# Høgskolen i Narvik - hovedside

Høgskolen i Narvik - hovedside

Høgskolen i Narvik

Sivilingeniørutdanningen

Eksamen i Faget STE6236 ELASTISITETSTEORI

Klasse: 4.ID

Dato: 26.02.2010

Tid: Kl. 09.00–12.00

Tillatte hjelpemidler under eksamen:

Kalkulator

Kopi av Boken “Mechanics of Aircraft Structures”

Kopi av transparenter “Elastisitetsteori”

Forelesningsnotater

Engelsk/Norsk, Norsk/Engelsk ordbok

Faglig kontaktperson under eksamen:

Professor II Gregory A. Chechkin

tel. 76 96 61 06

———————————————————————————

Narvik 2010

1. The state of stress in a wedge (see Figure 1) is uniform

and given by

σxx = 2 MPa τxy = 1 MPa τxz = 0 MPa

σyy = 2 MPa τyz = 0 MPa σzz = 5 MPa .

Assume that α = 45 ◦ and β = 45 ◦ .

Figure 1: Shape of a pyramid.

a) Find the three components of the stress vector t on the

inclined surface.

Solution. The unit normal to the inclined surface is

−→ 2

n = 0,

2 ,

2

2

and respectively

−→

t = ⎝

2 1 0

1 2 0

0 0 5

⎞⎛

⎠⎝

0

√ 2

2√ 2

2

⎠ =

√ 2

2

√ 2

5 √ 2

2

⎠.

b) Find the normal component σn of the stress vector on the

inclined surface.

Solution. By the definition

σn =

−→t

,

−→

n = 1 + 10 7

=

4 2 .

2

c) Find the principal stresses and the respective principal

directions for the stresses.

Solution. To find principal stresses we consider the equation

(2 − σ)

1

0

1

(2 − σ)

0

0

0

= 0

(5 − σ)

or equivalently

Hence,

and

and

and

For σ1 = 5 we have

⎪⎨

⎪⎩

For σ2 = 3 we have

⎪⎨

⎪⎩

For σ3 = 1 we have

⎪⎨

(1 − σ)(3 − σ)(5 − σ) = 0.

σ1 = 5, σ2 = 3, σ3 = 1.

−3n (1)

x + n (1)

y

⎪⎩

n (1)

x − 3n (1)

y

−n (2)

x

−→

n (1) = (0; 0; 1).

+ n (2)

y

n (2)

x − n (2)

y

n (3)

x

n (3)

x

0n (1)

z

2n (2)

z

−→

n (2)

1

= √2 ; 1

√ ; 0 .

2

+ n (3)

y

+ n (3)

y

4n (3)

z

−→

n (3)

1√2

= ; − 1

√ ; 0 .

2

3

= 0

= 0

= 0

= 0

= 0

= 0

= 0

= 0

= 0

Task 2. Calculate the strains and the stresses in an isotropic

body, if the displacements are

Solution. We have

u = x

z

v = x

y

w = y

z

cos (xy)

sin (yz)

cos (zx)

εxx = 1 xy

cos (xy) −

z z sin (xy), εyy = − x xz

sin (yz) +

y2 y

cos (yz),

εzz = y xy

cos (zx) −

z2 z sin (zx), γxy = − x2 1

sin (xy) + sin (yz),

z y

γxz = − x

z2 cos (xy) − y sin (zx), γyz = x cos (yz) + 1

cos (zx)

z

and respectively

E(1 − ν) 1 xy

σxx = cos (xy) − sin (xy) +

(1 + ν)(1 − 2ν) z z

+

(1 + ν)(1 − 2ν)

x

xz y xy

sin (yz) + cos (yz) + cos (zx) − sin (zx) ,

y2 y z2 z

E(1 − ν)

σyy = −

(1 + ν)(1 − 2ν)

x

xz

sin (yz) + cos (yz) +

y2 y

Eν 1 xy y xy

+

cos (xy) − sin (xy) + cos (zx) − sin (zx) ,

(1 + ν)(1 − 2ν) z z z2 z

E(1 − ν)

y xy

σzz = cos (zx) − sin (zx) +

(1 + ν)(1 − 2ν) z2 z

Eν 1 xy x xz

+

cos (xy) − sin (xy) − sin (yz) + cos (yz) ,

(1 + ν)(1 − 2ν) z z y2 y

τxy = G − x2

1

x

sin (xy) + sin (yz) , τxz = −G cos (xy) + y sin (zx) ,

z y z2

τyz = G x cos (yz) + 1

cos (zx) .

z

4

in Figure 2. Show that the Airy stress function

φ(x, y) = c1x 2 + c2xy + c3y 2

solves the problem. Find the constants c1, c2, c3.

Figure 2: Thin panel subjected to uniform loads.

Solution. The Airy stress function given by the formula, satisfies the equation

∆ 2 φ = 0

or

By the definition

consequently,

∂4φ ∂x4 + 2 ∂4φ ∂x2∂y2 + ∂4φ ∂y

4 = 0.

σxx = ∂2 φ

∂y 2 , σyy = ∂2 φ

∂x 2 , τxy = − ∂2 φ

∂x∂y ,

σxx = 2c3, σyy = 2c1, τxy = −c2.

It remains to consider the boundary conditions.

On I we have

−→ 1

n = ( √2 , 1 √ ),

2

−→ σ0

t = ( √2 , σ0

√2 ).

On II we have

−→ n = (0, 1),

5

−→ t = (σ0, 0).

On III we have

On IV we have

Hence, by the formula

tx

we deduce

−→ n = (− 1

√2 , 1 √ 2 ),

−→ n = (0, −1),

ty

=

σxx τxy

τxy σyy

−→ t = ( σ0

√2 , − σ0

√2 ).

−→ t = (−σ0, 0).

nx

ny

σxx = 0, σyy = 0, τxy = σ0.

and the respective Airy stress function is

φ(x, y) = −σ0xy.

Task 4. Consider two-point boundary value problem for

Bernoulli–Euler equation:

u (IV ) (x) = 1 for 0 < x < 3

u(0) = 0; u ′ (0) = 0;

u ′′ (3) = 1; u ′′′ (3) = 1.

Show that the solution u is also the solution of a variational

problem. Define the functional space (space formed by the set

of appropriate admissible functions) and derive the integral identity.

Solution. Let us introduce the set of admissible functions in the following

way:

V = {v ∈ C 2 [0, 3]; v(0) = 0, v ′ (0) = 0}.

Multiplying the equation by the test-function v ∈ V , integrating over [0, 3],

we obtain

3

u (IV ) 3

(x) v(x) dx = v(x) dx,

0

6

0

and finally integrating the left–hand side by parts twice and using the fact

that v(0) = 0, v ′ (0) = 0 and u ′′ (3) = 1, u ′′′ (3) = 1, we deduce

3

0

u (IV ) (x) v(x) dx = −

=

3

0

0

3

u ′′′ (x)v ′ (x) dx + u ′′′ (3)v(3) − u ′′′ (0)v(0) =

u ′′ (x)v ′′ (x) dx − u ′′ (3)v ′ (3) + u ′′ (0)v ′ (0) + v(3) =

=

3

which is valid for any function v ∈ V .

The variational formulation is:

Find u ∈ V such that

3

0

0

u ′′ (x)v ′′ (x) dx =

u ′′ (x)v ′′ (x) dx − v ′ (3) + v(3),

3

0

v(x) dx + v ′ (3) − v(3) ∀ v ∈ V.

The formulation of a minimization problem is:

Find u ∈ V such that F(u) ≤ F(v) for any v ∈ V , where

F(v) = 1

3

2

0

(v ′′ (x)) 2 dx −

7

3

0

v(x) dx − v ′ (3) + v(3).

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