A spring-sizing solution is effective requiring a simple root-finding algorithm such as Newton's Method or MathCAD.

1. NCEES Excercise 444 Polynomial-root Solution.mcdx
444.
SITUATION:
The front end of a car weighs 1200 pounds and is to have a natural frequency of 2 Hz.
Two (2) steel helical springs are to be used to support the front end of the vehicle.
The mean helix diameter is to be 4-1/2 inches.
The frame of the vehicle can deflect dynamically 1-1/2 inches relative to the axle before bottoming out
on a rubber bumper.
The maximum allowable spring shear stress is to be 95 ksi and fatigue effects are to be neglected.
The spring is to have square ends.
REQUIREMENTS:
(a) Determine the required minimum wire diameter for the spring.
(b) Determine the number of active spring coils.
(c) Determine the minimum free length of the spring.
Reference
"Engineering Examinations Principles and Practice of Engineering (PE) Sample Examination Group"
1984 ©
Solution
1.1 Determine the stiffness of the suspension system
Natural frequence of car-spring system: f ≔ 2 ⋅ Hz
The mean helix:
D ≔ 4.5 ⋅ in
Dynamic Deflection: δD ≔ 1.5 ⋅ in
The front end of a car weighs *:
w ≔ 1200 ⋅ lb
Shear strength of the spring: τ ≔ 95 ⋅ ksi
‾‾‾ ‾‾‾‾‾
gc ⋅ ke
k
ω ＝ ―＝ ――
m
w
where
lb
gc ≡ g ⋅ ――
lbf
rad
ω ≔ 2 ⋅ π ⋅ f = 12.57 ――
sec
* The soluton erroneously states 12000 lb but the calculation uses the correct weigh, 12000 lb
Julio C. Banks, PE, CGC
Page 1 of 5

2. NCEES Excercise 444 Polynomial-root Solution.mcdx
Solve for the stiffness of the spring system:
2 ⎛w⎞
lbf
ke ≔ ω ⋅ ⎜― = 490.8 ――
⎟
in
⎝ gc ⎠
Solve for the stiffness of each spring. Since the springs are in series (undergo identical
displacement) then ke ＝ n ⋅ k where n is the number of discrete springs
Number of springs:
n≔2
Maximum Shear stress in the spring:
Let
τmax ≔ τ = 95 ksi
1
lbf
k ≔ ― ke = 245.4 ――
⋅
n
in
∴
Let
8
F⋅D
τmax ＝ ― KW ⋅ ――
⋅
3
π
d
2
π τmax
a ＝ ― ――D
⋅
⋅
8
F
1 ⎛g⎞
Fst ≔ ― ⎜― ⋅ w = 600 lbf
⋅
n ⎝ gc ⎟
⎠
FD ≔ k ⋅ δD = 368.1 lbf
F ≔ Fst + FD = 968.1 lbf
1.1 Determine the required spring index, C
It can be shown that equations 1 through 4 produces the polynomial Eq. 14
Shear stress:
8
F⋅D
τmax ＝ ― KW ⋅ ――
⋅
3
π
d
(1)
Wahl factor:
4 ⋅ C − 1 0.615
KW ＝ ――― ――
+
4⋅C−4
C
(2)
D
C＝―
d
(3)
Where:
4 ≤ C ≤ 12
Solve for wired diameter, d, from Eq. 3
D
d＝―
C
Julio C. Banks, PE, CGC
(4)
Page 2 of 5

3. NCEES Excercise 444 Polynomial-root Solution.mcdx
The spring index, C, is calculated from equation 14 using a root-finding solution such as
Newton's Method of MathCAD
a3 ＝ 0.365
a2 ＝ −0.615
(12)
a0 ＝ λ
3
(11)
a1 ＝ −λ
4
(10)
(13)
2
(14)
C + a3 ⋅ C + a2 ⋅ C + a1 ⋅ C + a0 ＝ 0
Since the solution is the coefficients are dimensionless, then the answer is also dimensionless
and the coefficients being of homogeneous dimension, unity (1) we can store them in an
array.
2
π τmax
λ ≔ ― ――D = 780.3
⋅
⋅
8
F
(11)
(12)
a0 ≔ λ = 780.3
3
(10)
a1 ≔ −λ = −780.3
4
a3 ≔ 0.365
a2 ≔ −0.615
Let
(13)
2
C + a3 ⋅ C + a2 ⋅ C + a1 ⋅ C + a0 ＝ 0
4 ≤ C ≤ 12
(14)
Solve Eq. 13 with a polynomial solver where the objective function, O (x) , is the polynomial
itself.
Let
4
3
2
O (C) ≔ C + a3 ⋅ C + a2 ⋅ C + a1 ⋅ C + a0
C ≔ root (O (x) , x , 4 , 12) = 8.74
Check the root (solution): O (C) = −4.4 ⋅ 10
Julio C. Banks, PE, CGC
−12
Page 3 of 5

4. NCEES Excercise 444 Polynomial-root Solution.mcdx
1.2 Determine the spring-wire diameter
Post-Process:
D
d ≔ ― 0.515 in
=
C
0.52 − 0.515
ε ≔ ――――= 1.0 1%
0.515
2.0 Determine the total number of turns in the spring-coil
G⋅D
Λ ＝ ――
8⋅k
(19)
Λ
n ＝ ――
4
C
(20)
3
Assuming a shear modulus of steel to be G ≡ 12 ⋅ 10 ⋅ ksi
4
G⋅D
Λ ≔ ―― 2.751 ⋅ 10
=
8⋅k
Λ
The number of active turns is n ≔ ――
= 4.7
4
C
For "Square-end" condition given, add 1.5 turns. Therefore, n' ≔ n + 1.5 = 6.2
3.0 Determine the minimum free-length
The minimum free length is the wire diameter, d, times the number of turns plus the
deflections.
Let δn ≔ n' ⋅ d = 3.193 in
Fst
δS ≔ ――
= 2.445 in
k
δD = 1.500 in
δF ≔ δn + δS + δD = 7.1 in
Julio C. Banks, PE, CGC
Page 4 of 5

5. NCEES Excercise 444 Polynomial-root Solution.mcdx
Summary
It is an interesting problem to ask the student to show that the combination of equations 1
through 4 and 6 produces the polynomial C, Eq. 14. Also, that Eq. 16 can be expressed in
terms of as D and C as given in Eq. 20.
It has been shown that equations 1, through 4 can be combined
8
F⋅D
τmax ＝ ― KW ⋅ ――
⋅
3
π
d
(1)
4 ⋅ C − 1 0.615
KW ＝ ――― ――
+
4⋅C−4
C
(2)
Where:
D
C＝―
d
(3)
Post Process:
D
d＝―
C
(4)
2
π τmax
λ ＝ ― ――D
⋅
⋅
8
F
(6)
a3 ≔ 0.365
(10)
a2 ≔ −0.615
(11)
a1 ≔ −λ
(12)
a0 ≔ λ
(13)
Shear stress:
Wahl factor:
4
3
2
C + a3 ⋅ C + a2 ⋅ C + a1 ⋅ C + a0 ＝ 0
4 ≤ C ≤ 12
4 ≤ C ≤ 12
(14)
G⋅D
Λ ＝ ――
8⋅k
Λ
n ＝ ――
4
C
Julio C. Banks, PE, CGC
(19)
(20)
Page 5 of 5

6. ENGINEERING
EXAMINATIONS
'1ulio C c:B~nr;
PRINCIPLES AND PRACTICE
OF
ENGINEERING
(PE)
Sample Examination
GROUP I
Chemical, Civil/Sanitary/Structural,
Electrical, Mechanical
Published by
The National Council of Engineering Examiners
The questions and materials contained in this publication are designed to acquaint the
reader with typical NCEE examinations. No representation is made or intended as to
future questions, c<?ntent, or subject matter.

7. NATIONAL COUNCIL
OF
ENGINEERING EXAMINERS
Copyright ©, 1984
by
The National Council of Engineering Examiners.
All rights reserved. This book, or parts thereof, may not
be reproduced in any form without prior written permission.
First Printing - January 1984
PRINTED IN THE UNITED STATES OF AMERICA
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