plugin chapter6

Chapter 6

Failure Prediction for Static Loading

Section 6.2

6.1 Given that the stress concentration factor is 2.81 for a machine element made of steel with a modulusof elasticity of 207 GPa, find the stress concentration factor for an identical machine element made ofaluminum instead of steel. The modulus of elasticity for aluminum is 69 GPa.

Solution: Since the elastic stress concentration is entirely determined from the geometry of the ma-chine element, the stress concentration factor will remain the same. Therefore, Kc = 2.81.

6.2 A flat part with constant thickness b is loaded in tension as shown in Fig. 6.3(a). The height changesfrom 50 to 100 mm with a radius r = 10 mm. Find how much higher a load can be transmitted throughthe bar if the height increases from 50 to 87 mm and the radius decreases from 10 to 4 mm. Ans. 42%higher load.

Notes: To answer this question, one must compare the stress concentration factors for the two cases.The stress concentration factors are obtained from Figure 6.3(a).

Solution: Referring to the sketch in Figure 6.3(a), in the first case, H = 100 mm, h = 50 mm andr = 10 mm for the first case. Therefore, H/h = 100 mm/50 mm = 2 and r/h = 10 mm/50 mm =0.2. From Figure 6.3(a), Kc for this case is 1.8. For the second case, H/h = 100/87 = 1.15 andr/h = 4 mm/87 mm = 0.046. Therefore, From Figure 6.3(a), Kc is around 2.8. The load that can betransmitted depends on the maximum stress. Therefore, for the first case:

Sy = KcP

bh= 1.8

P1

b(0.050 m)→ P1 =

Syb(0.050 m)1.8

= 0.02778Syb

For the second case:

Sy = KcP

bh→ P2 =

Syb(0.087 m)2.2

= 0.03954Syb

The ratio of the two forces is:P2

P1=

0.03954Syb

0.02778Syb= 1.42

6.3 A flat steel plate axially loaded as shown in sketch a has two holes for electric cables. The holes aresituated beside each other and each has a diameter d. To make it possible to draw more cables, thetwo holes are replaced with one hole having twice the diameter 2d, as shown in sketch b. Assume thatthe ratio of diameter to width is d/b = 0.2 for the two-hole plate. Which plate will fail first?

117

118 CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

d

d

b

(a)

2d b

(b)

Sketches a and b, for Problem 6.3

Notes: The exact hole locations have not been specified, so some variation may occur with assumeddimensions. Also, the two-hole case does not correspond to a particular chart in Fig 6.1 through 6.6.However, it is possible to obtain a reasonable solution from the existing data. For a critical application,more advanced approaches, such as finite element analysis, would be necessary. See also Prob. 6.6.

Solution: It will be assumed that the top and bottom halves of the plate are symmetric about thecenterline, and that the holes are placed in the center of each half of the plate. The locations wherethe largest stress could occur are A and B in the sketch.

Considering the top half of the problem leads to:

d

Here the diameter to width ratio is 0.4, so Kc is around 2.2 from Fig. 6.2 (a). Note that this is truefor point A but not point B in the figure above.

d/2

Because of St. Venants Principal, we must be concerned about the stress concentrations interactingbetween the two holes. For B, take a section through the hole diameters to yield:

Here H = b/2, h = b/2 − d, r/h = (d/2)/(b/2 − d) = 0.1b/(0.5b − 0.2b) = 0.33. Therefore, Kc is justunder 2.0 from Fig. 6.4(a). Therefore, the larger stress concentration is Kc = 2.2 and point A is moreimportant than point B. For the single hole, Figure 6.2 (a) gives a stress concentration of Kc = 2.2(d/b = 0.4). Therefore, either design is expected to fail at the same stress.

6.4 A 5 mm thick 100 mm wide AISI 1020 steel rectangular plate has a central elliptical hole 6 mm inlength transverse to the applied stress and 2 mm in diameter along the stress. Determine the appliedload that causes yielding at the edge of the hole.

Notes: This is a straightforward problem that requires Eq. (6.2) to obtain the solution.

Solution: Note that the width is much larger than the elliptical hole width, so Eq. (6.2) can beapplied. Note that the dimensions in Eq. (6.2) are half-widths, so we need to use b = 3 mm and a = 1mm. Equation (6.2) gives

Kc = 1 +2a

b= 1 +

2(1)3

= 1.667

119

From the inside front cover for AISI 1020 steel, Sy = 295 MPa. Therefore,

σmax = Sy = Kcσave = KcP

bh

∴ P =Sybh

Kc=

(295 MPa)(0.094 m)(0.005 m)1.667

= 83, 170 N = 83.17 kN

6.5 A round bar has a fillet with r/d = 0.15 and D/d = 1.5. The bar transmits both bending moment andtorque. A new construction is considered to make the shaft stiffer and stronger by making it equallythick on each side of the fillet or groove. Determine whether that is a good idea.

Notes: Figures 6.5 and 6.6 are used to obtain the solution.

Solution: For r/d = 0.15 and D/d = 1.5, the stress concentrations for bending is just over 1.5 forbending (from Fig. 6.5 (b)) and about 1.25 for torsion (from Fig. 6.5(c)). If instead of a fillet thebar became a groove, with the same root diameter, then the stress concentrations are obtained fromFig. 6.6 (b) and (c). The new stress concentration factors are around 1.65 for bending and 1.325 fortorsion. Since the stress concentration factors are higher for the proposed redesign, it is not a goodidea.

6.6 A 10-inch wide plate loaded in tension contains a 2 inch long, 1/2 inch wide slot. Estimate the stressconcentration by:

(a) Approximating the slot as an ellipse that is inscribed within the slot.

(b) Obtaining the stress concentration at the edge of the slot by taking a section through the slot andapproximating the geometry as a rectangular plate with a groove.

Notes: Neither of these is an accurate capture of the stress concentration geometry. Students shouldbe encouraged to conduct a literature search to obtain the stress concentration for this case or toconduct a finite element analysis to obtain the factor.

Solution:

(a) If we assume the slot is an ellipse, and we take the member to be infinitely wide, we can writefrom Eq. (6.2),

Kc = 1 +2(0.25)

1= 1.5

(b) The geometry can be approximated as in Fig. 6.4a, with H = 5 in., h = 4 in., r = 0.5 in. Thus,H/h = 1.25 and r/h = 0.125, so that Kc = 2.5.

Note that part a) assumes a very large width exists, which is not the case here. Also in part b), thearea under consideration must be reduced, while it is not reduced in part a), so that the results areactually slightly closer.

6.7 A machine has three circular shafts, each with fillets giving stress concentrations. The ratio of filletradius to shaft diameter is 0.1 for all three shafts. One of the shafts transmits a tensile force, onetransmits a bending torque, and one transmits torsion. Because they are stressed exactly to the stresslimit (ns = 1), a design change is proposed doubling the notch radii to get a safety factor greater than1. How large will the safety factors be for the three shafts if the diameter ratio is 2 (D/d = 2)? Ans.ns,tension = 1.21, ns,bending = 1.19, ns,torsion = 1.17.

Notes: Figure 6.5 is used to solve this problem.


Solution: For the shaft under tension, Fig 6.5a is used to obtain the stress concentration factors. ForD/d = 2 and r/d = 0.1, Kc = 2.0, and corresponds to the original case. If r is doubled, then r/d = 0.2and Kc = 1.65. Since the original design was fully stressed, the new safety factor is

ns =2

1.65= 1.21

For the shaft under bending, Fig. 6.5b is used to obtain the stress concentration factors. For D/d = 2and r/d = 0.1, Kc = 1.7, and corresponds to the original case. If r is doubled, then r/d = 0.2 andKc = 1.43. Since the original design was fully stressed, the new safety factor is

ns =1.71.43

= 1.19

For the shaft under torsion, Fig. 6.5(c) is used to obtain the stress concentration factors. For D/d = 2and r/d = 0.1, Kc = 1.43, and corresponds to the original case. If r is doubled, then r/d = 0.2 andKc = 1.22. Since the original design was fully stressed, the new safety factor is

ns =1.431.33

= 1.17

Therefore, the lowest safety factor is ns = 1.17 and corresponds to the torsion-loaded shaft.

6.8 The shaft shown in sketch c is subjected to tensile, torsional, and bending loads. Determine theprincipal stresses at the location of stress concentration. Ans. σ1 = 52.99 MPa, σ2 = 0, σ3 = -12.27MPa.

150 mm

120 mm

D = 45 mm

d = 30 mm

r = 3 mm

100 N-m1000 N

500 N

Sketch c, for Problem 6.8

Notes: This problem can be easily solved through the principal of superposition. The stress concen-tration factors are obtained from Figures 6.5 (a) and (b).

Solution: The rod will see normal stresses due to axial loads and bending, and a shear stress due totorsion. Note that the shear stress due to shear is zero at the extreme fibers where the stresses arelargest. The critical location is at the bottom where the bending and axial stresses are both tensile.Assign the x-axis to the rod axis. The normal stress is given by:

σc = Kc1P

A+ Kc2

Mc

I= (1.9)

1000 Nπ

4(0.03 m)2

+ (1.65)(500 N)(0.120 m)(0.015 m)

π

64(0.030 m)4

= 40.0 MPa

where the stress concentration factors of 1.9 and 1.65 are obtained from Figure 6.5 (a) and (b). Theshear stress is

τ = KcTc

J= (1.4)

(100 Nm)(0.015 m)π

32(0.030 m)4

= 26.45 MPa

121

Equation (2.16) gives the principal stresses.

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=40.0 MPa

2±

√(26.45 MPa)2 +

(40.0 MPa

2

)2

or σ1 = 53.0 MPa and σ2=-12.7 MPa. Note that the shear stress is very small compared to the normalstress; we could have taken σx as a principal direction.

6.9 A steel plate with dimensions shown in sketch d is subjected to 150-kN tensile force and 300-N-mbending moment. The plate is made of AISI 1080 steel and is kept at 20◦C. A hole is to be punchedin the center of the plate. What is the maximum diameter of the hole for a safety factor of 1.5? Ans.d = 120 mm.

dM

M

P235 mm

5 mm

1000 mm

Sketch d, for Problem 6.9

Notes: Equation (3.16) gives the allowable stress in bending. The normal stress is the sum of thebending stress and the axial normal stress, and is equated to the allowable stress. This gives anequation in terms of the hole diameter and the stress concentration factors in tension and bendingwhich can be solved iteratively.

Solution: From the inside front cover, AISI steel has a yield strength of Sy = 380 MPa. Therefore,the allowable stress is given by Eq. (3.16) as:

σall = 0.6Sy = 0.6(380 MPa) = 228 MPa

However, since the safety factor is 1.5, the allowable stress for this problem is 228 MPa/1.5= 152 MPa.The stress associated with axial tension is (see Fig. 6.2a):

σa =KcaP

(b− d)h=

Kca(150 kN)(0.235 m− d)(0.025 m)

=6 MN/m

0.235 m− dKca

The stress associated with bending is (see Fig. 6.2b):

σb =6KcbM

(b− d)h2=

6Kcb(300 Nm)(0.235 m− d)(0.025 m)2

=2.88 MN/m0.235 m− d

Kcb

Therefore, the maximum stress is:

σmax = σa + σb =(6 MN/m)Kca + (2.88 MN/m)Kcb

0.235 m− d

This should be equated to the maximum allowable stress, of σmax = 152 MPa. Note that Kca and Kcb

are functions of only d, since the other variables needed for their definition are fixed. Therefore, thisequation can be iteratively solved. Note that we can re-write the equation as:

(6 MN/m)Kca + (2.88 MN/m)Kcb

0.235 m− d= 152 MPa


∴ d = 0.235− (6 MN/m)Kca + (2.88 MN/m)Kcb

152 MPaAssume d = 100 mm, so that d/b = (100 mm)/(235 mm) = 0.426, and d/h = 100/25 = 4. From Fig.6.2 (a), Kca = 2.2 and From Fig. 6.2 (b), Kcb = 1.46. Therefore, this equation would predict

d = 0.235 m− (6 MN/m)(2.2) + (2.88 MN/m)(1.46)152 MPa

= 0.120 m

Therefore, the initial value was too small. If we now use d = 0.120 m = 120 mm, then d/h = 120/25 =4.8 and d/b = 120/235 = 0.511. This gives Kca = 2.16 and Kcb = 1.4. Therefore,

d = 0.235 m− (6 MN/m)(2.16) + (2.88 MN/m)(1.4)152 MPa

= 0.123 m

This is close to the assumed value, and closer agreement between assumed and calculated values isdifficult because of the resolution of Figs. 6.2a and 6.2b.

Section 6.3

6.10 A Plexiglas plate with dimensions 1 m × 1 m × 10 mm is loaded by a nominal tensile stress of 55 MPain one direction. The plate contains a small crack perpendicular to the load direction. At this stresslevel a safety factor of 2 against crack propagation is obtained. Find how much larger the crack canget before it grows catastrophically. Ans, a2 = 4a1.

Notes: Equation (6.4) is used to solve this problem.

Solution: For the first case, Eq. (6.4) gives:

Kci

2= Y σnom

√πa1 → Kci = 2Y σnom

√πa1

For the second case, the safety factor would be unity so that:

Kci = Y σnom√

πa2

Substituting for Kci:Y σnom

√πa2 = 2Y σnom

√πa1

∴ a2 = 22a1 = 4a1

Therefore, the crack can be 300% larger before catastrophic failure occurs.

6.11 A pressure container is made of AISI 4340 steel. The wall thickness is such that the tensile stress inthe material is 1100 MPa. The dimensionless geometry correction factor Y = 1 for the given geometry.Find how big the largest crack can be without failure if the steel is tempered

(a) At 260◦C Ans. 1.31 mm.

(b) At 425◦C Ans. 4.00 mm.

Notes: The material properties as a function of temper temperature is obtained from Table 6.1.Equation 6.4 is used to solve the problem.

Solution: The nominal stress is given as σnom = 1100 MPa, and Y = 1. The critical crack length isderived from Eq. (6.4):

Kci = Y σnom

√πa → a =

1π

(Kci

Y σnom

)2

Note that a is one-half the crack length.

123

(a) From Table 6.1, at a temper of 260◦C, the fracture toughness is 50 MPa-m1/2. Therefore,

a =1π

(50 MPa

√m

(1)(1100 MPa)

)2

= 6.577× 10−4 m = 0.6577 mm

Note that a is one-half the crack length, so that the critical crack length is 1.31 mm.

(b) At a temper of 425◦C, Kci = 87.4 MPa√

m. Therefore,

a =1π

(87.4 MPa

√m

(1)(1100 MPa)

)2

= 0.002009× 10−4 m = 2.009 mm

The critical crack length is 4.00 mm.

6.12 Two tensile test rods are made of AISI 4340 steel tempered at 260◦C and aluminum alloy 2024-T351.The dimensionless geometry correction factor Y = 1. Find how high a stress each rod can sustain ifthere is a crack of 2-mm half-length in each of them. Ans. AISI 4340: σ = 631 MPa.

Notes: The fracture toughness for these materials is obtained from Table 6.1 on page 232. Thenominal stress that can be sustained is then given by Eq. (6.4).

Solution: From Table 6.1 on page 232, Kci for AISI 4340 is 50.0 MPa-m1/2. For Al 2024-T351, Kci

is 36 MPa-m1/2. Therefore, the stress in the steel is given by Eq. (6.4) as:

Kci = Y σnom

√πa → σnom =

Kci

Y√

πa=

50.0 MPa(1)

√π(0.002)

= 631 MPa

For the Al 2024-T351,

Kci = Y σnom

√πa → σnom =

Kci

Y√

πa=

36.0 MPa(1)

√π(0.002)

= 454 MPa

6.13 A plate made of titanium alloy Ti-6Al-4V has the dimensionless correction factor Y = 1. How largecan the largest crack in the material be if it still should be possible to plastically deform the plate intension? Ans. 1.488 mm.

Notes: To plastically deform the plate, the nominal stress must exceed the yield strength of thematerial. Therefore, Eq. (6.4) solves the problem.

Solution: From Table 6.1 on page 232, the fracture toughness for Ti-6Al-4V varies from 44-66 MPa-m1/2. Also, the yield strength is 910 MPa. To plastically deform the material, σnom > Sy, or, fromEq. (6.4),

Kci = Y σnom

√πa → σnom =

Kci

Y√

πa≥ Sy

Solving for a,Kci

Y√

πa≥ Sy → a =

1π

(Kci

Y Sy

)2

=1π

(Kci

(1)(910 MPa)

)2

Since Kci has a value between 44-66 MPa-m1/2, then a has a range of a ≤ 0.744−1.67 mm. Therefore,if the largest crack is below 0.744 mm in half-length (1.488 mm in length), then the nominal stress willbe larger than the yield strength.

6.14 A Plexiglas model of a gear has a 1-mm half-length crack formed in its fillet curve (where the tensilestress is maximum). The model is loaded until the crack starts to propagate. Y = 1.5. How muchhigher a load can a gear made of AISI 4340 steel tempered to 425◦C carry with the same crack andthe same geometry? Ans. σsteel/σplexiglass = 87.4


Notes: Equation (6.4) is used to solve the problem with data from Table 6.1. This solution makessure that the steel does not plastically deform before catastrophic crack propagation occurs.

Solution: From Table 6.1 on page 232, the fracture toughnesses for plexiglass and steel are 1.0 and87.4 MPa-m1/2, respectively. For a gear, the bending stress is directly proportional to the applied load,so for a constant crack size a and correction factor Y the load possible is directly proportional to Kci.Therefore, Kci for steel is 87.4 times larger than for plexiglass, so it may be possible to support a load87.4 times larger.

However, it is possible that the steel will plastically deform at a lower stress than that needed topropogate the crack. With a = 1 mm = 0.001 m, Eq. (6.4) predicts the nominal stress level as:

Kci = Y σnom

√πa → σnom =

Kci

Y√

πa=

87.4 MPa√

m(1.5)

√π(0.001 m)

= 1040 MPa

Table 6.1 gives the yield strength for AISI 4340 tempered to 425◦C as 1420 MPa. Therefore, thenominal stress level needed for crack propagation is not sufficient to cause plastic deformation, and aload 87.4 times larger than for plexiglass can be carried.

6.15 A pressure vessel made of aluminum alloy 2024-T351 is manufactured for a safety factor of 2.5 guardingagainst yielding. The material contains cracks through the wall thickness with a crack half-length lessthan 3 mm. Y = 1. Find the safety factor when considering crack propagation.

Notes: The safety factor guarding against crack propagation is obtained from the ratio of the fracturetoughness of the material to the stress intensity factor calculated by Eq. (6.4).

Solution: From Table 6.1 on page 232, Sy = 325 MPa and Kci = 36 MPa-m1/2. The safety factorguarding against yielding is 2.5, therefore the nominal stress is σnom = 130 MPa. The stress intensityfactor is therefore calculated from Equation (6.4) as

Ki = Y σnom

√πa = (1)(130 MPa)

√π(0.003 m) = 12.62 MPa

√m

The safety factor against crack propagation is therefore

ns =Kci

Ki=

36 MPa√

m12.62 MPa

√m

= 2.85

Since the safety factor for yielding is lower than the safety factor guarding against crack propagation,the safety factor for the pressure vessel is still 2.5.

6.16 The clamping screws holding the top lid of a nuclear reactor are made of AISI 4340 steel temperedat 260◦C. They are stressed to a maximum level of 1250 MPa during a pressurization test beforestarting the reactor. Find the safety factor guarding against yielding and the safety factor guardingagainst crack propagation if the initial cracks in the material have Y = 1 and a = 1 mm. Also, do thecalculations for the same material but tempered to 425◦C. Ans. AISI 4340 tempered at 260◦C: ns =0.714.

Notes: This problem is similar to the previous problem. Equation (6.4) is used to solve this problem.

Solution:

(a) AISI 4340 Tempered at 260◦CFrom Table 6.1 on page 232, Sy = 1640 MPa and Kci = 50 MPa-m1/2. The safety factor againstyielding is therefore

ns =Sy

σ=

1640 MPa1250 MPa

= 1.31

125

From Equation (6.4), the stress intensity factor is

Ki = Y σnom

√πa = (1)(1250 MPa)

√π(0.001 m) = 70.06 MPa

√m

The safety factor guarding against crack propagation is therefore

ns =Kci

Ki=

50 MPa√

m70.06 MPa

√m

= 0.714

Since the safety factor is less than 1, the bolts will fail.

(b) AISI 4340 Tempered at 435◦C From Table 6.1 on page 232, Sy = 1420 MPa and Kci = 87.4MPa-m1/2. Using the same equations, the safety factor against yielding is ns = 1.14, and thesafety factor against crack propagation is ns = 1.25. Therefore, the bolts will not crack.

6.17 A glass tube used in a pressure vessel is made of aluminum oxide (sapphire) to make it possible toapply 30-MPa pressure and still have a safety factor of 2 guarding against fracture. For a soda-limeglass of the same geometry only 7.5-MPa pressure can be allowed if a safety factor of 2 is maintained.Find the size of the cracks the glass tube can tolerate at 7.5-MPa pressure and a safety factor of 2. Y= 1 for both tubes. Ans. Sapphire: 2a < 75.2 µm, Glass: 2a < 65.6 µm.

Notes: Material properties are obtained from Table A.3. As is shown in Chapter 9, the stress isproportional to the pressure. Equation (6.4) is used to solve this problem.

Solution: From Table A.3 on page 901, the fracture strength of soda-lime glass is 69 MPa. Thestresses in the tube are directly proportional to the pressure, so the fracture strength of the aluminumoxide tube is:

Sfa

pa=

Sfs

ps→ Sfs =

Sfspa

ps=

(69 MPa)(30 MPa)7.5 MPa

= 276 MPa

Note that this is on the low end of the fracture strength values given in Table A.3 for aluminum oxide(Al2O3).

For a safety factor of 2, the applied stress is 276/2=138 MPa. From Table 6.1 on page 232, using alow value of fracture toughness for Al2O3, use Kcia = 3.0 MPa-m1/2. From Eq. (6.4),

Ki =Kci

2= Y σnom

√πa

∴ a =1π

(Kci

2Y σnom

)2

=1π

(3.0 MPa

√m

2(1)(138 MPa)

)2

= 3.76× 10−5 m = 37.6 µm

For the soda lime glass, the lowest value of fracture toughness is Kci = 0.7 MPa-m1/2. The appliedstress is σnom = 69 MPa/2 = 34.5 MPa. The stress intensity factor equation is

Ki =Kci

2= Y σnom

√πa

∴ a =1π

(Kci

2Y σnom

)2

=1π

(0.7 MPa

√m

2(1)(34.5 MPa)

)2

= 3.28× 10−5 m = 32.8 µm

Therefore, the largest crack in the aluminum oxide must be less than 2a or 75.2 µm, while for sodalime glass the largest crack must be smaller than 65.6 µm.

6.18 A stress optic model used for demonstrating the stress concentrations at the ends of a crack is madeof polymethylmethacrylate. An artificially made crack 100 mm long is perpendicular to the loading


direction. Y = 1. Calculate the highest tensile stress that can be applied to the model withoutpropagating the crack. Ans. σnom = 1.78 MPa.

Notes: Material properties are obtained from Table 6.1 and Table A-4. Equation (6.4) is used to solvethis problem.

Solution: From Table 6.1 on page 232, the critical stress intensity factor for polymethylmethacrylateis Kci = 1.0 MPa-m1/2. From Equation (6.4) the stress when the crack propagates catastrophically is:

Kci = Y σnom

√πa → σnom =

Kci

Y√

πa=

1.0 MPa√

m(1)

√π(0.1 m)

= 1.78 MPa

From Table A-4 on page 902, the ultimate strength is between 48 and 76 MPa. Therefore, the crackwill propagate at a stress far lower than the ultimate strength of the material.

6.19 A passengerless airplane requires wings that are lightweight and the prevention of cracks more than 2mm long. The dimensionless geometry correction factor Y is usually 2.15 for a safety factor of 2.

(a) What is the appropriate alloy for this application? Ans. Either Aluminum 2024-T351 or Alloysteel 4340 tempered at 425 ◦C.

(b) If Y is increased to 3.5, what kind of alloy from Table 6.1 should be used? Ans. Al 2020-T351.

Notes: This problem solution is restricted to the materials in Table 6.1. Equation (6.4) is needed toobtain the solution.

Solution:

(a) From Eq. (6.4),Kci = Y σnom

√πa

a is one-half the crack length, so a is set equal to 1 mm or 0.001 m for this case. Recognizing thesafety factor is 2, we assign σnom = Sy/2. Therefore, Eq. (6.4) yields

Kci = YSy

2√

πa → Kci

Sy=

Y

2√

πa = 2.1512

√π(0.001 m) = 0.060

Note from Table 6.1:

Material Sy Kci Kci/Sy

Al 2024-T351 325 36 0.110Al 2024-T651 505 29 0.057Steel 4340 Tempered at 260◦C 1640 50 0.0305Steel 4340 Tempered at 425◦C 1420 87.4 0.0615Ti-6Al-4V 910 44-66 0.048-0.0725

From these calculations, either Al 2024-T351 or Steel 4340 tempered at 425◦C would be suitable.Ti-6Al-4V may be acceptable depending on the quality of the material.

(b) If Y = 3.5, then from Eq. (6.4),

Kci

Sy=

Y

2√

πa = 3.512

√π(0.001 m) = 0.098

For this condition, only Al 2024-T351 would be suitable.

Section 6.6

127

6.20 The anchoring of the cables carrying a suspension bridge are made of cylindrical AISI 1080 steel bars210 mm in diameter. The force transmitted from the cable to the steel bar is 3.5 MN. Calculate thesafety factor range guarding against yielding based on the allowable stressed from Eq. (3.13). Ans.1.69 < ns < 2.25

Notes: The material property is obtained from the inside front cover. Equation (3.13) gives a rangefor allowable stresses in tension.

Solution: From the inside front cover, the yield strength for AISI 1080 is Sy = 380 MPa. The stressin the steel bar is

σ =P

A=

3.5 MNπ

4(0.21 m)2

= 101 MPa

Since the safety factor is ns = σall/σ, Eq. (3.13) gives

0.45Sy ≤ σall ≤ 0.60Sy

0.45Sy

σ≤ ns ≤

0.60Sy

σ

0.45(380 MPa)101 MPa

≤ ns ≤0.60(380 MPa)

101 MPaTherefore the safety factor is in the range of 1.69 ≤ ns ≤ 2.25.

6.21 The arm of a crane has two steel plates connected with a rivet that transfers the force in pure shear.The rivet is made of AISI 1040 steel and has a circular cross section with a diameter of 25 mm. Theload on the rivet is 20 kN. Calculate the safety factor. Ans. ns = 3.44

Notes: Equation (3.14) gives the allowable stress in shear.

Solution: The yield strength of AISI 1040 steel is obtained from the inside front cover as Sy = 350MPa. From Eq. (3.14),

τall = 0.4Sy = 0.4(350 MPa) = 140 MPa

The shear stress on the rivet is

τ =P

A=

20 kN(π

4(0.025 m)2

) = 40.74 MPa

Therefore, the safety factor is:

ns =τall

τ=

140 MPa40.74 MPa

= 3.44

Section 6.7

6.22 A machine element is loaded so that the principal normal stresses at the critical location for a biaxialstress state are σ1 = 20 ksi and σ2 = -15 ksi. The material is ductile with a yield strength of 60 ksi.Find the safety factor according to

(a) The maximum-shear-stress theory (MSST) Ans. ns = 1.714

(b) The distortion-energy theory (DET) Ans. ns = 1.97

Notes: Equation (6.6) is used to obtain the safety factor for the MSST. Equation (6.11) gives thesafety factor for the DET after the von Mises stress is calculated from Eq. (6.9). If the stress is biaxial,then one principal stress is zero.


Solution: First of all, since the stress state is biaxial, then one normal stress is zero. Therefore,the three principle stresses are properly referred to as σ1 = 20 ksi, σ2 = 0 and σ3 = −15 ksi, sinceσ1 ≥ σ2 ≥ σ3. For the maximum shear stress theory, Eq. (6.6) gives the safety factor as:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

60 ksi(20 ksi + 15 ksi)

= 1.714

The von Mises stress is obtained from Eq. (6.9) as:

σe =1√2

[(σ2 − σ1)

2 + (σ3 − σ1)2 + (σ3 − σ2)

2]1/2

=1√2

[(−20 ksi)2 + (−35 ksi)2 + (15 ksi)2

]1/2

or σe = 30.4 ksi. Therefore, the safety factor is, from Eq. (6.11),

σe =Sy

ns→ ns =

Sy

σe=

60 ksi30.4 ksi

= 1.97

6.23 A bolt is tightened, subjecting its shank to a tensile stress of 80 ksi and a torsional shear stress of 50ksi at a critical point. All of the other stresses are zero. Find the safety factor at the critical point bythe DET and the MSST. The material is high-carbon steel (AISI 1080). Will the bolt fail because ofthe static loading? Ans. ns,DET = 0.47, ns,MSST = 0.43.

Notes: Equations (2.16), (6.6), (6.10), and (6.11) are used to solve this problem.

Solution: From the inside front cover, the yield stress for AISI 1080 steel is 55 ksi. Directions arearbitrary; let’s refer to the tensile stress as σx = 80 ksi and the shear stress as τxy = 50 ksi. Since allother stresses are zero, Eq. (2.16) gives the principal stresses as

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=80 ksi

2±

√(50 ksi)2 +

(80 ksi

2

)2

or σ1 = 104 ksi, σ2 = −24 ksi. Note that the other stresses are zero, so the principal stress out of theplane of the normal and shear stresses is zero. Putting the stresses in the proper order (σ1 ≥ σ2 ≥ σ3),we assign them the values σ1 = 104 ksi, σ2 = 0 ksi, σ3 = −24 ksi. From Eq. (6.6),

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

55 ksi104 ksi− (−24 ksi)

= 0.43

which is the safety factor for the maximum shear stress theory. Equation (6.9) gives

σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(104 ksi− 0)2 + (104 ksi + 24 ksi)2 + (0 + 24 ksi)2

]1/2

= 118 ksi

From Eq. (6.11),

σe =Sy

ns→ ns =

Sy

σe=

55 ksi118 ksi

= 0.47

Since the safety factor is less than one for both cases, both cases predict failure.

6.24 A torque is applied to a piece of chalk used in a classroom until the chalk cracks. Using the maximum-normal-stress theory (MNST) and assuming the tensile strength of the chalk to be small relative to itscompressive strength, determine the angle of the cross section at which the chalk cracks. Ans. 45◦.

129

Notes: Given the loading condition, the angle of the largest tensile stress is obtained from Eq. (2.15).Based on the MNST, failure will occur at this angle.

Solution: For pure torque, the stress state is τxy = τ and σx = σy = 0. The angle of the largesttensile stress, φσ, is given by Eq. (2.15) as:

tan 2φσ =2τxy

σx − σy=

2τxy

0=∞

Therefore, φσ = 45◦. The chalk will crack along a 45◦ angle from its circumference.

6.25 A cantilevered bar 500 mm long with square cross section has 25-mm sides. Three perpendicular forcesare applied to its free end, a 1000 N force is applied in the x direction, a 100 N force is applied in they direction, and an equivalent force of 100 N is applied in the z direction. Calculate the equivalentstress at the clamped end of the bar by using the DET when the sides of the square cross section areparallel with the y and z directions.

Notes: The stresses are largest at the corners, where the total stress is the sum of two bending stressesand the axial stress. The effective stress is obtained from Eq. (6.9).

Solution: The moment of inertia for the cross section is:

I =bh3

12=

a4

12=

(0.025 m)4

12= 3.255× 10−8 m4

The cross sectional area is A = a2 = (0.025 m)2 = 6.25 × 10−4 m2. Since the bar is cantilevered,the loading is a combined situation of two bending moments and one axial load. The perpendicularmoments are M1 = M2 = Fl = (100 N)(0.5 m) = 50 Nm. The axial load is 1000 N. Therefore, themaximum stress occurs at a corner of the cross section, and is the sum of the stresses due to the threeloads. Therefore,

σx =M1c

I+

M2c

I+

P

A= 2

(50 Nm)(0.0125 m)3.255× 10−8 m4

+1000 N

6.25× 10−4 m2= 40.00 MPa

There is no stress in the y or z directions. Also, at the outside edge of the bar, the shear stress is zero(see page 169). Therefore, σ1 = 40.00 MPa, σ2 = σ3 = 0. From Eq. (6.9), the von Mises stress is

σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2=

1√2

[(40 MPa)2 + (40 MPa)2 + (0)2

]1/2

This is evaluated as σe = 40 MPa.

6.26 A shaft transmitting torque from the gearbox to the rear axle of a truck is unbalanced, so that acentrifugal load of 500 N acts at the middle of the 3-m-long shaft. The AISI 1040 tubular steel shafthas an outer diameter of 70 mm and an inner diameter of 58 mm. Simultaneously, the shaft transmitsa torque of 6000 N-m. Use the DET to determine the safety factor guarding against yielding. Ans. ns

= 1.196.

Notes: The moment must be determined, which then allows for calculation of the bending stress. Thetorque results in a shear stress; this combined stress state is then transformed to obtain the principalstresses. Equations (6.9) and (6.11) are then used to solve the problem.

Solution: From the inside front cover, the yield strength of AISI 1040 steel is Sy = 350 MPa. Themoment of inertia for the shaft is:

I =π

64(d4

o − d4i

)=

π

64[(0.070 m)4 − (0.058 m)4

]= 6.23× 10−7 m4


Similarly, J = 1.246 × 10−6 m4. For a simply supported shaft, the maximum moment occurs at thecenter of the shaft and has the value M = Pl/4 = (500 N)(3 m)/4 = 375 Nm. Therefore, the bendingstress is obtained from Eq. (4.48) as:

σx =Mc

I=

(375 Nm)(0.035 m)6.23× 10−7 m4

= 21.07 MPa

The shear stress due to the torque is given by Eq. (4.34) as:

τxy =Tc

J=

(6000 Nm)(0.035 m)1.246× 10−6 m4

= 168.5 MPa

Also, σy = σz = τzx = τyz = 0. Therefore, from Eq. (2.16),

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=21.07 MPa

2±

√(168.5 MPa)2 +

(21.07 MPa

2

)2

Therefore, σ1 = 179.4 MPa, σ2 = 0, and σ3 = −158.3 MPa. Note that the principal stresses have beenrenumbered so that σ1 ≥ σ2 ≥ σ3. Equation (6.9) gives the effective stress as

σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(179.4 MPa− 0)2 + (179.4 MPa + 158.3 MPa)2 + (158.3 MPa)2

]1/2

= 292.6 MPa

From Equation (6.11),

σe =Sy

ns→ ns =

Sy

σe=

350 MPa292.6 MPa

= 1.196

6.27 The right-angle-cantilevered bracket used in Problem 5.30, sketch w, has a concentrated force of 1000N and a torque of 300 N-m. Calculate the safety factor. Use the DET and neglect transverse shear.Assume that the bracket is made of AISI 1040 steel and use the following values: a = 0.5 m, b = 0.3m, d = 0.035 m, E = 205 GPa, and ν = 0.3. Ans. ns=1.76.

Notes: The stresses must be determined using the approach described in Chapter 4. From the stressstate, the principal stresses are determined. Equation (6.6) gives the safety factor for the MaximumShear Stress Theory, and Eqs. (6.9) and (6.11) give the safety factor for the Distortion-Energy Theory.

Solution: From the inside front cover, Sy = 350 MPa for AISI 1040 steel. The moment of inertia ofthe bracket cross section is:

I =π

64d4 =

π

64(0.035 m)4 = 7.366× 10−8 m4

Similarly, J = 1.4732× 10−7 m4. The maximum stress for the bracket occurs at the wall (x = a). Theloading is a bending moment and a torque. The moment is due to the applied torque T and the loadP , and is

M = Pa + T = (1000 N)(0.5 m) + 300 Nm = 800 Nm

Therefore, the bending stress is obtained from Eq. (4.48) as:

σx =Mc

I=

(800 Nm)(0.0175 m)7.366× 10−8 m4

= 190 MPa

At the wall, there is a torque of T = Pb = (1000 N)(0.3 m) = 300 Nm. The shear stress due to thetorque is given by Eq. (4.34) as:

τxy =Tc

J=

(300 Nm)(0.0175 m)1.4732× 10−7 m4

= 35.6 MPa

131


σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=190 MPa

2±

√(35.6 MPa)2 +

(190 MPa

2

)2

Therefore, σ1 = 196 MPa, σ2 = 0, and σ3 = −6.45 MPa. Note that the principal stresses have beenrenumbered so that σ1 ≥ σ2 ≥ σ3. From Eq. (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

350 MPa196 MPa + 6.45 MPa

= 1.73

Equation (6.9) gives the effective stress as

σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(196 MPa− 0)2 + (196 MPa + 6.45 MPa)2 + (6.45 MPa)2

]1/2

= 199 MPa

From Equation (6.11), the safety factor for DET is:

σe =Sy

ns→ ns =

Sy

σe=

350 MPa199 MPa

= 1.76

6.28 A 100-mm-diameter shaft is subjected to a 10 kN-m steady bending moment, an 8 kN-m steady torque,and a 150-kN axial force. The yield strength of the shaft material is 600 MPa. Use the MSST and theDET to determine the safety factors for the various types of loading. Ans. ns = 4.28.

Notes: This is similar to problems 6.28 and 6.29, but now a stress due to the axial force must beincluded. From the stress state, the principal stresses are determined. Equation (6.6) gives the safetyfactor for the Maximum Shear Stress Theory, and Eqs. (6.9) and (6.11) give the safety factor for theDistortion-Energy Theory.

Solution: The moment of inertia of the shaft cross section is:

I =π

64d4 =

π

64(0.10 m)4 = 4.909× 10−6 m4

Similarly, J = 9.817 × 10−6 m4. The area of the cross section is πd2/4 = 0.00785 m2. The bendingstress is obtained from Eq. (4.48) as:

σx =Mc

I=

(10 kNm)(0.05 m)4.909× 10−6 m4

= 101.8 MPa

The normal stress due to the axial load is

σx =P

A=

150 kN0.00785 m2

= 19.10 MPa

Therefore, the maximum normal stress is σx = 101.8 MPa+19.10 MPa = 120.9 MPa. The shear stressdue to the torque is given by Eq. (4.34) as:

τxy =Tc

J=

(8 kNm)(0.05 m)9.817× 10−6 m4

= 40.7 MPa


σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=120.9 MPa

2±

√(40.7 MPa)2 +

(120.9 MPa

2

)2


Therefore, σ1 = 133.3 MPa, σ2 = 0, and σ3 = −12.42 MPa. Note that the principal stresses have beenrenumbered so that σ1 ≥ σ2 ≥ σ3. From Eq. (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

600 MPa133.3 MPa + 12.42 MPa

= 4.12


σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(133.3 MPa− 0)2 + (133.3 MPa + 12.42 MPa)2 + (12.42 MPa)2

]1/2

= 140 MPa

From Eq. (6.11), the safety factor for DET is:

σe =Sy

ns→ ns =

Sy

σe=

600 MPa140 MPa

= 4.28

6.29 Use the MSST and the DET to determine the safety factor for 2024 aluminum alloys for each of thefollowing stress states:

(a) σx = 10 MPa, σy = -60 MPa Ans. ns,MSST = 4.64.

(b) σx = σy = τxy = -30 MPa Ans. ns,DET = 5.42.

(c) σx = −σy = 20 MPa, and τxy = 10 MPa Ans. ns,MSST = 7.27.

(d) σx = 2σy = -70 MPa, and τxy = 40 MPa Ans. ns,DET = 3.53.

Notes: This problem does not require determination of the stresses as in Problems 6.28 through 6.30,but uses the same approach. From the stress state, the principal stresses are determined. Equation(6.6) gives the safety factor for the Maximum Shear Stress Theory, and Equations (6.9) and (6.11) givethe safety factor for the Distortion-Energy Theory

Solution: From Table 6.1, the yield strength for 2024-T351 is Sy = 325 MPa.

(a) For σx = 10 MPa, σy = −60 MPa, note that there are no shear stresses. Therefore, we candirectly write the principal stresses as σ1 = 10 MPa, σ2 = 0 MPa and σ3 = −60 MPa. Note thatthe principal stresses have been renumbered so that σ1 ≥ σ2 ≥ σ3. From Eq. (6.6), the safetyfactor for MSST is:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

325 MPa10 MPa + 60 MPa

= 4.64


σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(10 MPa− 0)2 + (10 MPa + 60 MPa)2 + (60 MPa)2

]1/2

= 65.57 MPa

From Eq. (6.11), the safety factor for DET is:

σe =Sy

ns→ ns =

Sy

σe=

325 MPa65.57 MPa

= 4.96

133

(b) For σx = σy = τxy = −30 MPa, Eq. (2.16) gives

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=−30 MPa− 30 MPa

2±

√(30 MPa)2 + (0)2

Therefore, σ1 = 0 MPa, σ2 = 0 MPa and σ3 = −60 MPa. Note that the principal stresses havebeen renumbered so that σ1 ≥ σ2 ≥ σ3. From Eq. (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

325 MPa0 MPa + 60 MPa

= 5.42


σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(0 MPa)2 + (60 MPa)2 + (60 MPa)2

]1/2

= 60 MPa


σe =Sy

ns→ ns =

Sy

σe=

325 MPa60 MPa

= 5.42

(c) For σx = −σy = 20 MPa and τxy = 10 MPa, Eq. (2.16) gives

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=20 MPa− 20 MPa

2±

√(10 MPa)2 + (20 MPa + 20 MPa)2

Therefore, σ1 = 22.36 MPa, σ2 = 0 MPa and σ3 = −22.36 MPa. Note that the principal stresseshave been renumbered so that σ1 ≥ σ2 ≥ σ3. From Eq. (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

325 MPa22.36 MPa + 22.36 MPa

= 7.27


σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(22.36 MPa− 0)2 + (22.36 MPa + 22.36 MPa)2 + (22.36 MPa)2

]1/2

= 38.73 MPa


σe =Sy

ns→ ns =

Sy

σe=

325 MPa38.73 MPa

= 8.39

(d) For σx = 2σy = −70 MPa, and τxy = 40 MPa, Eq. (2.16) gives

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=−70 MPa− 35 MPa

2±

√(40 MPa)2 + (−70 MPa + 35 MPa)2


Therefore, σ1 = 0 MPa, σ2 = −8.84 MPa and σ3 = −96.16 MPa. Note that the principal stresseshave been renumbered so that σ1 ≥ σ2 ≥ σ3. From Eq. (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns→ ns =

Sy

σ1 − σ3=

325 MPa0 MPa + 96.16 MPa

= 3.38


σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2

=1√2

[(8.84 MPa)2 + (96.16 MPa)2 + (−8.84 MPa + 96.16 MPa)2

]1/2

= 92.06 MPa


σe =Sy

ns→ ns =

Sy

σe=

325 MPa92.06 MPa

= 3.53

6.30 Four different stress elements, each made of the same material, are loaded as shown in sketches e, f ,g, and h. Use the MSST and the DET to determine which element is the most critical. Ans. Sketch eis most critical.

21 MPa

21 MPa

7.5 MPa

28.5 MPa

30 MPa

30 MPa

10 MPa

30 MPa

(e) (f)

(g) (h)

Sketches e, f , g, and h, for Problem 6.30

Notes: Equations (2.16), (6.6), (6.10), and (6.11) are used to solve this problem.

Solution:

(e) σ1 = 21 MPa, σ2 = 0, σ3 = −21 MPa. Therefore, from Eq. 6.6: σ1 − σ3 = 42 MPa. Also, fromEq. 6.9,

σe =1√2

[(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2

]1/2= 36 MPa

(f) σ1 = 28.5 MPa, σ2 = 0, σ3 = −7.5 MPa. Thus, σ1 − σ3 = 36 MPa. Also,

σe =1√2

[(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2

]1/2= 33 MPa

135

(g) σ1 = 30 MPa, σ2 = 30 MPa, σ3 = 0. Thus, σ1 − σ3 = 30 MPa. Also σe = 30 MPa.

(h) σx = 30 MPa, σy = 0, τxy = −10 MPa. Therefore, from Equation (2.16), σ1 = 33 MPa, σ2 = 0,σ3 = −3.02 MPa. Therefore, σ1 − σ3 = 36.02 MPa, σe = 31.6 MPa.

This shows that the stress state in (e) is the largest.

6.31 The rod shown in sketch i is made of AISI 1040 steel and has two 90◦ bends. Use the MSST and theDET to determine the minimum rod diameter for a safety factor of 2 at the most critical section.

300 mm

750 mm

y′

yx

z

z′

x′

50 mm

1500 N100 N

800 N

Sketches i, for Problem 6.31

Notes: Recognizing that the critical section is at the wall, the component stresses can be expressedas functions of the rod diameter. Applying MSST or DET gives an expression that can be solved ford.

Solution: The yield strength for AISI 1040 is obtained from the inside front cover as 350 MPa. Thecritical section is at the wall; the rod is slender so transverse shear effects will be ignored. The 800 Nload causes a torque equal to T1 = 800 N(0.75 m)=600 Nm, and bending moment Mx1 = 800 N(0.3 m)= 240 Nm. The 100 N load causes axial normal stress, a bending moment Mz1 = 100 N(0.75m)=75Nm and a bending moment Mx2 = 100 N(0.05m)=5 Nm, which is in the opposite direction as Mx1.The 1500 N load causes torque equal to T2 = −1500 N(0.05m)=75 Nm (in the opposite direction asT1) and a bending moment Mz2 = 1500 N(0.3 m)= 450 Nm, which is in the opposite direction as Mz1.Therefore, the bar sees the following: Mx = Mx1 −Mx2 = 235 Nm, Mz = 450 Nm-75 Nm= 375 Nm,and T = T1 − T2 = 600 Nm-75 Nm= 525 Nm. Therefore, the moment at the wall is

M =√

M2x + M2

z =√

(235 Nm)2 + (375 Nm)2 = 442.5 Nm

The normal stress is therefore

σ =Mc

I+

P

A=

100 Nπ

4d2

+(442.5 Nm)(d/2)

π

64d4

=127.3 N

d2+

4507 Nmd3

=(127.3 N)d + 4507 Nm

d3

The shear stress is:

τ =Tc

J=

(525 Nm)(d/2)π

32d4

=2674 Nm

d3


Equation (2.16) gives

σ1, σ2 =σx + σy

2±

√τ2xy +

(σx − σy

2

)2

=1d3

{[(63.65 N)d + 2253.5 Nm]±

√(2674 Nm)2 + [(63.65 N)d + 2253.5 Nm]2

}It can be shown that unless d is in the thousands of meters, that one of these stresses will be positive,and the other negative. Therefore, for MSST, Eq. (6.6) is:

σ1 − σ3 =Sy

ns=

350 MPa2

= 175 MPa

Substituting for the stresses and solving yields d = 0.0341 m. Therefore, a 35 mm diameter cross-sectionis a good design designation. For DET, Equations (6.9) and (6.11) give:

σe =1√2

[(σ1 − σ2)2 + (σ1 − σ3)2 + (σ2 − σ3)2

]1/2=

1√2

[(σ1)2 + (σ1 − σ3)2 + (σ3)2

]1/2= 175 MPa

This is solved numerically as d = 0.0287 m. Therefore, a 35 mm diameter cross section is still accept-able.

6.32 The shaft shown in sketch j is made of AISI 1020 steel. Determine the most critical section by usingthe MSST and the DET. Dimensions of the various diameters shown in sketch j are d = 30 mm, D =45 mm, and d2 = 40 mm.

40 mm40 mm

40 mm

Dd

r = 6 mmr = 9 mm

T = 50 N-m

10 kN

1 kNd2

Sketches j, for Problem 6.32

Notes: This problem requires the incorporation of stress concentration effects into the componentstresses before determining the principal stresses.

Solution:

(a) Fillet. First, considering the location of stress concentration 80 mm from the wall:

J =π

32d4 =

π

32(0.040 m)4 = 7.95× 10−8 m4

I =J

2= 3.98× 10−8 m4

A = πd2/4 = 7.07× 10−4 m2

Also, from statics, V = 1 kN, M = 40 Nm, N = 10 kN, T = 50 Nm. The bottom locationis critical, since the bending and tensile stresses are additive at this location. Also, there is noshear stress due to shear at the extreme location. The stress concentration due to bending isobtained from Fig. 6.5(b) as 1.4, while for tension it is Kc = 1.55 from Fig. 6.5(a). The stressconcentration for torsion is Kc = 1.2 from Fig. 6.5(c). Therefore, σ1 = 458 MPa and σ2 = −28MPa.

137

(b) Groove For the location 40 mm from the wall, N = 10 kN, M = 80 Nm T = 50 Nm,Kc (bending) = 1.45, Kc (tension) = 1.5, Kc (torsion) = 1.2 (all from Fig. 6.6). Therefore,σ1 = 31.2 MPa and σ2 = −0.76.

This means that the critical location is 80 mm from the wall, as the stresses are higher.

plugin chapter6

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