![Figure [2-28]](fig0228.GIF)
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When two cylindrical parts are assembled by shrinking or press-fitting one part upon another, a contact pressure is created between the two parts. The stresses resulting from this pressure may easily be determined with the equations of the preceding section.
Figure 2-28 shows two cylindrical members which have been assembled with a shrink fit. A contact pressure p exists between the members at radius b, causing radial stresses sigma_r = -p in each member at the contacting surfaces. From equation 2-59, the tangential stress at the outer surface of the inner member is
sig_it = -p * (b^2 + a^2) / (b^2 - a^2) [2-61]
In the same manner, from equation 2-57, the tangential stress
at the inner surface of the outer member is
sig_ot = p * (c^2 + b^2) / (c^2 - b^2) [2-62]
These equations cannot be solved until the contact
pressure is known. In obtaining a shrink fit, the
diameter of the male member is make larger than the
diameter of the female member. The difference in these
dimensions is called the interference and is the
deformation which the two member must experience.
Since these dimensions are usually known, the
deformation should be introduced in order to evaluate
the stresses. Let
del_o = increase in radius of hole and
del_i = decrease in radius of inner cylinder
The tangential strain in the outer cylinder at the inner
radius is
epsilon_ot = change in circumference / original circumference
or
epsilon_ot = (2*pi*(b+del_o)-2*pi*b)/(2*pi*b)
= del_o / b
and so
del_o = b*epsilon_ot [a]
but since
epsilon_ot = sig_ot / E_o - mu*sig_or / E_o
then, from equations [2-61] and [2-62], we have
del_o = b*p/E_o * ((c^2+b^2)/(c^2-b^2)+mu_o) [b]
This is the increase in radius of the outer cylinder.
In a similar manner the decrease in radius of the
inner cylinder is found to be
del_i = -b*p/E_i * ((b^2+a^2)/(b^2-a^2)-mu_i) [c]
Then the total deformation del is
del = del_o - del_i =
(b*p/E_o * ((c^2+b^2)/(c^2-b^2)+mu_o)) +
b*p/E_i * ((b^2+a^2)/(b^2-a^2)-mu_i) [2-63]
This equation can be solved for the pressure p when the
interference del is given. If the two members are of the
same material, E_o = E_i = E and mu_o = mu_i = mu and
the relation simplifies to
p = E*del/b *
((c^2-b^2)*(b^2-a^2)/(2*b^2*(c^2-a^2))) [2-64]
Substitution of this value of p in equations [2-61] and [2-62]
will then five the tangential stresses at the inner surface of the
outer cylinder and at teh outer surface of the inner cylinder.
In addition, equation [2-63] or [2-64] can be employed to obtain
the value of p for use in the general equations (equations
[2-53] and [2-54]) in order to obtain the stress at any
point in either cylinder.
In addition to the assumptions both stated and implied by the development, it is necessary to assume that both members have the same length. In the case of a hub which has been press-fitted to a shaft, this assumption would not be true, and there would be an increased pressure at each end of the hub. It is customary to allow for this condition by the employment of a stress-concentration factor. The value of this factor depends upon the contact pressure and the design of the female member, but its theoretical value is seldom greater than 2.
A tube, such as a gun barrel, has nominal dimensions of i inch ID x 2 inch OD, over which a second tube having nominal dimensions of 2 inch ID x 3 inch OD is to be shrink-fitted. The material is steel. It is desired to fit these two members together to cause a stress of sig_t = 10 kpsi at the inner surface of the outer menber.
(a) Find the required original dimensions of the
members.
(b) Determine the resulting stress distributions.
SOLUTION: (a) Using equation [2-62], we solve for the
contact pressure. Thus
p=sig_ot*(c^2-b^2)/(c^2+b^2)
=(10000)*(1.5^2-1^2)/(1.5^2+1^2)
=3850 psi
Solving equation [2-64] for the deformation, we obtain
del=b*p/E*2*b^2*(c^2-a^2)/((c^2-b^2)*(b^2-a^2))
=(1)*(3850)/(30E6)*2*(1^2)*(1.5^2-.5^2)/
((1.5^2-1^2)*(1^2-.5^2))
=.000548 inch
The dimensions selected are as follows:
Outside diameter of inner member = 2.00000 inch
Inside dimension of outer member = 2.00000 - 2*(0.000548)
= 1.998904 inch
(b) We shall determine the stress distribution in the
outer member first. This is a cylinder subjected to an
internal pressre of p_i=3850 psi and an external pressure of
p_o=0 psi. Equations [2-55] and [2-56] apply. In these
equations a = 1 inch and b = 1.5 inch, and they are to be
solved for various values of r between 1 and 1.5 inch.
By using r = 1.1 inch, a sample calculation is as
follows:
sig_t=a^2*p_i/(b^2-a^2)*(1+b^2/r^2)
=1^2*3850/(1.5^2-1^2)*1+1.5^2/1.1^2)
=8810 psi
sig_r=a^2*p_i/(b^2-a^2)*(1-b^2/r^2)
=1^2*3850/(1.5^2-1^2)*1-1.5^2/1.1^2)
=-2650 psi
These, and other results obtained in the same manner, are shown
in Table [2-2],
Tangential and Radial stress in the Outer Member --------- ------------------------ -------------------- Radius, r Tangential Stress, sig_t Radial Stress, sig_r inch psi psi --------- ------------------------ -------------------- 1.0 10000 -3850 1.1 8810 -2650 1.2 7900 -1740 1.3 7180 -950 1.4 6630 -460 1.5 6160 0 --------- ------------------------ --------------------Equations [2-53] and [2-54] are used for the inner member. When p_i=0 and p_o=p, these equations reduce to
sig_t=-p*b^2/(b^2-a^2)*(1+a^2/r^2) [2-65]
sig_r=-p*b^2/(b^2-a^2)*(1-a^2/r^2) [2-66]
Upon applying this set of equations to this example,
a=.5 inch, b=1 inch, and r varies from .5 to 1 inch.
The contact pressure is 3850 psi as before. The results
are shown in Table [2-3], and the stress distribution for
both members is plotted in Figure [2-29].
Tangential and Radial stress in the Inner Member --------- ------------------------ -------------------- Radius, r Tangential Stress, sig_t Radial Stress, sig_r inch psi psi --------- ------------------------ -------------------- 0.5 -10300 0 0.6 -8720 -1570 0.7 -7750 -2520 0.8 -7160 -3130 0.9 -6730 -3560 1.0 -6420 -3850 --------- ------------------------ --------------------
![Figure [2-29]](fig0229.GIF)