Part 1-Mechanics Reviewramesh/courses/ME423/... · 2019-08-06 · Prof. Ramesh Singh Part...
Transcript of Part 1-Mechanics Reviewramesh/courses/ME423/... · 2019-08-06 · Prof. Ramesh Singh Part...
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Prof. Ramesh Singh
Mechanics Review
• Concept of stress and strain, True and engineering
• Stresses in 2D/3D, Mohr’s circle (stress and strain) for 2D/3D
• Elements of Plasticity
• Material Models
• Yielding criteria, Tresca and Von Mises
• Invariants of stress and strain
• Levy-Mises equations
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Prof. Ramesh Singh
Outline
• Stress and strain
– Engineering stresses/strain
– True stresses/strain
• Stress tensors and strain tensors
– Stresses/strains in 3D
– Plane stress and plane strain
• Principal stresses
• Mohr’s circle in 2D/3D
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Prof. Ramesh Singh
#1 - Match
a) Force equilibrium
b) Compatibility of
deformation
c) Constitutive
equation
1) δT = δ1 + δ2
2) SF=0; SM=0
3) s = E e
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Prof. Ramesh Singh
Steps of a Mechanics Problem
• O Read and understand problem
• 1 Free body diagram
• 2 Equilibrium of forces
– e.g. ΣF=0, SM=0.
• 3 Compatibility of deformations
– e.g. δT = δ1 + δ2
• 4 Constitutive equations
– e.g. s = E e
• 5 Solve
F F
A
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Prof. Ramesh Singh
Key Concepts
Load (Force), P, acting over area, A,
gives rise to stress, s.
Engineering stress: s = P/Ao
(Ao = original area)
True stress: st = P/A
(A = actual area)
P PA
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Prof. Ramesh Singh
Deformation
• Quantified by strain, e or e
• Engineering strain: e = (lf-li)/li
• True Strain: e = ln(lf/li)
• Shear strain: g = a/b
lilf
a
b
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Prof. Ramesh Singh
True and Engineering strains
e = (lf-li)/li
e = (lf/li) – 1
(lf/li) = e + 1
ln(lf/li) = ln(e + 1)
e = ln(e + 1)
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Prof. Ramesh Singh
3-D Stress State in Cartesian Plane
There are two subscripts in any stress component:
Direction of normal vector of the plane (first subscript)
s x x and t x y
Courtesy: http://www.jwave.vt.edu/crcd/kriz/lectures/Anisotropy.html
Direction of action (second subscript)
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Prof. Ramesh Singh
Stress Tensor
zzzyzx
yzyyyx
xzxyxx
stt
tst
tts
333231
232221
131211
sss
sss
sss
Mechanics Notation Expanded Tensorial Notation
It can also be written as sij in condensed form and is a second
order tensor, where i and j are indices
The number of components to specify a tensor
• 3n , where n is the order of matrix
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Prof. Ramesh Singh
Symmetry in Shear Stress
• Ideally, there has to be nine components
• For small faces with no change in stresses
• Moment about z-axis,
• Tensor becomes symmetric and have only six components, 3
normal and 3 shear stresses
xzzx
zyyz
yxxy
yxxy
Similarly
yzxxzy
tt
tt
tt
tt
=
=
=
=
,
)()(
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Prof. Ramesh Singh
Plane Stress
Only three components of stress
http://www.shodor.org/~jingersoll/weave4/tutorial/Figures/sc.jpg
𝜎𝑧𝑧 = 0𝜏𝑥𝑧 = 𝜏𝑦𝑧 = 0
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Prof. Ramesh Singh
Plane Strain (1)
One pair of faces has NO strain
– each cross-section has the same
strain
Material in a groove
𝜀𝑧𝑧 = 0𝜏𝑥𝑧 = 𝜏𝑦𝑧 = 0
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Prof. Ramesh Singh
2-D Stresses at an Angle
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Prof. Ramesh Singh
Shear Stress on Inclined Plane
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Prof. Ramesh Singh
Transformation Equations
Use q and 90 +q for x and y directions, respectively
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Prof. Ramesh Singh
Principal StressesFor principal stresses txy=0
= 0
2
2
2
2
2
22cos
2
2sin
2
2tan
xy
yx
yx
p
xy
yx
xy
p
yx
xy
p
tss
ss
q
tss
tq
ss
tq
+
−
−
=
+
−=
−=
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Prof. Ramesh Singh
Principal Stresses• Substituting values of sin2qp and cos2qp in
transformation equation we get principal stresses
2
2tan
plane, stress principal maximum The
22tan
02cos2sin2
stress,shear maximum of Angle
22
2
2
2,1
yx
xy
p
xy
yx
s
xy
xyxy
s
xy
yxyx
d
d
d
d
ss
tq
t
ss
q
qtss
t
q
tssss
s
−=
−
−=
=
+
−=
+
−
+=
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Prof. Ramesh Singh
Maximum Shear Stresses
2
2
max2
in of valuengsubstituti
45
9022
2cot2tan
xy
yx
xys
ps
ps
ps
tss
t
tq
+
−=
=
=
−=
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Prof. Ramesh Singh
Equations of Mohr’s Circle
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Prof. Ramesh Singh
Mohr’s Circle
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Prof. Ramesh Singh
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Prof. Ramesh Singh
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Prof. Ramesh Singh