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AME 30363 Mechanical Engineering Design Homework 1 · Solutions 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.

r = 3 mm

D = 45 mm

100 N-m 1000 N d = 30 mm

150 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) + Kc2 = (1.9) π + (1.65) = 40.0 MPa π A I 2 (0.03 m) (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 τ = Kc

(100 Nm)(0.015 m) Tc = (1.4) = 26.45 MPa π J (0.030 m)4 32

Equation (2.16) gives the principal stresses. s   σx + σy σx − σy 2 2 σ1 , σ 2 = ± τxy + 2 2 s   40.0 MPa 40.0 MPa 2 2 = ± (26.45 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. 6.12 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. 1

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 2024T351, 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 50.0 MPa p √ = 631 MPa = Y πa (1) π(0.002)

→

σnom =

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

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

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. 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: Sy Sy 60 ksi σ1 − σ3 = = 1.714 → ns = = ns σ1 − σ3 (20 ksi + 15 ksi) The von Mises stress is obtained from Eq. (6.9) as: i1/2 1/2 1 h 1  σe = √ (σ2 − σ1 )2 + (σ3 − σ1 )2 + (σ3 − σ2 )2 = √ (−20 ksi)2 + (−35 ksi)2 + (15 ksi)2 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

6.32 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 2

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 I=

J = 3.98 × 10−8 m4 2

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 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. (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. 7.3 The shaft shown in sketch a rotates at high speed while the imposed loads remain static. The shaft is machined from ground, high-carbon steel (AISI 1080). If the loading is sufficiently large to produce a fatigue failure after 1 million cycles, where would the failure most likely occur? Show all necessary computations and reasoning. P

3 in.

3 in. 2 in.

2 in.

1 1–4 in.

1 in.

x

1 in.

1 in.

y 1

P/2

1

r = –– 16 in.

r = –8 in.

1

r = –8 in.

P/2

Sketch a, for Problem 7.3 Notes: : The failure is expected to be due to bending, since the stresses are completely reversing. One must therefore investigate all likely locations to determine the maximum normal stress, including the effects of stress concentrations. Solution: Note the locations a, b, c, and d in the figure, which are the probable failure locations, either because of stress concentration (a, c, d) or because the loading is largest (b). Upon further inspection, the difference between shoulder a and d is the radius. Since 3

the radius is smaller for the left shoulder than the right one, it will have a higher stress concentration factor, and the right shoulder doesn’t have to be considered. In terms of the analysis to be followed, there are no differences between the locations in the shaft in terms of manufacturing method, size effects, temperature effects, reliability, etc, except that the fatigue stress concentration factor must be calculated. Also, from the inside front cover, AISI 1080 steel has an ultimate strength of 89 ksi. (a) Left Shoulder. Taking a section at point a, the moment is M = (P/2)(1 in.), with the diameter equal to 1 in. The stress concentration is obtained from Fig. 6.5b, using D/d = 1.25 in./1 in. = 1.25 and r/d = (1/16 in.)/1 in. = 0.0625 so that Kc = 1.75. From Fig. 7.9, qn is about 0.75. Therefore, from Eq. (7.19), Kf = 1 + (Kc − 1)qn = 1 + (1.75 − 1)(0.75) = 1.56 The maximum stress at a is: σ = Kf

Mc (0.5P )(1 in.)(0.5 in.) = (1.56) = (7.94 in.−2 )P π I 4 (1 in.) 64

(b) Midpoint of Shaft. At the midpoint, the bending moment is M = (P/2)(3 in.) = (1.5 in.)P . The diameter of the shaft is 1.25 in. at this location. There is no stress concentration here, so the maximum stress is given by Eq. (4.48) as: σ=

Mc [(1.5 in.)P ] (1.25 in./2) = = (7.82 in.−2 )P π I 4 (1.25 in.) 64

(c) Groove. Taking a section at the groove, the moment is M = (P/2)(2 in.) = (1 in.)P . Although the diameter is not specified at this location, it is reasonable to approximate it as 1 in. Kc is obtained from Fig. 6.6b, using D/d = 1.25 in./1 in. = 1.25 and r/d = (1/8 in.)/1 in. = 0.125, so that Kc = 1.7. From Fig. 7.9, qn is around 0.85. Therefore, from Eq. (7.19), Kf = 1 + (Kc − 1)qn = 1 + (1.7 − 1)(0.85) = 1.595 The maximum stress is σ = Kf

Mc [(1 in.)P ] (0.5 in. = (1.595) = (16.2 in.−2 )P π I 4 (1 in.) 64

(d) Summary. Since the stress is largest at the groove, this is the most likely fatigue failure location. 7.4 For each of the AISI 1040 cold-drawn steel bars shown in sketches b and c determine (a) The static tensile load causing fracture, (b) The alternating (completely reversing) axial load ±P that would be just on the verge of producing eventual fatigue failure.

4

r = 2.5 mm

P

25 mm

P

25 mm

25 mm

30 mm 25 mm r = 2.5 mm

P

P (b)

(c)

Sketches b and c, for Problem 7.4 Notes: The discussion on in Sec. 6.2.1 suggests that stress concentrations are relieved by plastic deformation under static loads; certainly this mild steel will relieve the stresses so that they are ignored. The stress concentration is not ignored for the fatigue failure. Also, a surface correction factor kf is calculated from Eq. (7.21). Solution: For AISI 1040 steel, the inside front cover gives Sut = 75 ksi = 520 MPa. Since the material is ductile, stress concentration are ignored (see discussion in Sec. 6.2.1). Therefore, the critical load for either bar is: P = σA = (520 MPa)(0.025 m)2 = 325 kN For a fatigue loading, and the bar without a notch, we note from Eq. (7.7) for axial loading: Se0 = 0.45Su = 0.45(520 MPa) = 234 MPa The bar may be cold drawn, but the notch must be machined; either case gives, from Table 7.3, that e = 4.51 MPa, and f = −0.265. Therefore, from Eq. (7.21), f kf = eSut = 4.51(520)−0.265 = 0.86

The size factor is ks = 1 for axial loading. All other factors are ignored. Therefore, from Equation (7.16), Se = kf ks kr kt km Se0 = (0.86)(1)(234 MPa) = 201 MPa For the notched bar, from Fig. 6.4a for H/h = 30/25 = 1.2 and r/h = 2.5/25 = 0.1, we get Kc = 2.35. From Fig. 7.9 we get for Sut = 520 MPa and r = 2.5 mm that qn = 0.75 (approx). Therefore, Eq. (7.19) gives Kf = 1 + (Kc − 1)qn = 1 + (2.35 − 1)(0.75) = 2.01 For the unnotched bar, there is no stress concentration, so we set the applied stress equal to the endurance limit: σall = Pall /A = Se

→

Pall = Se A = (201 MPa)(0.025 m)2 = 126 kN

For the notched bar, the endurance limit must be reduced by the factor Kf : σall =

P Se = A Kf

→

P =

Se A (201 MPa)(0.025 m)2 = = 62.5 kN Kf 2.01

Note that slightly different answers are acceptable; numbers read off charts can be slightly different, and there can be small discrepancies between using Eq. (7.21) versus Fig. 7.10.

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