Solid Mechanics Dr.NGUYEN T Hoang 1 1 SOLID MECHANICS SOLID MECHANICS Dr. NGUYEN The Hoang Dr. NGUYEN The Hoang Email: Email: [email protected][email protected]2 WHO AM I ? Educational and professional background: Educational and professional background: • PhD in Solid Mechanics and Material Science, University of Poitiers/ENSMA (Ecole Nationale Superieure de Mecanique et d’Aerotechnique), France, 2005 • Master in Engineering Mechanics, LMPM/University of Poitiers/ENSMA, France, 2002 • Bachelor of Engineering in Materials and Structures for Aeronautics and Transportation Vehicles, ENSMA, France, 2002 • Bachelor in Aeronautical Engineering, First Class Honours, HoChiMinh City University of Technology, Vietnam, 2001 • Deputy Head of Department of Aeronautical Engineering, HoChiMinh City University of Technology, Vietnam, 2005-2007
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Stress distribution near a hole in a plateloaded in tension.
Geometric irregularities (changes
in cross-sections) are a must in
most of machine components:
Shoulders for bearings, Key slots
for mounting gears and pulleys,
threads, and splines. Any change
in cross-section alters the stress
distribution and increases the
stress.
Discontinuities are called stress raisersand areas where they occur are called stress concentration
Crack Tip StressCrack Tip Stress
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1. Basic elasticity1.1 Stress
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1. Basic elasticity1.1 Stress
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1. Basic elasticity1.1 Stress
Stress matrixStress matrix
[ ]xx xy xz
yx yy yz
zx zy zz
σ τ τ
σ = τ σ τ τ τ σ
Or: in the form of vectorOr: in the form of vector
[ ]
xx
yy
zz
xy
yz
xz
σ σ σ
σ = τ
τ
τ
Direct
stresses
Shear
stresses
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1. Basic elasticity1.1 Stress
Average normal
stress
Compression
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1. Basic elasticity1.1 Stress
Average normal stressTraction / tension
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1.1.1. 1.1.1. Direct / Normal Stress
Direct Stress = Applied Force (P)
Cross Sectional Area (A)
� Units (SI): N/m2 (Pa), kPa, MPa, GPa
� US units: Force (P) in pounds (lb) or kilopounds (kip); Cross section (A) in square inches (in2) � Stress: pounds per square inch (psi) or kilopound per square inch (ksi)
� http://www.convertunits.com/SI-units.php
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1.1.2. Shear Stress
� Shear stresses are produced by equal
and opposite parallel forces not in line.
� The forces tend to make one part of the material slide over the other part.
� Shear stress is tangential to the area
over which it acts.
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1.1.2 Shear StressExample: A BEAM
Suppose: uniform / section
shear load:
shear stress:
Equilibrium
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1.1.2. Shear StressEx: Some Engineering Applications
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1.1.2. Shear StressEx: Engineering Application
F: shear force
(internal force)
P = P’
Shear stress
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Example 1Example 1Problem:
• 2 members are joined by a glue at
angle θ� Which stresses generated on inclined interface plane?� Calculate the stresses?
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1.1.3. Safety factor1.1.3. Safety factor
A: initial cross areaUltimate tensile strength
Working
load/design
load
Ultimate normal load
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Example 1 (cont.)Example 1 (cont.) Problem:
• 2 members are joined by a glue at
angle θ• Ultimate stresses:� Determine range of angles θ, if
safety factors: shear stress =4.27,
normal stress=5.28
U
U
22 MPa
11MPa
σ =
τ =
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ExampleExample Problem:
• 2 members are joined by a glue at
angle θ• Ultimate stresses:� Determine range of angles θ, if
safety factors: shear stress =4.27,
normal stress=5.28
U
U
22 MPa
11MPa
σ =
τ =
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Example 2
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1.1. Basic elasticityBasic elasticity
1.2. Equilibrium 1.2. Equilibrium
Forces applied:
+ Surface forces
+ Body forces
(gravitational,
inertial �per
unit of volume �
X, Y, Z )
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1.1. Basic elasticityBasic elasticity
1.2. Equilibrium 1.2. Equilibrium
Taking moments about an axis
through the centre of the element parallel to the zaxis
(Forces) equlibrium of the
element in directions x,y,z:
The equations of equilibrium must be
satisfied at all interior points in a
deformable body under a 3D force system.
Eqs. (1)
Eqs. (2)
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EqsEqs (1) demonstration(1) demonstration
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EqsEqs (1) demonstration(1) demonstration
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EqsEqs (2) demonstration(2) demonstration
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EqsEqs (2) demonstration(2) demonstration
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1. Basic elasticity1.2. Equilibrium: Plane stress
structural components are
fabricated from thin material sheet
Plane stress (2D case)
Equilibrium (2D plane stress)
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1.1. Basic elasticityBasic elasticity
1.3. Boundary Conditions1.3. Boundary Conditions
Equilibrium (3D) ���� ONLY 3 equations for 6 unknowns of stresses �
“Statically Indeterminate problems”
� we NEED: BOUNDARY CONDITIONS
Cosines:
l = dy/ds
m = dx/ds
3D
2D
Summations of forces in the
directions X,Y give:
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Demonstration:Demonstration:
2D
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1.1. Basic elasticityBasic elasticity
1.4. Principal Stresses1.4. Principal Stresses
Plane stress (2D)
forces are ignored
We want to
find stresses on plane (AB)..
..Element (ECD) is in
equilibrium
Direct stress
Shear stress
(1.8)
(1.9)
Both vary withθθθθ
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DemonstrationDemonstrationPlane stress (2D)
..Element (ECD) is in equilibrium
Direct stress
Shear stress
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1.1. Basic elasticityBasic elasticity
1.4. Principal Stresses1.4. Principal Stresses
� MAX or MIN
Student do by themselves
2 solutionsShear stress = 0 (comparing with
Equation 1.9)and / 2θ θ + π
τ
(1.8)
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1.1. Basic elasticityBasic elasticity
1.4. Principal Stresses1.4. Principal Stresses
maximum or major principal stress
2 solutions
minimum or minor principal stress
Shear stress = 0
2 principle stresses (σI, σII) on 2 perpendicular principal
planes (on which shear stresses =0)
τ
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1.1. Basic elasticityBasic elasticity
1.4. Principal Stresses1.4. Principal Stresses
And how about maximum SHEAR stress ??? ���� students answer
2 SOLUTIONS:
Remark:
NB: the planes of maximum shear stress are inclined at 45" to the principal planes
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1.1. Basic elasticityBasic elasticity
1.5. Mohr1.5. Mohr’’s Circle of Stresss Circle of Stress
The state of stress at a point in a deformable body may
be determined graphically by Mohr's circle of stress
�Q1
� Q2
� centre C
� rotate Q1
angle 2 θ�
( )ngiven , ?θ⇒ σ τ
( )nQ ,′ σ τ
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1.1. Basic elasticityBasic elasticity
1.5. MOHR1.5. MOHR’’S CIRCLE OF STRESSS CIRCLE OF STRESS
RADIUS:
CENTRE (C):
Principle stresses:
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1.1. Basic elasticityBasic elasticity
1.5. MOHR1.5. MOHR’’S CIRCLE OF STRESSS CIRCLE OF STRESS
MAX/MIN. NORMAL STRESSES
���� Where it is?
MAX./MIN. SHEAR STRESSES
���� Where they are ???
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1.1. Basic elasticityBasic elasticity
1.5. MOHR1.5. MOHR’’S CIRCLE OF STRESSS CIRCLE OF STRESS
EXAMPLEEXAMPLE
Direct stresses of 160 N/mm2, tension, and 120 N/mm2,
compression, are applied at a particular point in an elastic
material on two mutually perpendicular planes.
The principal stress in the material is limited to 200 N/mm2,
tension.
Calculate:+ the allowable value of shear stressat the point on the given planes
+ the value of the other principal stress
+ the maximum value of shear stress
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1.1. Basic elasticityBasic elasticity
1.5. MOHR1.5. MOHR’’S CIRCLE OF STRESSS CIRCLE OF STRESS
EXAMPLE Direct stresses of 160 N/mm2, tension, and 120 N/mm2,
compression, are applied at a particular point in an elastic
material on two mutually perpendicular planes.
The principal stress in the material is limited to 200 N/mm2,
tension.
UNKNOWN: θ…MOHR CIRCLE ?
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1.1. Basic elasticityBasic elasticity
1.5. MOHR1.5. MOHR’’S CIRCLE OF STRESSS CIRCLE OF STRESS
EXAMPLE
�OσT
� OP1 = 160
� OP2 = -120
� C: MIDPOINT OF P1P2
� OB = 200
� MOHR’S CIRCLE: WITH RADIUS = CB
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1. Basic elasticity1.6. STRAINS
Longitudinal / direct strains direct
stresses σchanges in lengthchanges in length
Shear strainsshear stresses changes in angle changes in angle
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1.6.1 Direct / Normal Strain1.6.1 Direct / Normal Strain
� loads applied to a body � deformation will occur � dimension
change
/ Lε = δ
A bar � subjected to axial tensile
loading force, then tensile strain is:
NB:
� strain is dimensionless
� Compressive strain = - /L
� Strain is positive for an increase in dimension and negative for a reduction in dimension.
δ
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1.6.2 Shear Strain1.6.2 Shear Strain
P Q
S R
F
D D’
A B
C C’
L
x
φ
Shear strain is the distortion produced by shear stress on
an element or rectangular block as above. The shear strain, (gamma) is given as:
elements OA, OB and OC at a point Oin a deformable body
subjected
to forces
(at O)
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DemonstrationDemonstration
(direct strains)
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DemonstrationDemonstration
(direct strains)
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1.1. Basic elasticityBasic elasticity
1.6.3. STRAIN: general case1.6.3. STRAIN: general case
DIRECT STRAINS SHEAR STRAINS
Eqs (1.18) and (1.20) are derived on the assumption that the displacementsinvolved are small. Normally these linearized equations are adequate for most types of structural problem but in cases where deflections are large,
for example types of suspension cable etc., the full, non-linear, large deflection equations must be employed.