Week 5

Chapter 4 Modelling and Solution4.5 Similarity and Difference in Plane Stress and Plane Strain (Page 130-132)

Similarity: In both plane stress and plane strain problems, Deformation occurs in a plane x-y, all independent stresses, strains and displacements are the functions of coordinate x and y only.

Hooke’s Law: Observe difference of the Hooke’s laws: Plane Stress Plane Strain

σ zz 0 σ zz=ν (σ xx+σ yy )≠0

ε zz ε zz=−νE(σ xx+σ yy)≠0 0

xx Eε xx=σ xx−νσ yy ( E1−ν2 )ε xx=σ xx−( ν

1−ν )σ yy

yy Eε yy=σ yy−νσ xx ( E1−ν2 )ε yy=σ yy−( ν

1−ν )σ xx

xy Eε xy=(1+ν )σ xy ( E1−ν2 )ε xy=[1+( ν

1−ν )]σ xy

E* E∗¿E E∗¿ E1−ν2

ν∗¿ν ν∗¿ ν1−ν

UnifiedHookes’ Law {E∗ε xx=σ xx−ν∗σ yy

E∗ε yy=σ yy−ν∗σ xx

E∗ε xy=(1+ν∗)σxy

Replacing Solution MethodsFrom plane stress to plane strain solution: This indicates that once we have obtained the solution of a plane-stress problem, we can actually obtain the solution to the corresponding plane-strain problem provided that the elastic constants E and of solution of the plane-stress

problem are replaced by

E1−ν2

and

ν1−ν , respectively.

From plane strain to plane stress solution: Similarly, by replacing the E and in the solution

of a plane-strain problem by

E(1+2 ν )(1+ν )2 and

ν1+ν , we can also obtain the solution to the

plane stress problems. This is because in plane strain:

For the Poisson’s ratio: νPstrain¿ = ν

1−ν , νPstrain¿ (1−ν )=ν νPstrain

¿ −ν Pstrain¿ ν=ν

νPstrain¿ =(1+νPstrain

¿ )ν ν=

νPstrain¿

(1+ν Pstrain¿ )

For Young’s modulus: EPstrain

¿ = E1−ν2

E=EPstrain¿ (1−ν2 )

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Week 5

E=EPstrain¿ [1−( νPstrain

¿

(1+ν Pstrain¿ ))

2]=EPstrain¿ [ (1+ν Pstrain

¿ )2

(1+ν Pstrain¿ )2

−νPstrain¿2

(1+ν Pstrain¿ )2 ]=EPstrain

¿ [ 1+2 ν Pstrain¿

(1+νPstrain¿ )2 ]

Key DifferencesDespite the abovementioned similarity, plain stress differs from plain strain. Plane stress concerns a thin plate with stress-free surfaces normal to z-axis and

Plane stress:{σ zz=0 ¿ {w≠0 ¿ ¿¿¿

Plant strain concerns a very long cylindrical component

Plane strain:{ε zz=0 ¿ {w=0 ¿ ¿¿¿

Example 4.2 A thin circular plate of radius R is subjected to a uniform pressure p1 on its edge. The two place surfaces are free, as illustrated in Figure below. The in-plane (xy-plane) displacements

in the plate are found to be u=−p1

1−νE

x and

v=−p11−ν

Ey

. Find the corresponding displacements in long circular bar under a uniform pressure p2. Assume that the length of the bar is much longer than its radius R. Solution: From the plane stress solution to a plane strain

solution: Using E∗¿ E

1−ν2 to replace E and and

ν∗¿ ν1−ν to

replace , pressure p2 to replace p1:

u=−p21−ν

Ex=−p2

1−( ν1−ν )

( E1−ν2 )

x=−p2(1−ν2)−ν (1+ν )

Ex=−p2

(1+ν )(1−2 ν )E

x

Similarly: v=−p2

(1+ν )(1−2 ν )E

y

4.6 Superposition Principles (Page 133-135)Complex problems can be often broken down as a number of simple problems. We can do so where materials and deformation are linear, where the problem is independence of loading history. We used this idea in Mechanics of Solids I last year. For example, we can resolve the combined loading problem in axial tension, torsion and bending.

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Week 5

x

y

x

y

Fig. 4.6 Superposition principle

Solution under concentrated force

Solution under distributed force

Integration

Fig. 4.7 A foundation under arbitrarily distributed normal pressure.

Example 4.3 The stress components in a plane stress elastic plates under an in-plane bending are known as σ yy=σ xy=0 and σ xx=6 ay . On the other hand, those in the same plate under in-plane

shearing as shown, σ xx=σ yy=0 and σ xy=−a . Find the stress in the same elastic plate under the combined loading.Solution: Known the stress functions in the two different loading cases.

x

y

-a-a6ay

6ay

Thus the stress functions can be added directly as per the “superposition principle”.σ xx=σ xx , case1+σ xx , case2=6 ay+0=6 ay .σ yy=σ yy , case1+σ yy , case 2=0σ xy=σ xy , case1+σ xy , case2=−a

4.7 Solution Approaches and SkillsIntroductionAfter define the B.C., one should solve for three groups of unknowns:

Displacement: u,v,w

Strain ε xx , ε yy , εzz , ε xy , ε yz , ε zx ,

Stress σ xx , σ yy , σ zz , σ xy ,σ yz , σ zx ,It is however impossible to solve for these unknowns altogether. We often have to solve one or two groups first. As such we have four different methods: displacement method, strain method, stress method and mixed method.

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Week 5

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21

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E

E

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Fig. 4.8 Flowchart of displacement method (replace stress and strain by displacement).Displacement MethodUnknowns: u, v, wProcedure: Other two sets of the unknown variables must be eliminated from the equations. Thus we replace strain and stress in displacements, which can be done as follows:

We derive (refer to Tutorial Question #3, Week 5)

{( λ+μ )∂ I 1

ε

∂ x +μ ∇2u+ ρf x=ρax ¿ {( λ+μ )∂ I 1

ε

∂ y +μ∇2 v+ρf y=ρa y ¿ ¿¿¿

where Laplace operator: ∇2= ∂2

∂ x2+ ∂2

∂ y2+ ∂2

∂ z2 and

I 1ε=∂ u

∂ x+∂ v∂ y

+∂w∂ z

After obtain u, v, w, one can calculate strain by using strain-displacement equation and then calculate the stress by using Hooke’s law. Note that the solution must satisfy the boundary conditions. Stress Method

Unknowns: σ xx , σ yy , σ zz , σ xy ,σ yz , σ zx ,Procedure: Solve for stress component first and then strains and displacements.

Strain Method

Unknowns: ε xx , ε yy , εzz , ε xy , ε yz , ε zx ,Procedure: Solve for strain component first and then stresses and displacements.

4.8 Problem 1: Solution to Cylinder under Internal and External PressureIntroductionIt is convenient to use cylindrical coordinate system for many engineering problem which involves in circular geometry (e.g. Fig. 4.8).

Cylindrical coordinate system

4

po

pi

r

zr

Ri

Ro

Fig. 4.8 Pressurised cylinder

Week 5

Similar to Cartesian coordinate system, cylindrical system consists of 3 independent coordinates: (r, , z) as shown in Fig. 4.9.

[σrr τrθ τ rz

τθr σ θθ τθz

τ zr τ zθ σ zz]

Equilibrium equations in 3D cylindrical system (can be derived by replacing coordinate):

{∂σ rr

∂r+1

r∂σ rθ

∂θ+∂σ rz

∂ z+

σrr−σθθ

r+ρf r=ρar ¿{∂ σrθ

∂r+1

r∂σ θθ

∂θ+∂σ θz

∂ z+

2σ rθ

r+ ρf θ= ρaθ ¿¿¿¿

In 2D: (σ zz=0 ,σ rz=0 , σθz=0 ) {∂σ rr

∂r+1

r∂σ rθ

∂θ+

σrr−σθθ

r+ρf r=ρar ¿¿¿¿

Strain-Displacement relations: Normal:

{εrr=∂u∂r ¿ {εθθ=

1r∂ v∂θ

+ur ¿ ¿¿¿

Shear:

{εrθ=12 (∂ v

∂r+1

r∂u∂θ

−vr ) ¿ {εθz=

12 (1r ∂w

∂θ+∂v∂ z ) ¿¿¿¿

Hooke’s Law in 2D:

{σ rr=E(1+ν )(1−2ν ) [(1−ν )εrr+ν ( εθθ+ε zz )] ¿{σθθ=

E(1+ν )(1−2 ν ) [(1−ν )εθθ+ν (ε zz+ε rr )]¿ {σ zz=

E(1+ν )(1−2 ν ) [(1−ν )ε zz+ν ( εrr+εθθ) ]¿ ¿¿¿

Displacement MethodStep 1: Check the Boundary conditions:

At r=Ri : σ rr=−pi , σ rθ=0 , σ rz=0

At r=Ro : σ rr=−po , σ rθ=0 , σ rz=0Step 2 Analysis: The deformation is axisymmetric and under plane strain. So the deformation is independent of coordinate z and . Thus the circumferential and axial displacement v and w vanish, and displacements can be expressed as:

5

drdz

d

dV=rddrdzFrom cylindrical to Cartesiandx=drdy=rddz=dz

Fig. 4.9 Cylindrical coordinate system and stress tensor

po

pi

r

zr

Ri

Ro

Week 5

{u=u(r )¿ {v=0 ¿ ¿¿ ¿Step 3 Strain – Displacement relation:

{εrr=∂u∂r ¿ {εθθ=

1r∂ v∂θ

+ur=1

r0+u

r=u

r ¿ ¿¿¿

{εrθ=12 (∂ v

∂r+1

r∂u∂θ

−vr )=1

2r∂ u(r )∂ θ

=12 r

0=0 ¿ {εθz=12 (1r ∂w

∂θ+∂ v∂ z )=1

2 (1r

0+0)=0 ¿ ¿¿¿Step 4 Apply Hooke’s law:

{σrr=E(1+ν )(1−2 ν ) [(1−ν )∂u

∂r+ν (u

r+0 )]=E

(1+ν )(1−2ν ) [(1−ν )∂u∂r

+νur ] ¿ {σ θθ=

E(1+ν )(1−2ν ) [(1−ν )u

r+ν (0+∂u

∂r)]=E

(1+ν )(1−2ν ) [(1−ν )ur+ν∂u

∂ r ] ¿ {σ zz=E(1+ν )(1−2 ν ) [(1−ν )0+ν(∂u

∂ r+u

r)]=Eν

(1+ν )(1−2 ν ) [∂ u∂ r

+ur ] ¿ ¿¿¿

Step 5: Equilibrium Equations

{∂σ rr

∂r +1r∂σ rθ

∂θ +∂σ rz

∂ z +σrr−σθθ

r =0¿ {∂σ rθ

∂r +1r∂ σθθ

∂ θ +∂ σθz

∂ z +2σ rθ

r =∂0∂r +

1r∂σθθ (r )∂θ +

∂ 0∂ z +

2×0r =0 ¿ ¿¿¿

The second and third equations are satisfied automatically. The first equation is:∂σ rr

∂ r+ 1

r∂σ rθ

∂θ+∂σ rz

∂ z+

σrr−σθθ

r=∂ σrr

∂r+1

r∂ 0∂ θ

+∂ 0∂ z

+σ rr−σ θθ

r=∂ σrr

∂r+

σ rr−σ θθ

r=0

Substitution of Hooke’s law into the above equation of

∂σ rr

∂ r+

σrr−σθθ

r=0

∂∂ r {E(1+ν )(1−2 ν ) [(1−ν )∂ u

∂ r+νu

r ]}+1r {(E

(1+ν )(1−2ν ) [(1−ν )∂u∂ r

+ν ur ])−(E

(1+ν )(1−2 ν ) [(1−ν )ur+ν ∂u

∂r ])}=0

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Week 5

LHS=∂∂ r {[(1−ν )∂u

∂r+ν u

r ]}+1r {[(1−ν )∂u

∂r+ν u

r ]−[(1−ν )ur+ν ∂u

∂r ]}=∂∂r {[(1−ν )∂u

∂ r+ν (1r ) (u )]}+1

r {∂u∂r

−2ν∂ u∂ r

−ur+2 ν u

r }={(1−ν )∂

2 u∂ r2 +(−νu

r 2 +νr∂ u∂ r )}+{1r ∂u

∂r −2 νr

∂u∂ r −

ur 2 +2 ν u

r2 }¿(1−ν )∂

2u∂ r2 −ν u

r2 +νr∂u∂r

+1r∂u∂r

−2 νr

∂u∂r

−ur2 +2 νu

r 2

¿(1−ν )∂2u

∂ r2 +1r (1−ν )∂ u

∂ r −(1−ν )ur2 =0

Thus LHS=∂2 u

∂r2 +1r∂u∂ r

− ur2 =0

Step 6: Solve for this linear and static ordinary differential equationThus its solution can be assumed as (Displacement Method)

u=c1 r+c21r

(in which c1 and c2 are constants to be determined by using B.C.)Step 7: plug this trial function (solution) into the Strain – Displacement equations

{εrr=∂u∂r

=∂∂r [c1 r+c2

1r ]=c1−c2

1r2 ¿ {εθθ=

1r∂ v∂θ

+ur=u

r=1

r [c1 r+c21r ]=c1+c2

1r2 ¿ ¿¿¿

σ rr=E(1+ν )(1−2 ν ) [(1−ν )∂u

∂ r+ν u

r ]=E(1+ν )(1−2 ν ) [(1−ν )(c1−c2

1r 2 )+ν

r (c1r+c21r )]

¿ E(1+ν )(1−2 ν ) [c1−c2

1r2 −c1ν+c2

νr 2 +

νr

c1 r+c2νr2 ]

¿E(1+ν )(1−2 ν ) [c1−

c2

r2 (1−2ν )]=[Ec1

(1+ν )(1−2ν ) ]−1r2 [Ec2

(1+ν ) ]=A−Br2

Similarly, we can have: σ θθ=A+ B

r2

where A=

Ec1

(1+ν )(1−2 ν ), B=

Ec2

(1+ν ) . Now the question is how to determine A and B.Note that since we used all the equations, the solution should satisfy all the equations.

Step 8: Apply B.C. to determine the constants

{σrr (r=Ri )=A−BR i

2 =−pi ¿¿¿¿

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Week 5

which leads to: A=

pi Ri2−po Ro

2

Ro2−Ri

2 and

B=( pi−po )R i

2 Ro2

Ro2−Ri

2

From A and B we can calculate c1 and c2:

c1=(1+ν )(1−2 ν )

EA=

(1+ν )(1−2 ν )E ( pi Ri

2−po Ro2

Ro2−R i

2 )c2=

(1+ν )E

B=(1+ν )E ( ( pi− po ) Ri

2 Ro2

Ro2−Ri

2 )Step 9: Calculate all the functions

Displacements: {u=c1r+c2

1r=(1+ν )(1−2ν )E ( pi Ri

2−po Ro2

Ro2−Ri

2 )r+(1+ν )E ( ( pi−po) Ri

2Ro2

Ro2−Ri

2 )1r¿¿¿¿

Strains:

{εrr=c1−c21r2 =

(1+ν )(1−2 ν )E ( pi Ri

2−po Ro2

Ro2−Ri

2 )−(1+ν )E (( pi−po )R i

2 Ro2

Ro2−Ri

2 )1r2 ¿ {εθθ=c1r+c21r2 =

(1+ν )(1−2 ν )E ( pi Ri

2−po Ro2

Ro2−R i

2 )+(1+ν )E ( ( pi− po) Ri

2 Ro2

Ro2−Ri

2 )1r2 ¿ ¿¿¿

Stresses:

{σrr (r )=A−Br2 =

pi Ri2−po Ro

2

Ro2−R i

2 −( pi−po )R i

2 Ro2

Ro2−Ri

2 (1r2 ) ¿ {σθθ(r )=A+Br2 =

pi Ri2−po Ro

2

Ro2−R i

2 +( p i− po) Ri

2 Ro2

Ro2−Ri

2 (1r2 ) ¿{σ zz (r )=2ν (pi Ri

2−po Ro2

Ro2−R i

2 ) ¿¿¿¿Plane Stress Problem:

Replacing E and by E∗¿ E (1+2 ν )

(1+ν )2 and ν∗¿ ν

(1+ν ) , we can further obtain the solution to the corresponding plane stress problems.

up−strain=(1+ν∗)(1−2 ν∗)

E∗¿ ( pi Ri2−po Ro

2

Ro2−Ri

2 )r+ (1+ν∗)E∗¿ ( ( p i−po) Ri

2 Ro2

Ro2−Ri

2 ) 1r¿¿

up−stress=

[1+( ν1+ν )] [1−2( ν

1+ν )]( E(1+2 ν )(1+ν )2 ) ( pi Ri

2−po Ro2

Ro2−R i

2 )r+ [1+( ν1+ν )]

( E(1+2 ν )(1+ν )2 )(

( p i−po ) Ri2 Ro

2

Ro2−Ri

2 ) 1r

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Week 5

Plant stress

u=[ 1+2 ν

1+ν ][ 1−ν1+ν ]

( E(1+2 ν )(1+ν )2 ) ( pi Ri

2−po Ro2

Ro2−Ri

2 )r+[ 1+2 ν

1+ν ](E (1+2 ν )

(1+ν )2 ) (( pi−po )R i

2 Ro2

Ro2−R i

2 ) 1r

Plane strain Plane stress

Fig. 4.10 Pressurised cylinder with plane strain and plane stress

Displacement: {u=(1−ν )

E ( pi Ri2−po Ro

2

Ro2−R i

2 )r+(1+ν )E ( ( pi−po) Ri

2 Ro2

Ro2−Ri

2 )1r ¿ {v=0 , ¿¿¿¿

Strains:

{εrr=(1−ν )E ( p i R i

2−po Ro2

Ro2−Ri

2 )−(1+ν )E (( p i−po) Ri

2 Ro2

Ro2−Ri

2 )1r2 ¿ {εθθ=(1−ν )E (pi Ri

2− po Ro2

Ro2−Ri

2 )+(1+ν )E (( p i− po) Ri

2 Ro2

Ro2−Ri

2 )1r2 ¿ ¿¿¿

Stresses:

{σ rr(r )=pi Ri

2− po Ro2

Ro2−R i

2 −( pi−po )R i

2 Ro2

Ro2−Ri

2 (1r2 )¿ {σθθ (r )=p i R i

2−po Ro2

Ro2−Ri

2 +( pi−po ) Ri

2 Ro2

Ro2−R i

2 (1r 2 )¿ ¿¿¿Remarks:

σ rr , σθθ are independent on material properties. The cylinder made of any materials will have the same stress values and thus if strength is the major concern, one should select the highest strength material.

However, the displacement and strains are dependent on material properties. If the stiffness is the main concern, a higher E modulus material should be chosen.

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Week 5

When po=0 , one have {σrr (r )=

pi Ri2

Ro2−R i

2 −p i R i

2 Ro2

Ro2−Ri

2 (1r2 ) ¿ ¿¿¿Since r≤Ro , the radial stress σ rr≤0 (always negative) and σ θθ≥0 (always positive).

Thus: σ θθ≤σ zz=0≤σrr . As all shear stresses are zero, thus the principal stresses are: σ 1=σθθ , σ2=σ zz=0 , σ 3=σrr

4.9 Saint-Venant Principle In the cantilever beam problem, some observed some difference of stress contours as shown in Fig. 4.11.

Saint Venant observed that in pure bending of a beam conforms a rigorous solution only when the external forces applied at the ends of beams are distributed over the end is the same as internal stress distribution, i.e. linear distribution.

Saint Venant Principle:If the force acting on a small portion of the surface of an elastic body are replaced by another statically equivalent system of forces acting on the same portion of the surface, such redistribution of loading produces substantial change in stress locally but has a negligible effect on the stress at a distances which are large in comparison with a linear dimensions of the surface on which the force are changed”.

Two key assumptions:(1) very small loading area compared with the whole dimension. The affected area will

be much smaller than the unaffected area Aunaffected>>Aaffected. e.g in the tensile bar as shown in Fig 4.12, L>>a, in which the affected area will take roughly: za.

(2) Force replaced must be statically equivalent. The replacement must not change either the resultant force or resultant couple.

For example the slender bar is stretched in different ways as below, where one can approximately define the affected and unaffected areas.

Tensile testIn the tensile test, the way of holding a specimen has no effect on the stress and deformation in the middle region of the specimen. In test code requires a sufficient length of the specimen to avoid the end effect on the testing result. It is an application of Saint-Venant’s principle.

10

x

y

0

L

P

FEA

Theoretical solutionFig. 4.11 Saint Venant Principles

Affectedzone

Affectedzone

unaffected zone

a

L

z z

Fig. 4.12 Affected and unaffected areas

Week 5

F

F

Y

Elastic Yielding Hardening Necking

StandardSpecimen Yield stress

Ultimatestress u

Fracturestress f

Plastic behavior

Elasticbehavior

plProportionallimit

Four-point bendingThe better positioning of strain gauge should be in a far field as shown below to get more stable and reliable testing result.

M(x)

Betterposition

worseposition

Strain gauge testing

Affect zone

x

Cantilever beam in FEA The end force can be applied in different way, which only affects a small area as shown.

-1000

-1000

-2000-2000N/m

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