Chapter 6.5, Problem 1P

To determine

Analyze the $2\times 4$ studs for combined action of compression and bending.

Answer to Problem 1P

The $\underset{\_}{2\times 4}$ stud with 24 inch spacing is adequate.

Explanation of Solution

Given information:

Height of the studs is 9 ft.

The size of the studs is $2\times 4$.

Use Douglas fir-larch, No. 1 grade.

Wind load on the surface is 17 psf and the gravity load on the wall is $400\text{lb/ft}$.

Calculation:

Refer Table 5.1a, “Reference design values for visually graded lumber of Douglas fir-larch” in the textbook for the values of bending $\left(F_b\right)$, compression parallel to grain $\left(F_c\right)$, and the modulus of elasticity for stability $\left(E_{\min }\right)$.

For grade No. 1 Douglas fir-larch, the value of $F_b$ is 1,200 psi, the value of $F_c$ is 1,000 psi, and the value of $E_{\min }$ is 580,000 psi.

Refer Table 5.2, “Size adjustment factors (CF) for Dimension lumber, Decking and Timber” in the textbook for the adjustment factor values of $F_b$ and $F_c$.

For grade No. 1 and the stud of $2\times 4$, the adjustment factor for $F_b$ is 1.5 and the adjustment factor for $F_c$ is 1.15.

Refer Table 5.3, “Applicability of adjustment factors for sawn lumber, ASD” in the textbook for the applicable adjustment factor.

For ASD and load duration, the adjustment factor applicable is load duration factor $\left(C_D\right)$.

Refer Table 5.4, “Adjustment factors for design values for structural lumber due to load duration, $C_D{}^a$” in the textbook for the adjustment factor $\left(C_D\right)$.

For wind load duration, the adjustment factor is 1.6.

Let assume the wall surface braces the $2\times 4$ studs on their weak axis. So the critical value of thickness is $d=3.5\text{in}\text{.}$

Determine the ratio of unbraced length of stud $\left(L\right)$ to the stud thickness (d) as shown below:

Substitute 9 ft for $L$ and 3.5 in. for d.

$\begin{array}{c}\frac{L}{d}=\frac{9\text{ft}}{3.5\text{in}\text{.}}\\ =\frac{9\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}}{3.5\text{in}\text{.}}\\ =30.86\end{array}$

Find the buckling stress $\left(F_{cE}\right)$ using the relation:

$F_{cE}=\frac{0.822E_{\min }}{{\left(L/d\right)}^2}$ (1)

Substitute 580,000 psi for $E_{\min }$ and 30.86 for $\frac{L}{d}$ in Equation (1).

$\begin{array}{c}{F}_{cE}=\frac{0.822\left(580,000\right)}{{30.86}^2}\\ =500.62\text{psi}\end{array}$

Find the modified design value for compression $\left(F_c^{*}\right)$ by multiplying $F_c$ with the adjustment factor 1.15 as follows:

$F_c^{*}=1.15F_c$

Substitute 1,000 psi for $F_c$.

$\begin{array}{c}{F}_c^{*}=1.15\times 1,000\text{psi}\\ =1,150\text{psi}\end{array}$

Find the ratio $\frac{F_{cE}}{F_c^{*}}$ as shown below:

Substitute $500.62\text{psi}$ for $F_{cE}$ and 1,150 psi for $F_c^{*}$.

$\begin{array}{c}\frac{F_{cE}}{F_c^{*}}=\frac{500.62}{1,150}\\ =0.435\end{array}$

Refer Figure 6.2, “Column stability factor as a function of $\frac{F_{cE}}{F_c^{*}}$” in the textbook for the column stability factor $C_p$.

For $\frac{F_{cE}}{F_c^{*}}$ of 0.435, the value of $C_p$ is 0.37.

Refer Table A.8 “Properties of sections” in the textbook for area and section modulus of $2\times 4$ studs.

The area $A=5.25\text{in}{\text{.}}^2$ and the section modulus $S=3.063\text{in}{\text{.}}^3$.

Find the stud compression capacity using the relation:

$P=\left(F_c^{*}\right)\left(C_p\right)\left(A\right)$ (2)

Substitute 1,150 psi for $F_c^{*}$, 0.37 for $C_p$, and $\text{5}\text{.25}\text{in}{\text{.}}^2$ for $A$ in Equation (2).

$\begin{array}{c}P=\left(1,150\right)\left(0.37\right)\left(5.25\right)\\ =2,233.9\text{lb}\end{array}$

Find the load P for stud spacing of 24 in. and the gravity load of 400 lb/ft.

$\begin{array}{c}P=\left(24\text{in}\text{.}\right)\left(400\text{lb/ft}\right)\\ =\left(24\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(400\text{lb/ft}\right)\\ =800\text{lb}\end{array}$

The above value shows that the gravity load alone is not a critical concern.

Consider the combined loading:

Find the modified design value for compression $\left(F_c^{*}\right)$ by multiplying $F_c$ with the adjustment factor 1.6 as follows:

$F_c^{*}=1.15F_c$

Substitute 1,000 psi for $F_c$.

$\begin{array}{c}{F}_c^{*}=1.6\times 1,000\text{psi}\\ =1,600\text{psi}\end{array}$

Find the ratio $\frac{F_{cE}}{F_c^{*}}$ as shown below:

Substitute $500.62\text{psi}$ for $F_{cE}$ and 1,600 psi for $F_c^{*}$.

$\begin{array}{c}\frac{F_{cE}}{F_c^{*}}=\frac{500.62}{1,600}\\ =0.313\end{array}$

Refer Figure 6.2, “Column stability factor as a function of $\frac{F_{cE}}{F_c^{*}}$” in the textbook for the column stability factor $C_p$.

For $\frac{F_{cE}}{F_c^{*}}$ of 0.313, the value of $C_p$ is 0.28.

For the combined loading,

$F_c^{\text{'}}=C_p{F}_c^{*}$

Substitute 0.25 for $C_p$ and 1,600 psi for $F_c^{*}$.

$\begin{array}{c}{F}_c^{\text{'}}=0.25\times 1,600\text{psi}\\ =400\text{psi}\end{array}$

Determine the computed compressive stress $\left(f_c\right)$ using the relation:

$f_c=\frac{P}{A}$

Substitute 800 psi for P and $\text{5}\text{.25}{\text{in}}^2$ for $A$.

$\begin{array}{c}{f}_c=\frac{P}{A}\\ =\frac{800}{5.25}\\ =152.38\text{psi}\end{array}$

Wind load is 17 psf. Therefore the load w is,

$\begin{array}{c}w=0.75\left(17\text{psf}\right)\\ =12.75\text{psf}\end{array}$

Find the moment (M) for the wind load as follows:

$M=\frac{w{L}^2}{8}\times d$

Here, L is the height of studs and d is the stud distance.

Substitute 12.75 psf for w, 9 ft for L, and 24 in. for d.

$\begin{array}{c}M=\left(\frac{\left(12.75\text{psf}\right){\left(\text{9}\text{ft}\right)}^2}{8}\right)\left(24\text{in}\text{.}\right)\\ =\left(\frac{\left(12.75\text{psf}\right){\left(\text{9}\text{ft}\right)}^2}{8}\right)\left(24\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\\ =258.19\text{lb-ft}\end{array}$

Determine the computed bending stress $\left(f_b\right)$ using the relation:

$f_b=\frac{M}{S}$

Substitute $258.19\text{lb-ft}$ for M and $\text{3}\text{.063}{\text{in}}^3$ for S.

$\begin{array}{c}{f}_b=\frac{258.19\text{lb-ft}}{\text{3}\text{.063}{\text{in}}^3}\\ =\frac{258.19\text{lb-ft}\left(\frac{12\text{in}\text{.}}{1\text{ft}}\right)}{\text{3}\text{.063}{\text{in}}^3}\\ =1,011.52\text{psi}\end{array}$

Determine the ratio $\frac{f_c}{f_{cE}}$:

Substitute $152.38\text{psi}$ for $f_c$ and $\text{500}\text{.62}\text{psi}$ for $f_{cE}$.

$\begin{array}{c}\frac{f_c}{f_{cE}}=\frac{152.38}{\text{500}\text{.62}}\\ =0.304\end{array}$

Apply the code formula for the interaction as follows:

${\left(\frac{f_c}{F_c^{\text{'}}}\right)}^2+\frac{f_b}{F_b\left(1-\frac{f_c}{F_{cE}}\right)}\le 1$

Substitute $152.38\text{psi}$ for $f_c$, $400\text{psi}$ for $F_c^{\text{'}}$, $1,011.52\text{psi}$ for $f_b$, $1.6\left(1200\text{psi}\right)$ for $F_b$, and 0.304 for $\frac{f_c}{f_{cE}}$.

$\begin{array}{l}{\left(\frac{152.38}{400}\right)}^2+\frac{1,011.52}{\left(16\times 1,200\right)\left(1-0.304\right)}\le 1\\ 0.145+0.076\le 1\\ 0.221<1\end{array}$

The condition satisfied. Therefore, the stud is adequate.

Therefore, the $\underset{\_}{2\times 4}$ stud with 24 inch spacing is adequate.