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Example: 2.7 A hollow circular nylon pipe (see Fig 1-23) supports a load P = 7800 N, which is uniformly distributed around a cap plate at the top of the lower pipe. A second load Pg is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are d, = 51 mm, d, = 60 mm, d, = 57 mm, and d = 63 mm, respectively. The upper pipe has a length L,= 350 mm; the lower pipe length is L2 = 400 mm. Neglect the self-weight of the pipes. (a) Find Pg so that the tensile stress in upper part is 14.5 MPa. What is the resulting stress in the lower part? (b) If P, remains unchanged, find the new value of P, so that upper and lower parts have same tensile stress.

Example: 2.7 A hollow circular nylon pipe (see Fig 1-23) supports a load P = 7800 N, which is uniformly distributed around a cap plate at the top of the lower pipe. A second load Pg is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are d, = 51 mm, d, = 60 mm, d, = 57 mm, and d = 63 mm, respectively. The upper pipe has a length L,= 350 mm; the lower pipe length is L2 = 400 mm. Neglect the self-weight of the pipes. (a) Find Pg so that the tensile stress in upper part is 14.5 MPa. What is the resulting stress in the lower part? (b) If P, remains unchanged, find the new value of P, so that upper and lower parts have same tensile stress.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Example: 2.7 A hollow circular nylon pipe (see Fig 1-23) supports a load P = 7800 N, which is uniformly distributed around a cap plate at the top of the lower pipe. A second load Pg is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are d, = 51 mm, d, = 60 mm, d, = 57 mm, and d = 63 mm, respectively. The upper pipe has a length L,= 350 mm; the lower pipe length is L2 = 400 mm. Neglect the self-weight of the pipes. (a) Find Pg so that the tensile stress in upper part is 14.5 MPa. What is the resulting stress in the lower part? (b) If P, remains unchanged, find the new value of Pe so that upper and lower parts have same tensile stress. (c) Find the tensile loads in part (b) pipe stress analysis be 3.56 mm and is 7.63 mm. Two-tier hanging e segments for the egment is known to bottom of the pipe L1 = 350 mm L2=400 mm Ps -----
Transcribed Image Text:Example: 2.7 A hollow circular nylon pipe (see Fig 1-23) supports a load P = 7800 N, which is uniformly distributed around a cap plate at the top of the lower pipe. A second load Pg is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are d, = 51 mm, d, = 60 mm, d, = 57 mm, and d = 63 mm, respectively. The upper pipe has a length L,= 350 mm; the lower pipe length is L2 = 400 mm. Neglect the self-weight of the pipes. (a) Find Pg so that the tensile stress in upper part is 14.5 MPa. What is the resulting stress in the lower part? (b) If P, remains unchanged, find the new value of Pe so that upper and lower parts have same tensile stress. (c) Find the tensile loads in part (b) pipe stress analysis be 3.56 mm and is 7.63 mm. Two-tier hanging e segments for the egment is known to bottom of the pipe L1 = 350 mm L2=400 mm Ps -----
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