Question 3.5.4: The space truss ABCDEFGH in Fig. 3.5-4(a) is in the form of ...

(a) The truss can be constructed in accordance with the rule for the construction of simple space trusses: starting with the tetrahedron AFHE, we can establish the other joints successively, as follows:
joint G from joints A, F, and H;
joint D from joints A, H, and G;
joint B from joints A, F, and G;
joint C from joints G, B, and D.
Hence, the truss ABCDEFGH is both stable and determinate.
(b) Let S_1, S_2, S_3, . . .,  S_{18} represent the tensile forces in members 1, 2, . . ., 18 respectively. At joint D, all members except member 1 lie in the yz plane. Hence, by Rule 1,

S_1=\underline{0}

Similarly,

S_2=\underline{0}

To determine the forces in members 3, 4, and 5, we note that the length of the line CE is l√3, where l is the length of an edge of the ‘cubic’ truss. Hence the angle α which CE makes with CG (Fig. 3.5-4(b)) is given by

\cos \alpha =\frac{l}{l√3}=\frac{1}{√3}

By symmetry, CE makes this same angle with CG, CB, and CD. That is, the line of action of P makes an angle \alpha=cos^{- 1} (1√3) with each of members 3, 4, and 5. Making use of symmetry, let

S_3=S_4=S_5=S

For equilibrium of joint C in the direction of P, 3S cos α = P

S=\frac{P}{3\cos \alpha}=0.577P    since     \cos \alpha =\frac{1}{√3}

i.e.

S_3=S_4=S_5=\underline{0.577P}

By similar reasoning,

S_6=S_7=S_8=\underline{0.577P}

Applying \sum{P_x}=0 to joint B,

-S_9\cos 45°-S_3=0

whence

S_9=-\frac{1}{\cos 45º}S_3=-\frac{1}{\cos 45°}\left(0.577P\right)

=-\underline{0.815P}            (i.e. compressive)

By symmetry,

S_{11}=-\underline{0.815P}

Applying \sum{P_z}=0 at joint B,

-S_{10}-S_9\cos 45°=0

Then

S_{10}=-\left(-0.815P\right)\cos 45°=\underline{0.577P}

By symmetry,

S_{12}=\underline{0.577P}

Applying \sum{P_z}=0 at joint F,

S_{13}\cos 45°+S_{10}=0

Hence

S_{13}=-\frac{1}{\cos 45°}\left(0.577P\right)=-\underline{0.815P}

By symmetry,

S_{14}=\underline{0.815P}

Applying \sum{P_x}=0 at joint H,

S_{15}\cos 45°+S_{14}\cos 45°+S_7=0

Then

S_{15}\cos 45°=-\left(-0.815P\right)\cos 45°-0.577P
=0

Hence

S_{15}=\underline{0}

With S_{15} = 0, then member 16 is the only member at joint F not lying on a plane normal to the x-axis. Hence (Rule 1)

S_{16}=\underline{0}

By symmetry,

S_{17}=\underline{0}

To determine S_{18}, let its components in the coordinate directions be X_{18},  Y_{18}, and Z_{18} respectively.
Consider equilibrium of joint A. \sum{P_x}=0 requires that

X_{18}-S_{14}\cos 45°=0

Then

X_{18}=S_{14}\cos 45°=-0.815P \cos 45°
=-0.577P

\sum{P_y}=0 requires that

Y_{18}-S_{13}\cos 45°=0

Then

Y_{18}=S_{13}\cos 45°=-0.815P \cos 45°
=-0.577P

\sum{P_z}=0 requires that

Z_{18}-S_8-S_{13}\cos 45°-S_{14}\cos 45°=0
Z_{18}=-0.577P+0.815\cos 45°+0.815P\cos 45°=0
Z_{18}=-0.577P

the magnitude of S_{18} is

√\left(X^{2}_{18}+Y^{2}_{18}+Z^{2}_{18} \right)=P

Since S_{18} has negative components in all coordinate directions, it must be tensile, i.e

S_{18}=\underline{P}

This example demonstrates that, even though the general method of joints would mean solving 3j – 6 = 3 x 8 – 6 = 18 simultaneous equations in this case, in the actual solution above we did not even have to solve two simultaneous equations at any time.