav 2m
1m
1m
2m
2m
MENENTUKAN BESARNYA REAKSI PERLETAKAN ∑MA = 0 -RBV.6 + RBH.2 – 12.4 – 20.4 + 10.1 - 5.2 = 0 -RBV.6 + RBH.2 = -32 ………………………………………………………………………………………..(1)
∑MB = 0 -RAV.6 + RAH.2 –5.8 – 10.5 – 20.2 – 12.2 = 0 -RAV.6 - RAH.2 = 154 ………………………………………………………………………………………..(2)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 19
∑Ms kanan = 0 -RBV.4 + RBH.4 + 12. 2 +20.2 = 0 -RBV.4 - RBH.4 = -64………………………………………………………………………………………..(3)
∑Ms kiri = 0 RAV.2 - RAH.6 – 5.4 – 10.1 = 0 RAV.2 - RAH.6 = 30…………………………………………………………………………………………..(4)
ELIMINASI PERSAMAAN (1) DAN (3) -RBV.6 + RBH.2 = -32
×2
-RBV.12 + RBH.4 = -64
-RBV.4 - RBH.4 = -64
×1
-RBV.4 - RBH.4 = -64 -16 RBV
= -128
RBV
= 8 kN
+
Substitusi nilai RBV ke persamaan 1 -RBV.6 + RBH.2
= -32
→ -6(8) + RBH.2 = -32 → RBH = 8 kN
Eliminasi Pers. 2 dan Pers 4 -RAV.6 - RAH.2 = -154
×1
-RAV.6 - RAH.2
= -154
RAV.2 - RAH.6 = 30
×3
RAV.6 - RAH.18
= 90
16 RAH RAH
-
= 154 = 4 kN
Substitusi nilai RAH ke Persamaan 2 -RAV.6 - RAH.2 = -154 → -6(RAV) – 2(4) = 154 → RAV = 27 kN Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 20
DIAGRAM BENDA BEBAS
R1 = 30 kN
P = 5 kN C
q1 = 5 kN/m
10
27
4
E 4
5 24
34
C 22
S
D
C q2 = 3 kN/m
Rbh = 8 B Rbv = 8 Rah = 4
A
Rav = 27
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 21
Tabel Gaya-Gaya Dalam:
Daerah Interval A–C 0 – 6
Momen (Mx) Mx = –RAH . x
Geser (Vx) Vx = – RAH
= –4x
Normal (Nx)
VA-C = –4 kN –27 kN
x = 0, MA = 0 x = 3, MC = – 24 kN m E–C
0 – 2
Mx = –P . x
Vx = – P
= –5x
VE-C = – 5 kN
x = 0, ME = 0
0
x = 2, MC = – 10 kN m C–D
0 – 6
Vx = 22 – 5x
Mx = 22x – 0,5qx2 – 34
x = 0, VB = 22 kN
= 22x – 2,5x2 – 34
x = 5, VD = – 8 kN
x = 0, MA = – 34 kNm x = 2, Ms = 0 x = 6, MD = 8 kN m
Jika Lintang sama dengan nol maka
–4 kN
Momen Mak.
Mmax = dMx/dx = 0 = 22 – 5x = 0
x = 4,4 m
Mmax = 22x – 2,5x2 – 34
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 22
= 14,4 kN B–D
0 – 4
Mx = RBH . x – 0,5qx2
Vx = – 8 + 3x
= 8x – 1,5x2
x = 0, VB = – 8 kN
x = 0, MB = 0 x = 4, VD = 4 kN x = 4, MD = 8 kN m Jika Lintang sama dengan nol maka
Mmax = dMx/dx = 0 = –8 + 3x = 0 22
x = 2,67 m
–8 kN
Momen Mak.
Mmax = 8x – 1,5x2 = 10,67 kN
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 23
+
D –+
E C
– 5
S
4 –
E 4
C
S
D
8 4,4
– 8
– 4
27
– B
8
A
– B
A
34 10 E
– 24
– C
8 S
+
D 8
– 4,4
14, 10,67 4 + 2,6 7 B
A
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 24
PELENGKUNG TIGA SENDI
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 25
SOAL 1
P=20 kN P = 10 kN 5m
S
C
D
Rbh 15 m
Rah
Rbv
10 m
25 m
Rav
10 m 10 m
20 m
25 m
100 m
Pertanyaan : Hitung GGD dititik C dan D SOLUSI : Y= 15 =
( )(
(
)
)(
)(
)
15L2 = 8000(L-100) 3L2 - 1600L + 160.000 =0 L1,2 =
(
) √(
)
( )(
)
L1 = 400 m (tidak memenuhi) L2 = 133,3 m Reaksi Perletakan Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 26
∑MA = 0 -Rbv(100)+Rbh(15)+20(56,665)+10(36,665) = 0 -100 Rbv+15Rbh
=-1499,95
…................1)
∑MA = 0 Rav(100)+Rah(15) -10(63,335)-20(43,335) = 0 100Rav + 15 Rah
= 1500,05 ………………….2)
∑Ms kiri = 0 Rav(66,665)+Rah (20)-10(30)-20(10) = 0 66,665Rav +20Rah = 500 ……………………3) Pers(2) dan (3) 100 Rav+15Rah = 1500,05
×4
400Rav+60Rah
= 6000,2
66,665Rav+20Rah = 500
×3
199,995Rav+60 Rah
= 1500
200,005Rav
= 4500,2
Rav = 22,5 kN (↑) 2250+15Rah
=1500,05
15 Rah
= -749,95
Rah =-49,997 kN (←) Rah = 49,997 kN(→) ∑V =0 → Rav+Rbv – 10-20 = 0 22,5+Rbv – 30 =0 Rbv ∑H =0 → Rah –Rbh 49, 997+Rbh
= 7,5 kN =0 =0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 27
Rbh
= 49,997 kN (←)
Gaya Gaya Dalam Y=
(
) (
)
(
)
(
Y=
)
(
Y=
)
Y = 0.5998x – 0.0045x2 X = 35 > Y = 15.48
X = 35 →
tan θ = 0,4423 → θ = 23.86 0 sin θ = 0,404 cos θ = 0,91 Vx = Rav – 10
= 22,5 – 10 = 12,5 kN (↓)
Hx = Rah
= 49,997 kN = 49,997 kN (←)
SFc
= V cos θ – H sin θ = (12,5)( 0,91) – (49,997)( 0,404) = 11.375 – 20.198 = - 8.823 kN
NFc
= V sin θ + H cos θ = (12,5)( 0,404)+( 49,997)( 0,91)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 28
= 50.547 kN
Mc
= Rav (35) – Rah(15.48) – 10(10) = 22.5(35) – 49.997 (15.48) - 100 = - 86.45 kNm
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 29
SOAL 2 Q1 = 3 kN/m
S C 5m
Q2 = 1,5 kN/m Rah
A
Rav Y =10 m Rbh
30 m
20 m
B
Rbv
HITUNG GGD DI TITIK C Q1 = 3 kN/m V = 33,15 kN H = 107,625 kN
C Q2 = 1,5 kN/m
86,125 kN 20 m
Kumpulan Soal Mekanika Rekayasa 1 dan 2
93,15 kN
Page 30
SOLUSI ∑MA = 0 → (3)(50)(25) - (1,5)(5)(2,5) + (1,5)(10)(5) + Rbh(10) + Rbv(50)
=0
3750 – 18,75 + 75 + 10 Rbh -50 Rbv
=0
10 Rbh -50 Rbv
= 3806,25
…………………….(1)
∑MB = 0 → (3)(50)(25) – (1,5)(15)(15/2 ) + Rah(10) + Rav(50)
=0
-3750 – 168,75 + 10 Rah + 50 Rav
=0 10 Rah + 50 Rav
= 3198,75
……..………………(2)
PERSAMAAN LENGKUNG PARABOLA, Y = 10, X = 50, h = 15 Y
=
10
=
( )(
(
)(
)
)(
)
10L2 = 3000 L – 150.000 L2 = 300L + 15000 = 0 L1,2 =
(
) √(
) ( )( )
( )(
)
L1 = 63,40 m L2 = 236,61 m (tidak memenuhi)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 31
∑Ms kanan = 0 (3) (31,7)(15,85) + (1,5)(15)(15/2) – Rbv(31,7) + Rbh(15)
=0
1507,335 + 168,75 – 31,7 Rbv + 15 Rbh
=0
31,7 Rbv – 15 Rbh
= 1676,085 ………………………………………….(3)
PERSAMAAN (1) DAN (3) 50Rbv – 10 Rbh = 3806,25
×3
150 Rbv – 30 Rbh
= 11418,75
31,7 Rbv – 15 Rbh = 1676,085
×2
63,4 Rbv – 30 Rbh
= 3352,17
86,6
Rbv
= 8066,58
Rbv
= 93,15 kN (↑)
50 Rbv – 10 Rbh
= 3806,25
50(93,15) – 10 Rbh
= 3806,25
4657,5 – 3806,25
= 10 Rbh
Rbh
= 85,125 kN
Untuk x = 20, h = 15 m, L = 63,4 m ( )(
)
Y
=
Y
=
Y
= 0,95x – 0,01 x2
(
)( )(
)
X = 20 →
tan θ = 0,55 → θ = 28,81 0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 32
sin θ = 0,48 cos θ = 0,88 Vx = Rbv – 3(x)
= 93,15 – 3(20) = 33,15 kN (↓)
Hx = Rbh + (1,5)(15)
= 85,125 + 22,5 = 107, 625 kN (→)
SFc
= V cos θ – H sin θ = (33,15)(0,88) – (107,625)(0,48) = 29,172 – 51,66 = -22,488 kN
NFc
= V sin θ + H cos θ = (33,15)(0,48)+(107,625)(0,88) = 110,622 kN
Mc
= Rbv (x) - Rbh(y) – 0,5(Q2)y2 – 0,5(Q1)x2 = 93,15(20) – 85,125(15) – 0,5 (1,5) (15)2 – 0,5 (3) (20)2 = 182,625 kNm
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 33
SOAL 3 P = 10 kN P = 5 kN D
S
B C
4m
Rbh 6m
A
Rah Rbv
Rav 5m
5m
8m 15 m 40 m
Pertanyaan : Hitung GGD titik C (8 meter dari A) dan titik D (15 meter dari A) SOLUSI : h = 10 m x = 40 m Y=6m Y
( )(
=
6 =
(
)(
)
)(
)
6 L2 – 1600L + 64000 = 0 L1,2 = L1,2 =
(
) √(
)
( )(
)
( ) √
)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 34
L1 = 217,6607 m (tidak mungkin) L2 = 49,0059 m Reaksi Perletakan
∑MB = 0 Rav(40) – Rah(6) – 5(35) – 10(30) = 0 40 Rav – 6Rah – 475 = 0
∑MS kiri = 0 Rav (L/2) – Rah.10 – 5(L/2-5) – 10(L/2 – 10) = 0 Rav (24,5030) – Rah.10 – 97,515 – 145,03 = 0 24,5030 Rav– 10.Rah – 97,515 – 242,545 = 0
……………………………(1)
……………………………(2)
ELIMINASI PERSAMAAN (1) DAN (2) : 40 Rav – 6Rah – 475 =0 24,5030 Rav– 10.Rah – 97,515 – 242,545 = 0
×10 ×6
400Rav – 60Rah 147,018 Rav – 60 Rah 252,982 Rav Rav
= 4750 = 1455,27 = 3294,73 = 13,0236 kN(↑)
Substitusi Rav ke persamaan (1) 40 Rav – 6Rah – 475 = 0 40(13,0236) – 6(Rah) – 475 = 0 Rah
= 7,6573 kN (→)
∑H = 0 Rah + Rbh = 0 Rah = - Rbh = -7,6573 kN (→)
∑MA = 0 -Rbv.40 + Rbh.6 + 10.10 + 5.5 = 0 -Rbv.40 +(-7,6573)6 + 125 =0 Rbv = 1,9764 kN (↑)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 35
∑V = 0 Rav + Rbv -10 - 5 = 0 13,0236 + 1,9764 – 15 = 0 …..ok!!
GAYA-GAYA DALAM Untuk h = 10 m , L = 49,0059 m Persamaan Parabola :
Y
( )(
=
Y =
(
)
) (
)
(
Y=
)
Untuk x = 8 m, maka y = Y= (
( )
( ) )
Y = 5.4693 m Titik c, untuk x = 8 m dari A, maka : = X=8 →
(
)
= 0,54790
tan θ = 28,7976 sin θ = 0,4817 cos θ = 0,8763 Titik C (8; 5,4693) Vx = Rav – 5 = 13, 0236 – 5 = 8,0236 kN (↓) Hx = Rah = 7,6573 kN (←) Gaya Lintang (SFx) SFx = Vcos θ – H sin θ = (8,0236)( 0,8763) – (7,6573)( 0,4817) = 3,3426 kN ≈ 3 kN Gaya Normal (NFx) Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 36
NFx = -(Vcos θ + H sin θ) = - ((8,0236)( 0,8763) + (7,6573)( 0,4817)) = - 8,4762 kN
Momen di titik C Mc = Rav 8 - Rah. 5,4639 – 5.3 = 13,0236. 8 – (7,6573)(5,4639) – 15 = 47,3501 kNm Titik D Untuk x =15 m, maka Y =
(
)
(
= X=15 m →
(
=
(
) )=
) (
(
))
m
tan θ = 0,3166 → θ = 17,56780 sin θ =0,3018, cos θ = 0,9534 Titik D (15; 8,4959) Vx = Rav – 5 – 10 = 13,0236 – 15 = -1,9764 kN (↓) Hx = Rah = 7,6573 kN (←) o Gaya Lintang (SFx) SFx = Vcos θ – H sin θ = (-1,9764)(0,9534) – (7,6573)( 0,3018) = -4,1953 kN ≈ -4 kN o Gaya Normal (NFx) NFx = -( Vcos θ + H sin θ) = -((-1,9764)( 0,9534)+(7,6573)(0,3018)) = -1,8843 +2,3110 = -0,4267 kN o Momen di Titik D MD = Rav 15 - Rah.(8,4959) – 5(10) – 10(5) MD = (13,0236)15 – (7,6573) .(8,4959) – 50 – 50 MD = 30.298 kNm
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 37
SOAL 4 q = 2 t/m’
SS D
2m A
Rah
q = 1 t/m’
C
4m Rav
B
Rbh
Rbv 2m
5m 20 m
Pertanyaan : Hitung GGD pada titik C (5 meter dari B) dan titik D (2 meter dari A) Penyelesaian : Untuk x = 20 m, y = 4 m, h = 6 m ( )(
Y= 4=
( )(
)
)(
)
4L2 – 480 L + 9600 = 0 L1,2 = L1,2 =
(
) √(
√
) ( )
( )(
)
)
L1 = 94,6410 m (tidak mungkin) L2 = 25,3590 m Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 38
Reaksi Perletakan ∑MA = 0 (2)(20)(10) + (1) (4)(2) – (1)(2)(1) - Rbv.20 + Rbh.4 = 0 400 + 8 – 2 - Rbv.20 + Rbh. = 0 - Rbv.20 + Rbh.4 + 406
=0
Rbv.20 - Rbh.4 – 406
= 0 ……………………….……………………………(1)
∑MB = 0 -(2)(20)(10) – (1)(6)(3) + Rav.20 + Rah.4 = 0 -400 – 18 + Rav.20 + Rah.4 = 0 Rav.20 + Rah.4 - 418 = 0……………………………………………………………..…(2) ∑Ms kanan = 0 -Rbv. 12,6795 + Rbh.6 +(2)(12,6795)(6,3398) + (1)(6)(3) = 0 -12,6795 Rbv +6 Rbh + 178,7710 = 0………………………………………..……(3) ∑Ms kiri = 0 -Rav.7,3025 – Rah.2 –(2)(7,3205)(3,6603) = 0 7,3205 Rav – 2Rah – 53,5905 = 0…………………………………………………..(4) ELIMINASI PERSAMAAN (1) DAN (3) 20 Rbv – 4Rbh = 406 -12,6795 Rbv + 6 Rbh = -178,7710
×6 ×4
120 Rbv - 24Rbh = 243,6 -50,718 Rbv + 24Rbh = -715,084 + 69,282 Rbv = 1720,916 Rbv = 24,8393 t (↑)
Substitusi Rbv = 24,8393 t ke persamaan (1) 20 Rbv – 4 Rbh = 406 20(24,8393) – 4Rbh = 406 Rbh = 22,6965 t (←)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 39
Eliminasi (2) dan (4) : Rav.20 + Rah.4 - 418 = 0 7,3205 Rav – 2Rah – 53,5905 = 0
×1 ×2
Rav.20 + Rah.4 14,641 Rav – 4 Rah
= 418 = 107,181 +
34,641 Rav
= 525,81
Rav
= 15,1607 t (↑)
Substitusi Rav = 15,1607 t ke (2) 2Rav + 4 Rah
= 418
20(15,1607) + 4 Rah
= 418
Rah
= 28,6965 t (→)
∑V=0 Rav + Rbv – 40 = 0 15,1607 + 24,8393 – 40 = 0……………….ok!
∑H=0 Rah – Rbh – 6 = 0 28,6965 – 22,6965 – 6 = 0………………….ok!
MENENTUKAN GAYA-GAYA DALAM Untuk h = 6 m, L = 25,3590 m → Y =
( )(
)
( )( )(
)
( (
Untuk titik c, x = 5 m dari B → y { } X=5m→{ }
(
( )
) ( ) )
) (
( )) = 0,5372
tan θ = 0,53720 → θ = 29,8213o sin θ = 0,4937 cos θ = 0,8676 Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 40
Titik C (5; 3,7990) dari B Vx = Rbv – qx =24,8393 – (2)(5) = 14, 8393 t (↓) Hx = Rbh +q.y = 22,6965 +(1)(3,7990) = 26,4955 t (→) Gaya Lintang (SFx) SFx = V cos θ – H sinθ = (14,8393)(0,4973) + (26,4955)(0,8676) = 30,3671 t Mc
= Rbv. 5 – Rbh. 3,7990 +(2)(5)(2,5) = 24,8393 (5) – 22,6965(3,7990) +25 = 62,9725 tn
Untuk titik D, x = 2m dari A, h = 2m , L = 25,3590 m, x = 2 m Y= =
( )(
( )( )(
)
(
)
)
Untuk x = 2m , y =
(
{ } {
}
( ) (
( ) ) = 0,5812 m )
(
( ))
tan θ = 0,26570 → θ = 14,8797o sin θ = 0,2567 cos θ = 0,9665
TITIK D (2 ; 0,5812) Vx = Rav – q.x = 15,1607 – 2(2) = 11,1607 t (↓) Hx = Rah = 28,6965 t (←)
GAYA LINTANG (SFx) SFx = Vcos θ – H sin θ = (11,1607(0,9665))-(28,6965(0,2567)) = 3,4204 t ≈ 3t
Gaya Normal (NFx) NFx= -(Vsin θ + Hcos θ) = -((11,1607(0,2567)) +((28,6965)(0,9665)) = -30,6001 t MD = RAv.2 – Rah. 0,5812 – (2)(2)(1) = 15,1607(2) – 28,6965(0,5812) – 4 = 19,1128 tm UNTUK TITIK A (0,0) DARI A
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 41
X=0→y=0{ }
(
→
)
Tan θ = 0,3155 → θ = 17,5105 sin θ = 0,3009 cos θ = 0,9537
TITIK A (0,0) Vx = Rav = 15,1607 t (↓) Hx = Rah = 28,6965 t (←)
Gaya Lintang (SFx) SFx = Vcos θ - Hsin θ (15,1607)(0,9537) – (28,6965)(0,3009) = 5,8240 t ≈ 6 t
Gaya Normal (NFx) NFx = -(Vsin θ + H cos θ) = -((15,1607) (0,3009) + (28,6965)(0,9537) = -31,9297 t M dititik A → Ma = 0
Untuk titik S (12,6795;6) dari B { }
=
(
( )(
))
Tan θ = 0 → sin θ = 0, cos θ = 1 S = (12,6795;6) Vx = Rbv – q.x = 24,8393 –(2)(12,6795) = -0,5197 t (↓) Hx = Rbh + q.x = 22,q.y = 22.,6965 +(1)(6) = 28,6965 t (→)
Gaya Lintang (SFx) SFx = Vcos θ – Hsin θ = (-0,5197.1) – (28,6965.0) =-0,5197 t Gaya Normal (NFx) NFx = Vsin θ + Hcos θ = (-0,5197.0) – (28,6965.1) = 28,6965 t ntuk titik B (0,0) dari B ( )( )) = { } ( Tan θ = 0,9464 → θ = 43,4226 Sin θ = 0,6874 Cos θ = 0,7263 Titik B (0,0) Vx = Rbv = 24,8393 t (↓) HX = Rbh = 22,6965 t (→) Gaya Lintang (SFx) SFx = Vcos θ – Hsin θ = (24,8393)(0,7263) – (22,6965)(0,6874) =2,4392 t Gaya Normal (NFx)
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 42
NFx = Vsin θ + Hcos θ = (-0,5197)(0,6874) – (28,6965)(0,7263) = 28,6965 t
SOAL 5
Pertanyaan : Tentukan Reaksi perletakan soal berikut ini : q = 2 kN/m P1 = 5kN S C
10 m
P2 = 4 kN
B
A 20 m
10 m
Persamaan dasar parabola yang digunakan adalah :
( )(
10 m
)
Dimana : Y = tinggi titik yang ditinjau dari tumpuan H = tinggi puncak parabola dari tumpuan X = jarak mendatar dari tumpuan terdekat L = jarak mendatar dua tumpuan
SOLUSI : ( )(
Y:
)
=
(
)(
)(
)
=
(
)
∑MA = 0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 43
(-Rbv.40)-(4.7,5) + (5.30) + (q.20.10) -40 Rbv – 30 +150 + 400 -40 Rbv Rbv
=0 =0 = -520 = 13 kNm
∑MB = 0 (-Rav.40) – (q.20.30) – (P1.10) – (P2.7,5) (-Rav.40) – 1200 – 50 – 30 40 Rav Rav
=0 =0 = 1280 = 32 kN
∑Ms kanan = 0 (-Rbv.20) + (Rbh.10) + (4.2,5) + (5.10) (-20.13) + (10 Rbh) + 10 + 50 -260 + 10 Rbh + 60 Rbh
=0 =0 =0 = 20 kN
∑Ms kiri = 0 (-Rav.20) - (Rah.10) - (q.20.10) (-20.32) - (10 Rah) – 400 -10 Rah Rah
=0 =0 =0 = 24 kN
∑V = 0 Rav + Rbv = 40 +5 32 + 13 = 45 ………..ok!
∑H = 0 Rah - Rbh = 4 24 – 20 = 4 …………..ok!
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 44
RANGKA BATANG
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 45
SOAL 1 5 kN
5 kN F9
6
5 F7
4m
F8
F6 F5
3
1 kN
4 1 kN
F3
F2
F4
4m
2
1
1 kN
F1
Rah Rav
Rbh Rbv
3m
Pertanyaan : Hitung gaya gaya batang yang terjadi pada struktur rangka batang tegrambar di bawah ini menggunakan metode kesetimbanan titik :
m = 2j – r
M= jumlah batang
9 = 2(6) – 3
J = jumlah titik buhul
9 = 9….(statis tentu)
R = jumlah reaksi perletakan
1. REAKSI PERLETAKAN
∑MA = 0 (-Rbv.3) + (5.3) + (-1.8) +(-1.4)
=0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 46
-3Rbv – 15 – 8 – 4 Rbv
∑MB = 0 (-Rav.3) + (-5.3) + (-1.8) +(-1.4) Rav
∑V = 0 Rav + Rbv = 0 5 + 5 = 0…………………………...ok!
∑H = 0 Rah = 1+1+1 Rah = 3…………………………...ok!
=0 =1
=0 =9
II. GAYA DI TITIK BUHUL BUHUL 1 F2 4m
α
Rah
F1
3m
∑V = 0
∑H = 0
Rav + F2 = 0
Rah + F1 = 0
9 + F2 = 0
3 + F1 = 0
F2 = -9 kN
F1 = -3 kN
Rav
BUHUL 2 F3
F3y
F4
F3x 3 m
F3x = F3 sin α = F3 F3y
5m
F3x
4m
F3y = F3 cos α = F3
α
F1
∑V=0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 47
Rbv + F4 + F3y = 0 1 + F4 + 3,33 = 0 F4 + 3,66 = 0
→ F4 = - 3,66
∑H=0 F1 + F3x + 1 = 0 -3 + F3x + 1 = 0 F3 = 3,333
BUHUL 3
F6 F3x = F3 cos α = F3
F5 F3x
F2
F3y = F3 sin α = F3
F3
F3y ∑V=0
F2 + F3y – F6 = 0 -9 + 3,33 = 0 → F6 = - 6,33
∑H=0 F3x + F5 = 0 → F5 = - 1,99
BUHUL 5 5 kN F9 F7x F6
F7y o
F7x = F7 cos α = F7 F7y = F7 sin α = F7
F7
∑V=0
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 48
F6 + F7y + 5 = 0 -6,33 + F7 . + 5= 0 → F7 = 1,66 o
∑H=0 F7x + F9 = 0 → F9 = - 0,99
BUHUL 6 5 kN 1 kN
F9
F8 o
∑V=0 F8 + 5 = 0 F8 = -5
BUHUL 4 (kontrol)
F8 F7
F7y
F7x = F7 cos α = 1,66 F7y = F7 sin α = F7 o
F7x
=0,99
= 1,33
∑V=0
F6 + F7y - F4 = 0
F5
-5 +1,33 – (-3,36) = 0 → ok o ∑H=0 F4
F5 + 1 + F7x = 0 → ok
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 49
SOAL 2
Kumpulan Soal Mekanika Rekayasa 1 dan 2
Page 50
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