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The given frame is subjected to the applied loads P and Q. Note that members ABC and CDE are pin-connected at C. a) Draw the FBDs of all members. A b) Determine the force in each link (AF, BG, DG, and EH). F Neglect the weight of the members. Parameters * 2 0 * D A B C 1 2 10 2 1 3 P = 100+ (ECD) x 0.1 Q = 100+ (BAC) x 0.1 8 = 30 x (P/Q) P = 100 + (534) x 0.1 = 153.4 kN Q=100+ (312) x 0.1 = 121.3 kN e = 30 x (153.4/121.3) = 37.94° 14 E5 E 2a B 3 e 1 HO (0/ HE D 8 a 2a

The given frame is subjected to the applied loads P and Q. Note that members ABC and CDE are pin-connected at C. a) Draw the FBDs of all members. A b) Determine the force in each link (AF, BG, DG, and EH). F Neglect the weight of the members. Parameters * 2 0 * D A B C 1 2 10 2 1 3 P = 100+ (ECD) x 0.1 Q = 100+ (BAC) x 0.1 8 = 30 x (P/Q) P = 100 + (534) x 0.1 = 153.4 kN Q=100+ (312) x 0.1 = 121.3 kN e = 30 x (153.4/121.3) = 37.94° 14 E5 E 2a B 3 e 1 HO (0/ HE D 8 a 2a

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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Using frame not joint

The given frame is subjected to the applied loads P and Q. Note that members ABC and CDE are pin-connected at C. a) Draw the FBDs of all members. A b) Determine the force in each link (AF, BG, DG, and EH). F Neglect the weight of the members. Parameters * 2 0 D A B C 1 2 10 2 1 3 P= 100+ (ECD) x 0.1 Q = 100+ (BAC) x 0.1 e = 30 x (P/Q) P = 100 + (534) x 0.1 = 153.4 kN Q=100+ (312) x 0.1 = 121.3 kN e = 30 x (153.4/121.3) = 37.94° 14 E5 E 2a B 3 OG HO HE D 8 a 2a
Transcribed Image Text:The given frame is subjected to the applied loads P and Q. Note that members ABC and CDE are pin-connected at C. a) Draw the FBDs of all members. A b) Determine the force in each link (AF, BG, DG, and EH). F Neglect the weight of the members. Parameters * 2 0 D A B C 1 2 10 2 1 3 P= 100+ (ECD) x 0.1 Q = 100+ (BAC) x 0.1 e = 30 x (P/Q) P = 100 + (534) x 0.1 = 153.4 kN Q=100+ (312) x 0.1 = 121.3 kN e = 30 x (153.4/121.3) = 37.94° 14 E5 E 2a B 3 OG HO HE D 8 a 2a
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