ZPQ'R= = LPQ'R= = MI i O rad
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- An clement of material in plane strain (see figure) is subjected to strains ex= 480 × 10-6, Ey= 70 × l0-6, and yxy= 420 × l0-6. Determine the following quantities: (a) the strains for an element oriented at an angle 0 = 75°, (b) the principal strains, and (c) the maximum shear strains. Show the results on sketches of properly oriented elements.A strain rosette (see figure) mounted on the surface of an automobile frame gives the following readings: gage A,310 × 10-6:gage B,180 × l0-6; and gage C. -160 × 10-6. Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.The strains for an element of material in plane strain (see figure) are as follows: x = 480 ×10-6. y = 140 × l0-6, and xy = —350 x 10”. Determine the principals strains and maximum shear strains, and show these strains on sketches of properly oriented elements.
- The normal strain in the 45n direction on the surface of a circular tube (sec figure) is 880 × 10 when the torque T = 750 lb-in. The tube is made of copper alloy with G = 6.2 × 106 psi and y = 0.35. If the outside diameter d2of the tube is 0.8 in., what is the inside diameter dt? If the allowable normal stress in the tube is 14 ksi, what is the maximum permissible inside diameter d?Solve the preceding problem if the cross- sectional dimensions are b = 1.5 in. and h = 5.0 in., the gage angle is ß = 750, the measured strains are = 209 × 10-6 and B = -110 × 10, and the material is a magnesium alloy with modulus E = 6.0 X 106 psi and Poisson’s ratio v = 0.35.A cube of cast iron with sides of length a = 4.0 in. (see figure) is tested in a laboratory under triaxialsire.ss. Gages mounted on the testing machine show that the compressive strains in the material arc a = -225 X l06and,ay = 37.5 X l0_. Determine the following quantities: (a) the norm al stresses i. r,.. and acting on the x, y, and z faces of the cube; (b) the maximum shear stress r in the material; (C) the change ..W in the volume of the cube: (d) the strain energy U stored in the cube; (e) the maximum value of s when the change in volume must be limited to O.O28%; and (f) the required value of when the strain energy must be 38 in.-lb. (Assume £ = 14,000 ksi and v = 0.25.)
- -7 A steel tube (G = 11.5 x 106 psi) has an outer diameter d2= 2.0 in. and an inner diameter dt=1,5 in. When twisted by a torque 7", the tube develops a maximum normal strain of 170 x 10-6. What is the magnitude of the applied torque T?A circular aluminum tube subjected to pure torsion by torques T(sec figure) has an outer radius r2equal to 1.5 times the inner radius r1. (a) If the maximum shear strain in the tube is measured as 400 × 10-6 rad, what is the shear strain y1at the inner surface? (b) If the maximum a1lo-abk rate of twist is 0.125 °/m and the maximum shear strain is to be kept at 400 × 10-6 rad by adjusting the torque T, that is the minimum required outer radius ( r2)Min?A punch for making a slotted hole in ID cards is shown in the figure part a. Assume that the hole produced by the punch can be described as a rectangle (12 mm X 3 mm) with two half circles (r = 1.5 mm) on the left and the right sides. If P = 10 N and the thickness of the ID card is 1 mm, what is the average shear stress in the card?
- - 7.2-26 The strains on the surface of an experiment al device made of pure aluminum (E = 70 GPa. v = 0.33) and tested in a space shuttle were measured by means of strain gages. The gages were oriented as shown in the figure. and the measured strains were = 1100 X 106, h = 1496 X 10.6, and = 39.44 X l0_. What is the stress o in the x direction?Assume that the normal strains x and y , for an clement in plane stress (see figure) are measured with strain gages. (a) Obtain a formula for the normal strain x in the : direction in terms x , y and Poisson’s ratio v. (b) Obtain a formula for the dilatation L’ in terms of x , y and Poisson’s ratio v.An aluminum tube has inside diameter dx= 50 mm, shear modulus of elasticity G = 27 GPa, v = 0.33, and torque T = 4.0 kN · m. The allowable shear stress in the aluminum is 50 MPa, and the allowable normal strain is 900 X 10-6. Determine the required outside diameter d2 Re-compute the required outside diameter d2, if allowable normal stress is 62 MPa and allowable shear strain is 1.7 X 10-3.