Chapter 3, Problem 3.8.1P

To determine

(a)

The section for loads given loads using load and resistance factor design (LRFD) method.

Answer to Problem 3.8.1P

The section for loads given loads using load and resistance factor design (LRFD) method is $\text{WT}5\times 11$.

Explanation of Solution

Given data:

Length of the connection is $11\text{in}$.

Spacing of truss in the roof system is $\text{12}\text{feet}\text{6}\text{inches}$.

Snow load is $20\text{psf}$ of horizontal projection.

Weight of roofing is $12\text{psf}$.

Section for the purlins is $\text{MC}8\times 8.5$.

Weight of the truss is $1000\text{lb}$.

Calculation:

Calculate the snow load.

$\begin{array}{c}S=20\text{psf}\times 10\text{ft}\times 12.5\text{ft}\\ =2500\text{lb}\end{array}$

Calculate the load due to purlins.

$\begin{array}{c}{D}_{\text{p}}=\left(8.5\text{lb}/\text{ft}\right)\times \left(12.5\text{ft}\right)\\ =106.25\text{lb}\end{array}$

Calculate the weight of the truss.

$\begin{array}{c}{D}_{\text{t}}=\frac{1000\text{lb}}{3}\\ =333.33\text{lb}\end{array}$

Calculate the slant height of the roof.

$\begin{array}{c}l=\sqrt{{\left(3\text{ft}\right)}^2+{\left(30\text{ft}\right)}^2}\\ =\sqrt{9{\text{ft}}^2+900{\text{ft}}^2}\\ =30.15\text{ft}\end{array}$

Calculate the weight of the roof.

$\begin{array}{c}{D}_{\text{r}}=\left(12\text{psf}\right)\times \left(10\text{ft}\right)\times \left(\frac{30.15\text{ft}}{30\text{ft}}\right)\times \left(12.5\text{ft}\right)\\ =1\text{507}\text{.5}\text{lb}\end{array}$

Write the expression to calculate the total dead load.

$D=D_{\text{t}}+D_{\text{r}}+D_{\text{p}}$ ...... (I)

Here, total dead load is $D$, load due to truss is $D_{\text{t}}$, load due to roofing is $D_{\text{r}}$ and load due to purlins is $D_{\text{p}}$.

Substitute $333.33\text{lb}$ for $D_{\text{t}}$, $106.25\text{lb}$ for $D_{\text{p}}$ and $1507.5\text{lb}$ for $D_{\text{r}}$ in Equation (I).

$\begin{array}{c}D=333.33\text{lb}+106.25\text{lb}+1507.5\text{lb}\\ =1947.08\text{lb}\end{array}$

Calculate the factored load using following load combination.

$P_{\text{u}}=1.2D+1.6S$ ...... (II)

Here, factored load is $P_{\text{u}}$, total dead load is $D$ and snow load is $S$.

Substitute $1947.08\text{lb}$ for $D$ and $2500\text{lb}$ for $S$ in Equation (II).

$\begin{array}{c}{P}_{\text{u}}=1.2\times \left(1947.08\text{lb}\right)+1.6\times \left(2500\text{lb}\right)\\ =2336.5\text{lb}+4000\text{lb}\\ =\left(6336.5\text{lb}\right)\left(\frac{1\text{kips}}{{10}^3\text{lb}}\right)\\ =6.34\text{lb}\end{array}$

Write the expression to calculate the exterior joint load.

$P_{\text{ux}}=1.2\left(\frac{D_{\text{t}}}{2}+\frac{D_{\text{r}}}{2}+D_{\text{p}}\right)+1.6\left(\frac{S}{2}\right)$ ...... (III)

Here, load on the exterior joint is $P_{\text{ux}}$.

Substitute $333.33\text{lb}$ for $D_{\text{t}}$, $1507.5\text{lb}$ for $D_{\text{r}}$, $106.25\text{lb}$ for $D_{\text{p}}$ and $2500\text{lb}$ for $S$ in Equation (III).

$\begin{array}{c}{P}_{ux}=1.2\left(\frac{333.33\text{lb}}{2}+\frac{1507.5\text{lb}}{2}+106.25\text{lb}\right)+1.6\left(\frac{2500\text{lb}}{2}\right)\\ =1.2\times 1026.66\text{lb}+1.6\times 1250\text{lb}\\ =3232\text{lb}\times \frac{1\text{kips}}{{10}^3\text{lb}}\\ =3.23\text{kips}\end{array}$

Consider the free body diagram of the truss shown below.

Figure-(1)

Write the expression to calculate the moment about point $\text{A}$.

$\begin{array}{l}\sum M_{\text{A}}=0\\ \left(P_{\text{u}}\times 10\text{ft}\right)+\left(P_{\text{u}}\times 20\text{ft}\right)+\left(P_{\text{ux}}\times 30\text{ft}\right)-\left(R_{\text{bx}}\times 3\text{ft}\right)=0\end{array}$ ....... (IV)

Substitute $6.34\text{kips}$ for $P_{\text{u}}$ and $3.23\text{kips}$ for $P_{\text{ux}}$ in Equation (IV).

$\begin{array}{l}\left(6.34\text{kips}\times 10\text{ft}\right)+\left(6.34\text{kips}\times 20\text{ft}\right)+\left(3.23\text{kips}\times 30\text{ft}\right)-\left(R_{\text{bx}}\times 3\text{ft}\right)=0\\ 63.4\text{kips}\cdot \text{ft}+126.8\text{kips}\cdot \text{ft}+96.9\text{kips}\cdot \text{ft}-\left(R_{\text{bx}}\times 3\text{ft}\right)=0\\ R_{\text{bx}}\times 3\text{ft}=287\text{kips}\cdot \text{ft}\\ R_{\text{bx}}=\frac{287\text{kips}\cdot \text{ft}}{3\text{ft}}\end{array}$

Solve further.

$R_{\text{bx}}=95.67\text{kips}$

Consider joint $\text{B}$ as shown in figure-(1).

Write the expression for summation of forces acting in the horizontal direction.

$\begin{array}{l}\sum F_{\text{x}}=0\\ R_{\text{Bx}}-F_{\text{BC}}\cos \theta =0\end{array}$ ...... (V)

Here, summation of all horizontal forces is $\sum F_{\text{x}}$, horizontal reaction at point $\text{B}$ is $R_{\text{Bx}}$ and force in member $\text{BC}$ is $F_{\text{BC}}$.

Substitute $95.67\text{kips}$ for $R_{\text{Bx}}$ in Equation (V).

$\begin{array}{l}95.67\text{kips}-F_{\text{BC}}\times \frac{30}{30.15}=0\\ F_{\text{BC}}=\frac{\left(95.67\text{kips}\times 30.15\right)}{30}\\ F_{\text{BC}}=96.15\text{kips}\end{array}$

Write the expression to calculate the required area.

$A_{\text{g}}=\frac{F_{\text{BC}}}{0.9f_{\text{y}}}$ ...... (VI)

Here, gross area is $A_{\text{g}}$ and yield strength is $f_{\text{y}}$.

Substitute $96.15\text{kips}$ for $F_{\text{BC}}$ and $36\text{ksi}$ for $f_{\text{y}}$ in Equation (VI).

$\begin{array}{c}{A}_{\text{g}}=\frac{96.15\text{kips}}{0.9\times 36\text{ksi}}\\ =2.96{\text{in}}^2\end{array}$

Write the expression to calculate the required area.

$A_{\text{e}}=\frac{F_{\text{BC}}}{0.75f_{\text{u}}}$ ...... (VII)

Here, effective area is $A_{\text{e}}$ and ultimate strength is $f_{\text{u}}$.

Substitute $96.15\text{kips}$ for $F_{\text{BC}}$ and $58\text{ksi}$ for $f_{\text{u}}$ in Equation (VII).

$\begin{array}{c}{A}_{\text{e}}=\frac{96.15\text{kips}}{0.75\times 58\text{ksi}}\\ =2.21{\text{in}}^2\end{array}$

Calculate the effective length of the truss.

$\begin{array}{c}l=10\text{ft}\times \frac{30.15\text{ft}}{30\text{ft}}\\ =10.05\text{ft}\end{array}$

Calculate the radius of gyration.

$r_{\min }=\frac{l}{300}$ ...... (VIII)

Substitute $10.05\text{ft}$ for $l$ in Equation (VIII)

$\begin{array}{c}{r}_{\min }=\frac{\left(10.05\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{300}\\ =0.402\end{array}$

Use section $\text{WT}4\times 9$.

$\begin{array}{l}{A}_{\text{g}}=3.24{\text{in}}^2\\ r_{\min }=1.33\text{in}\end{array}$

Write the expression to calculate reduction factor.

$U=1-\frac{\overline{x}}{l}$ ...... (IX)

Here, reduction factor is $U$, distance between the centroid and the plane is $\overline{x}$ and length is $l$.

Substitute $1.07$ for $\overline{x}$ and $11\text{in}$ for $l$ in Equation (IX).

$\begin{array}{c}U=1-\frac{1.07\text{in}}{11\text{in}}\\ =1-0.0972\\ =0.9027\end{array}$

Write the expression to calculate the effective area for the section.

$A_{\text{e}}=U{A}_{\text{g}}$ ...... (X)

Substitute $0.9027$ for $U$ and $3.24{\text{in}}^2$ for $A_{\text{g}}$ in Equation (X).

$\begin{array}{c}{A}_e=0.9027\times 3.24{\text{in}}^{\text{2}}\\ =2.92{\text{in}}^2\end{array}$

Conclusion:

Since the gross area, net area and radius of gyration for this greater than the calculated value,

$\begin{array}{l}{A}_{\text{g}}=3.24{\text{in}}^2>2.96{\text{in}}^2\\ r_{\text{min}}=1.33\text{in}>0.402\text{in}\\ A_{\text{e}}=2.92{\text{in}}^2>2.21{\text{in}}^2\end{array}$

To determine

(b)

The section for loads given loads using allowable strength design (ASD) method.

Answer to Problem 3.8.1P

The section for loads given loads using allowable strength design (ASD) method is $\text{WT}5\times 11$.

Explanation of Solution

Given data:

Length of the connection is $11\text{in}$.

Spacing of truss in the roof system is $\text{12}\text{feet}\text{6}\text{inches}$.

Snow load is $20\text{psf}$ of horizontal projection.

Weight of roofing is $12\text{psf}$.

Section for the purlins is $\text{MC}8\times 8.5$.

Weight of the truss is $1000\text{lb}$.

Calculation:

Calculate the ultimate load using the following load combination.

$P=D+S$ ...... (XI)

Here, ultimate load is $P$, dead load is $D$ and snow load is $S$.

Substitute $1947.08\text{lb}$ for $D$ and $2500\text{lb}$ for $S$ in equation (XI).

$\begin{array}{c}P=1947.08\text{lb}+2500\text{lb}\\ =\left(4447.08\text{lb}\right)\left(\frac{1\text{kips}}{{10}^3\text{lb}}\right)\\ =4.45\text{lb}\end{array}$

Write the expression to calculate the exterior joint load.

$P_{\text{ux}}=\frac{D_{\text{t}}}{2}+\frac{D_{\text{r}}}{2}+D_{\text{p}}+\frac{S}{2}$ ...... (XII)

Here, load on the exterior joint is $P_{\text{ux}}$.

Substitute $333.33\text{lb}$ for $D_{\text{t}}$, $1507.5\text{lb}$ for $D_{\text{r}}$, $106.25\text{lb}$ for $D_{\text{p}}$ and $2500\text{lb}$ for $S$ in Equation (XII).

$\begin{array}{c}{P}_{\text{ux}}=\left(\frac{333.33\text{lb}}{2}\right)+\left(\frac{1507.5\text{lb}}{2}\right)+\left(106.25\text{lb}\right)+\left(\frac{2500\text{lb}}{2}\right)\\ =\left(2276.665\text{lb}\right)\left(\frac{1\text{kips}}{{10}^3\text{lb}}\right)\\ =2.27\text{kips}\end{array}$

Consider the free body of the truss as shown below.

Figure-(2)

Write the expression to calculate the moment about point $\text{A}$.

$\begin{array}{l}\sum M_{\text{A}}=0\\ \left(P\times 10\text{ft}\right)+\left(P\times 20\text{ft}\right)+\left(P_{\text{ux}}\times 30\text{ft}\right)-\left(R_{\text{bx}}\times 3\text{ft}\right)=0\end{array}$ ....... (XIII)

Substitute $4.45\text{kips}$ for $P_{\text{u}}$ and $2.27\text{kips}$ for $P_{\text{ux}}$ in Equation (XIII).

$\begin{array}{l}\left(4.45\text{kips}\times 10\text{ft}\right)+\left(4.45\text{kips}\times 20\text{ft}\right)+\left(2.27\text{kips}\times 30\text{ft}\right)-\left(R_{\text{bx}}\times 3\text{ft}\right)=0\\ 44.5\text{kips}\cdot \text{ft}+89\text{kips}\cdot \text{ft}+68.1\text{kips}\cdot \text{ft}-\left(R_{\text{bx}}\times 3\text{ft}\right)=0\\ R_{\text{bx}}\times 3\text{ft}=201.6\text{kips}\cdot \text{ft}\\ R_{\text{bx}}=\frac{201.6\text{kips}\cdot \text{ft}}{3\text{ft}}\end{array}$

Solve further.

$R_{\text{bx}}=67.2\text{kips}$

Consider joint $\text{B}$ as shown in figure-(1).

Write the expression for summation of forces acting in the horizontal direction.

$\begin{array}{l}\sum F_{\text{x}}=0\\ R_{\text{Bx}}-F_{\text{BC}}\cos \theta =0\end{array}$ ...... (XIV)

Here, summation of all horizontal forces is $\sum F_{\text{x}}$, horizontal reaction at point $\text{B}$ is $R_{\text{Bx}}$ and force in member $\text{BC}$ is $F_{\text{BC}}$.

Substitute $67.2\text{kips}$ for $R_{\text{Bx}}$ in Equation (XIV).

$\begin{array}{l}67.2\text{kips}-F_{\text{BC}}\times \frac{30}{30.15}=0\\ F_{\text{BC}}=\frac{67.2\text{kips}\times 30.15}{30}\\ F_{\text{BC}}=67.54\text{kips}\end{array}$

Write the expression to calculate the required area.

$A_{\text{g}}=\frac{F_{\text{BC}}}{0.6f_{\text{y}}}$ ...... (XV)

Here, gross area is $A_{\text{g}}$ and yield strength is $f_{\text{y}}$.

Substitute $67.54\text{kips}$ for $F_{\text{BC}}$ and $36\text{ksi}$ for $f_{\text{y}}$ in Equation (XV).

$\begin{array}{c}{A}_{\text{g}}=\frac{67.54\text{kips}}{0.6\times 36\text{ksi}}\\ =3.13{\text{in}}^2\end{array}$

Write the expression to calculate the required area.

$A_{\text{e}}=\frac{F_{\text{BC}}}{0.5f_{\text{u}}}$ ...... (XVI)

Here, effective area is $A_{\text{e}}$ and ultimate strength is $f_{\text{u}}$.

Substitute $67.54\text{kips}$ for $F_{\text{BC}}$ and $58\text{ksi}$ for $f_{\text{u}}$ in Equation (XVI).

$\begin{array}{c}{A}_{\text{e}}=\frac{67.54\text{kips}}{0.5\times 58\text{ksi}}\\ =2.33{\text{in}}^2\end{array}$

Calculate the effective length of the truss.

$\begin{array}{c}l=\left(10\text{ft}\right)\times \left(\frac{30.15\text{ft}}{30\text{ft}}\right)\\ =10.05\text{ft}\end{array}$

Calculate the radius of gyration.

$r_{\min }=\frac{l}{300}$ ...... (XVII)

Substitute $10.05\text{ft}$ for $l$ in Equation (XVII).

$\begin{array}{c}{r}_{\min }=\frac{\left(10.05\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{300}\\ =0.402\end{array}$

Use section $\text{WT}4\times 9$.

$\begin{array}{l}{A}_{\text{g}}=3.24{\text{in}}^2\\ r_{\min }=1.33\text{in}\end{array}$

Write the expression to calculate reduction factor.

$U=1-\frac{\overline{x}}{l}$ ...... (XVIII)

Here, reduction factor is $U$, distance between the centroid and the plane is $\overline{x}$ and length is $l$.

Substitute $1.07$ for $\overline{x}$ and $11\text{in}$ for $l$ in Equation (XVIII).

$\begin{array}{c}U=1-\frac{1.07\text{in}}{11\text{in}}\\ =1-0.0972\\ =0.9027\end{array}$

Write the expression to calculate the effective area for the section.

$A_{\text{e}}=U{A}_{\text{g}}$ ...... (XIX)

Substitute $0.9027$ for $U$ and $3.24{\text{in}}^2$ for $A_{\text{g}}$ in Equation (XIX).

$\begin{array}{c}{A}_e=0.9027\times 3.24{\text{in}}^{\text{2}}\\ =2.92{\text{in}}^2\end{array}$

Conclusion:

Since the gross area, net area and radius of gyration for this greater than the calculated value,

$\begin{array}{l}{A}_{\text{g}}=3.24{\text{in}}^2>3.13{\text{in}}^2\\ r_{\text{min}}=1.33\text{in}>0.402\text{in}\\ A_{\text{e}}=2.92{\text{in}}^2>2.33{\text{in}}^2\end{array}$