Cee235b 2014 ExampleProblems02 (2)

7/23/2019 Cee235b 2014 ExampleProblems02 (2)

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cee235b: Finite Element Analysis of Structures

Example Problems #2 Prof. E. Taciroglu

Problem 1. The bulk modulus Kis defined as the ratio of the volumetric stress to the volumetricstrain. Using stress-strain relationship given below compute Kas function of the Youngs modulusCand Poissons ratio .

Sij = C

(1 +)(1 2)Ekkij+

C

1 +Eij (1)

Answer: The volumetric stress Sv and volumetric strain Ev are defined as

Sv =1

3tr(S) =

Sii

3 =

S11+S22+S333

Ev = tr(E) = Eii= E11+E22+E33

We can compute Sv as a function ofEv using (1).

Sii = Ekkii+ 2Eii= (3+ 2)Eii (note: i and k are dummy indices)

= C

(1 +)(1 2), and =

C

2(1 +)

Sv = (+2

3 )Eii= (+

2

3 )Ev

K= Sv

Ev=+

2

3

Problem 2. The strain-stress relationship for a three dimensional isotropic, linearly elastic materialcan be expressed in the form

Eij = 12

(Sij 1 +

Skkij)

where the subscripts i, j, and k range over the values 1, 2, and 3. Find equivalent expressions forthe strain-stress relationship that already reflect the plane strain condition E33 = E13 = E23 = 0.

Answer: To find equivalent strain-stress relationship for the plane strain condition, use E33 = 0to get S33.

E33 = 1

2(S33

1 +(S11+S22+S33)) = 0

S33 = (S11+S22)

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Now, we can compute E11, E22, and E12 as functions ofS11, S22, and S22.

E11 = 1

2(S11 (S11+S22))

E22 =

1

2 (S22 (S11+S22))

E12 = 1

2S12

E = 1

2(S S) for , ,= 1, 2 (1)

Problem 3. Arrange the stress-strain relationship for the plane strain condition in the followingform

S11S22S12

=

D11 D12 D13D21 D22 D23D31 D32 D33

E11E22E12

Answer: stress-strain relationships for three dimensional isotropic, linearly elastic material is

Sijei ej =Ekkijei ej+ 2Eijei ej for i ,j,k= 1, 2, 3

For the plane strain condition, E33 = E23 = E13 = 0. As a result

S11 = (E11+E22) + 2E11 = (+ 2)E11+E22

S22 = (E11+E22) + 2E22 = E11+ (+ 2)E22

S12 = 2E12

Arranging the above equations in the matrix form

S11S22S12

=

+ 2 0 + 2 00 0 2

E11E22E12

Problem 4. Consider the displacement map u(z) for a sphere of unit radius, given by the explicitexpression

u(z) = (z.z)z

where is a very small constant of the motion. Assume that the material is isotropic and linearlyelastic with material constantsand. Compute the body forcebrequired to maintain equilibrium.Compute the traction forces tthat must be acting on the surface of the sphere.

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Answer:

Compute the body force b required to maintain equilibrium.

u=

zj (z

kzk

zi)

ei e

j

=(2zkkjzi+zkzkij)ei ej

=(2zjzi+zkzkij)ei ej

=(2z z + (z.z)I)

The linearized strain tensor can be given by

E=1

2(u + (u)T) = (2z z + (z.z)I)

tr(E) = (2zizi+zkzkii) = 5(z.z)

The stress tensor can be obtained from the strain tensor as

S= tr(E)I + 2E= 5(z.z)I + 2(2z z + (z.z)I) = (5+ 2)(z.z)I + 4(z z)

The body force required for equilibrium is

b= divS= .S=

zj((5+ 2)(zkzk)ij+ 4zizj)ei

= (2(5+ 2)zkkjij+ 4ijzj+ 4jjzi)ei

= (10+ 20)z

Compute the traction forces t that must be acting on the surface of the sphere. Note thatthe radial vector z is normal to the surface of the sphere and is of unit length since the radiusof the sphere is 1.

Sz= (5+ 2)(z.z)z + 4(z z)z= (5+ 6)(z.z)z

Problem 5. Consider the rod of unit length and modulus C(x) that varies as shown. The rodis fixed at the left end, is free at the right end, and is subjected to a linearly varying body forceb(x) as shown. Consider the following displacement map: u(x) = a(x3 + 2x2 3x) where a is someconstant. Is the displacement map a solution to the given problem? Why or why not?

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Answer: For the given map to be a solution to the problem, it has to satisfy the governingdifferential equation and boundary condition.

(Cu(x)) +b(x) = 0, u(0) = 0, Cu(1) = 0

Substituting,

u(x) = a(3x2 + 4x 3), u(x) = a(6x+ 4)

(Cu(x)) +b(x) = C u(x) +Cu(x) +b(x)

=a(2 x2)(6x+ 4) +a(2x)(3x2 + 4x 3) + 15x

=a(12x3 12x2 + 18x+ 8) + 15x = 0

u(0) = 0, Cu(1) = 4a = 0

Therefore, the given map is not a solution to the problem.

Problem 6. Consider the rod of length l = 1 and constant modulus C = 1. The rod is re-strained by an elastic spring of modulus k = 1 at each end and rests on an elastic foundation,also with modulus k = 1. The rod is subjected to a quadratically varying body force as shown.The displacement u(x), positive in x direction, is governed by the following differential equationuu= 1x2. What are the boundary conditions for this problem? Is the following displacementfunction a classical solution to this problem?

u(x) = 1 +x2 2ex 1

2ex

Why or why not? If it is not then modify it so that it is.

Answer:

What are the boundary conditions for this problem?

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Is the given displacement function a classical solution to this problem? The solution has tosatisfy the differential equation and boundary conditions.

u(x) = 2x 2ex +1

2ex, u(x) = 2 2ex

1

2ex

Substituting in the differential equation

u(x) u(x) = (2 + 2ex 1

2ex) (1 +x2 2ex

1

2ex) = 1 x2

Therefore, the solution satisfies the differential equation for all values of x. Checking theboundary conditions:

u(0) u(0) = (2 +1

2) (1 2

1

2) = 0 (OK)

u(1) +u(1) = (2 2e+ 1

2e) (2 2e

1

2e) = 4(1 e) = 0

The boundary condition at x = 1 is not satisfied. Thus the given function is not a solution

to the problem. We can modify the function to

u(x) = 1 +x2 2

eex

1

2ex

u(x) = 2x2

eex +

1

2ex, u(x) = 2

2

eex

1

2ex

u(x) u(x) = 1 x2

u(0) u(0) = 0

u(1) +u(1) = 0

The modified function is the exact solution to the problem.

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Cee235b 2014 ExampleProblems02 (2)

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