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Jumat, 05 November 2010

MOMEN INERSIA (CONTOH SOAL DAN PENYELESAIANNYA)

Contoh soal

Tentukan momen inersia dari gambar di atas!
Penyelesaian.
Cari titik berat.
Penampang I
A = b x h
= 15 x 10
= 150 cm2
x = ½ b
= ½ . 15
= 7,5 cm
y = ½ h + 15
= ½ . 10 + 15
= 20 cm
Penampang II
A = b x h
= 5 x 15
= 75 cm2
x = ½ b
= ½ . 5
= 2,5 cm
y = ½ h
= ½ . 15
= 7,5 cm
Dari data di atas dibuat tabel.

x ̅ = (∑Axi)/(∑A)
= 1312,5/225
= 5,833 cm
y ̅ = (∑Ayi)/(∑A)
= 3562,5/225
= 15,833 cm
Momen inersia
Ix1 = 1/12 . b . h3 + A1 (y1 - y ̅)2
= 1/12 . 15 . 103 + 150 (20 – 15,833)2
= 1250 + 2604,583
= 3854,583 cm4
Ix2 = 1/12 . b . h3 + A2 (y2 - y ̅)2
= 1/12 . 5 . 153 + 75 (7,5 – 15,833)2
= 1406,25 + 5207,917
= 6614,167 cm4
∑Ix = Ix1 + Ix2
= 3854,583 +6614,167
= 10468,75 cm4

Iy1 = 1/12 . h . b3 + A1 (x1 - x ̅)2
= 1/12 . 10 . 153 + 150 (7,5 – 5,833)2
= 2812,5 + 416,833
= 3229,333 cm4
Iy2 = 1/12 . h . b3 + A2 (x2 - x ̅)2
= 1/12 . 15 . 53 + 75 (2,5 – 15,833)2
= 156,25 + 833,167
= 989,417 cm4
∑Iy = Iy1 + Iy2
= 3229,333 + 989,417
= 4218,75 cm4

Ixy = A1 (x1 - x ̅) (y1 - y ̅) + A2 (x2 - x ̅) (y2 - y ̅)
= 150 (7,5 – 5,833)(20 – 15,833) + 75 (2,5 – 5,833)(7,5 - 15,833)
= 150 (1,667)(4,167) + 75 (-3,333)(-8,333)
= 1041,958 + 2083,042
= 3125 cm4

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