Question 8.13: A fabricated beam is made by riveting square aluminum bars t...
A fabricated beam is made by riveting square aluminum bars to a vertical plate, as shown in Figure 8–23. The bars are 20 mm square. The plate is 6 mm thick and 200 mm high. The rivets can withstand 800 N of shearing force across one cross section. Determine the required spacing of the rivets if a shearing force of 5 kN is applied.
Objective Specify a suitable spacing for the rivets.
Given Shearing force = 5 kN. Fsd = 800 N/rivet.
Beam shape and dimensions in Figure 8–23.
Analysis Use the guidelines for specifying the spacing of fasteners.
Results Step 1. I is the moment of inertia of the entire cross section:
I = \frac{6(200)^{3}}{12}+4[\frac{20^{4}}{12}+(20)(20)(90)^{2}]
I = 17.0 \times 10^{6} mm^{4}
Step 2. Q is the product of Ayp for the area outside the section where the shear is to be calculated. In this case, the partial area A_{p} is the 20 mm square area to the side of the web.
For the beam in Figure 8–23,
Q = A_{p} \bar{y} (for one square bar)
Q = (20)(20)(90) mm³ = 36 000 mm³
Step 3. Then for V = 5 kN,
Q = \frac{VQ}{I} = \frac{(5 \times 10^{3} N)(36 \times 10^{3} mm³)}{17.0 \times 10^{6} mm^{4}} = 10.6 N/mm
Thus a shearing force of 10.6 N is to be resisted for each millimeter of length of the beam.
Step 4. Since each rivet can withstand 800 N of shearing force, the maximum spacing is
s_{max} = \frac{F_{sd}}{q} = \frac{800 N}{10.6 N/mm} = 75.5 mm
Step 5. Specify a spacing of s = 75 mm.