Question 3.5.1: A space truss ABCDEF is hinged to a vertical wall in the zx ...
A space truss ABCDEF is hinged to a vertical wall in the zx plane at A, B, C, and D (Fig. 3.5-1 (a)). Joints A, B, F, and E lie in the horizontal xy plane, with BA along the x-axis and BF along the y-axis. Joints B, C, and D lie on the vertical z-axis.
The horizontal angles between bars 2, 3, and 5, and the angles which bars 1 and 6 make with the vertical are as shown. Determine all the bar forces due to a vertical force P at joint E.
At joint F, all bars except bar 1 lie in the horizontal plane. Hence, by Rule 1, the force in bar 1 is zero, i.e.
S_{1}=\underline{0}
At joint E, S_{6}, the force in bar 6 is immediately determined from the condition \sum{P_{z}} = 0:
S_{6 } \cos 60° = P
Then
S_{6} =\underline{2P} (compressive)
The horizontal component of S_{6} is
H_{6} = S_{6} \sin 60° = 1.73P
From Fig. 3.5-1(b), it is clear that
S_{4} = H_{6} \cos 60° = \underline{0.87P} (tensile)
S_{5} = H_{6} \sin 60° = \underline{1.5P} (tensile)
By symmetry (Fig. 3.5-1(b)),
S_{3} = H_{6} = \underline{1.73P} (compressive)
S_{2} = S_{4} = \underline{0.87P} (tensile)
In this example, we do not even have to solve more than one equation at a time.
