Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
MODULE NO. 12
(Helical Spring)
BAGASALA, RHEN LAWRENCE M.
SAN JOAQUIN, EULLA V
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
PROBLEM NO.1
Figure 01:
Situation:
The helical spring shown is axially loaded with a compression force P equal to 5 kN. The
mean diameter of the spring is 100 mm and the wire used is 10 mm as indicated in the
figure.
1. What is the shear stress at A?
2. What is the shear stress at B?
3. On diameter AB, locate the point of zero stress measured from C.
SOLUTION:
T= 0.05P = 0.05(5)
T=0.25 kN*m
Shear stress due to direct force
�
�1 = =
�
5(1000)
�(52)
�1 = 63.66 MPa
Shear stress due to torque
�2 =
��
�
=
0.25 (10002 )(5)
1
�(54 )
2
�2 =1273.24 MPa
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
Part 1 - Shear stress at A (maximum shear stress)
�� =�1+ �2
�� =63.66 + 1273.24
�� =1336.90 MPa
Or you can use the formula for helical spring
16��
�
τ= ��3 (1+4� )
16(5000)(50)
���� =
�(103)
10
(1+4(50))
���� =1336.90 MPa
Part 2 - Shear stress at B
�� =τ2−τ1
�� =1273.24−63.66
�� =1209.58 MPa
Part 3 - At the point of zeroshear, ������� = �������−�ℎ���
��
�
=�1
0.25 (10002)(�)
1
�(54 )
2
= 63.66
x=0.25 mm from point C
Another way to solve for x is by ratio and proportion
�
�1
�
=�
2
�
63.66
=
x=0.25 mm from point C
5
1273.24
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
PROBLEM NO.2
As shown in Figure 02, a homogeneous 50-kg rigid block is suspended by the three springs
whose lower ends were originally at the same level. Each steel spring has 24 turns of 10mm-diameter on a mean diameter of 100 mm, and G = 83 GPa. The bronze spring has 48
turns of 20-mm-diameter wire on a mean diameter of 150 mm, and G = 42 GPa. Compute
the maximum shearing stress in each spring.
Figure 02:
SOLUTION:
ΣFV=0
ΣM1=0
�1 +�2 +�3 =490.5 → Equation (1)
�2 (1)+�3 (3)=490.5(1.5)
�2− �1
1
=
�2 = 1�
�3−�1
3
2
�
3 3+3 1
64 �2 (503)(24)
83000(104 )
3
830
3
166
�2
�2
1 64�3(753)(48)
42000(204 )
=3
9
1
= 8960 �3 + 415 �1
9
1
= 1792 �3 + 83 �1
+
2 64�1 (503 )(24)
3 83000(104)
→ Equation (3)
�2 +3�3 =735.75
→ Equation (2)
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
From Equation (1)
�1 =490.5 −�2 −�3
3
166
3
166
5
166
�2
�2
�2
9
1
= 1792 �3 + 83 (490.5
9
981
= 1792 �3 + 166
9
= 166
−
1045
148736
From Equation (2)
�2 =735.75−3�3 =
−
1
83
−�2 −�3 )
�2 −
1
83
�3
�3 → Equation (4)
2943
− - 3�3
4
Substitute P2 to Equation (4)
5 2943
(
166
4
1045
981
1045
− - 3�3 ) = ( 166 - 148736 �3 )
5
(148736 �3 -
166
�3 = 195.01N
981
)�3 = 166 -
14715
664
�2 =735.75−3(195.01)=150.72N
�1 =490.5−150.72−195.01=144.77N
���� =
16��
��3
�
(1+4� )
For steel at left:
���� =
16�(144.77)(50)
���� =
16�(150.72)(50)
���� =
16�(195.01)(75)
�(103 )
For steel at right:
�(103 )
10
[ 1 + 4(50)] = 38.709 MPa
10
[ 1 + 4(50)] = 40.300 MPa
For phosphor bronze:
�(203 )
20
[ 1 + 4(75)] = 9.932 MPa
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
PROBLEM NO. 3
Determine the maximum shearing stress and elongation in a bronze helical spring composed
of 20 turns of 1.0-in.-diameter wire on a mean radius of 4 in. when the spring is supporting a
load of 500 lb. Use Eq. (3-10) and G = 6 × 106 psi.
SOLUTION:
τmax =
Where:
16�� 4m−1
��3
(4m−4 +
0.615
m
→ Equation (3-10)
)
P = 500 lb; R = 4 in
d = 1 in; n = 20 turns
m = 2R/d = 2(4)/1 = 8
τmax =
16(1500)(4) 4(8)−1
�(13 )
(4(8)−4 +
0.615
τmax= 12060.3 psi = 12.1 ksi
δ=(
64��3 �
��4
δ= 6.83 in
)=
64(1500)(43)20
(6�106) (14 )
8
)
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
PROBLEM NO. 4
Two steel springs arranged in series as shown in Fig. supports a load P. The upper spring has
12 turns of 25-mm-diameter wire on a mean radius of 100 mm. The lower spring consists of
10 turns of 20-mm diameter wire on a mean radius of 75 mm. If the maximum shearing
stress in either spring must not exceed 200 MPa, compute the maximum value of P and the
total elongation of the assembly. Use Eq. (3-10) and G = 83 GPa. Compute the equivalent
spring constant by dividing the load by the total elongation.
SOLUTION:
τmax =
16�� 4m−1
��2
(4m−4 +
For Spring (1)
200 =
0.615
m
16�(100) 4(8)−1
[4(8)−4 +
�(253)
P = 5182.29N
→ Equation (3-10)
)
0.615
8
]
For Spring (2)
200 =
16�(75) 4(7.5)−1
�(203 )
[4(7.5)−4 +
P = 33498.28 N
0.615
7.5
]
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
Use
P=3498.28 N
Total elongation:
δ = δ1+δ2
δ= (
δ=
64��3 �
��4
)1 + (
64��3 �
64(3498.28)(1003)12
��4
+
83000(254 )
δ=153.99 mm
)2
64(3498.28)(753)10
83000(204 )
Equivalent spring constant, ����������� :
�
����������� = � =
3498.28
153.99
����������� =22.72N/mm
PROBLEM NO. 5
A helical spring is fabricated by wrapping wire 3/4 in. in diameter around a forming cylinder
8 in. in diameter. Compute the number of turns required to permit an elongation of 4 in.
without exceeding a shearing stress of 18 ksi. Use Eq. (3-9) and G = 12 × 106 psi.
SOLUTION:
τmax =
16��
18000 =
��3
d
( 1 + 4R )
16�(4)
3
4
�( )3
P=356.07lb
[1+
δ=
16��3 �
4=
64(356.07)(43)�
4
3
4
4(4)
��4
3
(12�106)( )4
→ Equation (3-9)
n=10.41 say 10 turns
]
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering
PROBLEM NO. 6
Determine the maximum shearing stress and elongation in a helical steel spring composed
of 20 turns of 20-mm-diameter wire on a mean radius of 90 mm when the spring is
supporting a load of 1.5 kN. Use Eq. (3-10) and G = 83 GPa.
SOLUTION:
τmax =
Where:
16��
�(203 )
4(m)−1
(4(m)−4 +
0.615
m
→ Equation (3-10)
)
P = 1.5 kN = 1500 N; R = 90 mm
d = 20 mm; n = 20 turns
m = 2R/d = 2(90)/20 = 9
τmax =
16(1500)(90) 4(9)−1
�(203 )
τmax=99.87MPa
δ=(
64��3 �
��4
)=
δ=105.4mm
(4(9)−4 +
64(1500)(903 )20
83000(204 )
0.615
9
)
Republic of the Philippines
Laguna State Polytechnic University
Province of Laguna
College of Engineering