Page 391 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown.cls
Brown˙C09
373
MACHINE ENERGY
The spring rate (k) is also called the stiffness, and the reciprocal of the
Terminology.
stiffness is called the compliance or the flexibility ( f ), which has units of length per unit
force. Typically, the compliance is used in place of stiffness when electrical circuit theory
is used to simulate the dynamic response of mechanical systems, such as in automatic
feedback controls.
Consider the following example where the mean diameter (D), the force (F), and the
maximum shear stress (τ max ) are given and the wire diameter (d) is to be determined.
U.S. Customary SI/Metric
Example 2. Determine the wire diameter (d) Example 2. Determine the wire diameter (d)
for a cylindrical helical spring under static for a cylindrical helical spring under static
conditions, where conditions, where
D = 1in D = 2.5 cm = 0.025 m
F = 50 lb F = 225 N
8
5
τ max = 100 kpsi = 1 × 10 lb/in 2 τ max = 700 MPa = 7 × 10 N/m 2
solution solution
Step 1. As the loading is static, substitute the Step 1. As the loading is static, substitute the
given information in Eq. (9.11) to give given information in Eq. (9.11) to give

d 8FD d 8FD
τ max = + 1 τ max = + 1
2D πd 3 2D πd 3

5 2 d 8 2 d
1 × 10 lb/in = + 1 7 × 10 N/m = + 1
2(1in) 2(0.025 m)
(8)(50 lb)(1in) (8)(225 N)(0.025 m)
× 3 × 3
πd πd
Step 2. Solve for the cube of the wire diameter Step 2. Solve for the cube of the wire diameter
3
3
(d ) in the expression in step 1. (d ) in the expression in step 1.
d (8)(50 lb)(1in) d (8)(225 N)(0.025 m)
3 3
d = + 1 d = + 1
2
2
5
2 π(1 × 10 lb/in ) 0.05 π(7 × 10 N/m )
8
d 1 3 d 1 3

= + 1 in = + 1 7 m
2 785 2 4.9 × 10
Step 3. Multiply the expression in step 2 Step 3. Multiply the expression in step 2
7
throughout by (2 × 785) and move all terms throughout by (2 × 4.9 × 10 ) and move all
to the left hand side to give the cubic equation terms to the left hand side to give the cubic
(without units) as equation (without units) as
3
7
3
(1,570)d − d − 2 = 0 (9.8 × 10 )d − d − 2 = 0
Step 4. Solve the cubic equation in step 3 by Step 4. Solve the cubic equation in step 3 by
trial and error. Start with a guess of (0.25) to trial and error. Start with a guess of (0.006) to
determine the wire diameter (d) to engineering determine the wire diameter (d) to engineering
accuracy. accuracy.
3
7
3
(1,570)d − d − 2 = 0 (9.8 × 10 )d − d − 2 = 0
?
?
7
3
3
(1,570)(0.25) − (0.25) − 2 = 0 (9.8 × 10 )(0.006) − (0.006) − 2 = 0
? ?
(24.53) − (0.25) − 2 = 0 (21.168) − (0.006) − 2 = 0
22.28 > 0 (guess too high) 19.162 > 0 (guess too high)