codySTR
Structural
- Dec 28, 2017
- 32
Hi all,
Using NDS2015, I'm calculating the connection capacity of a bolted knife plate connection in a heavy timber truss. The truss is to be constructed using 6"x6" nominal Douglas-Fir Larch truss members with 1/2" steel knife plates (i.e. an incision is made in the timber to sandwich and hide the steel - bolts are slid through the three "members" at the connection) at the joints/connections. The yield limit equations for double shear per Table 12.3.1A require determination of "ls" and "lm", the side member dowel bearing length and main member dowel bearing length. This determination is what is confusing me. I'm considering the 0.5" steel knife plate as my main member and the remaining material cut out of the 6x6 Doug-Fir as my two side members. Determining "lm" is easy: 1/2". However, what would you use for "ls"? Should "ls" be taken as 2.5", i.e. the thickness of ONE of the side members (that is: 0.5*[5.5" dressed lumber thickness - 0.5" steel])? Or should "ls" be taken as 5", i.e. the COMBINED thickness of both side members (that is: 5.5" wood - 0.5" steel)? A lot of my confusion stems from the dimensions and leaders used in Figure 12B and Figure 12C, which poorly indicate the correct values to use for "ls" in my opinion.
The choice here has a serious implication: Using the thickness of BOTH side members to determine "ls" provides TWICE the capacity compared to using the thickness of just one side member.
Thoughts?
Note: previous version of NDS use similar/equivalent formulae.
Using NDS2015, I'm calculating the connection capacity of a bolted knife plate connection in a heavy timber truss. The truss is to be constructed using 6"x6" nominal Douglas-Fir Larch truss members with 1/2" steel knife plates (i.e. an incision is made in the timber to sandwich and hide the steel - bolts are slid through the three "members" at the connection) at the joints/connections. The yield limit equations for double shear per Table 12.3.1A require determination of "ls" and "lm", the side member dowel bearing length and main member dowel bearing length. This determination is what is confusing me. I'm considering the 0.5" steel knife plate as my main member and the remaining material cut out of the 6x6 Doug-Fir as my two side members. Determining "lm" is easy: 1/2". However, what would you use for "ls"? Should "ls" be taken as 2.5", i.e. the thickness of ONE of the side members (that is: 0.5*[5.5" dressed lumber thickness - 0.5" steel])? Or should "ls" be taken as 5", i.e. the COMBINED thickness of both side members (that is: 5.5" wood - 0.5" steel)? A lot of my confusion stems from the dimensions and leaders used in Figure 12B and Figure 12C, which poorly indicate the correct values to use for "ls" in my opinion.
The choice here has a serious implication: Using the thickness of BOTH side members to determine "ls" provides TWICE the capacity compared to using the thickness of just one side member.
Thoughts?
Note: previous version of NDS use similar/equivalent formulae.
