Chapter 2, Problem 20P

An assembly of three springs connected in series as shown in Fig. P2.20 has an equivalent stiffness k given by

$k=\frac{k_1{k}_2{k}_3}{k_2{k}_3+k_1{k}_3+k_1{k}_2}$ (a) Suppose $k_2=6$ lb/in., $k_3=k_1+8$ lb/in., and the equivalent stiffness is $k=2$ lb/in. Substitute these values into equation (2.66) to obtain the following quadratic equation:

$4k_1^2+8k_1-96=0$ (2.67)

(b) Solve equation (2.67) for k ₁ by each of the following method: (i) factoring, (ii) quadratic formula, and (iii) completing the square. For each ca. determine the value of’ k ₃ corresponding to the only physical coalition for k ₁.