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**Unformatted text preview: **4/15/2020 Assignment Print View Score: 0/100 Points 0 % The ﬁgure shows the free-body diagram of a connecting-link portion having stress concentration at three
sections. The dimensions are r = 0.25 in, d = 0.40 in, h = 0.50 in, w1 = 3.50 in, and w2 = 3.0 in. The forces
F ﬂuctuate between a tension of 1 kip and a compression of 3.2 kip. Neglect column action and ﬁnd the
least factor of safety if the material is cold-drawn AISI 1018 steel. 1/27 4/15/2020 1. Assignment Print View Award: 0 out of 6.66 points What are the factors of safety against fatigue and yielding considering the impact of the fillet?
The factor of safety against fatigue is 0
The factor of safety against yielding is 1 .
. What are the factors of safety against fatigue and yielding considering the impact of the fillet?
The factor of safety against fatigue is
The factor of safety against yielding is 9.266 ± 2% .
25.313 ± 2% . Explanation:
For axial loading at the fillet, the theoretical stress concentration factor Kt is 1.85. This value is
determined as follows:
D
d r
d = = 3.5 mm
3 mm = 1.17 0.25 mm
3.00 mm = 0.083 Neuber’s constant, which is used to calculate the notch sensitivity, is (refer to Figure –A15–5)
‾ = 0.246 - 3.08 (10
√a −3 ) (64 kpsi) + 1.51 (10 −5 2 )(64 kpsi) - 2.67 (10 −8 3 )(64 kpsi) ‾‾
‾ = 0.1037 √in
√a q = 1
1 + a
√
√r 1 q = 0.1037 √in 1 + √ 0.25 in q = 0.82 2/27 4/15/2020 Assignment Print View
Kf = q (Kt - 1) + 1 Kf = 0.82 (1.85 - 1) + 1 Kf = 1.69 σmax = Kf Fmax
w2 h σmax = 1.69 1 kip
(3.0 in)(0.5 in) σmax = 1.127 kpsi σmin = Kf Fmin
w2 h σmin = 1.69 −3.2 kip
(3.0 in)(0.5 in) σmin = -3.635 kpsi The mean and alternating stresses are calculated as follows:
σm = σm = 1
2 1
2 (σmax + σmin ) (1.127 kpsi + (-3.635 kpsi)) σm = -1.254 kpsi ≅ 0 kpsi σa = ∣
∣ 1 σa = ∣
∣ 1 2 2 (σmax - σmin )∣
∣ (1.127 kpsi - (-3.635 kpsi))∣
∣ σa = 2.381 kpsi To determine the factor of safety against yielding, the largest stress, whether it is compressive or
tensile, is compared to the yield strength.
σ = max ∣
(∣ 1 kip
(3.0 in)(0.5 in) ∣
∣
, ∣
∣ -3.2 kip
(3.0 in)(0.5 in) ∣
∣) σ = 2.133 kpsi 3/27 4/15/2020 Assignment Print View
ny = ny = Sy
σ 54 kpsi
2.133 kpsi ny = 25.313 The endurance limit for cold-drawn AISI 1018 is
′
Se = 0.5 (Sut ) ′ Se = 0.5 (64 kpsi) ′ Se = 32 kpsi For a cold-drawn surface finish, the surface modification factor is
−0.217 ka = 2(64 kpsi)
ka = 0.8111 The size modification factor for axial loading is
kb = 1 The loading modification factor for axial loading is
kc = 0.85 Se = 0.8111 × 1 × 0.85 × 32 kpsi = 20.06192 kpsi The factor of safety against fatigue using the modified Goodman criterion is as follows. If the
midrange stress component is compressive, its value is set to 0.0.
1 nf = σa ( Se + ) ( σm ) Sut 1 nf = 2.381 kpsi ( 0 kpsi + 22.06192 kpsi ) ( 64 kpsi ) nf = 9.266 4/27 4/15/2020 2. Assignment Print View Award: 0 out of 6.66 points What is the factor of safety against yielding and fatigue considering the stress concentration around
the hole.
The factor of safety against yielding is 2
The factor of safety against failure is 3 .
. What is the factor of safety against yielding and fatigue considering the stress concentration around
the hole.
The factor of safety against yielding is
The factor of safety against failure is 26.156 ± 2% .
6.813 ± 2% . Explanation:
d
w1 = 0.4 in
3.5 in = 0.11 Kt = 2.68 The value of the notch sensitivity q will not change since the material and mode of loading are the
same. (Refer to Figure A–15–1.)
q = 0.81 Kf = q (Kt - 1) + 1 Kf = 0.81 (2.68 - 1) + 1 Kf = 2.36 The expressions for the maximum and minimum stresses experienced around the hole are as follows:
5/27 4/15/2020 Assignment Print View
σmax = Kf Fmax
h(w - d)
1 1 kip σmax = 2.36 0.5 in(3.5 in - 0.4 in) σmax = 1.542 kpsi σmin = Kf Fmin
h(w1 - d) σmin = 2.36 −3.2 kip
0.5 in(3.5 in - 0.4 in) σmin = -4.934 kpsi The midrange and alternating stresses are
σm = σm = 1
2 1
2 (σmax + σmin ) (1.542 kpsi + -4.934 kpsi) σm = -1.696 kpsi ≅ 0 kpsi σa = ∣
∣ 1 σa = ∣
∣ 1 2 2 (σmax - σmin )∣
∣ (1.542 kpsi - -4.934 kpsi)∣
∣ σa = ∣3.238∣ kpsi The factor of safety against yielding compares the largest stress with the yield strength:
σ = max 1 kpi ( 0.5 in(3.5 in - 0.4 in) , ∣
∣ −3.2 kpi
0.5 in (3.5 in - 0.4 in) ∣
∣) σ = 2.065 kpsi ny = ny = Sy
σ 54 kpsi
2.065 kpsi ny = 26.156 6/27 4/15/2020 Assignment Print View The factor of safety against fatigue using the modified Goodman failure criterion is (the midrange
stress component being set to 0.0 if it is compressive)
1 nf = σa ( Se + ) ( σm ) Sut 1 nf = 3.238 kpsi ( 0 kpsi + 22.06192 kpsi ) ( 64 kpsi ) nf = 6.813 The ﬁgure shows the free-body diagram of a connecting-link portion having stress concentration at three
sections. The dimensions are d = 0.40 in, h = 0.50 in, w1 = 3.50 in, and w2 = 3.0 in.
The forces F ﬂuctuate between a tension of 3 kip and a compression of 9.6 kip. The material is cold-drawn AISI 1018 steel. The following information can be used to determine the ﬁllet radius: The fatigue strength Se for this application is 24.4 kpsi. The notch sensitivity q is assumed to be 0.81. The fatigue stress concentration factor for the hole, Kf, is 2.36. The following information is needed to determine the fatigue stress concentration for the ﬁllet: D
d = 3.5 in
3 in D
d = 1.17 r
d = x
3.00 7/27 4/15/2020 3. Assignment Print View Award: 0 out of 6.66 points What is the factor of safety against fatigue around the hole?
The factor of safety is 4 . What is the factor of safety against fatigue around the hole?
The factor of safety is 2.545 ± 2% . Explanation:
The expressions for the maximum and minimum stresses experienced around the hole are as follows:
σmax = Kf Fmax
h(w - d)
1 σmax = 2.36 3 kip
0.5 in(3.5 in - 0.4 in) σmax = 4.566 kpsi σmin = Kf Fmin
h(w1 - d) σmin = 2.36 −9.6 kip
0.5 in(3.5 in - 0.4 in) σmin = -14.611 kpsi The midrange and alternating stresses are
σm = σm = 1
2 1
2 (σmax + σmin ) (4.566 kpsi + (-14.611 kpsi)) σm = -5.023 kpsi 8/27 4/15/2020 Assignment Print View
σa = 1 (σmax - σmin ) 2 σa = ∣
∣ 1
2 (4.566 kpsi - (-14.611 kpsi))∣
∣ σa = 9.589 kpsi The factor of safety against fatigue using the modified Goodman failure criterion is (the mid range
stress component being set to 0.0 if it is compressive)
nf 1 = hole σa ( nf + ) σm ( Sut ) 1 = 9.589 kpsi hole ( nf Se 24.4 kpsi ) = 2.545 hole 9/27 4/15/2020 4. Assignment Print View Award: 0 out of 6.66 points What are the values of the maximum and minimum stresses at the fillet (not including the adjustment
for Kf)?
The maximum stress value is 5 kpsi. The minimum stress value is 5 kpsi. What are the values of the maximum and minimum stresses at the fillet (not including the adjustment
for Kf)?
The maximum stress value is 2 ± 2% kpsi. The minimum stress value is -6.4 ± 2% kpsi. Explanation:
σmax = σmax = Fmax
w2 h 3 kip
(3.0 in)(0.5 in) σmax = 2 kpsi σmin = σmin = Fmin
w2 h −9.6 kip
(3.0 in)(0.5 in) −0.107 kb = 1.24(de ) σmin = -6.4 kpsi 10/27 4/15/2020 5. Assignment Print View Award: 0 out of 6.66 points What is the value of mean stress and alternating stress?
The value of the mean stress is 3 kpsi. The value of the alternating stress is 3 kpsi. What is the value of mean stress and alternating stress?
The value of the mean stress is -5.192 ± 2% kpsi. The value of the alternating stress is 9.912 ± 2% kpsi. Explanation:
σm = σm = 1
2 1
2 Kf (σmax + σmin ) × 2.36 × (2 kpsi + (-6.4 kpsi)) σm = -5.192 kpsi σa = σa = 1
2 1
2 Kf (σmax - σmin ) × 2.36 × (2 kpsi - (-6.4 kpsi)) σa = 9.912 kpsi 11/27 4/15/2020 6. Assignment Print View Award: 0 out of 6.66 points What is the value of Kf for the fillet if the factor of safety at the fillet and the hole are the same?
The value of Kf is 3 . What is the value of Kf for the fillet if the factor of safety at the fillet and the hole are the same?
The value of Kf is 0.967 ± 2% . Explanation:
nf hole 1 = K ( f σa Se K + )
( f σm Sut ) 1 Kf = 9.912 kpsi 2.545 ( 0 kpsi + 24.4 kpsi 64 kpsi ) Since the mean stress is negative it is considered to be zero.
Kf = 0.967 12/27 4/15/2020 7. Assignment Print View Award: 0 out of 6.66 points Assuming the notch sensitivity factor q to be 0.81, what is the value of the theoretical stress
concentration factor?
The value of the theoretical stress concentration factor is 3 . Assuming the notch sensitivity factor q to be 0.81, what is the value of the theoretical stress
concentration factor?
The value of the theoretical stress concentration factor is 0.96 ± 2% . Explanation:
1 Kf = n σa
fhole ( + Se σm Sut ) 1 q (Kt - 1) + 1 = nf σa hole Kt = ( + Se σm
Sut 1 ( 9.912 kpsi 2.545 ( 24.4 kpsi - 1 0 kpsi + 64 kpsi ) ) ) 1
0.81 + 1 Kt = 0.9597 13/27 4/15/2020 Assignment Print View A part is loaded with a combination of bending, axial, and torsional stresses such that the following
stresses are created at a particular location: Bending: Completely reversed, with a maximum stress of 43 MPa
Axial: Constant stress of 13 MPa
Torsion: Repeated load, varying from 0 MPa to 52 MPa Assume the varying stresses are in phase with each other. The part contains a notch such that Kf,bending
= 1.4, Kf,axial = 1.1, and Kf,torsion = 2.0. The material properties are Sy = 300 MPa and Su = 400 MPa. The
completely adjusted endurance limit is found to be Se = 200 MPa. 14/27 4/15/2020 8. Assignment Print View Award: 0 out of 6.66 points Determine the alternating and midrange stresses for all three types of stresses.
For bending stress, σmean = 0.9
For axial stress, σmean = 1 MPa and σalternating = 0 MPa and σalternating = 11 For torsional stress, τmean = 1 MPa. MPa. MPa and τalternating = 1 MPa. Determine the alternating and midrange stresses for all three types of stresses.
For bending stress, σmean =
For axial stress, σmean = 0 ± 2% MPa and σalternating = 13 ± 2% MPa and σalternating = For torsional stress, τmean = 26 ± 2% MPa and τalternating = 43 ± 2% MPa.
0 ± 2% MPa.
26 ± 2% MPa. Explanation:
For a completely reversed stress, like that of bending in this case,
σm = σm = σmax + σmin
2 43 MPa - 43 MPa
2 σm = 0 MPa σa = σa = σmax - σmin
2 43 MPa + 43 MPa
2 σa = 43 MPa For a constant stress, like that of the axial stress,
15/27 4/15/2020 Assignment Print View
σm = σm = σmax + σmin
2 13 MPa + 13 MPa
2 σm = 13 MPa σmax - σmin σa = 2 13 MPa - 13 MPa σa = 2 σa = 0 MPa For the repeated torsional loading, the alternating and mean shear stresses are
τm = τm = τ max + τ min
2 52 MPa + 0
2 τm = 26 MPa τm = τa = τ max + τ min
2 52 MPa - 0
2 τa = 26 MPa 16/27 4/15/2020 9. Assignment Print View Award: 0 out of 6.66 points What are the values of the mean and alternating von Mises stresses?
The value of the mean stress is n/r MPa. The value of the alternating stress is n/r MPa. What are the values of the mean and alternating von Mises stresses?
The value of the mean stress is 91.195 ± 2% MPa. The value of the alternating stress is 108.333 ± 2% MPa. Explanation:
2
2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
‾
′
σm = + 3(kf × τmean )
√ ((kfb × σbmean ) + (kfa × σamean ))
t 2
2
′
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
σm = √ ‾
((1.4 × 0) + (1.1 × 13 MPa)) + 3(2.0 × 26 MPa)‾ ′ σm = 91.195 MPa 2
2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
‾
′
σa = k
× σb ) + (kf × σa )) + 3(kf × τalt )
√ (( fb
alt
alt
a
t 2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
0
‾
2
′
σa = (1.4 × 43 MPa) + (1.1 × )) + 3(2.0 × 26 MPa) √(
0.85 ′ σa = 108.333 MPa 17/27 4/15/2020 10. Assignment Print View Award: 0 out of 6.66 points What is the factor of safety against fatigue using the Goodman failure criterion?
The factor of safety against fatigue is n/r . What is the factor of safety against fatigue using the Goodman failure criterion?
The factor of safety against fatigue is 1.299 ± 2% . Explanation:
1 nf = ′
σa ( Se + )
( ′
σm
Sut ) 1 nf = 108.333 MPa ( 200 MPa + )
( 91.195 MPa 400 MPa ) nf = 1.299 18/27 4/15/2020 11. Assignment Print View Award: 0 out of 6.66 points What is the factor of safety against yielding?
The factor of safety against yielding is 1 . What is the factor of safety against yielding?
The factor of safety against yielding is 1.504 ± 2% . Explanation:
The maximum stress can be determined as follows:
′ ′ ′ σmax = σm + σa ′ σmax = 91.195 MPa + 108.333 MPa ′ σmax = 199.528 MPa ny = ny = Sy
′ σmax 300 MPa
199.528 MPa ny = 1.504 19/27 4/15/2020 12. Assignment Print View Award: 0 out of 6.66 points Estimate the number of cycles, using the Walker criterion to find the equivalent completely reversed
stress. Be sure to check for yielding.
The number of cycles is 3 . Estimate the number of cycles, using the Walker criterion to find the equivalent completely reversed
stress. Be sure to check for yielding.
The number of cycles is 324709673 ± 2% . Explanation:
Finite life is predicted. To use the Walker criterion for estimating an equivalent completely reversed
stress.
Estimating the material fitting parameter for steels,
γ = −0.00002Sut + 0.8818 = −0.0002 (400 MPa) + 0.8818 = 0.8018 ′ 1−γ ′ σar = (σm + σa ) γ σa σar = (91.1948 MPa + 108.333 MPa) 1−0.8018 (108.333 MPa) 0.8018 σar = 122.2734 MPa Since the value obtained is off the chart, use f = 0.9.
2 a = (f Sut ) b = − b = − N = ( = Se 1
3 1
3 log 200 MPa = 648 f Sut ( log ( Se ) 0.9×400 MPa
200 MPa ) = −0.08509 1 σar
a (0.9(400 MPa)) ) b 1 N = ( 122.273 MPa
648 ) −0.08509 = 324709673 cycles 20/27 4/15/2020 Assignment Print View In the ﬁgure shown, shaft A, made of AISI 1020 hot-rolled steel, is welded to a ﬁxed support and is
subjected to loading by equal and opposite forces F via shaft B. A theoretical stress-concentration factor
Kts of 1.6 is induced in the shaft by the 1/8-in weld ﬁllet. The length of shaft A from the ﬁxed support to
the connection at shaft B is 2 ft. The load F cycles from 525 to 700 lbf. 21/27 4/15/2020 13. Assignment Print View Award: 0 out of 6.66 points Determine the endurance limit for this material as well as the value after correction for surface finish,
sizing, and loading.
The endurance limit for the material is 1 kpsi. The endurance limit for the material after correction for surface finish, sizing, and loading is 2 kpsi. Determine the endurance limit for this material as well as the value after correction for surface finish,
sizing, and loading.
The endurance limit for the material is 27.5 ± 2% kpsi. The endurance limit for the material after correction for surface finish, sizing, and loading is
13.0606 ± 2% kpsi. Explanation:
The endurance limit for the cold-drawn AISI 1040 steel is
′ Se = 0.5Sut ′ Se = 0.5 (55 kpsi) ′ Se = 27.5 kpsi For a hot-rolled surface finish, the surface modification factor is
−0.650 ka = 11(55 kpsi) ka = 0.8131 The equivalent diameter is
de = 0.37 (0.875 in) 22/27 4/15/2020 Assignment Print View
de = 0.324 in The size modification factor is
kb = 0.879(d) −0.107 −0.107 kb = 0.879(0.324 in) kb = 0.99 The loading modification factor for torsion is
kc = 0.59 Sse = ka kb kc (0.5Sut ) Sse = (0.8131) (0.99) (0.59) (0.5 × 55 kpsi) Sse = 13.0606 kpsi 23/27 4/15/2020 14. Assignment Print View Award: 0 out of 6.66 points What are the values of the midrange and alternating shear stresses at the critical section (including
the effects of stress concentration)?
The value of the midrange shear stress is 3
The value of the alternating shear stress is 5 kpsi.
kpsi. What are the values of the midrange and alternating shear stresses at the critical section (including
the effects of stress concentration)?
The value of the midrange shear stress is 13.783 ± 2% kpsi. The value of the alternating shear stress is 1.969 ± 2% kpsi. Explanation:
The theoretical stress concentration factor for torsional loading for this application, Kts, is 1.60.
Neuber’s constant for torsion is
‾ = 0.190 - 2.51 (10
√a −3 ) (55 kpsi) + 1.35 (10 −5 2 )(55 kpsi) - 2.67 (10 −8 3 )(55 kpsi) ‾‾
‾ = 0.08934 √in
√a q = 1
1 + a
√
√r 1 q = 0.08934 √in 1 + √ 0.125 in q = 0.79828 Kf s = q (Kt - 1) + 1 24/27 4/15/2020 Assignment Print View
Kf s = 0.8 (1.6 - 1) + 1 Kf s = 1.48 Tmax = 1400 in·lb Tmin = (525 lb) × 2 in Tmin = 1050 in·lb τmax = Kf s 16T max
π(0.875) τmax = 1.48 3 16 × 1400 in·lb ( π(0.875) 3 )( 1
1000 ) τmax = 15.752 in·lb τmin = Kf s 16T min
π(0.875) τmin = 1.48 3 16 × 1050 in·lb ( π(0.875) 3 )( 1
1000 ) τmin = 11.814 in·lb τm = τm = 1
2 1
2 (τmax + τmin ) (15.752 in·lb + 11.814 in·lb) τm = 13.783 in·lb τa = τa = 1
2 1
2 (τmax - τmin ) (15.752 in·lb - 11.814 in·lb) τa = 1.969 in·lb 25/27 4/15/2020 15. Assignment Print View Award: 0 out of 6.76 points What is the factor of safety against yielding and fatigue at infinite life?
The factor of safety against yielding is n/r . The factor of safety against fatigue using the modified goodman method is 2
The factor of safety against fatigue using the Gerber method is n/r . . What is the factor of safety against yielding and fatigue at infinite life?
The factor of safety against yielding is 0.952 ± 2% . The factor of safety against fatigue using the modified goodman method is
The factor of safety against fatigue using the Gerber method is 1.906 ± 2% . 2.191 ± 2% . Explanation:
Using the maximum shear stress theory to check for yielding,
Sy ny = ( 2 ) τ max 30 kpsi ny = ( ) 2 15.752 kpsi ny = 0.952 The factor of safety against fatigue, obtained with the modified Goodman failure theory (Ssu =
0.67Sut), since this application loading creates only shear stresses is
1 nf = τa ( Sse + ) ( τm
Ssu ) 1 nf = 1.969 kpsi ( 13.783 kpsi + 13.0606 kpsi ) ( 0.67 × 55 kpsi ) 26/27 4/15/2020 Assignment Print View
nf = 1.906 The Gerber fatigue failure theory gives the following factor of safety against fatigue: nf = nf = 1
2 ( τm 2 ) τa
Sse 2
‾‾‾‾‾‾‾‾‾‾‾‾
‾
2τ m Sse
−1 + 1 + √
( Ssu τa ) )
( 2 36.9 kpsi 1
2 Ssu ( 13.783 kpsi ) 1.969 kpsi ( 13.0606 kpsi 2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
‾
2 × 13.783 kpsi × 13.0606 kpsi
−1 + 1 + )(
(
) )
√
36.9 kpsi × 1.969 kpsi nf = 2.191 27/27 ...

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- Fall '08
- STAFF
- Shear Stress, Stress concentration, σm, KF, least factor of safety