Question

10–29. determine y, which locates the centroidal axis x for the cross-sectional area of the t-beam, and then find the moments of inertia ix and iy.

Answer 1

The centroidal axis (y) for the cross-sectional area of the T-beam is located at
`\( y = (h_1^2b_1 - h_2^2b_2)/(h_1b_1 - h_2b_2) \)`, where
`\( h_1 \)` and
`\( b_1 \)` are the height and width of the upper flange, and
`\( h_2 \)` and
`\( b_2 \)` are the height and width of the stem. The moments of inertia
`\( I_x \)` and
`\( I_y \)` are calculated based on the centroidal axis.

Step-by-step explanation:

In the given T-beam, the centroidal axis (y) is determined using the formula
`\( y = (h_1^2b_1 - h_2^2b_2)/(h_1b_1 - h_2b_2) \)`, where
`\( h_1 \)` and
`\( b_1 \)` are the dimensions of the upper flange, and
`\( h_2 \)` and
`\( b_2 \)` are the dimensions of the stem. This formula represents the centroidal axis as the weighted average of the individual centroids of the flange and the stem, considering their respective dimensions.

To calculate the moments of inertia
`\( I_x \)` and
`\( I_y \)`, the parallel axis theorem is often applied. The formula for
`\( I_x \) is \( I_x = I_(x1) + A_1d_(1y)^2 + I_(x2) + A_2d_(2y)^2 \)`, where
`\( I_(x1) \)` and
`\( I_(x2) \)` are the moments of inertia of the flange and the stem about their own centroidal axes, and
`\( A_1 \)` and
`\( A_2 \)` are their respective areas. The distances
`\( d_(1y) \)` and
`\( d_(2y) \)` are the distances between their centroidal axes and the overall centroidal axis (y). A similar formula is used for
`\( I_y \)`, considering the distances along the x-axis.

In summary, the centroidal axis and moments of inertia are crucial parameters in structural engineering, influencing the beam's behavior under different loading conditions. These calculations help engineers design and analyze T-beams for optimal performance and safety.

Answer 2

The value of y that locates the centroidal axis is 52.5 mm

How to determine the value of y?

Given data:

Area
`A_1` = 20 mm x 150 mm = 3000 mm²

Area
`A_2` = 150 mm x 20 mm = 3000 mm²

Calculation of Centroid (Ye):

Centroid of A1 (Ye₁) = 150 mm / 2 = 75 mm

Centroid of A2 (Yc₂) = 150 mm + 20 mm / 2 = 160 mm

Centroid of the combined area (Yc):

`\[ Yc = (A_1 * Yc_1 + A_2 * Yc_2)/(A_1 + A_2) \]`

`\[ Yc = (3000 * 75 + 3000 * 160)/(3000 + 3000) \]`

`\[ Yc = (225000 + 480000)/(6000) \]`

`\[ Yc = (705000)/(6000) \]`

`\[ Yc = 117.5 \, \text{mm} \]`

The value of y locating the centroidal axis:

`\[ y = 150 + 20 - 117.5 = 52.5 \, \text{mm} \]`

Therefore, the value of y that locates the centroidal axis is 52.5 mm.

See image below for missing part of the question.