Chapter 4, Problem 4.1PFS

To determine

The lightest section of W10 which support the working loads.

Answer to Problem 4.1PFS

The lightest section is $\text{W}10\times 33$ for LRFD and ASD.

Explanation of Solution

Given:

The specified minimum yield stress is $F_y=50\text{ksi}$, the specified minimum tensile strength is $F_u=65\text{ksi}$ ,working tensile loads are $P_D=120\text{k}$ and $P_E=160\text{k}$, the length of the member is $18\text{ft}$ and the diameter of bolt is $\frac{3}{4}\text{in}$.

Concept Used:

The formula to calculate the ultimate load.

$P_{\text{u}}=1.2P_D+1.0P_E$

The formula to calculate the allowable load.

$P_{\text{a}}=P_D+0.7P_E$

The formula to calculate the minimum gross area are

For LRFD,

$\min A_g=\frac{P_{\text{u}}}{{\varphi }_t{F}_y}$

And,

$\begin{array}{c}\min A_g=\min A_n+\text{estimated}\text{area}\text{of}\text{holes}\\ \text{=}\frac{P_u}{{\varphi }_t{F}_{u}U}+n\left(d+\frac{1}{8}\right)t_f\end{array}$

The formula to calculate minimum $r$.

$\min r=\frac{L}{300}$

The formula to calculate the nominal tensile strength.

$P_n=F_y{A}_g$

The formula to calculate the net area.

For ASD,

$\begin{array}{c}{A}_n=A_g-\text{estimated}\text{area}\text{of}\text{holes}\\ \text{=}{A}_g-n\left(d+\frac{1}{8}\right)t_f\end{array}$

The formula to calculate the effective area.

$A_e=U{A}_n$

Calculate the slenderness ratio.

$\frac{L}{r_{\min }}<300$

Calculation:

Calculate the ultimate load.

$\begin{array}{c}{P}_{\text{u}}=1.2P_D+1.0P_E\\ =1.2\left(120\text{k}\right)+1.0\left(160\text{k}\right)\\ =304\text{k}\end{array}$

Calculate the allowable load.

$\begin{array}{c}{P}_{\text{a}}=P_D+0.7P_E\\ =120\text{k}+0.7\left(160\text{k}\right)\\ =232\text{k}\end{array}$

Calculate the minimum gross area.

$\begin{array}{c}\min A_g=\frac{P_{\text{u}}}{{\varphi }_t{F}_y}\\ =\frac{304\text{k}}{0.9\left(50\text{ksi}\right)}\\ =6.76{\text{in}}^2\end{array}$

Assume reduction co-efficient $U$ is $0.85$ and the thickness of the flange $t_f$ is $0.44\text{in}$.

Calculate the minimum gross area.

$\begin{array}{c}\min A_g=\min A_n+\text{estimated}\text{area}\text{of}\text{holes}\\ \text{=}\frac{P_u}{{\varphi }_t{F}_{u}U}+n\left(d+\frac{1}{8}\right)t_f\\ =\frac{304\text{k}}{0.75\times 65\times 0.85}+4\left(\frac{3}{4}+\frac{1}{8}\right)\left(0.44\right)\\ =8.88{\text{in}}^2\end{array}$

Calculate the minimum $r$.

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

Take section $\text{W}10\times 33$.

From ASTM, the area of the section is $A=9.71{\text{in}}^2$, the minimum radius of gyration is $r=1.94\text{in}$, the breadth of the flange is $b_f=7.96\text{in}$, the depth is $d=9.73\text{in}$ and the thickness of the flange is $t_f=0.435\text{in}$.

Check for gross section yielding.

Calculate the nominal tensile strength.

$\begin{array}{c}{P}_n=F_y{A}_g\\ =50\left(9.71\right)\\ =485.5\text{k}\end{array}$

For LRFD,

Calculate the design tensile strength.

$\begin{array}{c}{\varphi }_t{P}_n=0.9\left(485.5\text{k}\right)\\ =436.95\text{k}>304\text{k}\left(\text{ok}\right)\end{array}$

For ASD,

Calculate the allowable design strength.

$\begin{array}{c}\frac{P_n}{{\Omega }_t}=\frac{485.5\text{k}}{1.67}\\ =290.72\text{k}>232\text{k}\left(\text{ok}\right)\end{array}$

Check for tensile rupture strength.

$\begin{array}{c}{b}_f\ge \frac{2}{3}d\\ 7.96\text{in}\ge \frac{2}{3}\left(9.73\text{in}\right)\\ 7.99\text{in}\ge 6.48\text{in}\end{array}$

Calculate the net area.

$\begin{array}{c}{A}_n=A_g-\text{estimated}\text{area}\text{of}\text{holes}\\ \text{=}{A}_g-n\left(d+\frac{1}{8}\right)t_f\\ =9.71{\text{in}}^2-4\left(\frac{3}{4}\text{in}+\frac{1}{8}\right)\left(0.435\text{in}\right)\\ =8.18{\text{in}}^2\end{array}$

Take the reduction coefficient $U=0.90$.

Calculate the effective area.

$\begin{array}{c}{A}_e=U{A}_n\\ =0.90\left(8.18{\text{in}}^2\right)\\ =7.36{\text{in}}^2\end{array}$

Calculate the nominal tensile strength.

$\begin{array}{c}{P}_n=F_u{A}_e\\ =65\left(7.36\right)\\ =478.4\text{k}\end{array}$

For LRFD,

Calculate the design tensile strength.

$\begin{array}{c}{\varphi }_t{P}_n=0.75\left(478.4\text{k}\right)\\ =358.8\text{k}>304\text{k}\left(\text{ok}\right)\end{array}$

For ASD,

Calculate the allowable design strength.

$\begin{array}{c}\frac{P_n}{{\Omega }_t}=\frac{478.4\text{k}}{2}\\ =239.2\text{k}>232\text{k}\left(\text{ok}\right)\end{array}$

Calculate the slenderness ratio.

$\begin{array}{c}\frac{L}{r_{\min }}<300\\ \frac{\left(18\text{ft}\times \frac{12\text{in}}{1\text{ft}}\right)}{1.94}<300\\ 111.34<300\left(\text{ok}\right)\end{array}$

Conclusion:

Therefore, the lightest section is $\text{W}10\times 33$ for LRFD and ASD.