contoho soal mekanika rekayasa IV( soal no 3)
contoho soal mekanika rekayasa IV( soal no 3)
Soal Nomor 3
Diketahui : struktur statis tak tentu seperti tergambar.
Ditanya : selesaikan struktur di atas menggunakan Metode Cross
Catatan : XYZ adalah 3 digit NPM Terakhir Mahasiswa
(Misal: G1B013XYZ) à X, Y, dan Z = 0, 0, dan 6
Penyelesaian
Anggapan Bahwa B dan C “Pinned”
Ø Menentukan nilai stiffness factor
|
Stifness Faktor |
||
|
AD |
- |
|
|
DA |
4(4EI)/3 |
5.3333 |
|
DB |
3(2EI)/8 |
0.7500 |
|
DE |
4(5EI)/9 |
2.2222 |
|
BD |
- |
|
|
ED |
4(5EI)/9 |
2.2222 |
|
EC |
3(EI)/7.5 |
0.4000 |
|
CE |
- |
|
Ø Menentukan nilai distribusi Faktor
|
Distribusi Faktor |
||
|
Joint A |
- |
|
|
Joint D |
0.6421 |
1.0000 |
|
|
0.0903 |
|
|
|
0.2676 |
|
|
Joint B |
|
|
|
Joint E |
0.8475 |
1.0000 |
|
|
0.1525 |
|
|
Joint C |
|
|
Ø Menentukan Nilai Fixed and Momen
|
FIX and MOMEN |
|
|
AD |
1.5450 |
|
DA |
-1.5450 |
|
DB |
9.0234 |
|
DE |
13.7391 |
|
BD |
0.0000 |
|
ED |
-24.0159 |
|
EC |
2.8125 |
|
CE |
0.0000 |
Ø Tabel Metode Cross
|
Joint |
A |
D |
B |
E |
C |
|||
|
Batang |
AD |
DA |
DB |
DE |
BD |
ED |
EC |
CE |
|
DF |
- |
0.6421 |
0.0903 |
0.2676 |
- |
0.8475 |
0.1525 |
- |
|
FEM |
1.5450 |
-1.5450 |
9.0234 |
13.7391 |
-24.0159 |
2.8125 |
||
|
BALANCE |
-13.6246 |
-1.9160 |
-5.6769 |
17.9690 |
3.2344 |
|||
|
CO |
-6.8123 |
8.9845 |
-2.8385 |
|||||
|
BALANCE |
-5.7693 |
-0.8113 |
-2.4039 |
2.4055 |
0.4330 |
|||
|
CO |
-2.8847 |
1.2027 |
-1.2019 |
|||||
|
BALANCE |
-0.7723 |
-0.1086 |
-0.3218 |
1.0186 |
0.1833 |
|||
|
CO |
-0.3862 |
0.5093 |
-0.1609 |
|||||
|
BALANCE |
-0.3270 |
-0.0460 |
-0.1363 |
0.1364 |
0.0245 |
|||
|
CO |
-0.1635 |
0.0682 |
-0.0681 |
|||||
|
BALANCE |
-0.0438 |
-0.0062 |
-0.0182 |
0.0577 |
0.0104 |
|||
|
CO |
-0.0219 |
0.0289 |
-0.0091 |
|||||
|
BALANCE |
-0.0185 |
-0.0026 |
-0.0077 |
0.0077 |
0.0014 |
|||
|
CO |
-0.0093 |
0.0039 |
-0.0039 |
|||||
|
BALANCE |
-0.0025 |
-0.0003 |
-0.0010 |
0.0033 |
0.0006 |
|||
|
CO |
-0.0012 |
0.0016 |
-0.0005 |
|||||
|
BALANCE |
-0.0011 |
-0.0001 |
-0.0004 |
0.0004 |
0.0001 |
|||
|
CO |
-0.0005 |
0.0002 |
-0.0002 |
|||||
|
BALANCE |
-0.0001 |
0.0000 |
-0.0001 |
0.0002 |
0.0000 |
|||
|
CO |
-0.0001 |
|||||||
|
TEM |
-8.7346 |
-22.1043 |
6.1323 |
15.9720 |
0.0000 |
-6.7003 |
6.7003 |
0.0000 |
Ø TEM hasil perhitungan di plotkan ke struktur dan selesaikan dengan cara mekanika rekayasa 1
Ø Menghitung shear force, Momen, dan Normal Force
· Bentang A – D
· Shear force
∑FV = 0
-RAV – 2.06X – SFx = 0
-1.3665 – 2.06X – SFx = 0
SFx = (-1.3665 – 2.06X) t
|
X (m) |
SF (t) |
|
0 |
-1.3665 |
|
0.5 |
-2.3965 |
|
1 |
-3.4265 |
|
1.5 |
-4.4565 |
|
2 |
-5.4865 |
|
2.5 |
-6.5165 |
|
3 |
-7.5465 |
· Bending Momen
∑Mx = 0
8.7346 – RAV (X) – 2.06X (1/2 X) – Mx = 0
8.7346 – 1.3665X – 1.03X2 – Mx = 0
Mx = (8.7346 – 1.3665X – 1.03X2) tm
|
X (m) |
Mx (tm) |
|
0 |
8.7346 |
|
0.5 |
7.79385 |
|
1 |
6.3381 |
|
1.5 |
4.36735 |
|
2 |
1.8816 |
|
2.5 |
-1.11915 |
|
3 |
-4.6349 |
· Normal Force
∑FH = 0
5.2849 + NFx = 0
NFx = -5.2849 t
|
X (m) |
NF (t) |
|
0 |
-5.2849 |
|
0.5 |
-5.2849 |
|
1 |
-5.2849 |
|
1.5 |
-5.2849 |
|
2 |
-5.2849 |
|
2.5 |
-5.2849 |
|
3 |
-5.2849 |
· Bentang D – B (potongan 1)
· Shear Force
∑FV = 0
RDV – SFx = 0
SFx = RDV
SFx = 3.3915 t
|
X (m) |
SF (t) |
|
0 |
3.3915 |
|
0.5 |
3.3915 |
|
1 |
3.3915 |
|
1.5 |
3.3915 |
|
2 |
3.3915 |
|
2.5 |
3.3915 |
|
3 |
3.3915 |
|
3.5 |
3.3915 |
|
4 |
3.3915 |
|
4.5 |
3.3915 |
|
5 |
3.3915 |
· Bending Momen
∑Mx = 0
-6.1323 + RDV (X) – Mx = 0
-6.1323 + 3.3915X – Mx = 0
Mx = (-6.1323 + 3.3915X) tm
|
X (m) |
Mx (tm) |
|
0 |
-6.1323 |
|
0.5 |
-4.43655 |
|
1 |
-2.7408 |
|
1.5 |
-1.04505 |
|
2 |
0.6507 |
|
2.5 |
2.34645 |
|
3 |
4.0422 |
|
3.5 |
5.73795 |
|
4 |
7.4337 |
|
4.5 |
9.12945 |
|
5 |
10.8252 |
· Normal Force
∑FH = 0
15.4442 + NFx = 0
NFx = -15.4442 t
|
X (m) |
NF (t) |
|
0 |
-15.4442 |
|
0.5 |
-15.4442 |
|
1 |
-15.4442 |
|
1.5 |
-15.4442 |
|
2 |
-15.4442 |
|
2.5 |
-15.4442 |
|
3 |
-15.4442 |
|
3.5 |
-15.4442 |
|
4 |
-15.4442 |
|
4.5 |
-15.4442 |
|
5 |
-15.4442 |
· Bentang D – B (potongan 2)
· Shear Force
∑FV = 0
RDV – P –SFx = 0
3.3915 – 7 – SFx = 0
SFx = -3.6085 t
|
X (m) |
SF (t) |
|
5 |
-3.6085 |
|
5.5 |
-3.6085 |
|
6 |
-3.6085 |
|
6.5 |
-3.6085 |
|
7 |
-3.6085 |
|
7.5 |
-3.6085 |
|
8 |
-3.6085 |
· Bending Momen
∑Mx = 0
-6.1323 + RDV (X) – P (X-5) – Mx = 0
-6.1323 + 3.3915X – 7 (X-5) – Mx = 0
Mx = (28.8677 – 3.6085X) tm
|
X (m) |
Mx (tm) |
|
5 |
10.8252 |
|
5.5 |
9.02095 |
|
6 |
7.2167 |
|
6.5 |
5.41245 |
|
7 |
3.6082 |
|
7.5 |
1.80395 |
|
8 |
-0.0003 |
· Normal Force
∑FH = 0
15.4442 + NFx = 0
NFx = -15.4442 t
|
X (m) |
NF (t) |
|
5 |
-15.4442 |
|
5.5 |
-15.4442 |
|
6 |
-15.4442 |
|
6.5 |
-15.4442 |
|
7 |
-15.4442 |
|
7.5 |
-15.4442 |
|
8 |
-15.4442 |
· Bentang D – E (Potongan 1)
· Shear force
∑FV = 0
RDV – SFx = 0
SFx = RDV
SFx = 7.8977 t
|
X (m) |
SF (t) |
|
0 |
7.8977 |
|
0.5 |
7.8977 |
|
1 |
7.8977 |
|
1.5 |
7.8977 |
|
2 |
7.8977 |
|
2.5 |
7.8977 |
|
3 |
7.8977 |
|
3.5 |
7.8977 |
|
4 |
7.8977 |
|
4.5 |
7.8977 |
· Bending Momen
∑Mx = 0
-15.9720 + RDV (X) – Mx = 0
-15.9720 + 7.8977X – Mx = 0
Mx = (-15.9720 + 7.8977X) tm
|
X (m) |
Mx (tm) |
|
0 |
-15.972 |
|
0.5 |
-12.0232 |
|
1 |
-8.0743 |
|
1.5 |
-4.12545 |
|
2 |
-0.1766 |
|
2.5 |
3.77225 |
|
3 |
7.7211 |
|
3.5 |
11.66995 |
|
4 |
15.6188 |
|
4.5 |
19.56765 |
· Normal Force
∑FH = 0
15.9720 + NFx = 0
NFx = -15.9720 t
|
X (m) |
NF (t) |
|
0 |
-1.8934 |
|
0.5 |
-1.8934 |
|
1 |
-1.8934 |
|
1.5 |
-1.8934 |
|
2 |
-1.8934 |
|
2.5 |
-1.8934 |
|
3 |
-1.8934 |
|
3.5 |
-1.8934 |
|
4 |
-1.8934 |
|
4.5 |
-1.8934 |
· Bentang D – E (Potongan 2)
· Shear force
∑FV = 0
RDV - P – (4.06X – 18.27) –SFx = 0
7.8977 – 4.6 -4.06X + 18.27 – SFx = 0
SFx = (21.5677 – 4.06X) t
|
X (m) |
SF (t) |
|
4.5 |
3.2977 |
|
5 |
1.2677 |
|
5.5 |
-0.7623 |
|
6 |
-2.7923 |
|
6.5 |
-4.8223 |
|
7 |
-6.8523 |
|
7.5 |
-8.8823 |
|
8 |
-10.9123 |
|
8.5 |
-12.9423 |
|
9 |
-14.9723 |
· Bending Momen
∑Mx = 0
-15.9720 + RDV (X) – P (X-4.5) – (4.06X – 18.27) (1/2(X-4.5)) – Mx = 0
-15.9720 + 7.8977X – 4.6 (X-4.5) – (4.06X – 18.27) (1/2(X-4.5)) – Mx = 0
-15.9720 + 7.8977X – 4.6X + 20.7 – 2.03 X2 + 18.27X – 41.1075 – Mx = 0
Mx = (-36.3795 + 21.5677X – 2.03 X2) tm
X (m)
|
|
Mx (tm) |
|
4.5 |
19.56765 |
|
5 |
20.709 |
|
5.5 |
20.83535 |
|
6 |
19.9467 |
|
6.5 |
18.04305 |
|
7 |
15.1244 |
|
7.5 |
11.19075 |
|
8 |
6.2421 |
|
8.5 |
0.27845 |
|
9 |
-6.7002 |
· Menentukan Momen Maksimum
Mx = (- 36.3795 + 21.5677X - 2.03X2) tm
=
0 = 21.5677 – 4.06X
X = 21.5677/4.06
X = 5.3122 m
Mmax = (- 36.3795 + 21.5677X - 2.03X2)) tm
= (- 36.3795 + 21.5677(5.3122) - 2.03(5.3122)2)tm
= 20.9069 tm
· Normal Force
∑FH = 0
15.9720 + NFx = 0
NFx = -15.9720 t
|
X (m) |
NF (t) |
|
4.5 |
-1.8934 |
|
5 |
-1.8934 |
|
5.5 |
-1.8934 |
|
6 |
-1.8934 |
|
6.5 |
-1.8934 |
|
7 |
-1.8934 |
|
7.5 |
-1.8934 |
|
8 |
-1.8934 |
|
8.5 |
-1.8934 |
|
9 |
-1.8934 |
· Bentang E – C (potongan 1)
· Shear force
∑FV = 0
REV – SFx = 0
SFx = REV
SFx = 1.8934 t
|
X (m) |
SF (t) |
|
0 |
1.8934 |
|
0.5 |
1.8934 |
|
1 |
1.8934 |
|
1.5 |
1.8934 |
|
2 |
1.8934 |
|
2.5 |
1.8934 |
|
3 |
1.8934 |
|
3.5 |
1.8934 |
|
3.75 |
1.8934 |
· Bending Momen
∑Mx = 0
-6.7003 + REV (X) – Mx = 0
-6.7003 + 1.8934X – Mx = 0
Mx = (-6.7003 + 1.8934X) tm
|
X (m) |
Mx (tm) |
|
0 |
-6.7003 |
|
0.5 |
-5.7536 |
|
1 |
-4.8069 |
|
1.5 |
-3.8602 |
|
2 |
-2.9135 |
|
2.5 |
-1.9668 |
|
3 |
-1.0201 |
|
3.5 |
-0.0734 |
|
3.75 |
0.39995 |
· Normal Force
∑FH = 0
14.9723 + NFx = 0
NFx = -14.9723 t
|
X (m) |
NF (t) |
|
0 |
-14.9723 |
|
0.5 |
-14.9723 |
|
1 |
-14.9723 |
|
1.5 |
-14.9723 |
|
2 |
-14.9723 |
|
2.5 |
-14.9723 |
|
3 |
-14.9723 |
|
3.5 |
-14.9723 |
|
3.75 |
-14.9723 |
· Bentang E – C (potongan 2)
· Shear force
∑FV = 0
REV – P – SFX = 0
1.8934 – 2 – SFx = 0
SFx = 1.8934 – 2
SFx = -0.1066 t
|
X (m) |
SF (t) |
|
3.75 |
-0.1066 |
|
4 |
-0.1066 |
|
4.5 |
-0.1066 |
|
5 |
-0.1066 |
|
5.5 |
-0.1066 |
|
6 |
-0.1066 |
|
6.5 |
-0.1066 |
|
7 |
-0.1066 |
|
7.5 |
-0.1066 |
· Bending Momen
∑Mx = 0
-6.7003 + REV (X) – P (X-3.75) – Mx = 0
-6.7003 + 1.8934X – 2 (X-3.75) – Mx = 0
-6.7003 + 1.8934X – 2X + 7.5 – Mx = 0
Mx = (0.7997 – 0.1066X) tm
|
X (m) |
Mx (tm) |
|
3.75 |
0.39995 |
|
4 |
0.3733 |
|
4.5 |
0.32 |
|
5 |
0.2667 |
|
5.5 |
0.2134 |
|
6 |
0.1601 |
|
6.5 |
0.1068 |
|
7 |
0.0535 |
|
7.5 |
0.0002 |
· Normal Force
∑FH = 0
14.9723 + NFx = 0
NFx = -14.9723 t
|
X (m) |
NF (t) |
|
3.75 |
-14.9723 |
|
4 |
-14.9723 |
|
4.5 |
-14.9723 |
|
5 |
-14.9723 |
|
5.5 |
-14.9723 |
|
6 |
-14.9723 |
|
6.5 |
-14.9723 |
|
7 |
-14.9723 |
|
7.5 |
-14.9723 |
Ø Hasil Analisis