• Author / Uploaded
• Gerard Mercado

Citation preview

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 modulus of elasticity of 207 GPa, find the stress concentration factor for an identical machine element made of aluminum 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 machine 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 changes from 50 to 100 mm with a radius r = 10 mm. Find how much higher a load can be transmitted through the 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 and r = 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 and r/h = 4 mm/87 mm = 0.046. Therefore, From Figure 6.3(a), Kc is around 2.8. The load that can be transmitted depends on the maximum stress. Therefore, for the first case: Sy = K c

P P1 = 1.8 bh b(0.050 m)

→

P1 =

Sy b(0.050 m) = 0.02778Sy b 1.8

For the second case: Sy = K c

P bh

→

P2 =

Sy b(0.087 m) = 0.03954Sy b 2.2

The ratio of the two forces is: P2 0.03954Sy b = = 1.42 P1 0.02778Sy b 6.3 A flat steel plate 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 to draw more cables, the two holes are replaced with one hole having twice the diameter 2d, as shown in sketch b. Assume that the 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 2d

b

b

d

(a)

(b)

Sketches a and b, for Problem 6.3 Notes: The exact hole locations have not been specified, so some variation may occur with assumed dimensions. 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 the centerline, and that the holes are placed in the center of each half of the plate. The locations where the 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 true for 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 interacting 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 Fig. 6.4(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 5 mm thick 100 mm wide AISI 1020 steel rectangular plate has a central elliptical hole 6 mm in length transverse to the applied stress and 2 mm in diameter along the stress. Determine the applied load 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 be applied. Note that the dimensions in Eq. (6.2) are half-widths, so we need to use b = 3 mm and a = 1 mm. Equation (6.2) gives 2a 2(1) Kc = 1 + =1+ = 1.667 b 3

119

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

∴P =

P bh

(295 MPa)(0.094 m)(0.005 m) Sy bh = = 83, 170 N = 83.17 kN Kc 1.667

6.5 A round bar has a fillet with r/d = 0.15 and D/d = 1.5. The bar transmits both bending moment and torque. A new construction is considered to make the shaft stiffer and stronger by making it equally thick 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 for bending (from Fig. 6.5 (b)) and about 1.25 for torsion (from Fig. 6.5(c)). If instead of a fillet the bar became a groove, with the same root diameter, then the stress concentrations are obtained from Fig. 6.6 (b) and (c). The new stress concentration factors are around 1.65 for bending and 1.325 for torsion. Since the stress concentration factors are higher for the proposed redesign, it is not a good idea. 6.6 A 10-inch wide plate loaded in tension contains a 2 inch long, 1/2 inch wide slot. Estimate the stress concentration 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 and approximating the geometry as a rectangular plate with a groove. Notes: Neither of these is an accurate capture of the stress concentration geometry. Students should be encouraged to conduct a literature search to obtain the stress concentration for this case or to conduct 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 write from Eq. (6.2), 2(0.25) Kc = 1 + = 1.5 1 (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), the area under consideration must be reduced, while it is not reduced in part a), so that the results are actually slightly closer. 6.7 A machine has three circular shafts, each with fillets giving stress concentrations. The ratio of fillet radius to shaft diameter is 0.1 for all three shafts. One of the shafts transmits a tensile force, one transmits a bending torque, and one transmits torsion. Because they are stressed exactly to the stress limit (ns = 1), a design change is proposed doubling the notch radii to get a safety factor greater than 1. 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.

120

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

Solution: For the shaft under tension, Fig 6.5a 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, then r/d = 0.2 and Kc = 1.65. Since the original design was fully stressed, the new safety factor is ns =

2 = 1.21 1.65

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

1.7 = 1.19 1.43

For the shaft under torsion, Fig. 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, then r/d = 0.2 and Kc = 1.22. Since the original design was fully stressed, the new safety factor is ns =

1.43 = 1.17 1.33

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. Ans. σ1 = 52.99 MPa, σ2 = 0, σ3 = -12.27 MPa.

D = 45 mm

r = 3 mm 100 N-m 1000 N

150 mm

d = 30 mm 120 mm

500 N

Sketch c, for Problem 6.8 Notes: This problem can be easily solved through the principal of superposition. The stress concentration 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. Assign the x-axis to the rod axis. The normal stress is given by: σc = Kc1

P Mc 1000 N (500 N)(0.120 m)(0.015 m) + (1.65) = 40.0 MPa + Kc2 = (1.9) π π A I (0.03 m)2 (0.030 m)4 4 64

where the stress concentration factors of 1.9 and 1.65 are obtained from Figure 6.5 (a) and (b). The shear stress is Tc (100 Nm)(0.015 m) τ = Kc = 26.45 MPa = (1.4) π J (0.030 m)4 32

121

Equation (2.16) gives the principal stresses. σ1 , σ 2

s

=

σx + σy ± 2

=

40.0 MPa ± 2



2 τxy

+

σx − σ y 2

2

s



(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 normal stress; 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-m bending moment. The plate is made of AISI 1080 steel and is kept at 20◦ C. A hole is to be punched in the center of the plate. What is the maximum diameter of the hole for a safety factor of 1.5? Ans. d = 120 mm. M

d

5 mm M P

1000 mm

235 mm

Sketch d, for Problem 6.9 Notes: Equation (3.16) gives the allowable stress in bending. The normal stress is the sum of the bending stress and the axial normal stress, and is equated to the allowable stress. This gives an equation in terms of the hole diameter and the stress concentration factors in tension and bending which 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 =

Kca P Kca (150 kN) 6 MN/m = = Kca (b − d)h (0.235 m − d)(0.025 m) 0.235 m − d

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

6Kcb M 6Kcb (300 Nm) 2.88 MN/m = = Kcb (b − d)h2 (0.235 m − d)(0.025 m)2 0.235 m − d

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, this equation can be iteratively solved. Note that we can re-write the equation as: (6 MN/m)Kca + (2.88 MN/m)Kcb = 152 MPa 0.235 m − d

122

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

(6 MN/m)Kca + (2.88 MN/m)Kcb 152 MPa Assume 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 −

d = 0.235 m −

(6 MN/m)(2.2) + (2.88 MN/m)(1.46) = 0.120 m 152 MPa

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) = 0.123 m 152 MPa

This is close to the assumed value, and closer agreement between assumed and calculated values is difficult 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 MPa in one direction. The plate contains a small crack perpendicular to the load direction. At this stress level a safety factor of 2 against crack propagation is obtained. Find how much larger the crack can get 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 = Y σnom πa1 2

→

√ 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 = 22 a1 = 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 in the 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 is derived from Eq. (6.4):  2 √ 1 Kci Kci = Y σnom πa → a = π Y σnom 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, 1 a= π



2 √ 50 MPa m = 6.577 × 10−4 m = 0.6577 mm (1)(1100 MPa)

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, 1 a= π



√ 2 87.4 MPa m = 0.002009 × 10−4 m = 2.009 mm (1)(1100 MPa)

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 if there 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. The nominal 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 =

→

σnom =

Y

Kci 50.0 MPa √ p = = 631 MPa πa (1) π(0.002)

For the Al 2024-T351, √ Kci = Y σnom πa

Kci 36.0 MPa √ p = = 454 MPa Y πa (1) π(0.002)

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 to plastically deform the plate in tension? Ans. 1.488 mm. Notes: To plastically deform the plate, the nominal stress must exceed the yield strength of the material. 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 MPam1/2 . Also, the yield strength is 910 MPa. To plastically deform the material, σnom > Sy , or, from Eq. (6.4), √ Kci ≥ Sy Kci = Y σnom πa → σnom = √ Y πa Solving for a, Kci √ ≥ Sy Y πa

→

a=

1 π



Kci Y Sy

2 =

1 π



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 will be 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 tensile stress is maximum). The model is loaded until the crack starts to propagate. Y = 1.5. How much higher a load can a gear made of AISI 4340 steel tempered to 425◦ C carry with the same crack and the same geometry? Ans. σsteel /σplexiglass = 87.4

124

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

Notes: Equation (6.4) is used to solve the problem with data from Table 6.1. This solution makes sure 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 87.4 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 load 87.4 times larger. However, it is possible that the steel will plastically deform at a lower stress than that needed to propogate the crack. With a = 1 mm = 0.001 m, Eq. (6.4) predicts the nominal stress level as: √

Kci = Y σnom πa

→

σnom

√ Kci 87.4 MPa m p = = 1040 MPa = √ Y πa (1.5) π(0.001 m)

Table 6.1 gives the yield strength for AISI 4340 tempered to 425◦ C as 1420 MPa. Therefore, the nominal 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 vessel made of aluminum alloy 2024-T351 is manufactured for a safety factor of 2.5 guarding against yielding. The material contains cracks through the wall thickness with a crack half-length less than 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 fracture toughness 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 factor guarding against yielding is 2.5, therefore the nominal stress is σnom = 130 MPa. The stress intensity factor is therefore calculated from Equation (6.4) as p √ √ Ki = Y σnom πa = (1)(130 MPa) π(0.003 m) = 12.62 MPa m The safety factor against crack propagation is therefore √ 36 MPa m Kci √ = 2.85 ns = = Ki 12.62 MPa m 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 tempered at 260◦ C. They are stressed to a maximum level of 1250 MPa during a pressurization test before starting the reactor. Find the safety factor guarding against yielding and the safety factor guarding against crack propagation if the initial cracks in the material have Y = 1 and a = 1 mm. Also, do the calculations 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◦ C From Table 6.1 on page 232, Sy = 1640 MPa and Kci = 50 MPa-m1/2 . The safety factor against yielding is therefore Sy 1640 MPa ns = = = 1.31 σ 1250 MPa

125

From Equation (6.4), the stress intensity factor is p √ √ Ki = Y σnom πa = (1)(1250 MPa) π(0.001 m) = 70.06 MPa m The safety factor guarding against crack propagation is therefore √ Kci 50 MPa m √ = 0.714 ns = = Ki 70.06 MPa m 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 = 87.4 MPa-m1/2 . Using the same equations, the safety factor against yielding is ns = 1.14, and the safety 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 to apply 30-MPa pressure and still have a safety factor of 2 guarding against fracture. For a soda-lime glass 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 is proportional 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. The stresses in the tube are directly proportional to the pressure, so the fracture strength of the aluminum oxide tube is: Sf a Sf s = pa ps

→

Sf s =

Sf s pa (69 MPa)(30 MPa) = = 276 MPa ps 7.5 MPa

Note that this is on the low end of the fracture strength values given in Table A.3 for aluminum oxide (Al2 O3 ). For a safety factor of 2, the applied stress is 276/2=138 MPa. From Table 6.1 on page 232, using a low value of fracture toughness for Al2 O3 , use Kcia = 3.0 MPa-m1/2 . From Eq. (6.4), √ Kci = Y σnom πa 2  2  2 √ 1 Kci 1 3.0 MPa m ∴a= = = 3.76 × 10−5 m = 37.6 µm π 2Y σnom π 2(1)(138 MPa) Ki =

For the soda lime glass, the lowest value of fracture toughness is Kci = 0.7 MPa-m1/2 . The applied stress is σnom = 69 MPa/2 = 34.5 MPa. The stress intensity factor equation is √ Kci = Y σnom πa 2  2  2 √ 1 Kci 1 0.7 MPa m ∴a= = = 3.28 × 10−5 m = 32.8 µm π 2Y σnom π 2(1)(34.5 MPa) Ki =

Therefore, the largest crack in the aluminum oxide must be less than 2a or 75.2 µm, while for soda lime 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 made of polymethylmethacrylate. An artificially made crack 100 mm long is perpendicular to the loading

126

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

direction. Y = 1. Calculate the highest tensile stress that can be applied to the model without propagating 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 solve this problem. Solution: From Table 6.1 on page 232, the critical stress intensity factor for polymethylmethacrylate is Kci = 1.0 MPa-m1/2 . From Equation (6.4) the stress when the crack propagates catastrophically is: √

Kci = Y σnom πa

→

σnom

√ Kci 1.0 MPa m p = √ = = 1.78 MPa Y πa (1) π(0.1 m)

From Table A-4 on page 902, the ultimate strength is between 48 and 76 MPa. Therefore, the crack 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 cracks more than 2 mm 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 Alloy steel 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 to obtain 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 the safety factor is 2, we assign σnom = Sy /2. Therefore, Eq. (6.4) yields Kci = Y

Sy √ πa 2

→

Kci Y√ 1p = πa = 2.15 π(0.001 m) = 0.060 Sy 2 2

Note from Table 6.1: Material Al 2024-T351 Al 2024-T651 Steel 4340 Tempered at 260◦ C Steel 4340 Tempered at 425◦ C Ti-6Al-4V

Sy 325 505 1640 1420 910

Kci 36 29 50 87.4 44-66

Kci /Sy 0.110 0.057 0.0305 0.0615 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 Y√ 1p πa = 3.5 π(0.001 m) = 0.098 = Sy 2 2 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 bars 210 mm in diameter. The force transmitted from the cable to the steel bar is 3.5 MN. Calculate the safety 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 range for allowable stresses in tension. Solution: From the inside front cover, the yield strength for AISI 1080 is Sy = 380 MPa. The stress in the steel bar is 3.5 MN P = π σ= = 101 MPa A (0.21 m)2 4 Since the safety factor is ns = σall /σ, Eq. (3.13) gives 0.45Sy ≤ σall ≤ 0.60Sy 0.45Sy 0.60Sy ≤ ns ≤ σ σ 0.60(380 MPa) 0.45(380 MPa) ≤ ns ≤ 101 MPa 101 MPa Therefore 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. The load 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 = 350 MPa. From Eq. (3.14), τall = 0.4Sy = 0.4(350 MPa) = 140 MPa The shear stress on the rivet is τ=

20 kN P  = 40.74 MPa = π A (0.025 m)2 4

Therefore, the safety factor is: ns =

τall 140 MPa = = 3.44 τ 40.74 MPa

Section 6.7 6.22 A machine element is loaded so that the principal normal stresses at the critical location for a biaxial stress 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 the safety 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.

128

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

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, since σ1 ≥ σ2 ≥ σ3 . For the maximum shear stress theory, Eq. (6.6) gives the safety factor as: σ1 − σ3 =

Sy ns

→

ns =

Sy 60 ksi = = 1.714 σ 1 − σ3 (20 ksi + 15 ksi)

The von Mises stress is obtained from Eq. (6.9) as: i1/2 1/2 1 h 1  2 2 2 = √ (−20 ksi)2 + (−35 ksi)2 + (15 ksi)2 σe = √ (σ2 − σ1 ) + (σ3 − σ1 ) + (σ3 − σ2 ) 2 2 or σe = 30.4 ksi. Therefore, the safety factor is, from Eq. (6.11), σe =

Sy ns

→

ns =

Sy 60 ksi = = 1.97 σe 30.4 ksi

6.23 A bolt is tightened, subjecting its shank to a tensile stress of 80 ksi and a torsional shear stress of 50 ksi at a critical point. All of the other stresses are zero. Find the safety factor at the critical point by the DET and the MSST. The material is high-carbon steel (AISI 1080). Will the bolt fail because of the 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 are arbitrary; let’s refer to the tensile stress as σx = 80 ksi and the shear stress as τxy = 50 ksi. Since all other stresses are zero, Eq. (2.16) gives the principal stresses as s s   2 2 σ − σ 80 ksi 80 ksi σx + σy x y 2 + σ1 , σ 2 = ± τxy = ± (50 ksi)2 + 2 2 2 2 or σ1 = 104 ksi, σ2 = −24 ksi. Note that the other stresses are zero, so the principal stress out of the plane 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), σ 1 − σ3 =

Sy ns

→

ns =

Sy 55 ksi = = 0.43 σ1 − σ 3 104 ksi − (−24 ksi)

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

= = =

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (104 ksi − 0)2 + (104 ksi + 24 ksi)2 + (0 + 24 ksi)2 2 118 ksi

From Eq. (6.11), σe =

Sy ns

→

ns =

Sy 55 ksi = = 0.47 σe 118 ksi

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 maximumnormal-stress theory (MNST) and assuming the tensile strength of the chalk to be small relative to its compressive 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 largest tensile stress, φσ , is given by Eq. (2.15) as: tan 2φσ =

2τxy 2τxy =∞ = σx − σy 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 forces are applied to its free end, a 1000 N force is applied in the x direction, a 100 N force is applied in the y direction, and an equivalent force of 100 N is applied in the z direction. Calculate the equivalent stress at the clamped end of the bar by using the DET when the sides of the square 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 bending stresses and the axial stress. The effective stress is obtained from Eq. (6.9). Solution: The moment of inertia for the cross section is: I=

a4 (0.025 m)4 bh3 = = = 3.255 × 10−8 m4 12 12 12

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 perpendicular moments are M1 = M2 = F l = (100 N)(0.5 m) = 50 Nm. The axial load is 1000 N. Therefore, the maximum stress occurs at a corner of the cross section, and is the sum of the stresses due to the three loads. Therefore, σx =

M1 c M2 c P (50 Nm)(0.0125 m) 1000 N + + =2 + = 40.00 MPa I I A 3.255 × 10−8 m4 6.25 × 10−4 m2

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 1/2 1/2 1  1  σe = √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 = √ (40 MPa)2 + (40 MPa)2 + (0)2 2 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 a centrifugal load of 500 N acts at the middle of the 3-m-long shaft. The AISI 1040 tubular steel shaft has an outer diameter of 70 mm and an inner diameter of 58 mm. Simultaneously, the shaft transmits a 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. The torque results in a shear stress; this combined stress state is then transformed to obtain the principal 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 = 350 MPa. The moment of inertia for the shaft is:
 π 4 π  do − d4i = (0.070 m)4 − (0.058 m)4 = 6.23 × 10−7 m4 I= 64 64

130

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

Similarly, J = 1.246 × 10−6 m4 . For a simply supported shaft, the maximum moment occurs at the center of the shaft and has the value M = P l/4 = (500 N)(3 m)/4 = 375 Nm. Therefore, the bending stress is obtained from Eq. (4.48) as: (375 Nm)(0.035 m) Mc = = 21.07 MPa I 6.23 × 10−7 m4

σx =

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

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

Also, σy = σz = τzx = τyz = 0. Therefore, from Eq. (2.16), s s  2  2 σx + σ y σ − σ 21.07 MPa 21.07 MPa x y 2 2 σ1 , σ 2 = ± τxy + = ± (168.5 MPa) + 2 2 2 2 Therefore, σ1 = 179.4 MPa, σ2 = 0, and σ3 = −158.3 MPa. Note that the principal stresses have been renumbered so that σ1 ≥ σ2 ≥ σ3 . Equation (6.9) gives the effective stress as σe

= = =

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (179.4 MPa − 0)2 + (179.4 MPa + 158.3 MPa)2 + (158.3 MPa)2 2 292.6 MPa

From Equation (6.11), σe =

Sy ns

→

ns =

Sy 350 MPa = = 1.196 σe 292.6 MPa

6.27 The right-angle-cantilevered bracket used in Problem 5.30, sketch w, has a concentrated force of 1000 N 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 = 0.3 m, 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 stress state, the principal stresses are determined. Equation (6.6) gives the safety factor for the Maximum Shear 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 of the bracket cross section is: π 4 π 4 I= d = (0.035 m) = 7.366 × 10−8 m4 64 64 Similarly, J = 1.4732 × 10−7 m4 . 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 the load P , and is M = P a + T = (1000 N)(0.5 m) + 300 Nm = 800 Nm Therefore, the bending stress is obtained from Eq. (4.48) as: σx =

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

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

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

131

Also, σy = σz = τzx = τyz = 0. Therefore, from Eq. (2.16), s s  2  2 σx + σy σ − σ 190 MPa 190 MPa x y 2 2 σ1 , σ 2 = ± τxy + = ± (35.6 MPa) + 2 2 2 2 Therefore, σ1 = 196 MPa, σ2 = 0, and σ3 = −6.45 MPa. Note that the principal stresses have been renumbered so that σ1 ≥ σ2 ≥ σ3 . From Eq. (6.6), the safety factor for MSST is: σ1 − σ3 =

Sy ns

→

ns =

Sy 350 MPa = = 1.73 σ 1 − σ3 196 MPa + 6.45 MPa

Equation (6.9) gives the effective stress as σe

= = =

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (196 MPa − 0)2 + (196 MPa + 6.45 MPa)2 + (6.45 MPa)2 2 199 MPa

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

Sy ns

→

ns =

Sy 350 MPa = 1.76 = σe 199 MPa

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 the DET 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 be included. From the stress state, the principal stresses are determined. Equation (6.6) gives the safety factor for the Maximum Shear Stress Theory, and Eqs. (6.9) and (6.11) give the safety factor for the Distortion-Energy Theory. Solution: The moment of inertia of the shaft cross section is: π 4 π I= d = (0.10 m)4 = 4.909 × 10−6 m4 64 64 Similarly, J = 9.817 × 10−6 m4 . The area of the cross section is πd2 /4 = 0.00785 m2 . The bending stress is obtained from Eq. (4.48) as: σx =

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

The normal stress due to the axial load is σx =

P 150 kN = = 19.10 MPa A 0.00785 m2

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

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

Also, σy = σz = τzx = τyz = 0. Therefore, from Eq. (2.16), s s  2  2 120.9 MPa σ x + σy σ − σ 120.9 MPa x y 2 + σ1 , σ 2 = = ± (40.7 MPa)2 + ± τxy 2 2 2 2

132

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

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

Sy ns

→

ns =

Sy 600 MPa = = 4.12 σ 1 − σ3 133.3 MPa + 12.42 MPa

Equation (6.9) gives the effective stress as σe

= = =

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (133.3 MPa − 0)2 + (133.3 MPa + 12.42 MPa)2 + (12.42 MPa)2 2 140 MPa

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

Sy ns

→

ns =

Sy 600 MPa = = 4.28 σe 140 MPa

6.29 Use the MSST and the DET to determine the safety factor for 2024 aluminum alloys for each of the following 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) give the 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 can directly write the principal stresses as σ1 = 10 MPa, σ2 = 0 MPa and σ3 = −60 MPa. Note that the principal stresses have been renumbered so that σ1 ≥ σ2 ≥ σ3 . From Eq. (6.6), the safety factor for MSST is: σ 1 − σ3 =

Sy ns

→

ns =

Sy 325 MPa = = 4.64 σ 1 − σ3 10 MPa + 60 MPa

Equation (6.9) gives the effective stress as σe

= = =

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (10 MPa − 0)2 + (10 MPa + 60 MPa)2 + (60 MPa)2 2 65.57 MPa

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

Sy ns

→

ns =

Sy 325 MPa = = 4.96 σe 65.57 MPa

133

(b) For σx = σy = τxy = −30 MPa, Eq. (2.16) gives s  2 q σx + σy σx − σy −30 MPa − 30 MPa 2 2 ± (30 MPa)2 + (0) σ1 , σ 2 = ± τxy + = 2 2 2 Therefore, σ1 = 0 MPa, σ2 = 0 MPa and σ3 = −60 MPa. Note that the principal stresses have been renumbered so that σ1 ≥ σ2 ≥ σ3 . From Eq. (6.6), the safety factor for MSST is: σ1 − σ3 =

Sy ns

→

ns =

Sy 325 MPa = = 5.42 σ1 − σ 3 0 MPa + 60 MPa

Equation (6.9) gives the effective stress as σe

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (0 MPa)2 + (60 MPa)2 + (60 MPa)2 2 60 MPa

= = =

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

Sy ns

→

ns =

Sy 325 MPa = = 5.42 σe 60 MPa

(c) For σx = −σy = 20 MPa and τxy = 10 MPa, Eq. (2.16) gives s  2 σx + σy σx − σy 2 + σ1 , σ 2 = ± τxy 2 2 q 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 stresses have been renumbered so that σ1 ≥ σ2 ≥ σ3 . From Eq. (6.6), the safety factor for MSST is: σ1 − σ3 =

Sy ns

→

ns =

Sy 325 MPa = = 7.27 σ 1 − σ3 22.36 MPa + 22.36 MPa

Equation (6.9) gives the effective stress as σe

= = =

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  √ (22.36 MPa − 0)2 + (22.36 MPa + 22.36 MPa)2 + (22.36 MPa)2 2 38.73 MPa

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

Sy ns

→

ns =

Sy 325 MPa = = 8.39 σe 38.73 MPa

(d) For σx = 2σy = −70 MPa, and τxy = 40 MPa, Eq. (2.16) gives s  2 σ x + σy σx − σy 2 σ1 , σ 2 = ± τxy + 2 2 q −70 MPa − 35 MPa 2 = ± (40 MPa)2 + (−70 MPa + 35 MPa) 2

134

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

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

Sy ns

→

ns =

Sy 325 MPa = = 3.38 σ1 − σ 3 0 MPa + 96.16 MPa

Equation (6.9) gives the effective stress as σe

1/2 1  √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 2 1/2 1  = √ (8.84 MPa)2 + (96.16 MPa)2 + (−8.84 MPa + 96.16 MPa)2 2 = 92.06 MPa =

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

Sy ns

→

ns =

Sy 325 MPa = = 3.53 σe 92.06 MPa

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 e is most critical. 21 MPa

7.5 MPa

21 MPa

(e)

28.5 MPa

(f) 10 MPa 30 MPa

30 MPa

30 MPa (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, from Eq. 6.9, 1/2 1  = 36 MPa σe = √ (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 2 (f) σ1 = 28.5 MPa, σ2 = 0, σ3 = −7.5 MPa. Thus, σ1 − σ3 = 36 MPa. Also, 1/2 1  σe = √ (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 33 MPa 2

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 the DET to determine the minimum rod diameter for a safety factor of 2 at the most critical section. z 300 mm x

y z′

750 mm

50 mm 100 N

1500 N x′

y′ 800 N

Sketches i, for Problem 6.31 Notes: Recognizing that the critical section is at the wall, the component stresses can be expressed as functions of the rod diameter. Applying MSST or DET gives an expression that can be solved for d. Solution: The yield strength for AISI 1040 is obtained from the inside front cover as 350 MPa. The critical section is at the wall; the rod is slender so transverse shear effects will be ignored. The 800 N load 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)=75 Nm 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 as T1 ) 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=

p p Mx2 + Mz2 = (235 Nm)2 + (375 Nm)2 = 442.5 Nm

The normal stress is therefore σ=

Mc P 100 N (442.5 Nm)(d/2) 127.3 N 4507 Nm (127.3 N)d + 4507 Nm + = + = π + = π 2 3 I A d d d3 d2 d4 4 64

The shear stress is: τ=

Tc (525 Nm)(d/2) 2674 Nm = = π 4 J d3 d 32

136

CHAPTER 6. FAILURE PREDICTION FOR STATIC LOADING

Equation (2.16) gives σ1 , σ 2

= =

s  2 σx + σy σx − σ y 2 ± τxy + 2 2   q 1 2 2 [(63.65 N)d + 2253.5 Nm] ± (2674 Nm) + [(63.65 N)d + 2253.5 Nm] d3

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: σ 1 − σ3 =

Sy 350 MPa = = 175 MPa ns 2

Substituting for the stresses and solving yields d = 0.0341 m. Therefore, a 35 mm diameter cross-section is a good design designation. For DET, Equations (6.9) and (6.11) give: 1/2 1/2 1  1  σe = √ (σ1 − σ2 )2 + (σ1 − σ3 )2 + (σ2 − σ3 )2 = √ (σ1 )2 + (σ1 − σ3 )2 + (σ3 )2 = 175 MPa 2 2 This is solved numerically as d = 0.0287 m. Therefore, a 35 mm diameter cross section is still acceptable. 6.32 The shaft shown in sketch j is made of AISI 1020 steel. Determine the most critical section by using the MSST and the DET. Dimensions of the various diameters shown in sketch j are d = 30 mm, D = 45 mm, and d2 = 40 mm. r = 9 mm d2

D

r = 6 mm d 1 kN 10 kN

40 mm

40 mm

T = 50 N-m 40 mm

Sketches j, for Problem 6.32 Notes: This problem requires the incorporation of stress concentration effects into the component stresses before determining the principal stresses. Solution: (a) Fillet. First, considering the location of stress concentration 80 mm from the wall: J=

π 4 π d = (0.040 m)4 = 7.95 × 10−8 m4 32 32 J = 3.98 × 10−8 m4 2 A = πd2 /4 = 7.07 × 10−4 m2 I=

Also, from statics, V = 1 kN, M = 40 Nm, N = 10 kN, T = 50 Nm. The bottom location is critical, since the bending and tensile stresses are additive at this location. Also, there is no shear stress due to shear at the extreme location. The stress concentration due to bending is obtained from Fig. 6.5(b) as 1.4, while for tension it is Kc = 1.55 from Fig. 6.5(a). The stress concentration for torsion is Kc = 1.2 from Fig. 6.5(c). Therefore, σ1 = 458 MPa and σ2 = −28 MPa.

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.