Chapter 22 Materials Selection and Design Considerations

Since d � 0.170
in. for this spring,
22.6 One Commonly Employed Steel Alloy • W99
Computation of the mean stress tm is made using Equation 8.14 modified to
the shear stress situation as follows:
(22.21)
It now becomes necessary to determine the minimum and maximum shear stresses
for the spring, using Equation 22.17.The value of may be calculated from Equations
22.17 and 22.13 inasmuch as the minimum is known (i.e., in.). A
shear modulus of 11.5 � 10 psi (79 GPa) will be assumed for the steel; this is the
room-temperature value, which is also valid at the 80�C service temperature. Thus,
is just
6
tmin dc dic � 0.060
t min
Now may be determined taking dc � dmc � 0.135 in. as follows:
t max
t min � d icGd
pD 2 K w
� dicGd c 1.60 aD
2 pD d b
�0.140
d
� c 10.060 in.2111.5 � 106 psi210.170 in.2
p11.062 in.2 2
� 41,000 psi 1280 MPa2
tmax � dmcGd c 1.60 aD
2 pD d b
�0.140
d
� c 10.135 in.2111.5 � 106 psi210.170 in.2
p11.062 in.2 2
� 92,200 psi 1635 MPa2
Now, from Equation 22.21,
t m � t min � t max
2
TS � 1169,000210.170 in.2 �0.167
� 227,200 psi 11570 MPa2
t m � t min � t max
2
41,000 psi � 92,200 psi
� � 66,600 psi 1460 MPa2
2
(22.22a)
(22.22b)
The variation of shear stress with time for this valve spring is noted in Figure 22.10;
the time axis is not scaled, inasmuch as the time scale will depend on engine speed.
Our next objective is to determine the fatigue limit amplitude (tal) for this
tm � 66,600 psi (460 MPa) using Equation 22.19 and for te and TS values of
45,000 psi (310 MPa) and 227,200 psi (1570 MPa), respectively. Thus,
66,600 psi
� 145,000 psi2c1 �
10.6721227,200 psi2
� 25,300 psi 1175 MPa2
d
tal � te c 1 � tm 0.67TS d
1.062 in.
dc1.60 a
0.170 in. b
�0.140
d
1.062 in.
dc1.60 a
0.170 in. b
�0.140
d