Question

The mass density of the material of a $\text{T}$ joint is $2\text{kg/m}$. The dimensions of this joint are shown in the figure.


What is the approximate moment of inertia ($\text{MI}$) of the $\text{T}$ joint about an axis passing through their common joining point and perpendicular to the plane of structure?

• A. $120{\text{kg-m}}^2$
• B. $240{\text{kg-m}}^2$
• C. $362{\text{kg-m}}^2$
• D. $426{\text{kg.m}}^2$

Answer 1

The correct option is C $362{\text{kg-m}}^2$

The mass density of the material of the T joint,
$\lambda =2\text{kg/m}$
Length of section $AB$:
$L_{AB}=5\text{m}$
Mass of section $AB$
$\therefore m_{AB}=\lambda \times L_{AB}=2\times 5=10\text{kg}$

Similarly, $L_{CD}=8\text{m}$
$m_{CD}=\lambda \times L_{CD}=2\times 8=16\text{kg}$
$\Rightarrow$ For section $AB$ the axis $Y{Y}^{\prime }$ is analogous to axis passing through mid point of a rod.
$\therefore$ $\text{MI}$ of section $AB$ about the axis,
${\text{MI}}_{AB}=\frac{m_{AB}\times L_{AB}^2}{12}$
$\Rightarrow {\text{MI}}_{AB}=\frac{10\times 5^2}{12}=\frac{10\times 25}{12}=\frac{125}{6}{\text{kg-m}}^2$

$\Rightarrow$ The axis $Y{Y}^{\prime }$ is analogous to the axis passing through one end of a rod.
$\therefore {\text{MI}}_{CD}=\frac{m_{CD}\times L_{CD}^2}{3}=\frac{16\times 8^2}{3}$

$\Rightarrow {\text{MI}}_{\text{CD}}=\frac{1024}{3}{\text{kg-m}}^2$

Net $\text{MI}$ of $\text{T}$ joint is:
${\text{MI}}_{Net}={\text{MI}}_{AB}+{\text{MI}}_{CD}$
$\therefore {\text{MI}}_{Net}=\frac{125}{6}+\frac{1024}{3}\approx 362{\text{kg-m}}^2$