TABLE 4 Data for Space Truss (Fig. 12)
Member AB AC BC BD BB'
dx, ft (m) O (O) 12 (37) 12 (3.7) 12 (3.7) O (O)
4,ft(m) 7(2.1) 7(21) 0(0) 10(3.0) 0(0)
dz, ft (m) 4 (1.2) 4 (12) O (O) 4 (1.2) 8 (2.4)
d, ft. (m) 8.06 (2.5) 14.46 (4.4) 12.00 (3.7) 16.12 (4.9) 8 (2.4)
Fx, Ib (N) 0(0) 1,250(5,560) 1,723(7,664) 1,723(7,664) 0(0)
Fy, Ib (N) 1,436 (6,367) 729 (3,243) O (O) 1,436 (6,367) O (O)
Fz,lb(N) 821(3,652) 417(1,655) 0(0) 574(2,553) 1,395(6,205)
F 5 Ib(N) +1,653 (+7,353) +1,506 (+6,699)-!,723 (-7,664) +2,315 (+10,297) -1,395 (-6,205)
- Compute the vertical reactions
Consider the equilibrium of joint D and the entire truss when you are computing the verti-
cal reactions. In all instances, assume that an unknown internal force is tensile. Thus, at
joint D: ^Fx = -H 1 + 2BDx = O; BDx = 1723-lb (7664-N) tension; BDy = 1723(10/12) =
1436 Ib (6387 N); likewise, ^Fy = V 1 - 2BDy = V 1 - 2(1436) - O; F 1 - 2872 Ib (12,275 N).
For the entire truss, 2Fy = V 1 + V 2 - 4330 = O; V 2 = 1458 Ib (6485 N).
The z components of the reactions are not required in this solution. Thus, the remain-
ing calculations for BD are BD 2 = 1723(4/12) = 574 Ib (2553 N); BD = 1723(16.12/12) =
2315 Ib (10,297 N).
- Compute the unknown forces by using the equilibrium of a joint
Calculate the forces AC and BC by considering the equilibrium of joint C. Thus ^Fx =
0.5H 2 + ACx + BC=Q, Eq. a; 2Fy = 0.5F 2 - ACy = O, Eq. b. From Eq. b, ACy = 729-lb
(3243-N) tension. Then ACx = 729(12/7) = 1250 Ib (5660 N). From Eq. a, BC = 1723-lb
(7664-N) compression. Then AC 2 = 729(4/7) = 417 Ib (1855 N); AC = 729(14.46/7) =
1506 Ib (6699 N).
- Compute another set of forces by considering joint equilibrium
Calculate the forces AB and BB' by considering the equilibrium of joint B. Thus ^Fy =
BDy - ABy = O; ABy = 1436-lb (6387-N) tension; AB 2 = 1436(4/7) = 821 Ib (3652 N); AB =
1436 (8.06/7) - 1653 Ib (7353 N); 2FZ = -AB 2 - BD 2 - BB' = O; BB' = 1395-lb (6205-N)
compression.
All the internal forces are now determined. Show in Table 4 the tensile forces as posi-
tive, and the compressive forces as negative.
- Check the equilibrium of the first joint considered
The first joint considered was A. Thus ^Fx = -2ACx + 2500 =-2(1250) + 2500 = O, and
^Fy = 2ABy + 2ACy - 4330 = 2(1436) + 2(729) - 4330 = O. Since the summation offerees
for both axes is zero, the joint is in equilibrium.
- Check the equilibrium of the second joint
Check the equilibrium of joint B by taking moments of the forces acting on this joint with
respect to the axis through A parallel to the x axis (Fig. 12c). Thus %MAx = -IBB' + IBD 2
- 4BDy = -7(1395) + 7(574) + 4(1436) = O.
- Check the equilibrium of the right-hand part of the structure
Cut the truss along a plane parallel to the yz plane. Check the equilibrium of the right-
hand part of the structure. Now ^Fx = -2BDx + 2BC-2ACx + 2500 = -2(1723) + 2(1723)
- 2(1250) + 2500 - O, and ^Fy = 2BDy + 2ACy - 4330 = 2(1436) + 2(729) - 4330 = O. The
calculated results are thus substantiated in these equations.
