CH_06

Page 6-1

Chapter 6 • Failure Prediction for Static Loading

6.1 Given that the stress concentration factor is 3.12 for a machine element made of steelwith a modulus of elasticity of 207GPa, find the stress concentration factor for anidentical machine element made of aluminum instead of steel. The modulus of elasticityfor aluminum is 69GPa.

Solution:Since the elastic stress concentration is entirely determined from the geometry of the machineelement, the stress concentration factor will remain the same. Therefore, Kc=3.12.

6.2 A flat part with constant thickness b is loaded in tension as shown in Fig. 6.3(a). Theheight changes from 50 to 100mm with a radius r=10mm. Find how much higher a loadcan be transmitted through the bar if the height increases from 50 to 70 mm and theradius decreases from 10 to 3mm.

Notes: To answer this question, one must compare the stress concentration factors for the twocases. The stress concentration factors are obtained from Figure 6.3(a) on page 233.

Solution:Referring to the sketch in Figure 6.3(a), H=100mm, h=50mm and r=10mm for the first case.Therefore, H/h=100mm/50mm=2 and r/h=10mm/50mm=0.2. From Figure 6.3(a), Kc for this caseis 1.8. For the second case, H/h=100/70=1.43 and r/h=3mm/70mm=0.043. Therefore, FromFigure 6.3(a), Kc is around 2.8.

The load that can be transmitted depends on the maximum stress. Therefore, for the firstcase:

σmax = Kc1P1

bh1; P1 = bh1σmax

Kc1= b 50mm( )σ max

1.8For the second case:

σmax = Kc 2P2

bh2; P2 =

bh2σmax

Kc2=

b 70mm( )σ max

2.8The ratio of the two forces is:

P1

P2=

b 50mm( )σ max

1.8

b 70mm( )σ max

2.8

= 50mm( )2.870mm( )1.8

= 1.111

Therefore the load carrying capacity decreases even though the thickness of the bar increasedfrom 50 to 70 mm, mainly because the radius of the notch decreased.

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

Page 6-2

Notes: The exact hole locations have not been specified, so some variation may occur withassumed dimensions. Also, the two-hole case does not correspond to a particular chart in Figure6.1 through 6.6. However, it is possible to obtain a reasonable solution from the existing data. Fora critical application, more advanced approaches, such as finite element analysis, would benecessary. See also Problem 6.6.

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

Considering the top half of the problem leads to:Here the diameter to width ratio is 0.4, so Kc isaround 2.2 from Figure 6.2(a). Note that this istrue for point A but not point B in the figureabove.

Because of St. Venant’s Principal, we must be concerned about the stress concentrationsinteracting between 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 just under 2.0 from Figure6.4(a).

Therefore, the larger stress concentration is Kc=2.2 and point A is more important 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 load-carrying beam is loaded with a bending moment M. The minor beam height is hand the width in the perpendicular direction is b. This beam, which carries a balcony onthe wall of a house, is welded together with a beam in the house structure. That beam hasa major height H. Because of manufacturing problems the radius r of the connectionscannot be made larger than rmax but can be made smaller, down to zero. Find how theload-carrying beam should be placed on the house beam to give maximum strength.

Notes: Design recommendations arise from studying Figure 6.3 (b).

Solution:From Figure 6.3 (b), the stick-out H/h should be small to give a low stress concentration factor.Also, the radius to height ratio should be as large as possible. The optimum configuration is tohave the small beam flush with the top side of the large beam which gives zero stressconcentration.

Page 6-3

6.5 A round bar has a fillet with r/d=0.15 and D/d=1.5. The bar transmits both bendingmoment and torque. A new construction is considered to make the shaft stiffer andstronger by making it equally thick on each side of the fillet or groove. Determinewhether 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 for bending (fromFigure 6.5 (b)) and about 1.25 for torsion (from Figure 6.5(c)). If instead of a fillet the bar becamea groove, with the same root diameter, then the stress concentrations are obtained from Figure 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 rectangular plate has a central hole. The width of the plate is b=50mm and the diameterof the hole is d=10mm. The plate is axially loaded with a force P=1500N, and the minorheight of the plate is h=5mm. Is it possible to find the stress concentration factor for theplate without using Figure 6.2?

Notes: Figure 6.2 can obviously be used to solve the problem, and serves as a useful comparisonfor the alternate technique. The problem can be approximated by Figure 6.4 if the plate issectioned. This solution compares the results of both approaches.

Solution:First of all, if Figure 6.2(a) is used with d/b=(10mm)/(50mm)=0.2, Kc=2.5. An alternativeapproach is to cut the beam in half across the center of the hole, then reassembling the twosections so the holes are on the exterior, thereby approximating the loading case in Figure 6.4(a).Note that there are differences; there are not two matching fillets on opposite sides, and theloading is complicated by the presence of a Poisson effect in the original geometry. However, thetwo should be fairly close. For the new, sectioned geometry, r=d/2=5mm, H=b=50mm, h=(b-d)=(50mm-10mm)=40mm. Therefore, H/h=(50mm)/(40mm)=1.25, andr/h=(5mm)/(40mm)=0.125. From Figure 6.4(a), Kc is around 2.3. The error between the two casesis:

%error =2.5 − 2.3

2.5×100% = 8%

6.7 A machine has three circular shafts, each with fillets giving stress concentrations. Theratio of fillet radius to shaft diameter is 0.1 for all three shafts. One of the shafts transmitsa tensile force, one transmits a bending torque, and one transmits torsion. Because theyare stressed exactly to the stress limit (ns=1), a design change is proposed doubling thenotch radii to get a safety factor greater than 1. How large will the safety factors be forthe three shafts if the diameter ratio is 2 (D/d=2)?

Notes: Figure 6.5 is used to solve this problem.

Page 6-4

Solution:For the shaft under tension, Figure 6.5(a) is used to obtain the stress concentration

factors. For D/d=2 and r/d=0.1, Kc=2.0, and corresponds to the original case. If r is doubled, thenr/d=0.2 and 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, Figure 6.5(b) is used to obtain the stress concentrationfactors. For D/d=2 and r/d=0.1, Kc=1.7, and corresponds to the original case. If r is doubled, thenr/d=0.2 and Kc=1.43. Since the original design was fully stressed, the new safety factor is

ns =1.7

1.43= 1.19

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

ns =1.43

1.33= 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 the principal stresses at the location of stress concentration.

Notes: This problem can be easily solved through the principal of superposition. The stressconcentration 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 to torsion.Note that the shear stress due to shear is zero at the extreme fibers where the stresses are largest.The critical location is at the bottom where the bending and axial stresses are both tensile. Assignthe x-axis to the rod axis. The normal stress is given by:

σx = Kc1P

A+ Kc2

Mc

I= 2.0( ) 10kN

π4

0.03m( )2+ 1.75( ) 50kN( ) 0.130m( ) 0.015m( )

π64

0.030m( )4= 4315MPa

where the stress concentration factors of 2.0 and 1.75 are obtained from Figure 6.5 (a) and (b).The shear stress is

τ = KcTc

J= 1.45( ) 100Nm( ) 0.015m( )

π32

0.030m( )4= 27.4MPa

Equation (2.16) gives the principal stresses.

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=4315MPa

2± 27.4MPa( )2 +

4315MPa

2

2

Page 6-5

or σ1=4315MPa and σ2=-0.17MPa. Note that the shear stress is very small compared to thenormal stress; we could have taken σx as a principal direction.

6.9 A steel plate with dimensions shown in sketch d is subjected to a 150kN tensile force and300Nm bending moment. The plate is made of AISI 1080 steel and is kept at 20°C. Ahole is to be punched in the center of the plate. What is the maximum diameter of thehole for a safety factor of 1.5?

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=380MPa. Therefore, theallowable stress is given by Equation (3.16) as:

σall=0.6Sy=0.6(380MPa)=228MPaHowever, since the safety factor is 1.5, the allowable stress for this problem is 228MPa/1.5=152MPa. The stress associated with axial tension is (see Figure 6.2(a)):

σa =Kca P

b − d( )h=

Kca 150kN( )0.235m − d( ) 0.025m( )

=6MN / m

0.235m − dKca

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

σb =6KcbM

b − d( )h2 =6Kcb 300Nm( )

0.235m − d( ) 0.025m( )2 =2.88MN / m

0.235m −dKcb

Therefore, the maximum stress is:

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

0.235m − dThis should be equated to the maximum allowable stress, of σmax=152MPa. Note that Kca and Kcb

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

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

0.235m − d= 152MPa; d = 0.235m −

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

152MPaAssume d=100mm, so that d/b=(100mm)/(235mm)=0.426, and d/h=100/25=4. From

Figure 6.2 (a), Kca=2.2 and From Figure 6.2 (b), Kcb=1.46. Therefore, this equation would predict

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

152MPa= 0.120m

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

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

152MPa= 0.123m

This is close to the assumed value, and closer agreement between assumed and calculated valuesis difficult because of the resolution of Figures 6.2(a) and 6.2(b).

Page 6-6

6.10 A Plexiglass plate with dimensions 1m x 1m x 1cm is loaded by a nominal tensile stressof 55MPa in one direction. The plate contains a small crack perpendicular to the loaddirection. At this stress level a safety factor of 1.3 against crack propagation is obtained.Find how much larger the crack can get before it grows catastrophically.

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

Solution:For the first case, Equation (6.4) gives:

Kci

1.3= Yσnom πa1 ; Kci = 1.3Yσnom πa1

For the second case, the safety factor would be unity so that:Kci = Yσnom πa2

Substituting for Kci:

Yσnom πa2 = 1.3Yσnom πa1 ; a2 = 1.32a1 = 1.69a1Therefore, the crack can be 69% larger before catastrophic failure occurs.

6.11 A pressure container is made of AISI 4340 steel. The wall thickness is such that thetensile stress in the material is 1100MPa. The dimensionless geometry correction factorY=1 for the given geometry. Find how big the largest crack can be without failure if thesteel is tempereda) at 260°Cb) at 425°C

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=1100MPa, and Y=1. The critical crack length is derived fromEq. (6.4):

Kci = Yσnom πa;a =1

πKci

Yσnom

2

From Table 6.1, at a temper of 260°C, the fracture toughness is 50MPam1/2. Therefore, thecritical crack length is 1.3 mm. At a temper of 425°C, The critical crack length is 4.0mm. (Note:a is one-half the crack length.)

6.12 Two tensile test rods are made of AISI 4340 steel tempered at 260°C and aluminum alloy2024-T351. The dimensionless geometry correction factor Y=1. Find how high a stresseach rod can sustain if there is a 2-mm crack half length in each of them.

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 Equation (6.4).

Solution:

Page 6-7

From Table 6.1 on page 232, Kci for AISI 4340 is 50.0MPam1/2

. For Al 2024-T351, Kci is36MPam

1/2. Therefore, the stress in the steel is given by Equation (6.4) as:

Kci = Yσnom πa;σnom =Kci

Y πa=

50.0MPa

1( ) π 0.002( )= 631MPa

For the Al 2024-T351,

Kci = Yσnom πa;σnom =Kci

Y πa=

36.0 MPa

1( ) π 0.002( )= 454MPa

6.13 A plate made of titanium alloy Ti-6Al-4V has the dimensionless correction factor Y=1.How large can the largest crack in the material be if it still should be possible toplastically deform the plate in tension?

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

Solution:From Table 6.1 on page 232, the fracture toughness for Ti-6Al-4V varies from 44-66 MPam

1/2.

Also, the yield strength is 910MPa. To plastically deform the material, σnom>Sy, or, from Equation(6.4),

Kci = Yσnom πa; σnom =Kci

Y πa≥ Sy

Solving for a,

Kci

Y πa≥ Sy ; a =

1

πKci

YSy

2

=1

πKci

1( ) 910 MPa( )

2

Since Kci has a value between 44-66MPam1/2

, then a has a range of a≤0.744-1.67mm. Therefore,if the largest crack is below 0.744mm in half-length (1.488mm in length), then the nominal stresswill be larger than the yield strength.

6.14 A plexiglass model of a gear has a 1-mm crack half-length formed in its fillet curve(where the tensile stress is maximum). The model is loaded until the crack starts topropagate. Y=1.5. How much higher a load can a gear made of AISI 4340 steel temperedto 425°C carry with the same crack and the same geometry?

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 and 87.4MPam

1/2, 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 toKci. Therefore, Kci for steel is 87.4 times larger than for plexiglass, so it may be possible tosupport a load 87.4 times larger.

However, it is possible that the steel will plastically deform at a lower stress than thatneeded to propogate the crack. With a=1mm=0.001m, Equation (6.4) predicts the nominal stresslevel as:

Page 6-8

Kci = Yσnom πa; σnom =Kci

Y πa=

87.4MPa m

1.5( ) π 0.001m( )=1040MPa

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

6.15 A pressure container made of aluminum alloy 2024-T351 is manufactured for a safetyfactor of 2 guarding against yielding. The material contains cracks through the wallthickness with a crack half-length less than 3mm. Y=1. Find the safety factor whenconsidering crack propagation.

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

Solution:From Table 6.1 on page 232, Sy=325MPa and Kci=36MPa-m

1/2. The safety factor guarding against

yielding is 2, therefore the nominal stress is one-half the yield strength, or σnom=162.5MPa. Thestress intensity factor is therefore calculated from Equation (6.4) as

Ki = Yσnom πa = 1( ) 162.5MPa( ) π 0.003m( ) =15.78MPa mThe safety factor against crack propagation is therefore

ns = Kci

Ki= 36MPa m

15.78MPa m= 2.28

Since the safety factor for yielding is lower than the safety factor guarding against crackpropagation, the safety factor for the pressure vessel is still 2.

6.16 The clamping screws holding the top lid of a nuclear reactor are made of AISI 4340 steeltempered at 260°C. They are stressed to a maximum level of 1250MPa during apressurization test before starting the reactor. Find the safety factor guarding againstyielding and the safety factor guarding against crack propagation if the initial cracks inthe material have Y=1 and a=1mm. Also, do the calculations for the same material buttempered to 425°C.

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

Solution:I. AISI 4340 Tempered at 260°CFrom Table 6.1 on page 232, Sy=1640MPa and Kci=50MPa-m

1/2. The safety factor against

yielding is therefore

ns =Sy

σ=

1640MPa

1250MPa= 1.31

From Equation (6.4), the stress intensity factor is

Ki = Yσ nom πa = 1( ) 1250MPa( ) π 0.001m( ) = 70.06 MPa mThe safety factor guarding against crack propagation is therefore

Page 6-9

ns = Kci

Ki= 50MPa m

70.06MPa m= 0.714

Since the safety factor is less than 1, the bolts will fail.II. AISI 4340 Tempered at 435°CFrom Table 6.1 on page 232, Sy=1420MPa and Kci=87.4MPa-m

1/2. Using the same equations, the

safety factor against yielding is ns=1.14, and the safety factor against crack propagation isns=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 itpossible to apply 30MPa pressure and still have a safety factor of 2 guarding againstfracture. For a soda-lime glass of the same geometry only 7.5 MPa pressure can beallowed if a safety factor of 2 is to be maintained. Find the size of the cracks the glasstube can tolerate at 7.5MPa pressure and a safety factor of 2. Y=1 for both cases.

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 69MPa. The stresses inthe tube are directly proportional to the pressure, so the fracture strength of the aluminum oxidetube is:

S fa

pa=

Sfs

ps;S fa =

S fs pa

ps= 69MPa( ) 30MPa( )

7.5MPa= 276MPa

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

For a safety factor of 2, the applied stress is 276/2=138MPa. From Table 6.1 on page232, using a low value of fracture toughness for Al2O3, use Kcia=3.0MPa-m

1/2. From Equation

(6.4),

Ki = Kci

2= Yσnom πa ; a = 1

πKci

2Yσnom

2

= 1π

3.0MPa m2 1( ) 138MPa( )

2

= 3.76 ×10−7 m = 37.6µm

For the soda lime glass, the lowest value of fracture toughness is Kci=0.7MPa-m1/2

. The appliedstress is σnom=69MPa/2=34.5MPa. The stress intensity factor is

Ki = Kci

2= Yσnom πa ; a = 1

πKci

2Yσnom

2

= 1π

0.7MPa m2 1( ) 34.5MPa( )

2

= 3.28×10−7m = 32.8µm

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

6.18 A stress optic model used for demonstrating the stress concentrations at the ends of acrack is made of polymethylmethacrylate. An artificially made crack 100mm long isperpendicular to the loading direction. Y=1. Calculate the highest tensile stress that can beapplied to the model without propagating the crack.

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

Page 6-10

Solution:From Table 6.1 on page 232, the critical stress intensity factor for polymethylmethacrylate isKci=1.0MPa-m

1/2. From Equation (6.4) the stress when the crack propagates catastrophically is:

Kci = Yσnom πa; σ nom =Kci

Y πa=

1.0MPa m

1( ) π 0.1m( )= 1.78MPa

From Table A-4 on page 902, the ultimate strength is between 48 and 76 MPa. Therefore, thecrack will 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 cracksmore than 2mm long. The dimensionless geometry correction factor Y is usually 1.5 for asafety factor of 2. What is the appropriate alloy for this application? If Y is increased to4.5, what kind of alloy from Table 6.1 should be used?

Notes: Equation (6.4), combined with material properties from Table 6.1, allow solution of thisproblem.

Solution:To maintain a safety factor of 2, set σnom=Sy/2. Therefore, Equation (6.4) gives:

Kci = Yσnom πa = 1.5( )Sy

2π 0.002m( ) = 0.0594 m( )Sy

From Table 6.1 on page 232, the following data is obtained:Material Kci/Sy (m

1/2)

Aluminum alloy, 2020-T351Aluminum alloy, 7075-T651

Alloy steel 4340, tempered at 260°CAlloy steel 4340, tempered at 425°C

Titanium alloy Ti-6Al-4V

0.1110.05740.03050.06150.0483

Therefore, either Aluminum alloy 2020-T351 or Alloy steel 4340 tempered at 425°C would work.For weight savings, the aluminum alloy is probably the better choice.

If Y=4.5, Equation (6.4) gives:

Kci = Yσnom πa = 4.5( )Sy

2π 0.002m( ) = 0.178 m( )Sy

None of the materials in Table 6.1 would be acceptable for this case.

6.20 The anchoring of the cables carrying a suspension bridge are made of cylindrical AISI1080 steel bars 210mm in diameter. The force transmitted from the cable to the steel baris 3.5 MN. Calculate the safety factor range guarding against yielding.

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

Solution:From the inside front cover, the yield strength for AISI 1080 is Sy=380MPa. The stress in the steelbar is

Page 6-11

σ =P

A=

3.5MNπ4

0.21m( )2= 101MPa

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

0.45Sy ≤ σall ≤ 0.60Sy ; 0.45 Sy

σ≤ ns ≤

0.60Sy

σ;

0.45 380MPa( )101MPa

≤ ns ≤0.60 380MPa( )

101MPaTherefore 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 inpure shear. The rivet is made of AISI 1040 steel and has a circular cross section with adiameter of 25mm. The load on the rivet is 10kN. Calculate the safety factor.

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. FromEquation (3.14),

tall=0.4Sy=0.4(350MPa)=140MPaThe shear stress on the rivet is

τ =P

A=

10kNπ4

0.025m( )2

= 20.37MPa

Therefore, the safety factor is:

ns =τall

τ=

140MPa

20.37MPa= 6.87

6.22 A telescope stands on four feet, each carrying a load of 200kN. The feet are made of AISI1020 steel, and the floor is a thick plate of AISI 1080 steel on a concrete base. Calculatehow large the diameter of the feet needs to be if they are circular and flat. The safetyfactor is 2.5.

Notes: This problem is straightforward. Equation (3.16) gives the allowable stress.

Solution:From the inside front cover, the yield strength of AISI 1020 steel is Sy=295MPa, and for AISI1080 the yield strength is 380MPa. Since the floor is stronger than the feet, the floor doesn’t haveto be analyzed - it will have a higher safety factor than the feet.

From Equation (3.16), the allowable stress is σall=0.9Sy=0.9(295MPa)=265.5MPa. Sincethe safety factor is 2.5, the design stress is σd=σall/ns=(265.5MPa)/2.5=106.2MPa. Therefore, thediameter of the feet is obtained from:

σ =P

A=

4P

πd2 ;d =4P

πσ=

4 200kN( )π 106.2MPa( )

= 0.049m = 49mm

Page 6-12

6.23 A machine element is loaded so that the principal normal stresses at the critical locationfor a biaxial stress state are σ1=20ksi and σ2=-15ksi. The material is ductile with a yieldstrength of 60ksi. Find the safety factor according toa) The maximum shear stress theory (MSST)b) The distortion-energy theory (DET)

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 Equation (6.9). If the stressis 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 threeprinciple stresses are properly referred to as σ1=20ksi, σ2=0 and σ3=-15ksi, since σ1≥σ2≥σ3. Forthe maximum shear stress theory, Equation (6.6) gives the safety factor as:

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

60ksi

20ksi +15ksi( )= 1.714

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

σe =1

2σ2 − σ1( )2 + σ3 − σ1( )2 + σ3 −σ2( )2[ ]1 / 2

=1

2−20ksi( )2 + −35ksi( )2 + 15ksi( )2[ ]1 / 2

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

σe =Sy

ns; ns =

Sy

σe=

60ksi

30.4ksi= 1.97

6.24 A glass bottle is filled with a liquid having a compressibility similar to (~1% lower than)the compressibility of the glass. To what depth within an ocean can the bottle be loweredwithout cracking.

Notes: Equations (6.9) is used to solve the problem.

Solution.If the pressure in the glass is hydrostatic, then σ1=σ2=σ3=-p, and Equation (6.10) gives

σe =12

σ2 −σ1( )2 + σ3 −σ1( )2 + σ3 −σ2( )2[ ]1 / 2=

12

−p + p( )2 + − p + p( )2 + − p + p( )2[ ]1 / 2= 0

Therefore, there is no limit to the depth that the bottle can be lowered (and is also the reasonrocks are not pulverized at the bottom of the ocean or inside the earth’s crust).

6.25 A bolt is tightened, subjecting its shank to a tensile stress of 80ksi and a torsional shearstress of 50ksi at a critical point. All the other stresses are zero. Find the safety factor tatthe critical point by the DET and the MSST. The material is high-carbon steel (AISI1080). Will the bolt fail because of the static loading?

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

Solution:

Page 6-13

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

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=80ksi

2± 50ksi( )2 +

80ksi

2

2

or σ1=104ksi, σ2=-24ksi. 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=104ksi, σ2=0ksi, σ3=-24ksi. From Equation (6.6),

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

55ksi

104ksi − −24ksi( )= 0.43

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

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

= 1

2104ksi − 0( )2 + 104ksi + 24ksi( )2 + 0 + 24ksi( )2[ ]1 / 2

= 118ksi

From Equation (6.11),

σe =Sy

ns; ns =

Sy

σe=

55ksi

118ksi= 0.47

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

6.26 A torque is applied to a piece of chalk used in a classroom until the chalk cracks. Usingthe maximum normal stress theory (MNST) and assuming the tensile strength of thechalk to be small relative to its compressive strength, determine the angle of the crosssection at which the chalk cracks.

Notes: Given the loading condition, the angle of the largest tensile stress is obtained fromEquation (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 largest tensile stress, φσ, isgiven by Equation (2.15) as:

t a n 2φσ =2τ xy

σ x − σy=

2τ xy

0= ∞

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

6.27 A cantilevered bar 500mm long with square cross section has 25mm sides. Threeperpendicular forces are applied to its free end, a 100-N force is applied in the y and z-directions, and a force of 1000N is applied in the x-direction (bar axis). Calculate theequivalent stress at the clamped end of the bar by using the DET when the sides of thesquare cross section are parallel with the y and z directions.

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

Solution:

Page 6-14

The moment of inertia for the cross section is:

I =bh3

12=

a4

12=

0.025m( )4

12= 3.255 × 10−8 m4

The cross sectional area is A=a2=(0.025m)

2=6.25x10

-4m

2. 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=(100N)(0.5m)=50Nm. The axial load is 1000N. Therefore, themaximum stress occurs at a corner of the cross section, and is the sum of the stresses due to thethree loads. Therefore,

σx =M1c

I+

M2c

I+

P

A= 2

50Nm( ) 0.0125m( )3.255 × 10−8m4 +

1000N

6.25 ×10−4 m2 = 40.00MPa

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

σe =1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

=1

240MPa( )2 + 40MPa( )2 + 0( )2[ ]1 / 2

This is evaluated as σe=40MPa.

6.28 A shaft transmitting torque from the gearbox to a rear axle of a truck is unbalanced, sothat a certrifugal load of 500N acts at the middle of the 3m-long shaft. The AISI 1040tubular steel shaft has an outer diameter of 70mm and an inner diameter of 58mm.Simultaneously, the shaft transmits a torque of 6000Nm. Use the DET to determine thesafety factor guarding against yielding.

Notes: The moment must be determined, which then allows for calculation of the bending stress.The torque results in a shear stress; this combined stress state is then transformed to obtain theprincipal stresses. 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=350MPa. The moment ofinertia for the shaft is:

I =π64

do4 − di

4( ) =π64

0.070m( )4 − 0.058m( )4( ) = 6.23 ×10−7 m4

Similarly, J=1.246x10-6m

4. For a simply supported shaft, the maximum moment occurs at the

center of the shaft and has the value M=Pl/4=(500N)(3m)/4=375Nm. Therefore, the bendingstress is obtained from Equation (4.48) as:

σx =Mc

I=

375Nm( ) 0.035m( )6.23 ×10−7m4 = 21.07MPa

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

τxy =Tc

J=

6000Nm( ) 0.035m( )1.246 × 10−6m4 =168.5MPa

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

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=21.07MPa

2± 168.5MPa( )2 +

21.07MPa

2

2

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

Page 6-15

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ3( )2[ ]1 / 2

= 1

2179.4MPa − 0( )2 + 179.4 MPa + 158.3MPa( )2 + 158.3MPa( )2[ ]1 / 2

=292.6 MPa

From Equation (6.11),

σe =Sy

ns; ns =

Sy

σe=

350MPa

292.6MPa= 1.196

6.29 The right-angle-cantilevered bracket used in Problem 5.30, sketch w, has a concentratedforce of 1000N and a torque of 300Nm. Calculate the safety factor. Use the DET andneglect transverse shear. Assume that the bracket is made of AISI 1040 steel and use thefollowing values: a=0.5m, b=0.3m, d=0.035m, E=205GPa, and ν=0.3.

Notes: The stresses must be determined using the approach described in Chapter 4. From thestress state, the principal stresses are determined. Equation (6.6) gives the safety factor for theMaximum Shear Stress Theory, and Equations (6.9) and (6.11) give the safety factor for theDistortion-Energy Theory.

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

I =π64

d4 =π64

0.035m( )4 = 7.366 × 10−8m4

Similarly, J=1.4732x10-7m

4. The maximum stress for the bracket occurs at the wall (x=a). The

loading is a bending moment and a torque. The moment is due to the applied torque T and theload P, and is

M=Pa+T=(1000N)(0.5m)+300Nm=800NmTherefore, the bending stress is obtained from Equation (4.48) as:

σx =Mc

I=

800Nm( ) 0.0175m( )7.366 ×10−8m4 = 190MPa

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

τxy =Tc

J=

300Nm( ) 0.0175m( )1.4732 × 10−7 m4 = 35.6MPa

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

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=190MPa

2± 35.6MPa( )2 +

190 MPa

2

2

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

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

350MPa

196MPa + 6.45MPa=1.73

Equation (6.9) gives the effective stress as

Page 6-16

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

= 1

2196MPa − 0( )2 + 196MPa + 6.45MPa( )2 + 6.45MPa( )2[ ]1/ 2

= 199MPa

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

σe =Sy

ns; ns =

Sy

σe=

350MPa

199MPa=1.76

6.30 A 10-cm diameter shaft is subjected to a 10kNm steady bending moment, an 8kN-msteady torque, and a 150kN axial force. The yield strength of the shaft material is600MPa. Use the MSST and the DET to determine the safety factors for the various typesof loading.

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 thesafety factor for the Maximum Shear Stress Theory, and Equations (6.9) and (6.11) give thesafety factor for the Distortion-Energy Theory.

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

I =π64

d4 =π64

0.10m( )4 = 4.909 × 10−6 m4

Similarly, J=9.817x10-6m

4. The area of the cross section is πd

2/4=0.00785m

2. The bending stress

is obtained from Equation (4.48) as:

σx =Mc

I=

10kNm( ) 0.05m( )4.909 × 10−6 m4 =101.8MPa

The normal stress due to the axial load is

σx =P

A=

150kN

0.00785m2 = 19.10MPa

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

τxy =Tc

J=

8kNm( ) 0.05m( )9.817 ×10−6m4 = 40.7MPa

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

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=120.9MPa

2± 40.7MPa( )2 +

120.9 MPa

2

2

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

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

600MPa

133.3MPa + 12.42MPa= 4.12

Equation (6.9) gives the effective stress as

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

= 1

2133.3MPa − 0( )2 + 133.3MPa + 12.42MPa( )2 + 12.42 MPa( )2[ ]1 / 2

=140MPa

Page 6-17

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

σe =Sy

ns; ns =

Sy

σe=

600MPa

140MPa= 4.28

6.31 Use the MSST and the DET to determine the safety factor for 2024 aluminum alloys foreach of the following stress states:a) σx=10MPa, σy=-60MPab) σx=σy=τxy=-30MPac) σx=-σy=20MPa and τxy=10MPad) σx=2σy=-70MPa, and τxy=40MPa

Notes: This problem does not require determination of the stresses as in Problems 6.28 through6.30, 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) give the safety factor for the Distortion-Energy Theory.

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

For σx=10MPa, σy=-60MPa, note that there are no shear stresses. Therefore, we candirectly write the principal stresses as σ1=10MPa, σ2=0MPa and σ3=-60MPa. Note that theprincipal stresses have been renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factorfor MSST is:

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

325MPa

10MPa + 60MPa= 4.64

Equation (6.9) gives the effective stress as

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

= 1

210MPa − 0( )2 + 10MPa + 60MPa( )2 + 60MPa( )2[ ]1 / 2

=65.57MPa

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

σe =Sy

ns; ns =

Sy

σe=

325MPa

65.57MPa= 4.96

For σx=σy=τxy=-30MPa, Equation (2.16) gives

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=−30MPa − 30MPa

2± 30MPa( )2 + 0( )2

Therefore, σ1=0MPa, σ2=0MPa and σ3=-60MPa. Note that the principal stresses have beenrenumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

325MPa

0MPa + 60MPa= 5.42

Equation (6.9) gives the effective stress as

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

= 1

20 + 60MPa( )2 + 60MPa( )2[ ]1/ 2

= 60MPa

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

Page 6-18

σe =Sy

ns; ns =

Sy

σe=

325MPa

60MPa= 5.42

For σx=-σy=20MPa and τxy=10MPa, Equation (2.16) gives

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=20MPa − 20MPa

2± 10MPa( )2 +

20MPa + 20MPa

2

2

Therefore, σ1=22.36MPa, σ2=0MPa and σ3=-22.36MPa. Note that the principal stresses havebeen renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

325MPa

22.36MPa + 22.36MPa= 7.27

Equation (6.9) gives the effective stress as

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ3( )2[ ]1 / 2

= 1

222.36MPa − 0( )2 + 22.36 MPa + 22.36 MPa( )2 + 22.36MPa( )2[ ]1 / 2

= 38.73MPa

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

σe =Sy

ns; ns =

Sy

σe=

325MPa

38.73MPa= 8.39

For σx=2σy=-70MPa, and τxy=40MPa, Equation (2.16) gives

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=−70MPa − 35MPa

2± 40MPa( )2 +

−70MPa + 35MPa

2

2

Therefore, σ1=0MPa, σ2=-8.84MPa and σ3=-96.16MPa. Note that the principal stresses havebeen renumbered so that σ1≥σ2≥σ3. From Equation (6.6), the safety factor for MSST is:

σ1 − σ3 =Sy

ns; ns =

Sy

σ1 − σ3=

325MPa

0MPa + 96.16MPa= 3.38

Equation (6.9) gives the effective stress as

σe = 1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ 3( )2[ ]1 / 2

= 1

28.84MPa( )2 + 96.16MPa( )2 + −8.84MPa + 96.16MPa( )2[ ]1 / 2

= 92.06MPa

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

σe =Sy

ns; ns =

Sy

σe=

325MPa

92.06MPa= 3.53

6.32 Four different stress elements, each of the same material, are loaded as shown in sketchese,f,g,h. Use the MSST and the DET to determine which element is the most critical.

Page 6-19

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

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

σe =1

2σ1 − σ2( )2 + σ2 − σ3( )2 + σ3 −σ1( )2[ ]1 / 2

= 36MPa

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

σe =1

2σ1 − σ2( )2 + σ2 − σ3( )2 + σ3 −σ1( )2[ ]1 / 2

= 33MPa

(g) σ1=30MPa, σ2=30MPa, σ3=0. Thus, σ1-σ3=30MPa. Also σe=30MPa.(h) σx=30MPa, σy=0, τxy=-10MPa. Therefore, from Equation (2.16), σ1=33MPa, σ2=0, σ3=-3.02MPa. Therefore, σ1-σ3=36.02MPa, σe=31.6 MPa.This shows that the stress state in (e) is the largest.

6.33 The rod shown in sketch i is made of AISI 1040 steeland has two 90° bends. Use the MSST and the DET todetermine the minimum rod diameter for a safety factor of 2 atthe most critical section.

Notes: Recognizing that the critical section is at the wall, thecomponent stresses can be expressed as functions of the roddiameter. Applying MSST or DET gives an expression that canbe solved for d.

Solution:The yield strength for AISI 1040 is obtained from the inside front cover as 350MPa. The criticalsection is at the wall; the rod is slender so transverse shear effects will be ignored. The 8kN loadcauses a torque equal to T1=8kN(0.75m)=6kNm, and bending moment Mx1=8kN(0.3m)=2.4kNm.The 10kN load causes axial normal stress, a bending moment Mz1=10kN(0.75m)=7.5kNm and abending moment Mx2=10kN(0.05m)=500Nm, which is in the opposite direction as Mx1. The 75kN

Page 6-20

load causes torque equal to T2=-75kN(0.05m)=3.75kNm (in the opposite direction as T1) and abending moment Mz2=75kN(0.3m)=22.5kNm, which is in the opposite direction as Mz1.Therefore, the bar sees the following: Mx=Mx1-Mx2=1.9kNm, Mz=22.5kNm-7.5kNm=15kNm, andT=T1-T2=6kNm-3.75kNm=2.25kNm. Therefore, the moment at the wall is

M = Mx2 + Mz

2 = 1.9kNm( )2 + 15kNm( )2 = 15.12kNmThe normal stress is therefore

σ =Mc

I+

P

A=

10kNπ4

d2+

15.12kNm( ) d / 2( )π64

d4=

12.73 kN

d2 +154kNm

d3 =12.73kN( )d + 154kNm

d3

The shear stress is:

τ =Tc

J=

2.25kNm( ) d / 2( )π32

d4=

11.46kNm

d 3

Equation (2.16) gives

σ1 ,σ2 =σ x + σ y

2± τ xy

2 +σ x − σy

2

2

=1

d 3 6.365kN( )d + 77kNm[ ] ± 11.46kNm( )2 + 6.365kN( )d + 77kNm( )2

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

σ1 − σ3 =Sy

ns=

350MPa

2= 175MPa

Substituting for the stresses and solving yields d=0.0964m. Therefore, a 0.10m or 100mmdiameter cross section is a good design designation. For DET, Equations (6.9) and (6.11) give:

σe =1

2σ1 − σ2( )2 + σ1 − σ3( )2 + σ2 −σ3( )2[ ]1 / 2

= 1

2σ1( )2 + σ1 − σ3( )2 + σ3( )2[ ]1 / 2

=Sy

ns= 175MPa

This is solved numerically as d=0.0963m. Therefore, a 0.10m or 100mm diameter cross section isstill acceptable.

6.34 The shaft shown in shetch j is made of AISI 1020 steel. Determine the most criticalsection by using the MSST and the DET. Dimensions of the various diameters shown in sketch jare d=30mm, D=45mm, and d2=40mm.

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

Page 6-21

Solution:First, considering the location of stress concentration 80 mm from the wall:

J =π32

d4 =π32

0.040m( )4 = 7.95 × 10−8 m4

I =J

2= 3.98 × 10−8m4 ,

A=πd2/4=7.07x10

-4m

2.

Also, from statics, V=10kN, M=0.4kNm, N=100kN, T=500Nm. The bottom location is critical,since the bending and tensile stresses are additive at this location. Also, there is no shear stressdue to shear at the extreme location. The stress concentration due to bending is obtained fromFig. 6.5(b) as 1.4, while for tension it is Kc=1.55 from Fig. 6.5(a). The stress concentration fortorsion is Kc=1.2 from Fig. 6.5(c). Therefore, σ1=458MPa and σ2=-28MPa.

For the location 40 mm from the wall, N=100kN, M=0.8kNm T=500Nm,Kc(bending)=1.45, Kc(tension)=1.5, Kc(torsion)=1.2 (all from Fig. 6.6). Therefore, σ1=312MPaand σ2=-7.6.

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