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Why is it compression

Why is it compression

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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Why is it compression?

STEPZ Bolution part 9 Drau FBD PL Boo lb 600 Jo 20 20 20 400lb Reニnoo.b 1000 lb Considey jeiw-by O-tan () = 63. 43' R FEG E Fy-0, Ffo sin(63.93)= -800 6343 FFg= m894.465lb -GOoeb FEG Efx-0 FFy6OS63, 43 +FE O=-600 80olb FEG =-600- (-894.465)0s FEG = 6343 200eb Considler jont F EFy-, (FeF ナFFy)sme393こ 63A 63 143 FDF FEF ニ +894.46516 I Fx =0 FDF+ FEplOs63.43= Ffofos63.43 FEP FDF = (Ffn- FEF) tel63.43 FF4=894.465 lb (compresin) FRF= 2 & FFO Cos 63. 43 FEG= 2004 (compressjon) FRE= 900blcompretsion) FDF = 2x (-894. 465-) Col 63.43 FDF800 lb ) methodof section Cconsider lett side of-seution) 400X304 9 FFX20 Rex 1D 1400x 1D 12000 + FCE X20 . FCE = 100 lb(Tensio )
Transcribed Image Text:STEPZ Bolution part 9 Drau FBD PL Boo lb 600 Jo 20 20 20 400lb Reニnoo.b 1000 lb Considey jeiw-by O-tan () = 63. 43' R FEG E Fy-0, Ffo sin(63.93)= -800 6343 FFg= m894.465lb -GOoeb FEG Efx-0 FFy6OS63, 43 +FE O=-600 80olb FEG =-600- (-894.465)0s FEG = 6343 200eb Considler jont F EFy-, (FeF ナFFy)sme393こ 63A 63 143 FDF FEF ニ +894.46516 I Fx =0 FDF+ FEplOs63.43= Ffofos63.43 FEP FDF = (Ffn- FEF) tel63.43 FF4=894.465 lb (compresin) FRF= 2 & FFO Cos 63. 43 FEG= 2004 (compressjon) FRE= 900blcompretsion) FDF = 2x (-894. 465-) Col 63.43 FDF800 lb ) methodof section Cconsider lett side of-seution) 400X304 9 FFX20 Rex 1D 1400x 1D 12000 + FCE X20 . FCE = 100 lb(Tensio )
Given STEP1 800lb 600lb 20 20: C 201E 20 400lb て RE ニ 1400 (^) Royx ニ 6oo b (a) Roy)= 80elb(か) Lusing method of joints) DeromPne> FFy, FEG, FRE O FCE (wing method of section)
Transcribed Image Text:Given STEP1 800lb 600lb 20 20: C 201E 20 400lb て RE ニ 1400 (^) Royx ニ 6oo b (a) Roy)= 80elb(か) Lusing method of joints) DeromPne> FFy, FEG, FRE O FCE (wing method of section)
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