Help with statics question.?
As shown, a truss is loaded by the forces p1= 491 lb and p2= 210 lb and has the dimension a= 10.6 ft
What is Fbc?
Thanks for the help
What is the force found between section BC?
- WoodsmanLv 79 years agoFavorite Answer
It might be useful to know the loads at A and E. Summing the moments around point A,
P2 * a + Fe * 3a - P1 * a = 0
a cancels; Fe = (P1 - P2)/3 = (491lb - 210lb)/3 = 93.7 lb
Then Fa = P1 - Fe = 491lb - 93.7lb = 379.3 lb
Now cut the truss vertically just to the right of B. The vertical forces must sum to 0. There is no shear in pinned members, so the only vertical forces in the cut members in the LHS of the truss are in CG. Then the vertical load in CG (assumed in compression) is such that
Fcg * sinΘ + Fa - P1 = 0
where sinΘ = sin(arctan(a/½a)) = sin(arctan2) = 0.894, so
0.894Fcg = P1 - Fa = 491lb - 379.3lb = 111.7 lb
Fcg = 125 lb
Since the sign is +ve, compression was the correct assumption.
Now we can sum the moments around B on the LHS of the truss.
Fcg * cos(arctan2) * a + Fgh * a - Fa * a = 0
Fgh = Fa - 0.447Fcg = 379.3lb - 0.447*125lb = 323 lb
and now we can simply consider the horizontal forces on the LHS of the truss.
Assume BC is in tension:
Fbc - Fgh - Fcg*cos(arctan(2) = 0
Fbc = Fgh + Fcg*cos(arctan(2)) = 323lb + 125lb * 0.447 = 379 lb
Lots of ways to get here.