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HW 4 - 471 SECTION 7.6 Triaxial Stress Problem 7.6-5 An...

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471 SECTION 7.6 Triaxial Stress Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is subjected to stresses U = 5200 psi (tension), u y = -4750 psi x (compression), and U = -3090 psi (compression). It is also known z that the normal strains in the x and y directions are Ex = 713.8 X 10- 6 (elongation) and E = -502.3 X 10- 6 (shortening). y What is the bulk modulus K for the aluminum? Solution 7.6-5 Triaxial stress (bulk modulus) -- x t u y Probs. 7.6-5 and 7.6·6 U x = 5200 psi u y = -4750 psi Substitute numerical values and rearrange: U = -3090 psi E: = 713.8 X 10- 6 (713.8 X 10- 6 ) E = 5200 + 7840 v (I) z x E: = -502.3 X 10 6 (-502.3 X 10- 6 ) E = -4750 - 2110 v (2) y Units: E = psi Find K. Solve simultaneously Eqs. (I) and (2): U x v Eq. (7-53a): ex = E - li(u y + u z ) E = 10.801 X 10 6 psi v = 0.3202 u y v Eq. (7-61): K = E = 10.0 X 10 6 psi - Eq. (7-53b): ey = E - li(u z + u x ) 3(1 - 2v) Problem 7.6-6 Solve the preceding problem if the material is nylon subjected to compressive stresses U = -4.5 MPa, u = -3.6 MPa, x y and U = -2.1 MPa, and the normal strains are Ex =" -740 X 10- 6 and z E = -320 X 10 6 (shortenings). y Solution 7.6-6 Triaxial stress (bulk modulus) U = -4.5 MPa u y = -3.6 MPa U. x = -2.1 MPa E: = -740 X 10- 6 x E: - = -320 X 10- 6 v Find K. U x v Eq. (7-53a): ex = E - li(u y + u) u y v Eq. (7-53b): ey = E - E(u z + u x ) Substitute numerical values and rearrange: (-740 X 10- 6 ) E = -4.5 + 5.7 v (I) (-320 X 10- 6 ) E = -3.6 + 6.6 v (2) Units: E = MPa Solve simultaneously Eqs. (I) and (2): E = 3,000 MPa = 3.0 OPa v = 0.40 E Eq. (7-61): K = = 5.0 OPa - 3(1 - 2v) Problem 7.6-7 A rubber cylinder R of length Land F cross-sectional area A is compressed inside a steel cylinder S by a force F that applies a uniformly distributed pressure to the rubber (see figure). (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel cylinder is rigid when compared to the rubber.) (b) Derive a formula for the shortening 8 of the rubber cylinder.

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37.5" SECTION 7.7 Plane Strain 477 Problem 7.7·5 An element of material subjected to plane strain (see y figure) has strains as follows: Ex = 220 X 10- 6 , E y = 480 X 10- 6 , and 'Y = 180 X 10- 6 . _1 X)' , Calculate the strains for an element oriented at an angle f} = 50° lOy and show these strains on a sketch of a properly oriented element. 1. ' .. '~y . . , . i f~ EJ ------" x Probs. 7.7-5 through 7.7-10 O~I___1Exf._ Solution 7.7·5 Element in plane strain e = 220 X 10- 6 e = 480 X 10- 6 y / = 180 X 10- 6 y X)' ex + e y ex - e y yry . ex = ---' + --- cos 213 + - Sill 213 '22 2 'Yx v ex - lOy yry _"_, = ---- sin 213 + - cos 213 222 y~ 239 X 10- 6 x FOR f} = 50°: e = 461 X 10- 6 'V = 225 X 10- 6 X1 I.1Wl = 239 X 10- 6 e y, Problem 7.7-6 Solve the preceding problem for the following data: 10 = 420 X 10- 6 E = -170 X 10 6 = 310 X 10- 6 and f} = 37.5°. x ' y , 'V I xy , Solution 7.7-6 Element in plane strain e = 420 X 10- 6 e = -170 X 10- 6 x y 'Y = 310 X 10- 6 y xy ex + e y ex - e y 'Y xy lOXI = --- + --- cos 213 + - sin 213 2 2 2 'YXlYI ex - ey .
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