What are the reactions and forces in a pin-jointed framework?

[MMCS]

ƩF_x = 0
ƩF_y = 0
ƩMoments about a point = 0

A_y + B_y - 2KN = 0

Momments about A = -(2KN * 2.4) + 1*ReactionC = 0
-4.8KN = reactionC



Sub into first eq

A_y +(-4.8) -2 = 0
A_y = 6.8

To find out the reaction in member AB

I have

6.8 - T_ab Cos(tan^-10.8/1) = 0
T_ab = 6.8/Cos(tan^-10.8/1) = 8.738

I have the answer to be 2.56

 

[PhanthomJay]

What about the horizontal component of the reaction force at C? What can you say about the horiz and vert components of the reaction force at A?

 

[MMCS]

What about the horizontal component of the reaction force at C? What can you say about the horiz and vert components of the reaction force at A?

The x and y at A should equal the support AB
The x values for A and C i am unsure how to work out, i have only dealt with problems where one support is a roller support

 

[PhanthomJay]

The x and y at A should equal the support AB

You mean the vector sum of the x and y components of the reaction force at A should be equal in magnitude to the force in member AB?

The x values for A and C i am unsure how to work out, i have only dealt with problems where one support is a roller support

Truss members are 2 force members. Since no other member besides AB frame into support A, then the total resultant reaction force at support A must be in the direction of member AB.

I note that you have posted this question in the engineering section also. You should probably respond in that thread instead of this one, and avoid double posting in the future thanks.

 

[MMCS]

If the reaction force at A is equal to the magnitude of vector AB, then i have an answer of 6.8KN, but the answer is 2.56KN can you point out the mistake in my working

 

[PhanthomJay]

The reaction force at A acts in the direction of member AB. Sum moments about C to solve for it or its components.

 

[MMCS]

Ok i realize where my confusion was on this problem thanks for your help