Design Procedure 6.1: Fracture Mechanics Applied to Design
6.7 Failure Prediction for Multiax- ial Stress State
6.7.1 Ductile Materials
As discussed in Section 3.2, most metals and thermoplastic polymers are considered to beductile. There are exceptions, however: metal castings are not as ductile as wrought or cold- worked parts and can thus behave in a brittle manner. Ductile materials typically have the same tensile strength as compres- sive strength and are not as susceptible to stress raisers as are brittle materials. For the purposes of this text, a ductile ma- terial is considered to have failed when it yields. Although in some applications a small amount of plastic deformation may be acceptable, this is rarely the case in machinery ele- ments. Two theories of yield criteria are presented next: the maximum-shear-stress theory and the distortion-energy the- ory.
Maximum-Shear-Stress Theory
The maximum-shear-stress theory (MSST) was first pro- posed by Coulomb [1773] but was independently developed by Tresca [1868] and is therefore often called theTresca yield criterion. Tresca noted that plastically deforming platinum exhibited bright shear bands under small strains, indicating that metals deformed under shear in all circumstances and that the shear was highly localized on well-defined planes.
His observations led to the MSST, which states that a part subjected to any combination of loads will fail (by yielding or fracturing) whenever the maximum shear stress exceeds a critical value. The critical value can be determined from stan- dard uniaxial tension tests. Experimental evidence verifies
Centerline of cylinder and hexagon
View along axis of cylinder σ1
σ1 σ2
σ3 σ3
σ2
MSST DET
Figure 6.11: Three-dimensional yield locus for MSST and DET.
that the MSST is a good theory for predicting yielding of duc- tile materials, and it is a common approach in design. If the nomenclatureσ1≥σ2 ≥σ3is used for the principal stresses in accordance with Eq. (2.18), the maximum shear stress the- ory predicts yielding when
σ1−σ3=Sy
ns, (6.8)
whereSyis the yield strength of the material andns is the safety factor.
For a three-dimensional stress state, the maximum- shear-stress theory provides an envelope describing the stress combinations that cause yielding, as illustrated in Fig. 6.11.
The curve, defined by a yield criterion, is known as ayield lo- cus. Any stress state in the interior of the yield locus results in elastic deformation. Points outside the yield locus are not possible because such stress states would cause yielding in the solid before these stresses could be attained. Stress states outside the yield locus cannot be supported by the material.
If a situation arises that would increase the material strength (such as strain rate effects or work hardening), the yield locus expands, so that fracture may not necessarily be the result.
It is helpful to present the yield criterion in a plane stress circumstance, for which there will be two principal stresses in the plane as well as a principal stress equal to zero perpendic- ular to the plane. Figure 6.12 graphically depicts failure pre- diction in the plane stress state by the maximum-shear-stress theory. The principal stresses used in the figure are labeled as σ1andσ2, but the ordering of normal stresses (σ1≥σ2≥σ3) is not being enforced. Remember that the principal stress out- side the page is zero. In the first quadrant, whereσ1 andσ2
are by definition positive, this means that the value ofσ3 in Eq. (6.8) would be zero and that yielding would occur when- everσ1 orσ2reached the uniaxial yield strength,Sy. In the second quadrant, where σ1 is negative and σ2 is positive, Eq. (6.8) would result in a line as shown in Fig. 6.12. The third and fourth quadrants of the curve follow the same reasoning in their development.
Distortion-Energy Theory
Thedistortion-energy theory(DET), also known as thevon Mises criterion, postulates that failure is caused by the elas-
Failure Prediction for Multiaxial Stress State 145
Shear diagonal 45°
+ σ1
+ σ2
– σ2
Sy
Sy
–Sy – σ1
–Sy
Figure 6.12: Graphical representation of maximum-shear- stress theory (MSST) for biaxial stress state (σz= 0).
tic energy associated with shear deformation in the material.
This theory is valid for ductile materials and predicts yield- ing under combined loading and works well (although the differences between the DET and the MSST are small).
The DET can be derived mathematically in a number of ways, but one of the more straightforward is based on the hypothesis that yielding occurs when the root mean shear stress exceeds a critical value. Mathematically, this can be expressed as
(σ1
−σ2)2+ (σ2−σ3)2+ (σ3−σ1)21/2
=Constant.
(6.9) Since in uniaxial yielding,σ1=Syandσ2=σ3= 0, the con- stant in Eq. (6.9) is evaluated as√
2Sy. Eq. (6.9) then becomes
√1 2
(σ1
−σ2)2+ (σ2−σ3)2+ (σ3−σ1)21/2
=Sy. (6.10)
Incorporating the safety factor, Eq. (6.10) becomes
σe= 1
√2 (σ1
−σ2)2+ (σ2−σ3)2+ (σ3−σ1)21/2
= Sy
ns, (6.11) whereσe is thevon Mises stress. For a biaxial stress state, assumingσ3= 0, the von Mises stress is given by
σe= σ2
1+σ22−σ1σ21/2
= σ2
x+σy2−σxσy+ 3τxy2 1/2 . (6.12) The DET yield locus is shown in Fig. 6.11 for a three- dimensional stress state and in Fig. 6.13 for plane stress load- ing (biaxial stress state). Compared to the MSST, the DET has the advantage that the yield criterion is continuous in its first derivative, an important consideration for applications in plasticity.
+ σ1
+ σ2
– σ2
– σ1
Syt Syt
Syt
Syt 45°
–Syt
–Syt –0.577 Syt
–Syt
–Syt 0.577 Syt
–0.577 Syt
0.577 Syt
Shear diagonal Figure 6.13: Graphical representation of distortion-energy theory (DET) for biaxial stress state (σz= 0).
Example 6.6: Determination of von Mises Stress
Given:In the rear-wheel suspension of the Volkswagen Bee- tle, the spring motion was provided by a torsion bar fastened to an arm on which the wheel was mounted, as depicted in Fig. 6.14. The torque in the torsion bar was created by a 2500-N force acting on the wheel from the ground through a 300-mm lever arm. Because of space limitations, the bear- ing holding the torsion bar was situated 100 mm from the wheel shaft. The diameter of the torsion bar was 28 mm.
Find:The stresses in the torsion bar at the bearing by using the distortion-energy theory.
Solution: The stresses acting on the torsion bar are a shear stress from torsion and a perpendicular tensile/compressive stress from bending, resulting in a plane stress loading (see Example 4.13). Using Eq. (4.33) gives the shear stress from torsion as
τ= T c
J = 2500(0.3)(0.014)32
π(0.028)4 Pa= 174.0MPa.
Using Eq. (4.45) gives the tensile stress from bending as σ= M c
I =2500(0.1)(0.014)64
π(0.028)4 Pa= 116.0MPa.
From Eq. (2.16), the principal normal stresses then are σ1, σ2 = σx+σy
2 ±
r
τxy2 +σx−σy
2 2
= 116.0 2 ±
s
(174.0)2+ 116.0
2 2
, orσ1 = 241.4MPa andσ2 =−125.4MPa. From Eq. (6.12), the von Mises stress is
σe = σ2
1+σ22−σ1σ20.5
= (241.4)2
+ (−125.4)2−241.4(−125.4)0.5
= 322.9MPa.
100 mm
Torsion bar BearingArm 2500 N
Wheel (top view)
300 mm
Figure 6.14: Rear wheel suspension used in Example 6.6.
Example 6.7: Yielding of a Ductile Bar
Given:A round, cantilevered bar made of a ductile material is subjected to a torque applied to the free end.
Find: Determine when yielding will occur by using (a) the MSST and (b) the DET.
Solution:Figure 6.15 shows the cantilevered bar, the stresses acting on an element, and a Mohr’s circle representation of the stress state. Since the goal is to determine stresses at yielding, the safety factor will be taken asns= 1. The prin- cipal stresses areσ1=τ1andσ3=−τ1.
(a) Using Eq. (6.8) the MSST predicts failure if
|σ1−σ3|= 2τmax=Sy
ns, or
τmax= 0.5Sy. (a) (b) Using Eq. (6.11) yields
σe = σ2
1−σ1σ2+σ22
1/2
= τ2
1 −τ1(−τ1) + (−τ1)21/2
=√ 3τ1
= √
3τmax=Sy
ns, or
τmax= 1
√3Sy= 0.577Sy. (b) Equations (a) and (b) show that the MSST and the DET are in fairly good agreement. This circumstance, that is, a loading of pure shear, results in the greatest difference between the MSST and the DET, suggesting that both theories will give close to the same results.
x y
T z
(a)
(b) τ1
(c) τ1
τ
σ1
σ3 0 σ
τ2
σ2
Figure 6.15: Cantilevered, round bar with torsion applied to free end (used in Example 6.7). (a) Bar with coordinates and load; (b) stresses acting on an element; (c) Mohr’s circle rep- resentation of stresses.
Example 6.8: Yield Criteria Applied to Design
Given:A round, cantilevered bar, similar to that considered in Example 6.7, is subjected not only to torsion but also to a transverse load at the free end, as shown in Fig. 6.16a. The bar is made of a ductile material having a yield stress of 350 MPa. The transverse force is 2000 N and the torque is 100 Nm applied to the free end. The bar is 150 mm long and a safety factor of 2 is assumed. Transverse shear can be neglected.
Find:Determine the minimum diameter to prevent yielding by using both (a) the MSST and (b) the DET.
Solution:Figure 6.16b shows the stress element on the top of the bar at the wall. Note that, in this example,σz= 0, so that the element encounters plane or biaxial stress. The critical section occurs at the wall. By using Eqs. (4.45) and (4.33) the normal and shear stresses can be written as
σx=M c
I =P l(d/2)
πd4/64 =32P l πd3 , τxy=T c
J = T(d/2) πd4/32 =16T
πd3.
Failure Prediction for Multiaxial Stress State 147
x y
d T
P z
Element
(a) l
(b) τxz dz dx
σx σx σx
(c) τ1 = τmax τ/d3
σ1 σ3 σ2
σ/d3 –800 3200
τ2
1600
Figure 6.16: Cantilevered, round bar with torsion and trans- verse force applied to free end (used in Example 6.8). (a) Bar with coordinates and loads; (b) stresses acting on element at top of bar and at wall; (c) Mohr’s circle representation of stresses.
From Eq. (2.16), the principal normal stresses in biaxial stress can be written as
σ1, σ2 = σx
2 ± r
σx
2 2
+τxz2
= 16P l πd3 ±
s 16P l
πd3
2 +
16T πd3
2 . Therefore,
σ1, σ2= 16 πd3
P l
±p
(P l)2+T2 . Substituting forPandlresults in
σ1, σ2= 16 πd3
h(2000)(0
.150)±p
(2000)2(0.150)2+ (100)2
i . Therefore,
σ1= 3140 d3
and σ2=−82.6 d3
.
Note that the stresses are in the wrong order; to ensureσ1≥ σ2 ≥ σ3, they are rearranged so thatσ1 =3140/d3,σ2 = 0, andσ3=−82.6/d3.
From Eq. (2.19), the maximum and principal shear stresses can be written as
τ1, τ2=± s
τxy2 +(σx−σz)2
4 ,
or
τmax = τ1= 16 πd3
p(P l)2+T2
= 16 πd3
p(2000)2(0.150)2+ (100)2. Therefore,
τmax=τ1=1610 d3 .
(a) Using Eq. (6.8), the MSST predicts that failure will be avoided if
|σ1−σ3|= 2τ1= 2τmax<Sy
ns
. Therefore,
3140 d3 +82.6
d3
<350×106
2 ,
ord= 0.0210 m = 21.0 mm.
(b) This is a plane stress state, since one of the principal stresses is zero. Therefore, Eq. (6.12) yields
σe =
"
3140 d3
2 +
3140 d3
82.6 d3
+
82.6 d3
2#1/2
= 3180 d3
.
Thus, using Eq. (6.11) the DET predicts that failure occurs when
σe=Sy
ns
, 26,950
d3 =50,000 2 .
Henced = 20.9mm. Note that both theories give approxi- mately the same solution.