Question 6.5: Compute the moment of inertia of the tee shape in Figure 6–1...
Compute the moment of inertia of the tee shape in Figure 6–12 with respect to its centroidal axis.
Objective Compute the moment of inertia.
Given Shape shown in Figure 6–12
Analysis Use the General Procedure listed in this section. As step 1, divide the tee shape into two parts, as shown in Figure 6–13. Part 1 is the vertical stem and part 2 is the horizontal flange.
Results The following table summarizes the complete set of data used to compute the total moment of inertia with respect to the centroid of the tee shape. Some of the data are shown also in Figure 6–13. Comments are given here to show the manner of arriving at certain data:
| Part | A_{i} | y_{i} | A_{i} y_{i} | I _{i} | d_{i} | A_{i} d_{i}^{2} | I_{i}+A_{i} d_{i}^{2} |
| 1 | 4.0 mm² | 2.0 mm | 8.0 mm³ | 5.333 cm^{4} | 0.75 cm | 2.25 cm^{4} | 7.583 cm^{4} |
| 2 | 2.0 mm² | 4.25 mm | 8.50 mm³ | 0.042 cm^{4} | 1.50 cm | 4.50 cm^{4} | 4.542 cm^{4} |
| A_{T} = 6.0 mm² | Σ (A_{i}y_{i}) = 16.5 mm³ | I_{T} = 12.125 cm^{4} | |||||
| \bar{Y} = \frac{∑(A_{i}y_{i})}{A_{T}}= \frac{16.5 mm^{3}}{6.0 mm^{3}} = 2.75 mm | |||||||
Steps 2 and 3. The first three columns of the table give data for computing the location of the centroid using the technique shown in Section 6–3. Distances, y_{i} , are measured upward from the bottom of the tee. The result is \bar{Y} = 2.75 cm.
Step 4. Both parts are simple rectangles. Then, the values of I are
I_{1} = bh^{3}/12 = (1.0)(4.0)^{3}/12 = 5.333 cm^{4}
I_{2} = bh^{3}/12 = (4.0)(0.5)^{3}/12 = 0.042 cm^{4}
Step 5. Distances, d_{i} , from the overall centroid to the centroid of each part
d_{1} = \bar{Y} – y_{1} = 2.75 cm – 2.0 cm = 0.75 cm
d_{1} = y_{2} -\bar{Y} = 4.25 cm – 2.75 cm = 1.50 cm
Step 6. Transfer term for each part
A_{1}d_{1}^{2} = (4.0 cm^{2})(0.75 cm)^{2} = 2.25 cm^{4}
A_{2}d_{2}^{2} = (4.0 cm^{2})(0.75 cm)^{2} = 2.25 cm^{4}
Step 7. Total moment of inertia
I_{T} = I_{1} + A_{1}d_{1}^{2} + I_{2} + A_{2}d_{2}^{2}
= 5.333 cm^{4} + 2.25 cm^{4} + 0.042 cm^{4} + 4.50 cm^{4}
I_{T} = 12.125 cm^{4}
Comment Notice that the transfer terms contribute over half of the total value to the moment of inertia.
