Engineering, 04.02.2021 02:50 bigb290

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Answer from: graceshort7orepzg

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

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;

$\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}$

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;

$\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552$

The maximum deflection of the beam, $\delta _{max}$ = 2.55552 inches

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