module 3 - plane elasticity theory

43

CHAPTER 2

PLANE ELASTICITY THEORY

2.1 Elastic Constants

Four elastic constants can be defined when isotropic materials are stressed elastically. Thefour constants apply to both metallic and non-metallic materials provided the stresses producereversible, proportional strains. Because elastic strains are normally small, three differentelastic moduli may be calculated from the ratio between engineering stress and straindepending upon the manner of the applied loading. These moduli define the stiffness of agiven material structure and are related to the strength of the interatomic forces bonding thatmaterial. Since these forces are controlled by interatomic spacing, the elastic modulus of amaterial is relatively insensitive to changes in its microstructure arising from alloying andheat treatment. However, the moduli will fall with increasing temperature as the forcesbetween atoms decrease with the increase in their spacing due to thermal expansion.

2.1.1 The Modulus of Elasticity (Young's Modulus)

Thomas Young (1773-1829) defined his elastic modulus E under uniaxial stress (see Fig.2.1). Most metallic materials are linear in their load-displacement relationship. That is, whenthe axial load W is elastic, the axial displacement, x, will fully recover upon unloading.

Engineering strain, e = x/l

Figure 2 .1 Uniaxial stress

For a bar of length l and section area A, it follows that the ratio between the engineering

stress, a = W /A, and strain, e = x/l, is a constant for a particular material. This ratio is themodulus of elasticity and is defined as

E = ale= (W 1) / (Ax) (2.1)

44 MECHANICS OF SOLIDS AND STRUCTURES

2.1.2 Poisson's Ratio

Simon-Denis Poisson (1781-1840) identified a further elastic constant v under the uniaxialstress state in Fig. 2.1. Poisson's ratio refers to the constant ratio between either of the lateralstrains, e2 or e3, and the axial strain e,. Poisson's ratio is a therefore a dimensionless elasticconstant and normally lies in the range t/4 to 1/3 for metallic materials. To make the ratiopositive, a minus sign must accompany the lateral strain within the definition. The sign of thelateral strain is always opposite to that of the axial strain. Each of the lateral and axial strainsis a diagonal component of the infinitesimal strain tensor, e;j. Thus, in Fig. 2.1, if x is theincrease in length l under tension and y is the contraction in the width dimension (or diameter)w, then

v = - e2 /e, = - (l Y)/(wx) (2.2)

Equation (2.2) is also valid for compression when x and y are respectively the decrease inlength and the increase in width. It follows that the two lateral strains may also be found fromE for a given uniaxial stress or. That is, from combining egs(2.1) and (2.2),

e2 = e3 = - ve, = - valE (2.3)

2.1.3 The Modulus of Rigidity (Shear Modulus)

In the elastic region, the shear displacement x (see Fig. 2.2) increases linearly with thetangential shear force F. The distortion produced is referred to the angular change in the rightangle g5 rad. The modulus of rigidity, G, is identified with the constant ratio between theshear stress r= F/A and the shear strain y= tang Provided 0 is small we may write tan - =xll (rad) and hence G becomes

G = r/ y= (Fl )l(Ax) (2.4)

In taking the reference length l to be the block height , we see from egs(2. 1) and (2.4) how ashear mode of deformation corresponds to a uniaxial mode of deformation.

x

Engineering shear strain , y = tan

Figure 2.2 Shear deformation

PLANE ELASTICITY THEORY 45

2.1.4 Bulk Modulus

Stress and strain are also linearly related when deformation is produced by hydrostatic stress.The bulk modulus , K, defines the ratio between the mean or hydrostatic stress , or., andvolumetric strain , &V/V. That is,

K = om /(dV/V) (2.5a)

Hydrostatic stress arises directly when a solid is subjected to mutually perpendicular equalstresses, typical of fluid pressure (see Fig. 2.3).

as

Figure 2.3 Hydrostatic stress

Equation (2.5a) remains positive irrespective of whether om is tensile or compressive becausedV will change sign accordingly . Assuming hydrostatic tension of a unit cube (1 = 1), theincrease in length of each side is x = e x l = e, and the volume change is 6 V = strainedvolume - initial volume . This gives , ignoring e2 and e3 terms,

dVIV=(l+e)3 -1=3e

Then, from eq(2.5a),

K= o,/(3e) (2.5b)

When the normal stresses are unequal the mean stress is defined from egs(1.21a) as

am=V3 o, =V3(a,+ a2+ o33) (2.6a)

which causes elastic compressibility only. Note that the remaining deviatoric stresscomponents (see Fig. 1.11 c) o,' = o, - or., a2' = 02 - om and a3' = 03 - om are responsible fordistortion without volume change. With associated length changes (strains), e„ e, and e3,in the sides of a unit cube (V = 1),

dV/V=(1 +e,)( 1 +e2)(1 +e3) - 1dVIV e,+e2+e3 = e,a (2.6b)


Using the summation convention, eq(2.5a) becomes

K= 1/3( OI+ o2+03)I(el +e2+e3)

= 1/3 or,lekk = Uml(3em) (2.6c)

Typical room-temperature elastic constants for engineering materials are given in Table 2.1.

Table 2.1 Elastic Constants at 20°C

Material E (GPa) G (GPa) v K (GPa)

Aluminium 70.3 26.1 0.345 75.6Al-Cu Alloy 75 28.5 0.31 65.8Brass 103.5 39 0.33 101.5Bronze 117 44.8 0.31 102.6Carbon Steel 207 81 0.28 157Cast Iron 113.4 45 0.26 50.3Chromium 279 115.5 0.21 160.3Copper 115.2 43 0.34 107.5Iron 211 82 0.29 167.5Nickel 200 76 0.31 175.4Stainless (18/8) 200 77 0.30 167Ni-Cr Steel 206 82 0.26 143Titanium 112.6 42 0.34 104Tungsten 400 157 0.27 290Nimonics 210 80 0.31 184Concrete 10 4.2 0.20 13.5Glass 72.1 29.3 0.23 58Quartz 73 31 0.17 37Tungsten Carbide 534.4 219 0.22 318.1Timber 4 1.3 0.50Rubber 0 .003 0.001 0.50

2.2 Relationship Between Elastic Constants

Engineers have now become used to working with four elastic constants E, G, K and v, butthey are not independent constants. Gabriel Lame (1795-1870) first showed that, provideda solid is isotropic, three independent elastic constants were sufficient to describe all possiblemodes of deformation. Thus, we should expect a relationship to exist between any three ofour four engineering constants.

2.2.IE,Gand v

This relationship is established from the stress state under pure shear. Figure 2.4a shows thatshear stresses, rA„ and rB, , resulting from the shear force on opposite faces, cannot exist


without complementary shear upon each adjacent face.

B C

(a) A DrAD 45°

Figure 2.4 Analysis of stress and forces under pure shear

OF, = r

02 = - r (b)

The complementary shear stresses, rAB and rcD, are found from taking moments about pointA. For a thickness, t, moment equilibrium gives

rBCx(BCxt)xAB= rtDx(CDxt)xAD

. rcD= rBC

and since rAD = rBC then TAD = rD. Similarly, taking moments about C gives rAB = rAD. Thisshows that all shear stresses are the same, which will be simply labelled r hereafter. Thecorresponding shear strain is y= r/G. To determine the states of stress and strain followinga rotation in the axes, we apply egs(1.35) and (1.40) to give

0 1.1, 0 1,2,

02.1, 02.2,

e1.1. e1.2.

E 2.1. a 2,2,

cos a

- sine

cos a

- sine

sine 0 T cos e - sin e (2.7a)

cos e

sine

i 0

0

sin

Y /2

s cos e

cos s - sin 9 (2.7b)

cos e Y /2 0 i l sine cos e

where the primed axes define the rotation shown in Fig. 1.15. In particular, when B= 45°,eq(2.7a,b) yields coincident principal stress and strain systems (see Fig. 2.4b)

0

- T

et 0 Y/2 0

also =0 e2 0 -y/2

(2.8a,b)

Thus pure shear is equivalent to a major principal tensile stress o, and a minor principalcompressive stress o2, each having a magnitude equal to r. Correspondingly, one principalstrain e, is tensile and the other e2 is compressive, each with magnitude y/2. Both stresses

contribute to the strain in each principal direction. One gives a direct strain o/E and the other

a lateral strain - vo/E:

e, _ (1/E)(o, - vo2) (2.9a)

e2 = (1/E)(u2 - vu,) (2.9b)

Setting o, = r, 02 - r, with e, = y12 and e2 = - yl2 reduces both eqs(2.9a,b) to


y12=(1+v)r/E

and, since G = r/yfrom eq(2.4), it follows from eq (2.9c) that

E=2G(1+v)

2.2.2 E, K and v

(2.9c)

(2.10)

Consider the total strain e, for any one direction under hydrostatic stressing in Fig. 2.3. Thismean strain is the sum of three contributions: the direct tensile strain, om/E, and two lateralcompressive strains produced by the remaining stresses, which are each -vom /E, fromeq(2.3). The total strain is

e= om/E - v om/E - vom/E _ (om/E)(1- 2v)

Substituting eq(2.11) into eq (2.6c) for when e= e, = e2 = e3

K=oml(3em) =1/3 am /[(om/E)(1 - 2v )]

from which,

E=3K(1 - 2v)

To connect the mean stress and strain we combine eqs(2.6c) and (2.12a)

ekk=(1 - 2v)o,,/E

2.2.3 Relationships Between G, K, E and G, K, v

(2.11)

(2.12a)

(2.12b)

Further relationships between G, K and v are found by eliminating E and v between egs(2.10and 2.12a). These are

G=3KE/(9K-E)=3K(1-2v)/[2( 1+v)] (2.13a)

From eq(2.6c) and (2.13a), the mean stress and strains become connected through G and v

em = om(1 - 2v)/[2G (1 + v)] (2.13b)

where em = V3 e, and om = okk

2.3 Cartesian Plane Stress and Plane Strain

Many problems which are plane in nature may be analysed using a reduced form of thegeneral theory of elasticity. Specifically, in Cartesian co-ordinates x,y,z, plane stress refersto a body with small thickness in the z-direction. For the thin plate in Fig. 2.5a, the stress inthe plane of the plate is two dimensional, while the stress through the thickness (ok) is zero.Plane strain, on the other hand, refers to a body with a large dimension in the z - axial


direction (see Fig . 2.5b) and this gives either zero axial strain if the ends are constrained orsome constant value (say e.).

rri

(a)

Figure 2.5 Plane stress and plane strain in Cartesian co-ordinates

2.3.1 Stress-Strain (Constitutive) Relations

The non-zero stresses ox, ay and= 0 and the strains become

ex=(ox - voy)/E

ey=(ay - vox)/E

eZv(ox+oy)/E

yxy= rxy/G=2 (1 +v)r,,yIE

rxy may be functions of x and y but not z. In plane stress oZ

(2.14a)

(2.14b)

(2.14c)

(2.14d)

where the through -thickness strain eZ is not zero . The plane strain constitutive relations are

ex = [Ux - v (ay + o= )]IE (2.15a)

ey = [oy v (ax + of )]/E (2.15b)

ez=[o - v ( ax+0y )]/E (2.15c)

yxy=rte, /G=2(1+v) rxy/E (2.15d)

Since e, =e„ is constant, then from eq (2.15c), aZ = e,, E + v (ox + ay ). It follows from this thatplane sections will remain plane when the stress sum (ox + ay) is a constant . Moreover, sinceax , oy and rxy are each functions of x and y only , the following analyses of equilibrium andcompatibility will apply to both plane stress and plane strain . The theory of equilibrium andcompatibility , appropriate to plane polar co -ordinates , will be derived later from transformingthe Cartesian co-ordinate relationships.


2.3.2 Equilibrium

Let the stresses vary over the elemental dimensions ox and dy in the manner illustrated inFigs 2.6a and b. These act together, as shown in Fig. 2.5a,b, but, for clarity, the x and y forcesare separated. Body forces, due to self-weight, are ignored.

L

l aY+

a' by

By

aTS * _8X

by

'Y ax

yA

Qx

by

atT_- y, byy

Bar,^-- IN. a,+-bx

U` (a)vx (b

Figure 2.6 Stress variation in the x and y directions

Assuming unit thickness in Fig . 2.6a, vertical force equilibrium gives

[ci +(ao,.lay)6y- o^.]8x+[r,,+(arxYlax)ox- r ]Ey=o

ao,.lay+arxy.lax=0

and, from Fig . 2.6b, for horizontal force equilibrium

[ox +(a oxlax)ox- ox]oy +[ryx +(arYx lay)oy- ryx]U.[ =0

a oxlax+aryxlay= o

ax

► x (b)

(2.16a)

(2.16b)

Equations(2.16a,b) are plane equilibrium equations. They apply to plane stress states in whichthere are x and y variations in the stress components from point to point.

2.3.3 Strain-Displacement Relations

In Chapter 1 it was shown that initially unstrained perpendicular line elements &x and 6ystretch and rotate whilst translating into a new position under stress (see Fig. 1.12a). Forplane deformation the u and v displacements defining the shift in the origin from 0 to 0' aredifferent functions of the x, y co-ordinates. Plane sections remain plane when all points in thebody displace only within the x, y plane, i.e. u and v do not depend upon z. The displacementgradients in egs(1.23a,b) and (1.24a), found from the partial differentiation of u and v withrespect to x and y, define the normal and shear strain components. In confirmation of this, thenormal strain components identify with the corresponding length changes in 8x and 6y

Ex= {[cSx +(aulax)&] - ox}l&x=aulax

e,,= {[ay+(avlay) oy] - 6y 116y = May

)

(2.17a)

(2.17b)

The small angular distortions are approximated by


ex,. = [(au / ay)6y]1oy = aulaye,x = [(av / ax)6x]/8x = Max

Their sum gives the net change in the right angle, which , when expressed in radians , definesthe engineering shear strain

Yx,. = aulay + Max (2.17c)

2.3.4 Compatibility

It is seen from eqs(2.17a,b and c) that the three strains ex, e,. and yx ,, depend upon the twodisplacements u and v. The relationship between the strains expresses a condition ofcompatibility . This condition ensures that the material deforms as a continuum without adiscontinuity arising in the strain distribution . A mathematical expression for straincompatibility is found by eliminating u and v between eqs(2 . 17 a-c).

a2exlay2 +a2elax2= a2yx,.laxay (2.18)

Equation (2.18) may further be written in terms of the stress components . With plane stress,for example , substituting from eqs (2.14a,b,d) leads to

ae(o- vo,)/ay2+ae(o,,-vox)/axe=2(1 +v)a2r /ax ay (2.19a)

The corresponding plane strain relationship is, from eqs(2 . 15a,b,d)

ae[ox(1 - v2)- v o,(1 +v)- ve,EI/aye

+ae[o.(1-v2)-vox(l+v)-ve„EJ/axe=2(1+v)a21x,,/ax ay (2.19b)

in which the following substitution for oz has been made from eq(2.15c):

oz=Ee,, +v(ox+o,.)

2.3.5 The Biharmonic Equation and Stress Components

The requirements that both equilibrium and compatibility are satisfied in plane stress andplane strain are met by combining egs(2 . 16a,b) and (2.19a,b). For plane stress egs(2 . 16a and2.19a) lead to

a20,x lay2- va2o,.lay2+a2o,.lax2- va2ox /axe=- 2(1+v)a2Q.lay2 (2.20a)

and, from eqs(2.16b) and (2.19a),

a2ox/ay2- va2q./ay2+a2o,,/axe- va2ox 1axe=- 2(1 +v)a2ox /axe (2.20b)

Subtracting eq(2.20a) from (2.20b),

- 2(1 +v)a2oxlax2+2(1 +v)a2o,,/ay2=0

ago. lay2 - a2ox l aze = 0 (2.21a)


Adding eqs (2.20a and b),

2[ a2oX l ay2 - v a2a, l ay2+ a2a,,l ax2 - v a2aXl ax2 1 2(1 + v)( a2a,./ ay2+ ago / ax2 )a2 as / axe + a2 ax l ay2 + a2 o.l ay2 + a2a,, l 3x 2 = 0(a2 /ax2 +a2/ay2)(a+o)=o (2.21b)

It is seen that the equilibrium eq(2.21a ) is satisfied when as and oy are defined from a stressfunction 0_ 0(x, y) (G.B.Airy 1862) in the following manner

ax =a2, /ay2 and a,.= a20 /ax2 (2.22a,b)

and, from either eq(2.16a) or eq(2.16b), the shear stress is given by

arxy lax=- aa,lay=- a3p /ax2 ayr^y = - a2 ax ay (2.22c)

The result of combining eq(2.21b) with eqs(2.22a and b) may be written as a biharmonicequation in any one of the following three forms:

(all axZ+ a2Iay2)(a2Olay2+a2p/ax2)=0 (2.23a)

(a2/ax2+a2/ ay2 )25=0 (2.23b)

V2(V20) =V40=0 (2.23c)

where 'del-squared' is defined as V 2 = a2 / ax2 + al l ay2. The reader should check that identicaleqs(2 .22) and (2.23) apply to plane strain when eqs(2.16a,b ) are combined with eq (2.19b).

2.3.6 Body Forces

The biharmonic equation V'q = 0 applies to plane stress and plane strain only in the absenceof body forces. When the latter are present , e.g. in the form of components X and Y of selfweight , the equilibrium equations (2.16a,b) are modified to

aoy.lay+arx,lax+Y=0

aaxlax+ar,,lay+X=0

(2.24a)

(2.24b)

where X and Y are the Cartesian body force components/unit volume . Individual solutionsmay be derived by combining eqs(2 .24a,b ) with the appropriate compatibility condition. Aclosed solution is possible in which the body forces are derivatives of a potential function Sa(x, y)

x = - au lax, Y = - au lay (2.25a,b)

Combining eqs(2 .24a,b) and (2 .25a,b)

a (oy - o )lay + a ;,, lax = o (2.26a)a (ax - u )lax + a r, lay = o (2.26b)


A stress function O(x, y) will satisfy eqs (2.26a ,b) as follows:

ox = Q + ago /ay2, ay = 0 + a2Olax2, r,, = - alolax ay (2.27a,b)

In plane strain the compatibility condition (2.19b) remains unaltered. Combining this witheqs(2 .26a,b) and noting that e„ is either zero or a constant leads to

a4,0 laX4 +2 a4 / Iax 2 ay2 + a40lay4 + [(1 - 2v / (1 + v )](a2Q lax2 + a2L /aye) = 0 (2.28a)

V4o +[(1 - 2v)/(1 +v)JV2Q=0 (2.28b)

Functions 0 (x, y) and 0 (x, y ) which satisfy eq(2.28b ) will provide stress components(2.27a,b) which must also be made to match the boundary conditions.

Strictly, when body forces are present in plane stress , additional compatibility conditionsaccompany eq(2.19a). These arise from the dependence of the through -thickness strain eZupon the in-plane stresses

a2e,lax2=a2e,lay2=a2e,laxay=0 (2.29a)

which shows that eL should obey an equation of the form

ez= a, +a2X+a3y (2.29b)

where a ,, a2 and a3 are constants . Note , however, that eq(2.29b) will not be upheld whenegs(2 . 19a) and (2.26a ,b) are combined to give a single equation to be satisfied by 0 and 0in plane stress

a 4.0 lax4 + 2 a4,0 lax2 ay2 + a 4 ' /ay4 + (I - v )(a 2U /axe + a 2U /ay2) = 0 (2.30a)V4.0 + (1 - v)V2Q =0 (2.30b)

This approximate solution is acceptable for thin plates . Otherwise , eq(2.29a) requires threeadditional equations which 0 and 0 should satisfy:

2 a 2 U /axe + a 2 (v 20)/axe = o

2a2a /ay2+a2(v2o)/ay2=0

2 a2Qlaxay +a2(V20)laxay =0

In the absence of body forces (X = Y = 0) we see again from eqs(2.28b) and (2.30b) that bothplane stress and plane strain simplify to v4q5= 0.

2.4 Cartesian Stress Functions

If a function 0 (x, y) can be found that satisfies V 40= 0, it will satisfy both equilibrium andcompatibility in plane stress and plane strain . The stresses are found from 0 (x, y) using Airy'seqs(2 .22a,b,c). Constants appearing in the function 0 must meet a final requirement to satisfythe boundary conditions for a particular problem . That is, the internal stress distribution,which will in general vary throughout the volume of the body, must be made to contain theknown forces , moments and torques applied to its boundary.


2.4.1 Polynomial Functions

The stress functions of most common use in plate and beam problems are taken from thepolynomial function

0(x, y) =ax2 + bxy+cy2 + dx3 +ex2y +fxy2+ gy3 +hx4 + ix3y+jx2y2+kxy3 + 1y4+.....

Linear and constant terms are excluded since these disappear with the second derivative. Thefollowing examples illustrate the usefulness of certain combinations of these terms.

Example 2.1 Show that the plane displacement system u = ax2 y2, v = byx3 is compatible.

The following strain components are found from eqs(2.17a-c):

ex = au lax = 2axy2

ey= avlay=bx3yx,.=aulay+avlax=2ax2y+3byx2

Applying the left- and right-hand sides of eq(2.18) separately, we see that

a2exlay2 + a2e,,lax2 = a (4 axy) lay + a (3bx2 )lax = 4ax + 6bx

a2 yx,, lax ay = a (4axy + 6bxy)lay = 4ax + 6bx

Example 2.2 Derive the displacements corresponding to the following stresses in plane strainox = c(y2 + 2x), oy = - cx2 and zxy = - 2cy when e„ = 0. Show that the corresponding strainsare compatible.

Clearly, the given stress expressions satisfy the equilibrium equations(2.16a,b). Since e0, _0, we have, from eq(2.15c), oZ = v (ox + oy). Then, from eqs(2.15a,b,d),

ex au lax=(l +v)[o(1 - v)- v oy]/E

=(1+v)[c(y2+2x)(1 - v)+vcx2]/E

u=(1 +v)[c(y2x+x2)(1 - v)+vcx3l3]/E+f(y) (i)

ey=avlay=(l+v)[oy(1-v)-vox]/E

- (1 +v)[cx2(1 - v)+vc(y2+2x)]/E

v=- (1 +v)[cx2y(1 - v)+vc(y3/3+2xy)]/E+g(x) (ii)

yx,. =au lay+av lax= 2(l +v )sxy /E

:. au lay + av lax =- 4 cy (1 + v )l E (iii)

Equations (i), (ii) and (iii) clearly satisfy the compatibility condition given in equation (2.18).Substituting eqs(i) and (ii) into (iii),

(l + v )2cyx (1 - v )/ E + f'(y) - (l + v )[2cyx (1 - v) + 2v cy]l E + g'(x) 4cy(l + v )/ E

f'(y)- 2vcy(1 +v)lE+g'(x)=- 4cy(1 +v)/E


from which it follows that g'(x) = 0 and

f'(y)=2vcy( l+v)/E-4cy ( 1+v)/E=2cy(1+v)(v-2)/E

f (y) = cy2 ( 1 +v )(v - 2)/ E

The displacement expressions are then, from eqs(i) and (ii)

u=(1 +v)[c(y2x+x2)(1 - v)+vcx3/3]/E+cy2(1+v)(v- 2)/E

v=- (1 +v)[cx2y(1 - v)+vc(y3/3+2xy)]/E

Constants of integration have been omitted here as no boundary conditions are given. Notethat u and v will differ for plane stress where oZ = 0.

Example 2 .3 The square plate a x a in Fig. 2.7 is subjected to the plane stresses shown.Determine, for the given boundary conditions, the general expressions for the u, vdisplacements at any point (x, y) in the plate. What are the displacements for a pointoriginally at the origin?

U=0

Y A Axa,=-

a

.4

(a. a)

rry = B

au_0

ayu=0

Figure 2.7 Plane stress plate

fx

The given stresses obviously satisfy the equilibrium eqs(2.16a,b). For plane stress,eqs(2.14a,b,d) and (2.17a-c) give

ex =aulax=(ox- voy)IE=A(1 - vx/a)/Eu = A [x - vx2l(2a)]/E + f (y) + P (i)

ey=avlay =( a,- vox )/E=A(x/a- v)/Ev=A(xyla - vy)IE+g(x)+Q (ii)

Yxy=au lay + avlax = 2(1 +v)ry/E=2B (1 +v)/E

Substituting (i) and ( ii) in (iii) gives

f ' (y) + Ay/(aE) + g'(x) = 2B(1 + v )/E

Since the right-hand side is a constant , say a+ X13= 2B (1 + v )/ E, it follows that

.f'(y) + Ay/(aE) = a f(y) = ay - Ay 2 1(2aE)

MECHANICS OF SOLIDS AND STRUCTURES

g'(x)=Q, g(x)=fix

Hence eqs( i) and (ii) become

u=A[x - vx2/(2a)]/E+ay-Ay2/(2aE)+P (iv)

v = A(xy la - v y) /E + ,6x + Q (v)

The following boundary conditions apply to eqs(iv) and (v)

u=Oatx =O, y=a,:. 0 = a a - Aa 2 /(2aE) + P, P = Aal(2E) - a a

v=Oatx=a, y=0..0=ia+Q, Q=-,6a

au1ay=0atx=0, y=0:. a=0, 6=2B(1 +v)/E

Finally, eqs(iv) and (v) become

u = Ax [I - vx/(2a)]/E -A y2 1(2aE) + Aal(2E)

v = Ay(xla - v )/E + 2B (1 + v )(x - a)/E

which give the displacements at the origin (x = 0, y = 0)

u = Aal(2E) and v = - 2Ba(1 + v )/E

Example 2.4 The rectangular section bar in Fig. 2.8 is line loaded as shown. Assuming planestrain conditions, derive the stresses and displacements from the stress function 0= Ax 3. Ifall the points along the z-axis are fixed, determine the displacements beneath the load.

Figure 2.8 Plane strain bar


Since 0 satisfies V45= 0, the stress components are, from eqs (2.22a-c)

Oy = a2.0 lax2 = a2 (Ax3 )lax2 = a (3Ax2 )lax = 6Ax

Qx = rxy. =0

Now, at the boundary , F is the resultant of the internal oy distribution. That is , for unitthickness in the z - direction,

dx=6A f axdx= 3Aa2F= f oy oo

A = F/(3a2) and c = 2Fx lag

Alternatively, A may be found from moment balance

2aF l3 = f "a x dx = 6A f x2 dx, A = F/(3a2 )f f

Since e„ = 0 and ox = 0, eq(2 . 15c) gives oZ = voy. Then eqs(2.15a and b ) become

ex=au lax= - v(1 +v)oylE=- 2Fv(1 +v)x/(a2E)

e),=avlax=(1- v2) oy/E=2F(1 - v2)x/(a2E)

Integrating these leads to the u, v displacements

u=-Fv(1+v )x2/(a2E)+f(y)+C, (i)

v = 2F(1- v 2 )xy /(a2E) +g(x) + C2 (ii)

Since rxy = 0, egs(2.15d and 2.17c) become

Yxy= aulay + avlax = o

Substituting eqs(i ) and (ii ) into (iii) gives

f (y) + 2F ( 1- v 2) y 1(a 2E) + g'(x) = 0

:. g'(x) = 0 and f '(y) + 2F(1 - v 2 )y /(a2E) = 0

.f (y) _ - F (l - v 2) y2/(a2E)

Hence eqs(i) and (ii) become

u = - Fv (1 + v )x2/(a2E) - F(1 - v 2) y2 /(a2E) + C, (iv)

v = 2F (1 - V2 )xyl(a2E) + C2 (v)

where C , = C2 = 0 since u = v = 0 for x = y = 0. At the load (x = 2a/3, y = h) egs(iv) and (v)give

u=- F[h2 ( 1 - v2)/a2- 4v(1 +v)/9]/E

v = 4Fh (1 - v2)l(3aE)


Example 2.5 Determine the stresses and displacements from the plane stress function:Ax2 + Bxy + Cy2. Apply these to the plate in Fig. 2.9.

= 2A

o,= 2C

Figure 2.9 Thin plate under uniform stressing

Clearly 0 satisfies V4O= 0. The stress components are, from eqs(2.22a-c)

ox = a2olay2 = a (Bx + 2C y)lay = 2Coy= a20 lax2 = a(2Ax + By)lax=2Aray. a2O /ax ay = a(2Ax + B y)lay= - B

That is, the function here provides a solution to a thin plate under the action of uniformstresses along its sides (see Fig. 2.9). The displacements follow from eqs(2.14 and 2.17)

es= aulax= ( ox- voy )/E=2(C- vA)/E

u=2(C- vA)xlE +f (y) + Q, (i)

ey.= av /ay= (oy - vox)/E=2 (A - v C)/E

v = 2(A - v C )y/ E + g(x) + Q2 (ii)

ny, = aulay + avlax = 2 (1 + v ) rte, IE = - 2(1 + v )B/E (iii)

Substituting eqs(i) and (ii) into (iii) leads to

f'(y)+g'(x)= - 2 (1 +v)B/E= a+,13

Since the derivatives are the respective constants a and f3, then f (y) = ay and g(x) =,6x andeqs(i ) and (ii) become

u=2(C- vA)x/E+ay+Q,

v = 2(A - v C )ylE +)6x + Q2

Example 2.6 Show that the stress function 0 = Dxy3 + Bxy provides the stress distribution forthe cantilever of depth 2h and of unit thickness in Fig. 2.10, when it is loaded by aconcentrated force P at its free end.


Figure 2 .10 Plane cantilever with concentrated end load

Equations (2.27a-c) gives the stress components

ox = a2q5 /ay2 = a(3Dx y2 + Bx)lay = 6Dxy

a. = a20 /axe = a(Dy' + By)lax = 0

(i)

(ii)rx,. = - a2Olax ay = - a(3Dxy2 + Bx)lax = - (3Dy2 + B)

The boundary conditions are

(a) rx,. = 0 for y = ± h, since there is no shear stress along the free edges.

- (B + 3Dh2) = 0 B = - 3Dh2

(b) P is the resultant of the internal rxy distribution acting at the free end. That is

P= f tirx,.dy3D f ti (h2-y2)dy

= 3D h2 y - y3 /3 Ih

= 4Dh3, D = PI(4h3)h

From eqs( i) and (ii)

ox=Pxyl(2h3/3)_(Px)yl I=My/I

rx,. = 3P(h2 - y2 )/(4h3) = [P1(21 )I (h 2 - y2)

which are distributed in the manner shown in Fig. 2.10. Note that the stresses agree with theseparate applications of bending and shear flow theory. In the stress function approach, x andnot z is used to denote the length of a plane stress beam within the x, y plane. Then y boundsthe depth and z bounds the thickness.

2.4.2 Sinusoidal Function

If a stress function of the form

0 = f (y) sin (n,rx IL) = f (y) sin (ax) (2.31 a)


is to satisfy V4O= 0, the following condition is found

d4 f (y)ldy4 - 2a2 d 2.f (y)/dy2 + ce4f (y) = 0

The solution is

f (y) _ (A + By) exp (ay) + (C +Dy) exp (- ay)= E cosh ay + F sinh ay + Hy cosh ay + Jy sinh ay (2.31b)

where a = n z IL. The function will apply directly to sinusoidal edge loadings where theparticular boundary conditions enable the constants E, F, H and J to be found.

Example 2 .7 A long thin strip of depth 2h is subjected to a normal stress distribution psin(ax/L) along its longer edges (see Fig. 2.11). Determine the constants in eq(2.3 I b) and thestress state along the x - axis.

p sin

(0,0)

2L

Figure 2.11 Sinusoidal wave edge loading

Here a =n /L since n = 1. The plane stress components are found from eqs(2.22a-c) andegs(2.31 a,b),

oy.= a20/az2 =- [Ecosh ay+ Fsinh ay+Hy cosh ay +Jysinh ay]a2 sin ax (i)

r^y _ - a2' lax ay = - [a (E sinh ay + F cosh ay) + H (ya sinh ay + cosh ay)

+ J (yacosh ay + sinh ay)]acos ax (ii)

ox = a2() lay2 = [a2 (E cosh ay + Fsinhay) + Ha (ya cosh ay + 2 sinh ay)

+ Ja (ya sinh ay + 2 cosh ay)] sin ax (iii)

In eq (i) the boundary condition is oy = p sin ax (tensile) for y = ± h and all x. This gives

- a2 [E cosh ah + F sinh ah + Hh cosh ah + Jh sinh ah] = p

- a2 [E cosh ah - F sinh ah - Hh cosh ah + Jh sinh ah] = p

By addition and subtraction,

E = - (p + Ja2h sinh ah)/(a2 cosh ale) (iv)F = - (Hh) cosh ah / sinhat (v)


In eq(ii) the boundary condition is rte, = 0 for y = ± h and all x. Thus,

- [a (E Binh ah + F cosh ah) + H (ha sinh ah + cosh ah) + J (ha cosh ah + sinh ah)] = 0

- [a (- E sinh alt + F cosh adz) + H (hasinh ah + coshah) - J(hacosh ah + sinh ah)] = 0

By addition and subtraction,

H = - aF (coshah) l (ha sinhah + cosh adz) (vi)

J = - aE (sinh ah) / (hacoshah + sinhah) (vii)

Substituting eq(iv) into (vii) leads to

J ap (sinh adz) / [a2(sinhah coshah + adz)]

E _ - p (sinh ah + alt cosh adz) / [a2 (sinh alt cosh ah + adz)]

Substituting eq(v) into (vi) gives F = H = 0, which is a consequence of the common amplitude

p for each edge distribution. When y = 0 the stress state along the x axis is, from eqs (i) - (iii)

oy, = - Eat sin ax= p sinax (sinhah + alt cosh adz) / (sinhah coshah + ah)

ox = (a2E + 2Ja) sin ax= p sin ax [- (sinhah + alt coshah) / (sinhah coshah + adz)

+ 2 sinhah / (sinhah coshah + ah)]

rte,=- (Fa+H) acosar=0

2.4.3 Fourier Series

(a) Full Range SeriesWhen the edge loading is not sinusoidal it may still be represented by a sinusoidal seriesfunction g(x) of period 2L. The stress function in egs(2.31a,b) then becomes

0 (x, y) = [E coshay +F sinhay + Hy cosh ay + Jy sinh ay] g(x)

where

g(x) = a012 + a, sin (rrx/L) + a2 sin (2rrx/L) + a3 sin (3rrx/L)

+ b, cos (rrx/L) + b2 cos (2rrx/L) + b3 cos (3 rrx/L).....

00

=a„/2+E [ a,, sin (nzx/L)+ b„cos (nrrx/L)]n•I

(2.32)

(2.33a)

If ^&(x) describes the actual edge loading, the coefficients a0, a„ and b„ in eq(2.33a) are found

by identifying {/r(x) with each part of g (x) and integrating over the range ± L, as follows:

L

f L [a„/2 + a„sin (nlrxlL) sin (nnxlL) + b,,cos (nrrxlL) cos (nrrxIL)]dx

= f L ^&(x) dx+ f L 0,(x) sin (nrrx IL) dx + f L j&(x) cos (nrrx IL) dx


This gives

fL (a„/2)dx = f L,;& (x) dx

a„_(l/L) f LV,(x)dx (2.33b)

f L a„ sin (nirxlL) sin (nnx/L)dx= f L r&(x) sin (nirx/L) dx

La„L = f L O(x) sin (n,rx/L)dx

a„ = (1/L) E L Or(x) sin (n;rxIL)dx

f Lb„cos(nirx/L) cos (nirx/L)dx= f L j&(x) cos (n;rxIL)dx

Lb„ L = f L r/r(x) cos (nitxlL)dx

b,, = (I IL) f L r& (x) cos (n irx/L)dx

(2.33c)

(2.33d)

Note that when x lies outside the range -L < x s L, g(x) in eqs(2.33a) will not , in general,represent r/r(x) unless Or(x) is itself periodic.

Example 2.8 A rectangular wave loading of amplitude p and period 2L is applied normallyto the longer edges of a thin plate of depth 2h, as shown in Fig. 2.12. Determine the Fourierseries loading function g(x) in eq(2.33a). If L = 4h, determine ox and or, at x = L/2, y = 0 usingthe result from Example 2.7.

Figure 2.12 Rectangular wave

The loading function is

for Osx<L

for Lsxs2L

From eqs(2.33b - d),

ao (1/L)[ f L (p)dx + f L^ (- p)dx)


=(1/L)[IPxIo -Ipxlu ]=0

a,, = (1/L)[ f ,LP sin(n^rxIL)dx + f ^(- p) sin(nrrx/L)dx ]

= (I/ L)[Lpl(n)r)][ I - cos(nnx/L) IL + I cos(nnrx/L) 12L ]

= [pl(n7r)][(1- cos n,r) + (cos 2nir- cos n,r)]

a,, =[2pl(nir)](1- cos nn) for n= 1, 3, 5... , a. = 0 for n = 2, 4, 6...

b,, = (1/L)[ f, L p cos(n rrx/L)dx+ fig(- p) cos(n ,rxIL)dx]

= (1/L)[Lpl(n;r)][ I sin(nirx/L) IL - I sin(nrrx/L) 12L ]

= [pl(n;r)][O - 01 = 0

Equation (2.33a) becomes

g(x) = F, [2p/(nir)](1 - cos nor) sin(nirx/L) for n = 1, 3, 5...n-1

= (4 p/ir)[sin(irxlL) + (1/3)sin(3 rrx/L) + (1/5)sin(5 nx/L) +... ]

Now, since the functions x and y in eq(2.32) are separable , the hyperbolic function in y maybe taken from the previous example where similar boundary conditions were applied. It issufficient to take n = 1 within g(x), when it follows for y = 0, that the plane stresses are

o,, = (4plrr)[ sin (irx/L) + (1/3) sin (3,rx/L) + ( 1/5) sin (5rrx/L) +... I

x [ sink (,rh/L) + (irh/L) cosh (rrh/L)] / [ sinh (,rh/L) cosh (,rh/L) + irh/L]

aX = (4p/rr)[sin (rrx/L) + ( 1/3) sin (31rxIL) + ( 1/5) sin(5;rx/L) +... I

x [ - sinh(irh/L) - (rr h/L) cosh (rr h/L) + 2 sinh (n h/L)]l [ sinh (ir h1L) cosh (rr h/L) + n h/L]

Taking x = L/2, hIL =1/4

o,, = (4plir ) [ I - 1/3 + 1/5 -in + 1/9 - 1/11...]1.9801 / 1.835 = 1.036paX = (4p /rr )[0.75431 [ - 1.9801 + 1 .73731/ 1.835 = - 0.127p

(b) Odd and Even FunctionsFor an odd loading function r/r(- x) Or (x), b„ = 0 and the Fourier representation g(x)

contains sine terms only. For an even loading function r&(- x) = - r&(x), which gives a,, = 0,

resulting in a Fourier cosine series. In addition, when r&(x) = - f/r(x + L), as in this example,

the Fourier series will contain odd harmonics only. When Or(x) = O(x + L), the series appearsin even harmonics. Simplified Fourier sine or cosine series may be applied over the half-range 0 < x s L. These are

g(x) _ a,, sin ax (2.34a)n

where a= nrr/L, and


La,, = (2/L) r&(x) sin ax dx

g(x) = a,, + b„ cos ax (2.34b)h-1

a,,=(1/L) foLr/r(x)dx

Lb,, = (2/L) f

0O(x) cos ax dx

Beyond the range of L, the sine series will extend 0,(x) by alternating it in positive andnegative y with period L about the x - axis. The cosine series duplicates 0,(x) with period Lalong the x - axis. Half-range series' representations of odd and even functions are identicalto the full series representation of each function.

Example 2 .9 Find the expression g(x) for the edge loading in Fig. 2.13.

Y

P

Figure 2.13 Triangular edge loading

Here g(x) will follow from the half-range sine series, since r/r (x) alternates about x withperiod L. The loading function becomes

12' IL for 0sxsL12^r (x)

2p(1-x/L) for L/2sxsL

Equations (2.34a,b) may be written as

a,, = (2/L) fo n (2pxlL ) sin ax dx + (2/L) fLa 2p (1 - x/L) sin ax dx

Using the following result from integrating by parts:

f x (sinax) dx = (1/a2) sin ax - (x/a) cos ax

eq(i) becomes

a„ = (4p/L2) I (1/a2) sin ax - (xla) cosax X0,2- [4pl(La)] I cosax

ILn

- (4p/L2) I (1 /a2) sinax - (x /a) cosaxI Ln

(i)

_ (4p1L2)[(1/a2) sin(n;r/2) - [Ll(2a)] cos(n;r/2)] - [4p1(La)][cos(nrr) - cos(nrrl2)]

- (4p1L2)([( l/a2) sin(n,r) - (Lla) cos(n,r )] - [( 1/a2) sin(nn'/2)- [L/(2a)]cos(n,r/2)])


= [8pl(L2a2 )] sin(n ;r /2) = [8pl(z2 n2 )] sin (n ,r l2)

g(x) _ Fla„ sin(n ,rx/L) = (8p/ir2) E (1/ n 2) sin(n ,r /2) sin (n,rx/L)n - I n •I

_ (8p/,r2)[sin (nxIL) - (1/32) sin (3,rx/L) + (1/52) sin (5;rx/L) - .... ]

2.5 Cylindrical Plane Stress and Plane Strain

Many plane problems are better represented by cylindrical co-ordinates (r,0). For this it isnecessary to re-express the constitutive relations, the strain-displacement relations,equilibrium, compatibility and the stress functions in terms of r and 9. Fortunately, thenumber of derivations in the theory may be reduced by transforming the Cartesian co-ordinates' relations. Firstly, we should note that the constitutive relations are identical in formto the Cartesian eqs(2.14 and 2.15) where r and 0 now replace x and y respectively (z is acommon co-ordinate).

2.5.1 Strain-Displacement Relations

Strain components er, eB and y, are found from the distortion occurring to the cylindrical lineelements or x rOO, as point 0 displaces to 0' with radial and tangential components u andv respectively (see Fig. 2.14).

au 60

Figure 2.14 Plane deformation in cylindrical co-ordinates

The radial displacement of OP is (au/ar)6 r. Hence the radial strain is

e,= [ar+(aular)ar- ar]/Or=aular (2.35a)

For the tangential (hoop) strain in OQ, the increase in length of r88 is due to both the rateof change of v with respect to 0 and the change in radius from r to r + u as 0 displaces to 0'.This gives


eo = [(r+ u)8B+ (avlao)80 - r60]1(r60)eo=ulr+ (1/r)avlao (2.35b)

The true change in the angle LPOQ defines the shear strain . This must omit the rigid rotation(vlr rad) which also contributes to the new position O'P'. That is,

Yro= (avlar)8r18r - vlr+ (au 1ao)[80/(r80)JYro = avlar - vlr + (I/ r)(au la 0)

2.5.2 Equilibrium

(2.35c)

Equilibrium equations are found from radial and tangential force balance for the cylindricalelement in Figs 2.15a,b. Note that the force components lying in each direction have beenseparated in (a) and (b) for clarity. The hoop ao, radial ar and shear rro stresses vary acrossthe element in the manner shown when taking positive r and 0 along the centre-line directionsindicated. Since the inclinations of ao and rro vary with 00, these must be resolved parallelto r and 0. The following derivations apply in the absence of body forces, assuming unitthickness of the element.

(a)

Figure 2.15 Stress variation on a plane cylindrical element

For radial equilibrium along the centre-line r in Figs 2.15a,b,

(ar+8raarlar)(r+8r)80- or r80-r,8rcos(8012)+(rer+80arorla0)8rcos(80/2)

- ao 8r sin(80/2) - (ao +80aaola0)8r sin(80/2) = 0

Now when 80-4 0, cos (80/2) 4 1 and sin (80/2)' 80/2 rad. This gives

(ar+ 8raarlar)(r+ 8r)80- orr80- rr9 Or+(rro+ 80arrola0)8r

- aoOr80/2 - ( a0+ 80aaola0)8roe12 = 0

where rro = r, Neglecting products of infinitesimals leads to

(b)

ar or8B + ror8Baarlar + 80 8r arrola0- a0Ord0 12 - a0Or80/2 = 0


Dividing through by (r86 8r) gives the first equilibrium equation,

(o,.- oo)lr + (1/r) arrola6+ aorlar = 0 (2.36a)

For equilibrium along the tangential centre - line 0 in Figs 2.15a,b,

(zro+ 6 raz olar)(r+ 6r)86- zrorO6 +(zor+ 86 azorla6) a rsin(86 /2)

+ rroOrsin(86/2) +(°B+ 86aoo1a6)6 rcos(66 /2) - o'6rcos(86/2)=0

and, with similar reductions, this leads to the second equilibrium equation

(2/r) zro + (1/r) aoola 6+ a zrolar = 0

2.5.3 Bi-harmonic Equation

(2.36b)

Figures 2.15a,b show that the Cartesian and cylindrical co-ordinates are related by r2 = x2 +y2, x = r cos 6, and y = r sin B. These relationships are used to derive the first and secondderivatives

alax = (ar/ax) afar + (a 6 /ax) ala 6

= cos6 (alar) - ( 1/r) sin6(ala6)

a2/ax2 = [cos6(alar) - (1/r) sino (alaO)][cos0 (alar) - ( 11r) sin6 (ala6)]

= cos0 (alar)[cos0 (alar)-( 1/r) sin0 (ala0)]

- (1/r) sin6 (ala6)[cos6 (alar) - ( 1/r) sin6 (ala6)]= cos26(a2/are) + ( 1/ r2) sinecos6 (a/a6) + (1/r) sin26(alar)

+ (1/ r2) sin20(a2/a02) + (1/ r2) sinocos6(a/a6)

alay = (arlay) alar+ (a0lay) ala6

= sin0 (alar) + (1/r) cos0(ala0)

a2/ay2 = [sin6 (a/ar) + (1/ r) cos6 (ala6)][sin6 (alar) + (1/ r) cos6(ala6 )]= sin6 (alar)[sin6 (alar) + (1/ r) cos6 (ala6)]

+ (1/ r) cos6(ala6 )[sin6 (alar) + (1/ r) cos6 (ala6)]

= sin26 (a2/are) - (1/r2) sinecos6(a /a6) + (1/r) cos26(alar)+ (1/ r2) cos26(a2/a62) - (1/ r2) sinocos6(a/a6)

Adding eqs(2.37a and b), the biharmonic eq(2.23b) is transformed to

[ a2 /ar2 + (1/ r)(alar) + (1/ r2)(a2 /a62)]2q= 0

(2.37a)

(2.37b)

(2.38a)

If we let V2 = [ ], the LH side of eq(2.38a ) becomes [V 2 ] [V 2 ] O , which is abbreviated to

V4(r,0).0 =0 (2.38b)


2.5.4 Stress Component Functions

When 0= 0 in Figs 2.15a,b, we see that ox = o, and oY = oo. Equations (2.37a,b) may then beused to transform or, and oy (eqs 2 .22a and b) into o, and oo . The shear stress rro will followfrom either of the equilibrium eqs(2 . 36a or b ). This gives three plane polar stress functions:

or= (1/r)aelar+ (1/r2)a2,0 la02 (2.39a)

00 = a2(plar2 (2.39b)

rro - a [(11 r)(aplas)]lar_ (1 / r2 )aOlaO- (11r) (a2.lar aa) (2.39c)

The reader should check that eqs(2.39a-c) satisfy the two equilibrium equations (2.36a,b).

2.5.5 Body Forces

Apart from self-weight, an example of a radial body force R is the centrifugal force producedin a rotating disc or cylinder. This force derives from a potential Sa (r) as R = au /ar, so thatthe equilibrium eqs(2.36a,b) are modified to

(o,- oo)lr+(1/r)ar,olaO+aorlar +aolar =0 (2.40a)

(2/r) r,0 + (1/ r) a oola B+a r,olar = 0 (2.40b)

Equations (2.40a,b) are satisfied by the stress functions

or = (l/r)aq51ar + (1/r2 )a2-olaa2 (2.41 a)

op = a2q5 lar2+r ae lar (2.41 b)

rro= - a [(1/r)(ab1a0)]1ar (2.41 c)

At this stage a distinction is made between plane stress and plane strain as their constitutiverelations differ. For cylindrical plane stress (e.g. a thin disc), 0 and 0 in eqs(2.39a-c) mustsatisfy

v40(B, r)+(1 - v)V2SZ(r)=0

For cylindrical plane strain (a cylinder), 0 and 0 in egs (2.41a-c) must satisfy

V40(B,r)+[(1 -2v)/(1+v)]V2Q(r)=0

(2.42a)

(2.42b)

The expressions (2.42a,b) are identical to their Cartesian counterparts (2.28b) and (2.30b).However, the meaning of V 2 will now differ between the first and second terms ofeqs(2.42a,b). They are respectively

V4 = (v2)2 = [a2/ ar2 + (1/ r) (afar) + (1/ r2) (a 2/a 9 2)]2V 2

= a2 /ar2 + (I/ r) (alar)

These forms may be employed for the analysis of cylindrical bodies rotating about an axis ofsymmetry (see Chapter 3).


2.6 Polar Stress Functions

There is a number of useful stress functions 0 satisfying eq(2.38b). These may depend uponSand r alone or together, so that the following three classes of function are considered.

2.6.1 Stress Functions in 0

The function (0) is independent of r. Thus al ar = 0, and the biharmonic eq(2.38a)

becomes

[a2lar2+(1/r)(aalar)+(1/r2)(a2/a92)][(1/r2)a201a02]=0

which simplifies to

d'0/d0'+4d2O/d02=0

The solution , which is obtained from a substitution of the form N exp(m0), where N and

in are constants , has the final form

0= A cos20+ B sin20+CO (2.43)

where A , B and C are constants . The corresponding stress functions eqs(2 .39a-c) gives oo =

0 with

or= (1/ r2) d2tb /d02 (2.44a)

ro= (l/ r2) d4'ldO (2.44b)

Any part of eq(2.43) is a valid stress function . The following functions are of interest.

(a) Torsion of a DiscThe simple function 0= CO applies to the torsion of a thin disc or rotating wheel. Here oo =o, = 0 and eq(2.44b) gives the shear stress as

rre= Clr2

If the torque is applied to the axis of a solid disc of radius R, then C is found from the

equilibrium condition

T=2 7r f rrer2dr=2,rC fodr=2rrCRR R

which gives C = TI (2 nrR) and z = TI (2 it R r'). When a torque is applied to the axis of athin annulus with inner radius R; and outer radius R. (see Fig. 2. 16a) the equilibrium

condition becomes

R RoT= 21r f °rror2dr=2rrC f dr=2>rC(R0-R;)

Ri Ri

which gives C = T l [2ir (R„- R,)] and co = T l [21r r2 (R0- R;)]. In each case z varies


inversely with r2. For the hollow tube rre varies from a maximum value at R. to a minimumat R,,, as shown in Fig. 2.16b.

Figure 2.16 Hollow disc under axial torque

(b) Wedge with Tip MomentThe following example shows how the function q5= B sin20+ CO can be applied to a wedgewith a moment applied to its tip.

Example 2.10 Determine the stress distribution in the body of a wedge of unit thickness andapex angle 2a when a moment M is applied at its tip in the direction of 0 negative (see Fig.2.17)

Figure 2.17 Wedge with tip moment

Applying eqs(2.44a,b) with 0= B sin2a+ Ca yields the radial and shear stress components,

ar_(-4B/r2)sin2B (i)

ra = (l/ r ')(2B cos 20+ C) (ii)

The constants B and C are found from the two conditions: (a) rro = 0 for 0 = ± a in eq(ii)which gives

C= - 2B cos 2a


and (b) M = f : r,o r2 d6. Substituting from eq(ii) and (iii), we find the constants

B = M l [2(sin 2a - 2a cos2a)]

C = - M cos2a / (sin2a - 2a cos2a )

Substituting for B and C in eqs(i) and (ii) prescribes the stresses in the body of the wedge as

or - 2M sin20/ [r2(sin2a- 2acos2a)] (iv)

rro = (M / r2)[(cos20- cos2a) / [(sin2a- 2acos2a)] (v)

Equations (iv) and (v) show (see Fig . 2.17) that or, is zero along the centre line of the wedgeand reaches a maximum in tension and compression along the outer edges of magnitude

o,=±2M1[r2(1- 2a cot 2q)]

In contrast, the shear stress is zero at the edges and with a maximum value on the centre-line

r,o = (M / r 2)[(l - cos 2a) l (sin 2a - 2a cos 2a)]

Both stress components diminish rapidly with the square of radii away from the wedge tip.

2.6.2 Stress Functions in r

The stress function 5=5(r) refers to a body with symmetry about the z - axis. All points in

the r, B plane must displace radially. Thus v = av/ar = au/ar = 0 and eqs (2.35a-c) give r,o =

0 with the strains

e, = duldr, eo = ulr (2.45a,b)

These two strains depend upon a single displacement u. A compatibility condition , expresses

this dependence as

er = duldr = d (reo)ldr = r deoldr + eo

:. e, - eo = r deoldr

Since 0 is independent of 0, then aplaB= 0 and eq(2.38a) reduces to

[d2/dr2 + (11r) d/dr][d2O/dr2 + (1/r) dO/dr] = 0

which gives

d°Oldr°+ (2/r) d30/dr3 - (1/r2)d20/dr2+ (1/r3) dO/dr= 0 (2.46a)

Following a change of variable, with the substitution r = exp(t), the solution to eq(2.46a) is

0= A In r+ Br 2 In r+ C r 2 (2.46b)

The stress function eqs(2.39a-c) yield rre = 0 and


or = (11r) dO /dr (2.47a)

a0 = d2O ldr2 (2.47b)

where 0 may be the sum of any combination of right-hand terms within eq(2.46b). Thefollowing three functions are of practical interest.

(a) Equi-biaxial StressingApplying eqs(2.47a,b) to the function 45 = C r 2 gives equal stresses ar = o0 = 2C. Thesedescribe uniform plane stress states (i) within the wall of a thin-walled spherical vessel underpressure and (ii ) in a thin plate subjected to co -planar, concurrent forces . Other functions thatfollow employ the C r2 term to match boundary conditions for axi -symmetric bodies.

(b) Thick Cylinder Under PressureTaking the stress function 0 = A In r + C r 2, eqs(2.47a,b) supply radial and hoop stresscomponents as

ar = 2C + A / r2 and o0 = 2C - A l r2 (2.48a,b)

These are the Lame stresses (Gabriel Lame 1795-1870). They exist in the wall of a long,thick cylinder subjected to internal (and external) pressure. The axial stress or, depends uponthe end condition of the cylinder. The ends are said to be open when an internal pressure iscontained by bore pistons. The cylinder wall in then unstressed axially (oo = 0). When closedends contain the pressure, or, is found from a horizontal force equilibrium equation betweenthe forces exerted by the axial stress in the cylinder wall and the internal pressure acting uponthe end plates:

uu,r(r„2 - r;2)=p;r r2

.: az=pr2/(r„2- r.2) (2.49)

If the cylinder is built in to rigid end closures , its natural extension will be prevented . A planestrain condition may then be applied in which the axial strain eZ is set either to zero (seeExample 2 . 11) or to a constant value to find the axial wall stress.

Example 2.11 Determine the constants A and C in the Lame equations, given that the cylinderis subjected to internal pressure only and that the axial strain is zero. Plot the distributionof or, a0 and aZ and derive a general expression for the radial displacement.

The boundary conditions are or = - p; for r = r; and or = 0 for r = r0 . Substituting intoeq(2.48a) leads to the two simultaneous equations:

- pi= 2C+ A I r,2 (i)0=2C+A/rn2 (ii)

Subtracting (ii) from (i) gives

A=-pr,?r,,2/( r„2- r2 ), 2C=pr;2/(r„2- r2)

Substituting for A and 2C into eqs (2.48a and b)


o1. =pr,2(1- r02/r2)l (r02- r,2) (iii)

00 = pr,? (1 + r„2l r 2) l (r„2 - r,2 ) (iv)

When eZ = 0

o:=v(or+ oo)=2vpr;2/(r„2- r,2) (v)

Equation (v) differs slightly from the closed-end cylinder expression (2.49) but both giveconstant or, (independent of r). The stresses in eqs(iii)-(v) are distributed as shown in Fig.2.18.

Figure 2.18 Stress distributions within the wall of a thick cylinder under internal pressure

At r = r, the radial stress or is maximum in compession and the hoop stress ae is a maximumin tension. The radial displacement follows from eq(iii)-(v) and either of eqs(2.45a or b):

u=re,= r[ae - v(or+ oo) ]IE

u=r[ae(1- V2) - var(1 +v)]/E

=prr;2[(1- v2)(1 +r„2/r2) - v(1 +v)(1- r„2/r2)] / [E(r,,2- r2) ]

=pr,2[(1- v - 2v2)r+(1+v)r„2/rI /[E(r„2- r,?)]

(c) Bending of a Curved BeamTaking the complete function 45 from eq(2.46b), the following radial and tangential stressesare found from eq(2.47a,b):

ar=A/r2+ B(1+2lnr)+2C (2.50a)

ao=-Alr2+ B(3+2lnr )+2C (2.50b)

Golovin (1881) recognised that egs(2.50a,b) are the exact solutions to the stresses in a curvedbeam under pure bending (see Fig. 2.19 and reference p.75). To find constants A, B and C,we let a hogging moment M be applied to a curved rectangular beam of unit thickness andwith inner and outer radii r, and r. respectively. The first boundary condition is that of zerostress, normal to the outer surfaces, ar = 0 for r = r; and r = re, . From eqs (2.47a) and (2.50a)

or=[(1/r)dOldr]^ =Al r,2+B(1+2lnr,)+2C=0 (2.51a)

ar=[(1/r)d0ldr], =A/r02+B(1+21nre,)+2C=0 (2.51b)


The second boundary condition refers to end moment equilibrium and uses eq (2.47b)

M = f oo r dr = f (d2O/dr2 )r dr = I r dp /dr I'° - f (d, /dr) dr (2.52a)

It follows from eqs(2.51a and b), that the first term is zero . Integrating eq(2.52a) andsubstituting into eq (2.46b),

M= -I^I;° =-[ AIn (r„lr,)+B (r021nr„-r21nr,)+C(r„2-r,2)] (2.52b)

The constants A, B and C are found from the simultaneous solution to egs(2.51 a,b) and(2.52b). This results in the final expression for the radial and hoop stresses

or

oe

4M[(rrr2Ir2) In(ro/r, ) + ro In (r/ro) + r+ In (r,lr)]

(r0-r+ 2)2 - 4rzr2 [In(ro/r;)]2

4M[-(r2ro /r2)In(ro / rj) . roIn ( r/ro) . r? In(r1/r ) + (ro - ri)]

(r0-r2 )2 - 4ror?[ In (ra/rr)]2

(2.53a)

(2.53b)

Equation (2.53b) is the bending stress with the distribution shown in Fig. 2.19. The positionof the neutral axis can be found from setting oB = 0 in eq(2.53b). Equations (2.53a,b) satisfyboth compatibility and the boundary conditions and will provide exact solutions to thestresses everywhere in the beam, provided that the applied moments correspond to the or,,end-distribution, as shown. Even if this end condition is not met exactly, St Venant's principle(Saint-Venant 1797-1886) assures us that the solution will be accurate at a distance equal tothe beam depth beyond the ends. The principal cannot guarantee similar accuracy forapproximate solutions to this problem where they do not satisfy the three fundamentalrequirements of elasticity theory. These are equilibrium of forces, compatibility of strain anda match to the boundary conditions.

Figure 2.19 Bending of curved bar


2.6.3 General Function, 0 = 0 (9, r)

(a) Point Normal Force on Semi-infinite PlateOne of the most useful functions to satisfy eq(2.38b) is that attributed to Boussinesq (1885)'and Flambert (1892). Their function has the following form:

0= CrOsin0 (2.54)

Applying eqs(2.39a-c) to eq(2.54) gives the stresses as oa = rre = 0 and

or= (1/ r) aOlar+ (1/ r2)a2p/ae2

= (C/r) O sinO+ (1/ r2 )(alae )[Cr sin6 + CrO cos0]

= (2C / r) cos O 2.55a)

This radial stress arises from applying an outward point force normally to the straight edgeof a large (semi-infinite) plate, as shown in Fig. 2.20a.

(a) Imo' / \ I L\'I (b)

Figure 2.20 Radial stress under a point force on a semi-infinite plate

For a given radius, or varies with cosO as shown, taking a zero value for 0= ± it/2 and amaximum value 2C/r for 0= 0. The constant C, in eq(2.55a), is found from the horizontalforce equilibrium equation within a plate of unit thickness

P= f or (r d19) cosO

P=2C f nrz cos20 d6=CI0+thsin2B nn =Crrn/2 -nt2

In practice, it is more common for a compressive force to act in a line across a plate ofthickness t (see Fig. 2.20b). Then P is replaced by - Pit to give - Pit = Cn when eq(2.55a)supplies a compressive radial stress

or= - [2Pl(ntr)]cos0 (2.55b)

Consider the circle of diameter d, lying tangential to y at P, in Fig.2.20a.

a see Timoshenko S. P. and Goodier J. N., Theory of'Elacticity, 2nd edition , McGraw-Hill, 1984.


Since r = d cos 0, eq(2.55a) becomes

or = 2C l d = 2Pl(,r d) = constant (2.55c)

Note also that the maximum shear stress in the plate is given by r, = (or - oo )/2.Substituting from eq (2.55c), with oe= 0, gives r,,, = orl2 = Pl( ird). This means that both orand r,,, are constant around the given tangent circle , each being inversely proportional to itsdiameter is not the stress component rr0). The theory is readily confirmed from pointloading the straight edge of a birefringent transparent polymer and viewing the image underpolarised light . The photoelastic method reveals the tangent circles as isochromatic fringes,each of a constant shear stress value. An example of an isochromatic fringe pattern is shownon the front cover of this book.

Example 2 .12 Determine the displacement expressions in the body of a thin, semi-infiniteplate when a compressive point force is applied normal to the straight edge.

Referring to Fig. 2.20b, the displacements follow from combining the plane stress constitutiverelations egs(2.14a,b,d) with the polar strain displacement relations (2.35a,b,c),

er=aular=(or - voo)IE

ea=ulr+(1/r)avlao= (a0- vor)/E

Yro= rro/G=avlar - vl r+(l/r)aula9

Substituting oo = rro = 0 and or = - [2Pl(7r tr)] cos 0, from eq (2.55b), gives

aular= - [2Pl(,rEtr)]cos6 (i)ulr+ (11r) av1a6= [2vPl( 7rEtr)] cosO (ii)av lar - v / r + (1/r) aula6= 0 (iii)

Integrating eq(i) leads to

u = - (2Pl(,r tE)] (cos 6) In r +f (0)

Substituting (iv) in (ii)

- [2P/(,rtE)] (cos 0 ) In r + f (6) + av/a0= [2vPl( ,rtE)] cos6avla0= [2vP/(;rtE)] cos6+ [2P/(,rtE)] (cos 0 ) In r - f (6)

v = f { [2vP/(irtE)] cos6+ [2Pl(;rtE)] (cos6 ) In r - f (6)} d6

= [2vP/( 7rtE)] sin6+ [2P/(,rtE)] (sin 0 ) In r - f f (6) d6+g(r) (v)

From eqs(iv) and (v), the partial derivatives of u and v are

au1a6 = [2Pl(;rtE)} (sin 0 ) In r + f '(0) (vi)avlar = [2P/( n tEr)] sin 0+ g'(r) (vii)

Substituting eqs(v), (vi) and (vii) into (iii) gives

f'(6)+[2Pl(,rtE)]( l -v)sin6+rg '(r)+ f f(0)d6- g(r)=0


The right-hand side is zero , so the sum of the separate functions in r and Oequals zero:

rg'(r) - g (r) = 0 or dglg = drlr

..ing=lnr+lnA , g(r)=A,r

and

f'(e) + [2P1( n.tE)](1 - v) sin0+ f f (0) de= 0

d2f/d02 +f=- [2Pl( ntE)](1 - v)cose

The solution to eq(ix) is

f (6) =A2 sine+A, cose- (1 - v )[P/(ntE)] 0sine (x)

The general u, v displacement expressions are found by substituting eqs(viii) and (x) intoeqs(iv) and (v). This gives

u =A2 sin0+A, cosh- [2Pl(ntE)] (cosh ) In r - (1 - v )[P/( ntE)]0sin0 (xi)v = (1 + v)[Pl(ntE)] sin6+ [2Pl(ntE)] (sine) In r - (1 - v )[P/(ntE)]ecose

+ A, r + A2 cos O - A, sin O

Since v = 0 when 0= 0 along the x - axis, then, from eq(xii), A2 = - A, r. This condition canonly be satisfied by A2 = A, = 0 for all r. Because deformation is localised, it is assumed thatu = 0 at some radius r„ from P along the x - axis (where 0= 0°). Thus, from eq(xi),

u=0=A3 - [2P/(ntE)] In rA, = [2P/(ntE)] In r„

Substituting for A, , A2 and A, into eqs(xi) and (xii), the plate displacements at a point 0 andr s r0 are given by

u = [Pl(ntE)][2 In (r°l r) cos0- (1 - v )0sin0] (xiii)

v = [Pl(ntE)][(1 + v) sin0+ 2 (sin0) In (r/r„) - (1 - v )0cos0] (xiv)

When 0= 0°, eqs(xiii) and (xiv) give v = 0 and u = [2P/(ntE)]ln(rjr). This shows that thedisplacement under P falls rapidly to zero as r -► r„ . The fact that both a, and u becomeinfinite for r = 0 indicates that localised plasticity would occur if it were possible to achievea true point force. Yielding is less likely to occur in practice where the force is distributedover a finite area. Any yielding that does occur at the load point will result only in a localdisturbance to the predicted internal elastic stress distribution.

(b) Tangential Force on Semi-infinite PlateThe radial stress expression eq(2.55a) also applies when a tangential shear force is appliedto the straight edge of a semi-infinite plate. This requires that 0 be measured from an x - axisaligned with the force direction, as shown in Fig. 2.21a.

(c) Point Force Applied to Finite BodiesEquation (2.55a) may also be applied to point loading of finite bodies when the constant Cis re-defined. For example, this equation supplies the radial stress distribution within the bodyof a wedge when an inclined compressive force is applied to its tip. Figure 2.21b shows awedge of unit thickness, apex angle 2a and inclination 8 of its compressive tip force P.


Figure 2.21 Further applications of the Boussinesq function

By taking 0 anti-clockwise from the force line, the horizontal equilibrium equation becomes

f (a 4)PIt+J o,(rdO)cos0=0

(a-P)(a .0) 2

..PIt=-2CJ cosBdo(a -P)

C 10+ t/2 sin 26 (Cc +P) C (2a+ sin2acos2Q)(a - P)

C= - PI [t (2a+ sin2acos2i)]:. or _ - (2P cos 0) / [ r t (2a+ sin2acos2,0)] (2.56)

The wedge stresses arising from horizontal and vertical tip forces will be found bysubstituting ,6= 0 and ,3=ir/2 respectively in eq(2.56). When these loads are combined, theo, distributions corresponding to each load applied separately may be superimposed to givea net stress distribution.

Example 2 .13 The wedge in Fig. 2.21b has an apex angle of 30° and supports a verticaldownward force of 10 kN. If the wedge thickness is 20 mm, find the maximum stress on aradius of 50 mm. Show that the simple theory of bending will provide 75% of the maximumradial stress when 2a:5 100.

In this case we set 8= 90° =n/2, in eq(2.56), to give

o, _ - (P cos O )/ [ r t (a - sina cosa )] (i)

Since 0 is measured from the force line, the wedge interior is defined by 75° s 0 s 105°. Onthe bottom edge 0= 75° and on the top edge 0= 105°. In addition P = 10 x 10' N, t = 20mm, r = 50 mm and a= 15° in eq(i). The radial stress on the bottom edge becomes: o, = -219.4 MPa in compression and on the top edge o, = 219.4 MPa in tension. Also when 0=90°, o, = 0 along the wedge axis.

The radial stress distribution resembles that obtained from bending a cantilever beam ofrectangular section (see Fig. 2.10). Indeed, we may apply bending theory to a slightly taperedcantilever beam to provide an approximation to these maxima. Let x define the wedge axisin Fig. 2.21b so that a point on the sloping side has co-ordinates Q (x, y). When X13= 90°, thebending stress at Q is

o= My/I (ii)

where the bending moment M = Px and the second moment of area for the transverse plane


containing Q is I = t (2 y) 3/ 12. Substituting into (ii) gives

o= 3Px / (2 t y2) (iii)

When a s 5°, we may write

y2=r2-x2=(r-x)(r+x)=2x(r-x) (iv)

Substituting eq(iv) into (iii) gives

a=3Pl[4t(r-x)] (v)

When is small, sina= a, so that eq(i) approximates to a tensile bending stress,

o,j - Pcos (90+a)/[rta ( l - cosa)]=P/[rt( I-cosa)] (vi)

and setting cos a= xlr in eq(vi) gives

o,=PI[t(r-x)] (vii)

Comparing eqs(v) and (vii) shows that bending theory approximates to 75% of the maximumtensile radial stress. The same approximation holds for the compressive side where negative

signs accompany eqs(v) and (vii).

(d) Disc Under Diametral CompressionConsider a disc of diameter d and thickness t subjected to diametral compressive force P, as

shown in Fig. 2.22a. A stress function of the following form applies:

i= CrOsinO+ Dr'

Equations (2.39a-c) give the stress components as r,B = 0

o, _ (2C /r) cos h+ 2D and or,=2D

(2.57)

(2.58a,b)

The term in D superimposes a uniform biaxial tension upon Boussinesq's radial stress field.This is necessary to account for the interruption caused to the stress field by the discboundary. Neither a normal stress nor a tangential shear stress can exist around the boundary.This condition can be matched when eqs(2.58a,b) superimpose to give oe = a, = 0. Referringto Fig. 2.22a, the disc diameter is d = r/cosO= r'/cosh'.

The combined stress state at any point A on the boundary, resulting from each force P, isgiven by the sum of eqs(2.58a,b). This gives an equibiaxial tension as shown:

o,=oo=(2Cld)+4D (2.58c)

The resolution in oo and or normal and parallel to the boundary at A, will give two directstresses of equal magnitude with no shear stress. It follows from eq(2.58c) that the normalstress will be zero when D = - C/(2d). Hence, for any point (r,O) in the disc, the stresses

eqs(2.58a,b) due to a single force are


o,= (2C/r) cos9 - CId, oo= - Cld (2.59a,b)

The constant C is found by applying the horizontal equilibrium condition across a sectionaligned with the y - axis

P+ f oxtdy=0

where x is aligned with horizontal force line.

2D 2CId+2D

Figure 2.22 Stresses o, and ore in a disc under compression

(2.60)

At point B in Fig. 2.22b, ox is the sum of the the radial and hoop stresses resulting from eachforce resolved into the x direction. The stress transformations for the left half of the disc are

Qx = a. COS29+ op sin20or, = o, sin2B+ Oro cos20

r,,y=th( o,- oo)sin20

For the right half of the disc, B in egs(2.61a-c) is replaced by - 0, to give

ox = o, COS20 + op sin20

o, = o, sin 20+ oe cos29

r=- '/2(o,- oe)sin29

(2.61 a)(2.61 b)(2.61 c)

(2.62a)(2.62b)(2.62c)

The sum of eqs(2.61c) and (2.62c) confirms that Erx,, = 0 along an axis of symmetry.Substituting L.ox from eqs(2.61a) and (2.62a) into eq(2.60),

P + 2 t f (o, cos 29+ oo sin20)dy = 0 (2.63a)

Since y = (d /2) tan 0, then dy/d9= (d /2) sec29. Substituting for dy and o, and oo fromeq(2.59a,b) into (2.63a) leads to

n

P + 2C t f0

4 (4 cos 29 - sec29) d9= 0 (2.63b)

Equation (2.63b) gives C = - PI(7rt). Thus, for the y - axis,


ox = - [2P l(rrdt )](4 cos2B- 1) cos2B+ (2Pl;rdt) sin20 (2.64a)

oY = - [2Pl(rrdt)](4 cos2B- 1) sin2B+ (2P1 dt) cos2B (2.64b)

Finally , substituting cos0= d /3(d 2 + 4y2 ), eqs(2.64a,b) lead to

ox = - 2P (3d 2 + 4y2)(d 2 - 4y2) / [(,rdt)(d 2 + 4 y2 )] (2.65a)

oY = 2P (d 2 - 4 y/ [(n dt )(d 2 + 4 y2 )] (2.65b)

Equations (2.65a,b) show that or = ay = 0 at the boundary (y = d /2). At the disc centre (y =0), these attain their maxima ox = - 6Pl(rrdt) and oY = 2Pl(7rdt ), in compression and tensionrespectively.

(e) Stress ConcentrationsA function of the following form

0(r,0)=f(r)cos20

will satisfy eq(2.38b) when f (r) is written as

f(r)=Ar2+Br4+Clr2+D

(2.66a)

(2.66b)

in which A, B, C and D are constants. Equation (2.66b) may be applied to the uniformlystressed plate with the small central hole, of radius a, shown in Fig. 2.23a.

S

Figure 2.23 Stress distributions around a hole in a uniformly stressed plate


The full effect that the hole has in concentrating stress is determined by superimposing theLame stress function (2.46b). This is written as

0= 0 (r)=Elnr+Fr2 (2.67)

where E and F are constants. Using St Venant's principle, it may be assumed that at anyradius b >> a, the stress distribution is not affected by the hole. At radius b, a uniform radialLame pressure S/2 appears as a result of transforming the horizontally applied stress, S. Nowat the position (b, 0) shown, the radial and shear stress components are

ar = S cos20= (S /2) (1 + cos 20) (2.68a)rro = - (S /2) sin 20 (2.68b)

The accompanying tangential shear stress component (2.68b) and that part of eq(2.68a) whichis or= (S /2) cos20are derived from eq(2.66a) and the stress functions (2.39a,c). It followsthat for r s b, the stress components or, ae and zre are the result of summing the stressfunctions (2.66a) and (2.67). That is,

0(r,B)=(Ar2+Br4+C1r2+D)cos20+(Elnr+Fr2) (2.69a)

for which eqs(2 . 39a-c ) supply the stress components . The boundary conditions foreqs(2 . 68a,b ) are (i ) Cr = rro = 0 for r = a and (ii) or = S12 and rro = - (S/2) sin2 0 for r = b.Applying conditions (i) and (ii) to an infinitely large plate , where alb ,., 0, leads to A = - S/4,B = 0, C = - a4S /4, D = a2S /2, E = - a2S /2 and F = S /4. Equation(2.69a) becomes

0 (r, 61) = (S/4)[r2 - 2a2ln r - (r - a 2/r) 2 cos 20] (2.69b)

From eqs(2.39a,b,c),

or=(S/2)[(1 - a2/r2)+( 1 - 4a2/r2 +3a4/r4)cos20] (2.70a)

vo= (S/2)[( 1 +a2/ r2 ) - ( 1 + 3a4/ r4 ) cos 20] (2.70b)

rro= - (S/2)( 1+2a2/r2 - 3a4/r4) sin20 (2.70c)

Equations (2.70a,b,c) are particularly useful for determining the stress concentration factor.

On a transverse y - axis, passing through the centre of the hole, where 0= n/2 and 0= 3;r/2,

they give rro = 0, and

ao=(S/2) (2+a2/r2+3a4/r4) (2.71 a)ar=(S/2) (3a2/r2) (1 - a2/r2) (2.71b)

Clearly when r z a, ao = 3S, ar = 0 and when r 2! 10a, ao = S, ar = 0. The manner in whichstress is distributed according to egs(2.7la,b), within intermediate radii a < r < 10a, is shownin Fig. 2.23a. The two distributions agree with experimental data given for a 10 s.w.g., 200mm wide L72 duralumin plate with a 50.8 mm central hole. They confirm a threefoldmagnification (the stress concentration factor) in the axial stress for two transverse points Aand B on the hole boundary. In Fig. 2.23b, the hoop stress distribution around the hole isfound by substituting r = a in eq(2.70b). This shows that the stress changes from tensile tocompressive at the four singular points (P, Q, R and S) where ao = 0. Equation (2.70b) showsthat these points occur at 0=1/2 cos -'(0.5) = 30°, 150°, 210° and 330°.


(a) / I I I (b)

Figure 2.24 Hole in a plate under uniform shear

The solutions (2.70a-c) may be extended to pure shear loading of a plate with a hole, shownin Fig . 2.24a. For this, the equivalent stress system given in Fig . 2.24b is employed. It followsfrom eqs(2.70a-c) that the stress components under the uniform compressive stress (- S) forthe y - direction are found by replacing B with 0 - ,r/2. This gives upon factorisation,

ar = (- S /2) (1 - a 2/ r2)11 + (I - 3a 2/ r2) cos [2(0- is/2)] } (2.72a)

a,=( - S/2) ((I +a 2 / r2) - (1 + 3a 4/ r4) cos [2(0- ,r/2)] } (2.72b)

rro= - ( - S/2)(1 - a2/ r2) (1 + 3a2/ r2) sin [2(B- fr/2)] (2.72c)

Adding eqs(2.70) and (2.72) gives the stress state under pure shear

ar=S(1+3a4/r4-4a2/r2)cos20 (2.73a)

ao= - S(1+3a41r4)cos20 (2.73b)

rro=- S(1 - 3a4/r4+2a2/r2)sin20 (2.73c)

Figures 2.25a,b show the manner in which eqs(2.73a and b) are distributed along they (0=n /2) and x (0 = 0°) axes respectively. Where these axes intersect the hole (r = a), ao ismagnified by a factor of 4 in both the x and y directions. Setting r = a in eq(2.73b), the hoopstress around the hole is ao = - 4S cos26. This reveals unstressed points for 0= 45°, 135°,225° and 315° in Fig. 2.25c.

The principal of superposition can again be applied to the general case of in-plane biaxialstressing of a plate with a hole under under S,t and Sy (see Fig. 2.26a). Setting 0= 0- 90° with

S = Sy in eqs(2.70a-c) gives ar, ao and rro for Sy acting alone. Adding these to eqs(2.70a-c),for when S. acts alone, gives the stress components for the combined loading

ar=t/2(Sx+Sx)(1 - a2/ r2)+th(SX- Sy)(1 - 4a2/r2+3a4/r4)cos20 (2.74a)

ao = t/2 (Sx + S}.)(1 + a 2/ r2) - t/z(SS - S,,)(1 + 3a 4/ r4) cos 28 (2.74b)

rro = - t/2 (S., - Sy)(1 + 2a2/ r2 - 3a4/ r4) sin 20 (2.74c)


(c)

Figure 2.25 Stress distributions around a hole in a plate under pure shear

Equation (2.74b) allows the stress concentration at the hole (r = a) to be determined for givenratios K = S,,ISX. For 0= 0° this gives ao/S,t = 3K - 1 and for 0=z/2, .7o/Sx = 3 - K. Thelinear dependence of a normalised stress concentration (oo/SS) upon K is shown in Fig. 2.26b.This includes the previous cases of uniaxial tension when K = 0 and also pure shear when K= - 1. As K increases beyond 3, the stress concentration increases positively for 0= 0° underbiaxial tension and decreases negatively for 0= ,r12.

s

S.

-

Lt.-2 -I I 2 3 4

0=a/2

(a) / I - (b)B=0

Fig. 2 .26 Stress concentrations for a plate with hole under biaxial tension

Example 2.14 Determine the thickness of reinforcement necessary around a circular windowin a pressurised fuselage in order to eliminate the stress concentration

In a thin-walled tubular fuselage , the pressure produces biaxial stressing in the ratio S},/ SS =2. In the absence of reinforcement , the greatest stress concentration would be 5 (see Fig.2.26b). To eliminate this , the hole bead reinforcement must serve to reduce stresses to those

PLANE ELASTICITY THEORY as

values existing away from the hole, i .e for r >> a in eqs(2 .74a,b,c). That is,

(i)o, = (Ss/2)(3 - cos 20)oo = (SX l2)(3 + cos 29) (ii)rre = (SX/2) sin 20 (iii)

Let the reinforcement consist of a circular section bead bonded to the hole at radius a. Figure2.27a shows that the radial and shear forces exerted by the fuselage plate (thickness t) on anadjacent element of this bead are or(a89 )t and rre(a89)t respectively. The inner surface isforce free. Also shown are the variations in the tensile force T and transverse shear force Q

carried by the bead. These are (dT/d 0) 80 and (dQld 0) 60 respectively in the direction 0-positive.

(a)

Figure 2.27 Forces on a reinforcing bead

Radial force equilibrium gives

o,t (a89) + (dQ/d9)89cos (80/2) - [T + (dT/d9)89]sin(89/2) - T sin (89/2) = 0

As 89-► 0 we may set cos (80/2) -> 89/2 rad and sin (89/2) -> 0 to give

orta+dQld9- T=0

(b)

(iv)

where small products have been ignored . Tangential force equilibrium (Fig. 2.27a) leads to

rrota+dTld9+Q=O

Combining the two equilibrium equations (iv) and (v) leads to a differential equation

d2T/d02 + T = to (or - drre /d 0)

Substituting for or and rre from (i) and (iii ) into (vi) gives

d2T/d02 + T= (3 S.tal2)(1 - cos 20)

The solution to eq(vii) is

T=A cos; 9+ B sin 0+ (3Ss ta/2)[1 +1/scos 20]

(v)

The constants A and B are found from a horizontal force equilibrium condition across the


vertical section, i.e. 2T= 2aSxt when 0= ±n/2 (see Fig. 2.27b). Substituting into eq(viii)gives aSt = B + Sxta, from which B = 0. Similarly, for vertical force equilibrium across thehorizontal section (6= 0, ;r), 2T= 2aSy t = 2a(2Sx) t. Equation (viii) gives 2aSt = A + 2aSxt, from which A = 0. Thus eq(viii) simplifies to

T= (3Stal2)[ 1 + 1/3 cos 261 (ix)

A final compatibility condition is required to match the hoop strain in the sheet with that inthe bead . The sheet is under biaxial tensile stresses ao and a, and the bead is under a directtensile force T. The corresponding strains are equalised when , for a bead of area A,,

Tl(A,,E)_(a0- var)IE (x)

Substituting from eqs (i) and (ii) and (viii ) into (ix)

[3Sxt al(2A,,E)][1 + 1/3 cos 26] = [Sx/(2E)][3(1 - v) + (1 + v) cos 26]

This shows that the cross-sectional area of the bead must vary with B, according to

A,,=at(3+cos26)/[3(1 - v)+(1 +v)cos26] xi)

For 6= 0°, A,, = 2at l(2 - v) and for 0= ir12, A = atl(1 - 2v ). This shows that by taking v= 1/9 for an aluminium alloy bead, its area increases from 6at/5 to 3at as 0 increases from 0°to 90°. Symmetry holds in the remaining quadrants. The volume of material required is

n /2V=4 J

r

0A,,ad0

Substituting eq(xi) into eq(xii), the bead volume becomes

n/2

V=4f [a2t(3 + cos 20)d9]

J [3(1-v)+(l+v)cos200

It is left as an exercise for the reader to evaluate eq(xiii) numerically, using Simpson's rule.This shows that V - 10 a2 t. Since the volume of the disc removed is nag t, then 10 /it timesthe disc volume must be introduced as a bead to remove the stress concentration.

EXERCISES

Equilibrium and Compatibility in Cartesian Co-ordinates

2.1 Show that the following strains and displacements are all compatible in plane stress : (i) es = axy,e, = by3, y, = c - dye , (ii) ex = q (x2 + y2 ), e,. = qy2, y, = 2gxy and (iii ) u = ax2y2, v = byx3.

2.2 Analysis of plane strain (e. = y,,, = yx = 0) shows that non-zero strains can be represented by:ex=a+b(x2+y2)+x"+y°, e,,=c+d(x2+y2)+x°+y°, yxv=e+fxy( x2+y2- g2)

where a, b, c , d, e, f and g are constants . Show that these are only possible iff = 4 and b + d + 2g2 =0. Hence derive the displacements from the strains.Answer: u = Ax + b(x3/3 + y2x) + x5/5 + y°x + C3 y + C1, v = cy + d(x2y + y3/3) + x4y + y5/5 + Cox + C2


2.3 Find the conditions for which the shear strain yxv = Ax2y + Bxy + Cx2 + Dy is compatible with the

displacements u = ax2y2 + bxy2 + cx2y and v = ax2y + bxy.

Answer : B=a(a+b),A=2a,D=b,C=c

2.4 A thin plate is subjected to the following in-plane stress field components : ox = ay' + bx2 y - cx,

o, = dy' - e, r_ = fxy2 + gx2 y - h. What are the constraints on the constants a, b, c, d, e, f, g and h so

that the stress field satisfies both equilibrium and compatibility?Answer: a = 2f /3, b = - f, d = - f 13, c = g = 0, e, f and h are unconstrained

2.5 The stress system in a plate consists of direct tensile stresses oz = cyx2, o,, = cy' /3 together with a

shear stress rn,. Determine rx,,.

Answer: r_ cxy 2

2.6 The general expressions for the two direct stress components of a plane stress field have the formox = ax 3 + bx 2 y + cxy 2 + dy', oa = kx' + lx 2 y + na y' + ny', where a , b, c, d, k , 1, in, n are constants.Determine the necessary relationship for these constants if the stress distribution is to be compatible.For the special case where a = c = d = k = m = 0, determine expressions for the compatible stress

components o, o,, and r_. (CEI)Answer : (6a + 3k + c)(3a - b) - (m - 3a)(2b + I + 3d) = 0; ou = 3x2y; o,, = y3 - 6x2y, rx,, = 2x' - 3xy2

2.7 A square plate of side length a and with sides along the co -ordinate axes is fixed in position at the

origin and with the side along the y - axis fixed in direction ( au/ay = 0). If the stresses in the plate are

ox = Ay/a, o,, = A with a consistent value of shear stress, find expressions for this shear stress and thegeneral displacements . What are the displacements existing at the corner (a,a)? (IC)

Answer: rx,, = K, u = Axyl(aE) - v AVE, v = Ay/E - v Aye/(2aE) + 2Kx(1 + v)IE - Axe 1(2aE) and at

(a, a), u = Aa(1 - v )IE, v = Aa(1 - v)/(2E) + 2Ka(1 + v )lE

2.8 Derive general expressions for the displacements at any point (x, y) in the cantilever of Fig. 2.10.

Find the constants and identify the terms in the final displacement expressions for each of the following

boundary conditions : (i) u = v = 0 and aulay = 0 at (1, 0), (ii) u = v = 0 and av/ax = 0 at (1, 0).

Answer: (i) u = [Pl(2EI )](x2y - 12y + v y'/3) - Py31(6GI ), v = [P/(EI )](12x12 - vxy2/2 - x3/6-

0/3) + [Ph2 /(2GI )](x - 1); (ii) u = [P1(2EI)](x2 y - 12y + v y'13) - [Py3 /(6GI )](3h2/y2 - 1), v =

[P/(El )](I2x/2 - vxy2/2 - x3 /6 - 1' /3) + [Ph2/(2GI )](x - 1). Terms with E are deflections due to

bending . Terms with G are deflections due to shear

Cartesian Stress Functions

2.9 Derive the stress function supplying the stress components ox = cy, ov = cx, rx,, k, where c and

k are constants.Answer: c(x3 + y3 )/6 + kxy

2.10 Find the condition for which the stress function 0 = Axy° + Bx3 y2 is valid.

Answer: A + B = 0

2.11 The stress function 0 = Axe + Bxy + C? provides the direct and shear stresses in a thin rectangular

plate . One comer, which lies at the origin , is position - and direction-fixed . Show that the displacements

for the diagonally opposite corner (a, b) are : u = 2a(C - v A)/E, v = 2b(A - v C)/E.

2.12 If 0 in Exercise 2.11 applies to a thick plate in plane strain , where e , = 0, determine the

displacements at the comer (a, b).Answer : u = 2x(1 + v)[C (l - v)- Av ]/E, v = 2y(1 + v)[A (1 - v) - Cv]/E

2.13 Show that the stress function 0 = Dx3 matches in-plane bending of a thin plate of width 2a underend moments M (see Fig. 2.28). Taking the thicknesses to be unity, derive the displacements. Given


that the plate centre is position fixed, show that the displacements at the point (a,0) are given byu = - Mva2 /(2EI)andv=0.

Answer: u = - M (v x2 + y2 )/(2EI) + C, y + C2, v = Mxyl(EI) + C3x + C, where C; are constants

v

(010)

a,

(a, 0)Ilb

xx

Figure 2 .28 Figure 2.29

2.14. Show that the stress function 0 = Dx3 applies to linearly varying edge loading of the plate shownin Fig . 2.29. F is the total force applied to the plate of width a. Taking the thicknesses to be unity,derive the displacements . Given that the plate origin (0,0) is position fixed show that the displacementsat point (a, 0) are u = - v FIE, v = 0.

Answer: u = - F (v x2 + y2)/(Ea2) + C, y + C2 , v = 2Fxyl(Ea2) - C3x + C4 where C, are constants

2.15 Derive the stress components from the stress function 0 = Axy3 + Bxy + Cy3, given that the stressresultants are a shear force F and a bending moment M at the section x = 0 of a beam. The latter hasconstant I and depth ± h relative to the neutral x-axis . How could these'stress distributions be madeto represent those for an encastre beam carrying a central concentrated load?

Answer: ox = FxylI + My/I, rx,, = F(h2 - x2)/(2I )

2.16 The stress function 0 = Ax2 + Bx2y + Cx2y3 provides the exact solution to the distribution of stressin a beam of depth 2h with unit thickness when it is simply supported over a length 2L and carries auniformly distributed load q/unit length . With the origin at the beam centre , derive the stresses andcompare these with those from simple bending theory. What is the vertical displacement at the centre?

Answer: ax = [gyl(21)](x2 - 12 + 2h2/5 - 2)2/3), or, = [q/(2n](y2/3 - yh2 - 2h3/3),

rte. = [gxl(21)](h2 - y2), v= [5ql/(24E1)][1 + [12h2/(5l2)l(v /2 + 4/5)]

2.17 Determine the state of stress represented by the stress function 0 = Axe - By3 when applied to athin square plate (a x a) with origin at the centre of the left vertical side. Check that both equilibriumand compatibility are satisfied. Show the variation of stress on the plate surfaces. (CEI)

2.18 Show that the stress function 0 _ [q/(8c3 )][x2 (y3 - 3c2 y + 2c3) - (y3 /5)(y2 - 2c2 )] is valid anddetermine the problem it solves when applied to the region: y = ± c, x = 0 on the side x - positive.

2.19 A cantilever beam with rectangular cross-section occupies the region - a s z s a, - h s y s h and0 s x s 1. The end x = I is built-in and a concentrated force P is applied at the free end (x = 0) in the+ y direction. Given that the non-zero stress components are ox = Cxy and rr, = A + Bye show that theconstants become 2B + C = 0 and that C = 3Pl(4ah3) when the stresses satisfy force and momentequilibrium at the free end.

2.20 If, for the cantilever in Exercise 2.19, a = 2 mm, h = 4 mm, l = 10 mm and P = I kN determine(i) the Airy stress function from which the stresses derive, (ii) the principal stresses at the point Q(5,2,0)and (iii) the normal and shear stresses at Q on an oblique plane whose edge is given by y - x + 3 = 0.(Hint: Use Table 1.1 in Chapter 1 for appropriate plane stress transformation equations.)


2.21 Establish the relationship between those constants in the quartic stress function 0 = Ax° + Bx3y+ Cx2y2 + Dxy3 + Ey4 if the function is to be valid . Determine further reltionships between theconstants , from a consideration of known resultant normal forces X and Y and shear force S acting onthe sides of a square plate of side length a and unit thickness , when the plate lies in the positive x, yquadrant with one comer at the origin . Answer : 3A + C + 3E = 0, X = (2C + 3D + 4E)d 3, y = (4A +3B + 2C)a3, S = - (B + 2C + 3D)a3 = (3B + 2C + D)a3, B = D

2.22 A rectangular strip of thickness t is simply supported at its ends (x = ± l) and carries on its upper

edge (y = + d) a uniformly distributed load, w per unit length , acting in the plane of the strip , as shown

in Fig . 2.30. The expression 0 = [3w/(4 t)] { [y3 /(30d3)](5x2 - y2 - 512 + 2d2) - [X21(6d)1(3y + 2d)])

is proposed as an Airy stress function . Show that g5 satisfies the compatibility condition and determineexpressions for the direct and shear stresses in the strip . Investigate whether the direct stress expression

satisfies the boundary equilibrium conditions. (CEI)YI q

Figure 2.30 Figure 2.31

2.23 The rectangular plate shown in Fig . 2.31 is of uniform thickness t and is built in at the end x = 0.It is loaded along the upper edge y = c by a tangential shear flow q . The lower edge of the plate is not

loaded . The expression 0 = [q/(4 t)][xy - xy2Ic - xy3 IcZ + 1 y2 / c + I y3 / c2 ] is proposed as an

appropriate stress function . To what extent does 0 satisfy boundary conditions and compatibility? (CEI)

2.24 Show that the stress function 0 = axy12 - axy3 /(2c2) + aby3 /(2c2) is biharmonic and that it maybe used to supply the stresses in the cantilever of Fig . 2.32. Is the condition that the cantilever isunloaded at its free end met by the function?

Q

0

-a

P/2

-L L

P/2


2.25 The function 0 = - xy3 [PI(4a3) + QI(4a2L)] + y3 [PU(4a3) + Q/(4a2)1 - xy2QI(4aL) + y2QI(4a) +

xy[3P/(4a) + Q1(4L)] is proposed for the loaded beam in Fig. 2.33. Check that this satisfies V40 = 0and determine the component stresses. To what extent are the boundary conditions satisfied in relationto the ratio Ua?

2.26 A cantilever beam of length l and depth 2h carries a downward acting uniformly distributed loadw/unit area along its top and bottom faces under plane strain conditions . Using the stress function 0= Px2 y + Qx2 y3 + Rya + Sys , find the constants from a consideration of the known conditions existingat the beam surface . Take the origin of co-ordinates to lie at the centre of the free end. (CEI)


2.27 The stress function 0 = [3F1(2c)1[xy - xy3/(3c2)] + Qy3/(4c3) is to be used to determine the stressfield in a cantilever beam of length L, unit thickness and depth 2c, subjected to a given loading.Determine the nature of that loading. (CEI)

Fourier Series Loading Functions

2.28 Identify full and half-range series for the edge loadings shown in Figs 2.34a-d and derive theappropriate Fourier functions g(x) in eqs(2 .33) and (2.34).Answer: (a) g(x) = (2pl ir) sin (,rxll) - [2p/(2,r )] sin (2.r x/1) + [2p/(3,r )] sin (3 irxll) - ...

(b) g(x) = (2pl it )Q [ - cos (ir 1/L) + (U I ;r ) sin ('r 1/L)] sin (irx/l)

+ (1/2 cos (22r lIL) + [U(4 hr)] sin (27r 11L)) sin (2,rx/L) ... I

(c) g(x) =p/2 - (4p/,r2)[cos (,rxll) + ( 1/32) cos (3,rx/1) + (1/52) cos (5;rxll) ... ](d) g(x) = 3p/2 - (2p/ r) [sin (nxll) + ( 1/3) sin (3zrz11) + ( 1/5) sin (5,rxll) ... ]

YA.

P

(a)fop^

Y

2p

P

Figure 2.34

I

(d)

21

2.29 Show that the full range Fourier series for the saw-tooth edge loading in Fig. 2.35 is given byg(x) = 5p/4 + (2p/1r2 )[sin (m/1) + (1/32) sin (3,rx/l) + (1/52) sin (5,rx/1) + ... ]. Hence find the planestresses ox and or, at the origin in terms of p.

2p

p

Ix x+21

I 21

IL=h 2L=2h

I T

I I I

I

I I

I I



2.30 The stress function in egs (2.31 a,b ) provides the solution to the problem of a long flat strip with

its axis in direction x, under a sinusoidal edge loading of wavelength 2L. Use this stress function to find

the normal stress ov at the origin (see Fig . 2.36) in a strip of depth 2h, loaded with a uniform pressure

p under guides which are equally spaced along its edges , for the case when h = L. (IC)

Cylindrical Co-ordinates

2.31 For a particular plane strain problem the strain-displacement equations in cylindrical co-ordinates

(0,z) are e, = du/dr, ee = uJr, e,= y, = re,= y, = 0. Show that the appropriate compatibility equation

in terms of stress is rvdo,/dr - r(1 - v )do8/dr+ o, - ae= 0. State the nature of a problem represented

by the above equations. (CEI)

2.32 A thick-walled cylinder with inner and outer radii r , and r„ respectively is pressurised both

internally (p) and externally (p). Find the expressions for the constants A and C in the corresponding

stress function io = A In r + C r2.Answer: A = - (pi - p„) r„2 /[(r„ Ir )2 - 11, 2C = [pi - p° (r0l ri )2 ]1[(r.1 r i)2 - I ]

2.33 Determine the hoop stress at the inner and outer radii of a curved bar of 45 mm mean radius and

45 mm square cross -section when it is subjected to a pure moment of 300 Nm, inducing tension along

the top edge.Answer: 54.6 MPa, - 86.2 MPa

2.34 The stress function 0 = CO prescribes the shear stress distribution in an annular disc of thicknesst and inner radius r, under torsion. If the torque is transmitted to the disc by a keyed shaft of radius r,show that there is no discontinuity in shear stress at r, when t = r.14.

2.35 Examine how to modify the stress distributions in the body of the wedge in Fig. 2.17 when the tipmoment is produced by a force applied vertically to the wedge axis at a distance s from the wedge tip.

2.36 Find the maximum radial stress at a radius of 75 mm when the 30° wedge in Fig. 2.2lb is loadedat its tip with 15 kN compression across its 5 mm thickness, with an inclination of 5° to the axis.

Answer : - 78.77 MPa on the load line

2.37. A very large plate of thickness t is subjected to a membrane tensile stress o„ applied around itsperiphery. If a small circular hole is cut from the middle of the plate, show , from the Lame equations,

that the maximum stress at the edge of the hole is 2c r..

2.38 The stress function 0 = [qr2/(22r )] ('h sin 20 - 0) prescribes the stresses in a semi-infinite platewhen a normal pressure q is applied over one half of the plate straight edge. Taking the origin at the

load end with 0 measured anti-clockwise from unloaded edge, show that the stresses at radius r in the

plate are ar = - (q/r)('/2 sin 20+0) and r,9 = - [q/(2ir)](cos20 - 1).

2.39 A hole of radius a in an infinite plate is loaded with an internal horizontal pressure p. Taking theorigin of co-ordinates at the hole centre and assuming that the outer boundary is stress free, derive

expressions for the polar stress components or, oB and r,8 in the body of the plate and hence find the

maximum stress concentration at the hole. (IC)

2.40 A window port in a large diameter , thin-walled pressurized cylinder is idealised as a large flat platecontaining a hole with normal boundary stresses in the ratio S,/S, = 3/2 applied to the plate boundary.

If the hole is to be free from stress concentration , what thickness of hole reinforcement is necessary?

2.41 Match the stress function 0 = (Ar3 + B/r + Cr + Dr In r) f (0) to each of the thin cantilever arcs

in Figs 2.37a,b taking f (0) = sing and f (0) = cos0 respectively. Examine the local distributions ofstress at the fixed and free ends and show their effects on the stress predictions from the stress function.


Answer: = F sin 0 [r3 - a2b2 /r - 2(a2 + b2) r In r]l (2[(a 2 - b2) +(a 2 + b2) In (b/a)] }and 0 = - l p cos 0 1[8(a + b)]) [r3 + a3 (a + 2b)l r + 4abr In r]

(a)

F

P

Figure 2.37

(b)

2A2 Employ the Boussinesq function 0 = Cr8 sin0 to obtain separate solutions for the stresses at pointA in a semi-infinite plate of thickness t, due to two concentrated forces which constitute a couple Pa(see Fig. 2.38). Resolve forces at A to show that the stresses in the 8 direction are given by o, =[2P/(rrt)][ - (1/r) cos0+ (l/r,) cos0, costa ], oe= [2P/(,rtr,)] cosO, sin2a, rre= [P/(,r tr,)] cosO, sin2a.Show that when a -► 0 and the couple Pa is replaced by a moment M, the stresses are reduced to or =[2M/(,r tr2 )] sin 20, ore = 0 and rie = [2Ml(,r tr2 )] cos2B.


2.43 In Fig. 2.39 a linear distribution of loading is applied to the edge of a semi-infinite plate. Use theBoussinesq function and plane stress transformations for the given co-ordinates to derive integralexpressions for the Cartesian stress components at a point P(x, y) in the body of the plate.(Hint : Box = - [26pl(ir r)] cos3B, do,, = - [26 p/(fr r)]cosO sin20 and drx,, = - [26 p/(ir r)] sin8 cos20

module 3 - plane elasticity theory

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