ch5-1 - 5 Stresses in Beams (Basic Topics) Longitudinal...

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Longitudinal Strains in Beams Problem 5.4-1 Determine the maximum normal strain e max produced in a steel wire of diameter d 5 1/16 in. when it is bent around a cylindrical drum of radius R 5 24 in. (see figure). Solution 5.4-1 Steel wire 5 Stresses in Beams (Basic Topics) 285 d R R 5 24 in. From Eq. (5-4): Substitute numerical values: e max 5 1 / 16 in. 2(24 in.) 1 1 / 16 in. 5 1300 3 10 2 6 5 d / 2 R 1 d / 2 5 d 2 R 1 d e max 5 y r d 5 1 16 in. d R Cylinder Problem 5.4-2 A copper wire having diameter d 5 3 mm is bent into a circle and held with the ends just touching (see figure). If the maximum permissible strain in the copper is e max 5 0.0024, what is the shortest length L of wire that can be used? Solution 5.4-2 Copper wire d = diameter L = length d 5 3 mm « max 5 0.0024 From Eq. (5-4): L min 5 p d e max 5 p (3 mm) 0.0024 5 3.93 m e max 5 y r 5 d / 2 L / 2 p 5 p d L L 5 2 p r Ê r 5 L 2 p d r
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Problem 5.4-3 A 4.5 in. outside diameter polyethylene pipe designed to carry chemical wastes is placed in a trench and bent around a quarter- circular 90° bend (see figure). The bent section of the pipe is 46 ft long. Determine the maximum compressive strain e max in the pipe. Solution 5.4-3 Polyethylene pipe 286 CHAPTER 5 Stresses in Beams (Basic Topics) 90 ° Angle equals 90º or p /2 radians, r 5 r 5 radius of curvature e max 5 p d 4 L 5 p 4 ¢ 4.5 in. 552 in. 5 6400 3 10 2 6 r 5 L p / 2 5 2 L p Ê e max 5 y r 5 d / 2 2 L / p r 5 radius r d L Problem 5.4-4 A cantilever beam AB is loaded by a couple M 0 at its free end (see figure). The length of the beam is L 5 1.5 m and the longitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm. Calculate the radius of curvature r , the curvature k , and the vertical deflection d at the end of the beam. Solution 5.4-4 Cantilever beam A B M 0 d L L 5 length of beam L 5 1.5 m « max 5 0.001 k 5 1 r 5 0.01333 m 2 1 r 5 y e max 5 75 mm 0.001 5 75 m y 5 75 mm Ê e max 5 y r A C B M 0 d L r r u 0 L 5 length of 90º bend L 5 46 ft 5 552 in. d 5 4.5 in. L 5 2 p r 4 5 p r 2 Assume that the deflection curve is nearly flat. Then the distance BC is the same as the length L of the beam. u 5 arcsin 0.02 5 0.02 rad d 5 r (1 2 cos u ) 5 (75 m)(1 2 cos (0.02 rad)) 5 15.0 mm N OTE : which confirms that the deflection curve is nearly flat. L d 5 100, sin u 5 L r 5 1.5 m 75 m 5 0.02
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SECTION 5.4 Longitudinal Strains in Beams 287 Problem 5.4-5 A thin strip of steel of length L 5 20 in. and thickness t 5 0.2 in. is bent by couples M 0 (see figure). The deflection d at the midpoint of the strip (measured from a line joining its end points) is found to be 0.25 in. Determine the longitudinal normal strain e at the top surface of the strip. Solution 5.4-5 Thin strip of steel t M 0 M 0 d L 2 L 2 L 5 20 in. t 5 0.2 in. d 5 0.25 in. The deflection curve is very flat (note that L / d 5 80) and therefore u is a very small angle.
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ch5-1 - 5 Stresses in Beams (Basic Topics) Longitudinal...

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