Question 1: A steel machine part is statically loaded and has a yield strength of 320 MPa. For each of the following stress states find the factor of safety using each of the three static failure theories. a) σ x = 60 MPa σ y = -30 MPa σ z = -20 MPa τ xy = 40 MPa b) σ x = 70 MPa τ xy = 30 MPa c) Solution: Steel is a ductile material so we will use the ductile static failure theories. First the principal stresses for the given stress state should be calculated. (refer to Tutorial 2 - Question 1) ( ) ( ) σ ⋅ τ − τ − τ − σ σ + + σ σ + σ σ + σ ⋅ σ + σ + σ − σ 2 zx 2 yz 2 xy z y z x y x 2 z y x 3 ( ) 2 xy z 2 zx y 2 yz x zx yz xy z y x 2 τ σ − τ σ − τ σ − τ τ τ + σ σ σ − = 0 3 2 2 1 3 I I I − σ + σ − σ = 0 a) Inserting the known stresses to the given eqn. I 1 = 10 I 2 = -4000 I 3 = 68000 68000 4000 10 2 3 − σ ⋅ − σ ⋅ − σ = 0 Recall the roots of the equation provides the principal stresses. Solving and arranging; σ 1 = 75.21 MPa σ 2 = - 20 MPa σ 3 = - 45.21 MPa Factor of safety for each failure theories : i) Maximum Normal Stress Theory: (Theory states that failure occurs if any of the principal stresses exceeds the yield strength of the material.) n S y max = σ => 21 . 75 320 n = => n = 4.26 ii) Maximum Shear Stress Theory: (Theory states that yielding starts whenever the maximum shear stress at any point becomes equal to the maximum shear stress in a tension test specimen of the same material when that specimen starts yielding) n 2 S y max = τ 2 ) 21 . 45 ( 21 . 75 2 3 1 max − − = σ − σ = τ = 60.21 MPa Static Failure σ MPa ( ) 40 − 30 0 30 60 − 0 0 0 10 − :=
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Question 1: A steel machine part is statically loaded and has a yield strength of 320 MPa. For each of the following stress states find the factor of safety using each of the three static failure theories. a) σx = 60 MPa σy = -30 MPa σz = -20 MPa τxy = 40 MPa b) σx = 70 MPa τxy = 30 MPa c) Solution: Steel is a ductile material so we will use the ductile static failure theories. First the principal stresses for the given stress state should be calculated. (refer to Tutorial 2 - Question 1)
( ) ( ) σ⋅τ−τ−τ−σσ++σσ+σσ+σ⋅σ+σ+σ−σ 2zx
2yz
2xyzyzxyx
2zyx
3
( )2xyz
2zxy
2yzxzxyzxyzyx 2 τσ−τσ−τσ−τττ+σσσ− = 0
32
21
3 III −σ+σ−σ = 0 a) Inserting the known stresses to the given eqn.
I1 = 10 I2 = -4000 I3 = 68000
68000400010 23 −σ⋅−σ⋅−σ = 0 Recall the roots of the equation provides the principal stresses. Solving and arranging; σ1 = 75.21 MPa σ2 = - 20 MPa σ3 = - 45.21 MPa Factor of safety for each failure theories :
i) Maximum Normal Stress Theory: (Theory states that failure occurs if any of the principal stresses exceeds the yield strength of the material.)
n
Symax =σ =>
21.75320n = => n = 4.26
ii) Maximum Shear Stress Theory: (Theory states that yielding starts whenever the maximum shear stress at any point becomes equal to the maximum shear stress in a tension test specimen of the same material when that specimen starts yielding)
n2
Symax =τ
2)21.45(21.75
231
max−−=
σ−σ=τ = 60.21 MPa
Static Failure
σ MPa( )
40−
30
0
30
60−
0
0
0
10−
:=
21.602
320n⋅
= => n = 2.66 (minimum)
iii) Distortion Energy Theory: (Theory states that yielding occurs whenever the distortion energy in a unit volume reaches the distortion energy in the same volume corresponding to the yield strength in tension or compression)
Factor of safety for each failure theories : i) Maximum Normal Stress Theory:
(max. in compression) n
)S(S ycymax
==σ =>
6.81320
−−=n => n = 3.92 (minimum)
ii) Maximum Shear Stress Theory:
n2
Symax =τ = 35.8 MPa
8.352
320⋅
=n => n = 4.47
iii) Distortion Energy Theory:
the von Mises stress 2/12
312
322
21
2)()()(
σ−σ+σ−σ+σ−σ=σ′ =
nSy =>
2/1222
2)106.81()6.814.18()4.1810(
320
+−++−++−=n => n = 4.72
Question 2: A steel LPG tank is shown in the figure. The wall thickness of the tank is 15 mm and has a yield strength of 340 MPa. The full weight of the tank is 6500 kg and the internal pressure is 3 MPa. Calculate the factor of safety of the tank according to the distortion energy theory.
(checking 201
501
75015
rt <== the tank can be treated as thin-walled pressure vessel)
B
4 m
1.5 m15
2)6.81(10
231
max−−−=−= σστ
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Solution: First the principal stresses should be calculated for both cylindrical and spherical sections. For cylindrical vessel: For point A:
tangential stress 157503
tPr
t⋅==σ = 150 MPa radial stress 3. −=−= pArσ MPa
longitudinal stress 1527503
t2Pr
⋅⋅==σ
� = 75 MPa
bending stress due to weight of the tank: (consider weight as a concentrated force which is a conservative assumption compared with the distributed weight assumption)
( ) ( ) 10444i
4o 10929.114701500
64DD
64I ⋅=−π=−π= mm4
4.210929.1
)15750(10766.6310
6
, ≅⋅
−⋅⋅==IMc
Abσ MPa
At the bottom of the tank the tensile stresses will be larger, so the bottom mid-point is critical. Recalling there will be no traverse shear stress due to weight at the bottom fiber, the axial stresses are to be taken as principal stresses. Arranging as 321 σ>σ>σ ;
150t1 =σ=σ MPa 5.77b2 =σ+σ=σ�
MPa 3r3 −=σ=σ MPa After calculating the stress state we can find the factor of safety using the distortion energy
theory: 2/12
312
322
21
2)()()(
σ−σ+σ−σ+σ−σ=σ′ =
nSy
2/1222
2)3150()35.77()5.77150(
340n
++++−=
For point B: tangential stress =tσ 150 MPa radial stre longitudinal stress =
�σ 75 MPa
5.210929.1
75010766.6310
6
, ≅⋅
⋅⋅==IMc
Bbσ Mpa
4000 mm
6500 kg
F F
=> n = 2.56
ss 0. =Brσ
Nmm 10766.632
400031883M
N 318832
81.96500F
6⋅=⋅=
=⋅=
z
x y
F = 1.5 i – k kN
φ30
φ45
R6
200
100
Comment: Since a thin walled cylinder is used, as it can be seen in above equations, bending moments at point A and B can be considered to be equal. Also, pressure in the tank is small which results in small radial stress at point A compared to longitudinal and tangential stresses. Therefore checking safety factor according to stress element at point A is sufficient for this problem. For the spherical cap: on spherical shells stresses in orthogonal directions are same:
1527503
t2Pr
t2l ⋅⋅==σ=σ=σ = 75 MPa
7521 =σ=σ MPa 3pr3 −==σ=σ MPa
2/1222
2)375()375()7575(
340n
++++−= n = 4.36
Factor of safety used for the production of the tank is 2.56 (the smaller of the two factors calculated above). Question 3: A cast iron structure is loaded as shown in the figure. The material has Sut = 325 MPa and Suc = 912 MPa. Find the factor safety of the structure using brittle failure theories at the points A and B (Coulomb-Mohr and Modified Mohr).
Solution: The maximum bending moment on the shoulder is to be calculated using My (in N.m).
For machine elements made of brittle materialsstress concentrations should be considered. Theneck for this case is critical.
Mx = Fz . 100 = 100000 N.mm My = Fx . 200 = 300000 N.mm T = Fx . 100 = 150000 N.mm Fz = 1000 N Fx = 1500 N
T
MyMx
Fz
x
y
A
B
Fx
σA
σA
σy= 0
σB
σy= 0
The transverse shear due to Fx at points A and B is zero.
As stated before, the stress concentrations should be considered on brittle elements. Certain fillets, notches, holes, grooves on the element should be checked as critical sections, as the stress concentrates around these sections.
For 5.13045
dD == and 2.0
306
dr == => Kt.axial = 1.57 (Fig E-1, Norton, pp.994)
You can also use the equations 12.c, 12.d, 12.e in pp. 274 of Norton to obtain the same result.
Question 4. The steel crankshaft is loaded statically as shown in figure. The steady force is counterbalanced by a twisting torque T and by reactions at A and B. The yield strength of the material is 420 MPa. If the factor of safety according to maximum shear stress theory is to be 2.0, what should be the minimum diameter of the crankshaft? (Note: In practice such problems are dealt with dynamic considerations. Here it is taken as a static example.) At point C, there is normal stress in axial directiondue to bending (max. moment). At point D, both axial stress due to bending and shear stress due to torsion exist.
At point C:
M = 1250 . 90 = 1.125 . 105 N.mm 2dc = 4d
64I π=
3
6
4
5
b d10146.1
d64
2d10125.1
IMc ⋅=
π
⋅⋅==σ MPa b1 σ=σ 032 =σ=σ
Recall , for max shear stress theory: 10522
4202
=⋅
==nS y
allτ MPa and τall = τmax
1052
0d
10146.1
2
3
6
31max =
−⋅
=σ−σ
=τ MPa => d3 = 5457 mm3
which gives d = 17.6 mm
FA = FB = 2500 / 2 = 1250 N T = 2500 . 45 = 1.125 . 105 N.mm Sections at points C and D should be checked.
90 90
48
45
BA
F = 2.5 kN
T
FA
C
DFB
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Note: In practice such problems are
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dealt with dynamic considerations. Here it is taken as a static example.)
At point D:
M = 1250 . 48 = 6 . 104 N.mm 2dc = 4d
64I π= J = 2 . I
3
5
4
4
10112.6
64
2106
dd
d
IMc
bending⋅=
⋅⋅== πσ MPa 3
5
4
5
1073.5
32
210125.1
dd
d
JTc
tortion⋅=
⋅⋅== πτ
2xy
2yx
max 2τ+
σ−σ=τ 105
22420
2=
⋅==
nS y
allτ MPa τall = τmax
105d
10494.6d
1073.5d2
10112.63
52
3
52
3
5
max =⋅=
⋅+
⋅=τ MPa => d3 = 6185 mm3
which gives, d = 18.36 mm Checking both points, point D found to be more critical. The minimum diameter of the shaft should be 18.36 mm. But it should be better to get used to accept preferred numbers in machine elements design, so it can be set as d = 20 mm.