1/1 SOE 1032 SOLID MECHANICS
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Website
www.ex.ac.uk~TWDavies/solid_mechanics
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course organisation,lecture notes,
tutorial problems,deadlines
Course book:J M Gere, Mechanics of
Materials, Nelson Thornes, 2003, £29.
1/2 TUTORIALS
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Three groups, A, B, C
Three sessions
Monday, Tuesday and Friday weeks
8, 9, 10, 11
1/3 LABORATORY SESSIONS
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ONE DEMONSTRATION
Week 10
Write up deadline Friday week 11
Hand in to 307 and get it date stamped.
1/4 LEARNING TRIANGLE
BOOK
ME
YOU
1/5 DEFORMATIONS to be studied
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Static objects
Extension or compression of a rod
under an axial load
Twisting of a rod by applied torque
Bending of a beam subjected to point
loads, uniformly distributed loads and
bending moments
1/6 BASIC SCIENCE USED
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Newton’s 3rd Law, equilibrium (STATICS)
Auxiliary relationships based on material
properties – e.g. Hooke’s Law
1/7 POSSIBLE MATERIALS
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NATURAL
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MANUFACTURED
1/8 NATURAL MATERIALS
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WOOD
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STONE
1/9 MANUFACTURED
MATERIALS
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METALS (examples used in this course)
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PLASTICS
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CONCRETE AND BRICK
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CERAMICS AND GLASS
1/10 SCOPE OF COURSE
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STRESS AND STRAIN IN SIMPLE
SYSTEMS
DEFORMATIONS IN TENSION,
COMPRESSION & TORSION
STRESS & BENDING IN BEAMS
MOHR’S CIRCLE
i.e. essential parts of Chapters 1 to 5
and 7 (see reading list).
1/11 NORMAL STRESS
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Axial force per unit X-sectional area
P/A =  N/m2 or Pa (like pressure)
Tensile stress (positive)
Compressive stress (negative)
eg m=100 kg held by rod of A = 1 cm2
g = 10ms-2
=P/A = mg/A = 1000/10-4 = 10 MPa
1/12 SHEAR STRESS
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TANGENTIAL FORCE PER UNIT AREA
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P/A =  N/m2 or Pa
1/13 NORMAL STRAIN
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Change in length  caused by normal
stress
 = /L (dimensionless)
Tensile strain
Compressive strain
eg  = 2 mm, L = 2 m, then  = 1
mm/m
or  = 0.1%
1/14 UNIAXIAL STRESS
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Conditions are that:
Deformation is uniform throughout the
volume (prismatic bar) which requires
that:
Loads act through the centroid
Material is homogeneous
See Section 1.2 in book
1/15 LINE OF ACTION OF AXIAL FORCES FOR
UNIFORM STRESS DISTRIBUTION
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Prismatic bar of arbitrary cross-section A
Axial forces P producing uniformly distributed
stresses  = P/A
1/16 BALANCE THE MOMENTS
1/17 Mechanical Properties
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Strength
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compression
tension
shear
Elasticity, plasticity, ductility,creep
Stiffness, flexibilty
Used to relate deformation to applied
force
1/18 Mechanical Testing
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Tensile test machine
1/19 Tensile test for mild steel
1/20 Nominal and true SS
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Nominal stress based on initial area
True stress based on necked area
Nominal strain based on initial length
True strain based on current length
Use nominal values when operating
within elastic limit
1/21 Test data – linear scale
1/22 TENSILE TEST
SPECIMEN
1/23 BRITTLE MATERIAL
1/24 CAST IRON
400
200
Tension
0
-0.03
-0.02
-0.01
Compression
-200
-400
-600
-800
-1000
0
0.01
1/24 STONE
Tension
Compression
Stress
Strain
5-200MPa
1/26 WOOD
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Not an isotropic or homogeneous
material
Stronger and stiffer along the grain
Stronger in tension than compression
Fibres buckle in compression
Very high strength/weight ratio
Very high stiffness/weight ratio
1/27 COMPRESSION TEST
1/28 Compression test - concrete
1/29 ELASTICITY
1/30 PLASTICITY
1/31 DUCTILITY
1/32 CREEP
1/33 LINEAR ELASTICITY
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STRAIGHT LINE PORTION OF STRESSSTRAIN CURVE
 = E. (Hooke’s Law)
E is the modulus of elasticity or Young’s
Modulus and is the slope of the curve
For stiff materials E is high (steel
200GPa)
For plastics E is low (1 to 10 GPa)
1/34 POISSON’S RATIO
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Tensile stretching of a bar results in
lateral contraction or strain (and v v)
For homogeneous materials axial strain
is proportional to lateral strain
Poisson’s Ratio = - (lateral/axial strain)
 = - (’ / )
For a bar in tension  is positive and ’
is negative, and v.v. for compression.
1/35 Poisson (1781-1840)
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Normal values 0.25-0.35 for metals
Concrete about 0.2
Cork about 0 (makes it a good stopper)
Auxetic materials have NEGATIVE 
1/36 Axial and lateral deformation
B
B..
= /L
L
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1/37 Volume change
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V1=L*B2 (original volume)
V2=L(1+  )*(B(1- * ))2 (final volume)
V2=L B 2(1+  -2 -22+ 22 +32)
V2L B 2(1+  -2)
1/38 DILATION
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V=L B 2 (1 -2) (change in volume)
V/V1=(1 -2) or (1 -2)/E = DILATION
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Max value of  is 0.5
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Since  is 1/4 to 1/3 then dilation is /3 to /2
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1/39 SUMMARY
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Stress - Force/Area
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Strain – (Extension or compression)/Length
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For some materials and subcritical loads
strain is proportional to stress (Hooke)
Change in length proportional to change in
width (Poisson). Shrinkage/expansion.
Characteristics determined by experiment