Steel: E = 200 GPa В V 50 kN 3 m C 2.5 m Brass: E = 105 GPa 9.18 The 36-mm-diameter steel rod ABC and a brass rod CD of the same diameter are joined at point C to form the 7.5-m rod ABCD. For the loading shown and neglecting the weight of the rod, deter- mine the deflection of (a) point C, (b) point D. V100 kN Fig. P9.18
Steel: E = 200 GPa В V 50 kN 3 m C 2.5 m Brass: E = 105 GPa 9.18 The 36-mm-diameter steel rod ABC and a brass rod CD of the same diameter are joined at point C to form the 7.5-m rod ABCD. For the loading shown and neglecting the weight of the rod, deter- mine the deflection of (a) point C, (b) point D. V100 kN Fig. P9.18
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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Question

Transcribed Image Text:A
2 m
Steel: E = 200 GPa
В
50 kN
3 m
C
2.5 m
Brass: E = 105 GPa
9.18 The 36-mm-diameter steel rod ABC and a brass rod CD of the
same diameter are joined at point C to form the 7.5-m rod ABCD.
For the loading shown and neglecting the weight of the rod, deter-
mine the deflection of (a) point C, (b) point D.
V100 kN
Fig. P9.18
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