Question

Could I please get help with this​

Answer 1

1.
`I_(xc)` = 7.161458
`\overline 3` in.⁴

`I_(yc)` = 36.661458
`\overline 3` in.⁴

Iₓ = 28.6458
`\overline 3` in.⁴

`I_y` = 138.6548
`\overline 3` in.⁴

2.
`I_(xc)` = 114.
`\overline 3` in.⁴

`I_(yc)` = 37.
`\overline 3` in.⁴

Iₓ = 457.
`\overline 3` in.⁴

`I_y` = 149.
`\overline 3` in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Step-by-step 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;

`I_(xc)` = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458
`\overline 3`

`I_(xc)` = 7.161458
`\overline 3` in.⁴

`I_(yc)` = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458
`\overline 3`

`I_(yc)` = 36.661458
`\overline 3` 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
`\overline 3`

Iₓ = 28.6458
`\overline 3` in.⁴

`I_y` = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548
`\overline 3`

`I_y` = 138.6548
`\overline 3` 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;

`I_(xc)` = b·h³/12 = 4 × 7³/12 = 114.
`\overline 3`

`I_(xc)` = 114.
`\overline 3` in.⁴

`I_(yc)` = h·b³/12 = 7 × 4³/12 = 37.
`\overline 3`

`I_(yc)` = 37.
`\overline 3` in.⁴

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

Iₓ = b·h³/3 = 4 × 7³/3 = 457.
`\overline 3`

Iₓ = 457.
`\overline 3` in.⁴

`I_y` = h·b³/3 = 2.5 × 5.5³/3 = 149.
`\overline 3`

`I_y` = 149.
`\overline 3` in.⁴

3. The deflection,
`\delta _(max)`, of a simply supported beam having a point load at the center is given as follows;

`\delta_(max) = (W * L^3)/(48 * E * I)`

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,
`I_(xc)` = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

`\delta_(max) = (750 * 264^3)/(48 * 10,000,000 * 11.25) = 2.55552`

The maximum deflection of the beam,
`\delta _(max)` = 2.55552 inches