10-105 Determine the moment of inertia of the compos-ite body shown in Fig. P10-105 with respect to the y-axisshown on the figure. The specific weiglıt of the material is175 lb/ft.3 in.8 in.6 in.6 in.6 in. .6 in.5 in.8 in.5 in.12 in.3 in.Fig. P10-105

Answered: 10-105 Determine the moment of inertia…

Related questions

Question

10-105 Determine the moment of inertia of the compos-
ite body shown in Fig. P10-105 with respect to the y-axis
shown on the figure. The specific weiglıt of the material is
175 lb/ft.
3 in.
8 in.
6 in.
6 in.
6 in. .
6 in.
5 in.
8 in.
5 in.
12 in.
3 in.
Fig. P10-105

Expert Solution

Concept and Principle: Moment of Inertia.

Moment of inertia it determines the torque needed for a desired angular acceleration about rotational axis. it depends on the mass of the body chosen and the axis in consideration. Mathematically moment of inertia I is defined as the ratio of the net angular momentum L of a system to its angular velocity ω around a principal axis that is,

$I = \frac{L}{ω}$

Parallel Axes Theorem:

The moment of inertia of a body about an axis parallel to the body passing through its center is equal to the sum of moment of inertia of body about the axis passing through the center and product of mass of the body times the square of distance between the two axes.

$I = I_{c} + {Mh}^{2} I : the moment of inertia of the body I_{c} : the moment of inertia about the center M : the mass of the body h^{2} : the square of the distance between the two axes$

Now the given problem is of a composite area. For such a system to find moment of inertia around and axis we need to,

Identify simple shaped subareas in the composite.
Determine the distance from centroid to each of those subareas.
Determine moment of inertia of each subarea
Apply Parallel Axes theorem on each subarea
Add or subtract (negative subareas) the moments of inertia from the last step.

Step 1: Selecting subparts

The lines in section 2 shows the hole

Calculation: Moment of inertia of section 1

Mass of section 1 is,

$\begin{array}{rcl} m & = & \frac{W}{g} \\ = & \frac{ρV}{g} — — — — \to (1) \end{array}$

Now volume V of section1 is given by,

$\begin{array}{rcl} V & = & bhL \\ b & = & \frac{12}{12} ft \\ h & = & \frac{8}{12} ft \\ L & = & \frac{3}{12} ft \\ V & = & (\frac{12}{12} ft) \times (\frac{8}{12} ft) \times (\frac{3}{12} ft) \\ = & 0.166 {ft}^{3} \end{array}$

Now substituting in eq(1)

$\begin{array}{rcl} m & = & \frac{ρV}{g} \\ ρ & = & 175 lb . {ft}^{- 3} \\ g & = & 32.2 \\ m & = & \frac{(175 lb . {ft}^{- 3}) \times (0.166 {ft}^{3})}{32.2} \\ = & 0.902 slugs \end{array}$

Now we calculate moment of inertia of section 1 along Y axis,

$\begin{array}{rcl} I_{y 1} & = & I_{yG} + {d_{y}}^{2} m \\ = & \frac{1}{12} m (b^{2} + h^{2}) + {(z_{G})}^{2} m \\ I_{yG} : Moment of inertia of section 1 at its centroid about y axis \\ z_{G} : Distance from centriod to z axis . \end{array}$

Now substituting,

$\begin{array}{rcl} m & = & 0.902 slugs \\ z_{G} & = & \frac{4}{12} ft \\ I_{y 1} & = & \frac{1}{12} m (b^{2} + h^{2}) + {(z_{G})}^{2} m \\ = & \frac{1}{12} \times (0.902 slugs) \times ({(\frac{12}{12} ft)}^{2} + {(\frac{8}{12} ft)}^{2}) + {(\frac{4}{12} ft)}^{2} \times (0.902 slugs) \\ = & 0.209 slug . {ft}^{2} \\ I_{y 1} = = 0.209 slug . {ft}^{2} \end{array}$

Calculation: Moment of inertia of section 2

Now volume V of section 2 is given by,

$\begin{array}{rcl} V & = & bhL \\ b & = & \frac{12}{12} ft \\ h & = & \frac{3}{12} ft \\ L & = & \frac{15}{12} ft \\ V & = & (\frac{12}{12} ft) \times (\frac{3}{12} ft) \times (\frac{15}{12} ft) \\ = & 0.3125 {ft}^{3} \end{array}$

Now substituting in eq(1)

$\begin{array}{rcl} m & = & \frac{ρV}{g} \\ ρ & = & 175 lb . {ft}^{- 3} \\ g & = & 32.2 \\ m & = & \frac{(175 lb . {ft}^{- 3}) \times (0.3125 {ft}^{3})}{32.2} \\ = & 1.69 slugs \end{array}$

Now we calculate moment of inertia of section 2 along Y axis,

$I_{y 2} = \frac{1}{12} m (b^{2} + h^{2}) + {(z_{G})}^{2} m$

Now substituting,

$\begin{array}{rcl} m & = & 1.69 slugs \\ z_{G} & = & \frac{1.5}{12} ft \\ I_{y 2} & = & \frac{1}{12} m (b^{2} + h^{2}) + {(z_{G})}^{2} m \\ = & \frac{1}{12} \times (1.69 slugs) \times ({(\frac{12}{12} ft)}^{2} + {(\frac{3}{12} ft)}^{2}) + {(\frac{1.5}{12} ft)}^{2} \times (1.69 slugs) \\ = & 0.1769 slug . {ft}^{2} \\ I_{y 2} = = 0.1769 slug . {ft}^{2} \end{array}$

Step by step

Solved in 8 steps with 1 images

10-105 Determine the moment of inertia of the compos- ite body shown in Fig. P10-105 with respect to the y-axis shown on the figure. The specific weiglıt of the material is 175 lb/ft. 3 in. 8 in. 6 in. 6 in. 6 in. . 6 in. 5 in. 8 in. 5 in. 12 in. 3 in. Fig. P10-105