Question 2.1.3: A security camera is to be mounted on a building in an alley......

The sketch on the left in Figure 2.6 suggests the beam–mass, transverse-vibration model of Figure 1.26, repeated in the middle of Figure 2.6, for modeling this system. The beam–mass model, in turn, suggests the spring–mass system on the right in Figure 2.6. The equation of motion is then given by equation (2.2) with the beam stiffness suggested in the figure or

m\ddot{x}(t)  +  kx(t)  =  F_{0}  \cos wt                   (2.2)

m\ddot{x}(t)  +  \frac{3El}{l³} x(t)  =  F_{0}  \cos wt

From strength of materials, the value of I for a rectangular beam is

I = \frac{bh³}{12}

Thus the natural frequency of the system is given by

Note that the length l is the quantity we need to solve for in the design.

Next, consider the expression for the maximum deflection of the response computed in Example 2.1.2. Requiring this amplitude to be less than 0.01 m yields the following two cases:

\left|\frac{2f_{0}}{w²_{n}  –  w²}\right| < 0.01 ⇒ (a) -0.01 < \frac{2f_{0}}{w²_{n}  –  w²} and (b) \frac{2f_{0}}{w²_{n}  –  w²} < 0.01

First, consider case (a), which holds for w²_{n}  –  w² < 0:

-0.01 < \frac{2f_{0}}{w²_{n}  –  w²} ⇒ 2f_{0} < 0.01w² – 0.01w²_{n} ⇒ 0.01w² – 2f_{0} > 0.01\frac{Ebh³}{4ml³} ⇒ l³ > 0.01\frac{Ebh3}{4m(0.01w²  –  2f_{0})} = 0.02 ⇒ l > 0.272 m

Next, consider case (b), which holds for w²_{n}  –  w² > 0:

\frac{2f_{0}}{w²_{n}  –  w²} < 0.01 ⇒ 2f_{0} < 0.01w²_{n}  –  0.01w²  ⇒ 2f_{0} + 0.01w² < 0.01\frac{Ebh³}{4ml³} ⇒ l³ < 0.01\frac{Ebh3}{4m(2f_{0}  +  0.01w²)} = 0.012 ⇒ l < 0.229 m

Here the value of E for aluminum is taken from Table 1.2 (7.1 × 10^{10} N/m²) and ω is changed to rad/s. To conserve material, case (b) is chosen. Given the constraint that the length of the bracket must be at least 0.2 m, 0.2 < l < 0.229 so the value of l = 0.22 is chosen for the solution.

Next, a couple of simple checks are performed to make sure the assumptions made in solving the problem are reasonable. First, compute the value of ω_{n} for this value of l to see that the solution is consistent with the inequality:

w²_{n}  –  w² = \frac{3Ebh³}{12ml³}  –  (20 \pi)² = (74.543)² – (20π)² = 1609 > 0

Thus case (b) is satisfied.

Note that the mass of the mounting bracket was neglected. It is always important to check assumptions. Using the density for aluminum given in Table 1.2, the mass of the bracket is

m = ρlbh = (2.7 × 10³) (0.22) (0.01) (0.01) = 0.149 kg

This is less than the mass of the camera (about 5%), so according to Example 1.4.4 it is reasonable to ignore the spring’s mass in these calculations.