M. Inersia
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Tugas Struktur Baja II
Menghitung Inersia Komposit: 1.
Jarak Sumbu X, diukur dari bawah:
Yt =
A A ⋅ YA + A B ⋅ YB +... + A E ⋅ YE At
Momen Inersia Segitiga:
I=
1 ⋅ b ⋅ h3 36
Momen Inersia Segiempat:
I=
1 ⋅ b ⋅ h3 12
AA = 1/2 × 8 × 6
YA = (1/3 × 6) + 20
IA = 1/36 × 8 × 63
AA = 24
YA = 22
IA = 48
AB = 8 × 5
YB = (1/2 × 5) + 15
IB = 1/12 × 8 × 53
AB = 40
YB = 17,5
IB = 83,333
AC = 4 × 5
YC = (1/2 × 5) + 10
IC = 1/12 × 4 × 53
AC = 20
YC = 12,5
IC = 41,667
AD = 6 × 5
YD = (1/2 × 5)
ID = 1/12 × 6 × 53
AD = 30
YD = 2,5
ID = 62,5
AE = 6 × 26
YE = (1/2 × 26)
IE = 1/12 × 6 × 263
YE = 13
IE = 8.788
AE = 156
+
At = 270
1
Tugas Struktur Baja II
Jarak Sumbu X, diukur dari bawah:
Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD + A E ⋅ YE At
Yt =
(24 ⋅ 22) + (40 ⋅ 17,5) + (20 ⋅ 12,5) + (30 ⋅ 2,5) + (156 ⋅ 13) 270
Yt =
528 + 700 + 250 + 75 + 2.028 270
Yt =
3.581 270
Yt = 13,263. AA = 24
AB = 40
AC = 20
AD = 30
AE = 156
a = YA – Yt
a2 = (8,737)2
a = 22 – 13,263
a2 = 76,336
a = 8,737 b = YB – Yt
b2 = (4,237)2
b = 17,5 – 13,263
b2 = 17,952
b = 4,237 c = Yt – YC
c2 = (0,763)2
c = 13,263 – 12,5
c2 = 0,582
c = 0,763 d = Yt – YD
d2 = (10,763)2
d = 13,263 – 2,5
d2 = 115,841
d = 10,763 e = Yt – YE
e2 = (0,263)2
e = 13,263 – 13
e2 = 0,069
AA × a2 = 1.832,060
AB × b2 = 718,099
AC × c2 = 11,642
AD × d2 = 3.475,241
AE × e2 = 10,787
e = 0,263
Inersia Komposit:
2
Tugas Struktur Baja II It
It
=(I A + A A ⋅a 2 ) + (I B + A B ⋅b 2 ) + (I C + A C ⋅c 2 ) + (I D + A D ⋅d 2 ) + (I E + A E ⋅e 2 )
= (48 + 1.832,060) + (83,333 + 718,099) + (41,667 + 11,642) + (62,5 + 3.475,241) + (8.788 + 10,787)
2.
It
= 1.880,060 + 801,433 + 53,309 + 3.537,741 + 8.798,787
It
= 15.071,330.
Jarak Sumbu X, diukur dari bawah:
Yt =
A A ⋅ YA + A B ⋅ YB +... + A D ⋅ YD At
Momen Inersia Segiempat:
I=
1 ⋅ b ⋅ h3 12
AA = 10 × 10
YA = (1/2 × 10) + 36
IA = 1/12 × 10 × 103
AA = 100
YA = 41
IA = 833,333
AB = 40 × 10
YB = (1/2 × 10) + 26
IB = 1/12 × 40 × 103
AB = 400
YB = 31
IB = 3.333,333
AC = 10 × 16
YC = (1/2 × 16) + 10
IC = 1/12 × 10 × 163
AC = 160
YC = 18
IC = 3.413,333
AD = 20 × 10
YD = (1/2 × 10)
ID = 1/12 × 20 × 103
YE = 5
IE = 1.666,667
AD = 200
+
At = 860 Jarak Sumbu X, diukur dari bawah: Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD At
3
Tugas Struktur Baja II
Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD At
Yt =
(100 ⋅ 41) + (400 ⋅ 31) + (160 ⋅ 18) + (200 ⋅ 5) 860
Yt =
4.100 + 12.400 + 2.880 + 1.000 860
Yt =
20.380 860
Yt = 23,698. AA = 100
AB = 400
AC = 160
AD = 200
a = YA – Yt
a2 = (17,302)2
a = 41 – 23,698
a2 = 299,370
a = 17,302 b = YB – Yt
b2 = (7,302)2
b = 31 – 23,698
b2 = 53,324
b = 7,302 c = Yt – YC
c2 = (5,698)2
c = 23,698 – 18
c2 = 32,463
c = 5,698 d = Yt – YD
d2 = (18,698)2
d = 23,698 – 5
d2 = 349,603
AA × a2 = 29.937,047
AB × b2 = 21.329,584
AC × c2 = 5.194,159
AD × d2 = 69.920,606
d = 18,698 Inersia Komposit: It
It
=(I A + A A ⋅a 2 ) + (I B + A B ⋅b 2 ) + (I C + A C ⋅c 2 ) + (I D + A D ⋅d 2 )
= (833,333 + 29.937,047) + (3.333,333 + 21.329,584) + (3.413,333 + 5.194,159) + (1.666,667 + 69.920,606)
It
= 30.770,380 + 24.662,917 + 8.607,492 + 71.587,272
It
= 135.628,062.
4
Tugas Struktur Baja II
3.
Jarak Sumbu X, diukur dari bawah:
Yt =
Momen Inersia Segiempat:
A A ⋅ YA + A B ⋅ YB +... + A D ⋅ YD At
I=
1 ⋅ b ⋅ h3 12
AA = 20 × 10
YA = (1/2 × 10) + 30
IA = 1/12 × 20 × 103
AA = 200
YA = 35
IA = 1.666,667
AB = 40 × 5
YB = (1/2 × 5) + 25
IB = 1/12 × 40 × 53
AB = 200
YB = 27,5
IB = 416,667
AC = 10 × 20
YC = (1/2 × 20) + 5
IC = 1/12 × 10 × 203
AC = 200
YC = 15
IC = 6.666,667
AD = 40 × 5
YD = (1/2 × 5)
ID = 1/12 × 40 × 53
YE = 2,5
IE = 416,667
AD = 200
+
At = 800 Jarak Sumbu X, diukur dari bawah: 5
Tugas Struktur Baja II
Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD At
Yt =
(200 ⋅ 35) + (200 ⋅ 27,5) + (200 ⋅ 15) + (200 ⋅ 2,5) 800
Yt =
7.000 + 5.500 + 3.000 + 500 800
Yt =
16.000 800
Yt = 20. AA = 200
AB = 200
AC = 200
AD = 200
a = YA – Yt
a2 = (15)2
a = 35 – 20
a2 = 225
a = 15 b = YB – Yt
b2 = (7,5)2
b = 27,5 – 20
b2 = 56,25
b = 7,5 c = Yt – YC
c2 = (5)2
c = 20 – 15
c2 = 25
c=5 d = Yt – YD
d2 = (17,5)2
d = 20 – 2,5
d2 = 306,250
AA × a2 = 45.000
AB × b2 = 11.250
AC × c2 = 5.000
AD × d2 = 61.250
d = 17,5 Inersia Komposit: It
It
=(I A + A A ⋅a 2 ) + (I B + A B ⋅b 2 ) + (I C + A C ⋅c 2 ) + (I D + A D ⋅d 2 )
= (1.666,667 + 45.000) + (416,667 + 11.250) + (6.666,667 + 5.000) + (416,667 + 61.250)
It
= 46.666,667 + 11.666,667 + 11.666,667 + 61.666,667
It
= 131.666,667.
6
Menghitung Inersia Komposit: 1.
Jarak Sumbu X, diukur dari bawah:
Yt =
A A ⋅ YA + A B ⋅ YB +... + A E ⋅ YE At
Momen Inersia Segitiga:
I=
1 ⋅ b ⋅ h3 36
Momen Inersia Segiempat:
I=
1 ⋅ b ⋅ h3 12
AA = 1/2 × 8 × 6
YA = (1/3 × 6) + 20
IA = 1/36 × 8 × 63
AA = 24
YA = 22
IA = 48
AB = 8 × 5
YB = (1/2 × 5) + 15
IB = 1/12 × 8 × 53
AB = 40
YB = 17,5
IB = 83,333
AC = 4 × 5
YC = (1/2 × 5) + 10
IC = 1/12 × 4 × 53
AC = 20
YC = 12,5
IC = 41,667
AD = 6 × 5
YD = (1/2 × 5)
ID = 1/12 × 6 × 53
AD = 30
YD = 2,5
ID = 62,5
AE = 6 × 26
YE = (1/2 × 26)
IE = 1/12 × 6 × 263
YE = 13
IE = 8.788
AE = 156
+
At = 270
1
Tugas Struktur Baja II
Jarak Sumbu X, diukur dari bawah:
Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD + A E ⋅ YE At
Yt =
(24 ⋅ 22) + (40 ⋅ 17,5) + (20 ⋅ 12,5) + (30 ⋅ 2,5) + (156 ⋅ 13) 270
Yt =
528 + 700 + 250 + 75 + 2.028 270
Yt =
3.581 270
Yt = 13,263. AA = 24
AB = 40
AC = 20
AD = 30
AE = 156
a = YA – Yt
a2 = (8,737)2
a = 22 – 13,263
a2 = 76,336
a = 8,737 b = YB – Yt
b2 = (4,237)2
b = 17,5 – 13,263
b2 = 17,952
b = 4,237 c = Yt – YC
c2 = (0,763)2
c = 13,263 – 12,5
c2 = 0,582
c = 0,763 d = Yt – YD
d2 = (10,763)2
d = 13,263 – 2,5
d2 = 115,841
d = 10,763 e = Yt – YE
e2 = (0,263)2
e = 13,263 – 13
e2 = 0,069
AA × a2 = 1.832,060
AB × b2 = 718,099
AC × c2 = 11,642
AD × d2 = 3.475,241
AE × e2 = 10,787
e = 0,263
Inersia Komposit:
2
Tugas Struktur Baja II It
It
=(I A + A A ⋅a 2 ) + (I B + A B ⋅b 2 ) + (I C + A C ⋅c 2 ) + (I D + A D ⋅d 2 ) + (I E + A E ⋅e 2 )
= (48 + 1.832,060) + (83,333 + 718,099) + (41,667 + 11,642) + (62,5 + 3.475,241) + (8.788 + 10,787)
2.
It
= 1.880,060 + 801,433 + 53,309 + 3.537,741 + 8.798,787
It
= 15.071,330.
Jarak Sumbu X, diukur dari bawah:
Yt =
A A ⋅ YA + A B ⋅ YB +... + A D ⋅ YD At
Momen Inersia Segiempat:
I=
1 ⋅ b ⋅ h3 12
AA = 10 × 10
YA = (1/2 × 10) + 36
IA = 1/12 × 10 × 103
AA = 100
YA = 41
IA = 833,333
AB = 40 × 10
YB = (1/2 × 10) + 26
IB = 1/12 × 40 × 103
AB = 400
YB = 31
IB = 3.333,333
AC = 10 × 16
YC = (1/2 × 16) + 10
IC = 1/12 × 10 × 163
AC = 160
YC = 18
IC = 3.413,333
AD = 20 × 10
YD = (1/2 × 10)
ID = 1/12 × 20 × 103
YE = 5
IE = 1.666,667
AD = 200
+
At = 860 Jarak Sumbu X, diukur dari bawah: Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD At
3
Tugas Struktur Baja II
Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD At
Yt =
(100 ⋅ 41) + (400 ⋅ 31) + (160 ⋅ 18) + (200 ⋅ 5) 860
Yt =
4.100 + 12.400 + 2.880 + 1.000 860
Yt =
20.380 860
Yt = 23,698. AA = 100
AB = 400
AC = 160
AD = 200
a = YA – Yt
a2 = (17,302)2
a = 41 – 23,698
a2 = 299,370
a = 17,302 b = YB – Yt
b2 = (7,302)2
b = 31 – 23,698
b2 = 53,324
b = 7,302 c = Yt – YC
c2 = (5,698)2
c = 23,698 – 18
c2 = 32,463
c = 5,698 d = Yt – YD
d2 = (18,698)2
d = 23,698 – 5
d2 = 349,603
AA × a2 = 29.937,047
AB × b2 = 21.329,584
AC × c2 = 5.194,159
AD × d2 = 69.920,606
d = 18,698 Inersia Komposit: It
It
=(I A + A A ⋅a 2 ) + (I B + A B ⋅b 2 ) + (I C + A C ⋅c 2 ) + (I D + A D ⋅d 2 )
= (833,333 + 29.937,047) + (3.333,333 + 21.329,584) + (3.413,333 + 5.194,159) + (1.666,667 + 69.920,606)
It
= 30.770,380 + 24.662,917 + 8.607,492 + 71.587,272
It
= 135.628,062.
4
Tugas Struktur Baja II
3.
Jarak Sumbu X, diukur dari bawah:
Yt =
Momen Inersia Segiempat:
A A ⋅ YA + A B ⋅ YB +... + A D ⋅ YD At
I=
1 ⋅ b ⋅ h3 12
AA = 20 × 10
YA = (1/2 × 10) + 30
IA = 1/12 × 20 × 103
AA = 200
YA = 35
IA = 1.666,667
AB = 40 × 5
YB = (1/2 × 5) + 25
IB = 1/12 × 40 × 53
AB = 200
YB = 27,5
IB = 416,667
AC = 10 × 20
YC = (1/2 × 20) + 5
IC = 1/12 × 10 × 203
AC = 200
YC = 15
IC = 6.666,667
AD = 40 × 5
YD = (1/2 × 5)
ID = 1/12 × 40 × 53
YE = 2,5
IE = 416,667
AD = 200
+
At = 800 Jarak Sumbu X, diukur dari bawah: 5
Tugas Struktur Baja II
Yt =
A A ⋅ YA + A B ⋅ YB + A C ⋅ YC + A D ⋅ YD At
Yt =
(200 ⋅ 35) + (200 ⋅ 27,5) + (200 ⋅ 15) + (200 ⋅ 2,5) 800
Yt =
7.000 + 5.500 + 3.000 + 500 800
Yt =
16.000 800
Yt = 20. AA = 200
AB = 200
AC = 200
AD = 200
a = YA – Yt
a2 = (15)2
a = 35 – 20
a2 = 225
a = 15 b = YB – Yt
b2 = (7,5)2
b = 27,5 – 20
b2 = 56,25
b = 7,5 c = Yt – YC
c2 = (5)2
c = 20 – 15
c2 = 25
c=5 d = Yt – YD
d2 = (17,5)2
d = 20 – 2,5
d2 = 306,250
AA × a2 = 45.000
AB × b2 = 11.250
AC × c2 = 5.000
AD × d2 = 61.250
d = 17,5 Inersia Komposit: It
It
=(I A + A A ⋅a 2 ) + (I B + A B ⋅b 2 ) + (I C + A C ⋅c 2 ) + (I D + A D ⋅d 2 )
= (1.666,667 + 45.000) + (416,667 + 11.250) + (6.666,667 + 5.000) + (416,667 + 61.250)
It
= 46.666,667 + 11.666,667 + 11.666,667 + 61.666,667
It
= 131.666,667.
6
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