lecture note for solid mechanics -...
TRANSCRIPT
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Lecture Note for Solid Mechanics- Analysis of Stress -
Prof. Sung Ho YoonDepartment of Mechanical EngineeringKumoh National Institute of Technology
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Text book : Mechanics of Materials, 6th ed.,
W.F. Riley, L.D. Sturges, and D.H. Morris, 2007.
Prerequisite : Knowledge of Statics, Basic Physics, Mathematics, etc.
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Analysis of stress
Application of equilibrium equations
Determine the forces on a structure by its support and internal forces on the parts of the structure
Objective of this chapter
Determine the internal effect of forces on the structure in order to understand the behavior of materials under the action of forces
Design considerations : safety and economy
Calculate the internal force intensity and deformation on the parts
By knowing the properties of the material, engineers can establish the most effective size and shape of the parts
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Normal stress under axial loading Axial force is defined as the force collinear with the centroidal axis of member
Stress (force intensity) =
Normal stress : Normal force divided by the area over which the force is distributed
- Distribution of internal force- develops on exposed cross section- has a resultant F that is normal to the exposed
cross section and is equal in magnitude of P- has a line of action collinear with the line of
action of P
- Average intensity of internal force
AF
avg + : tensile- : compressive
- Normal stress at a point on the cross section
AF
avg
AF
A
0
lim
areaforce
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
- Shear stress at a point on the transverse cross section
AV
avg
AV
A
0
lim
Single shear
Double shear
Shear stress in connections The bolted and pinned connections are used to introduce the concept of shear
stress
Shear stress : shear force divided by the area over which the force is distributed
- Average shear stress on the transverse cross section
AV
avg
- The shear stress can not be uniformly distributed over the area. Actual shear stress or maximum shear stress is different from the average stress.
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Bearing stress (compressive normal stress) Bearing stress occurs on the contact surface between two interacting members
and just names normal stress resulting from contact between two different bodies
AF
b
- For the case of connection on the surfaces of contact between the head of the bolts and the top plate
: annular area of the bolt or nut)(4
22io ddA
- For the case of connection where the shank of the bolt is pressed against the sides of the hole
: projected area (not actual contact area)dtA
: Average bearing stress
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Unit of stress Force per unit area (FL-2) psi or ksi (=1000psi) in English unit Pa (N/m2) or MPa (MN/m2 or N/mm2) in SI unit
(Example) Determine the normal stress in the bar (a) On a section 20 in. to the right of point A, (b) On a section 20 in. to the right section of point B, and (c) On a section 20 in. to the right of point C. Assume that the cross sectional area is 3 in2.
(Sol)
(a) kipFFF
kipFFF
kipFFF
CDCD
BCBC
ABAB
450811854
2708154
81081
ksiA
F
ksiA
F
ksiA
F
CDCD
BCBC
ABAB
0.153
45
0.93
27
0.273
81
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Stresses on an inclined plane in an axially loaded member
cossin
cos
AAPV
PN
n
- Normal stress is either max. or min. on the planes for which shear stress is zero. Also, shear stress is zero on the planes of max. or min. normal stress.
- Normal and shear stresses
2sin2
cossincossin
)2cos1(2
coscoscos 2
AP
AP
AP
AV
AP
AP
AP
AN
nn
nn
- Variations of normal and shear stresses
APAP
2
max
max
- Resultant F of internal force is equal to the applied load P
- Average total stress AF
Savg
at 0° and 180°
at 45° and 135°
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(Example) The block has a 200x100-mm rectangular cross section. Normal stress on plane a-a is 12.0MPa (compression). If Φ is 36°, determine (a) load P (b) shear stress on plane a-a (c) maximum normal and shear stresses.
(Sol)
)(695)10(7.694
)54(2cos1)10)(100)(200(2
)10(12
2cos12
3
6
6
CkNNP
P
A
P
A
N
nn
(a)
MPa
A
P
A
V
nn
52.16)54(2sin)10)(100)(200(2
)10()695(
2sin2
6
3
(b)
MPam
N
A
P
MPam
N
A
P
4.17)10(37.17)10)(100)(200(2
)10(695
2
7.34)10(74.34)10)(100)(200(
)10(695
2
6
6
3
max
2
6
6
3
max
(c)
2sin2
)2cos1(2
A
P
A
P
n
n
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Stress at a general point in an arbitrarily loaded member
AFS n
An
0
lim
- For any other plane through O the values of F and M could be different- General concept of stress state at a point is needed
- Stress vector or resultant stress
AFA
F
nt
An
nn
An
0
0
lim
lim
- Normal stress and shear stress are defined as
- Consider plane having an outward normal in the x-direction
x
x
xz
xy
x
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Stress vector itself depends on the orientation of the plane with which it is associated
Infinite number of planes can be passed through the point, resulting in an infinite number of stress vectors being associated with the point
Specification of stresses on three mutually perpendicular planes is sufficient to completely describe the state of stress at the point
- Normal stress : single subscript indicates the plane on which the stress acts
z
y
x
- Shear stress : first subscript indicates the plane on which the stress acts and second subscript indicates the coordinate axis to which the stress is parallel
zx
yz
xy
zz
yy
xx
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Two dimensional or plane stress
0 yzxzzyzxz
- Two parallel faces to be perpendicular to z-axis of the small element are assumed to be free of stress
- The only components of stress in two dimensional problem
- Two dimensional or plane stress problem
yxxyyx ,,
Thin plates where z-dimension of the body is small and the z-components of force are zero
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Stress transformation equation for plane stress
- Equation relating the normal and shear stresses n and nt on an arbitrary plane () and the known stresses x, y and xy on the reference planes
:0 nF
0sin)cos(cos)sin(
sin)sin(cos)cos(
dAdA
dAdAdA
xyyx
yxn
2222
22
212
21
222
sincos
sincoscoscossinsincos
xyyxyx
xyyx
xyyxn
:0 tF
0
sin)sin(cos)cos(cos)sin(sin)cos(
dAdAdAdAdA
yxxy
yxnt
222
22
cossin
)sin(coscossin)(
xyyx
xyyxnt
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
sign convention for stress transformation equation
Tensile normal stresses are positive; compressive normal stresses are negative. The sign of a normal stress is independent of the coordinate system.
Shear stress is positive if it points in the positive direction of the coordinate axis of the second subscript when it is acting on a surface whose outward normal is in a positive direction of the coordinate axis of the first subscript. The sign of a shear stress depends on the coordinate system.
Angle measured counterclockwise from the reference positive x-axis is positive.
(n, t, z) axes have the same order and form a right-hand coordinate system as (x, y, z) axes.
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(Example) At a point on the outside surface of a thin walled pressure vessel, stresses are 8000 psi(T) and zero shear on a horizontal and 4000 psi(C) and zero shear on a vertical plane. Determine the stresses at this point on plane b-b having a slope of 3 vertical to 4 horizontal.
(Sol)
)(368036800)13.53(sin)8000()13.53(cos)4000(
cossin2sincos22
22
Tpsipsi
xyyxn
)(57605760
0)13.53cos()13.53sin()8000()4000(
)sincos(cossin)( 22
Cpsipsi
xyyxnt
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(Example) Determine (a) normal and shear stresses on plane a-b, (b) normal and shear stresses on plane c-d, which is perpendicular to plane a-b, (c) show the stresses on planes a-b and c-d using a small element.
(Sol)
(a)
(b)
MPaMPaMPa yxxyyx 60,100,80
MPaMPa
CMPaMPa
xyyxnt
xyyxn
8.9578.95)]42(sin)42()[cos60()42cos()42sin()]100(80[
)sin(coscossin)()(3.6026.60
)42cos()42sin()60(2)42(sin)100()42(cos80
cossin2sincos
22
22
22
22
MPaMPa
TMPaMPa
xyyxnt
xyyxn
8.9578.95)]48(sin)48()[cos60()48cos()48sin()]100(80[
)sin(coscossin)()(3.4026.40
)48cos()48sin()60(2)48(sin)100()48(cos80
cossin2sincos
22
22
22
22
(c) Since planes a-b and c-d through the point are perpendicular, the stresses on these planes may be drawn on an element as figure (g).
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Principal stresses and maximum shear stress - plane stress Transformation equations for plane stress provide a means for determining the
normal stress and the shear stress on different planes through a point.
For design purposes, critical stresses at the point are the maximum normal and shear stresses
yx
xy
p
pp
xyyxn
xyyxyx
n
dd
22cos2sin
2tan
2cos22sin)(0
2sin2cos22
- The above relation gives the orientations of two principal planes.- Principal planes are normal to each other. (Two values of p differ by 180o)- A third principal plane for the plane stress state has outward normal in z-direction.
xyyxnt
yno
xynt
xno
at
at
90
0
Shear stress is zero on planes experiencing maximum and minimum values of normal stress.
Planes free of shear stress are the principal planes
Normal stresses on principal planes are principal stresses
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
yx
xy
p
pp
2
2cos2sin
2tan
0
22
3
2
2
2,1
zp
xyyxyx
pp
If xy and (x - y) have the same sign (tan2p is positive),
2p is between 0o and 90o with the other value 180o greaterp is between 0o and 45o with the other value 90o greater (p and p+90o)
Both sin2p and cos2p are positive
For a given set of stress values, there are two values of p differing by 180o
The above relation gives the orientations of two principal planes which are normal to each other. (Two values of p differ by 90o)
A third principal plane for the plane stress state has outward normal in z-direction.
2
2
2
2
2
2sin
2
22cos
xyyx
xyp
xyyx
yx
p
Both sin 2p and cos 2p are negative
2
2
2
2
2
2sin
2
22cos
xyyx
xyp
xyyx
yx
p
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
xy
yx
xyyxnt
xyyx
nt
dd
22cos
2sin2tan
2sin22cos)(0
2cos22sin2
2
2
21
2
2
pp
xyyx
p
There are two perpendicular planes of maximum in-plane shear stress, having the same magnitude but opposite signs
Maximum shear stress is equal in magnitude to one-half the difference between the two principal stresses
(a) Two tangents are negative reciprocals(b) Two angles 2p and 2 differ 90o, and p and differ 45o apart(c) Planes on which maximum in-plane shear stress occur are 45o from the principal planes
yx
xy
p
pp
2
2cos2sin
2tan
Maximum in-plane shear stress occurs on planes by the specified values of
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
2
22
2
2222
21
2
2
2
2
2
2
21
ppp
pxyyx
xyyxyx
xyyxyx
pp
There are two perpendicular planes of maximum in-plane shear stress, having the same magnitude but opposite signs
Maximum shear stress is equal in magnitude to one-half the difference between the two principal stresses
Relation between the principal stresses and the maximum in-plane shear stress
yxyx
xyyxyx
xyyxyx
pp
22
22222
2
2
2
21
Relation between the principal stresses and the normal stresses on the orthogonal planes
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(Example) (a) Determine principal stresses and maximum shearing stress (b) locate planes on which these stresses act and show the stresses on a complete sketch.
(Sol) (a) psipsipsi xyyx 4000,8000,10000
psi
Cpsipsi
Tpsipsi
zp
p
p
xyxxyx
pp
0
)(88508850
)(1085010850
98491000
)4000(2
)8000(10000
2
)8000(10000
22
3
2
1
2
2
2
2
2,1
psi98502
)8850(10850
2minmax
max
Maximum in-plane shear stress
psi
xyyx
p
9850
)4000(2
)8000()10000(
2
2
2
2
2
Maximum shear stress
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(b)
,02.78,98.11
,04.156,96.232
444.0)8000(10000
)4000(222tan
p
p
yx
xyp
)(1085010850
)98.11cos()98.11sin()4000(2)98.11(sin)8000()98.11(cos10000
cossin2sincos
98.11
1
22
22
Tpsipsi
when
p
xyyxn
p
)(88508850
)02.78cos()02.78sin()4000(2)02.78(sin)8000()02.78(cos10000
cossin2sincos
02.78
2
22
22
Cpsipsi
when
p
xyyxn
p
psi
Tpsipsi
when
xyyxnt
xyyxn
9850
)]02.33(sin)02.33()[cos4000(
)02.33cos()02.33sin()]8000(10000[
)sin(coscossin)(
)(10001000
)02.33cos()02.33sin()4000(2)02.33(sin)8000()02.33(cos10000
cossin2sincos
02.334598.11
22
22
22
22
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
psi
Tpsipsi
when
xyyxnt
xyyxn
9850
)]02.123(sin)02.123()[cos4000(
)02.123cos()02.123sin()]8000(10000[
)sin(coscossin)(
)(10001000
)02.123cos()02.123sin()4000(2)02.123(sin)8000()02.123(cos10000
cossin2sincos
02.1239002.33
22
22
22
22
Normal stresses on the perpendicular planes of maximum shear stresses are equal both in magnitude and sign
Shear stress have equal magnitude but opposite sign
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Mohr’s circle for plane stressGraphic interpretation of transformation equations for plane stress
22
2
2
22
2cos2sin2
2sin2cos22
xyyx
ntyx
n
xyyx
nt
xyyxyx
n
0,
2yxisCenter
22
2xy
yxRisRadius
How to draw a Mohr’s circle
Coordinates of each point on the circle represent the normal and shear stresses on one plane through the stresses point
Angular position of the radius to the point gives the orientation of the plane
Normal stresses are plotted as horizontal coordinates (tensile stresses plotted to the right of the origin)
Shear stresses are plotted as vertical coordinates (clockwise rotation of stress element plotted above -axis)
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
In the case of the x greater than y
1. Choose a set of x-y reference axes
2. Identify stresses and list them with proper sign3. Draw a set of coordinate axes4. Plot and label it point V (vertical plane)5. Plot and label it point H (horizontal plane)6. Draw line between V and H7. Establish center C and radius R8. Draw a circle
),( xyx ),( yxx
Procedures to draw a Mohr’s circle
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Mohr’s Circle for plane stress
2sin2sin2cos2cos
)22cos('
pp
p
CVCVOC
CFOCOF
V : stresses on the vertical planeH : stresses on the horizontal planeC : center of circleLine CV : vertical plane from which angle is
measured
Using
avgyx
xyp
yxp
OC
VVCV
CVCV
2)(
2sin
2)(2cos
'
'
2sin2cos22
2sin'2cos''
xyxxxx
VVCVOCOF
2sin2
2cos
2sin'2cos'
2sin2cos2cos2sin
)22sin('
xxxy
pp
p
CVVV
CVCV
CFFF
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
22
22
1
2
22
xyxx
p
xyxxxx
p
CBCA
CVOCCDOCOD
22tan yxxy
p
• If the two nonzero principal stresses have the same sign, the maximum shear stress at the point will not be in the plane of the applied stresses
• The angle between the vertical plane and one of the principal planes is p
• All angle on Mohr’s circle are twice the corresponding angles for the actual stressed body
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(Example 2-11) Determine (a) principal stresses and maximum shearing stress (b) normal and shearing stresses on a-a plane.
(Sol)
(a)
ksiCV
CHV
06.847
)0,1(:)4,6(:)4,8(:
22
0
)(06.706.706.81
)(06.906.906.81
3
2
1
zp
p
p
CksiksiOE
TksiksiOD
Since two principal stresses have opposite signs
ksiCBCAp 06.8max
The principal planes are represented by lines CD and CE
87.1474.292
5714.07
42tan
pp
p
or
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Two orthogonal surfaces are sufficient to completely specify the principal stresses One of the surfaces is required to completely specify the maximum shear stress
Analysis of stress
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Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(Homework)
(2-8), (2-18), (2-37), (2-58), (2-74), (2-93)