3. Find the force in members CE, DE, DF using the method of sections. The force in member CE is: kN [Tension:+, Compression: -] The force in member DE is: Kn[Tension:+, Compression: -] The force in member DF is: kN [Tension:+, Compression: -]

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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This is the whole question but i want specially the 3rd part
B
F
RK
E
G
F1 F2
AC, CE, EG, GI, IK, BD, DF, FH, HJ = 12.5 m, AZ=15 m, ZB=6.25 mF_1= 60 kN, F_2 = 60 kN
For the truss as shown and the loading indicated.
Show the free body diagram for the joint:
Olncluded in manual solution
ONot made
1.Find the reactions.
pe here to search
Transcribed Image Text:B F RK E G F1 F2 AC, CE, EG, GI, IK, BD, DF, FH, HJ = 12.5 m, AZ=15 m, ZB=6.25 mF_1= 60 kN, F_2 = 60 kN For the truss as shown and the loading indicated. Show the free body diagram for the joint: Olncluded in manual solution ONot made 1.Find the reactions. pe here to search
UNot made
Time left 0:40:32
1.Find the reactions.
Reaction at A is:
kN. (answer to the nearest whole number)
Reaction at K is:
kN. (answer to the nearest whole number)
2. Use the method of joints to find the forces in AB and AC.
The force in member AB is:
kN [Tension:+, Compression: -]
The force in member AC is:
kN [Tension:+, Compression: -]
3. Find the force in members CE, DE, DF using the method of sections.
The force in member CE is:
kN [Tension:+, Compression: -]
The force in member DE is:
Kn [Tension:+, Compression: -]
The force in member DF is:
kN [Tension:+, Compression: -]
Type here to search
hp
Transcribed Image Text:UNot made Time left 0:40:32 1.Find the reactions. Reaction at A is: kN. (answer to the nearest whole number) Reaction at K is: kN. (answer to the nearest whole number) 2. Use the method of joints to find the forces in AB and AC. The force in member AB is: kN [Tension:+, Compression: -] The force in member AC is: kN [Tension:+, Compression: -] 3. Find the force in members CE, DE, DF using the method of sections. The force in member CE is: kN [Tension:+, Compression: -] The force in member DE is: Kn [Tension:+, Compression: -] The force in member DF is: kN [Tension:+, Compression: -] Type here to search hp
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