Elasticity Part 6 ppsx

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. It was further shown that solution to each of these problem types could be conveniently handled using the Airy stress function approach. This scheme reduces the field equations to a single partial 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 the biharmonic equation in a particular domain of interest. Such a solution must also satisfy the given boundary conditions associated with the particular problem under study. Several general solution 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-dimensional problems using these methods. Our formulation and solution is conducted using both Cartesian and polar coordinate systems. In many cases we use MATLAB software to plot the stress and displacement field distributions in order to better understand the nature of the solution. Plane problems can also be solved using complex variable theory, and this powerful method is discussed in Chapter 10. 8.1 Cartesian Coordinate Solutions Using Polynomials We begin the solution to plane elasticity problems with no body forces by considering problems formulated in Cartesian coordinates. When taking boundary conditions into account, this formulation is most useful for problems with rectangular domains. The method is based on the inverse solution concept where we assume a form of the solution to the biharmonic equation @ 4 f @x 4 þ 2 @ 4 f @x 2 @y 2 þ @ 4 f @y 4 ¼ 0(8:1:1) and then try to determine which problem may be solved by this solution. The assumed solution form for the Airy stress function is taken to be a general polynomial of the in-plane coordin- ates, and this form can be conveniently expressed in the power series 139 Sadd / Elasticity Final Proof 5.7.2004 6:13pm page 139 TLFeBOOK f(x, y) ¼ X 1 m¼0 X 1 n¼0 A mn x m y n (8:1:2) where A mn are constant coefficients to be determined. This representation was given by Neou (1957), who proposed a systematic scheme to solve such plane problems. Using the stress-stress function relations (7.5.3) with zero body forces s x ¼ @ 2 f @y 2 , s y ¼ @ 2 f @x 2 , t xy ¼À @ 2 f @x@y (8:1:3) Note that in the Airy function form the three lowest-order terms with m þ n 1 do not contribute to the stresses and therefore are dropped. It is observed that second-order terms produce a constant stress field, third-order terms give a linear distribution of stress, and so on for higher-order polynomials. Terms with m þ n 3 automatically satisfy the biharmonic equation (8.1.1) for any choice of constants A mn . However, for higher-order terms with m þ n > 3, the constants A mn must be related in order to have the polynomial satisfy the biharmonic equation. For example, the fourth-order polynomial terms A 40 x 4 þ A 22 x 2 y 2 þ A 04 y 4 will not satisfy the biharmonic equa- tion unless 3A 40 þ A 22 þ 3A 04 ¼ 0. This condition specifies one constant in terms of the other two, thus leaving two constants to be determined by the boundary conditions. Considering the general case, substituting the series form (8.1.2) into the governing biharmonic equation (8.1.1) yields X 1 m¼4 X 1 n¼0 m(m À 1)(m À 2)(m À 3)A mn x mÀ4 y n þ 2 X 1 m¼2 X 1 n¼2 m(m À 1)n(n À 1)A mn x mÀ2 y nÀ2 þ X 1 m¼0 X 1 n¼4 n(n À 1)(n À 2)(n À 3)A mn x m y nÀ4 ¼ 0 (8:1:4) Collecting like powers of x and y, the preceding equation may be written as X 1 m¼2 X 1 n¼2 [(m þ 2)(m þ 1)m(m À 1)A mþ2, nÀ2 þ 2m(m À 1)n(n À 1)A mn þ (n þ 2)(n þ 1)n(n À 1)A mÀ2, nþ2 ]x mÀ2 y nÀ2 ¼ 0 (8:1:5) Because this relation must be satisfied for all values of x and y, the coefficient in brackets must vanish, giving the result (m þ 2)(m þ 1)m(m À 1)A mþ2, nÀ2 þ 2m(m À 1)n(n À 1)A mn þ (n þ 2)(n þ 1)n(n À 1)A mÀ2, nþ2 ¼ 0 (8:1:6) For each m,n pair, (8.1.6) is the general relation that must be satisfied to ensure that the polynomial grouping is biharmonic. Note that the fourth-order case (m ¼ n ¼ 2) was previously discussed. Because this method produces polynomial stress distributions, we would not expect the scheme to satisfy general boundary conditions. However, this limitation can be circum- Sadd / Elasticity Final Proof 5.7.2004 6:13pm page 140 140 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK 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 statically equivalent 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, this method has applications to problems of rectangular shape in which one dimension is much larger than the other. This would include a variety of beam problems, and we shall now consider three 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 simple change in elastic constants given in Table 7-1. Of course, for the case with zero body forces and traction 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 rectangu- lar beam under uniform tension T at each end, as shown in Figure 8-1. This problem could be considered the Saint Venant approximation to the more general case with nonuniformly distributed tensile forces at the ends x ¼Æl. 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 s x (Æl, y) ¼ T, s y (x, Æ c) ¼ 0 t xy (Æl, y) ¼ t xy (x, Æ 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’s boundaries, we are motivated to try a second-order stress function of the form f ¼ A 02 y 2 (8:1:8) and this gives the following constant stress field: s x ¼ 2A 02 , s y ¼ t xy ¼ 0(8:1:9) The first boundary condition (8.1.7) implies that A 02 ¼ T=2 and all other boundary conditions are identically satisfied. Therefore, the stress field solution to this problem is given by Continued x y T T 2l 2c FIGURE 8-1 Uniaxial tension problem. Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 141 Two-Dimensional Problem Solution 141 TLFeBOOK EXAMPLE 8-1: Uniaxial Tension of a Beam–Cont’d s x ¼ T, s y ¼ t xy ¼ 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 is calculated using Hooke’s law. Then the strain-displacement relations are used to determine various displacement gradients, and these expressions are integrated to find the individual displacements. Using this scheme, the in-plane displacement gradients are found to be @u @x ¼ e x ¼ 1 E (s x À ns y ) ¼ T E @v @y ¼ e y ¼ 1 E (s y À ns x ) ¼Àn T E (8:1:11) These results are easily integrated to get u ¼ T E x þ f (y) v ¼Àn T E y þ g(x) (8:1:12) where f(y) and g(x) are arbitrary functions of the indicated variable coming from the integration process. To complete the problem solution, these functions must be deter- mined, and this is accomplished using the remaining Hooke’s law and the strain- displacement relation for the shear stress and strain @u @y þ @v @x ¼ 2e xy ¼ t xy m ¼ 0 ) f 0 (y) þ g 0 (x) ¼ 0(8:1:13) This result can be separated into two independent relations g 0 (x) ¼Àf 0 (y) ¼ constant and integrated to get f (y) ¼À! o y þ u o g(x) ¼ ! o x þ v o (8:1:14) where ! o , u o , v o are arbitrary constants of integration. The expressions given by relation (8.1.14) represent rigid-body motion terms where ! o is the rotation about the z-axis and u o and v o are the translations in the x and y directions. Such terms will always result from the integration of the strain-displacement relations, and it is noted that they do not contribute to the strain or stress fields. Thus, the displacements are determined from the strain field only up to an arbitrary rigid-body motion. Additional boundary conditions on the displacements are needed to determine these terms explicitly. For example, if we agree that the center of the beam does not move and the x-axis does not rotate, all rigid-body terms will vanish and f ¼ g ¼ 0. Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 142 142 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK EXAMPLE 8-2: Pure Bending of a Beam As a second plane stress example, consider the case of a straight beam subjected to end moments as shown in Figure 8-2. The exact pointwise loading on the ends is not considered, and only the statically equivalent effect is modeled. Hence, the boundary conditions on this problem are written as s y (x, Æ c) ¼ 0, t xy (x, Æ c) ¼ t xy (Æ l, y) ¼ 0 ð c Àc s x ( Æl, y)dy ¼ 0, ð c Àc s x (Æ l, y) ydy ¼ÀM (8:1:15) Thus, the boundary conditions on the ends of the beam have been relaxed, and only the statically equivalent condition will be satisfied. This fact leads to a solution that is not necessarily valid near the ends of the beam. The choice of stress function is based on the fact that a third-order function will give rise to a linear stress field, and a particular linear boundary loading on the ends x ¼Æl will reduce to a pure moment. Based on these two concepts, we choose f ¼ A 03 y 3 (8:1:16) and the resulting stress field takes the form s x ¼ 6A 03 y, s y ¼ t xy ¼ 0(8:1:17) This field automatically satisfies the boundary conditions on y ¼Æc and gives zero net forces at the ends of the beam. The remaining moment conditions at x ¼Æl are satisfied if A 03 ¼ÀM=4c 3 , and thus the stress field is determined as s x ¼À 3M 2c 3 y, s y ¼ t xy ¼ 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 substituted into the strain-displacement relations and integrated yielding the result @u @x ¼À 3M 2Ec 3 y ) u ¼À 3M 2Ec 3 xy þ f (y) @v @y ¼ n 3M 2Ec 3 y ) v ¼ 3Mn 4Ec 3 y 2 þ g(x) (8:1:19) Continued x y M M 2l 2c FIGURE 8-2 Beam under end moments. Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 143 Two-Dimensional Problem Solution 143 TLFeBOOK EXAMPLE 8-2: Pure Bending of a Beam–Cont’d where f and g are arbitrary functions of integration. Using the shear stress-strain relations @u @y þ @v @x ¼ 0 )À 3M 2Ec 3 x þ f 0 (y) þ g 0 (x) ¼ 0(8:1:20) This result can again be separated into two independent relations in x and y, and upon integration the arbitrary functions f and g are determined as f (y) ¼À! o y þ u o g(x) ¼ 3M 4Ec 3 x 2 þ ! o x þ v o (8:1:21) Again, rigid-body motion terms are brought out during the integration process. For this problem, the beam would normally be simply supported, and thus the displacement conditions could be specified as v(Æ l,0)¼ 0 and u(Àl,0)¼ 0. This specification leads to determination of the rigid-body terms as u o ¼ ! o ¼ 0, v o ¼À3Ml 2 =4Ec 3 . We now wish to compare this elasticity solution with that developed by elementary strength of materials (mechanics of materials). Introducing the cross-sectional area moment of inertia I ¼ 2c 3 =3 (assuming unit thickness), our stress and displacement field can be written as s x ¼À M I y, s y ¼ t xy ¼ 0 u ¼À Mxy EI , v ¼ M 2EI [ny 2 þ x 2 À l 2 ] (8:1:22) Note that for this simple moment-loading case, we have verified the classic assumption from elementary beam theory, that plane sections remain plane. Note, however, that this will not be the case for more complicated loadings. The elementary strength of materials solution is obtained using Euler-Bernoulli beam theory and gives the bending stress and deflection of the beam centerline as s x ¼À M I y, s y ¼ t xy ¼ 0 v ¼ v(x,0)¼ M 2EI [x 2 À l 2 ] (8:1:23) Comparing these two solutions, it is observed that they are identical, with the exception of the x displacements. In general, however, the two theories will not match for other beam problems with more complicated loadings, and we investigate such a problem in the next example. Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 144 144 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK EXAMPLE 8-3: Bending of a Beam by Uniform Transverse Loading Our final example is that of a beam carrying a uniformly distributed transverse loading w along its top surface, as shown in Figure 8-3. Again, plane stress conditions are chosen, and we relax the boundary conditions on the ends and consider only statically equivalent effects. Exact pointwise boundary conditions will be specified on the top and bottom surfaces, while at the ends the resultant horizontal force and moment are set to zero and the resultant vertical force will be specified to satisfy overall equilibrium. Thus, the boundary conditions on this problem can be written as t xy (x, Æ c) ¼ 0 s y (x, c) ¼ 0 s y (x,Àc) ¼Àw ð c Àc s x (Æ l, y)dy ¼ 0 ð c Àc s x (Æ l, y)ydy ¼ 0 ð c Àc t xy (Æ l, y)dy ¼Çwl (8:1:24) 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 includ- ing second-, third-, and fifth-order terms: f ¼ A 20 x 2 þ A 21 x 2 y þ A 03 y 3 þ A 23 x 2 y 3 À A 23 5 y 5 (8:1:25) It is noted that the fifth-order term has been generated to satisfy the biharmonic equation. The resulting stress field from this stress function is given by s x ¼ 6A 03 y þ 6A 23 (x 2 y À 2 3 y 3 ) s y ¼ 2A 20 þ 2A 21 y þ 2A 23 y 3 t xy ¼À2A 21 x À 6A 23 xy 2 (8:1:26) Continued x y w 2c 2l wl wl FIGURE 8-3 Beam carrying uniformly transverse loading. Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 145 Two-Dimensional Problem Solution 145 TLFeBOOK 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 equations among the unknown coefficients A 20 , A 21 , and A 23 . Solving this system determines these constants, giving the result A 20 ¼À w 4 , A 21 ¼ 3w 8c , A 23 ¼À w 8c 3 (8:1:27) Using these results, it is found that the stress field will now also satisfy the fourth and sixth conditions in (8.1.24). The remaining condition of vanishing end moments gives the following A 03 ¼ÀA 23 (l 2 À 2 5 c 2 ) ¼ w 8c l 2 c 2 À 2 5  (8:1:28) This completes determination of the four constants in the trial Airy stress function, and the resulting stress field is now given by s x ¼ 3w 4c l 2 c 2 À 2 5  y À 3w 4c 3 x 2 y À 2 3 y 3  s y ¼À w 2 þ 3w 4c y À w 4c 3 y 3 t xy ¼À 3w 4c x þ 3w 4c 3 xy 2 (8:1:29) We again wish to compare this elasticity solution with that developed by elementary strength of materials, and thus the elasticity stress field is rewritten in terms of the cross- sectional area moment of inertia I ¼ 2c 3 =3, as s x ¼ w 2I (l 2 À x 2 )y þ w I y 3 3 À c 2 y 5  s y ¼À w 2I y 3 3 À c 2 y þ 2 3 c 3  t xy ¼À w 2I x(c 2 À y 2 ) (8:1:30) The corresponding results from strength of materials for this case is given by s x ¼ My I ¼ w 2I (l 2 À x 2 )y s y ¼ 0 t xy ¼ VQ It ¼À w 2I x(c 2 À y 2 ) (8:1:31) where the bending moment M ¼ w(l 2 À x 2 )=2, the shear force V ¼Àwx, the first moment of a sectioned cross-sectional area is Q ¼ (c 2 À y 2 )=2, and the thickness t is taken as unity. Comparing the two theories, we see that the shear stresses are identical, while the two normal stresses are not. The two normal stress distributions are plotted in Figures 8-4 Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 146 146 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK EXAMPLE 8-3: Bending of a Beam by Uniform Transverse Loading–Cont’d and 8-5. The normalized bending stress s x for the case x ¼ 0 is shown in Figure 8-4. Note that the elementary theory predicts linear variation, while the elasticity solution indicates 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 stress values is simply w/5, a result independent of the beam dimensions. For most common beam problems where l >> c, the bending stresses will be much greater than w, and thus the differences between elasticity theory and strength of materials will be relatively small. For example, the set of curves in Figure 8-4 for l=c ¼ 4 gives a maximum difference of about only 1 percent. Figure 8-5 illustrates the behavior of the stress s y , and the maximum difference between the two theories is given by w and this occurs at Continued σ x / w - Elasticity σ x / w - Strength of Materials l/c = 2 l/c = 4 l/c = 3 Dimensionless Stress Dimensionless Distance, y/c 10 5 0 - 5 - 10 15 - 15 - 0.5 0.5 - 1 10 FIGURE 8-4 Comparison of bending stress in beam Example 8-3. σ y / w - Strength of Materials σ y / w - Elasticity Dimensionless Distance, y / c Dimensionless Stress - 0.5 0 0.5 - 11 - 0.9 - 0.8 - 0.7 - 0.6 - 0.5 - 0.4 - 0.3 - 0.2 - 0.1 0 - 1 0.1 FIGURE 8-5 Comparison of s y stress in beam Example 8-3. Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 147 Two-Dimensional Problem Solution 147 TLFeBOOK 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 beam problems where l >> c. These results are generally true for beam problems with other transverse loadings. That is, for the case with l >> c, approximate bending stresses determined from strength of materials will generally closely match those developed from theory of elasticity. Next let us determine the displacement field for this problem. As in the previous examples, the displacements are developed through integration of the strain- displacement relations. Integrating the first two normal strain-displacement relations gives the result u ¼ w 2EI [(l 2 x À x 3 3 )y þ x( 2y 3 3 À 2c 2 y 5 ) þvx( y 3 3 À c 2 y þ 2c 3 3 )] þf (y) v ¼À w 2El [( y 4 12 À c 2 y 2 2 þ 2c 3 y 3 ) þv(l 2 À x 2 ) y 2 2 þ v( y 4 6 À c 2 y 2 5 )] þ g(x) (8:1:32) where f(y) and g(x) are arbitrary functions of integration. Using these results in the shear strain-displacement equation gives the relation w 2EI [l 2 x À x 3 3 þ x(2y 2 À 2c 2 5 ) þvx(y 2 À c 2 )] þf 0 (y) þ w 2EI vxy 2 þ g 0 (x) ¼À w 2mI x(c 2 À y 2 ) (8:1:33) This result can again be rewritten in a separable form and integrated to determine the arbitrary functions f (y) ¼ ! o y þ u o g(x) ¼ w 24EI x 4 À w 4EI [l 2 À ( 8 5 þ v)c 2 ]x 2 À ! o x þ v o (8:1:34) Choosing the conditions u(0, y) ¼ v(Æ l,0) ¼ 0, the rigid-body motion terms are found to be u o ¼ ! o ¼ 0, v o ¼ 5wl 4 24EI [1 þ 12 5 ( 4 5 þ n 2 ) c 2 l 2 ](8:1:35) Using these results, the final form of the displacements is given by u ¼ w 2EI [(l 2 x À x 3 3 )y þ x( 2y 3 3 À 2c 2 y 5 ) þvx( y 3 3 À c 2 y þ 2c 3 3 )] v ¼À w 2EI y 4 12 À c 2 y 2 2 þ 2c 3 y 3 þ v[(l 2 À x 2 ) y 2 2 þ y 4 6 À c 2 y 2 5 ]  À x 4 12 þ [ l 2 2 þ ( 4 5 þ n 2 )c 2 ]x 2  þ 5wl 4 24EI [1 þ 12 5 ( 4 5 þ n 2 ) c 2 l 2 ] (8:1:36) Sadd / Elasticity Final Proof 5.7.2004 6:14pm page 148 148 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK [...]... solution are given by Continued Two-Dimensional Problem Solution 167 TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 168 EXAMPLE 8-8: Wedge and Semi-Infinite Domain Problems–Cont’d y r b a q x FIGURE 8- 16 Wedge domain geometry sr ¼ 2a2 þ 2a6 y À 2a21 cos 2y À 2b21 sin 2y sy ¼ 2a2 þ 2a6 y þ 2a21 cos 2y þ 2b21 sin 2y try ¼ Àa6 À 2b21 cos 2y þ 2a21 sin 2y (8:4:20) Note that this general stress... FIGURE 8-8 160 Thick-walled cylinder problem FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 161 EXAMPLE 8 -6: Thick-Walled Cylinder Under Uniform Boundary Pressure–Cont’d From plane strain theory, the out-of-plane longitudinal stress is given by sz ¼ v(sr þ sy ) ¼ 2v 2 2 r1 p1 À r2 p2 2 À r2 r2 1 (8:4:4) Using the strain displacement relations (7 .6. 1) and... cylinder example Two-Dimensional Problem Solution 161 TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 162 8.4.1 Pressurized Hole in an Infinite Medium Consider the problem of a hole under uniform pressure in an infinite medium, as shown in Figure 8-10 The solution to this problem can be easily determined from the general case of Example 8 -6 by choosing p2 ¼ 0 and r2 ! 1 Taking these limits... established in Exercise 3-3 (or see Appendix B) Continued Two-Dimensional Problem Solution 163 TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 165 EXAMPLE 8-7: Infinite Medium with a Stress-Free Hole Under Uniform Far-Field Tension Loading–Cont’d a1 ¼0 a2 6a23 4a24 2a21 þ 4 þ 2 ¼ 0 a a 6a23 2a24 2a21 À 4 À 2 ¼ 0 a a T 2a21 ¼ À 2 T 2a2 ¼ 2 2a2 þ (8:4:14) This system is easily solved for the constants,... compression in the vertical This T T T 45Њ T T T T (a) Biaxial Loading T (b) Shear Loading FIGURE 8-15 Stress-free hole under biaxial and shear loading 166 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 167 far-field loading is equivalent to a pure shear loading on planes rotated 458 as shown in case (b) Thus, the solution to this case would apply to either... solved Two particular problems are now presented EXAMPLE 8-4: Beam Subject to Transverse Sinusoidal Loading Consider the simply supported beam carrying a sinusoidal loading along its top edge as shown in Figure 8 -6 qosin px/l y qol/p qol/p 2c x l FIGURE 8 -6 150 Beam carrying sinusoidal transverse loading FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page... r2 r2 (8:4:9) The maximum stress occurs at the boundary of the hole r ¼ r1 and is given by r1 p FIGURE 8-10 Pressurized hole in an infinite medium 162 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 163 T r1 FIGURE 8-11 T Stress-free hole under uniform biaxial far-field loading smax ¼ (sy )max ¼ sy (r1 ) ¼ 2T (8:4:10) and thus the stress concentration... displacements Two-Dimensional Problem Solution 159 TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 160 8.4 Polar Coordinate Solutions With the general solution forms determined, we now explore the solution to several specific problems of engineering interest, including cases with both axisymmetric and general geometries EXAMPLE 8 -6: Thick-Walled Cylinder Under Uniform Boundary Pressure The first... (a,q) / T 180 0 330 210 240 FIGURE 8-13 (8:4:17) 300 270 Variation of hoop stress around hole boundary in Example 8-7 Continued Two-Dimensional Problem Solution 165 TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 166 EXAMPLE 8-7: Infinite Medium with a Stress-Free Hole Under Uniform Far-Field Tension Loading–Cont’d Therefore, the stress concentration factor for this problem is 3, a result that... (1 963 ) A function f(x) periodic with period 2l can be represented on the interval (Àl, l) by the Fourier trigonometric series 1 X 1 npx npx an cos þ bn sin f (x) ¼ ao þ 2 l l n¼1 (8:2:25) ð 1 l npx dx, n ¼ 0, 1, 2, Á Á Á f (x) cos l Àl l ð 1 l npx bn ¼ f (x) sin dx, n ¼ 1, 2, 3, Á Á Á l Àl l (8:2: 26) where an ¼ Two-Dimensional Problem Solution 153 TLFeBOOK Sadd / Elasticity Final Proof 5.7.2004 6: 14pm . from this stress function is given by s x ¼ 6A 03 y þ 6A 23 (x 2 y À 2 3 y 3 ) s y ¼ 2A 20 þ 2A 21 y þ 2A 23 y 3 t xy ¼À2A 21 x À 6A 23 xy 2 (8:1: 26) Continued x y w 2c 2l wl wl FIGURE 8-3 Beam. The two normal stress distributions are plotted in Figures 8-4 Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 1 46 1 46 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK EXAMPLE 8-3: Bending. x 2 ) y 2 2 þ y 4 6 À c 2 y 2 5 ]  À x 4 12 þ [ l 2 2 þ ( 4 5 þ n 2 )c 2 ]x 2  þ 5wl 4 24EI [1 þ 12 5 ( 4 5 þ n 2 ) c 2 l 2 ] (8:1: 36) Sadd / Elasticity Final Proof 5.7.2004 6: 14pm page 148 148