Page 77 - E-modul Mekanika Teknik 2 Fase E/Kelas X SMK

Y

6 cm

b1
Zo (2,3)
X X
b2
y = 5 cm
2 cm

a1 a2

X = 2 cm Y
2 cm 4 cm

Gambar 50. Penentuan Letak Momen Kelembaman Latihan Soal Nomor 1 Momen
Inersia

Jarak Z1 & Z2 akan sumbu x dan y:
b1 = 5 cm – 3 cm = 2 cm
b2 = 3 cm – 1 cm = 2 cm
a1 = 2 cm – 1 cm = 1 cm
a2 = 4 cm – 3 cm = 1 cm
Momen Inersia akan sumbu s:
Ix = I2x + Fb 2
4
3
4
2
2
Ix1 = 1/12 . 2 cm . (6 cm) + 12 cm . (2 cm) = 36 cm + 48 cm = 84 cm 4
2
4
3
Ix2 = 1/12 . 6 cm . (2 cm) + 12 cm . (2 cm) = 4 cm + 48 cm = 56 cm 4
2
4
Jadi Ix = 140 cm 4
Momen Inersia akan sumbu y:
Iy = I2y + F.a 2
2
2
4
3
4
Iy1 = 1/12 . 6 cm . (2 cm) + 12 cm . (1 cm) = 4 cm + 12 cm = 16 cm 4
4
3
Iy2 = 1/12 . 2 cm . (6 cm) + 12 cm . (1 cm) = 36 cm + 12 cm = 48 cm 4
4
2
2
Jadi Iy = 64 cm
4

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