. The Bell System technical journal . ofthe surface of the cable. Suppose these two strains to be parallel withllie .r and y coordinate axes, respectively, and that one side of the squareis aligned, before straining, at the angle 0 willi the .v axis. This is illus-trated in Fig. 1. Let ex = maximum principal strainCy = minimum principal strain\x = I + ex \ = 1 ->r Cy 1 Cf. for example, Max Frocht, Photoelasticitj-, 1, p. 37, 1941. 523 524 THE BELL SYSTEM TECHNICAL JOURNAL, MAY 1952 If primes are used to refer to the strained state,K = Xu — Xb - •^a *^d *^c Xh - ~ Xa Xd Xc yb - 1 / / Va _ Vd
. The Bell System technical journal . ofthe surface of the cable. Suppose these two strains to be parallel withllie .r and y coordinate axes, respectively, and that one side of the squareis aligned, before straining, at the angle 0 willi the .v axis. This is illus-trated in Fig. 1. Let ex = maximum principal strainCy = minimum principal strain\x = I + ex \ = 1 ->r Cy 1 Cf. for example, Max Frocht, Photoelasticitj-, 1, p. 37, 1941. 523 524 THE BELL SYSTEM TECHNICAL JOURNAL, MAY 1952 If primes are used to refer to the strained state,K = Xu — Xb - •^a *^d *^c Xh - ~ Xa Xd Xc yb - 1 / / Va _ Vd - Vc Vb - Va yd - yc If Li and L2 are the lengths of the sides of the unstrained square, andLs and L4 the diagonals, (Li + A LiY = Xlixb - Xaf + kliub - yaf {Li + A L2)- = \l{xd - Xcf + \l{yd - ycf whence, if {Xb - Xaf = {yd - ycf = L\ cosi = LI cos^i(yb — yaf = {xd - Xcf = Li sin^0i = LI sm(j)i ^ + ^^ = L[,^\^^=L[,etc. U Li Li = Xx cos ^1 + X^ sm 01L2 = Xi sin 01 + Xj, cos 01 (la)(lb)(2a) (2b) (2c) (3a)(3b) ^b>yb. Fig. 1—Lines ab and cd, before the xy plane is strained by stretching (or com-pressing) in the x and y directions. PRINCIPAL STRAINS IN BUCKLED SURFACES 525 Henceforth, for clarity, suppose the subscript 1 to refer to the longerside of the parallelogram, 2 to the shorter side, 3 to the longer dia-gonal, and 4 to the shorter. *Si = original slope of Li = tan [ =^Si (4a) Ax S, = tan 0; = ^ ^o (4b) Vb - - Va Xb - - Xa Vd - - Vc *^d »^c Smce 01 - 02 = 90°, 82= -^ (5a) Oi /S2 = — Xj//Xi;Si (5a) By expansion and substitution from Equations (4) and (5), r (« +1) tan (.i,; - *;) = ^-^ ,^^l (6) - fe) Let ^ = 90° - (0i - 0o) then 1 - ftan (90° - (01 - 02)) = tan ^ = —-. ^^^^ (7) H^ + s. which is the shear between Li and Li . si) 2 {Sx + l/^Si) = tan 01 + cot 0i = ^^— (8) sin Z01 and substituting this in equation (7), . 2K\y tan ^p «^ 201 = -^^^r^ (9) 526 THE BELL SYSTEM TECHNICAL JOURNAL, MAY 1952 whence cos 201 = a/i - 4^
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