introduction to mechanics-i
TRANSCRIPT
Prof. Ramesh Singh
What is Bulk Deformation?
• Imparting changes in material (dM~0):– Geometry
• Force• Die
– Material properties and Mechanics• Strength• Hardness• Toughness• Yield Criteria• Theory of plasticity
Prof. Ramesh Singh
Part 1- 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• Strengthening mechanisms• Basic formulation of slip line in plane strain
Prof. Ramesh Singh
Part 2- Bulk Deformation
• Process description, videos and first order mathematical analysis– Plane strain forging– Cylindrical forging– Closed die forging– Rolling– Drawing– Tube drawing– Extrusion
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
Prof. Ramesh Singh
#1 - Match
a) Force equilibriumb) Compatibility of
deformationc) Constitutive
equation
1) δT = δ1 + δ2
2) ΣF=0; ΣM=0
3) σ = E e
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, ΣM=0.• 3 Compatibility of deformations
– e.g. δT = δ1 + δ2
• 4 Constitutive equations– e.g. σ = E e
• 5 Solve
F F
A
Prof. Ramesh Singh
Key ConceptsLoad (Force), P, acting over area,
A, gives rise to stress, σ.
Engineering stress: σ = P/Ao
(Ao = original area)
True stress: σt = P/A(A = actual area)
P PA
Prof. Ramesh Singh
Deformation
• Quantified by strain, e or ε
• Engineering strain: e = (lf-li)/li
• True Strain: ε = ln(lf/li)
• Shear strain: γ = a/b
lilf
a
b
Prof. Ramesh Singh
True and Engineering strains
e = (lf-li)/lie = (lf/li) – 1(lf/li) = e + 1ln(lf/li) = ln(e + 1)ε = ln(e + 1)
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)
σ x x and τ x y
Courtesy: http://www.jwave.vt.edu/crcd/kriz/lectures/Anisotropy.html
Direction of action (second subscript)
Prof. Ramesh Singh
Stress Tensor
zzzyzx
yzyyyx
xzxyxx
στττστττσ
333231
232221
131211
σσσσσσσσσ
Mechanics Notation Expanded Tensorial Notation
It can also be written as σij in condensed form and is a second order tensor, where i and j are indicesThe number of components to specify a tensor
• 3n , where n is the order of matrix
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
ττ
ττ
ττ
ττ
=
=
=
∆∆∆=∆∆∆
,
)()(
Prof. Ramesh Singh
Plane Stress
00
===
zxyx
zz
ττσ
Only three components of stress
http://www.shodor.org/~jingersoll/weave4/tutorial/Figures/sc.jpg
Prof. Ramesh Singh
2-D Stresses at an Angle
Prof. Ramesh Singh
Shear Stress on Inclined Plane
Prof. Ramesh Singh
Transformation Equations
Use θ and 90 +θ for x and y directions, respectively
Prof. Ramesh Singh
Principal StressesFor principal stresses τxy=0
= 0
22
22
2
22cos
2
2sin
2
2tan
xyyx
yx
p
xyyx
xyp
yx
xyp
τσσ
σσ
θ
τσσ
τθ
σστ
θ
+
−
−
±=
+
−±=
−=
Prof. Ramesh Singh
Principal Stresses• Substituting values of sin2θp and cos2θp in
transformation equation we get principal stresses
2
2tan
plane, stress principal maximum The
22tan
02cos2sin2
stress,shear maximum of Angle
222
2
2,1
yx
xyp
xy
yx
s
xyxyxy
s
xyyxyx
dd
dd
σστ
θ
τ
σσ
θ
θτϑσσ
ϑϑτ
θ
τσσσσ
σ
−=
−
−=
=
+
−=
′
+
−±
+=
Prof. Ramesh Singh
Maximum Shear Stresses
22
max 2
in of valuengsubstituti
45
9022
2cot2tan
xyyx
xys
ps
ps
ps
τσσ
τ
τθ
θθ
θθ
θθ
+
−=
′
±=
±=
−=
Prof. Ramesh Singh
Equations of Mohr’s Circle
Prof. Ramesh Singh
Mohr’s Circle
Prof. Ramesh Singh
Prof. Ramesh Singh
Prof. Ramesh Singh