Chapter 4.7, Problem 112P

A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using an allowable stress of 600 psi, determine the largest compressive load P that can be applied at the center of the top section of the timber column as shown if (a) the column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed, (d) planks 1, 2, and 3 are removed, (e) all planks are removed.

Fig. P4.112

To determine

Find the largest compressive load P that can be applied at the center of the top section of the timber column.

Answer to Problem 112P

The largest compressive load P is $\underset{\_}{19.20\text{kips}}$.

Explanation of Solution

Given information:

The compressive load P is $16\text{kips}$.

The allowable stress $\left(\sigma \right)$ is $600\text{psi}$.

The width $\left(b_p\right)$ of the planks is $1\text{in}\text{.}$

The depth $\left(d_p\right)$ of the planks is $4\text{in}\text{.}$

The width $\left(b_t\right)$ of the timber is $4\text{in}\text{.}$

The depth $\left(d_t\right)$ of the timber is $4\text{in}\text{.}$

Calculation:

Sketch the centric loading as shown in Figure 1.

Refer to Figure 1.

Find the area of the timber section using the relation:

$A={\left(bd\right)}_{\text{planks}}+4{\left(bd\right)}_{\text{timber}}$ (1)

Substitute $1\text{in}\text{.}$ for $b_p$, $4\text{in}\text{.}$ for $d_p$, $4\text{in}\text{.}$ for $b_t$, and $4\text{in}\text{.}$ for $d_t$ in Equation (1).

$\begin{array}{c}A=\left(4\times 4\right)+4\left(1\times 4\right)\\ =16+16\\ =32{\text{in}}^{\text{2}}\end{array}$

Calculate the largest compressive load P using the relation:

$\begin{array}{l}\sigma =\frac{P}{A}\\ P=\sigma A\end{array}$ (2)

Substitute $600\text{psi}$ for $\sigma$ and $32{\text{in}}^{\text{2}}$ for $A$ in Equation (2).

$\begin{array}{c}P=600\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\times 32\\ =0.600\times 32\\ =19.20\text{kips}\end{array}$

Thus, the largest compressive load P is $\underset{\_}{19.20\text{kips}}$.

To determine

Find the largest compressive load P that can be applied at the center of the top section of the timber column without plank 1.

Answer to Problem 112P

The largest compressive load P that can be applied at the center of the top section of the timber column without plank 1 is $\underset{\_}{11.68\text{kips}}$.

Explanation of Solution

Calculation:

Sketch the Eccentric loading as shown in Figure 2.

Find the area of the timber section using the relation:

$A={\left(bd\right)}_{\text{planks}}+3{\left(bd\right)}_{\text{timber}}$ (3)

Substitute $1\text{in}\text{.}$ for $b_p$, $4\text{in}\text{.}$ for $d_p$, $4\text{in}\text{.}$ for $b_t$, and $4\text{in}\text{.}$ for $d_t$ in Equation (3).

$\begin{array}{c}A=\left(4\times 4\right)+3\left(1\times 4\right)\\ =16+12\\ =28{\text{in}}^{\text{2}}\end{array}$

Refer to Figure 2.

Find the centroid $\left(\overline{y}\right)$ of the section as shown below:

$\overline{y}=\frac{\sum A\overline{y}}{A}$ (4)

Substitute $\left(1\times 4\right){\text{in}}^{\text{2}}$ for $\sum A$ and $2.5\text{in}\text{.}$ for $\overline{y}$ in Equation (4).

$\begin{array}{c}\overline{y}=\frac{1\times 4\times 2.5}{28}\\ =0.35714\text{in}\text{.}\end{array}$

Refer to Figure 2.

Find the moment of inertia $I_1$ using the relation:

$I_1=\frac{b_1{d}_1^3}{12}+\left(bd\right){\left(\overline{y}-y_1\right)}^2$ (5)

Substitute $6\text{in}\text{.}$ for $b_1$, $4\text{in}\text{.}$ for $d_1$, $0.35714\text{in}$ for $\overline{y}$, and 0 for $y_1$ in Equation (5).

$\begin{array}{c}{I}_1=\frac{6\times 4^3}{12}+\left(6\times 4\right){\left(0.35714-0\right)}^2\\ =32+3.06\\ =35.06{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia $I_2$ using the relation:

$I_2=\frac{b_2{d}_2^3}{12}+\left(bd\right){\left(y_2-\overline{y}\right)}^2$ (6)

Substitute $4\text{in}\text{.}$ for $b_1$, $1\text{in}\text{.}$ for $d_1$, $0.35714\text{in}$ for $\overline{y}$, and $2.5$ for $y_2$ in Equation (6).

$\begin{array}{c}{I}_2=\frac{4\times 1^3}{12}+\left(4\times 1\right){\left(2.5-0.35714\right)}^2\\ =0.33+18.366\\ =18.69{\text{in}}^{\text{4}}\end{array}$

Find the total moment of inertia as follows:

$I=I_1+I_2$ (7)

Substitute $35.06{\text{in}}^{\text{4}}$ for $I_1$ and $18.69{\text{in}}^{\text{4}}$ for $I_2$ in Equation (7).

$\begin{array}{c}I=35.06+18.69\\ =53.762{\text{in}}^{\text{4}}\end{array}$

Calculate the largest compressive load P that can be applied at the center of the top section of the timber column without plank 1using the relation:

$\begin{array}{l}\sigma =P\left(\frac{1}{A}+\frac{ec}{I}\right)\\ P=\frac{\sigma }{\left(\frac{1}{A}+\frac{ec}{I}\right)}\end{array}$ (8)

Here, e is the eccentricity, I is the moment of inertia, A is the area of cross section, and c is the distance between the centroid from extreme fibre.

Substitute $600\text{psi}$ for $\sigma$, $28{\text{in}}^{\text{2}}$ for A, $53.762{\text{in}}^{\text{4}}$ for I, $0.35714\text{in}\text{.}$ for e, and $2.35714\text{in}\text{.}$ for c in Equation (8).

$\begin{array}{c}P=\frac{600\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)}{\left(\frac{1}{28}+\frac{0.35714\times 2.35714}{53.762}\right)}\\ =\frac{0.6}{0.0357+0.0156}\\ =11.68\text{kips}\end{array}$

Thus, the largest compressive load P that can be applied at the center of the top section of the timber column without plank 1 is $\underset{\_}{11.68\text{kips}}$.

To determine

Find the largest compressive load P that can be applied at the center of the top section of the timber column without plank 1and 2.

Answer to Problem 112P

The largest compressive load P that can be applied at the center of the top section of the timber column without plank 1 and 2 is $\underset{\_}{14.40\text{kips}}$.

Explanation of Solution

Calculation:

Sketch the centric loading as shown in Figure 3.

Refer to Figure 3.

Find the area of the timber section using the relation:

$A=\left(b\right)\times d$ (9)

Substitute $6\text{in}\text{.}$ for $b$, and $4\text{in}\text{.}$ for d in Equation (9).

$\begin{array}{c}A=6\times 4\\ =24{\text{in}}^{\text{2}}\end{array}$

Calculate the largest compressive load P using the relation:

$\begin{array}{l}\sigma =\frac{P}{A}\\ P=\sigma A\end{array}$ (10)

Substitute $600\text{psi}$ for $\sigma$ and $24{\text{in}}^{\text{2}}$ for $A$ in Equation (10).

$\begin{array}{c}P=600\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\times 24\\ =0.600\times 24\\ =14.40\text{kips}\end{array}$

Thus, the largest compressive load P is $\underset{\_}{14.40\text{kips}}$.

To determine

Find the largest compressive load P that can be applied at the center of the top section of the timber column without plank , 2, and 3.

Answer to Problem 112P

The largest compressive load P that can be applied at the center of the top section of the timber column without plank 1, 2, and 3 is $\underset{\_}{\text{7}\text{.50}\text{kips}}$.

Explanation of Solution

Calculation:

Sketch the Eccentric loading as shown in Figure 4.

Refer to Figure 4.

Find the area of the timber section using the relation:

$A={\left(bd\right)}_{\text{timber}}+1{\left(bd\right)}_{\text{planks}}$ (11)

Substitute $4\text{in}\text{.}$ for $b_t$, $4\text{in}\text{.}$ for $d_t$, $4\text{in}\text{.}$ for $b_p$, and $1\text{in}\text{.}$ for $d_p$ in Equation (11).

$\begin{array}{c}A=\left(4\times 4\right)+1\left(4\times 1\right)\\ =16+4\\ =20{\text{in}}^{\text{2}}\end{array}$

Find the centroid $\left(\overline{x}\right)$ of the section as shown below:

$\begin{array}{c}\overline{x}=2.5-2\\ =0.5\text{in}\text{.}\end{array}$

Determine the moment of inertia (I) of eccentric section as follows:

$I=\frac{b{d}^3}{12}$ (12)

Substitute $4\text{in}\text{.}$ for b and $5\text{in}\text{.}$ for d in Equation (12).

$\begin{array}{c}I=\frac{4\times 5^3}{12}\\ =41.667{\text{in}}^{\text{4}}\end{array}$

Calculate the largest compressive load P that can be applied at the center of the top section of the timber column without plank , 2, and 3 using the relation:

$\begin{array}{l}\sigma =P\left(\frac{1}{A}+\frac{ec}{I}\right)\\ P=\frac{\sigma }{\left(\frac{1}{A}+\frac{ec}{I}\right)}\end{array}$ (13)

Here, e is the eccentricity, I is the moment of inertia, A is the area of cross section, and c is the distance between the centroid from extreme fibre.

Substitute $600\text{psi}$ for $\sigma$, $20{\text{in}}^{\text{2}}$ for A, $41.667{\text{in}}^{\text{4}}$ for I, $0.5\text{in}\text{.}$ for e, and $2.5\text{in}\text{.}$ for c in Equation (13).

$\begin{array}{c}P=\frac{600\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)}{\left(\frac{1}{20}+\frac{0.5\times 2.5}{41.667}\right)}\\ =\frac{0.6}{0.05+0.02999}\\ =7.5\text{kips}\end{array}$

The largest compressive load P that can be applied at the center of the top section of the timber column without plank 1, 2, and 3 is $\underset{\_}{\text{7}\text{.50}\text{kips}}$.

To determine

Find the largest compressive load P that can be applied at the center of the top section of the timber all columns are removed.

Answer to Problem 112P

The largest compressive load P that can be applied at the center of the top section of the timber all columns are removed is $\underset{\_}{\text{9}\text{.60}\text{kips}}$.

Explanation of Solution

Calculation:

Sketch the centric loading as shown in Figure 5.

Refer to Figure 5.

Find the area of the timber section using the relation:

$A=\left(b\right)\times d$ (14)

Substitute $4\text{in}\text{.}$ for $b$, and $4\text{in}\text{.}$ for d in Equation (14).

$\begin{array}{c}A=4\times 4\\ =16{\text{in}}^{\text{2}}\end{array}$

Calculate the largest compressive load P using the relation:

$\begin{array}{l}\sigma =\frac{P}{A}\\ P=\sigma A\end{array}$ (15)

Substitute $600\text{psi}$ for $\sigma$ and $16{\text{in}}^{\text{2}}$ for $A$ in Equation (15).

$\begin{array}{c}P=600\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\times 16\\ =0.600\times 16\\ =9.60\text{kips}\end{array}$

Thus, the largest compressive load P that can be applied at the center of the top section of the timber all columns are removed is $\underset{\_}{\text{9}\text{.60}\text{kips}}$.