Chapter 3, Problem 3.8.4P

To determine

(a)

The double angle section for a tension member of the roof truss using Load And Resistance Factor Design (LRFD) method.

Answer to Problem 3.8.4P

The double angle section to be selected for the bottom chord by LFRD method is $\text{2L6}\times \text{L6}\times \frac{\text{7}}{\text{16}}$

The double angle section to be selected for the web members by LFRD method is $\text{2L3}\times \text{L2}\times \frac{3}{\text{16}}$

Explanation of Solution

Given:

Metal deck load is $4\text{psf}$ of roof surface.

Build up roof load is $12\text{psf}$.

Purlins load is $6\text{psf}$ of roof surface.

Snow load is $18\text{psf}$ of horizontal projection.

Truss load is $5\text{psf}$ of horizontal projection.

The trusses are spaced at $25\text{ft}$.

Thickness of gusset plate is $\frac{3}{8}\text{in}$.

$\text{A572}\text{Grade}\text{50}$ steel.

Calculation:

The properties of $\text{A572}\text{Grade}\text{50}$ steel from the AISC steel manual are as follows.

• The yield strength $\left(F_{\text{y}}\right)$ is $50\text{ksi}$.
• The ultimate strength is $\left(F_{\text{u}}\right)$ is $65\text{ksi}$.

The truss diagram is shown below.

Figure (1)

Write the expression to calculate the area on which the load will act on the truss.

$A=2\times h\times l_{\text{t}}$ ...... (I)

Here, the area on which the load will act is $A$, the hypotenuse length of the truss is $h$ and the length of the truss is $l_{\text{t}}$.

Calculate the value of the hypotenuse length of the truss.

$\begin{array}{c}h=\sqrt{8^2+{40}^2}\text{ft}\\ =\sqrt{1664}\text{ft}\\ =40.79\text{ft}\end{array}$

Here, the hypotenuse height $EH$ of $\Delta AEH$ is $h$.

Substitute $40.79\text{ft}$ for $h$ and $25\text{ft}$ for $l_{\text{t}}$ in Equation (I).

$\begin{array}{c}A=2\times \left(40.79\text{ft}\right)\times \left(25\text{ft}\right)\\ =\text{2040}{\text{ft}}^2\end{array}$

Write the expression to calculate the total dead load.

$D=\left(MD+R+P\right)A+\left(T_{\text{w}}\times PA\right)$ ...... (II)

Here, the total dead load is $D$, the metal deck load is $MD$, the roof load is $R$, the purlin load is $P$, the truss load is $T_{\text{w}}$ and the projection area is $PA$.

Substitute $4\text{psf}$ for $MD$, $12\text{psf}$ for $R$, $6\text{psf}$ for $P$, $2040{\text{ft}}^{\text{2}}$ for $A$, $5\text{psf}$ for $T_{\text{w}}$ and $2000{\text{ft}}^{\text{2}}$ for $PA$ in Equation (II).

$\begin{array}{c}D=\left(\left(4\text{psf}+12\text{psf}+6\text{psf}\right)\times \left(2040{\text{ft}}^{\text{2}}\right)\right)+\left(5\text{psf}\times 2000{\text{ft}}^{\text{2}}\right)\\ =44880\text{lb}+10000\text{lb}\\ =54880\text{lb}\end{array}$

Write the expression to calculate the load per truss.

$DL=\frac{D}{N_{\text{T}}}$ ...... (III)

Here, the dead load per truss is $DL$, the number of trusses is $N_{\text{T}}$.

Substitute $54880\text{lb}$ for $D$ and $8$ for $N_{\text{T}}$ in Equation (III).

$\begin{array}{c}DL=\frac{54880\text{lb}}{8}\\ =6860\text{lb}\end{array}$

Write the expression to calculate the snow load per truss.

$S=\frac{SL\times PA}{N_{\text{T}}}$ ...... (IV)

Here, the snow load per truss is $S$ and the snow load is $SL$.

Substitute $18\text{psf}$ for $SL$, $8$ for $N_{\text{T}}$ and $2000{\text{ft}}^{\text{2}}$ for $PA$ in Equation (IV).

$\begin{array}{c}S=\frac{\left(18\text{psf}\right)\times \left(2000{\text{ft}}^{\text{2}}\right)}{8}\\ =4500\text{lb}\end{array}$

Write the expression to calculate the factored load by LRFD.

$P_{\text{n}}=1.2DL+1.6S$ ...... (V)

Here, the factored load is $P_{\text{n}}$.

Substitute $6860\text{lb}$ for $DL$ and $4500\text{lb}$ for $S$ in Equation (V).

$\begin{array}{c}{P}_{\text{n}}=\left(1.2\times 6860\text{lb}\right)+\left(1.6\times 4500\text{lb}\right)\\ =\left(15432\text{lb}\right)\left(\frac{1\text{kips}}{1000\text{lb}}\right)\\ =15.43\text{kips}\end{array}$

The free body diagram of join E is shown below.

Figure (2)

Here, the vertical reaction at support $E$ is $R$ and the force members are $F_{\text{EF}}$ and $F_{\text{DE}}$.

Write the expression to calculate the value of $R$.

$R=\frac{N_{\text{S}}\times P_{\text{n}}}{2}$ ...... (VI)

Here, the number of spans is $N_{\text{S}}$.

Substitute $8$ for $N_{\text{S}}$ and $15.43\text{kips}$ for $P_{\text{n}}$ in Equation (VI).

$\begin{array}{c}R=\frac{8\times 15.43\text{kips}}{2}\\ =61.72\text{kips}\end{array}$

Consider the forces in vertical equilibrium.

$R-\frac{V_{\text{H}}}{h}{F}_{DE}=0$ ...... (VII)

Here, the vertical height is $V_{\text{H}}$.

Substitute $8\text{ft}$ for $V_{\text{H}}$, $61.72\text{kips}$ for $R$ and $40.79\text{ft}$ for $h$ in Equation (VII).

$\begin{array}{l}61.72\text{kips}-\frac{8\text{ft}}{\left(\sqrt{8^2+{40}^2}\right)\text{ft}}{F}_{\text{DE}}=0\\ F_{\text{DE}}=\frac{61.72\text{kips}}{\frac{8\text{ft}}{\left(\sqrt{8^2+{40}^2}\right)\text{ft}}}\\ F_{\text{DE}}=314.7\text{kips}\end{array}$

Consider the force in horizontal equilibrium.

$F_{\text{DE}}\times \left(\frac{40\text{ft}}{40.79\text{ft}}\right)-F_{\text{FE}}=0$

Substitute $314.7\text{kips}$ for $F_{\text{DE}}$.

$\begin{array}{l}\left(314.7\text{kips}\right)\times \left(\frac{40\text{ft}}{40.79\text{ft}}\right)-F_{\text{FE}}=0\\ F_{\text{FE}}=308.61\text{kips}\end{array}$

Write the expression to calculate the gross area.

$A_{\text{g}}=\frac{F_{\text{FE}}}{0.9F_{\text{y}}}$ ...... (VIII)

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

Substitute $308.61\text{kips}$ for $F_{\text{FE}}$ and $50\text{ksi}$ for $F_{\text{y}}$ in Equation (VIII).

$\begin{array}{c}{A}_{\text{g}}=\frac{308.61\text{kips}}{0.9\times 50\text{ksi}}\\ =6.858{\text{in}}^{\text{2}}\end{array}$

Write the expression to calculate the effective area.

$A_{\text{e}}=\frac{F_{\text{FE}}}{0.75F_{\text{u}}}$ ...... (IX)

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

Substitute $308.61\text{kips}$ for $F_{\text{FE}}$ and $65\text{ksi}$ for $F_{\text{y}}$ in Equation (IX).

$\begin{array}{c}{A}_{\text{e}}=\frac{308.61\text{kips}}{0.75\times 65\text{ksi}}\\ =6.33{\text{in}}^{\text{2}}\end{array}$

Write the expression to calculate the radius of gyration.

$r_{\min }=\frac{L}{300}$ ...... (X)

Here, the length of the member is $L$ and the minimum radius of gyration is $r_{\min }$.

Substitute $25\text{ft}$ for $L$ in Equation (X).

$\begin{array}{c}{r}_{\min }=\frac{10\text{ft}\times \left(\frac{12\text{in}}{1\text{ft}}\right)}{300}\\ =0.4\text{in}\end{array}$

From the above calculated values try double angle section $\text{2L6}\times \text{L6}\times \frac{\text{7}}{\text{16}}$.

The properties for $\text{2L6}\times \text{L6}\times \frac{7}{\text{16}}$ section from AISC steel manual are as follows.

• The gross area $\left(A_{\text{g}}\right)$ is $10.2{\text{in}}^{\text{2}}$.
• The radius of gyration is $1.86\text{in}$.

The gross area of the section is greater than the required gross area of $6.858{\text{in}}^{\text{2}}$.

Thus, the section in terms of gross area is ok.

The radius of gyration of the section is greater than the required radius of gyration of $0.4\text{in}$.

Thus, the section in terms of radius of gyration is ok.

Write the expression to calculate the effective net area.

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

Here, the effective net area is $A_{\text{e}}$ and the shear lag factor is $U$.

Substitute $10.2{\text{in}}^{\text{2}}$ for $A_{\text{n}}$ and $0.85$ for $U$ in Equation (XI).

$\begin{array}{c}{A}_{\text{e}}=0.85\times 10.2{\text{in}}^{\text{2}}\\ =8.67{\text{in}}^{\text{2}}\end{array}$

The section’s effective area is greater than the required area of $6.33{\text{in}}^{\text{2}}$.

Thus, the section in terms of effective area is ok.

For the web members.

The maximum force occurs at $AH$.

Calculate the length of member of member $HG$.

$\begin{array}{c}L=\sqrt{8^2+{10}^2}\text{ft}\\ =\text{12}\text{.81}\text{ft}\end{array}$

Here, the length of the member $HG$ is $L$.

Consider the moment about $E$.

$\begin{array}{l}\frac{8}{12.81}{F}_{AH}\times \left(25\text{ft}\right)-\left(15.42\times 60\right)\text{k-ft}=0\\ F_{AH}=\left(\frac{925.2}{\frac{8\times 25}{12.81}}\right)\text{kips}\\ F_{AH}=\text{59}\text{.26}\text{kips}\end{array}$

Here, the force on the member $AH$ is $F_{AH}$.

Write the expression to calculate the gross area of the member.

$A_{\text{g}}=\frac{F_{\text{AH}}}{0.9F_{\text{y}}}$ ...... (XII)

Substitute $59.26\text{kips}$ for $F_{\text{AH}}$ and $50\text{ksi}$ for $F_{\text{y}}$ in Equation (XII).

$\begin{array}{c}{A}_{\text{g}}=\frac{59.26\text{kips}}{0.9\times 50\text{ksi}}\\ =1.32{\text{in}}^{\text{2}}\end{array}$

Write the expression to calculate the effective area.

$A_{\text{e}}=\frac{F_{\text{AH}}}{0.75F_{\text{u}}}$ ...... (XIII)

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

Substitute $59.26\text{kips}$ for $F_{\text{FE}}$ and $65\text{ksi}$ for $F_{\text{u}}$ in Equation (XIII).

$\begin{array}{c}{A}_{\text{e}}=\frac{59.26\text{kips}}{0.75\times 65\text{ksi}}\\ =1.22{\text{in}}^{\text{2}}\end{array}$

Calculate the radius of gyration.

Substitute $12.81\text{ft}$ for $L$ in Equation (X).

$\begin{array}{c}{r}_{\min }=\frac{\left(12.81\text{ft}\right)\times \left(\frac{12\text{in}}{1\text{ft}}\right)}{300}\\ =0.51\text{in}\end{array}$

From the above calculated values try double angle section $\text{2L3}\times \text{L2}\times \frac{\text{3}}{\text{16}}$.

The properties for $\text{2L6}\times \text{L6}\times \frac{7}{\text{16}}$ section from AISC steel manual are as follows.

• The gross area $\left(A_{\text{g}}\right)$ is $1.83{\text{in}}^{\text{2}}$.
• The radius of gyration is $0.577\text{in}$.

The gross area of the section is greater than the required gross area of $1.32{\text{in}}^{\text{2}}$.

Thus, the section in terms of gross area is ok.

The radius of gyration of the section is greater than the required radius of gyration of $0.512\text{in}$.

Calculate the effective area.

Substitute $1.83{\text{in}}^{\text{2}}$ for $A_{\text{n}}$ and $0.85$ for $U$ in Equation (XI).

$\begin{array}{c}{A}_{\text{e}}=0.85\times 1.83{\text{in}}^{\text{2}}\\ =1.55{\text{in}}^{\text{2}}\end{array}$

The effective area of the section is greater than the required effective area of $1.22{\text{in}}^{\text{2}}$.

Thus, the section in terms of effective area is ok.

Conclusion:

The double angle section $\text{2L6}\times \text{L6}\times \frac{\text{7}}{16}$ is safe in terms of gross area, the radius of gyration and the effective area for the bottom chord.

Thus, the double angle section to be selected for the bottom chord by LFRD method is $\text{2L6}\times \text{L6}\times \frac{\text{7}}{\text{16}}$.

The double angle section $\text{2L3}\times \text{L2}\times \frac{\text{3}}{16}$ is safe in terms of gross area, the radius of gyration and the effective area for the web members.

Thus, the double angle section to be selected for the web members by LFRD method is $\text{2L3}\times \text{L2}\times \frac{3}{\text{16}}$.

To determine

(b)

The double angle section for a tension member using Allowed Strength Design (ASD) method.

Answer to Problem 3.8.4P

The double angle section to be selected for the bottom chord by ASD method is $\text{2L6}\times \text{L6}\times \frac{\text{7}}{\text{16}}$.

The double angle section to be selected for the web members by ASD method is $\text{2L3}\times \text{L2}\times \frac{3}{\text{16}}$.

Explanation of Solution

Calculation:

Write the expression to calculate the factored load using ASD.

$P_{\text{n}}=DL+S$ ...... (XIV)

Substitute $6860\text{lb}$ for $DL$ and $4500\text{lb}$ for $S$ in Equation (XIV).

$\begin{array}{c}{P}_{\text{n}}=6860\text{lb}+4500\text{lb}\\ =\text{11360}\text{lb}\left(\frac{{10}^{-3}\text{kips}}{1\text{lb}}\right)\\ =11.3\text{kips}\end{array}$

Calculate the vertical reactions $R$.

Substitute $8$ for $N_{\text{S}}$ and $11.3\text{kips}$ for $P_{\text{n}}$ in Equation (VI).

$\begin{array}{c}R=\frac{8\times 11.3\text{kips}}{2}\\ =45.2\text{kips}\end{array}$

Calculate the value of force in the member $DE$.

Substitute $8\text{ft}$ for $V_{\text{H}}$, $45.2\text{kips}$ for $R$ and $40.79\text{ft}$ for $h$ in Equation (VII).

$\begin{array}{l}45.2\text{kips}-\frac{8\text{ft}}{40.79\text{ft}}{F}_{\text{DE}}=0\\ F_{\text{DE}}=\frac{45.2\text{kips}}{\frac{8\text{ft}}{40.79\text{ft}}}\\ F_{\text{DE}}=230.5\text{kips}\end{array}$

Consider the force in horizontal equilibrium.

$F_{\text{DE}}\times \left(\frac{40\text{ft}}{40.79\text{ft}}\right)-F_{\text{FE}}=0$

Substitute $230.5\text{kips}$ for $F_{\text{DE}}$.

$\begin{array}{l}230.5\text{kips}\times \left(\frac{40\text{ft}}{40.79\text{ft}}\right)-F_{\text{FE}}=0\\ F_{\text{FE}}=226.04\text{kips}\end{array}$

Write the expression to calculate the required gross area.

$F_{\text{FE}}=0.6F_{\text{y}}{A}_{\text{g}}$ ....... (XV)

Substitute $50\text{ksi}$ for $F_{\text{y}}$ and $226.04\text{kips}$ for $F_{\text{FE}}$ in Equation (XV).

$\begin{array}{l}226.04\text{kips}=0.6\times 50\text{ksi}\times A_{\text{g}}\\ A_{\text{g}}=\frac{226.04\text{kips}}{0.6\times 50\text{ksi}}\\ A_{\text{g}}=7.53{\text{in}}^{\text{2}}\end{array}$

Write the expression to calculate the net area.

$F_{\text{FE}}=0.5F_{\text{u}}{A}_{\text{n}}$ ...... (XVI)

Substitute $65\text{ksi}$ for $F_{\text{u}}$ and $226.04\text{kips}$ for $F_{\text{FE}}$ in Equation (XVI).

$\begin{array}{l}226.04\text{kips}=0.5\times 65\text{ksi}\times A_{\text{e}}\\ A_{\text{e}}=\frac{226.04\text{kips}}{0.5\times 65\text{ksi}}\\ A_{\text{e}}=6.95{\text{in}}^{\text{2}}\end{array}$

From the above calculated values try double angle section $\text{2L6}\times \text{L6}\times \frac{\text{7}}{\text{16}}$.

The properties for $\text{2L6}\times \text{L6}\times \frac{7}{\text{16}}$ section from AISC steel manual are as follows.

• The gross area $\left(A_{\text{g}}\right)$ is $10.2{\text{in}}^{\text{2}}$.
• The radius of gyration is $1.86\text{in}$.

The gross area of the section is greater than the required gross area of $7.53{\text{in}}^{\text{2}}$.

Thus, the section in terms of gross area is ok.

The radius of gyration of the section is greater than the required radius of gyration of $0.4\text{in}$.

Thus, the section in terms of radius of gyration is ok.

The section’s effective area is greater than the required area of $6.95{\text{in}}^{\text{2}}$.

Thus, the section in terms of effective area is ok.

For the web members.

The maximum force occurs in the member $AH$.

Consider the moment about $E$.

$\begin{array}{l}\frac{8}{12.81}{F}_{\text{AH}}\times 25\text{ft}-11.3\text{kips}\times 60\text{ft}=0\\ F_{\text{AH}}=\left(\frac{678}{\frac{8\times 25}{12.81}}\right)\text{kips}\\ F_{\text{AH}}=\text{43}\text{.43}\text{kips}\end{array}$

Write the expression to calculate the required gross area.

$F_{\text{AH}}=0.6F_{\text{y}}{A}_{\text{g}}$ ....... (XVII)

Substitute $50\text{ksi}$ for $F_{\text{y}}$ and $43.43\text{kips}$ for $F_{\text{AH}}$ in Equation (XVII).

$\begin{array}{l}43.43\text{kips}=0.6\times 50\text{ksi}\times A_{\text{g}}\\ A_{\text{g}}=\frac{43.43\text{kips}}{0.6\times 50\text{ksi}}\\ A_{\text{g}}=1.45{\text{in}}^{\text{2}}\end{array}$

Write the expression to calculate the net area.

$F_{\text{AH}}=0.5F_{\text{u}}{A}_{\text{n}}$ ...... (XVIII)

Substitute $65\text{ksi}$ for $F_{\text{u}}$ and $43.43\text{kips}$ for $F_{\text{AH}}$ in Equation (XVIII).

$\begin{array}{l}43.43\text{kips}=0.5\times 65\text{ksi}\times A_{\text{e}}\\ A_{\text{e}}=\frac{43.43\text{kips}}{0.5\times 65\text{ksi}}\\ A_{\text{e}}=1.33{\text{in}}^{\text{2}}\end{array}$

From the above calculated values try double angle section $\text{2L3}\times \text{L2}\times \frac{\text{3}}{\text{16}}$.

The properties for $\text{2L6}\times \text{L6}\times \frac{7}{\text{16}}$ section from AISC steel manual are as follows.

• The gross area $\left(A_{\text{g}}\right)$ is $1.83{\text{in}}^{\text{2}}$.
• The radius of gyration is $0.577\text{in}$.

The gross area of the section is greater than the required gross area of $1.32{\text{in}}^{\text{2}}$.

Thus, the section in terms of gross area is ok.

The radius of gyration of the section is greater than the required radius of gyration of $0.512\text{in}$.

The effective area of the section is greater than the required effective area of $1.33{\text{in}}^{\text{2}}$.

Conclusion:

The double angle section $\text{2L6}\times \text{L6}\times \frac{\text{7}}{16}$ is safe in terms of gross area, the radius of gyration and the effective area for the bottom chord.

Thus, the double angle section to be selected for the bottom chord by ASD method is $\text{2L6}\times \text{L6}\times \frac{\text{7}}{\text{16}}$.

The double angle section $\text{2L3}\times \text{L2}\times \frac{\text{3}}{16}$ is safe in terms of gross area, the radius of gyration and the effective area for the web members.