Overview: The horizontal shear does not occur along a continuous surface, as it would between components that were welded or glued wood blocks. Therefore, I cannot use
Instead, I will find the horizontal shear per unit length, q, in kips/in, using VQ/I. Then I will multiply this by the spacing s to find the shearing force on the bolts.
Here are the data from Appendix C:
| A (in2) | I (in4) |  |
tw (in) | |
| C 8 x 13.75 | 4.04 | 1.53 (Iy) | 0.553 | 0.303 |
| S 10 x 25.4 | 7.46 | 124 (Ix) | 5 | N/A |
Notice that I need Iy for the two C-sections, since they are turned 90o from the picture given in Appendix C?
To compute the moment of inertia of the entire cross section, I'll need d, the distance from the centroid of one of the C-sections to the overall centroid:
| section |  |
A (in2) | d (in) |  |
| C shape | 1.53 | 4.04 | 4.75 | 92.68 |
| C shape | 1.53 | 4.04 | 4.75 | 92.68 |
| S shape | 124 | 124 | ||
| total | 309.365 |
The moment of the area above the shearing plane is:
Therefore, the horizontal shear per unit length is
And the shearing force is 
.
But there are two bolts in line sharing this force, so the force on each bolt is 9.30/2 = 4.65 kips. The shearing force on a bolt is
