Moment of Inertia of a Rectangle

Click and drag the blue dots in the corner of the rectangle to change its shape and size.[br][br]Definitions:[br][br][math]I'_x[/math]The moment of inertia of the area about its centroidal x-axis, [i]x’[/i][br][br][math]I'_y[/math]The moment of inertia of the area about its centroidal y-axis, [i]y’[/i][br][br][math]I_x[/math]The moment of inertia of the area about the x-axis[br][br][math]I_y[/math]The moment of inertia of the area about the y-axis[br][br]

Fundamentally, the moment of inertia of a given area about a given axis tells you...

The ability of the area to resist bending about the axis The ability of the area to resist axial compression The ability of the area to resist shear forces The ability of the area to bend about the axis

If the height [i]h [/i]of the rectangle is doubled while the base [i]b[/i] stays the same, how will the moment of inertia about its centroidal [i]x’[/i] axis change?

It will stay the same It will increase by a factor of 8 It will increase by a factor of 4 It will double

If the base [i]b[/i] of the rectangle is tripled while the height [i]h[/i] stays the same, how will the moment of inertia about its centroidal [i]x’[/i] axis change?

It will increase by a factor of 4 It will increase by a factor of 3 It will stay the same It will double

If the height [i]h [/i]of the rectangle is doubled while the base [i]b[/i] stays the same, how will the moment of inertia about its centroidal [i]y’[/i] axis change? [br]

It will stay the same It will increase by a factor of 8 It will increase by a factor of 4 It will double

If the rectangle is translated one unit to the right with no change in [i]b[/i] or [i]h[/i], how will the moment of inertia about its centroidal [i]x’[/i] axis change?

It will increase by a factor of 4 It will increase by a factor of 8 It will stay the same It will double

For a rectangle with [i]b [/i]= 6 and [i]h[/i] = 4 mm, which is the correct expression for the moment of inertia about the centroidal [i]x’[/i] axis?[br]

[math]I'_x=\frac{1}{12}\cdot4\cdot6^3[/math] [math]I'_x=\frac{1}{12}\cdot6\cdot4^4[/math] [math]I'_x=\frac{1}{12}\cdot4\cdot6^4[/math] [math]I'_x=\frac{1}{12}\cdot6\cdot4^3[/math] [math]I'_x=\frac{1}{6}\cdot6\cdot4^3[/math]

For a rectangle with [i]b [/i]= 6 and [i]h[/i] = 4 mm, which is the correct expression for the moment of inertia about the centroidal y[i]’[/i] axis?[br]

[math]I'_y=\frac{1}{12}\cdot6\cdot4^3[/math] [math]I'_y=\frac{1}{12}\cdot6\cdot4^4[/math] [math]I'_y=\frac{1}{12}\cdot4\cdot6^4[/math] [math]I'_y=\frac{1}{12}\cdot4\cdot6^3[/math] [math]I'_y=\frac{1}{6}\cdot6\cdot4^3[/math]

For a square with [i]b [/i]= 4 and [i]h[/i] = 4 mm, which is the correct expression for the moment of inertia about the centroidal [i]x’[/i] axis?

[math]I'_x=\frac{1}{2}\cdot4\cdot4^3[/math] [math]I'_x=\frac{1}{12}\cdot4\cdot4^2[/math] [math]I'_x=\frac{1}{12}\cdot4\cdot4^3[/math] [math]I'_x=\frac{1}{6}\cdot4\cdot4^3[/math] [math]I'_x=\frac{1}{12}\cdot4^2[/math]

Which rectangle will have a greater moment of inertia about its centroidal [i]x’[/i] axis?[br][br] Rectangle A: [i]b[/i] = 3, [i]h[/i] = 4[br][br] Rectangle B: [i]b[/i] = 9, [i]h[/i] = 2[br]

Rectangle A Rectangle B Not enough information Both are equal Neither

Which rectangle will have a greater moment of inertia about its centroidal y[i]’[/i] axis?[br][br] Rectangle A: [i]b[/i] = 3, [i]h[/i] = 4[br][br] Rectangle B: [i]b[/i] = 9, [i]h[/i] = 2[br]

Both are equal Not enough information Rectangle A Neither Rectangle B

Parallel-axis Theorem

The moment of inertia of a given cross-sectional area about any axis other than its own centroidal axis, depends on which of the following terms?

All of the above The moment of inertia of the area about a parallel axis going through its centroid The perpendicular distance between the two axes The total cross-sectional area of the shape

Why does [i]I[sub]y [/sub][/i]increase when the square is translated to the right?

The moment of inertia about the centroidal y' axis increased The cross-sectional area of the shape increased The perpendicular distance between the y and y' axis increased [i]I[sub]y[/sub] [/i]depends on the location of the squares corners

If a square is translated 2 units in the +x direction and 4 units in the +y direction with no change in its cross-sectional area, will this have a greater effect on [i]I[sub]x [/sub][/i]or [i]I[sub]y[/sub][/i]?

[i]I[sub]x[/sub][/i] [i]I[sub]y[/sub][/i] Neither will be impacted Both will increase by the same amount

For the above rectangle, what is the correct expression for the moment of inertia of the area about the y-axis?

[math]I_y=\frac{1}{12}\cdot4\cdot6^3+4\cdot6\cdot5^2[/math] [math]I_y=\frac{1}{12}\cdot6\cdot4^3+4\cdot6\cdot5^2[/math] [math]I_y=\frac{1}{12}\cdot4\cdot6^3+4\cdot6\cdot2^2[/math] [math]I_y=\frac{1}{12}\cdot4\cdot6^3+4\cdot6\cdot6^2[/math]

For the above rectangle, what is the correct expression for the moment of inertia of the area about the -axis?

[math]I_x=\frac{1}{12}\cdot6\cdot4^3+4\cdot6\cdot6^2[/math] [math]I_x=\frac{1}{12}\cdot4\cdot6^3+4\cdot6\cdot6^2[/math] [math]I_x=\frac{1}{12}\cdot6\cdot4^3+4\cdot6\cdot4^2[/math] [math]I_x=\frac{1}{12}\cdot6\cdot4^3+4\cdot6\cdot5^2[/math]

Moment of Inertia of a Rectangle

Parallel-axis Theorem

Informação: Moment of Inertia of a Rectangle