%%%%%B--A (inclined at 55 degree)
%%%%/|%%/\
%%%/%|%/%%\Pull 3900N
%%/%%|/
%D---C
%|%%/|
%|%/%|
%|/%%|
%E
%|%%/|
%|%/%|
%|/%%|
%H---G
(All is Pin Joint)
Distance DC = FE = HG = AB = 2.7m
FH = FD = GE = EC = CB = 2 m
Find the supporting reaction of H, Member BD (Tension Force), Member CE, and Member CF?
To find the supporting reactions and tension forces in the given structure, we can analyze each joint and segment of the structure using the method of joints. This method involves applying the equations of equilibrium to each joint to solve for the unknown forces.
1. Supporting reaction at H:
Consider the joint at H. The forces acting on this joint are the horizontal and vertical reactions at H, as well as the tension force in member HG.
Since the joint is in equilibrium, the sum of the horizontal forces and vertical forces must be zero.
Horizontal forces: Hx = 0
Vertical forces: Vy - 3900N = 0 (vertical force upward is positive)
Therefore, the supporting reaction at H is Vy = 3900N.
2. Tension force in member BD:
Consider the joint at D. The forces acting on this joint are the vertical reaction at D, the tension force in member BD, and the tension force in member CD.
Since the joint is in equilibrium, the sum of the vertical forces must be zero.
Vertical forces: Tension in BD - Vy + 3900N = 0 (vertical force downward is positive)
Therefore, the tension force in member BD is equal to the supporting reaction at H: Tension in BD = 3900N.
3. Tension force in member CE:
Consider the joint at C. The forces acting on this joint are the tension force in member CE, the tension force in member CB, and the horizontal and vertical reactions at C.
Since the joint is in equilibrium, the sum of the horizontal and vertical forces must be zero.
Horizontal forces: Tension in CE - Hx = 0
Vertical forces: Tension in CB - Vy = 0
Therefore, the tension force in member CE is equal to Hx.
4. Tension force in member CF:
Consider the joint at C. The forces acting on this joint are the tension force in member CF and the horizontal reaction at C.
Since the joint is in equilibrium, the sum of the horizontal forces must be zero.
Horizontal forces: Tension in CF + Hx = 0
Therefore, the tension force in member CF is equal to the negative of Hx.
In summary, the supporting reaction at H is Vy = 3900N, the tension force in member BD is 3900N, the tension force in member CE is Hx, and the tension force in member CF is -Hx.