me311 machine design - fairfield...
TRANSCRIPT
ME311 Machine Design
W Dornfeld26Sep2019 Fairfield University
School of Engineering
Lecture 4: Stress Concentrations; Static Failure
Stress Concentration
We saw that in a curved beam, the stress was distorted from the uniform distribution of a straight beam, and higher at the inside radius.
A similar distortion happens near holes, notches, and changes in section:
The effect varies with the loading type: tension, bending, or torsion.
Stress Concentration“Stress Concentration Factors” relate the maximum stress to
the average stress in the minimum cross section of the part.
375.00.2
75.0==
bd
psiK
psi
nomc
nom
216096025.2
9603125.0
300
25.0)75.00.2(
300
max =×=×=
==
−
=
σσ
σ
Hamrock Fig. 6.2aHamrock’s plots tell
you the nominal stress.
Which has the bigger Kc?
150 100 150 100
r r
M M
=
=
=
=
cKh
r
rh
H
FigUse _______:.
=
=
=
=
cKh
r
rh
H
FigUse _______:.
Predicting Failure
σ
τ
100
50
σ
τ
60
80-40
We are given some material that is tensile tested and yields at Sy=100ksi.
It has a Mohr’s circle that looks like this:
We want to use this material in an application where it sees this loading:
It has a Mohr’s circle that looks like this:
80
40
y
x
τ=0
The Max Principal stress is less than 100ksi, but will it yield?
Failure Theory
Failure Theory addresses how to translate a real, multi-axial state of stress into something that can be compared with a simple uniaxial (tensile) test result.
For ductile materials, there are two prevailing theories:
1. Max Shear Stress Theory (MSST) or Tresca Theory
2. Distortion Energy Theory (DET) or von Mises criterion.
Max Shear Stress
MSST says "Yield occurs whenever the maximum shear stress in an element exceeds the max shear at yield in a tensile test"
In other words, the Mohr's circle lies within the shear bounds (upper and lower) of the tensile test.
σ
τ
100
50
Max Shear Stress
How did we do?
σ
100
50
τ
60
80-40
Doesn’t look good.
MSST predicts yielding when σ1 - σ3 ≥ Sy. (~Eqn. 6.8)
σ1 is Max principal and σ3 is Min principal.
In our case, σ1 - σ3 = [80 – (-40)] = 120 > Sy of 100.
Another way to say is that τmax ≤ Sy / 2
Max Shear Stress
Here is a representation of MSST in terms of limits to Max and Min principal stresses:
σσσσ 2
σσσσ 1
Syt
Syt
σ1σ2
σ1σ2
σ1σ2
σ1σ2
σ2σ1σ1&σ2
σ2σ1
Inside the “box” is safe.
This is the pure shear line, where
σ1 = -σ2.
Distortion Energy
DET observes that materials can take enormous hydrostatic pressures (e.g., rocks deep in the earth that see pressures well above their compressive strength) and not fracture – suggesting that it must be distortion that causes failure.
DET essentially computes the total strain energy and subtracts the volume change energy to get the distortion energy.
In terms of principal stresses:
where σe is the von Mises equivalent stress.
2
31
2
32
2
21 )()()(2
1σσσσσσσ −+−+−=e
Note that for hydrostatic loading, σ1=σ2=σ3, and σe=0!
Distortion Energy
In the more common case of biaxial principal stresses (σ3=0), this reduces to:
In our example, this is:
This exceeds Sy = 100ksi, so DET also predicts yielding.
ksi
e
8.105200,11
160032006400
)40()40)(80(8022
==
++=
−+−−=σ
2
221
2
1σσσσσ +−=e
Distortion Energy - Simplified
If you have only direct biaxial stress (σz=0), you can
calculate DET directly from σx, σy, and τxy without getting
the principal stresses:
And, if you have only bending and torsion (σy=0), this
reduces to:
(Hamrock buries this in Ex. 7.4 on page 170.)
2223 yxyxyxe τσσσσσ +−+=
223 yxxe τσσ +=
The Failure Theories, overlaid
MSST is a little more conservative than DET.
Brittle Theories
Max Normal Stress Theory
Brittle materials can take more compression than tension, so have different failure bounds.
Internal Friction Theory & Modified Mohr Theory
The Brittle theories use Ultimate Strengths because they fracture rather than yield, and both Compressive and Tensile strengths because they are different.
Remember the Crank Arm ?
1200 in
lb T
orq
ue
1800 in.lbMoment
300 lb Shear
1200 in
lb
To
rqu
e
300 lb Shear
A
BShearPure
psi
psiA
V
psiJ
Tr
psiI
Mc
BENDINGxzTOTAL
BENDING
xz
x
→
=−=
====
====
====
582,13
905325.1
1200
4)75.0(3
)300)(4(
3
4
487,1403106.0
450
32)75.0(
)375.0)(1200(
460,4301553.0
675
64)75.0(
)375.0)(1800(
2
4
4
τττ
π
τ
π
τ
π
σA
B
y
xz
At "A": Bending M = 1800 in.lb.Torsion T = 1200 in.lb.
At “B": Torsion T = 1200 in.lb.Shear V = 300 lb
C
Note that at location “C”, the two shears would ADD to 15,392 psi.
Stresses in a Crank Arm
The stresses at A occur at the same point, so can use Mohr.
psiJ
Tr
psiI
Mc
xz
x
487,14
460,43
==
==
τ
σ
A
y
xz
For Xstress = 43.5, Ystress = 0.0, XYShear = -14.5, Angle = -16.85, Stress1 =
47.85, Stress2 = 0.00, Stress3 =-4.39, ShearMax = 26.12
X,-T
Y,T
-30
-20
-10
0
10
20
30
-10 0 10 20 30 40 50 60
Face Stresses
Sh
ea
r S
tre
ss
The Mohr circle gives:σ1 = 47.846 ksiσ2 = 0.000 ksiσ3 = -4.386 ksiτMax = 26.116 ksi
What are the DET stresses at “A” on the Crank Arm?
psiJ
Tr
psiI
Mc
xz
x
487,14
460,43
==
==
τ
σA
At "A": Bending M = 1800 in.lb.Torsion T = 1200 in.lb.
The Mohr circle for “A” gave:σ1 = 47.846 ksiσ2 = 0.000 ksiσ3 = -4.386 ksiτMax = 26.116 ksi
ksi
yxyxyxe
18.504.25186.6298.1888
)487.14(30046.43322222
==+=
+−+=+−+= τσσσσσ
ksi
e
18.502518192102285
)39.4()39.4)(8.47(8.47222
331
2
1
==++=
−+−−=+−= σσσσσ
Using the Principal stresses:
Using the Direct stresses:
What are the MSST stresses at “A” on the Crank Arm?
psiJ
Tr
psiI
Mc
xz
x
487,14
460,43
==
==
τ
σA
At "A": Bending M = 1800 in.lb.Torsion T = 1200 in.lb.
The Mohr circle for “A” gave:σ1 = 47.846 ksiσ2 = 0.000 ksiσ3 = -4.386 ksiτMax = 26.116 ksi
Using the Principal stresses (our only option with MSST):
ksiMSST 24.52)39.4(85.4731 =−−=−= σσ
This number is slightly higher than the von Mises stress of 50.2 – consistent with the MSST being more conservative.
What are the MSST & DET stresses at “C” on the Crank Arm?
ShearPure
psi
psiA
V
BENDINGxzTOTAL
BENDING
→
=+=
==
392,15
9053
4
τττ
τ
C
At “C": Torsion T = 1200 in.lb.Shear V = 300 lb
“C” sees pure shear, with σx & σy both zero.
σ
τ
15.4
15.4
So σ1 = 15.4 & σ3 = -15.4
ksi
MSST
8.30)4.15(4.1531
=−−=
−= σσ
ksi
e
7.265.711)4.15(3
)4.15()4.15)(4.15(4.15
2
22
2
331
2
1
===
−+−−=
+−= σσσσσ
ksi
yxyxyxe
7.265.711)4.15(3
)4.15(3000
3
2
222
222
===
+−+=
+−+= τσσσσσ
DET using the Principal stresses:
DET using the Direct stresses:
Factors of SafetyThe Factor of Safety, ns, for a component is the ratio of the stress allowed by the component’s material, divided by the maximum stress predicted (or measured). We usually use the yield strength, Sy, as the allowable stress for ductile materials. See Hamrock’s Appendix A for Sy.
maxσ
Syns =
Using the Failure Theories, we will generally compute ns either by:
e
s
Syn
σ
=
For DET (von Mises)max
31
5.0
)(
τ
σσ
Syn
or
Syn
s
s
=
−
=
or
For MSST
Crank Arm Safety Factors
This assumes a material with Sy = 100ksi.Can you find a material with Sy = 100ksi?
C
A
99.12.50
100
91.124.52
100
==
==
ADET
AMSST
n
n
75.37.26
100
25.38.30
100
==
==
CDET
CMSST
n
n
Static Safety Factors
A bar of AISI 1040 Steel sees a bending and twisting load, resulting in a bending stress of 15 ksi and a shear stress of 6.5 ksi at its most highly stressed location.
Find the MSS and the DET factors of safety against yielding.
Calculate both FoS’s using both Direct stresses and Principal stresses.
Selecting Factors of SafetyHamrock gives guidelines for selecting Safety Factors in Section 1.4.1.
),(),,( EDnCBAnn sysxs ×=
Quality of materials,
processing, maintenance, & inspection
Environment (Load) control
Quality of analysis &
testDanger to humans
$$ impact
1.1 ≤ nsx ≤ 3.95; 1.0 ≤ nsy ≤ 1.6; so 1.1 ≤ ns ≤ 6.32