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Transcript of mechanics of solids week 1 lectures
Week 1
Mechanics of Solids II (MECH3361/5361)
1. Introduction COURSE OUTLINE
Aims: To learn how to analyse the behaviour of solid materials and structures subjected to stress and deformation in more complex scenarios.
Lecturers, Tutorials and Lab:Lectures: Monday: 11am-1pm and Wednesday: 11am-12pm (PNR Lect Theatre 1-
Farrell)Tutorials: 2-4pm Monday (Mech Tut Rm 1 & 2, PNR Drawing office 1), Wednesday
(Chemical Engineering Lecture Room 1), Friday (Mechanical Engineering Drawing Office)
Classroom activities: Run “Classroom activities” for attendance checking during lectures indefinitely.
Lab: 2-5pm each Tuesday, Thursday and Friday, s163, Mechanical Building
Learning suggestions:Reading text and lecture notes. Do more exercises. Participate in computer lab sessions, Prepare yourself before walk into tutorial class.
Assessments: A final examination at the end of the semester (50%) Four assignments: 20% (5% each) to be finished INDEPENDENTLY Two in-class quizzes: 20% (10% each) “Semi-open book” A laboratory experiment on the strain gauge technique and knowledge: 10%
(5% for quiz, and 5% for the group’s lab report). Each student is responsible to make sure that his/her name and student ID are presented in the submitted report. Fail to do so will lead to 0% of the lab report mark.Note: A late submission of each day will result in a mark reduction of 25% for both the assignments and lab report.
2. RELATION WITH OTHER UNITS
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Week 1
Engineering MechanicsENGG1802
Dynamics I
AMME2301Mechanicsof Solids I
AMME2500EngineeringDynamics II
Particle dynamics• displacement, • velocity, • acceleration
Non-deformableDeformable
Forces, MomentsFree body diagram
Stress, deflectionBar, shaft, beam, column (simple)
3U Maths/2U Physics in high school
MECH3361Mechanicsof Solids II
Deformable
Stress, Strain2D and 3D problems (complex)
AMME3500EngineeringDynamics III
Control
Non-deformable
Non-deformableNon-deformable
Rigid body dynamicsLinear/angular motion
3. MECHANICS OF SOLIDS I (AMME2301)Stress analysis:
Bending normal stress
M = -My/I
Total normal stress
=F/A -My/I
TorsionalLoad
(Torque T)
BendingLoad
(TransverseForce P)
CombinedLoads
StressDistributions
StressesStresses Produced by Each Load Individually
T
B
xA
D
B
N.A. xA
D
P
P
T
A
B
D N.A.
x
BAD
F
F
avg
Tensile average normal stressavg=F/A
Torsional shear stress
T = Tρ/J
Transverse shear stressV = VQ/It
T
B
A
D
C
C
BAD
M
B
A,CD
BA,C
D
B
A
D
C
B
A
D
CTotal shear
stress at N.A. = VQ/ItTρ/J
AxialLoad
(Force F)
y
y
N.A.
N.A.
N.A.
Q ( y )=∫y
y top yt ( y )dy= y ' A'
A’ is the top (or bottom) portion of the member’s cross-section
Deformation:Torsional angle of twist:
ϕ=TLGJ T-Torque, L=length, G=shear modulus, J=polar moment of inertia
Axial deformation (elongation):
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Week 1
δ= PLEA P=axial force, L=length, E-Young’s modulus, A=cross-sectional area
Bending deformation
Deflection: v=∬ M ( x )
EIdxdx+Cx+D
or EIv=∬M ( x ) dxdx+Cx+D
Slope: θ=dv
dx=∫ M (x )
EIdx+C
or EI θ=EI dv
dx=∫M ( x ) dx+C
M=bending moment, I=second moment of inertia, E=Young’s modulus
Biaxial Stress Systems and Mohr circles
+xx
+yy
xx
x
y
+xy
+yx
Upward in the right hand face
Tensile or outward direction
-xx
-yy
xx
x
y
-xy
-yx
Downward in the right hand face
Compressive orinward direction
s n
n ny y
x x
x yx y
m a x
1 12 2
m a x
2 = 2 p 1
2
2
yyxx
22
2 xyyyxxR
2 p 2
1 8 0 °
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Week 1
Mechanics of Solids II (MECH3361)
Chapter 1 Stress
1.1 Definition of Stress
External forces on a bodyConsider an element of continuous (no voids) and cohesive (no cracks, breaks and defects) material subjected to a number of externally applied loads as shown in Fig. 1.1a). It is supposed that the member is in equilibrium.
FFn
Ft
Cross section: A A
F1
F2
F3 F5
F4Free Body Diagram
F1
F2
t
n
(a) (b) Fig. 1.1 External and internal forces in a structural member
If we now cut this body, the applied forces can be thought of as being distributed over the cut area A as in Fig. 1.1b). Now if we look at infinitesimal regions A, we assume the resultant force in this infinitesimal area is F. In fact, F is also a distributed force. When A is extremely small, we can say that the distributed force F is nearly uniform. In other words, if we look at the whole sectioned area, we can say that the entire area A is subject to an infinite number of forces, where each one (of magnitude F) acts over a small area of size A. Now, we can define stress.
Definition: Stress is the intensity of the internal force on a specific plane passing through a point.
Mathematically, stress at a point can be expressed as
σ=Tn
= limΔA→0
ΔFΔA (1.1)
Dividing the magnitude of internal force F by the acting area A, we obtain the stress. If we let A approach zero, we obtain the stress at a point. In general, the stress could vary in the body, which depends on the position that we are concerning.
The stress is one of most important concepts that we introduced in mechanics of solids. Why? Design of structures is largely dependent on stress level for safety reasons.
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Normal and Shear StressAs we known, force is a vector that has both magnitude and direction. But in the stress definition, we only consider the magnitude of the force so far. Obviously, this may easily confuse us. Let’s still take patch A as an example. As we can see, force F is not perpendicular to the sectioned infinitesimal area A. If we only take the magnitude of the force into account, apparently, the stress may not reflect the real mechanical status at this point. In other words, we need to consider both magnitude and direction of the force.
Now let’s resolve the force F in normal (Fn) and tangential (Ft) directions of the acting area as Fig. 1.1b). The intensity of the force or force per unit area acting normally to section A is called Normal Stress, nn (sigma), and it is expressed as:
σ nn= limΔA→0
ΔFn
ΔA (1.2)If this stress “pulls” on the area it is referred as Tensile Stress and defined as Positive . If it “pushes” on the area it is called Compressive Stress and defined as Negative .
The intensity or force per unit area acting tangentially to A is called Shear Stress, nt (tau), and it is expressed as:
τ nt= limΔA→0
ΔF t
ΔA (1.3)
Unit of stress: N/m2 or Pa (Pascal). In engineering practice: KPa=103Pa, MPa=106Pa, GPa=109Pa are used generally.
1.2 Notation for StressesObviously, the elementary notation described above is not sufficiently flexible and convenient for use in general, because (1) the direction of surface A can change, and (2) there are infinite tangential directions on a specific surface. i.e. the normal stress nn can vary with the direction change of n and shear stress nt can be in any tangential direction of the surface.
Cartesian coordinate systemHowever, recall that we often solve the engineering problems under a reference coordinate system, for instance, a Cartesian coordinate system xyz as show in Fig. 1.2. Clearly, it will be convenient to discuss stress at a point of interest P on an infinitesimal plane through P with its external normal n in one of the directions along a reference coordinate.
For example, we can consider an infinitesimal sectional plane through P in coordinate z, where n is coincident with z. Then the traction can be resolved along these three axes as On the z-sectional plane:
Tn= lim
ΔA→0 ( ΔFΔA )
z=( dF
dA )z=( dF x+dF y+dF z
dA )z=( dFx
dA+
dF y
dA+
dF z
dA )z= (σ x i+σ y j+σ z k )z
kji zzzyzxn
Tz - sectional plane
Resolution direction(i.e.coordinate directions)
(1.4)The first suffix of a stress component indicates the direction of the sectional plane (z, here). The second denotes the direction of the stress component along the coordinate.
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Positive/negative planes:If normal is the same as coordinate direction, this plane is positive (Fig. 1.2b). Otherwise negative (Fig. 1.2c).Similarly, we can have other infinitesimal planes through P in the direction of other coordinates. Positive/negative z; Positive/negative y; Positive/negative x; totally six planes.
F1
F2
F3F5
F4
x
y
z
ox
y
z
oy
z
o
Z-planeP
zz
zyzx
x
y
z
ox
y
z
oy
z
ox
y
z
ox
y
z
oy
z
o
P P
n
n +
Normal in the same direction of coordinate (z)
Positivez-plane
Negativez-plane
Normal in the opposite direction to coordinate (z)
Fig. 1.2 Stress in Cartesian coordinate
Representation of infinitesimal cubeFor the sake of convenient presentation, we often use an infinitesimal cube formed by the six infinitesimal planes mentioned above as shown in Fig. 1.3. On each plane, we have one normal stress component and two shear stress components.
x
y
z
ox
y
z
oy
z
o
zzzy
zx
zzzy
zx yz
yy
yxxz
xy
xx
Fig. 1.3 Stress in infinitesimal cube
We can arrange the stress component in a form of matrix (or namely, tensor)
xx xy xz
yx yy yz
zx zy zz
x y zPlane normal to x
Plane normal to y
Plane normal to z
Direction of stress component (coordinate direction)
Thus from above, we know that the stress state at Point P should be expressed by nine stress components. In engineering, we denote them as “stress matrix” or “stress tensor”:
[σ xx σ xy σ xz
σ yx σ yy σ yz
σ zx σzy σzz]
or [σ xx τxy τ xz
τ yx σ yy τ yz
τ zx τ zy σ zz]
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Cylindrical coordinate system
[σrr τ rθ τ rz
τθr σ θθ τθz
τ zr τ zθ σ zz]
Example 1.1Draw the stress states at two different points in a machine component measured as
A=[16 18 018 17 −150 −15 19 ] B=[−19 20 0
20 −25 00 0 20 ]
Soln: To reflect the stress matrix to an infinitesimal cube, you can first determine the corresponding notation.
A=[16 18 018 17 −150 −15 19 ]=[σ xx σ xy σ xz
σ yx σ yy σ yz
σ zx σ zy σ zz]
Thus σ xx=16 , σ xy=18 , σ xz=0 , which means in the x-section (ie. front face, or namely, x-plane), the three components in x, y, and z directions can be determined. Thus, in the x-section (x-plane), draw positive 16 in x-direction, positive 18 in y-direction, and 0 in z-direction as shown in the front face of the left infinitesimal element. Similarly, y and z plane stresses can be drawn as shown:
x
y
z
ox
y
z
oy
z
o
19
15
18
16
1815
17
x
y
z
ox
y
z
oy
z
o
20
20
1920
25
x-section y-section
z-section z-section
y-sectionx-section
1.3 Sign of StressesIt is necessary to define positive and negative sense of stresses for convenience.
Positive direction of normal stress.Consider normal stress nn on an infinitesimal plane An, whose external normal is n, We define that nn is positive if its direction is in the normal direction of section, as in Fig. 1.4 (left). Otherwise, negative (right).
Tensile = positive Compressive= negative
x
y
z
o
xx
+
positivex
y
z
o
xx
negative
n n
Fig. 1.4 Sign of normal stress (left – the same direction as the normal direction of section plane; right – opposite to the normal direction of section plane)
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Positive direction of shear stress.For a shear stress nt in the infinitesimal plane An, where t is a tangential direction of An. In the positive plane (e.g. Fig. 1.5a, where the external normal n of An has the same direction as a coordinate axis x), the positive nt should have the same direction as coordinate axis t (i.e. y in Fig. 1.5a). In the negative plane, the positive nt should have the opposite direction to coordinate axis t (i.e. y in Fig. 1.5d).
x
y
z
o xy
+
positivex
y
z
o xy
negative
x
y
z
o
xy
+positive
x
y
z
o
xy
negative
n n
n n(Positive-section) (Positive-section)
(Negative-section) (Negative-section)
(a) (b)
(c) (d)
Fig. 1.5 Sign of shear stress
1.4 Symmetry of the stress matrix
Are they all independent? Or Can we use a smaller number of stress component to facilitate the description of stress state?
Let’s check the infinitesimal element shown before. If we cut the infinitesimal element in the middle, i.e. a z-section as shown in dashed line. We can have the sectional model on the right hand side, a 2D version of infinitesimal element.
x
y
z
ox
y
z
oy
z
o
zzzy
zx
zzzy
zx yz
yy
yxxz
xy
xxA B
CD
A
BC
D
x
y
x
y
yy
yy
xxxy
xy
yx
yx
x
yO
Fig. 1.6 Equilibrium of element
In the 2D element, the “moment equation” can be written as:
σ xy ( Δy ) ( Δz )[( 12
Δx)]+σxy ( Δy ) ( Δz )[( 12
Δx)]−σ yx ( Δx ) ( Δz )[( 12
Δy)]−σ yx ( Δx ) ( Δz )[( 12
Δy)]=0
∴σxy−σ yx=0
Thus ∴σxy=σ yx (Or: τ xy=τ yx )Similarly by checking equilibrium conditions in the yz- and xz planes, we can have
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Week 1
σ xz=σ zx (Or: τ xz=τ zx )σ yz=σ zy (Or: τ yz=τ zy )
Thus we have shown that the stress matrix or stress tensor is symmetrical, i.e. σ ij=σ ji (where i , j=x , y , z ) (1.5)
Only SIX independent stress components are needed to describe the stress state at a point.
zzzyzx
yzyyyx
xzxyxx
[σ xx
σ yx σ yy
σ zx σzy σ zz]
Example 1.2
(1) Draw an infinitesimal cube to show the stress tensor A :
σ A=[10 0 −400 −30 0
−40 0 10 ]Soln:
(2) Write the stress tensor from the stressed infinitesimal cube (note the signs of the shear stress are not given in the figure and you need to decide them):
xyz
o
2020
40
40
60
60
60
- y-plane
+x-plane
xyz
o
2020
40
40
60
60
60
+z-plane
Soln:Look at the planes in the infinitesimal cube. Obviously, the front y-section is a negative plane (its normal direction is opposite to y-positive). Other two faces shown are positive planes:
x-plane (positive): σ xx=−60(compression ) , σ xy=−20(opposite to y ), σ xz=−40 ( opp to z )
y-plane (negative): σ yx=−20(same as x ) , σ yy=60( tension or opp to y ), σ yz=0
z-plane (positive): σ zx=−40(opposite to x ) , σzy=0 , σ zz=60 ( tension or same as z ) So the stress tensor can be written as:
σ B=[−60 −20 −40−20 60 0−40 0 60 ]
9
xy
z
o
10
40(-)
40(-)
10
30(-)
+x-plane
+y-plane
+z-plane
Week 1
Recap
Stress σ=T
n
= limΔA→0
ΔFΔA
σ nn= limΔA→0
ΔFn
ΔA τ nt= lim
ΔA→0
ΔF t
ΔA
FFn
Ft
Cross section: A A
F1
F2
F3 F5
F4Free Body Diagram
F1
F2
t
n
F1
F2
F3F5
F4
x
y
z
ox
y
z
oy
z
o
Z-plane
P
Fx
Fy
Fz
Notation for Stresses Tn=( dF
dA )z=( dFx
dA+
dF y
dA+
dF z
dA )z=(σ x i+σ y j+σ z k )z
kji zzzyzxn
Tz - sectional plane
Resolution direction(i.e.coordinate directions)
x
y
z
ox
y
z
oy
z
o
zzzy
zx
zzzy
zx yz
yy
yxxz
xy
xx
xx xy xz
yx yy yz
zx zy zz
x y zPlane normal to x
Plane normal to y
Plane normal to z
Direction of stress component (coordinate direction)
Stress tensor:
[σ ]=[ σxx σ xy σ xz
σ yx σ yy σ yz
σ zx σ zy σ zz]
Convention of normal stress:In any plane (Positive or Negative): Tension: positive Compression: negative
x
y
z
ox
y
z
oy
z
o
xx
+
positivex
y
z
ox
y
z
oy
z
o
xx
negative
Convention of shear stress:In Positive plane: + shear stress follows the same direction of coordinate directionIn Negative plane: + shear stress follows the opposite direction of coordinate direction
x
y
z
ox
y
z
oy
z
o xy
+
positivex
y
z
ox
y
z
oy
z
o xy
negative
x
y
z
ox
y
z
oy
z
o
xy
+positive
x
y
z
ox
y
z
oy
z
o
xy
negative
n n
x
y
z
ox
y
z
oy
z
o xy
+
positivex
y
z
ox
y
z
oy
z
o xy
negative
x
y
z
ox
y
z
oy
z
o
xy
+positive
x
y
z
ox
y
z
oy
z
o
xy
negative
n n
Example 1.3(1) Draw infinitesimal cube to show the stress tensor [A]
[σ A ]=[10 20 −4020 −30 0−40 0 10 ]
10
x
y
oz
Week 1
Soln: Note that the coordinator is different from that in Ex 1.2.Step 1: Determine the convention of section planes (all + in the visible planes).
Step 2:
[σ A ]=[10 20 −4020 −30 0−40 0 10 ]=[ σ xx σ xy σ xz
σ yx σ yy σ yz
σ zx σ zy σ zz].
Step 3: In +x-plane: σ xx=10 , σ xy=20 , σ xz=−40
In +y-plane: σ yx=20 , σ yy=−30 , σ yz=0
In +z-plane: σ zx=−40 , σ zy=0 , σ zz=10(2) Write the stress tensor from the stressed infinitesimal cube (note the signs of the shear stress are not given in the figure and you need to decide them):Soln:Step 1: Determine the convention of section planes. Look at the visible planes in the infinitesimal cube. Obviously, the visible x-plane is a negative plane (its normal direction is opposite to x-positive). Similarly, other two faces shown are also negative sections as shown.Step 2: In – x-plane: σ xx=−60(compression ) , σ xy=20 (opposite to + y ), σ xz=−40 (same as +z )
In – y-plane: σ yx=20 (opposite to +x ), σ yy=60 ( tension ), σ yz=0
In – z-plane: σ zx=−40(same as +x ) , σ zy=0 , σ zz=60 ( tension)Step 3: write the stress tensor
σ B=[−60 20 −4020 60 0−40 0 60 ]
1.5 Stress TransformationStresses in any direction (2D) (Mechanics of Solids I)Cut a triangle in 2D infinitesimal element, leaving the left and bottom sides and a third side inclined at an angle from the vertical. Two of its surfaces have the normals in the opposite x and y directions; the third has a normal at an angle from the x axis, as in Fig. 1.6 (right).
xx
yy
yy
xx
x
y
x
y
xy
yx
Ac o s
Asin
tn
xy
yy
xx
nn
x
y
x
y
nt
A
yx
xx
yy
yy
xx
x
y
x
y
xy
yx
Ac o s
Asin
tn
xy
yy
xx
nn
x
y
x
y
nt
A
yx
x
y
ox
y
o
x’y’
x’y’
x’x’
y’y’
xx
x’y’
yy
xy
+90
Fig. 1.6 Stress in different direction Fig. 1.7 Stresses with coordinate rotation
It is now necessary to apply the equilibrium equations about the Normal n & Tangent t axes.
11
x
y
o
1040
30
+z-plane
+x-plane
+y-plane
z
10
40
2020
xy
zo 2020
40
40
60
60
60
– y-plane
– x-plane
xy
zo 2020
40
40
60
60
60
– z-plane
Week 1
∑ Fn=0=Aσnn−(σ yy A sinθ⏟F y
)sin θ−(σ xx A cos θ⏟F x
)cosθ−( τ yx A sinθ )⏟V x
cosθ−(τ xy A cosθ )⏟V y
sin θ
Since τ xy=τ yx , the above equation can be simplified to: σ nn=σ xx cos2 θ+σ yysin2θ+2 τ xycos θ sinθ (1.6a)
Using the following trigonometric functions:
cos2θ=12
(1+cos2 θ )
sin2θ=12
(1−cos 2θ )sin 2 θ=2 cosθ sin θ
we can obtain: σ nn=
(σ xx+σ yy )2
+(σ xx−σ yy )
2cos2 θ+τ xy sin 2θ
(1.6b)And in a similar way, by applying equilibrium in tangential (t) axis and using the trigonometric functions we can get: ∑ F t=0=Aσnt−(σ yy A sin θ⏟
F y
)cosθ+( σxx A cosθ⏟F x
)sin θ+(τ yx A sin θ )⏟V x
sin θ−( τ xy A cosθ )⏟V y
cosθ
σ nt=12 [2 σ yy sin θ cosθ ]− 1
2 [2 σ xx sinθ cos θ ]−τ yx sin2 θ+ τ xy cos2 θ
Use trigonometrics, we can have: cos2θ−sin2θ= 1
2(1+cos 2θ )−1
2(1−cos2 θ )=cos2 θ
τ tn=(σ yy−σ xx )
2sin 2θ+τxy cos2 θ
(1.7)Based on these above two equations, we can determine the stress in any plane.
Stresses with coordinate axis rotation (Mechanics of Solids II)Let’s consider 2D stress state undergoing coordinate rotation, from xoy to x’oy’ (Fig. 1.7). σ x ' x ' can be determined from the previous section where the coordinate x’ is coincide with normal n. Thus the stress in an inclined plane of can be calculated by
σ x ' x '=σ xx cos2 θ+σ yysin2 θ+2 σ xy cosθ sinθ (1.8)σ y ' y ' can be determined by viewing the inclined plane with angle of (+90)
σ y ' y '=σ xx cos2 (θ+90° )+σ yy sin2 (θ+90° )+2σ xy cos (θ+90° )sin (θ+90° )¿σ xx sin2 θ+σ yy cos2 θ+2 σ xy (−sinθ ) cosθ¿σ xx sin2 θ+σ yy cos2 θ−2 σ xy sin θ cosθ
Thus σ y ' y '=σ xxsin2 θ+σ yy cos2θ−2 σ xy sinθ cos θ (1.9)
Similarly σ x ' y '=τ x ' y '=
(σ yy−σ xx )2
sin 2θ+σ xy cos2θ
(1.10)
Remarks:Let’s add (1.8) to (1.9)σ x ' x '+σ y ' y '=σ xx (sin2θ+cos2θ )+σ yy ( sin2θ+cos2θ )+2 σ xy sin θ cosθ−2 σxy sin θ cosθ
Thus: σ x ' x '+σ y ' y '=σ xx+σ yy
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which means that the summation of two normal stress components is independent on the
rotation of coordinate system. We will show this again in 3D: σ xx+σ yy+σ zz=const .
Example 1.4
Rotate the following stress tensors about z-axis for =90o (units MPa).
[ σ ]=( 1 −1 0−1 2 00 0 3 ) .
Soln: Since the rotation is about z-axis and σ zx=σ zy=0 , we do not need to change z-directional stresses:
σ x ' x '=σ xx cos2θ+σ yysin2 θ+2 σ xy cosθ sinθ¿(1 )×cos2 90 °+(2 )×sin2 90°+2×(−1)×cos90 ° sin 90°=(1)×0+(2 )×1+2×(−1 )×0×1=2σ y ' y '=σ xx sin2 θ+σ yy cos2 θ−2 σ xy sinθ cos θ¿(1 )×sin2 90°+ (2)×cos2 90°−2×(−1 )×sin 90 °×cos 90 °=1
τ x ' y '=(σ yy−σ xx )2
sin 2θ+σ xycos 2θ
¿(2−1 )2
sin (2×90 ° )+ (−1 )cos (2×90 ° )=12
sin (180° )+(−1 ) cos (180° )=1
σ z ' z'=σ zz=3 (rotates about z-axis)
Thus:
[ σ ' ]=(2 1 01 1 00 0 3)MPa .
Principal stresses (Mechanics of Solids I)To find the maximum stress, mathematically, we can solve from Eq. (1.6b)∂ σnn
∂θ=0=∂
∂ θ [ (σ xx+σ yy )2
+(σ xx−σ yy )2
cos2 θ+σ xy sin 2 θ]¿−(σ xx−σ yy )sin 2θ+2 σ xy cos2 θ=0
∴ tan 2θ p=2σ xy
σxx−σ yy (1.11)
where the maximum normal stresses occur. We call such a maximum and minimum σ nn the
principal stresses. From Eq. 1.10, 2 σ x ' y '=−(σ xx−σ yy ) sin 2 θ+2σ xy cos2 θ , we can obtain 2 σ x ' y '=(∂ σnn /∂θ )=0 , meaning that when σ nn reaches its extrema (principal stresses)
on the plane, σ nt=0 . (In other words, if seeing a zero shear plane, this plane is a principal plane).In Eq. (1.11), there are two roots p1 and p2. (2p1 and 2p2 are 180 apart, thus p1 and p2 are 90 apart.), i.e.
Two roots: tan2 θp= tan (2θp+180° )=
2σ xy
σ xx−σ yy
Thus for p1
13
2
yyxx
xy
22
2 xyyyxx
2p
2p+180
Fig. 1.8 principal stress and principal plane
Week 1
{sin 2θ p1=σ xy /√ [(σ xx−σ yy )2 ]
2
+σxy2 ¿ ¿¿¿
(1.12)
For p2 (=p1+90)
{sin 2θ p2=−σ xy /√[(σ xx−σ yy )2 ]
2
+σ xy2 ¿ ¿¿¿
(1.13)Substituting the above two trigonometric relations into
σ nn=(σ xx+σ yy )
2+
(σ xx−σ yy )2
cos2θ+σ xy sin 2θ
we can have
σ nn=(σxx+σ yy )2
±(σ xx−σ yy )2
(σ xx−σ yy )2
√ [(σ xx−σ yy )2 ]
2
+σ xy2
±σ xy
2
√[ (σ xx−σ yy )2 ]
2
+σ xy2
¿(σ xx+σ yy )2
±[ (σ xx−σ yy )2 ]
2
+σ xy2
√[(σ xx−σ yy )2 ]
2
+σ xy2
=(σ xx+σ yy )2
±√[(σ xx−σ yy )2 ]
2
+σ xy2
σ 1,3=(σxx+σ yy )
2 ±√[ (σ xx−σ yy )2 ]
2
+σ xy2
(1.14)
Maximum shear stresses (Mechanics of Solids I)To find the maximum shear stress, mathematically, we can solve for∂ σnt
∂θ=0=∂
∂θ [(σ yy−σ xx )2
sin 2 θ+σ xy cos2 θ]¿−(σ xx−σ yy )cos2θ−2 σ xy sin 2θ=0
∴ tan 2θ s=−σ xx−σ yy
2σ xy (1.15)
There are two roots ∴ tan 2θ s=tan (2 θs+180 ° )=−
σ xx−σ yy
2σ xy
By comparison with the max normal stresses (principal stresses) orientation, each roots of 2s
is 90 from 2p. Thus the roots of s and p are 45 apart. The planes for max shear stress can be determined by orienting 45 from the principal plane.Ref to Fig. 1.9, we can have:
14
2
yyxx
xy
22
2 xyyyxx
2s
2s+180
Fig. 1.9 Maximum shear stress/shear plane
Week 1
sin 2θs=−( σ xx−σ yy
2 )/√[ (σ xx−σ yy )2 ]
2
+σ xy2
,
cos2θs=σ xy /√[ (σ xx−σ yy )2 ]
2
+σ xy2
Thus the maximum shear stress is calculated as follows:
(σnt )max=(σ yy−σ xx )
2sin 2θs+σ xy cos 2θs
(σnt )max=(σ yy−σ xx )
2
−( σ xx−σ yy
2 )
√[ (σxx−σ yy )2 ]
2
+σ xy2
+σ xy
σ xy
√[ (σ xx−σ yy )2 ]
2
+σ xy2
=√[ (σ xx−σ yy )2 ]
2
+σ xy2
(σnt )max=√[ (σ xx−σ yy )2 ]
2
+σ xy2
(1.16)
From the definition of the principal stresses, we have:{(σ nt )max ¿ ¿¿¿
(1.17)Example 1.5: Determine the principal and maximum shear stresses for the following stress
tensor:
[σ ]=[50 20 020 10 00 0 30 ]
Soln: Step 1: Draw the Mohr Circle Center of Mohr Circle:
c=(σ xx+σ yy )
2=50+10
2=30 MPa
Radius of Mohr Circle: R=√(σ xx−σ yy
2 )2
+τ xy2 =√(50−10
2 )2+ (20 )2=28. 28 MPa
Step 2: Determine the orientation of the principle stress:
tan2 θp=2 τxy
(σxx−σ yy )= 2×20
50−10=1 . 0
, ∴θ p1=
arctan (1 . 0 )2
=22. 5 ° and
∴θ p2=90°+θ p1=112. 5°Step 3: Compute the principal stresses and the maximum shear stress
σ 11
σ33
=(σ xx+σ yy )
2±√( σ xx−σ yy
2 )2
+τxy2 =c±R=30±28 .28=58 .28 MPa
1.72 MPa
So we can write: σ 11=58 .28 MPa , σ22=30 MPa , σ33=1 . 72MPa ( i . e . σ11≥σ22≥σ33)
15
yy=10xx=50
xy=20
max=28.28
11=58.2822=1.72
max
2
2 R=28.28
C = 30
0
Week 1
τ max=R=√( σ xx−σ yy
2 )2
+τ xy2 =√(50−10
2 )2+ (20 )2=28. 28 MPa
Step 4: Draw infinitesimal elements indicating magnitude and orientations Note that the principal stresses correspond to zero shear; but max shears do not correspond to zero normal stress
11=58.28MPa
22= 1.72MPa
= 22.5o30MPa
30MPa
= 22.5o
max=28.28MPa
Orientation of Principal Stresses Orientation of Maximum Shear Stress
Cylindrical Pressure Vessels for experiments (Mechanics of Solids I) This analysis will look at tubes with an internal pressure and closed ends. Let xx be the Axial Stress due to the pressure on the end walls, and = yy be the Hoop Stress due to the pressure acting on the curved surface.
xxP
t Sectioned plane
r
x
y
xx
yy
P
L
t
Fig. 1.10 FBD of axial section of vessel Fig. 1.11 FBD of circumferential section of the vessel
Axial StressLook at a FBD of the axial section as shown in Fig. 1.10 and check for the axial equilibrium.
∑ F x=0=−(πr2 )P+(2 π rt ) σ xx ie: Pπr2= (2 π rt )σ xx
which gives the equation for Axial Stress( or Longitudinal Stress):
σ xx=
Pr2 t (1.18)
Hoop Stress Look now at a FBD of the circumferential section as shown in Fig. 1.11.Equating the forces vertically gives: ∑ F y=0=−P× (2 r×L )+σθθ×2 ( L×t ) ∴2 σθθ ( Lt )=2rLPwhich simplifies to give the equation for Hoop Stress (or Circumferential Stress):
σ θθ=σ yy=Pr/ t (1.19)
Example 1.6 Determine the Principal stresses and maximum shear stresses and their orientation for a
pressurised vessel. Assume Pr / (2 t )=10 MPa .Soln:Step 1: Principal stresses: Since there is no shear stress, x and y are the principal directions so
16
Week 1
σ 1=σ yy=2 σ xx=20 and σ 2=σ xx=10 (note that σ 1≥σ2 ). (θp 1=0 , θ p2=90 ° )
Step 2: Shear stresses: {(σ nt )max ¿ ¿¿¿MPa
(or use the equation: (σnt )max=√[ (σ xx−σ yy )
2 ]2
+τ xy2 =√[ (10−20 )
2 ]2
+02=5 MPa)
Step 3: Max shear direction ∴ tan 2θs=tan (2 θ s+180 ° )=−
σ xx−σ yy
2 σ xy=−10−20
2×0=−∞
{2θ s1=270 ° ¿ ¿¿¿ ∴¿ {θs 1=135 ° ¿¿¿
17