V

Finally, the displacement components can be obtained by integrating the above equations

10.2.2.3 Example: Thick-Walled Cylinder

4i If we consider a circular cylinder with internal and external radii a and b respectively, subjected to internal and external pressures p^ and po respectively, Fig. 10.2, then the boundary conditions for the plane strain problem are

Saint Venant rrrrrrrm m m M m

Figure 10.2: Pressurized Thick Tube

42 These Boundary conditions can be easily shown to be satisfied by the following stress field

These equations are taken from Eq. 10.50, 10.51 and 10.52 with B = 0 and therefore represent a possible state of stress for the plane strain problem.

43 We note that if we take B =0, then uo = Ar'EB (1 — v2) and this is not acceptable because if we were to start at 0 = 0 and trace a curve around the origin and return to the same point, than 0 = 2n and the displacement would then be different.

44 Applying the boundary condition we find that

45 We note that if only the internal pressure p, is acting, then Trr is always a compressive stress, and Too is always positive.

46 If the cylinder is thick, then the strains are given by Eq. 10.53, 10.54 and 10.55. For a very thin r cylinder in the axial direction, then the strains will be given by

Err =

Î = ETrr -

(10.63-a)

Eee =

U = -f(T00 - vTrr) r E

(10.63-b)

Ezz =

Tz = V (Trr + Tee)

(10.63-c)

E Tr6

(10.63-d)

47 It should be noted that applying Saint-Venant's principle the above solution is only valid away from the ends of the cylinder.

10.2.2.4 Example: Hollow Sphere

48 We consider next a hollow sphere with internal and xternal radii ai and ao respectively, and subjected to internal and external pressures of pi and po, Fig. 10.3.

Figure 10.3: Pressurized Hollow Sphere

49 With respect to the spherical ccordinates (r,9,4>), it is clear due to the spherical symmetry of the geometry and the loading that each particle of the elastic sphere will expereince only a radial displacement whose magnitude depends on r only, that is ur = ur (r), uq = up = 0 (10.64)

10.2.2.5 Example: Stress Concentration due to a Circular Hole in a Plate

50 Analysing the infinite plate under uniform tension with a circular hole of diameter a, and subjected to a uniform stress ao, Fig. 10.4.

51 The peculiarity of this problem is that the far-field boundary conditions are better expressed in cartesian coordinates, whereas the ones around the hole should be written in polar coordinate system.

52 First we select a stress function which satisfies the biharmonic Equation (Eq. 10.23), and the far-field boundary conditions. From St Venant principle, away from the hole, the boundary conditions are given by:

Recalling (Eq. 10.19) that Txx = dyr, this would would suggest a stress function $ of the form $ = a0y2. Alternatively, the presence of the circular hole would suggest a polar representation of $. Thus, substituting y = r sin 9 would result in $ = a0r2 sin2 9.

Circular Hole Infinite Plate
Figure 10.4: Circular Hole in an Infinite Plate

53 Since sin2 9 = 2 (1 — cos 20), we could simplify the stress function into

Substituting this function into the biharmonic equation (Eq. 10.46) yields d 2

dr dr2

54 The general solution of this ordinary linear fourth order differential equation is f (r) = Ar2 + Br4 + C -1 + D

thus the stress function becomes

Using Eq. 10.41-10.43, the stresses are given by

1 d$ 1 32$ f 6C 4D\ Trr = ^ + ^ = — ^^ ^^ ) cos 29

55 Next we seek to solve for the four constants of integration by applying the boundary conditions. We will identify two sets of boundary conditions:

1. Outer boundaries: around an infinitely large circle of radius b inside a plate subjected to uniform stress oo, the stresses in polar coordinates are obtained from Eq. 9.35

Trr Tr0

TrQ Tqq

yielding (recalling that sin2 9 = 1/2sin29, and cos2 9 = 1/2(1 + cos29)).

(Trr)r=b = o0 cos2 9 =1 o0(1 + cos29) (Tre)r=b = 1 oo sin 29 (Tee)r=b = o0 (1 — cos 29)

For reasons which will become apparent later, it is more convenient to decompose the state of stress given by Eq. 10.72-a and 10.72-b, into state I and II:

(TroYrU

Where state I corresponds to a thick cylinder with external pressure applied on r = b and of magnitude a0/2. This problem has already been previously solved. Hence, only the last two equations will provide us with boundary conditions.

2. Around the hole: the stresses should be equal to zero:

56 Upon substitution in Eq. 10.70-a the four boundary conditions (Eq. 10.73-c, 10.73-d, 10.74-a, and 10.74-b) become

6C 4D\

1

(10.75-a)

"b4 + H2 )

= 2 a°

6C 2D)

1

(10.75-b)

- H2 )

= 2 a°

6C 4D)

(10.75-c)

a4 + a2 )

=0

6C 2D)

(10.75-d)

a4 - H2)

=0

57 Solving for the four unknowns, and taking a =0 (i.e. an infinite plate), we obtain:

58 To this solution, we must superimpose the one of a thick cylinder subjected to a uniform radial traction a0/2 on the outer surface, and with b much greater than a. These stresses were derived in Eqs. 10.60 and 10.61 yielding for this problem (carefull about the sign)

Thus, upon substitution into Eq. 10.70-a, we obtain

0 0

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