--------> PRODUCED BY STABMOT SOFTWARE <-------
The Joint at point C of the body/beam is a pin
connection for the entire system.
AC is one member/body and BC is one member/body
of the diagram given.
Solving for the force acting on the right side of the
beam/body.
Arch design/shape is the same as the way an
Aeroplane/Airplane Hangar looks and other buildings
such as Cathedral and Dome (Stadium Dome) .
Fx = 700 x Cos(46)
Fx = 700 x 0.694425881510006
Fx = 486.098117057004
Fx = 486.10N
Fy = 700 x sin(46)
Fy = 700 x 0.719564239723634
Fy = 503.695
Fy = 503.70N
The Horizontal Force Fx = 486.10N
The vertical Force Fy = 503.70N
--------------------> SESSION ENDED <--------------------
--------> PRODUCED BY STABMOT SOFTWARE <-------
The Joint at point C of the body/beam is a pin
connection for the entire system.
AC is one member/body and BC is one member/body
of the diagram given.
Take moment for the entire member ACB of the
diagram(Arch) to get the equation at point B.
Fx = 486.1N
Fx = -486.1N , Because Fx points to the left side of the
arch
Fy = -503.7N , Because Fy points downward.
Sum of the forces in Y direction is equal to zero (0).
^ + F(y) = 0
|
--> Ay + By - 503.7 - 1000 = 0 Eqn.(1)
--> Ay + By = 1503.7 Eqn.(1)
--> + F(X) = 0
--> Ax - Bx - 700 = 0 Eqn.(2)
--> Ax = 486.1 = 0 Eqn(2)
--> Ax - Bx = 486.1 Eqn.(2)
Sum of the moment at A is equla to zero.
From Left side/hand to the right is positive.
L--> + M(A) = 0
-9 x 1000 + 18 x By + 6 x 486.1 - 16 x 503.7 = 0
18By = 14142.6 Eqn.(3)
By = 14142.6 / 18 Eqn.(3)
By = 785.7
By = 785.7
By = 785.7N
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