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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Week 1

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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Week 1

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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Week 1

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

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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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Week 1

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