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6.5 Space Trusses
A space truss consists of members joined
together at their ends to form a stable 3D
structure
The simplest space truss is a tetrahedron,
formed by joined 6
members as shown
Any additional members
added would be redundant
in supporting force P
6.5 Space Trusses
Assumptions for Design
 The members of a space truss may be
 treated as two force members provided
 the external loading is applied at the joints
 and the joints consist of ball and socket
 connections
 If the weight of the member is to be
 considered, apply it as a vertical force, half
 of its magnitude applied at each end of the
 member
6.5 Space Trusses
Procedure for Analysis
Method of Joints
 To determine the forces in all the members of
 the truss
 Solve the three scalar equilibrium ∑Fx = 0, ∑Fy =
 0, ∑Fz = 0 at each joint
 The force analysis begins at a point having at
 least one unknown force and at most three
 unknown forces
 Cartesian vector analysis used for 3D
6.5 Space Trusses
Procedure for Analysis
Method of Sections
 Used to determine a few member forces
 When an imaginary section is passes through a
 truss and the truss is separated into two parts,
 the below equations of equilibrium must be
 satisfied
     ∑Fx = 0, ∑Fy = 0, ∑Fz = 0
     ∑Mx = 0, ∑My = 0, ∑Mz = 0
 By proper selection, the unknown forces can be
 determined using a single equilibrium equation
6.5 Space Trusses

Example 6.8
Determine the forces acting in the members
of the space truss. Indicate
whether the members are
in tension or compression.
6.5 Space Trusses
 Solution
 Joint A
r       r       v         r r             r
P = {−4 j }kN , FAB = FAB j , FAC = − FAC k
              r
r            rAE 
FAE = FAE        
             rAE 
              r        r          r
= FAE (0.577i + 0.577 j − 0.577 k )

   r
∑ F = 0;
r r         r     r
P + FAB + FAC + FAE = 0
    r       r       r             r             r             r
− 4 j + FAB j − FAC k + 0.577 FAE i + 0.577 FAE j − 0.577 FAE k = 0
6.5 Space Trusses
Solution
Joint A
∑ Fx = 0;− RB cos 45o + 0.707 FBE = 0
∑ Fy = 0;−4 + RB sin 45o = 0
∑ Fz = 0;2 + FBD − 0.707 FBE = 0


RB = FBE = 5.66kN (T )
FBD = 2kN (C )
To show
FDE = FDC = FCE = 0

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6161103 6.5 space trusses

  • 1. 6.5 Space Trusses A space truss consists of members joined together at their ends to form a stable 3D structure The simplest space truss is a tetrahedron, formed by joined 6 members as shown Any additional members added would be redundant in supporting force P
  • 2. 6.5 Space Trusses Assumptions for Design The members of a space truss may be treated as two force members provided the external loading is applied at the joints and the joints consist of ball and socket connections If the weight of the member is to be considered, apply it as a vertical force, half of its magnitude applied at each end of the member
  • 3. 6.5 Space Trusses Procedure for Analysis Method of Joints To determine the forces in all the members of the truss Solve the three scalar equilibrium ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 at each joint The force analysis begins at a point having at least one unknown force and at most three unknown forces Cartesian vector analysis used for 3D
  • 4. 6.5 Space Trusses Procedure for Analysis Method of Sections Used to determine a few member forces When an imaginary section is passes through a truss and the truss is separated into two parts, the below equations of equilibrium must be satisfied ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 ∑Mx = 0, ∑My = 0, ∑Mz = 0 By proper selection, the unknown forces can be determined using a single equilibrium equation
  • 5. 6.5 Space Trusses Example 6.8 Determine the forces acting in the members of the space truss. Indicate whether the members are in tension or compression.
  • 6. 6.5 Space Trusses Solution Joint A r r v r r r P = {−4 j }kN , FAB = FAB j , FAC = − FAC k r r  rAE  FAE = FAE    rAE  r r r = FAE (0.577i + 0.577 j − 0.577 k ) r ∑ F = 0; r r r r P + FAB + FAC + FAE = 0 r r r r r r − 4 j + FAB j − FAC k + 0.577 FAE i + 0.577 FAE j − 0.577 FAE k = 0
  • 7. 6.5 Space Trusses Solution Joint A ∑ Fx = 0;− RB cos 45o + 0.707 FBE = 0 ∑ Fy = 0;−4 + RB sin 45o = 0 ∑ Fz = 0;2 + FBD − 0.707 FBE = 0 RB = FBE = 5.66kN (T ) FBD = 2kN (C ) To show FDE = FDC = FCE = 0