Chapter 10.3, Problem 73P

(a)

To determine

Find the number of wood pieces.

Answer to Problem 73P

The number of wood pieces required to support the centric load is $\underset{\_}{3}$.

Explanation of Solution

Given information:

The effective length of the laminated column is $l_e=2.1\text{m}$.

The adjusted allowable stress for compression is ${\sigma }_C=7.7\text{MPa}$.

The adjusted modulus of elasticity is $E=5.4\text{GPa}$.

The centric load acting in the column is $P=52\text{kN}$.

The width of the laminated column is $b=150\text{mm}$.

The depth of each laminated column is $d=25\text{mm}$.

Calculation:

Find the cross sectional area (A) of the column using the equation.

$A=b\left(nd\right)$

Here, the width of the column is b, the number of pieces is n, and the depth of each piece is d.

Substitute 150 mm for b and 25 mm for d.

$\begin{array}{c}A=150\left(n\times 25\right)\\ =3,750n{\text{mm}}^2\end{array}$

Find the radio of the effective length to the dimension of the cross section $\left(\frac{L_e}{d}\right)$.

$\begin{array}{c}\frac{L_e}{d}=\frac{2.1\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{25n}\\ =\frac{84}{n}\end{array}$

Find the stress $\left({\sigma }_{CE}\right)$ using the equation.

${\sigma }_{CE}=\frac{0.822E}{{\left(\frac{L_e}{d}\right)}^2}$

Substitute 5.4 GPa for E and $\frac{84}{n}$ for $\frac{L_e}{d}$.

$\begin{array}{c}{\sigma }_{CE}=\frac{0.822\times 5.4\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{{\left(\frac{84}{n}\right)}^2}\\ =\frac{4,438.8n^2}{7,056}\end{array}$

Find the column stability factor $C_p$ using the Equation.

$C_p=\frac{1+\left({\sigma }_{CE}/{\sigma }_C\right)}{2c}-\sqrt{{\left[\frac{1+\left({\sigma }_{CE}/{\sigma }_C\right)}{2c}\right]}^2-\frac{{\sigma }_{CE}/{\sigma }_C}{c}}$

Here, the allowable stress for compression grain is ${\sigma }_C$.

For Glue laminated column, the value of c is $c=0.9$.

Substitute $\frac{4,438.8n^2}{7,056}$ for ${\sigma }_{CE}$, 7.7 MPa for ${\sigma }_C$, and 0.9 for c.

$\begin{array}{c}{C}_p=\frac{1+\left(\frac{\left(\frac{4,438.8n^2}{7,056}\right)}{7.7}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{\left(\frac{4,438.8n^2}{7,056}\right)}{7.7}\right)}{2\times 0.9}\right]}^2-\frac{\frac{\left(\frac{4,438.8n^2}{7,056}\right)}{7.7}}{0.9}}\\ =\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\end{array}$

Calculate the allowable stress $\left({\sigma }_{all}\right)$ using the relation.

${\sigma }_{all}={\sigma }_C{C}_P$

Substitute 7.7 MPa for ${\sigma }_C$ and $\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}$ for $C_p$.

${\sigma }_{all}=7.7\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]$

Calculate the allowable stress $\left({\sigma }_{\text{all}}\right)$ in terms of allowable load (P) using the equation.

${\sigma }_{\text{all}}=\frac{P}{A}$

Substitute 52 kN for P and $3,750n{\text{mm}}^2$ for A.

$\begin{array}{c}{\sigma }_{\text{all}}=\frac{52\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{3,750n}\\ =\frac{52,000}{3,750n}\end{array}$

Substitute $7.7\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]$ for ${\sigma }_{all}$.

$\begin{array}{l}7.7\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]=\frac{52,000}{3,750n}\\ 7.7n\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]=\frac{52,000}{3,750}\end{array}$

Solve the equation using Solver tool in Excel.

The number of piece required is $n=2.94\simeq 3\text{no's}$

Therefore, the number of wood pieces required to support the centric load is $\underset{\_}{3}$.

(b)

To determine

Find the number of wood pieces.

Answer to Problem 73P

The number of wood pieces required to support the centric load is $\underset{\_}{5}$.

Explanation of Solution

Given information:

The effective length of the laminated column is $l_e=2.1\text{m}$.

The adjusted allowable stress for compression is ${\sigma }_C=7.7\text{MPa}$.

The adjusted modulus of elasticity is $E=5.4\text{GPa}$.

The centric load acting in the column is $P=108\text{kN}$.

The width of the laminated column is $b=150\text{mm}$.

The depth of each laminated column is $d=25\text{mm}$.

Calculation:

Find the cross sectional area (A) of the column using the equation.

$A=b\left(nd\right)$

Substitute 150 mm for b and 25 mm for d.

$\begin{array}{c}A=150\left(n\times 25\right)\\ =3,750n{\text{mm}}^2\end{array}$

Find the radio of the effective length to the dimension of the cross section $\left(\frac{L_e}{d}\right)$.

$\begin{array}{c}\frac{L_e}{d}=\frac{2.1\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}}{25n}\\ =\frac{84}{n}\end{array}$

Find the stress $\left({\sigma }_{CE}\right)$ using the equation.

${\sigma }_{CE}=\frac{0.822E}{{\left(\frac{L_e}{d}\right)}^2}$

Substitute 5.4 GPa for E and $\frac{84}{n}$ for $\frac{L_e}{d}$.

$\begin{array}{c}{\sigma }_{CE}=\frac{0.822\times 5.4\text{GPa}\times \frac{\text{1,000}\text{MPa}}{\text{1}\text{GPa}}}{{\left(\frac{84}{n}\right)}^2}\\ =\frac{4,438.8n^2}{7,056}\end{array}$

Find the column stability factor $C_p$ using the Equation.

$C_p=\frac{1+\left({\sigma }_{CE}/{\sigma }_C\right)}{2c}-\sqrt{{\left[\frac{1+\left({\sigma }_{CE}/{\sigma }_C\right)}{2c}\right]}^2-\frac{{\sigma }_{CE}/{\sigma }_C}{c}}$

For Glue laminated column, the value of c is $c=0.9$.

Substitute $\frac{4,438.8n^2}{7,056}$ for ${\sigma }_{CE}$, 7.7 MPa for ${\sigma }_C$, and 0.9 for c.

$\begin{array}{c}{C}_p=\frac{1+\left(\frac{\left(\frac{4,438.8n^2}{7,056}\right)}{7.7}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{\left(\frac{4,438.8n^2}{7,056}\right)}{7.7}\right)}{2\times 0.9}\right]}^2-\frac{\frac{\left(\frac{4,438.8n^2}{7,056}\right)}{7.7}}{0.9}}\\ =\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\end{array}$

Calculate the allowable stress $\left({\sigma }_{all}\right)$ using the relation.

${\sigma }_{all}={\sigma }_C{C}_P$

Substitute 7.7 MPa for ${\sigma }_C$ and $\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}$ for $C_p$.

${\sigma }_{all}=7.7\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]$

Calculate the allowable stress $\left({\sigma }_{\text{all}}\right)$ in terms of allowable load (P) using the equation.

${\sigma }_{\text{all}}=\frac{P}{A}$

Substitute 108 kN for P and $3,750n{\text{mm}}^2$ for A.

$\begin{array}{c}{\sigma }_{\text{all}}=\frac{108\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{3,750n}\\ =\frac{108,000}{3,750n}\end{array}$

Substitute $7.7\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]$ for ${\sigma }_{all}$.

$\begin{array}{l}7.7\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]=\frac{108,000}{3,750n}\\ 7.7n\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}-\sqrt{{\left[\frac{1+\left(\frac{4,438.8n^2}{54,331.2}\right)}{2\times 0.9}\right]}^2-\frac{4,438.8n^2}{48,898.08}}\right]=\frac{108,000}{3,750}\end{array}$

Solve the equation using Solver tool in Excel.

The number of piece required is $n=4.27\simeq 5\text{no's}$

Therefore, the number of wood pieces required to support the centric load is $\underset{\_}{5}$.