Chapter 3, Problem 3.6.1P

To determine

(a)

The design of a single-angle tension member using Load Resistance Factor Design (LRFD).

Answer to Problem 3.6.1P

The single angle tension member using LRFD is $L8\times 4\times \frac{1}{2}$.

Explanation of Solution

Given:

The member is $18\text{ft}$ long with $1\text{in}$ diameter bolts.

Dead Load is $28\text{kips}$.

Live load is $84\text{kips}$.

Concept Used:

Write the expression for factored load.

$P_{\text{u}}=1.2DL+1.6LL$ …… (I)

Here, the factored load is $P_{\text{u}}$, dead load is $DL$ and live load is $LL$.

Write the expression for the required gross area.

$A_{\text{g}}=\frac{P_{\text{u}}}{0.9F_{\text{y}}}$ …… (II)

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

Write the expression for the required effective area.

$A_{\text{e}}=\frac{P_{\text{u}}}{0.75F_{\text{u}}}$ …… (III)

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

Write the expression for the effective area of the section.

$A_{\text{es}}=A_{\text{ns}}U$ …… (IV)

Here, the effective area of the section is $A_{\text{es}}$, the net area of the section is $A_{\text{ns}},$ and the area reduction factor is $U$.

Write the expression for the net area of the section.

$A_{\text{ns}}=A_{\text{gs}}-t\left(d_{\text{bolt}}+\frac{1}{8}\text{in}\right)$ …… (V)

Here, the gross area of the section is $A_{\text{gs}}$, thickness of the member is $t,$ and the diameter of bolt is $d_{\text{bolt}}$.

Calculation:

Calculate the factored load.

Substitute $28\text{kips}$ for $DL$ and $84\text{kips}$ for $LL$ in the Equation (I).

$\begin{array}{c}{P}_{\text{u}}=\left(1.2\times 28\text{kips}\right)+\left(1.6\times 84\text{kips}\right)\\ =\left(33.6+134.4\right)\text{kips}\\ =168\text{kips}\end{array}$

Calculate the required gross area of the tension member.

Substitute $168\text{kips}$ for $P_{\text{u}}$ and $36\text{ksi}$ for $F_{\text{y}}$ in the Equation (II).

$\begin{array}{c}{A}_{\text{g}}=\frac{168\text{kips}}{0.9\left(36\text{ksi}\right)}\\ =\frac{168}{32.4}{\text{in}}^2\\ =5.185{\text{in}}^2\end{array}$

Calculate the required net area of the tension member.

Substitute $168\text{kips}$ for $P_{\text{u}}$ and $58\text{ksi}$ for $F_{\text{u}}$ in the Equation (III).

$\begin{array}{c}{A}_{\text{e}}=\frac{168\text{kips}}{0.75\left(58\text{ksi}\right)}\\ =\frac{168}{43.5}{\text{in}}^2\\ =3.862{\text{in}}^2\end{array}$

Before choosing the sections calculate the minimum radius of gyration.

Write the expression for the radius of gyration.

$r_{\text{min}}=\frac{L}{300}$ …… (VI)

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

Substitute $18\text{ft}$ for $L$ in Equation (VI).

$\begin{array}{c}{r}_{\text{min}}=\frac{18\text{ft}}{300}\times \left(\frac{12\text{in}}{1\text{ft}}\right)\\ =\frac{18\times 12\text{in}}{300}\\ =0.72\text{in}\end{array}$

Choose the section that has parameters slightly above the required values.

Trial 1:

Choose the section $L8\times 4\times \frac{1}{2}$.

The properties of the section are,

$\begin{array}{l}{A}_{\text{gs}}=5.80{\text{in}}^2\\ r_{\text{mins}}=0.863\text{in}\end{array}$

Here, the gross area of the section is $A_{\text{gs}}$ and the minimum radius of gyration of the section is $r_{\min \text{s}}$.

Compare the gross area.

$\left(A_{\text{gs}}=5.80{\text{in}}^2\right)>\left(A_{\text{g}}=5.185{\text{in}}^2\right)$

Hence, the section is feasible.

Compare the radii of gyration.

$\left(r_{\text{mins}}=0.863\text{in}\right)\text{>}\left(r_{\text{min}}=0.72\text{in}\right)$

Hence, the section is feasible.

Calculate the net area of the section.

Substitute $5.80{\text{in}}^2$ for $A_{\text{gs}}$ and $1\text{in}$ for $d_{\text{bolt}}$ in the Equation (V).

$\begin{array}{c}{A}_{\text{ns}}=5.80{\text{in}}^2-\frac{1}{2}\left(1\text{in}+\frac{1}{8}\text{in}\right)\\ =5.80{\text{in}}^2-0.5\times 1.125{\text{in}}^2\\ =\left(5.80-0.563\right){\text{in}}^2\\ =5.233{\text{in}}^2\end{array}$

Calculate the effective area of the section.

Substitute $5.233{\text{in}}^2$ for $A_{\text{ns}}$ and $0.80$ for $U$ in the Equation (IV).

$\begin{array}{c}{A}_{\text{es}}=5.233\times 0.80\\ =4.19{\text{in}}^2\end{array}$

Compare the effective area required and that of the section.

$\left(A_{\text{es}}=4.19{\text{in}}^2\right)>\left(A_{\text{e}}=3.862{\text{in}}^2\right)$

Hence, the section is feasible for design.

Adopt the section $L8\times 4\times \frac{1}{2}$.

Conclusion:

Thus, the single angle tension member using LRFD is $L8\times 4\times \frac{1}{2}$.

To determine

(b)

The single angle tension member using Allowable Strength Design (ASD).

Answer to Problem 3.6.1P

The single angle tension member using ASD is $L8\times 4\times \frac{1}{2}$.

Explanation of Solution

Concept used:

Write the expression for factored load,

$P_{\text{u}}=DL+LL$ …… (VII)

Here, the factored load is $P_{\text{u}}$, dead load is $DL,$ and live load is $LL$.

Write the expression for the required gross area.

$A_{\text{g}}=\frac{P_{\text{u}}}{0.6F_{\text{y}}}$ …… (VIII)

Write the expression for the required effective area.

$A_{\text{e}}=\frac{P_{\text{u}}}{0.5F_{\text{u}}}$ …… (IX)

Write the expression for the effective area of the section.

$A_{\text{es}}=A_{\text{ns}}U$ …… (X)

Here, the effective area of the section is $A_{\text{es}}$, the net area of the section is $A_{\text{ns}},$ and the area reduction factor is $U$.

Calculation:

Calculate the factored load.

Substitute $28\text{kips}$ for $DL$ and $84\text{kips}$ for $LL$ in the Equation (VII).

$\begin{array}{c}{P}_{\text{u}}=28\text{kips}+84\text{kips}\\ =112\text{kips}\end{array}$

Calculate the required gross area of the tension member.

Substitute $168\text{kips}$ for $P_{\text{u}}$ and $36\text{ksi}$ for $F_{\text{y}}$ in the Equation (VIII).

$\begin{array}{c}{A}_{\text{g}}=\frac{112\text{kips}}{0.6\left(36\text{ksi}\right)}\\ =\frac{112}{21.6}{\text{in}}^2\\ =5.185{\text{in}}^2\end{array}$

Calculate the required net area of the tension member.

Substitute $112\text{kips}$ for $P_{\text{u}}$ and $58\text{ksi}$ for $F_{\text{u}}$ in the Equation (IX).

$\begin{array}{c}{A}_{\text{e}}=\frac{112\text{kips}}{0.5\left(58\text{ksi}\right)}\\ =\frac{112}{29}{\text{in}}^2\\ =3.862{\text{in}}^2\end{array}$

Before choosing the sections calculate the minimum radius of gyration.

Substitute $18\text{ft}$ for $L$ in Equation (VI).

$\begin{array}{c}{r}_{\text{min}}=\frac{18\text{ft}}{300}\times \left(\frac{12\text{in}}{1\text{ft}}\right)\\ =\frac{18\times 12\text{in}}{300}\\ =0.72\text{in}\end{array}$

Choose the section that has parameters slightly above the required values.

Trial 1:

Choose the section $L8\times 4\times \frac{1}{2}$.

The properties of the section are,

$\begin{array}{l}{A}_{\text{gs}}=5.80{\text{in}}^2\\ r_{\text{mins}}=0.863\text{in}\end{array}$

Here, the gross area of the section is $A_{\text{gs}}$ and the minimum radius of gyration of the section is $r_{\min \text{s}}$.

Compare the gross area.

$\left(A_{\text{gs}}=5.80{\text{in}}^2\right)>\left(A_{\text{g}}=5.185{\text{in}}^2\right)$

Hence, the section is feasible.

Compare the radii of gyration.

$\left(r_{\text{mins}}=0.863\text{in}\right)\text{>}\left(r_{\text{min}}=0.72\text{in}\right)$

Hence, the section is feasible.

Calculate the net area of the section.

Substitute $5.80{\text{in}}^2$ for $A_{\text{gs}}$ and $1\text{in}$ for $d_{\text{bolt}}$ in the Equation (V).

$\begin{array}{c}{A}_{\text{ns}}=5.80{\text{in}}^2-\frac{1}{2}\left(1\text{in}+\frac{1}{8}\text{in}\right)\\ =5.80{\text{in}}^2-0.5\times 1.125{\text{in}}^2\\ =\left(5.80-0.563\right){\text{in}}^2\\ =5.233{\text{in}}^2\end{array}$

Calculate the effective area of the section.

Substitute $5.233{\text{in}}^2$ for $A_{\text{ns}}$ and $0.80$ for $U$ in the Equation (IV).

$\begin{array}{c}{A}_{\text{es}}=5.233\times 0.80\\ =4.19{\text{in}}^2\end{array}$

Compare the effective area required and that of the section.

$\left(A_{\text{es}}=4.19{\text{in}}^2\right)>\left(A_{\text{e}}=3.862{\text{in}}^2\right)$

Hence, the section is feasible for design.

Adopt the section $L8\times 4\times \frac{1}{2}$.

Conclusion:

Thus, the single angle tension member using ASD is $L8\times 4\times \frac{1}{2}$.