The rain has started out here in Oregon and consequently, I have been spending my nights engineering the joints for a timberframe of my own design. I have been referencing the Guild's Timber Frame Joinery and Design Workbook extensively and in the process have developed a theory I would like to run up the collective flagpole.

Specifically, my theory involves calculating the allowable shear strength of mortised carrying members. Borrowing an example from the workbooks' article, Joint Engineering I (page 42), suppose we are calculating the shear between a floor joist and it's carrying member as a basic shear connection. I will assume the same design values and member sizing.

In this example, a floor frame of 5 x 6 floor joists are tenoned into an 8 x 10 chimney girt. The shear force at the end of each joist is calculated to be: V=700lbs. If the 5 x 6 joist uses a soffit tenon, 4 in of timber remains below the mortise in the 8 x 10 girt.

In this example, we calculate the horizontal shear force in the mortised member by hypothesizing that each joist is supported by a 4 x 4 cross section of the remaining chimney girt. I agree with this logic but would theorize that if we treat this 4 x 4 cross section as a stand alone beam, the 700 lb joist load would be distributed uniformly along a 5 inch section of the 4 x 4 beam. Therefore, 350 lbs of load would apply toward the shear at the one end of the 4 x 4, and 350 lbs of load would apply at the other end of the 4 x 4. This approach effectively cuts the resulting shear values in half for the mortised member... or treats the condition for failure as a double-shear problem, depending on how you look at it. In this example, the width of the joist (5 in), is greater than the height of the theoretical beam (4 in), so I thought solving for shear along two surfaces would be an acceptable solution to the problem.

This approach is key for me if I am to keep my principle rafters at 8 x 10 and still use drop-in purlins for ease-of-assembly, (with simpson straps across the top for tensile integrity). Do you think this is a reasonable method for calculating shear in mortised members? I would like to use this approach on the rest of my joint calculations but don't want to see my purlins falling through the rafter and cracking me in the head on some snowy day years from now. Thoughts from the floor?

Best regards,
Bart Popenoe

You are correct in calculating the shear in the chimnet girt at the end of the joist as a double shear problem. The wood under each side of the recieving mortise carries half of the 700 lbs of shear at the end of the joist. For the joint to fail, both sections of wood have to rip out. There is also extra protection in the fact that the girt does not end immediately past the mortise. Shear failure is a problem at the end of members because the cut end of the timber allows the wood fibers to move and seperate. The wood in the rest of the beam around an interior mortise helps restrain the shearing movement.