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Second Moment of Area.
Ixx = Second moment of Area about the x-x axis
Ixx = bh3/12 = 6.9*13^3/12 = 1263.275 mm4
Iyy = hb3/12 = 13*6.9^3/12 = 355.8848 mm4
Do outside (solid shaft)
Ixx = *D4/64 = Pi*13^4/64 = 1401.985 mm4
Do inside (the hole)
ID = 13-2*1.7 = 9.6
Ixx = *D4/64 = Pi*9.6^4/64 = 416.922 mm4
Net Ixx = 1401.985 - 416.922 = 985.063 mm4
Second Moment of Area
Tuesday, 26 July 2011
8:35 PM
Area-Moment Page 1
I = Ic + Ad2
I = The second moment of area about non-centroidal plane
Ic = The second moment of area about its own centroid
A = Area of region
d = Distance from centroid to new plane
Ic = bh3/12
= 18*4.9^3/12 = 176.4735 mm4
I = Ic + Ad2
= 176.4735 + (18*4.9)*(6.2^2)
= 3566.9 mm4
Split into 3 regions (elements)
E1 & E3: Top and bottom. E2 is middle web.
height = (h1-h2)/2 = (26-15)/2 = 5.5 mm
d1 = 15/2+5.5/2 = 10.25 mm
I1 = Ic1 + Ad2
Ic1 = bh3/12 = 14*5.5^3/12 = 194.1042 mm4
Ad2 = (14*5.5)*(10.25^2) = 8089.8125 mm4
I1 = 194.1042 + 8089.8125 = 8283.9167 mm4
E3 is same as E1.
E2:
I1 = Ic1 + Ad2
Ic2 = bh3/12 = 5.8*15^3/12 = 1631.25 mm4
Total I = I1+I2+I3
= 8283.917 + 8283.917 +1631.25 = 18199.08 mm4
Parallel Axis Theroem
Tuesday, 13 March 2012
5:24 PM
Area-Moment Page 2
Not so easy if the compound shape is not symmetrical…
For example, this "T" beam.
Where is the Neutral Plane (x----x)?
Need to find the centroid.
Well, at least the y coordinate
of the centroid, yc.
Can we split this shape into elements that have the
same neutral axis? Yes!
E1: I=bh3/12 = 14*26^3/12 = 20505.33 mm4
E2: I=bh3/12
I = 4.1*15^3/12 = 1153.125 mm4
Ixx = I1 - I2 - I3
= 20505.33 - 1153.125 - 1153.125
= 18199.08 mm4
b = (14-5.8)/2 = 4.1 mm
This is a great shortcut if you can split up into
elements that all have the same centroidal axis.
Area-Moment Page 3
yc = (Ay)/(A) xc = (Ax)/(A)
y1 = 412/2 = 206 mm
y2 = (615-412)/2 + 412 = 513.5 mm
A1 = 335*412 = 138020 mm2
A2 = 130*(615-412) = 26390 mm2
A1y1 = 138020*206 = 28432120
A2y2 = 26390*513.5 = 13551265
(Ay) = 28432120+13551265 = 41983385
(A) = 138020+26390 = 164410
= 255.3579 mm
yc = (Ay)/(A) = 41983385/164410
Element A y Ay x Ax
mm2 mm mm3 mm mm3
1 138020 206 28432120 167.5 23118350
2 26390 513.5 13551265 270 7125300
Total 164410 41983385 30243650
Centroid 255.35786 183.9526
The centroid is where this piece
of polycarbonate will balance.
Finding Centroid
Wednesday, 27 March 2013
12:06 PM
Area-Moment Page 4
= 255.3579 mm
yc = (Ay)/(A) = 41983385/164410
The centroid is where this piece
of polycarbonate will balance.
Element A y Ay Ic d Ad^2 I
mm2 mm mm3 mm4 mm mm4 mm4
1 138020 206 28432120 1952338907 49.35786 336244032 2288582939
2 26390 513.5 13551265 90625459.2 258.1421 1.759E+09 1849185570
Total 164410 41983385 NO! NO! 4137768509
Centroid 255.35786
4137.76851 E6mm4
I = Ic + Ad2
Finding Total Ixx
Wednesday, 27 March 2013
12:06 PM
Area-Moment Page 5
I = Ic + Ad2
I = The second moment of area about non-centroidal plane
Ic = The second moment of area about its own centroid
A = Area of region
d = Distance from centroid to new plane
yc = (Ay)/(A)
yc = (Ay)/(A)
Element 1: A1 = 16*4.2 = 67.2 mm2
y1 = 4.2/2 = 2.1 mm
A1y1 = 67.2*2.1 = 141.12 mm3
Element 2: A2 = 6.2*13 = 80.6 mm2
y2 = 4.2 + 13/2 = 10.7 mm
A2y2 = 80.6*10.7 = 862.42 mm3
= 1003.54 mm3
(Ay) = A1y1 + A2y2 = 141.12+ 862.42
(A) = 67.2+80.6 = 147.8 mm2
yc = (Ay)/(A) = 1003.54/147.8 = 6.7899mm
Table format
Element A y Ay x Ax
mm2 mm mm3 mm mm3
1 67.2 2.1 141.12 8 537.6
2 80.6 10.7 862.42 3.1 249.86
Total 147.8 n.a. 1003.54 n.a. 787.46
Centroid 6.7899 5.328
yc = (Ay)/(A) xc = (Ax)/(A)
Centroid
Finding Centroid for Composite Shape (p368)
Tuesday, 13 March 2012
5:38 PM
Area-Moment Page 6
I = Ic + Ad2
I = The second moment of area about non-centroidal
plane
Ic = The second moment of area about its own centroid
A = Area of region
d = Distance from centroid to new plane
yc = (Ay)/(A) xc = (Ax)/(A)
Table format (Centroid)
Element A y Ay
mm2 mm mm3
1 67.2 2.1 141.12
2 80.6 10.7 862.42
Total 147.8 X 1003.54
Centroid 6.7899
Working
Ic1 = 16*4.2^3/12 = 98.784 mm4
Ic2 = 6.2*13^3/12 = 1135.1167 mm4
d1 = yc - h1/2 = 6.7899-2.1 = 4.6899 mm
d2 = 10.7 - 6.7899 = 3.9101 mm
A1d1
2 = 67.2 * 4.6899^2 = 1478.0749 mm4
I1= 98.784 + 1478.07 = 1576.854 mm4
A2d2
2 = 80.6 *3.9101^2 = 1232.28 mm4
I2 = 1135.12 + 1232.28 = 2367.4 mm4
Total I = I1 + I2 = 1576.854 + 2367.4 = 3944.254 mm4
I = Ic + Ad2Now find Ixx.
Finding Ixx for Composite Shape (p368)
Tuesday, 13 March 2012
5:38 PM
Area-Moment Page 7
Element A y Ay Ic d Ad2 I
mm2 mm mm3 mm4 mm mm4 mm4
1 67.2 2.1 141.12 98.784 4.6899 1478.07 1576.85
2 80.6 10.7 862.42 1135.12 3.9101 1232.28 2367.4
Total 147.8 n.a. 1003.54 n.a. n.a. n.a. 3944.254
Centroid yc = 6.7899
Ixx
I = Ic + Ad2
Based on centroidal axis
Centroid
yc = (Ay)/(A)
Based on temporary axis
Area-Moment Page 8
Q13: If b1 = 14 , h1 = 4.2 , b2 =
3.9 and h2 = 25 mm. (a) Find Yc.
(Y coordinate of Centroid)
y1 = 4.2/2 = 2.1 mm
y2 = 25/2 = 12.5 mm
yc = (Ay)/(A)
A1= 14*4.2 = 58.8 mm2
A2 = 3.9*25 = 97.5 mm2
Element A y Ay
mm2 mm mm3
1 58.8 2.1 123.48
2 97.5 12.5 1218.75
Total 156.3 n.a. 1342.23
Centroid yc = 8.5875
A1 + A2 = 58.8+97.5 = 156.3
A1*y1 = 58.8*2.1 = 123.48
A2*y2 = 97.5*12.5 = 1218.75
A1y1+A2y2 = 123.48+1218.75=1342.23
Yc = Ay/A = 1342.23/156.3 = 8.5875
Element A y Ay Ic d Ad2 I
mm2 mm mm3 mm4 mm mm4 mm4
1 58.8 2.1 123.48 86.436 6.4875 2474.75 2561.19
2 97.5 12.5 1218.75 5078.125 3.9125 1492.50 6570.62
Total 156.3 n.a. 1342.23 n.a. n.a. n.a. 9131.81
Centroid yc = 8.5875
I = Ic + Ad2
Ic1 = 14*4.2^3/12 = 86.436
Ic2 = 3.9*25^3/12 = 5078.125
d1 = 8.5875-2.1 = 6.4875
d2 = 12.5 - 8.5875 = 3.9125
Ad12 = 58.8*6.4875^2 = 2474.75
Ad22 = 97.5*3.9125^2 = 1492.4965
I1 = 86.436+2474.75 = 2561.19
I2 = 5078.125+1492.4965 = 6570.6215
Ixx = 2561.19+6570.6215 = 9131.8115
Full Centroid Question
Tuesday, 26 March 2013
5:58 PM
Area-Moment Page 9

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Area moment

  • 1. Second Moment of Area. Ixx = Second moment of Area about the x-x axis Ixx = bh3/12 = 6.9*13^3/12 = 1263.275 mm4 Iyy = hb3/12 = 13*6.9^3/12 = 355.8848 mm4 Do outside (solid shaft) Ixx = *D4/64 = Pi*13^4/64 = 1401.985 mm4 Do inside (the hole) ID = 13-2*1.7 = 9.6 Ixx = *D4/64 = Pi*9.6^4/64 = 416.922 mm4 Net Ixx = 1401.985 - 416.922 = 985.063 mm4 Second Moment of Area Tuesday, 26 July 2011 8:35 PM Area-Moment Page 1
  • 2. I = Ic + Ad2 I = The second moment of area about non-centroidal plane Ic = The second moment of area about its own centroid A = Area of region d = Distance from centroid to new plane Ic = bh3/12 = 18*4.9^3/12 = 176.4735 mm4 I = Ic + Ad2 = 176.4735 + (18*4.9)*(6.2^2) = 3566.9 mm4 Split into 3 regions (elements) E1 & E3: Top and bottom. E2 is middle web. height = (h1-h2)/2 = (26-15)/2 = 5.5 mm d1 = 15/2+5.5/2 = 10.25 mm I1 = Ic1 + Ad2 Ic1 = bh3/12 = 14*5.5^3/12 = 194.1042 mm4 Ad2 = (14*5.5)*(10.25^2) = 8089.8125 mm4 I1 = 194.1042 + 8089.8125 = 8283.9167 mm4 E3 is same as E1. E2: I1 = Ic1 + Ad2 Ic2 = bh3/12 = 5.8*15^3/12 = 1631.25 mm4 Total I = I1+I2+I3 = 8283.917 + 8283.917 +1631.25 = 18199.08 mm4 Parallel Axis Theroem Tuesday, 13 March 2012 5:24 PM Area-Moment Page 2
  • 3. Not so easy if the compound shape is not symmetrical… For example, this "T" beam. Where is the Neutral Plane (x----x)? Need to find the centroid. Well, at least the y coordinate of the centroid, yc. Can we split this shape into elements that have the same neutral axis? Yes! E1: I=bh3/12 = 14*26^3/12 = 20505.33 mm4 E2: I=bh3/12 I = 4.1*15^3/12 = 1153.125 mm4 Ixx = I1 - I2 - I3 = 20505.33 - 1153.125 - 1153.125 = 18199.08 mm4 b = (14-5.8)/2 = 4.1 mm This is a great shortcut if you can split up into elements that all have the same centroidal axis. Area-Moment Page 3
  • 4. yc = (Ay)/(A) xc = (Ax)/(A) y1 = 412/2 = 206 mm y2 = (615-412)/2 + 412 = 513.5 mm A1 = 335*412 = 138020 mm2 A2 = 130*(615-412) = 26390 mm2 A1y1 = 138020*206 = 28432120 A2y2 = 26390*513.5 = 13551265 (Ay) = 28432120+13551265 = 41983385 (A) = 138020+26390 = 164410 = 255.3579 mm yc = (Ay)/(A) = 41983385/164410 Element A y Ay x Ax mm2 mm mm3 mm mm3 1 138020 206 28432120 167.5 23118350 2 26390 513.5 13551265 270 7125300 Total 164410 41983385 30243650 Centroid 255.35786 183.9526 The centroid is where this piece of polycarbonate will balance. Finding Centroid Wednesday, 27 March 2013 12:06 PM Area-Moment Page 4
  • 5. = 255.3579 mm yc = (Ay)/(A) = 41983385/164410 The centroid is where this piece of polycarbonate will balance. Element A y Ay Ic d Ad^2 I mm2 mm mm3 mm4 mm mm4 mm4 1 138020 206 28432120 1952338907 49.35786 336244032 2288582939 2 26390 513.5 13551265 90625459.2 258.1421 1.759E+09 1849185570 Total 164410 41983385 NO! NO! 4137768509 Centroid 255.35786 4137.76851 E6mm4 I = Ic + Ad2 Finding Total Ixx Wednesday, 27 March 2013 12:06 PM Area-Moment Page 5
  • 6. I = Ic + Ad2 I = The second moment of area about non-centroidal plane Ic = The second moment of area about its own centroid A = Area of region d = Distance from centroid to new plane yc = (Ay)/(A) yc = (Ay)/(A) Element 1: A1 = 16*4.2 = 67.2 mm2 y1 = 4.2/2 = 2.1 mm A1y1 = 67.2*2.1 = 141.12 mm3 Element 2: A2 = 6.2*13 = 80.6 mm2 y2 = 4.2 + 13/2 = 10.7 mm A2y2 = 80.6*10.7 = 862.42 mm3 = 1003.54 mm3 (Ay) = A1y1 + A2y2 = 141.12+ 862.42 (A) = 67.2+80.6 = 147.8 mm2 yc = (Ay)/(A) = 1003.54/147.8 = 6.7899mm Table format Element A y Ay x Ax mm2 mm mm3 mm mm3 1 67.2 2.1 141.12 8 537.6 2 80.6 10.7 862.42 3.1 249.86 Total 147.8 n.a. 1003.54 n.a. 787.46 Centroid 6.7899 5.328 yc = (Ay)/(A) xc = (Ax)/(A) Centroid Finding Centroid for Composite Shape (p368) Tuesday, 13 March 2012 5:38 PM Area-Moment Page 6
  • 7. I = Ic + Ad2 I = The second moment of area about non-centroidal plane Ic = The second moment of area about its own centroid A = Area of region d = Distance from centroid to new plane yc = (Ay)/(A) xc = (Ax)/(A) Table format (Centroid) Element A y Ay mm2 mm mm3 1 67.2 2.1 141.12 2 80.6 10.7 862.42 Total 147.8 X 1003.54 Centroid 6.7899 Working Ic1 = 16*4.2^3/12 = 98.784 mm4 Ic2 = 6.2*13^3/12 = 1135.1167 mm4 d1 = yc - h1/2 = 6.7899-2.1 = 4.6899 mm d2 = 10.7 - 6.7899 = 3.9101 mm A1d1 2 = 67.2 * 4.6899^2 = 1478.0749 mm4 I1= 98.784 + 1478.07 = 1576.854 mm4 A2d2 2 = 80.6 *3.9101^2 = 1232.28 mm4 I2 = 1135.12 + 1232.28 = 2367.4 mm4 Total I = I1 + I2 = 1576.854 + 2367.4 = 3944.254 mm4 I = Ic + Ad2Now find Ixx. Finding Ixx for Composite Shape (p368) Tuesday, 13 March 2012 5:38 PM Area-Moment Page 7
  • 8. Element A y Ay Ic d Ad2 I mm2 mm mm3 mm4 mm mm4 mm4 1 67.2 2.1 141.12 98.784 4.6899 1478.07 1576.85 2 80.6 10.7 862.42 1135.12 3.9101 1232.28 2367.4 Total 147.8 n.a. 1003.54 n.a. n.a. n.a. 3944.254 Centroid yc = 6.7899 Ixx I = Ic + Ad2 Based on centroidal axis Centroid yc = (Ay)/(A) Based on temporary axis Area-Moment Page 8
  • 9. Q13: If b1 = 14 , h1 = 4.2 , b2 = 3.9 and h2 = 25 mm. (a) Find Yc. (Y coordinate of Centroid) y1 = 4.2/2 = 2.1 mm y2 = 25/2 = 12.5 mm yc = (Ay)/(A) A1= 14*4.2 = 58.8 mm2 A2 = 3.9*25 = 97.5 mm2 Element A y Ay mm2 mm mm3 1 58.8 2.1 123.48 2 97.5 12.5 1218.75 Total 156.3 n.a. 1342.23 Centroid yc = 8.5875 A1 + A2 = 58.8+97.5 = 156.3 A1*y1 = 58.8*2.1 = 123.48 A2*y2 = 97.5*12.5 = 1218.75 A1y1+A2y2 = 123.48+1218.75=1342.23 Yc = Ay/A = 1342.23/156.3 = 8.5875 Element A y Ay Ic d Ad2 I mm2 mm mm3 mm4 mm mm4 mm4 1 58.8 2.1 123.48 86.436 6.4875 2474.75 2561.19 2 97.5 12.5 1218.75 5078.125 3.9125 1492.50 6570.62 Total 156.3 n.a. 1342.23 n.a. n.a. n.a. 9131.81 Centroid yc = 8.5875 I = Ic + Ad2 Ic1 = 14*4.2^3/12 = 86.436 Ic2 = 3.9*25^3/12 = 5078.125 d1 = 8.5875-2.1 = 6.4875 d2 = 12.5 - 8.5875 = 3.9125 Ad12 = 58.8*6.4875^2 = 2474.75 Ad22 = 97.5*3.9125^2 = 1492.4965 I1 = 86.436+2474.75 = 2561.19 I2 = 5078.125+1492.4965 = 6570.6215 Ixx = 2561.19+6570.6215 = 9131.8115 Full Centroid Question Tuesday, 26 March 2013 5:58 PM Area-Moment Page 9