In the seventeenth century it was experimentally established that if a direct axial external forceFis applied to a machine element, whether it is a traditional spring (see Chapter 14) or a straight cylindrical bar such as the one shown in Figure 2.1, changes in the length of the machine element are produced. Further, for a broad class of materials, a linearrelationship Figure 2.1
Straight cylindrical bar loaded by direct axial force.
A F
F y=f
lo lf
3See Hooke’s Law discussion of 5.2.
exists between the applied force, F, and the induced change in length, y, as long as the ma- terial is not stressed beyond its elastic range.3The elastic deflection, y, and the corresponding elastic strain, , produced by the aggregation of small changes in interatomic spacing within the material, are fully recoverable4as long the applied forces do not produce stresses that exceed the yield strengthof the material.
For any linear spring, the relationship between force and deflection may be plotted as shown in Figure 2.2, and expressed as
(2-1) wherekis called the spring constantorspring rate.
The change in length induced by the applied force is
(2-2) where is the original bar length with no force applied, is the bar length after external forceFhas been applied, and is force-induced elastic deformation, so defined to distin- guish it from temperature-induced elastic deformation, discussed later.
If, in a machine element, exceeds the design allowableaxial deformation, failure will occur. For example, if the blade-axial deformation of an aircraft gas turbine blade, caused by the centrifugal force field, exceeds the tip clearance gap, failure will occur because of force-induced elastic deformation.
Based on (2-1), the spring rate, , for the uniform bar shown in Figure 2.1 may be written as
(2-3) where is the spring rate, or stiffness, of the axially loaded bar. It should again be em- phasized that all realmachine parts and structural elements behave as springs because they all have finite stiffnesses. Thus the concept of spring rate is important not only when dis- cussing “traditional” springs, as in Chapter 14, but also when considering potential failure by force-induced elastic deformation, or when examining the consequences of load shar- ing, preloading, and/or residual stresses (see Chapter 4).
The uniaxial force-deflection plot of Figure 2.2 may be normalized by dividing the force coordinate by the original cross-sectional area and the deflection coordinate by original length . The resulting plot, shown in Figure 2.3, is the familiar engineering stress-strain diagram,5 with slope equal to Young’s modulus of elasticity, E. For the uniform bar shown in Figure 2.2, the engineering stress, , is
(2-4) where is the original cross-sectional area of the bar.
The engineering strain induced by the applied force is
(2-5) ef = df
lo Ao
s = F Ao
s lo
Ao kax
kax = F y = F
df
kax
df
df
lf
lo
y = df = lf - lo
F = ky e
Elastic Deformation, Yielding, and Ductile Rupture 29
4See ref. 2.
5Engineering stress and strainare based on original values of area and length, in contrast with true stressand true strain, which are based on instantaneous values of area and length.
From the engineering stress-strain curve of Figure 2.3,
(2-6) whereEis Young’s modulus of elasticity and is force-induced elastic strain.
Combining (2-3), (2-4), (2-5), and (2-6),
(2-7) where is the axial spring rate, a function of material and geometry. Thus if it were of interest, the force-induced elastic deformation could be easily calculated for the axial loading case, as
(2-8) If the limiting or allowable design value of deflection, , in a particular design sit- uation were exceeded by the design would be unacceptable; hence elastic deformation failure is predicted to occur if (FIPTOI)
(2-9) and redesign would be necessary if failure were predicted.
df-max Ú dallow
df,
dallow
df = F kax
df
kax
kax = AoE lo ef
s = Eef
Example 2.1 Force-Induced Elastic Deformation
An axially loaded straight steel bar with a rectangular cross section will fail to perform its design function if its length increases 0.20 mm or more. It is desired to operate with the stress level in the bar greater than one-half the yield strength, Syp. The bar, made of 1040 HR steel (Syp290 MPa), is 300 mm long.
a. What is the probable governing failure mode?
b. Is failure predicted to occur?
k lb in
Deflection,f(in)
Force,F(lb)
F=kf E lb
in2
Engineering strain, f=f/lo(in/in) =Ef
Engineering stress, =F/Ao(psi)
Figure 2.2
Force-deflection curve for linear elastic behavior.
Figure 2.3
Engineering stress-strain diagram for linear elastic behavior.
Temperature fluctuations may also produce dimensional changes and associated re- coverable elastic strains in a machine part. If temperature changes are slow enough so that significant gradients are not generated, and if no external constraints are imposed, the ma- chine part remains stress-free as temperature-induced elastic deformations are produced.
The change in length, , induced by the temperature difference , may be calculated as (2-10) where is the initial length of the part and ais the linear coefficient of thermal expansion for the material. From (2-10), then, the temperature-induced elastic strain (thermal strain) is (2-11) Temperature-induced strains are recoverable (elastic), normal (as opposed to shear), and usually close to linear functions of temperature over a wide temperature range (often a few hundred F). The temperature-induced elastic strain in a given direction, i, may be found from (2-11) as
(2-12) where is the thermal or temperature-induced elastic strain in the ith direction due to tem- perature change eti ¢®, and aiis the thermal expansion coefficient in the ith direction.
eti = ai¢®
et = dt
lo = a¢®
lo
dt = loa¢®
¢®
dt
Elastic Deformation, Yielding, and Ductile Rupture 31
Solution
a. The probable governing mode of failure is force-induced elastic deformation.
b. From (2-9), FIPTOI (failure is predicted to occur if) . Now,
Hence failure is predicted to occur since 0.21 0.20.
df-max = Fmax
kax
= smaxAo
aAoE lo
b
= Syplo
2E = 290* 106 (0.30)
21207* 1092 =0.00021 m (0.21 mm) df-max Ú dallow
Example 2.2 Temperature-Induced Elastic Deformation
The straight rectangular 1040 HR steel bar of Example 2.1 is to be evaluated for an en- tirely different application in which there are no applied forces at all but the temperature is to be increased by 300F. Again, in this new application, the bar will cease to properly perform its design function if its length increases 0.0060 inch or more. The linear coeffi- cient of thermal expansion for steel is approximately 6.3 106in/in/F.
a. What is the governing failure mode?
b. Is failure predicted to occur?
Solution
a. The governing failure mode is temperature-induced elastic deformation.
b. Following the logic leading to (2-9), FIPTOI (failure is predicted to occur if) From (2-10),
dt-max = loa¢® = 112.0000216.3 * 10-6213002 = 0.0227 inch dt-max Ú dallow
Hence, FIPTOI
Therefore failure ispredicted to occur.
It is worth noting that the change in length of this 12-inch steel bar produced by a temperature change is nearly three times the length change produced by an axial force cor- responding to a stress equal to half the yield strength of the steel. (Temperature-induced elastic strains are often large enough to be important design criteria.)
Theprinciple of superpositionmay be utilized when governing equations are linear.
Since force-induced elastic strain and temperature-induced elastic strain are both linear functions, the principle of superposition may be used to give the total strain in any given ith direction as
(2-13) where and are force- and temperature-induced components of strain in the ith direc- tion, respectively. This expression may be used to assess the consequences of constraint to thermal expansion. For example, if a prismatic steel bar of length were axially con- strained between rigid walls so that its total change in length were forced to remain zero, and the temperature increased by , equation (2-13) would require
(2-14) or
(2-15) Utilizing the uniaxial Hooke’s Law and (2-12), the temperature-induced stress (thermal stress), , in this fully constrained bar, would be
(2-16) a compressive stress induced in the steel bar because it attempts to increase in length as temperature increases. If the walls are not fully rigid (a more realistic case), the calcula- tions become more complicated, as discussed later in 4.7.
When the state of stress is more complicated than the uniaxial cases just discussed, it becomes necessary to calculate the elastic strains induced by the multiaxial state of stress in three mutually perpendicular directions, say x,y,z, through the use of the generalized Hooke’s Law equations given in 5.2. The resulting elastic strains, and , may be used to calculate the total force-induced elastic deformation of a member in any of the coordinate directions by integrating the strain over the member’s length in that direction.
Temperature-induced strains may be included as shown in (2-13). If the change in length of the member in any direction exceeds the design-allowable deformation in that direction, elastic deformation failure will occur.
If the externally applied forces produce stresses that exceed the yield strengthof a ductile material, the induced strains are not fully recovered upon release of the loads and permanent or plastic strains remain. Such permanent deformations usually (but not al- ways) render a machine part incapable of performing its intended function, whereupon failure by yielding is said to occur. For example, if the axially loaded bar of Figure 2.1 were to reach a stress , calculated from (2-1), that exceeds the material’s yield strength,s
efz
efx,efy, s = -Eai¢®
s
efi = -eti
0 = efi + eti
¢®
lo
eti
efi
ei = efi + eti
ei
300°
0.0227 Ú 0.0060 Example 2.2
Continues
, the design would be unacceptable; hence yielding failure would be predicted to occur if (FIPTOI)
(2-17) and redesign would be necessary.
For the case of uniaxial loadng, the onset of yielding may be accurately predicted to occur when the uniaxial maximum normal stress reaches a value equal to the yield strength of the material, as shown in (2-17) and Example 2.3. If the loading is more complicated, and a multiaxial state of stress is produced by the loads, the onset of yielding may no longer be predicted by comparing any one of the normal stress components with uniaxial material yield strength, not even the maximum principal normal stress. Onset of yielding for multiaxially stressed critical points in a machine or structure is more accurately pre- dicted through the use of a combined stress theory of failurethat has been experimentally validated for the prediction of yielding. The two most widely accepted theories for pre- dicting the onset of yielding under multiaxial states of stress are the distortion energy the- ory(also known as the octahedral shear stress theory, or the Huber– von–Mises–Hencky theory), and the maximum shearing stress theory. The use of these theories is discussed in 5.4.
If the externally applied forces are so large that a ductile material not only experiences plastic deformation but proceeds to separate into two pieces, the process is called ductile rupture. In most cases, such a separation renders a machine part incapable of performing its intended function, and failure by ductile rupture is said to occur. For example, if the axially loaded bar of Fig. 2.1 reaches a stress , calculated from (2-1), that attempts tos
s Ú Syp Syp
Elastic Deformation, Yielding, and Ductile Rupture 33
Example 2.3 Yielding
An axially loaded straight bar of circular cross section will fail to perform its design func- tion if applied axial loads produce permanent changes in length when the load is removed.
The bar has a diameter of 12 mm, a length of 150 mm, and is made of 1020 HR steel (Su379 MPa, Syp207 MPa; e25 percent in 50 mm). The required axial load for this particular case is 22 kN.
a. What is the governing failure mode?
b. Is failure predicted to occur?
Solution
a. Since the elongation in 50 mm is 25 percent, the material is ductile and the governing failure mode is yielding.
b. From (2-17), FIPTOI . Now,
Hence FIPTOI 195 207, Therefore, failure is notpredicted to occur. It must be noted, however, that failure is imminent, and as a practical matter a designer would nearly always redesign to lower the stress. Typically, a safety factorwould be utilized to determine the properly redesigned dimensions, as discussed in Chapter 6.
s = F A0
= F apd2
4 b
= 4F
pd2 = 4(22)
p(0.012)2 =194522 kN/m2 =195 MPA s Ú Syp
exceed the material’s ultimate strength, , the design would be unacceptable; hence fail- ure by ductile rupture would be predicted to occur if (FIPTOI)
(2-18) and redesign would be necessary.
s Ú Su
Su
Example 2.4 Ductile Rupture
The axially loaded straight cylindrical bar of Example 2.3 is to be used in a different application for which permanent deformations are acceptable but separation of the bar into two pieces destroys the ability of the device to perform its function. The required axial load for this application is to be 11,000 lb.
a. What is the governing failure mode?
b. Is failure predicted to occur?
Solution
a. From material specifications in Example 2.3, the 1020 HR material is ductile (e25 percent in 2 inches); hence the governing failure mode is ductile rupture.
b. From (2-18), FIPTOI
Now,
Also, from Example 2.3, 55,000 psi, hence FIPTOI
Therefore failure ispredicted to occur by ductile rupture, and redesign would be necessary.
56,120 Ú 55,000 Su
s = F
Ao = 11,000 pc10.50022
4 d
= 11,000
0.196 = 56,120 psi s Ú Su
Again, if the loading is more complicated, and a multiaxial state of stress is produced by the loads, ductile rupture may no longer be accurately predicted by comparing any of the normal stress components with the uniaxial ultimate strength, not even the maximum principal normal stress. A combined stress theory of failure is required for prediction of ductile rupture under multiaxial states of stress. The distortion energy theory is usually chosen for such cases, as discussed in 4.6.