. The strength of materials; a text-book for engineers and architects. I Stresses. y/Uulhide Tuhe^nsidc TuLe\^ R -^ -H Fig. 249.—Shrinkage. Stresses in Compound Tube. f ^ Cli - x (4) 526 THE STRENGTH OF MATERIALS We must have the same value of p for the junction where • n A. ^ — 1 ^ i. e. {a,. — a^) r^ = (6, — 6,.) (5) Next consider the circumferential strains at the the outer tube we have Unital circumferential strain = != -\-^^ (6) Similarly for inner tube Unital circumferential strain = — (^ + ~f j . .(7) The value of y is the same in each case, and in (6) / is as in(2) and in
. The strength of materials; a text-book for engineers and architects. I Stresses. y/Uulhide Tuhe^nsidc TuLe\^ R -^ -H Fig. 249.—Shrinkage. Stresses in Compound Tube. f ^ Cli - x (4) 526 THE STRENGTH OF MATERIALS We must have the same value of p for the junction where • n A. ^ — 1 ^ i. e. {a,. — a^) r^ = (6, — 6,.) (5) Next consider the circumferential strains at the the outer tube we have Unital circumferential strain = != -\-^^ (6) Similarly for inner tube Unital circumferential strain = — (^ + ~f j . .(7) The value of y is the same in each case, and in (6) / is as in(2) and in (7) as in (4). .*. Increase in circumference of outer tube 2 7rri 4(-j + >7P E Decrease in circumference of inner tube .. Difference in circumference of two tubes before heatingand shrinking on . •. Corresponding difference in radius_ difference in circumference 2 r= (from (5)) -^ (a, - aj Proportional difference in radius = -p, (a^ — a^). . (8) This will probably be made more clear by the followingnumerical example. THICK PIPES 527 Numerical Exam
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