Elasticity Part 6 ppsx

8 Two-Dimensional Problem SolutionThe previous chapter developed the general formulation for the plane problem in elasticity.This formulation results in two types of in-plane problems—pl

8 Two-Dimensional Problem Solution

The previous chapter developed the general formulation for the plane problem in elasticity.This formulation results in two types of in-plane problems—plane strain and plane stress Itwas further shown that solution to each of these problem types could be conveniently handledusing the Airy stress function approach This scheme reduces the field equations to a singlepartial differential equation, and for the case of zero body forces, this result was the biharmo-nic equation Thus, the plane elasticity problem was reduced to finding the solution to thebiharmonic equation in a particular domain of interest Such a solution must also satisfy thegiven boundary conditions associated with the particular problem under study Several generalsolution techniques were briefly discussed in Section 5.7 These include the use of power series

or polynomials and Fourier methods We now pursue the solution to several two-dimensionalproblems using these methods Our formulation and solution is conducted using both Cartesianand polar coordinate systems In many cases we use MATLAB software to plot the stress anddisplacement field distributions in order to better understand the nature of the solution Planeproblems can also be solved using complex variable theory, and this powerful method isdiscussed in Chapter 10

We begin the solution to plane elasticity problems with no body forces by consideringproblems formulated in Cartesian coordinates When taking boundary conditions into account,this formulation is most useful for problems with rectangular domains The method is based onthe inverse solution concept where we assume a form of the solution to the biharmonicequation

f(x, y)¼X1

m¼0

X1 n¼0

Terms withmþ n  3 automatically satisfy the biharmonic equation (8.1.1) for any choice

of constantsAmn However, for higher-order terms withmþ n > 3, the constants Amnmust berelated in order to have the polynomial satisfy the biharmonic equation For example, thefourth-order polynomial termsA40x4þ A22x2y2þ A04y4will not satisfy the biharmonic equa-tion unless 3A40þ A22þ 3A04¼ 0 This condition specifies one constant in terms of the othertwo, thus leaving two constants to be determined by the boundary conditions

Considering the general case, substituting the series form (8.1.2) into the governingbiharmonic equation (8.1.1) yields

X1 m¼4

X1 n¼0

m(m
1)(m
2)(m
3)Amnxm
4yn

þ 2X1 m¼2

X1 n¼2

m(m
1)n(n
1)Amnxm
2yn
2

þX1 m¼0

X1 n¼4

n(n
1)(n
2)(n
3)Amnxmyn
4¼ 0

(8:1:4)

Collecting like powers ofx and y, the preceding equation may be written as

X1 m¼2

X1 n¼2

vented by modifying boundary conditions on the problem using the Saint Venant principle This

is accomplished by replacing a complicated nonpolynomial boundary condition with a staticallyequivalent polynomial condition The solution to the modified problem would then be accurate

at points sufficiently far away from the boundary where adjustments were made Normally, thismethod has applications to problems of rectangular shape in which one dimension is muchlarger than the other This would include a variety of beam problems, and we shall now considerthree such examples Solutions to each of these problems are made under plane stress condi-tions The corresponding plane strain solutions can easily be determined by using the simplechange in elastic constants given in Table 7-1 Of course, for the case with zero body forces andtraction boundary conditions, the stress fields will be identical in either theory

EXAMPLE 8-1: Uniaxial Tension of a Beam

As a simple example, consider the two-dimensional plane stress case of a long lar beam under uniform tensionT at each end, as shown in Figure 8-1 This problemcould be considered the Saint Venant approximation to the more general case withnonuniformly distributed tensile forces at the endsx¼ l For such an interpretation,the actual boundary conditions are replaced by the statically equivalent uniform distri-bution, and the solution to be developed will be valid at points away from these ends.The boundary conditions on this problem may be written as

rectangu-sx( l, y) ¼ T, sy(x, c) ¼ 0

txy( l, y) ¼ txy(x, c) ¼ 0 (8:1:7)These conditions should be carefully verified by making reference to Figure 5-3.Because the boundary conditions specify constant stresses on each of the beam’sboundaries, we are motivated to try a second-order stress function of the form

and this gives the following constant stress field:

sx¼ 2A02, sy¼ txy¼ 0 (8:1:9)The first boundary condition (8.1.7) implies that A02¼ T=2 and all other boundaryconditions are identically satisfied Therefore, the stress field solution to this problem isgiven by

Continued

x

y

T T

2l 2c

FIGURE 8-1 Uniaxial tension problem.

EXAMPLE 8-1: Uniaxial Tension of a Beam–Cont’d

sx¼ T, sy¼ txy¼ 0 (8:1:10)Next we wish to determine the displacement field associated with this stress distribution.This is accomplished by a standard procedural technique First, the strain field iscalculated using Hooke’s law Then the strain-displacement relations are used todetermine various displacement gradients, and these expressions are integrated to findthe individual displacements Using this scheme, the in-plane displacement gradientsare found to be

@u

@yþ@v

@x¼ 2exy¼tx y

m ¼ 0 ) f0(y)þ g0(x)¼ 0 (8:1:13)This result can be separated into two independent relationsg0(x)¼
f0(y)¼ constantand integrated to get

do not contribute to the strain or stress fields Thus,the displacements are determinedfrom the strain field only up to an arbitrary rigid-body motion Additional boundaryconditions on the displacements are needed to determine these terms explicitly Forexample, if we agree that the center of the beam does not move and thex-axis does notrotate, all rigid-body terms will vanish andf ¼ g ¼ 0

EXAMPLE 8-2: Pure Bending of a Beam

As a second plane stress example, consider the case of a straight beam subjected to endmoments as shown in Figure 8-2 The exact pointwise loading on the ends is notconsidered, and only the statically equivalent effect is modeled Hence, the boundaryconditions on this problem are written as

The choice of stress function is based on the fact that a third-order function will giverise to a linear stress field, and a particular linear boundary loading on the endsx¼ lwill reduce to a pure moment Based on these two concepts, we choose

f¼ A03y3 (8:1:16)and the resulting stress field takes the form

sx¼ 6A03y, sy¼ txy¼ 0 (8:1:17)This field automatically satisfies the boundary conditions ony¼ c and gives zero netforces at the ends of the beam The remaining moment conditions atx¼ l are satisfied

ifA03¼
M=4c3, and thus the stress field is determined as

sx¼
3M2c3y, sy¼ txy¼ 0 (8:1:18)The displacements are again calculated in the same fashion as in the previous example.Assuming plane stress, Hooke’s law will give the strain field, which is then substitutedinto the strain-displacement relations and integrated yielding the result

@u

@x¼
3M2Ec3y) u ¼
3M

2Ec3xyþ f (y)

@v

@y¼ n 3M2Ec3y) v ¼ 3Mn

2l 2c

FIGURE 8-2 Beam under end moments.

EXAMPLE 8-2: Pure Bending of a Beam–Cont’d

where f and g are arbitrary functions of integration Using the shear stress-strainrelations

@u

@yþ@v

@x¼ 0 )
3M

2Ec3xþ f0(y)þ g0(x)¼ 0 (8:1:20)This result can again be separated into two independent relations inx and y, and uponintegration the arbitrary functionsf and g are determined as

f (y)¼
!oyþ uo

g(x)¼ 3M4Ec3x2þ !oxþ vo

(8:1:21)

Again, rigid-body motion terms are brought out during the integration process For thisproblem, the beam would normally be simply supported, and thus the displacementconditions could be specified asv( l, 0) ¼ 0 and u(
l, 0) ¼ 0 This specification leads

to determination of the rigid-body terms asuo¼ !o¼ 0, vo¼
3Ml2=4Ec3

We now wish to compare this elasticity solution with that developed by elementarystrength of materials (mechanics of materials) Introducing the cross-sectional areamoment of inertia I¼ 2c3=3 (assuming unit thickness), our stress and displacementfield can be written as

Comparing these two solutions, it is observed that they are identical, with the exception

of thex displacements In general, however, the two theories will not match for otherbeam problems with more complicated loadings, and we investigate such a problem inthe next example

EXAMPLE 8-3: Bending of a Beam by Uniform

Transverse Loading

Our final example is that of a beam carrying a uniformly distributed transverseloadingw along its top surface, as shown in Figure 8-3 Again, plane stress conditionsare chosen, and we relax the boundary conditions on the ends and consider onlystatically equivalent effects Exact pointwise boundary conditions will be specified

on the top and bottom surfaces, while at the ends the resultant horizontal forceand moment are set to zero and the resultant vertical force will be specified tosatisfy overall equilibrium Thus, the boundary conditions on this problem can bewritten as

Again, it is suggested that these conditions be verified, especially the last statement

Using the polynomial solution format, we choose a trial Airy stress function ing second-, third-, and fifth-order terms:

EXAMPLE 8-3: Bending of a Beam by Uniform

Transverse Loading–Cont’d

Applying the first three boundary conditions in the set (8.1.24) gives three equationsamong the unknown coefficientsA20, A21, andA23 Solving this system determines theseconstants, giving the result

A20¼
w

4, A21¼3w

8c, A23¼
w

8c3 (8:1:27)Using these results, it is found that the stress field will now also satisfy the fourth and sixthconditions in (8.1.24) The remaining condition of vanishing end moments gives thefollowing

A03¼
A23(l2
2

5c

2)¼w8c

l2

c2
25

l2

c2
25

y
3w4c3 x2y
2

txy¼
3w4cxþ3w4c3xy2

(8:1:29)

We again wish to compare this elasticity solution with that developed by elementarystrength of materials, and thus the elasticity stress field is rewritten in terms of the cross-sectional area moment of inertiaI¼ 2c3=3, as

sx¼ w2I(l

sy¼
w2I

2
x2)y

sy¼ 0

txy¼VQ

It ¼
w2Ix(c

2
y2)

(8:1:31)

where the bending momentM¼ w(l2
x2)=2, the shear forceV¼
wx, the first moment

of a sectioned cross-sectional area isQ¼ (c2
y2)=2, and the thicknesst is taken as unity.Comparing the two theories, we see that the shear stresses are identical, while the twonormal stresses are not The two normal stress distributions are plotted in Figures 8-4

EXAMPLE 8-3: Bending of a Beam by Uniform

Transverse Loading–Cont’d

and 8-5 The normalized bending stress sx for the casex¼ 0 is shown in Figure 8-4.Note that the elementary theory predicts linear variation, while the elasticity solutionindicates nonlinear behavior The maximum difference between the two theories exists

at the outer fibers (top and bottom) of the beam, and the actual difference in the stressvalues is simplyw/5, a result independent of the beam dimensions For most commonbeam problems wherel >> c, the bending stresses will be much greater than w, and thusthe differences between elasticity theory and strength of materials will be relativelysmall For example, the set of curves in Figure 8-4 for l=c¼ 4 gives a maximumdifference of about only 1 percent Figure 8-5 illustrates the behavior of the stress sy,and the maximum difference between the two theories is given byw and this occurs at

-1 0.1

FIGURE 8-5 Comparison of s y stress in beam Example 8-3.

EXAMPLE 8-3: Bending of a Beam by Uniform

Transverse Loading–Cont’d

the top of the beam Again, this difference will be negligibly small for most beamproblems wherel >> c These results are generally true for beam problems with othertransverse loadings That is, for the case withl >> c, approximate bending stressesdetermined from strength of materials will generally closely match those developedfrom theory of elasticity

Next let us determine the displacement field for this problem As in theprevious examples, the displacements are developed through integration of the strain-displacement relations Integrating the first two normal strain-displacement relationsgives the result

u¼ w2EI[(l

4
w4EI[l

u¼ w2EI[(l

EXAMPLE 8-3: Bending of a Beam by Uniform

c2

l2, and this term is caused by the presence of the shear force.For beams wherel >> c, this difference is very small Thus, we again find that for longbeams, strength of materials predictions match closely to theory of elasticity results.Note from equation (8.1.36), the x component of displacement indicates that planesections undergo nonlinear deformation and do not remain plane It can also be shownthat the Euler-Bernoulli relationM¼ EId

2v(x, 0)

dx2 used in strength of materials theory isnot satisfied by this elasticity solution Timoshenko and Goodier (1970) provide add-itional discussion on such differences

Additional rectangular beam problems of this type with different support and loading conditionscan be solved using various polynomial combinations Several of these are given in the exercises

A more general solution scheme for the biharmonic equation may be found by usingFouriermethods Such techniques generally use separation of variables along with Fourier series orFourier integrals Use of this method began over a century ago, and the work of Pickett (1944),Timoshenko and Goodier (1970), and Little (1973) provide details on the technique

In Cartesian coordinates, the method may be initiated by looking for an Airy stress function

of the separable form

f(x, y)¼ X(x)Y(y) (8:2:1)Although the functionsX and Y could be left somewhat general, the solution is obtained moredirectly if exponential forms are chosen asX¼ eax, Y¼ eby Substituting these results into thebiharmonic equation (8.1.1) gives

(a4þ 2a2b2þ b4)eaxeby¼ 0and this result implies that the term in parentheses must be zero, giving the auxiliary orcharacteristic equation

(a2þ b2)2¼ 0 (8:2:2)The solution to this equation gives double roots of the form

The general solution to the problem then includes the superposition of the zero root cases plusthe general roots For the zero root condition with b¼ 0, there is a fourfold multiplicity of theroots, yielding a general solution of the form

fb¼0¼ C0þ C1xþ C2x2þ C3x3 (8:2:4)while for the case with a¼ 0, the solution is given by

fa¼0¼ C4yþ C5y2þ C6y3þ C7xyþ C8x2yþ C9xy2 (8:2:5)Expressions (8.2.4) and (8.2.5) represent polynomial solution terms satisfying the biharmonicequation For the general case given by equation (8.2.3), the solution becomes

f¼ eibx[Aebyþ Be
byþ Cyebyþ Dye
by]

þ e
ibx[A0ebyþ B0e
byþ C0yebyþ D0ye
by] (8:2:6)The parameters Ci, A, B, C, D, A0,B0,C0, and D0 represent arbitrary constants to bedetermined from boundary conditions The complete solution is found by the super-position of solutions (8.2.4), (8.2.5), and (8.2.6) Realizing that the final solution must be real,the exponentials are replaced by equivalent trigonometric and hyperbolic forms, thus giving

f¼ sin bx[(A þ Cby) sinh by þ (B þ Dby) cosh by]

þ cos bx[(A0þ C0by) sinh byþ (B0þ D0by) cosh by]

þ sin ay[(E þ Gax) sinh ax þ (F þ Hax) cosh ax]

þ cos ay[(E0þ G0ax) sinh axþ (F0þ H0ax) cosh ax]

EXAMPLE 8-4: Beam Subject to Transverse Sinusoidal

be most useful for the case where l >> c Because the vertical normal stress has asinusoidal variation inx along y¼ c, an appropriate trial solution from the general case is

f¼ sin bx[(A þ Cby) sinh by þ (B þ Dby) cosh by] (8:2:9)The stresses from this trial form are

sx¼ b2sin bx[A sinh byþ C(by sinh by þ 2 cosh by)

þ B cosh by þ D(by cosh by þ 2 sinh by)]

sy¼
b2sin bx[(Aþ Cby) sinh by þ (B þ Dby) cosh by]

txy¼
b2cos bx[A cosh byþ C(by cosh by þ sinh by)

þ B sinh by þ D(by sinh by þ cosh by)]

(8:2:10)

Condition (8:2:8)2implies that

[A cosh byþ C(by cosh by þ sinh by)

þ B sinh by þ D(by sinh by þ cosh by)]y¼c¼ 0 (8:2:11)This condition can be equivalently stated by requiring that theeven and odd functions of

y independently vanish at the boundary, thus giving the result

A cosh bcþ D(bc sinh bc þ cosh bc) ¼ 0

B sinh bcþ C(bc cosh bc þ sinh bc) ¼ 0 (8:2:12)Solving for the constantsA and B gives

A¼
D(bc tanh bc þ 1)

B¼
C(bc coth bc þ 1) (8:2:13)and thus the vertical normal stress becomes

Continued

EXAMPLE 8-4: Beam Subject to Transverse Sinusoidal

Loading–Cont’d

sy¼
b2sin bx{D[by cosh by
(bc tanh bc þ 1) sinh by]

þ C[by sinh by
(bc coth bc þ 1) cosh by]} (8:2:14)Applying boundary condition (8:2:8)3to this result gives the relation betweenC and D

C¼
tanh bc bc
sinh bc cosh bc

Using these results, the remaining boundary conditions (8:2:8)1and (8:2:8)5,6will now besatisfied Thus, we have completed the determination of the stress field for this problem.Following our usual solution steps, we now wish to determine the displacements, andthese are again developed through integration of the strain-displacement relations.Skipping the details, the final results are given by

u¼
b

Ecos bx{A(1þ n) sinh by þ B(1 þ n) cosh by

þ C[(1 þ n)by sinh by þ 2 cosh by]

þ D[(1 þ n)by cosh by þ 2 sinh by]}
!oyþ uo

v¼
b

Esin bx{A(1þ n) cosh by þ B(1 þ n) sinh by

þ C[(1 þ n)by cosh by
(1 þ n) sinh by]

þ D[(1 þ n)by sinh by
(1
n) cosh by]} þ !oyþ vo

(8:2:19)

To model a simply supported beam, we choose displacement boundary conditions as

u(0, 0)¼ v(0, 0) ¼ v(l, 0) ¼ 0 (8:2:20)

EXAMPLE 8-4: Beam Subject to Transverse Sinusoidal

Loading–Cont’d

These conditions determine the rigid-body terms, giving the result

!o¼ vo¼ 0, uo¼b

E[B(1þ n) þ 2C] (8:2:21)

In order to compare with strength of materials theory, the vertical centerline displacement

is determined Using (8:2:19)2and (8:2:13)1, the deflection of the beam axis reduces to

v(x, 0)¼Db

E sin bx[2þ (1 þ n)bc tanh bc] (8:2:22)For the casel >> c, D
3qol5=4c3p5, and so the previous relation becomes

(8:2:23)The corresponding deflection from strength of materials theory is given by

Considering again the casel >> c, the second term in brackets in relation (8.2.23) can

be neglected, and thus the elasticity result matches with that found from strength ofmaterials

8.2.1 Applications Involving Fourier Series

More sophisticated applications of the Fourier solution method commonly incorporate Fourierseries theory This is normally done by using superposition of solution forms to enable moregeneral boundary conditions to be satisfied For example, in the previous problem the solutionwas obtained for a single sinusoidal loading However, this solution form could be used togenerate a series of solutions with sinusoidal loadings having different periods; that is,

b¼ bn¼ np=l, (n ¼ 1, 2, 3,    ) Invoking the principle of superposition, we can form alinear combination of these sinusoidal solutions, thus leading to a Fourier series representation

to a general transverse boundary loading on the beam

In order to use such a technique, we shall briefly review some basic concepts of Fourierseries theory Further details may be found in Kreyszig (1999) or Churchill (1963) A functionf(x) periodic with period 2l can be represented on the interval (
l, l) by the Fourier trigono-metric series

f (x)¼1

2aoþX1 n¼1

an¼1l