2692584306b

2692584306b - 524 CHAPTER 8 Applications of Plane Stress...

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Combined Loadings The problems for Section 8.5 are to be solved assuming that the structures behave linearly elastically and that the stresses caused by two or more loads may be superimposed to obtain the resultant stresses acting at a point. Consider both in-plane and out-of-plane shear stresses unless otherwise specified. Problem 8.5-1 A bracket ABCD having a hollow circular cross section consists of a vertical arm AB , a horizontal arm BC parallel to the x 0 axis, and a horizontal arm CD parallel to the z 0 axis (see figure). The arms BC and CD have lengths b 1 5 3.2 ft and b 2 5 2.4 ft, respectively. The outer and inner diameters of the bracket are d 2 5 8.0 in. and d 1 5 7.0 in. A vertical load P 5 1500 lb acts at point D . Determine the maximum tensile, compressive, and shear stresses in the vertical arm. Solution 8.5-1 Bracket ABCD 524 CHAPTER 8 Applications of Plane Stress z 0 y 0 A x 0 B C D P b 1 b 2 b 1 5 3.2 ft b 2 5 2.4 ft P 5 1500 lb C ROSS SECTION d 2 5 8.0 in. d 1 5 7.0 in. I 5 p 64 ( d 2 4 2 d 1 4 ) 5 83.203 in. 4 A 5 p 4 ( d 2 2 2 d 1 2 ) 5 11.781 in. 2 V ERTICAL ARM AB P 5 1500 lb M 5 P (distance BD ) 5 6,000 lb-ft 5 72,000 lb-in. M AXIMUM STRESSES occur on opposite sides of the vertical arm. M AXIMUM TENSILE STRESS 52 127.3 psi 1 3461.4 psi 5 3330 psi M AXIMUM COMPRESSIVE STRESS 3590 psi M AXIMUM SHEAR STRESS Uniaxial stress. t max 5 2 s c 2 2 5 1790 psi s c P A 2 M ( d 2 / 2) I 127.3 psi 2 3461.4 psi 1500 lb 11.781 in. 2 1 (72,000 lb-in.)(4.0 in.) 83.203 in. 4 s t P A 1 M ( d 2 / I 5 P Ï b 1 2 1 b 2 2 5 (1500 lb)(4.0 ft) A B C D P b 1 b 2 d 1 d 2 P M A B
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Problem 8.5-2 A gondola on a ski lift is supported by two bent arms, as shown in the figure. Each arm is offset by the distance b 5 180 mm from the line of action of the weight force W . The allowable stresses in the arms are 100 MPa in tension and 50 MPa in shear. If the loaded gondola weighs 12 kN, what is the mininum diameter d of the arms? Solution 8.5-2 Gondola on a ski lift SECTION 8.5 Combined Loadings 525 (a) (b) b d W W b 5 180 mm s allow 5 100 MPa (tension) t allow 5 50 MPa Find d min M AXIMUM TENSILE STRESS or ¢ p s t 4 W d 3 2 d 2 8 b 5 0 s t 5 W A 1 M S 5 4 W p d 2 1 32 Wb p d 3 S 5 p d 3 32 A 5 p d 2 4 W 5 12 kN 2 5 6 kN S UBSTITUTE NUMERICAL VALUES : 8 b 5 1.44 m 13,090 d 3 2 d 2 1.44 5 0( d 5 meters) Solve numerically: d 5 0.04845 m [ d min 5 48.4 mm M AXIMUM SHEAR STRESS Since t allow is one-half of s allow , the minimum diameter for shear is the same as for tension. t max 5 s t 2 (uniaxial stress) p s t 4 W 5 p s allow 4 W 5 p (100 MPa) 4(6 kN) 5 13,089.97 1 m 2 d W W b W W M 5 Wb
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Problem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.0 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 60 k and the torque (due to drilling) is 170 k-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe. Solution 8.5-3 Drill pipe for an oil well 526 CHAPTER 8 Applications of Plane Stress P 5 compressive force T 5 torque P 5 60 k T 5 170 k-in.
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2692584306b - 524 CHAPTER 8 Applications of Plane Stress...

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