How do I calculate forces in a truss cross section?

AI Thread Summary
To calculate the forces in a truss cross section, it is essential to recognize that each section supports half of the total load. When analyzing one of the cross sections, the loads should be divided in half, as only half of the truss is being considered. This approach ensures accurate force calculations for the beams within that specific cross section. Understanding the distribution of loads is crucial for effective structural analysis. Accurate calculations will lead to better design and safety of the truss system.
carlcla
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Homework Statement
Calculate the trusses in each beam of one of the cross sections
Relevant Equations
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I need to calculate the forces in each of the beams in one of the two equal cross sections of the truss attatched in the picture. The second picture is the cross section of the truss.

As you can see on the 3d picture of the truss the loads are attatched with equal distance inbetween the 2 "cross sections" that make up the truss. When calculating forces in one of the cross sections as in the 2. Picture, do i also need to devide the loads in half as I am only calculating for half the truss?
 

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carlcla said:
When calculating forces in one of the cross sections as in the 2. Picture, do i also need to devide the loads in half as I am only calculating for half the truss?
Yes, each truss is supporting half of the total hanging weight.
 
Thanks!
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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