Determine the product of inertia of the beams crosssectional

Determine the product of inertia of the beam\'s cross-sectional area with re to the x and y axes that have their origin located at the centroid, C.

Solution

Equation of shorter line is y = x+ 3

Equation of longer line is y = -x + 9

The shape can be divided into 2 smaller shapes;

a triangle between lines x = 3, y = 0 and y = -x + 9

a trapezium between lines x = 0, x = 3, y = 0 and y = x + 3

Ixy = Integral xy dA........where x and y are coordiantes of elemental area dA

For the triangle taking vertical elemental strips, x = x, y = (-x + 9)/2, dA = (-x + 9)dx

Ixy = Integral [x * (-x + 9)/2 * (-x + 9)dx].......x = 3 to 9

= Integral [(x³ - 18x² + 81x)/2 dx]......x = 3 to 9

= [x⁴/8 - 3x³ + 81x²/4]....x = 3 to 9

= (9⁴ - 3⁴)/8 - 3(9³ - 3³) + 81*(9² - 3²)/4

= 810 - 2106 + 1458

= 162 in⁴

For the trapezium taking vertical elemental strips, x = x, y = (x + 3)/2, dA = (x + 3)dx

Ixy = Integral [x * (x+3)/2 (x+3)dx ]..........x = 0 to 3

= Integral (x3 + 6x2 + 9x)/2 dx...........x = 0 to 3

= (x⁴/8 + x³ + 9x²/4)..........x = 0 to 3

= (3⁴ - 0)/8 + (3³ - 0) + 9*(3² - 0)/4

= 57.375 in⁴

Total product of inertia = 162 + 57.375 = 219.375 in⁴

Now we have to find centroid C.

xc = (Integral x dA) / A

yc = (Integral y dA) / A

For the triangle:

area A = 1/2*6*6 = 18 in²

xc = Integral (x *(-x + 9)dx ) / 18

xc = (-x³ /3 + 9x²/2)/18.............x = 3 to 9

xc = [(-9³ + 3³)/3 + 9*(9² - 3²)/2] /18

xc = 5 in

yc = Integral ((-x + 9)/2 *(-x + 9)dx ) / 18

yc = Integral (x² - 18x + 81) dx / 36.........x = 3 to 9

yc = (x³ /3 - 9x² + 81x) /36.......x = 3 to 9

yc = [(9³ - 3³) /3 - 9*(9² - 3²) + 81*(9-3)] / 36

yc = 2 in

For the trapezium:

area A = 1/2 *3* (3+6) = 13.5 in²

xc = Integral (x * (x + 3)dx) / 13.5.....x = 0 to 3

xc = (x³ /3 + 3x² /2) / 13.5......x = 0 to 3

xc = (3³ /3 + 3*3² /2) / 13.5

xc = 1.67 in

yc = Integral ((x + 3)/2 (x + 3)dx) / 13.5........x = 0 to 3

yc = Integral (x² + 6x + 9) /2 / 13.5..........x = 0 to 3

yc = (x³ /3 + 3x² + 9x) / 27..........x = 0 to 3

yc = (3³ /3 + 3*3² + 9*3) / 27

yc = 2.33 in

Total area A = 18 + 13.5 = 31.5 in²

Centroid of lamina Xc = [(xc*A)triangle + (xc*A)trapezium ] / A

Xc = [(5*18) + (1.67*13.5)] / 31.5 = 3.571 in

Centroid of lamina Yc = [(yc*A)triangle + (yc*A)trapezium ] / A

Yc = [(2*18) + (2.33*13.5)] / 31.5 = 2.143 in

By parallel axis theorem, Product of inertia about centroidal axes = Ixy - Xc*Yc*A

= 219.375 - 3.571* 2.143 * 31.5

= -21.67 in⁴