CVG2140 – Solution to Assignment No. 3 (Centroids & Moments of Inertia) Problem 1: For the section shown below, find the position of the centroid. (Note: solve the problem by taking the origin at the point "O")
1 2
4
5 3
A (mm2) A1 A2 A3 A4 A5 Σ
170×130=22,100 130×130/2=8,450 300×150=45,000 -×252=-1,963.5 -×252=-1,963.5 71,623
xi (mm)
yi (mm)
xi Ai 3
(mm ) 85 150+65=215 1,878,500 170+(130/3)=213.3 150+(130/3)=193.3 1,802,385 150 75 6,750,000 80 80 -157,080 300-80=220 80 -431,970 9,841,835
yi Ai (mm3) 4,751,500 1,633,385 3,375,000 -157,080 -157,080 9,445,725
x A 9,841,835 137.4 mm 71, 623 A y A 9, 445, 725 131.9 mm y 71, 623 A
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CVG2140 – Assignment 3
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Problem 2: For the section shown below, find the moments of inertia Ix, Iy, Jo, Ixy with respect to a coordinate system located at the centroid. (Dimensions are in mm). (a) Centroid
150 mm
A1 A2 A3 A4 A5 Σ
150 mm
A (mm2)
xi (mm)
yi (mm)
xi Ai (mm3)
yi Ai (mm3)
2,000 7,600 2,000 6,400 ×202/2=628.3 18,628.32
-180 0 180 0 0
70 10 70 100 180+(4×20/3)=188.5
-360,000 0 360,000 0 0 0
140,000 76,000 140,000 640,000 118,438 1,114,438.04
x A 0 0 mm A 18, 628.32 y A 1,114, 438 59.8 mm (from the bottom) y A 18, 628.32
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CVG2140 – Assignment 3
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(b) Moments of inertia The parallel axis theorem, which refers to the centroidal axes of each area, will be used, i.e., I xi I xci Ai d xi . 2
59.8 mm
150 mm
A1 A2 A3 A4 A5 Σ
A1 A2 A3 A4 A5 Σ
150 mm
I xci (mm4)
A (mm2)
d xi (mm)
I xi (mm4)
1,666,667 253,333 1,666,667 13,653,333 17,568
2,000 7,600 2,000 6,400 ×202/2=628.3 18,628.32
10.2 49.8 10.2 40.2 128.7
1,874,747 19,101,637 1,874,747 23,995,989 10,424,841 57,271,961
i I yc (mm4)
A (mm2)
d yi (mm)
I yi (mm4)
66,667 91,453,333 66,667 853,333 62,832
2,000 7,600 2,000 6,400 ×202/2=628.3 18,628.32
-180 0 180 0 0
64,866,667 91,453,333 64,866,667 853,333 62,832 222,102,832
Ix = 57,271,961 mm4, Iy = 222,102,832 mm4 Jo = Ix + Iy = 57,271,961 + 222,102,832 = 279,374,793 mm4 Since the figure is symmetric with respect to the y-axis, Ixy = 0.
CVG2140 – Assignment 3
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Problem 3: Consider the section shown below. For a coordinate system located at the centroid, calculate the maximum and minimum values of the moments of inertia Imax and Imin (obtained by rotating the axes) and the corresponding angles.
1
2
(a) Centroid
A1 A2 Σ
A (cm2)
xi (cm)
yi (cm)
xi Ai (cm3)
yi Ai (cm3)
500 200 700
5 20
25 5
2,500 4,000 6,500
12,500 1,000 13,500
x A 6,500 9.3 cm (from the left) 700 A y A 13,500 19.3 cm (from the bottom) y 700 A
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(b) Moments of inertia with respect to centroidal axes
A1 A2 Σ
A1 A2 Σ
I xci (cm4)
A (cm2)
d xi (cm)
I xi (cm4)
104,167 1,667
500 200 700
5.7 -14.3
120,412 42,565 162,976
i I yc (cm4)
A (cm2)
d yi (cm)
I yi (cm4)
4,167 6,667
500 200 700
-4.3 10.7
13,412 29,565 42,976
CVG2140 – Assignment 3
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i (cm4) I xcyc
A (cm2)
d xi (cm)
d yi (cm)
I xyi (cm4)
0 0
500 200 700
5.7 -14.3
-4.3 10.7
-12,255 -30,602 -42,857
A1 A2 Σ
(c) Principal moments of inertia and principal axes
tan 2
I xy Ix I y
42,857 0.714283 162,976 42,976
tan 1 0.714283 17.7° (counterclockwise from x-axis) 2
To find out the correspondence between principal moments of inertia and principal axes, the transformation equation is first used for = 17.7°:
I xc I yc
I xc I yc
cos 2 17.7 I xcyc sin 2 17.7 2 2 162,976 42,976 162,976 42,976 cos 2 17.7 42,857 sin 2 17.7 2 2 176, 710 cm 4
Ix'
Then it is used for = 17.7°+90° = 107.7°:
I xc I yc
I xc I yc
cos 2 107.7 I xcyc sin 2 107.7 2 2 162,976 42,976 162,976 42,976 cos 2 107.7 42,857 sin 2 107.7 2 2 29, 242 cm 4
Ix'
Therefore, Imax = I1 = 176,710 cm4 acting at 17.7° Imin = I2 = 29,242 cm4 acting at 107.7°
CVG2140 – Assignment 3
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2
1
107.7° 17.7°
CVG2140 – Assignment 3
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