Chapter 7, Problem 7.71RP

The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail along the sides C and the top D must resist if the nails are uniformly spaced at s=3 in. The beam is subjected to a shear of V=4.5 kip.

To determine

The shear force $\left(F_C\right)$ each nails along the side C.

The shear force $\left(F_D\right)$ each nails along the top D.

Answer to Problem 7.71RP

The shear force $\left(F_C\right)$ each nails along the side C is $\underset{\_}{197\text{lb}}$.

The shear force $\left(F_D\right)$ each nails along the top D is $\underset{\_}{1.38\text{kip}}$.

Explanation of Solution

Given information:

The shear force is $4.5\text{kip}$.

The uniform nail spacing is 3 in.

Calculation:

Sketch the diagram of the T section as shown in Figure 1.

Refer Figure 1,

The area of the beam is the sum of area of three rectangles 1, 2, and 3.

The dimensions of rectangle 1 as width $b_1=10\text{in}\text{.}$ and depth $d_1=1\text{in}\text{.}$

The dimensions of rectangle 2 as width $b_2=4\text{in}\text{.}$ and depth $d_2=2\text{in}\text{.}$

The dimensions of rectangle 2 as width $b_3=12\text{in}\text{.}$ and depth $d_2=1\text{in}\text{.}$

Find the value of area section 1 as shown below:

$a_1=b_1{d}_1$ (1)

Substitute 10 in. for $b_1$ and 1 in. for $d_1$ in Equation (1).

$\begin{array}{c}{a}_1=10\times 1\\ =10{\text{in}}^{\text{2}}\end{array}$

Find the value of area section 2 as shown below:

$a_2=b_2{d}_2$ (2)

Substitute 4 in. for $b_2$ and 2 in. for $d_2$ in Equation (2).

$\begin{array}{c}{a}_2=4\times 2\\ =8{\text{in}}^{\text{2}}\end{array}$

Find the value of area section 3 as shown below:

$a_3=b_3{d}_3$ (3)

Substitute 12 in. for $b_3$ and 1 in. for $d_3$ in Equation (3).

$\begin{array}{c}{a}_3=12\times 1\\ =12{\text{in}}^{\text{2}}\end{array}$

Calculate the centroid of $\overline{y}$ as shown in below:

$\overline{y}=\frac{a_1{y}_1+a_2{y}_2+a_3{y}_3}{a_1+a_2+a_3}$ (4)

Here, $a_1,a_2$, and $a_3$ area of section 1,2, and 3 and $y_1,y_2$ $y_3$ is the centroid of section 1,2, and 3.

Substitute $10{\text{in}}^{\text{2}}$ for $a_1$, $8{\text{in}}^{\text{2}}$ for $a_2$, $12{\text{in}}^{\text{2}}$ for $a_3$, $0.5\text{in}\text{.}$ for $y_1$, $2\text{in}\text{.}$ for $y_2$, and $7\text{in}\text{.}$ for $y_3$ in Equation (4).

$\begin{array}{c}\overline{y}=\frac{0.5\left(10\right)+2\times 8+7\times 12}{10+8+12}\\ =\frac{5+16+84}{30}\\ =3.5\text{in}\end{array}$

Sketch the diagram of $\overline{y}$ as shown in Figure 2.

Calculate the moment of inertia of the beam (I) as follows:

$\begin{array}{c}{I}_{NA}=I_1+I_2+I_3\\ =\left[\begin{array}{l}\frac{b_1{d}_1{}^3}{12}+a\left(\overline{y}-y_1\right)+\frac{b_2{d}_2{}^3}{12}+a\left(\overline{y}-y_2\right)\\ +\frac{b_1{d}_1{}^3}{12}+a\left(y_3-\overline{y}\right)\end{array}\right]\end{array}$ (5)

Refer to Figure 2:

The value of $y_1$ is $0.5\text{in}\text{.}$

The value of $y_2$ is $2\text{in}\text{.}$

The value of $y_3$ is $7\text{in}\text{.}$

Substitute 10 in. for $b_1$, 1 in. for $d_1$, 4 in. for $b_2$, $2\text{in}$ for $d_2$, 12 in. for $b_3$, and 1 in. for $d_2$ in Equation (5).

$\begin{array}{c}{I}_{NA}=\left[\begin{array}{l}\frac{\left(10\right)\left(1^3\right)}{12}+\left(10\right)\left(1\right){\left(3.50-0.5\right)}^2+\frac{2\times 4^3}{12}+2\times 4{\left(3.50-2\right)}^2\\ +\frac{1\times {12}^3}{12}+1\times 12{\left(7-3.50\right)}^2\end{array}\right]\\ =90.8333+28.6667+291\\ =410.5{\text{in}}^{\text{4}}\end{array}$

Calculate the first moment area $\left(Q_C\right)$ as shown in below:

$Q_C={{\overline{y}}^{\prime }}_1{A}^{\prime }$ (6)

Here, ${\overline{y}}^{\prime }$ is the distance between the neutral axis and centroid of the area under consideration and $A^{\prime }$ is the area of cross section.

Refer to Figure 2.

The value of ${{\overline{y}}^{\prime }}_1$ is $1.5\text{in}\text{.}$.

Substitute $1.5\text{in}\text{.}$ for ${{\overline{y}}^{\prime }}_1$ and $\left(4\times 1\right){\text{in}}^{\text{2}}$ for $A^{\prime }$ in Equation (6).

$\begin{array}{c}{Q}_C=1.5\times 4\times 1\\ =6.00{\text{in}}^{\text{3}}\end{array}$

Calculate the first moment area $\left(Q_D\right)$ as shown in below:

$Q_D={{\overline{y}}^{\prime }}_2{A}^{\prime }$ (7)

Here, ${\overline{y}}^{\prime }$ is the distance between the neutral axis and centroid of the area under consideration and $A^{\prime }$ is the area of cross section.

Refer to Figure 2.

The value of ${{\overline{y}}^{\prime }}_2$ is $3.50\text{in}\text{.}$.

Substitute $3.50\text{in}\text{.}$ for ${{\overline{y}}^{\prime }}_2$ and $\left(12\times 1\right){\text{in}}^{\text{2}}$ for $A^{\prime }$ in Equation (7).

$\begin{array}{c}{Q}_C=3.5\times 12\times 1\\ =42.0{\text{in}}^{\text{3}}\end{array}$

Show the formula for shear flow $\left(q_C\right)$ as follows:

$q_C=\frac{V{Q}_C}{I}$ (8)

Here, V is the shear force, I is the moment of inertia, and ${\overline{y}}^{\prime }$ is the distance between the neutral axis and centroid of the area under consideration.

Substitute $4.5\text{kip}$ for $V$, $6.00{\text{in}}^{\text{3}}$ for $Q_C$, and $410.5{\text{in}}^{\text{4}}$ for $I$ in Equation (8).

$\begin{array}{c}{q}_C=\frac{4.5\times {10}^3\times 6.00}{410.5}\\ =\frac{27,000}{410.5}\\ =65.773\text{lb}/\text{in}\text{.}\end{array}$

Show the formula for shear flow $\left(q_D\right)$ as follows:

$q_D=\frac{V{Q}_D}{I}$ (9)

Here, V is the shear force, I is the moment of inertia, and ${\overline{y}}^{\prime }$ is the distance between the neutral axis and centroid of the area under consideration.

Substitute $4.5\text{kip}$ for $V$, $42.0{\text{in}}^{\text{3}}$ for $Q_D$, and $410.5{\text{in}}^{\text{4}}$ for $I$ in Equation (9).

$\begin{array}{c}{q}_D=\frac{4.5\times {10}^3\times 42.0}{410.5}\\ =\frac{189,000}{410.5}\\ =460.414\text{lb}/\text{in}\text{.}\end{array}$

Calculate the shear force $\left(F_C\right)$ using the relation:

$F_C=q_{C}s$ (10)

Here, s is the spacing and $q_C$ is the shear flow at point C.

Substitute $65.773\text{lb}/\text{in}\text{.}$ for $q_C$ and 3 in. for s in Equation (10).

$\begin{array}{c}{F}_C=65.773\times 3\\ =197\text{lb}\end{array}$

Hence, the shear force $\left(F_C\right)$ resisted nail at C is $\underset{\_}{197\text{lb}}$.

Calculate the shear force $\left(F_D\right)$ using the relation:

$F_D=q_{D}s$ (11)

Here, s is the spacing and $q_D$ is the shear flow at point D.

Substitute $460.41\text{lb}/\text{in}\text{.}$ for $q_D$ and 3 in. for s in Equation (11).

$\begin{array}{c}{F}_D=460.41\text{lb}/\text{in}\text{.}\times 3\\ =1.38\text{kip}\end{array}$

Hence, the shear force $\left(F_D\right)$ resisted nail at D is $\underset{\_}{1.38\text{kip}}$.