True or False: As the axle rotates, the wheel moves a greater distance then the axle, so it increases the speed but also turns with less force.

Answer:

1.[tex]I_{xc}[/tex] = 7.161458[tex]\overline 3[/tex] in.⁴

[tex]I_{yc}[/tex] = 36.661458[tex]\overline 3[/tex] in.⁴

Iₓ = 28.6458[tex]\overline 3[/tex] in.⁴

[tex]I_y[/tex] = 138.6548[tex]\overline 3[/tex] in.⁴

2. [tex]I_{xc}[/tex] = 114.[tex]\overline 3[/tex] in.⁴

[tex]I_{yc}[/tex] = 37.[tex]\overline 3[/tex] in.⁴

Iₓ = 457.[tex]\overline 3[/tex] in.⁴

[tex]I_y[/tex] = 149.[tex]\overline 3[/tex] in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

[tex]I_{xc}[/tex] = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458[tex]\overline 3[/tex]

[tex]I_{xc}[/tex] = 7.161458[tex]\overline 3[/tex] in.⁴

[tex]I_{yc}[/tex] = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458[tex]\overline 3[/tex]

[tex]I_{yc}[/tex] = 36.661458[tex]\overline 3[/tex] in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458[tex]\overline 3[/tex]

Iₓ = 28.6458[tex]\overline 3[/tex] in.⁴

[tex]I_y[/tex] = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548[tex]\overline 3[/tex]

[tex]I_y[/tex] = 138.6548[tex]\overline 3[/tex] in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

[tex]I_{xc}[/tex] = b·h³/12 = 4 × 7³/12 = 114.[tex]\overline 3[/tex]

[tex]I_{xc}[/tex] = 114.[tex]\overline 3[/tex] in.⁴

[tex]I_{yc}[/tex] = h·b³/12 = 7 × 4³/12 = 37.[tex]\overline 3[/tex]

[tex]I_{yc}[/tex] = 37.[tex]\overline 3[/tex] in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457.[tex]\overline 3[/tex]

Iₓ = 457.[tex]\overline 3[/tex] in.⁴

[tex]I_y[/tex] = h·b³/3 = 2.5 × 5.5³/3 = 149.[tex]\overline 3[/tex]

[tex]I_y[/tex] = 149.[tex]\overline 3[/tex] in.⁴

3. The deflection, [tex]\delta _{max}[/tex], of a simply supported beam having a point load at the center is given as follows;

[tex]\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}[/tex]

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, [tex]I_{xc}[/tex] = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

[tex]\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552[/tex]

The maximum deflection of the beam, [tex]\delta _{max}[/tex] = 2.55552 inches