How did they solve the F(cd)

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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How did they solve the F(cd)

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- Bzll:5) + 2910 (3 -6)- 0
Bュ--2102
at- joint- C
- - 6984 N- Es (comprzession)
1.5
FEB 9
- FEB-Fac coso +FeoCoso = 0
Fee
6984 - Fac cus40+F costo = 0
Fac
→ Fac Cos 40 - Feb Cos 40° = 6984
& ZFy-0
- Fic Sing -FeD Sino = 0
9 Frc.= - FeD
putting
-- 1)
AC
in 1),
ー Cos 40°
2 Fep X 0.766
Feg Cos40°= 6984
6984
4558.75 N.
ラ Fp-
Fo - 4558-75N (compression)
FAc= 4558.75 N (Tension)
FAc IE=0, ED+ F, coso.=0
FAD
'Ac
→ FAD
Fac Cos 40
A
4558.75x Cos40
FAD= -3492:20N
at joint A
FAD
comprassin)
- 3492:20 N Ccompress
→ Fig +F Sing = 0
AC
4558-75 Sin4o
- 2930-31 N
-FAC
Sin40° =
AB
'AB
FaB- 2930-31N (compression)..
AB = 2930-31 N. (comphession
Ac= 4558.75 N (Teision)
AD = 3492: 20 N (compression)
Bc = 6984 N (Compression)
CD =4558.75 N
(compression).
Transcribed Image Text:91 1:10 AM O Answered: Determine .. bartleby.com = bartleby Q&A Engineering / Mechanical Engineeri... / Q&A Library / D... Determine the force in each member of ... ••• Try bartlek tutor today Get live help whenever you need from online tutors! - Bzll:5) + 2910 (3 -6)- 0 Bュ--2102 at- joint- C - - 6984 N- Es (comprzession) 1.5 FEB 9 - FEB-Fac coso +FeoCoso = 0 Fee 6984 - Fac cus40+F costo = 0 Fac → Fac Cos 40 - Feb Cos 40° = 6984 & ZFy-0 - Fic Sing -FeD Sino = 0 9 Frc.= - FeD putting -- 1) AC in 1), ー Cos 40° 2 Fep X 0.766 Feg Cos40°= 6984 6984 4558.75 N. ラ Fp- Fo - 4558-75N (compression) FAc= 4558.75 N (Tension) FAc IE=0, ED+ F, coso.=0 FAD 'Ac → FAD Fac Cos 40 A 4558.75x Cos40 FAD= -3492:20N at joint A FAD comprassin) - 3492:20 N Ccompress → Fig +F Sing = 0 AC 4558-75 Sin4o - 2930-31 N -FAC Sin40° = AB 'AB FaB- 2930-31N (compression).. AB = 2930-31 N. (comphession Ac= 4558.75 N (Teision) AD = 3492: 20 N (compression) Bc = 6984 N (Compression) CD =4558.75 N (compression).
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