Answers
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