Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 29)
29.
A pin-jointed tower truss is loaded as shown in the given figure. The force induced in the member DF is
Discussion:
14 comments Page 1 of 2.
Dhritiman said:
2 years ago
Force at member DF = (2*3)/4 = 1.5 Tension.
CIVi said:
4 years ago
The answer is correct.
taking reaction about F.
4xE= (2 x 3) + (2 x 6)
E = 18/4.
E = 4.5 (compression).
taking reaction about F.
4xE= (2 x 3) + (2 x 6)
E = 18/4.
E = 4.5 (compression).
Arashdeep Singh said:
7 years ago
Take a section from mid of CE to mid of DF and calculate moment about joint C.
Then, You will get a compressive force of 1.5 kN.
Then, You will get a compressive force of 1.5 kN.
Sharma said:
7 years ago
1.5 Tension.
The reaction at F will be 4.5 upward.
And taking vertical equilibrium we have;
3/5 FC - FD = 4.5.
Now, the force in FC will be compressive only and subtracting force FD is equal to reaction 4.5 which means force FD and reaction is in same direction i.e upward which means the force FD is a Tension force
The reaction at F will be 4.5 upward.
And taking vertical equilibrium we have;
3/5 FC - FD = 4.5.
Now, the force in FC will be compressive only and subtracting force FD is equal to reaction 4.5 which means force FD and reaction is in same direction i.e upward which means the force FD is a Tension force
Bharti Sharma said:
7 years ago
1.5 KN Compression.
Calculate horizontal and vertical reactions and then by the method of sections, take equilibrium of horizontal, vertical forces and moments.
Calculate horizontal and vertical reactions and then by the method of sections, take equilibrium of horizontal, vertical forces and moments.
Mike Edrad said:
8 years ago
1.5 KN Compression because it is towards the joint.
Maxp50 said:
8 years ago
Ans is C.
Calculate reactions now go with the method of section cutting member EF CF DF.
Now, do summation Fx=0 and Fy=0.
Force in DF=-1.5KN i.e. compression.
Calculate reactions now go with the method of section cutting member EF CF DF.
Now, do summation Fx=0 and Fy=0.
Force in DF=-1.5KN i.e. compression.
Immu azm said:
8 years ago
Reaction at E 4.5 downward and at F 4.5 upwards.
Immu azm said:
8 years ago
Answer is D.
Reaction Re=4.5 downward now apply zero force member rule force in member EF =0 ( force CE in member AND reaction at E are collinear force in non collinear member must be zero ).
Now cut the member X-X passing through CA,CD,FD. NOW TAKE moment about C we will get 4.5 * 4 = DF * 4.
Reaction Re=4.5 downward now apply zero force member rule force in member EF =0 ( force CE in member AND reaction at E are collinear force in non collinear member must be zero ).
Now cut the member X-X passing through CA,CD,FD. NOW TAKE moment about C we will get 4.5 * 4 = DF * 4.
Chintu said:
8 years ago
Taking moment about E we will get reaction at F as 4.5 upwards and hence 4.5 kN comp force in member DF.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers