cmats lect4-compatibility & shear stress and strain
TRANSCRIPT
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8/3/2019 CMats Lect4-Compatibility & Shear Stress and Strain
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Civil Engineering Materials 267Stresses in Materials
Lecture 4: Compatibility
Shear Stress and Strain
Kerri Bland
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Lecture 4 2
Civil Engineering Materials 267 - Stresses
References
P.P. Benham & R.J. Crawford, Mechanics of Engineering Materials, 1987,Longman Scientific & Technical
R.C. Hibbeler, Mechanics of Materials, 6th Ed., 2005, Prentice Hall/Pearson.
E.P. Popov, Engineering Mechanics of Solids, 2nd Ed., 1991, Prentice Hall.
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Lecture 4 3
Civil Engineering Materials 267 - Stresses
Principal of Compatibility
Strains must be compatible with any
internal or external restraints
Force applied
resulting in xx
y
FF x=F/A
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=Determine x, y and z,using general strainequations:
Effect of Restraining Mechanical Strain:
Civil Engineering Materials 267 - Stresses
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Lecture 4 4
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or externalrestraints
Force appliedresulting in xx
y
FFx=F/A
tEEE
tEEE
tEEE
yxz
z
zxy
y
zyxx
+=
+=
+=
If there is no restraint against strain in the y and z
direction (ie: it expands as it wants due to Poissons
ratio), then y and Z = 0, so:
E
E
E
xz
xy
xx
=
=
=
x
yy=0
Principal of Compatibility
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Lecture 4 5
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or externalrestraints
Force appliedresulting in x
x
y
FFx=F/A
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
If there is restraint against strain in the y direction then y = 0,
y 0 (ie: in order to prevent the strain that wants to occur(due to ) a restraining force or stress is applied to the elementby the restraint) (Assume no restraint
in z direction so Z
= 0)
xy
xy
y 0EE
=
==
x
y=0y0
x primary or applied stress
y secondary or derivative stress
Principal of Compatibility
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Lecture 4 6
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or externalrestraints
Force appliedresulting in xx
y
FFx=F/A
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
Also, if there is restraint against strain in the y direction then
x x/E(Assume no restraint in z direction so Z = 0)
( )2xx
xy
yxx
1E
and
EE
=
=
=
x
y=0y0
Principal of Compatibility
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Lecture 4 7
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or externalrestraints
x
y
FFx=F/A
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
( )2xx
xy
yxx
1E
and
EE
=
=
=x
y=0y0
steelfor1.1E
)1(
E
)1(E
EModulussYoung'Apparent
2
2x
x
x
x'x
=
=
=
So, for 100% restraint in the y or z direction (not both) we get:
Principal of Compatibility
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Lecture 4 8
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or externalrestraints
x
y
FFx=F/A
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
x
y=0y0
)1(
EEModulussYoung'Apparent
2
'x
=
For a reduced proportion of restraint (where 1.0> >0)in the y or z direction (not both) we get:
( )2xx
xy
yxx
1E
and
EE
=
=
=
Principal of Compatibility
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Lecture 4 9
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or externalrestraints
x
y
FFx=F/A
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
x
y=0y0
)21(
)1EEModulussYoung'Apparent
2
'x
=
For 100% restraint in the y and z directions (both) we get:
Principal of Compatibility
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Lecture 4 10
Civil Engineering Materials 267 - Stresses
Strains must be compatible with any internal or external restraints
x
y
Know: l, t, , A, EFind x, induced axial force:
Need 2 equations to solve:
Effect of Restraining Thermal Strain:
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
Heat t
(1)Compatibility:
x = thermal + mechanical strain = 0
=
000 === lll
Principal of Compatibility
(2) Axial force (AF) = xA
= -tEA
tEE
t
x
x
=
=+
0
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Lecture 4 11
Civil Engineering Materials 267 - Stresses
Principal of Compatibility Strains must be compatible with any internal or external restraints
x
y
Find st, br, induced axial force:
Need 3 equations to solve:
Effect of Restraining Thermal Strain:
tEEE
tEEE
tEEE
yxzz
zxy
y
zyxx
+=
+=
+=
Heat t
(2) & (3) Axial force (AF) = (xA)st=(xA)br
Solve forx.stand x.br, then AF
steel brass
lst
lbr
(1)Compatibility:
( ) ( )
0..
0:
00
=
++
+
=+
=
=
=+=
br
br
xst
st
x
brasssteel
brasssteeltotal
Et
Et
so
and
ll
ll
lll
l
lll
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Lecture 4 12
Civil Engineering Materials 267 - Stresses
Principal of Compatibility
Strains must be compatible with any internal orexternal restraints
Restraint problems are always statically
indeterminate and require knowledge of material
characteristics and member values (ie: crosssectional area, E, )
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Lecture 4 13
Civil Engineering Materials 267 - Stresses
Thermal strain example Concrete mass after initial set
Outer layers heat of hydration able to dissipate
cooled and set in fixed positionrelative to ambient temperature
Inner layers heat of hydration not able to
dissipate
heat builds up
wants to expand (=t) restrained due to outer fixed layers stresses induced
Compatibility
x
y
0
0
=+=
=+=
tEE
tEE
xy
y
yxx
cold
cold coldcold
cold
cold
coldcold
cold
coldheat
x
y
Two equations and two unknowns:
solve to find x
and y
.
Assuming no
restraint inthe z direction
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Lecture 4 14
Civil Engineering Materials 267 - Stresses
CompatibilityWorked exampleLong reinforced concrete wallGiven: Econcrete = 14 GPa
= 0.2 = 10-5/C
y= -2.5 MPax=?
y=?
x=?
t=100C
(a) (b)
(a) The compressive stress in an element of concrete at the bottom of areinforced concrete wall due to the self-weight of the concrete has been
calculated as -2.5 MPa.
Determine the normal stress in the x-direction due to the confinement of
the concrete in the longitudinal direction of the wall.
y= -2.5 MPa
x=?
x= 0 MPaEE
ityCompatibil
yx
yxx
5.02.0*5.2
00
:
===
+==
As a wall is relatively thin
(cf width and height)
assume that z = 0
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Lecture 4 15
Civil Engineering Materials 267 - Stresses
CompatibilityWorked example
(2)tEE
(1)tEE
ityCompatibil
xy
y
yxx
+==
+==
0
0
:
Long reinforced concrete wallGiven: Ecocnrete = 14 GPa
= 0.2 = 10-5/C
y= -2.5 MPax=?
y=?
x=?
t=100C
(a) (b)
(b) A small part of the wall is being heated such that the temperature rise is100C.
Determine the normal stresses in the x and y-directions due to the
confinement of the concrete in the longitudinal direction of the wall.
y= ?
x=?
x= 0
As a wall is relatively thin (cf width andheight) assume thatz= 0
y= 0
)5.175.17
)2.01(
100*10*10*14
)1(
0)1()1(
53
2
MPa(andMPa
tE
tE
(2)intosubstituteandby(1)Multiply
x
y
y
==
=
=
=++
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Lecture 4 16
Civil Engineering Materials 267 - Stresses
CompatibilityWorked exampleLong reinforced concrete wallGiven: Ecocnrete = 14 GPa
= 0.2 = 10-5/C
y= -2.5 MPax=?
y=?
x=?
t=100C
(a) (b)
(c) If parts a and b occur simultaneously, determine the normal stresses inthe x and y-directions
y= ?
x=?
x= 0
As a wall is relatively thin (cf width andheight) assume thatz= 0
y= 0
By superposition:
x = -0.5+(-17.5) = -18 MPa
y = -2.5+(-17.5) = -20 MPa
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Lecture 4 17
Civil Engineering Materials 267 - Stresses
CompatibilityExample:Steel rod
Brass tube
airHeat t
Find st, br, xCompatibility of strains:
(1)tE
tE
tube
br
br
br
rod
st
st
stx
+=
+=
( ) ( ) (2)0AA
0F
brbrstst
x
=+=
Equilibrium:
Solve (1) and (2) to find st and br, then use to find x.
tuberodassembly
tuberodassembly
lll
lll
==
==
x,assembly=x,rod=x,tube
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Lecture 4 18
Civil Engineering Materials 267 - Stresses
External Shear ForceExternal Shear Force
External Axial Tension Force
Internal Normal Stress Response
External Shear ForceExternal Shear Force
Internal ShearStress Response
(AverageNormal Stress)A
P=
A
Vav =
(Average
Shear Stress)
Similarly:
Shear Stress
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Lecture 4 19
Civil Engineering Materials 267 - Stresses
Examples of shear in bolts:
Top bolt (a, b, c, d) shows single shear plane
Bottom bolt (e, f, g, h) shows double shear planes
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Lecture 4 20
Civil Engineering Materials 267 - Stresses
Class examples:
1. Determine the average shear stress in each bolt in the following detail.
Assume M24 bolts are being used. (ie bolt diameter = 24mm)
A
A
30
0kN
a) Section AA
b) Section AA
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Lecture 4 21
Civil Engineering Materials 267 - Stresses
Class examples:
2. If we have an applied load of 540kN, what bolt size would be required
for the three bolt arrangement shown below? Assume that the
maximum possible shear stress in the bolts is 400MPa. (Standard bolt
sizes: M12, M16, M20, M24, M30, M36).
A
A
540kN
a) Section AA
b) Section AA
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Lecture 4 22
Civil Engineering Materials 267 - Stresses
xxusually denoted asx
Stress direction and type (ie normal/shear) changes depending on direction and
type of externally applied loads. There needs to be a consistent method of
describing the stresses so that the type and direction of stress can
be easily recognised.
x
z
y Typically adopt an orthogonal set of axes:
In this unit the x-axis will correspond with
the longitudinal axis of the member inquestion.
Warning: In later design units axes changeand the x and z axes swap so that thelongitudinal axis becomes the z axis.
Stress expressed as with two subscripts.1
st
Subscript: direction normal to the plane on which stress acts2nd Subscript: actual direction of the stress
xy xz
Normal & Shear Stresses
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Lecture 4 23
Civil Engineering Materials 267 - Stresses
Normal & Shear Stresses 3D In a 3D elemental cube, the normal and shear
stresses can be designated as follows:-
1st subscript = direction
normal to
the planeon which
stress acts
2nd subscript = actualdirection of
the stress
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Lecture 4 24
Civil Engineering Materials 267 - Stresses
The element is in equilibrium:-MA = 0
yx= xy Complementary or conjugate shearstresses are paired and in equilibrium. For conjugate shear stresses in 3D:-
xy=yx yz=zy zx=xz
Conjugate Shear Stresses
x
z
y
xyxyxy
yx
yx(Initial action)(Reaction)
(Reaction)
(Reaction)
Considering the x-y plane only (ie: in 2D):
A
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Lecture 4 25
Civil Engineering Materials 267 - Stresses
Shear Strain Consider a small
element subjectedto shear:-
Shear stress = = V/A (MPa)
Shear strain = = l/l = angular strain = change of an angle that
was formally 90
Shear strain = l/l = tan for small values.
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Lecture 4 26
Civil Engineering Materials 267 - Stresses
Visualising shear straindeformation in 2D
subscript of (shear strain)= subscript of stresses causing distortion
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Lecture 4 27
Civil Engineering Materials 267 - Stresses
For a ductile material, we can draw a -relationship, which is analogous to -
relationship.
This can be idealised:-
Slope = G = = shear modulusor modulus of rigidity
Shear Modulus G
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Lecture 4 28
Civil Engineering Materials 267 - Stresses
SummaryShear stress and Strain
Average Shear stress :
Shear strain (angular strain):
Shear modulus (modulus of rigidity):
(a measure of the shear stiffness of
a linear elastic material)
AVav =
l
l =
=G