§10.9 Thick Cylinders 225 compression.The outer tube will conversely be brought into a state of tension.If this compound cylinder is now subjected to internal pressure the resultant hoop stresses will be the algebraic sum of those resulting from internal pressure and those resulting from shrinkage as drawn in Fig.10.11;thus a much smaller total fluctuation of hoop stress is obtained.A similar effect is obtained if a cylinder is wound with wire or steel tape under tension(see §10.19). (a)Internal pressure (b）Shrinkage only (c)Combined shrinkage only and internal pressure Fig.10.11.Compound cylinders-combined internal pressure and shrinkage effects. (a)Same materials The method of solution for compound cylinders constructed from similar materials is to break the problem down into three separate effects: (a)shrinkage pressure only on the inside cylinder; (b)shrinkage pressure only on the outside cylinder; (c)internal pressure only on the complete cylinder(Fig.10.12). (a)Shrinkoge-internal (b)Shrinkage-external cylinder (c)Internal pressure- cylinder compound cylinder Fig.10.12.Method of solution for compound cylinders. For each of the resulting load conditions there are two known values of radial stress which enable the Lame constants to be determined in each case. i.e.condition (a)shrinkage-internal cylinder: At r=R1,,=0 At r=Re,a,=-p (compressive since it tends to reduce the wall thickness) condition (b)shrinkage-external cylinder: At r=R2,,=0 At r=R,,=-p condition (c)internal pressure-compound cylinder: Atr=R2,0,=0 At r=R1,a,=-P1
$10.9 Thick CyIinders 225 compression. The outer tube will conversely be brought into a state of tension. If this compound cylinder is now subjected to internal pressure the resultant hoop stresses will be the algebraic sum of those resulting from internal pressure and those resulting from shlinkflge as drawn in Fig. 10.11; thus a much smaller total fluctuation of hoop stress is obtained. A similar effect is obtained if a cylinder is wound with wire or steel tape under tension (see 410.19). (a) Internal pressure (b) Shrinkage only (c) Combined shrinkage only and internal pressure Fig. 10.1 1. Compound cylinders-combined internal pressure and shrinkage effects. (a) Same materials The method of solution for compound cylinders constructed from similar materials is to (a) shrinkage pressure only on the inside cylinder; (b) shrinkage pressure only on the outside cylinder; break the problem down into three separate effects: (c) internal pressure only on the complete cylilider (Fig. 10.12). (a) Shrinkage-internal (b) Shrinkage-aternol Cylw (c) IntnnaI pressurecylinder compound cylinder Fig. 10.12. Method of solution for compound cylinders. For each of the resulting load conditions there are two known values of radial stress which enable the Lame constants to be determined in each case. i.e. condition (a) shrinkage - internal cylinder: At r= R,, ur=O At r = R,, u, = -p (compressive since it tends to reduce the wall thickness) condition (b) shrinkage - external cylinder: At r= R,, u,=O At r = R,, u, = -p condition (c) internal pressure - compound cylinder: At r = R,, u, =O At r = R,, a, = -PI
226 Mechanics of Materials §10.10 Thus for each condition the hoop and radial stresses at any radius can be evaluated and the principle of superposition applied,i.e.the various stresses are then combined algebraically to produce the stresses in the compound cylinder subjected to both shrinkage and internal pressure.In practice this means that the compound cylinder is able to withstand greater internal pressures before failure occurs or,alternatively,that a thinner compound cylinder (with the associated reduction in material cost)may be used to withstand the same internal pressure as the single thick cylinder it replaces. Internal pressure Total Shrinkuge Internal Shrinkage pressure Total (a）Hoop stres5es (b)Rodiol stresses Fig.10.13.Distribution of hoop and radial stresses through the walls of a compound cylinder. (b)Different materials (Sees10.14.) 10.10.Compound cylinders-graphical treatment The graphical,or Lame line,procedure introduced in $10.8 can be used for solution of compound cylinder problems.The vertical lines representing the boundaries of the cylinder walls may be drawn at their appropriate 1/r2 values,and the solution for condition(c)of Fig.10.12 may be carried out as before,producing a single line across both cylinder sections (Fig.10.14a). The graphical representation of the effect of shrinkage does not produce a single line, however,and the effect on each cylinder must therefore be determined by projection of known lines on the radial side of the graph to the respective cylinder on the hoop stress side, i.e.conditions (a)and(b)of Fig.10.12 must be treated separately as indeed they are in the analytical approach.The resulting graph will then appear as in Fig.10.14b. The total effect of combined shrinkage and internal pressure is then given,as before,by the algebraic combination of the separate effects,i.e.the graphs must be added together,taking due account of sign to produce the graph of Fig.10.14c.In practice this is the only graph which need be constructed,all effects being considered on the single set of axes.Again,all values should be calculated from proportions of the figure,i.e.by the use of similar triangles. 10.11.Shrinkage or interference allowance In the design of compound cylinders it is important to relate the difference in diameter of the mating cylinders to the stresses this will produce.This difference in diameter at the
226 Mechanics of Materials $10.10 Thus for each condition the hoop and radial stresses at any radius can be evaluated and the principle of superposition applied, i.e. the various stresses are then combined algebraically to produce the stresses in the compound cylinder subjected to both shrinkage and internal pressure. In practice this means that the compound cylinder is able to withstand greater internal pressures before failure occurs or, alternatively, that a thinner compound cylinder (with the associated reduction in material cost) may be used to withstand the same internal pressure as the single thick cylinder it replaces. Toto I ... (a) Hoop stresses (b) Rodiol stresses Fig. 10.13. Distribution of hoop and radial stresses through the walls of a compound cylinder. (b) Diferent materials (See $10.14.) 10.10. Compound cylinders - graphical treatment The graphical, or Lame line, procedure introduced in 4 10.8 can be used for solution of compound cylinder problems. The vertical lines representing the boundaries of the cylinder walls may be drawn at their appropriate l/r2 values, and the solution for condition (c) of Fig. 10.12 may be carried out as before, producing a single line across both cylinder sections (Fig. 10.14a). The graphical representation of the effect of shrinkage does not produce a single line, however, and the effect on each cylinder must therefore be determined by projection of known lines on the radial side of the graph to the respective cylinder on the hoop stress side, i.e. conditions (a) and (b) of Fig. 10.12 must be treated separately as indeed they are in the analytical approach. The resulting graph will then appear as in Fig. 10.14b. The total effect of combined shrinkage and internal pressure is then given, as before, by the algebraic combination of the separate effects, i.e. the graphs must be added together, taking due account of sign to produce the graph of Fig. 10.14~. In practice this is the only graph which need be constructed, all effects being considered on the single set of axes. Again, all values should be calculated from proportions of the figure, i.e. by the use of similar triangles. 10.11. Shrinkage or interference allowance In the design of compound cylinders it is important to relate the difference in diameter of the mating cylinders to the stresses this will produce. This difference in diameter at the
§10.11 Thick Cylinders 227 Inner Outer Oufer Inner a)Interna cylinder cylinder cylinder cylinder pressure only 2 Hoop Radial stresses stresses b)Shrinkage only Shrinkage pressure Internal pressure Total only c)Combined shrinkage and internal pressure Shrinkage only Fig.10.14.Graphical (Lame line)solution for compound cylinders. "common"surface is normally termed the shrinkage or interference allowance whether the compound cylinder is formed by a shrinking or a force fit procedure respectively.Normally, however,the shrinking process is used,the outer cylinder being heated until it will freely slide over the inner cylinder thus exerting the required junction or shrinkage pressure on cooling. Consider,therefore,the compound cylinder shown in Fig.10.15,the material of the two cylinders not necessarily being the same. Let the pressure set up at the junction of two cylinders owing to the force or shrink fit be p. Let the hoop stresses set up at the junction on the inner and outer tubes resulting from the pressure p be (compressive)and a(tensile)respectively. Then,if δ，=radial shift of outer cylinder and ;radial shift of inner cylinder (as shown in Fig.10.15)
$10.11 Thick Cylinders 227 I II I 11 I- I I- II I Hoop 1 I stresses I u 10uler Inner I a)- pressurc cylinder 7 I - only I! IInternol pressure I I I I I I1 I I1 I I -<- Fig. 10.14. Graphical (Lam6 lirie) solution for compound cylinders. “common” surface is normally termed the shrinkage or interference allowance whether the compound cylinder is formed by a shrinking or a force fit procedure respectively. Normally, however, the shrinking process is used, the outer cylinder being heated until it will freely slide over the inner cylinder thus exerting the required junction or shrinkage pressure on cooling. Consider, therefore, the compound cylinder shown in Fig. 10.15, the material ofthe two cylinders not necessarily being the same. Let the pressure set up at the junction of two cylinders owing to the force or shrink fit be p. Let the hoop stresses set up at the junction on the inner and outer tubes resulting from the pressure p be uHi (compressive) and aHo (tensile) respectively. Then, if 6, = radial shift of outer cylinder ai = radial shift of inner cylinder (as shown in Fig. 10.15) and