a) struktur balok dan tiang b) portal kaku c) portal biasa ...€¦ · statika dan mekanika bahan 1...
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Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 111
BAB VI
Portal
Struktur portal yaitu struktur balok yang mempunyai kaki-kaki (Wesli,
2012). Pada portal yang terdiri dari balok dan tiang yang dibebani muatan
diatasnya akan timbul lenturan pada balok saja, dan akan meneruskan gaya-gaya
tersebut ke tiang berupa gaya normal. Balok yang demikian disebut balok
sederhana. Adapun gaya yang bekerja pada tiang pada umumnya berupa gaya
horisontal, tidak berpengaruh pada balok (Mulyati). Struktur portal dengan
kondisi seperti ini dapat diselesaikan dengan prinsip struktur statis tertentu dalam
artian hukum kesetimbangan (ΣM = 0, ΣV = 0, dan ΣH = 0) dapat digunakan.
Bentuk portal yang umumnya dipelajari berupa portal segi empat, segi banyak
atau lengkungan, yang dapat dilihat pada Gambar 6.1 berikut. Pada bab ini hanya
akan dibahas struktur portal segi empat.
Gambar 6.1 Struktur Portal (Mulyati)
a) Struktur balok dan tiang b) Portal kaku
c) Portal biasa d) Portal segi banyak
c) Portal lengkung
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 112
Berikut ini beberapa contoh soal dan penyelesaian untuk struktur portal.
Contoh Soal 1
Penyelesaian :
1. Reaksi Perletakan
ΣMB = 0
VA . 4 + P . L – q . L (½ . L) = 0
VA . 4 + 5 . 2,5 – 2 . 4 (½ . 4) = 0
VA = 0,875 ton ( )
ΣMA = 0
-VB . 4 + P . L + q . L (½. L) = 0
-VB . 4 + 5 . 2,5 + 2 . 4 (½. 4) = 0
VB = 7,125 ton ( )
ΣV = 0
VA + VB = q . L
0,875 + 7,125 = 2 . 4
8,0 ton = 8,0 ton (Oke!!!)
P = 5 ton1 m
2,5 m
4 m
q = 2 t/m
A B
D
C
E
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 113
ΣH = 0
HA + P = 0
HA = - P = - 5,0 ton ( )
= 5,0 ton ( )
2. Perhitungan Gaya-gaya Dalam
a) Bentang A – C ( 0 ≤ x ≤ 2,5 meter )
ΣMx = 0
HA . x – Mx = 0
Mx = 5x
x = 0 ; MA = 0 tm
x = 2,5 ; MC = 12,5 tm
ΣV = 0
- VA – Nx = 0
Nx = - 0,875
x = 0 ; NA = - 0,875 ton
x = 2,5 ; NC = - 0,875 ton
X
HA
VA
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 114
ΣH = 0
HA – Lx = 0
Lx = 5
x = 0 ; LA = 5,0 ton
x = 2,5 ; LC = 5,0 ton
b) Bentang C – D ( 0 ≤ x ≤ 1,0 meter )
ΣMx = 0
HA . (2,5 + x) – P . x – Mx = 0
Mx = 5 (2,5 + x) – 5x
= 12,5 + 5x – 5x
x = 0 ; MC = 12,5 tm
x = 1 ; MD = 12,5 tm
ΣV = 0
- VA – Nx = 0
Nx = - 0,875
x = 0 ; NC = - 0,875 ton
x = 1 ; ND = - 0,875 ton
2,5 m
HA
VA
P = 5 tonX
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 115
ΣH = 0
HA – P – Lx = 0
Lx = 5 – 5
= 0
x = 0 ; LC = 0 ton
x = 1 ; LD = 0 ton
c) Bentang D – E ( 0 ≤ x ≤ 4,0 meter )
ΣMx = 0
VA . x + HA . 3,5 – P . 1 – q . x (½ . x) – Mx = 0
Mx = 0,875 . x + 5 . 3,5 – 5 . 1 – 2 . x (½ . x)
= 12,5 + 0,875x – x2
x = 0 ; MD = 12,5 tm
x = 4 ; ME = 0 tm
ΣV = 0
VA – q . x – Lx = 0
Lx = 0,875 – 2x
x = 0 ; LD = 0,875 ton
x = 4 ; LE = - 7,125 ton
2,5 m
HA
VA
P = 5 ton
q = 2 t/m
Lx
Nx
Mx
X
1 m
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 116
ΣH = 0
- HA + P – Nx = 0
Nx = 5 – 5
= 0
x = 0 ; ND = 0 ton
x = 4 ; NE = 0 ton
terjadi Mmaks pada jarak Lx = 0
0,875 – 2x = 0 x = 0,438 meter
M maks = 12,5 + 0,875x – x2
= 12,5 + 0,875 (0,438) – (0,438)2
= 12,691 tm
d) Bentang E – B ( 0 ≤ x ≤ 3,5 meter )
ΣMx = 0
Mx = 0
x = 0 ; ME = 0 tm
x = 3,5 ; MB = 0 tm
X
VB
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 117
ΣV = 0
- VB – Nx = 0
Nx = - 7,125
x = 0 ; NE = - 7,125 ton
x = 3,5 ; NB = - 7,125 ton
ΣH = 0
Lx = 0
x = 0 ; LE = 0 ton
x = 3,5 ; LB = 0 ton
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 118
3. Gambar bidang Momen, Lintang dan Normal
P = 5 ton1 m
2,5 m
4 m
q = 2 t/m
A B
HA
VA
VB
0 tm 0 tm12,5 tm
12,5 tm
0 tm 0 tm
+
+
+
0 t 0 t
5 t
7,125 t
0,875 t
+
-
+
0 t 0 t
0 t 0 t
7,125 t0,875 t
0 t 0 t
- -
MOMEN
LINTANG
NORMAL
12,5 tm12,691 tm
D
C
E
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 119
Contoh Soal 2
Penyelesaian :
1. Reaksi Perletakan
ΣMB = 0
VA . 4 + q . L (½ . L) – P . L = 0
VA . 4 + 2 . 4 (½ . 4) – 2 . 2 = 0
VA = - 3,0 ton ( )
= 3,0 ton ( )
ΣMA = 0
-VB . 4 + P . L + q . L (½. L) = 0
-VB . 4 + 2 . 2 + 2 . 4 (½. 4) = 0
VB = 5,0 ton ( )
ΣV = 0
VA + VB = P
- 3,0 + 5,0 = 2
2,0 ton = 2,0 ton (Oke!!!)
4 m
2 m2 m
P = 2 ton
q = 2 t/m
A B
C ED
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 120
ΣH = 0
q . L + HA = 0
HA = - 2 . 4
= - 8,0 ton ( )
= 8,0 ton ( )
2. Perhitungan Gaya-gaya Dalam
a) Bentang A – C ( 0 ≤ x ≤ 4,0 meter )
ΣMx = 0
HA . x – q. x (½ x) – Mx = 0
Mx = 8 . x – 2 . x (½ x)
= 8x – x2
x = 0 ; MA = 0 tm
x = 4 ; MC = 16,0 tm
ΣV = 0
VA – Nx = 0
Nx = 3
x = 0 ; NA = 3,0 ton
x = 4 ; NC = 3,0 ton
X q = 2 t/m
LX
NX
MX
VA
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 121
ΣH = 0
HA – q . x – Lx = 0
Lx = 8 – 2x
x = 0 ; LA = 8,0 ton
x = 4 ; LC = 0 ton
b) Bentang C – D ( 0 ≤ x ≤ 1,0 meter )
ΣMx = 0
-VA . x + HA . 4 – q . 4 (½ . 4) – Mx = 0
Mx = - 3 . x + 8 . 4 – 2 . 4 (½ . 4)
= - 3x + 16
x = 0 ; MC = 16,0 tm
x = 2 ; MD = 10,0 tm
ΣV = 0
- VA – Lx = 0
Lx = - 3
x = 0 ; LC = - 3,0 ton
x = 2 ; LD = - 3,0 ton
4 m q = 2 t/m
C
VA
Lx
Nx
Mx
X
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 122
ΣH = 0
- HA + q . 4 – Nx = 0
Nx = - 8 + 2. 4
= 0
x = 0 ; NC = 0 ton
x = 2 ; ND = 0 ton
c) Bentang D – E ( 0 ≤ x ≤ 2,0 meter )
ΣMx = 0
- VA . (2 + x) + HA . 4 – P . x – q . 4 (½ . 4) – Mx = 0
Mx = - 3 (2 + x) + 8 . 4 – 2 . x – 2 . 4 (½ . 4)
= - 5x + 10
x = 0 ; MD = 10,0 tm
x = 2 ; ME = 0 tm
ΣV = 0
- VA – P – Lx = 0
Lx = - 3 – 2
= - 5
x = 0 ; LD = - 5,0 ton
x = 2 ; LE = - 5,0 ton
4 m
P = 2 ton
q = 2 t/m
C D
VA
Lx
Nx
Mx
X2 m
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 123
ΣH = 0
- HA + q . x – Nx = 0
Nx = - 8 + 2 . 4
= 0
x = 0 ; ND = 0 ton
x = 2 ; NE = 0 ton
d) Bentang E – B ( 0 ≤ x ≤ 4,0 meter )
ΣMx = 0
Mx = 0
x = 0 ; ME = 0 tm
x = 4 ; MB = 0 tm
ΣV = 0
- VB – Nx = 0
Nx = - 5
x = 0 ; NE = - 5,0 ton
x = 4 ; NB = - 5,0 ton
ΣH = 0
Lx = 0
x = 0 ; LE = 0 ton
x = 4 ; LB = 0 ton
X
VB
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 124
3. Gambar bidang Momen, Lintang dan Normal
4 m
2 m2 m
P = 2 ton
q = 2 t/m
VA
VB
0 tm 0 tm16 tm
16 tm
10 tm
0 tm 0 tm
0 t 0 t8 t
0 t 0 t
5 t
3 t
3 t
+
+
+
+
-
0 t
0 t
5 t-
0 t
0 t
A B
C ED
MOMEN
LINTANG
NORMAL
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 125
Contoh Soal 3
Penyelesaian :
1. Reaksi Perletakan
ΣMB = 0
VA . 4 + P1. L – P2 . L – P3 . L = 0
VA . 4 + 2 . 3 – 2 . 3 – 3 . 1 = 0
VA = 0,75 ton ( )
ΣMA = 0
-VB . 4 + P3 . L + P2 . L + P1 . L = 0
-VB . 4 + 3 . 3 + 2 . 1+ 2 . 3 = 0
VB = 4,25 ton ( )
ΣV = 0
VA + VB = P2 + P3
0,75 + 4,25 = 2 + 3
5,0 ton = 5,0 ton (Oke!!!)
P1 = 2 ton1 m
3 m
P2 = 2 ton P3 = 3 ton
2 m1 m 1 m
A B
D GE F
C
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 126
ΣH = 0
P1 + HA = 0
HA = - 2
= - 2,0 ton ( )
= 2,0 ton ( )
2. Perhitungan Gaya-gaya Dalam
a) Bentang A – C ( 0 ≤ x ≤ 3,0 meter )
ΣMx = 0
HA . x – Mx = 0
Mx = 2 . x
x = 0 ; MA = 0 tm
x = 3 ; MC = 6,0 tm
ΣV = 0
- VA – Nx = 0
Nx = - 0,75
x = 0 ; NA = - 0,75 ton
x = 3 ; NC = - 0,75 ton
X
HA
VA
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 127
ΣH = 0
HA – Lx = 0
Lx = 2
x = 0 ; LA = 2,0 ton
x = 3 ; LC = 2,0 ton
b) Bentang C – D ( 0 ≤ x ≤ 1,0 meter )
ΣMx = 0
HA (3 + x) – P1 . x – Mx = 0
Mx = 2 (3 + x) – 2 . x
= 6 + 2x – 2x
x = 0 ; MC = 6,0 tm
x = 1 ; MD = 6,0 tm
ΣV = 0
- VA – Nx = 0
Nx = - 0,75
x = 0 ; NC = - 0,75 ton
x = 1 ; ND = - 0,75 ton
3 m
HA
VA
P = 5 tonX
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 128
ΣH = 0
HA – P1 – Lx = 0
Lx = 2 – 2
= 0
x = 0 ; LC = 0 ton
x = 1 ; LD = 0 ton
c) Bentang D – E ( 0 ≤ x ≤ 1,0 meter )
ΣMx = 0
VA . x + HA . 4 – P1 . 1 – Mx = 0
Mx = 0,75 . x + 2 . 4 – 2 . 1
= 6 + 0,75 x
x = 0 ; MD = 6 tm
x = 1 ; ME = 6,75 tm
ΣV = 0
VA – Lx = 0
Lx = 0,75
x = 0 ; LD = 0,75 ton
x = 1 ; LE = 0,75 ton
P1 = 2 ton1 m
3 m
X
VA
Lx
Nx
Mx
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 129
ΣH = 0
- HA + P1 – Nx = 0
Nx = - 2 + 2
= 0
x = 0 ; ND = 0 ton
x = 1 ; NE = 0 ton
d) Bentang E – F ( 0 ≤ x ≤ 2,0 meter )
ΣMx = 0
VA . (1 + x) + HA . 4 – P1 . 1 – P2 . x – Mx = 0
Mx = 0,75 (1 + x) + 2 . 4 – 2 . 1 – 2 . x
= 6,75 – 1,25 x
x = 0 ; ME = 6,75 tm
x = 2 ; MF = 4,25 tm
ΣV = 0
VA – P2 – Lx = 0
Lx = 0,75 – 2
= - 1,25
x = 0 ; LE = - 1,25 ton
x = 2 ; LF = - 1,25 ton
P1 = 2 ton1 m
3 m
2
X1 m
VA
Lx
Nx
Mx
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 130
ΣH = 0
- HA + P1 – Nx = 0
Nx = - 2 + 2
= 0
x = 0 ; NE = 0 ton
x = 2 ; NF = 0 ton
e) Bentang F – G ( 0 ≤ x ≤ 1,0 meter )
ΣMx = 0
VA . (3 + x) + HA . 4 – P1 . 1 – P2 (2 + x) – P3 . x – Mx = 0
Mx = 0,75 . (3 + x) + 2 . 4 – 2 . 1 – 2 (2 + x) – 3 . x
= 4,25 – 4,25 x
x = 0 ; MF = 4,25 tm
x = 1 ; MG = 0 tm
ΣV = 0
VA – P2 – P3 – Lx = 0
Lx = 0,75 – 2 – 3
= - 4,25
x = 0 ; LF = - 4,25 ton
x = 1 ; LG = - 4,25 ton
P1 = 2 ton1 m
3 m
P2 = 2 ton P3 = 3 ton
2 m1 m X
VA
Lx
Nx
Mx
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 131
ΣH = 0
- HA + P1 – Nx = 0
Nx = - 2 + 2
= 0
x = 0 ; NF = 0 ton
x = 1 ; NG = 0 ton
f) Bentang G – B ( 0 ≤ x ≤ 4,0 meter )
ΣMx = 0
Mx = 0
x = 0 ; MG = 0 tm
x = 4 ; MB = 0 tm
ΣV = 0
- VB – Nx = 0
Nx = - 4,25
x = 0 ; NG = - 4,25 ton
x = 4 ; NB = - 4,25 ton
ΣH = 0
Lx = 0
x = 0 ; LG = 0 ton
x = 4 ; LB = 0 ton
X
VB
LX
NX
MX
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 132
3. Gambar bidang Momen, Lintang dan Normal
P1 = 2 ton1 m
3 m
P2 = 2 ton P3 = 3 ton
2 m1 m 1 m
HA
VA
VB
0 tm 0 tm
6 tm
6 tm
6 tm
0 tm 0 tm
6,75 tm
4,25 tm
0 t 0 t2 t
2 t
0 t 0 t
4,25 t0,75 t
0 t0 t
0,75 t
1,25 t
4,25 t
+
+
+
-
+
- -
A B
D GE F
C
MOMEN
LINTANG
NORMAL
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 133
Contoh Soal 4
Penyelesaian :
1. Reaksi Perletakan
ΣMB = 0
VA . 4,5 - P. L – q . L (½ . L + L) = 0
VA . 4,5 - 3 . 4 – 2 . 2,5 (½ . 2,5 + 2) = 0
VA = 6,278 ton ( )
ΣMA = 0
- VB . 4,5 - P. L + q . L (½ . L) = 0
- VB . 4,5 - 3 . 4 + 2 . 2,5 (½ . 2,5) = 0
VB = - 1,278 ton ( )
ΣV = 0
VA + VB = q . L
6,278 + (- 1,278) = 2 . 2,5
5 ton = 5 ton (Oke!!!)
P = 3 ton
2,5 m 2 m
q = 2 t/m
4 m
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 134
Konstruksi bentang A – D
ΣMD = 0
HA . 4 - VA . 2,5 – q . L (½ . L) = 0
HA . 4 - 6,278 . 2,5 – 2 . 2,5 (½ . 2,5) = 0
HA = 2,361 ton ( )
Konstruksi bentang B – D
ΣMD = 0
HB . 4 – VB . 2 = 0
HB . 4 + 1,278 . 2 = 0
HB = - 0,639 ton ( )
= 0,639 ton ( )
2. Perhitungan Gaya-gaya Dalam
a) Bentang A – C ( 0 ≤ x ≤ 4 meter )
ΣMx = 0
- HA . x – Mx = 0
Mx = - 2,361 . x
x = 0 ; MA = 0 tm
x = 4 ; MC = - 9,444 tm
X
LX
NX
MX
VA
HA
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 135
ΣV = 0
VA – Nx = 0
Nx = 6,278
x = 0 ; NA = 6,278 ton
x = 4 ; NC = 6,278 ton
ΣH = 0
- HA – Lx = 0
Lx = - 2,361
x = 0 ; LA = - 2,361 ton
x = 4 ; LC = - 2,361 ton
b) Bentang C – D ( 0 ≤ x ≤ 2,5 meter )
ΣMx = 0
VA . x – HA . L - q . x (½ . x) – Mx = 0
Mx = 6,278 . x – 2,361 . 4 - 2 . x (½ . x)
= 6,278 x – 9,444 - x²
x = 0 ; MC = - 9,444 tm
x = 2,5 ; MD = 0 tm
X
q = 2 t/m
Lx
Nx
Mx
4 m
VA
HA
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 136
ΣV = 0
VA – q . x – Lx = 0
Lx = 6,278 – 2 x
x = 0 ; LC = 6,278 ton
x = 2,5 ; LD = 1,278 ton
ΣH = 0
HA – Nx = 0
Nx = 2,361
x = 0 ; NC = 2,361 ton
x = 2,5 ; ND = 2,361 ton
c) Bentang B – E ( 0 ≤ x ≤ 4 meter )
ΣMx = 0
HB . x + Mx = 0
Mx = - 0,639 x
x = 0 ; MB = 0 tm
x = 4 ; ME = - 2,556 tm
X
LX
NX
MX
VB
HB
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 137
ΣV = 0
-VB + Nx = 0
Nx = 1,278
x = 0 ; NB = 1,278 ton
x = 4 ; NE = 1,278 ton
ΣH = 0
- HB – Lx = 0
Lx = - 0,639
x = 0 ; LB = - 0,639 ton
x = 4 ; LE = - 0,639 ton
d) Bentang E – D ( 0 ≤ x ≤ 2 meter )
ΣMx = 0
- VB . x + HB . 4 + Mx = 0
Mx = 1,278 . x – 0,639 . 4
= 1,278 x – 2,556
x = 0 ; ME = - 2,556 tm
x = 2 ; MD = 0 tm
P = 3 ton
X
Nx
Mx
Lx
VB
HB
4 m
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 138
ΣV = 0
HB – P + Nx = 0
Lx = - 0,639 + 3
x = 0 ; LE = 2,361 ton
x = 2 ; LD = 2,361 ton
ΣH = 0
- VB + Lx = 0
Nx = 1,278
x = 0 ; NE = 1,278 ton
x = 2 ; ND = 1,278 ton
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 139
3. Gambar bidang Momen, Lintang dan Normal
P = 3 ton
2,5 m 2 m
q = 2 t/m
4 m
0 tm 0 tm
9,444 tm
9,444 tm
2,556 tm0 tm 0 tm
2,556 tm
0 t 0 t
1,278 t
6,278 t
2,361 t
2,361 t
0 t 0 t
6,278 t
6,278 t
1,278 t
1,278 t
+ +
--
+
- -
-
-
MOMEN
LINTANG
NORMAL
+
2,361 t
A B
C D E
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 140
SOAL LATIHAN
1. Hitunglah reaksi perletakan, gaya-gaya dalam dari struktur statis tertentu
dibawah ini dan gambarkan bidang momen, lintang dan normal dengan
mencantumkan ordinat-ordinat penting!
2. Hitunglah reaksi perletakan, gaya-gaya dalam dari struktur statis tertentu
dibawah ini dan gambarkan bidang momen, lintang dan normal dengan
mencantumkan ordinat-ordinat penting!
2 m
2 m
P = 3 ton
2 m2 m
q = 3 t/m
P = 6 ton
2 m
2 m
4 m
q = 4 t/m
Noviyanthy H. MT.
Statika dan Mekanika Bahan 1 141
3. Hitunglah reaksi perletakan, gaya-gaya dalam dari struktur statis tertentu
dibawah ini dan gambarkan bidang momen, lintang dan normal dengan
mencantumkan ordinat-ordinat penting!
P = 4 ton
2 m
2 m
4 m
q = 3 t/m