Lecture 28 Centroidal Axis Section Example Problem

Lecture 28
Example Problem

Centroidal Axis & the Transfer Formula

Given:
a built-up bridge girder section.
Determine:
its centroidal axis x-x and Ixx.

Lecture 28

Solution:
For this problem, before the combined Moment of Inertia can be calculated, the centroidal axis, x-x must be found. When a Moment of Inertia problem gets complex, it can be very useful to utilize a chart such as the one illustrated above to keep track of the many bits of information necessary to solve. As the problem is worked below, fill in the chart given in the above illustration.

Begin by filling in known quantites. In this case, A and Ic for both pieces. The values for the wide flange beam are given in the illustration (A = 24.8 in^2; Ic = 2850 in^4) and those for the bar are easily calcu lated (A = 12 in^2; Ic = 1/12(12in)(1in)^3 = 1 in^4). The total area equals 24.8 in^2 + 12 in^2 = 36.8 in ^2.

Next find 'y' for each piece. 'y' will be the distance between the centroid of each area and a common axis. Use the bottom edge of the wide flange section. For the purposes of this explaination, '1' will signify the bar and '2' will signify the wide fl ange section. Therefore, y1 will equal the total height of the wide flange section plus half the height of the bar or

y1 = 26.75in + .5in = 27.25in

y2 will equal half the depth of the wide flange section or

y2 = 26.75in/2 = 13.375in

Now Ay can be calculated for each part as well as the whole.

A1y1 = 12in^2 (27.25in) = 327 in^3
A2y2 = 24.8in^2 (13.375in) = 331.7 in^3
ATyT = 327 in^3 + 331.7 in^3 = 658.7 in^3

From this, yT, the centroidal axis can be found by dividing ATyT by AT or

658.7 in^3/36.8 in^2 = 17.9 in

This means that yT is 17.9in away from the previous common axis which was the bottom of the wide flange beam.

Now the centroidal axis can be used to calculate the final variable, d, the absolute distance between the centroidal axis of the whole and the centroidal axis of each of the parts.

d1 = 27.5in - 17.9in = 9.35 in
d2 = 13.375in - 17.9in = 4.525 in

Next, calculate Ad^2 and apply the transfer formula to find Ixx.

Ixx = Ic1 + A1d1^2 + Ic2 + A2d2^2
Ixx = 1 in^4 + 12in^2 (9.35in)^2 +2850 in^4 + 24.8in^2 (4.525in)^2
Ixx = 1050 in^4 + 3358 in^4
Ixx = 4408 in^4

In conclusion, the centroidal axis, x-x, is located 17.9in above the bottom of the wide flange beam, and the moment of inertia of the composite area equals 4408 in^4.

Lecture 28 Example Problem

Lecture 28
Example Problem