The question is to calculate the forces ina ll members of the trussnby method of joints.
First i have calculated the reaction forces for A and B by taking the moment from both these points.
These came too: Ay=50N and By=30N
I do not know where to go from here, what point should i analyse first, i tried analysing point H however i am not sure how i should go about this, is F_hj equal to zero. and would CH equal to the force of AC.
Please help i am lost. i just need to know how to start, so maybe could you just help me solve two points and i can probably finish the rest. thank you for your time.
here is the figure:
http://www.flickr.com/photos/78859486@N04/7059400621/
First i have calculated the reaction forces for A and B by taking the moment from both these points.
These came too: Ay=50N and By=30N
I do not know where to go from here, what point should i analyse first, i tried analysing point H however i am not sure how i should go about this, is F_hj equal to zero. and would CH equal to the force of AC.
Please help i am lost. i just need to know how to start, so maybe could you just help me solve two points and i can probably finish the rest. thank you for your time.
here is the figure:
http://www.flickr.com/photos/78859486@N04/7059400621/
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Bars CH, HJ, LM, GM, and EK are all zero-force-members (F = 0). This is because if you look at node H for example, if CH had a nonzero force, there would be nothing to balance it out because the only other bar that influences that node is bar HJ which cannot have a y-component of force (force in a bar has to be along the direction of the bar). So those 5 bars can essentially be taken out of the picture (just pretend that they aren't there).
Now we can look at the simplified structure:
First of all, your reaction forces are not right. Retry your calculations and you should get Ay = 65N and By = 55N.
Let's start at node A. We know that Ay = 65N, so the force in bar AC = 65N by the joint at A.
Now move on the node C. Because bar CH has been removed, the only bar that has a y-component to balance the force from bar AC is bar CJ. So use trig to find that CJ = 65/sin(45) = 91.92N. Now we have to balance the x-component of the force from bar CJ with bar CD. Do the force balance correctly and you should get the force in bar CD = 65N.
Continue in this fashion. The order of nodes that I would recommend would be
A (shown)
C (shown)
J (solve JD and JK)
D (solve DJ then solve DE)
B
G
L
F
You never have to actually solve nodes E and K because by the time that you do the other nodes, all the bars will be solved already.
Hope this helps!
Now we can look at the simplified structure:
First of all, your reaction forces are not right. Retry your calculations and you should get Ay = 65N and By = 55N.
Let's start at node A. We know that Ay = 65N, so the force in bar AC = 65N by the joint at A.
Now move on the node C. Because bar CH has been removed, the only bar that has a y-component to balance the force from bar AC is bar CJ. So use trig to find that CJ = 65/sin(45) = 91.92N. Now we have to balance the x-component of the force from bar CJ with bar CD. Do the force balance correctly and you should get the force in bar CD = 65N.
Continue in this fashion. The order of nodes that I would recommend would be
A (shown)
C (shown)
J (solve JD and JK)
D (solve DJ then solve DE)
B
G
L
F
You never have to actually solve nodes E and K because by the time that you do the other nodes, all the bars will be solved already.
Hope this helps!