Chapter 12, Problem 12.1PFS

To determine

The LRFD design tensile strength and the ASD allowable tensile strength for given members.

Answer to Problem 12.1PFS

The LRFD design tensile strength is $219.1\text{k}$ .

The ASD allowable tensile strength is $146.1\text{k}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{size of plates}=1"\times 16"\\ \text{yield strength of steel},F_y=36\text{ksi}\\ \text{ultimate}\text{tensile}\text{strength}{F}_u=58\text{ksi}\\ \text{Bolt}\text{size}=\frac{7}{8}\text{in}\end{array}$

Formula used:

  $\begin{array}{l}{R}_n=F_{nv}\times A_b\times \text{number}\text{of}\text{bolts}\\ \text{where},R_n\text{is}\text{the}\text{shearing}\text{strength of bolts}\\ {\text{F}}_{nv}\text{is}\text{the}\text{nominal}\text{strength}\text{in}\text{bearing}\text{type}\text{connection}\\ A_b\text{is}\text{the}\text{area of bolts}\end{array}$

Given that the A325, $\frac{7}{8}\text{in}$ a bolt is used.

Calculate the area of the bolt as follows:

  $\begin{array}{c}{A}_b=\frac{\pi }{4}{\left(d\right)}^2\\ =\frac{\pi }{4}{\left(\frac{7}{8}\right)}^2\\ =0.60132{\text{in}}^{\text{2}}\end{array}$

From table 12.5, $F_{nv}=54\text{ksi}$

The shearing strength of the bolt is given by,

  $\begin{array}{c}{R}_n=F_{nv}\times A_b\times \text{number}\text{of}\text{bolts}\\ =\text{54}\times \text{0}\text{.60132}\times \text{9}\\ =292.2\text{k}\end{array}$

By LFRD design, the tensile strength is given as,

  $\begin{array}{c}{R}_n^{\text{'}}=\varphi R_n\\ =0.75\times 292.2\\ =219.1\text{k}\end{array}$

Hence, the tensile strength of LFRD is $219.1\text{k}$

By ASD design, the allowable tensile strength is given by,

  $\begin{array}{c}{R}_n^{\text{'}}=\frac{R_n}{\Omega }\\ =\frac{292.2}{2}\\ =146.1\text{k}\end{array}$

Hence, the allowable tensile strength by ASD design is $146.1\text{k}$

Conclusion:

The tensile strength by LFRD is $219.1\text{k}$ and the allowable tensile strength by ASD design is $146.1\text{k}$