Problem 6.13

The equation to be used is

The shearing stress is given. Q, I, and t are needed. Let's discuss where shear should be considered. If the horizontal shearing stress is greater at the weld, we would need

Actually, the weld cannot be made across the cross section, so the weld cross section is actually much smaller than 220 mm. However, no information is given about this. Let's assume that the weld cross section is of the same order as the web thickness.

If the horizontal shearing stress is greater at the web, we would have

The thickness t has been assumed to be the same, so let's consider the effect of Q. Since the allowable shearing stress is fixed, the use of a larger Q will result in a smaller allowable shearing force. So we must make the calculations at the web.

To compute Q, I need the centroid:

section		A_i (mm²)
top plate	15/2 + 15.7 + 112.8 = 136	(220)(15) = 3300	4.488 x 10⁵
top flange	15.7/2 + 112.8 = 120.65	(204)(15.7) = 3202.8	3.864 x 10⁵
half web	112.8/2 = 56.4	(112.8)(8.9) = 1003.92	56621.1
totals	N/A	7506.72	891.84

The overall centroid of the upper half is 891.84/7506.72 = 118.8 mm, taking the middle of the web as the reference axis. Q is the area sum times this centroid, or 8.918 x 10⁵ mm³.

The moment of inertia is of the entire cross section:

		d (mm)	A (mm²)	Ad² (mm³)	total I (mm⁴)
each plate			(15)(220) = 3300	6.104x10⁷	6.10987 x 10⁷
beam	104 x 10⁶	N/A	8580	N/A	104 x 10⁶

The total, after multiplying the plate moment of inertia by two, is 2.262 x 10⁸ mm⁴.

Finally,