What Are the Forces in Truss Members BH and HG?

AI Thread Summary
The discussion revolves around determining the forces in truss members BH and HG within a symmetrical plane truss. The reaction forces at supports A and E are both 15 kN, leading to calculations of forces along the x and y axes. The participant struggles with the next steps, particularly in identifying the forces at node G, which must balance the forces from the members meeting there. It is suggested to analyze one node at a time, focusing on the forces acting directly at that node rather than external loads. Understanding the balance of forces at node G is crucial for deriving the necessary equations to solve for the unknown forces.
sophiebeth100
Messages
12
Reaction score
0

Homework Statement



The right triangles of the plane truss are all the same. What are the forces in members BH and HG?

Diagram for question is below, or a link:
http://puu.sh/nsw1k/aad8a8b134.png

Homework Equations



n/a

The Attempt at a Solution



Reaction force at A + reaction force at E are equal, therefore both = 15kN.

It was after this point I became stuck.

I calculated the forces in the x axis:
ABcos60 + AHcos30 = 0
Then in the y axis:
ABcos30 + AHcos30 = 15kN

Not sure if the above is correct, and not sure how to continue with the question. Any help would be great!
 

Attachments

  • a.png
    a.png
    9.7 KB · Views: 423
Last edited by a moderator:
Physics news on Phys.org
You can get more equations by considering the balance of forces at each node. But that could generate a lot of equations and a lot of unknowns.
Using the symmetry, there are 7 unknowns.
It will help if you can find an easy one to start with. I suggest G. What do you notice there?
 
haruspex said:
You can get more equations by considering the balance of forces at each node. But that could generate a lot of equations and a lot of unknowns.
Using the symmetry, there are 7 unknowns.
It will help if you can find an easy one to start with. I suggest G. What do you notice there?

Perhaps that GC is equal to 12kN? Not entirely sure if that's right but if so I could work from there.
 
sophiebeth100 said:
Perhaps that GC is equal to 12kN? Not entirely sure if that's right but if so I could work from there.
No, that isn't right.
What forces act at G? What are their directions?
 
haruspex said:
No, that isn't right.
What forces act at G? What are their directions?
I wouldn't think there are any forces in the x-axis since the total force in the x-axis is zero.

Does the force at GC act upwards? Equal to the total downward force, so 30kN?
 
sophiebeth100 said:
I wouldn't think there are any forces in the x-axis since the total force in the x-axis is zero.

Does the force at GC act upwards? Equal to the total downward force, so 30kN?
In analysing forces in such interconnected systems, you need to focus on one node at a time, just thinking about what that node experiences directly. The only forces node G feels are from the three members that meet there, it doesn't 'know' anything about the externally applied loads. (Only nodes B, C, D feel those.)
The forces from the three members meeting at G must balance both vertically and horizontally. That allows you to write two equations involving the forces in those members. What equations do you get?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top