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Solution: The first part of the problem is to determine the growth equation Starting with dP -= kP , the general growth equation is P = Poekt dt

Finally, the strain is simply de ection per unit length, so the de ection of the rod is d = Le = FL EA

1 + [ f ( x) ]2

The spring rate follows directly from this expression, so for the axially loaded rod K= F EA = d L

Equivalent Linear and Rotary Spring Constants Spring constant (K = F/x)

1 + [6x 1/2 ]2

L Simply Supported Beam L/2

so that on switching,

( 1 + 36x) 1/2 dx 1 ( 1 + 36x) 3/2 54 1 1093/2 13/2 54 ( 109) 3/2 1 54

Spring constant (K1 = T/Dq)

338 d

Let us calculate the length of the graph of the function f ( x) = ( 1/2) ( ex + e x ) over the interval [1, ln 8]

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1143 Combinations of Springs If combinations of springs are connected by elements that are assumed to be massless, the combination can be replaced with a single spring of equivalent stiffness and free length The need for this often arises as a result of model reduction where a mass has been assumed to be zero (as with the suspension in Fig 113c) There are two possible ways to connect springs, known as parallel and series connections 11431 Springs in Parallel Two springs connected in parallel, as shown in Fig 1114a, can be replaced by the single spring in Fig 1114b to simplify modeling The spring constant for the equivalent spring can be found by realizing that two springs in parallel are constrained to have the same de ection Thus, if the two springs de ect by a distance d in response to an applied force F, the rst spring will produce a force

We calculate that f ( x) = ( 1/2) ( ex e x )

F1 = K1d while the second spring produces a force F2 = K2d The sum of these forces must equal the applied force F so F = F1 + F2 = ( K1 + K2 )d = Keqd Clearly, the equivalent spring stiffness is therefore Keq = K1 + K2

k = 026

(1148)

= = =

(1149)

(1150)

graph of y = sin x on the interval /4 x 3 /4

(1151)

Sometimes an arc length problem is more conveniently solved if we think of the curve as being the graph of x = g (y) Here is an example

The specific growth law for these deer in this habitat is P = Poe026t, Now solve for the initial number PO needed to produce the 10,000 population in 5 more years 10,000 = P0e026'5 P0e130= 367Po or PO = 2725 =

Since the equivalent stiffness is simply the sum of the two individual spring stiffnesses, if the two springs have very different stiffnesses, the equivalent stiffness will be dominated by the spring with higher stiffness 11432 Springs in Series Two springs connected end-to-end, as shown in Fig 1115a, are de ned to be connected in series These springs can also be replaced by a single equivalent spring in Fig 1115b The stiffness of the equivalent spring can be derived by realizing that the forces must sum to zero at the point where the springs connect for the spring to be in equilibrium As a result, the force exerted on each spring is the same F1 = F2 = F (1152)

The de ection in each spring is different with the total de ection of the system equal to the sum of the two spring de ections Taking this information together with the spring stiffnesses, d = d1 + d 2 = F1 F2 1 F 1 + = F + = K1 K2 Keq K1 K2 (1153)

Calculate the length of that portion of the graph of the curve 16x 2 = 9y3 between the points ( 0, 0) and ( 6, 4)

So 1 1 1 = + Keq K1 K2 (1154)

Then dx/dy = 9 y1/2 Now, reversing the roles of x and y in ( ) , we find 8 that the requested length is

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