Chapter 15, Problem 15.1PFS

To determine

The maximum end reaction that can be transferred through the A36 web angle connection by LRFD and ASD

Answer to Problem 15.1PFS

136.89 kips and 91.26 kips

Explanation of Solution

Given:

  $\begin{array}{l}A36\text{web}\text{angle}\\ F_y\text{for}W18\times 55\text{beam}=50\text{ksi}\\ \text{Diameter}\text{of}\text{bolts}=\frac{3}{4}\text{in}.\\ \text{Angle}\text{thickness}=\frac{5}{16}\text{in}.\\ l_{ev}=1\frac{1}{4}\text{in}.\\ l_{eh}=1\frac{1}{2}\text{in}.\end{array}$

Calculation:

For 4 rows of bolts, $\frac{5}{16}\text{in}\text{.}$ angle, $\frac{3}{4}\text{in}.$ and A325-N bolts with standard holes,

The bolt and angle available strength from Table 10-1, “All-Bolted Double-Angle Connections” in AISC manual by LRFD is

  $\varphi R_n=126\text{kips}$

The bolt and angle available strength by ASD is

  $\frac{R_n}{\Omega }=83.9\text{kips}$

For 4 rows of bolts having uncoped section and

  $\begin{array}{l}{l}_{ev}=1\frac{1}{4}\text{in}.\\ l_{eh}=1\frac{1}{2}\text{in}.\end{array}$

The beam web available strength per in. thickness from Table 10-1, “All-Bolted Double-Angle Connections” in AISC manual by LRFD is

  $V_n=351\text{kips/in}.$

The thickness of web of section $W18\times 55$ is

  $t_w=0.39\text{in}.$

The beam web available design strength is

  $\begin{array}{l}\varphi V_n=t_w{V}_n\\ \varphi V_n=0.39\times 351\\ \varphi V_n=136.89\text{kips}\end{array}$

The beam web available strength per in. thickness by ASD is

  $V_n=234\text{kips/in}.$

The beam web available design strength is

  $\begin{array}{l}\frac{V_n}{\Omega }=t_w{V}_n\\ \frac{V_n}{\Omega }=0.39\times 234\\ \frac{V_n}{\Omega }=91.26\text{kips}\end{array}$

The bolt and angle available strength controls both LRFD and ASD.

Conclusion:

The maximum end reaction in LRFD method is 136.89 kips and the maximum end reaction in ASD method is 91.26 kips.