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Find moment of inertia of angle ISA : 100 mm x 75 mm x 6 mm about the centroidal XX and YY axis.
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written 5.2 years ago by |
Solution:
a1 = 6 x 100 = $600mm^{2}$
a2 = 69 x 6 = $414mm^{2}$
x1 = 6/2 = 3mm.
x2 = 6 + 69/2 = 40.50mm
y1 = 100/2 = 50 mm
y2 = 6/2 = 3mm
$\bar{x} = \frac{a1x1 + a2x2}{a1 + a2} = 18.31 mm from (1) and (2)$
$\bar{y} = \frac{a1y1 + a2y2}{a1 + a2} = 30.81 mm from (1) and (2)$
$Ixx = (Ixx)_{1} + (Ixx)_{2}$
$= \big(\frac{6 x 100}{12}^{3} + 600(50-30.81)^{2}\big) + \big(\frac{69 x 6}{12}^{3} + 414(30.81-3)^{2}\big)$
$Ixx = 7.21 \times 10^{5} mm^{4} + 3.21 \times 10^{5} mm^{4}$
$Ixx = 10.42 \times 10^{5} mm^{4}$
$Iyy = (Iyy)_{1} + (Iyy)_{2}$
$= \big(\frac{6 x 100}{12}^{3} + 600(18.31-3)^{2}\big) + \big(\frac{69 x 6}{12}^{3} + 414(40.50 - 18.31)^{2}\big)$
$= 1.42 \times 10^{5} + 3.68 \times 10^{5}$
$Iyy = 5.1 \times 10^{5} mm^{4}$
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