CHAPTER 6 Space Trusses
INTRODUCTION •
•
A space truss consists of members joined together at their ends to form a stable threedimensional structures A stable simple space truss can be built from the basic tetrahedral, formed by connecting six members with four joints
INTRODUCTION •
•
A space truss consists of members joined together at their ends to form a stable threedimensional structures A stable simple space truss can be built from the basic tetrahedral, formed by connecting six members with four joints
INTRODUCTION
OBJECTIVES •
•
To determine the stability and determinacy of space trusses To determine member forces of space trusses using tension coefficient analysis
TYPE OF SPACE TRUSSES 1.
Simple Space Truss
This truss is constructed from a tetrahedron. The truss can be enlarged by adding three members.
TYPE OF SPACE TRUSSES 2.
Compound Space Truss
This truss is constructed by combining two or more simple truss.
TYPE OF SPACE TRUSSES 3.
Complex Space Truss
Complex truss is a truss that cannot be classified as simple truss or compound truss.
DETERMINACY & STABILITY •
Due to three dimensions, there will be three equations of equilibrium for each joint. (F x = 0; F y = 0; F z = 0)
•
•
The external stability of the space truss requires that the support reactions keep the truss in force and moment equilibrium. Generally, the least number of required reactions for stable and externally determinate is SIX
If r < 6
Unstable
If r > 6
Externally Indeterminate
DETERMINACY & STABILITY •
For internal determinacy, if m = number of members; j = number of joints; r = number of supports; therefore:
If m = 3 j + r Determinate
•
Stable and Internally
If m < 3 j + r
Unstable
If m > 3 j + r
Internally Indeterminate
Internal stability can sometimes be checked by careful inspection of the member arrangement.
DETERMINACY & STABILITY Internally m + r = 3 j m + r > 3 j m + r < 3 j
Determinate Truss Indeterminate Truss Unstable Truss
Externally r < 6 r = 6 r > 6
Unstable Truss Determinate if Truss is Stable Indeterminate Truss
TYPES OF SUPPORT
EXAMPLE 1
m = 3, j = 4, r = 9 Ball & Socket
m + r = 12 3 j = 12 m + r = 3 j
Determinate Truss
EXAMPLE 2 m = 15, j = 10, r = 15
Ball & Socket
m + r = 30 3 j = 30 m + r = 3 j
Determinate Truss
EXAMPLE 3 m = 13, j = 8, r = 12 m + r = 25 3 j = 24 m + r = 25 3 j = 24 Ball & Socket
Indeterminate Truss in the First Degree
ASSUMPTIONS FOR DESIGN •
•
•
The members are joined together by smooth pins (no friction – cannot resist moment)
All loadings and reactions are applied centrally at the joints The centroid for each members are straight and concurrent at a joint
Therefore, each truss member acts as an axial force member: If the force tends to elongate If the force tends to shorten
Tensile (T) Compressive (C)
ZERO FORCE MEMBERS THEOREM 1: If all members and external force except one member at a joint, (say, joint B) lie in the same plane, then, the force in member A is zero. Member A z
B P
x y
The force in member A is zero
ZERO FORCE MEMBERS THEOREM 2: If all members at a joint has zero force except for two members, (say member A and B), and both members ( A and B) do not lie in a straight line, then the force in member A and B are zero. Member B F By
Member A
F Bx F Ay
Both members A and B has zero force because both members do not lie in a straight line.
0
F Ax
0
0
ZERO FORCE MEMBERS If members A and B lie in a straight line, then, the forces in these members MIGHT NOT be zero. In fact, referring to the example below: F Ax = -F Bx F Ay = -F By
Member B Member A F Ay
F Bx F Ax
0 0
0
F By
ZERO FORCE MEMBERS THEOREM 3: If three members at a joint do not lie in the same plane and there is no external force at that joint, then the force in the three members is zero.
Member B
Three members connected at a joint has zero force. A plane can consists of two members, say member A and B. Thus, no force can balance the component of member C that is normal to the plane.
Member A
Member C
EXAMPLE 4 Identify the members of the space truss that has zero force.
Theorem 1: Joint L F 1= 0 and F 2 = 0 Theorem 3: Joint K F 3 = F 4 = F 5 = 0 and F 1= 0 Theorem 3: Joint M F 6 = F 7 = F 8 = 0 and F 2= 0 Theorem 3: Joint J F 9 = F 10 = F 11= 0 and F 5 = F 6= 0
EXAMPLE 5 Theorem 1: Joint F Members FE, FC, FD lie in a plane, except member FG. Thus, member FG has zero force.
Theorem 3: Joint F Members FE, FC, FD have zero force.
Theorem 3: Joint E Members ED, EA, EH have zero force.
ANALYSIS METHOD: TENSION COEFFICIENT •
To determine member forces
•
Based on 3D particle
equilibrium
L x = L cos : Ly = L cos ; Lz = L cos F x = F cos : F y = F cos ; F z = F cos
Therefore:
=
=
=
ANALYSIS METHOD: TENSION COEFFICIENT •
Tension Coefficient: t = F/L
F x = t Lx ; F y = t Ly ; F z = t Lz
•
If t positive (tension), if t negative (compression)
•
The actual force and length are given from: =
•
2 + 2 + 2
=
For 3D equilibrium:
F x = 0 ; F y = 0 ; F z = 0 tLx = 0 ; tLy = 0 ; tLz = 0
2 + 2 + 2
ANALYSIS METHOD: TENSION COEFFICIENT y
L, F
Lz , F z
x
Ly , F y
A
Lx , F x
z
The component of length, L, and force, F in the x, y, z direction
ANALYSIS METHOD: TENSION COEFFICIENT y B
F y A
F O
X
D
F x
F z E z
C
x
= = = = ∙ Where t = F /L t = tension coefficient
ANALYSIS METHOD: TENSION COEFFICIENT y B
F y A Y
O
F D
F x
F z E z
C
x
= = = = ∙
ANALYSIS METHOD: TENSION COEFFICIENT y B
F y A
F O
D
F x Z
F z E z
C
x
= = = = ∙
EXAMPLE 6 The space truss shown in the figure has roller and socket support at joint A, B, C and D i)
ii)
Determine the member force for all members at joint F and G
z y
x 40 kN 40 kN 10 kN
E
G
20 kN
40 kN 2m
F
Determine the reaction at support C
B A C 2m 4m
1m
D
EXAMPLE 6 – Solution 1. Start at Joint G Member Lx (m) Ly (m) Lz (m) GC 0 0 -2 GE -4 -2 0 0 -2 0 GF 10 0 -40 Force (kN)
L (m) 2 4.47 2
t (kN/m) F (kN) -20 -40 2.5 11.18 -2.5 -5
F x
= 0: -4t GE + 10 = 0
t GE=
F z
= 0: -2t GC 40 = 0
t GC =
-20 kN/m
t GF =
-2.5 kN/m
F y =
–
2.5 kN/m
0: -2t GE – 2t GF = 0 -2(2.5) – 2t GF = 0
EXAMPLE 6 – Solution 2. Move to Joint F Member FC FD FE FG Force (kN)
Lx (m) 0 0 -4 0 0
Ly (m) Lz (m) 2 -2 -1 -2 0 0 2 0 0 -40
L (m) 2.83 2.24 4
t (kN/m) -5 -15 0 -2.5
F x
= 0: -4t FE = 0
t FE =
F y
= 0: 2t FC – t FD + 2t FG = 0
2t FC – t FD – 5 =
F z
= 0: -2t FC – 2t FD – 40 = 0 .......(ii)
F (kN) -14.14 -33.6 0
0 kN/m
(i) + (ii): -3 t FD -45 = 0
t FD =
-15 kN/m
From (i): -2t FC – (-15) – 5 = 0
t FC =
-5 kN/m
0 ....... (i)
EXAMPLE 6 – Solution 3. Calculate the Reaction at Joint C Member
Lx (m) Ly (m) Lz (m)
CF CG Force (kN) F x = F y
0 0 RCx
-2 0 RCy
t (kN/m) F (kN)
2 2 RCz
-5 -20
0: RCx = 0 kN
= 0: -2t CF + RCy = 0 -2(-5) + RCy = 0
F z
L (m)
RCy =
-10 kN (
)
= 50 kN (
)
= 0: 2t CF + 2t CG + RCz = 0 2(-5) + 2(50) + RCz = 0
RCz
EXAMPLE 7
Slotted roller constraint in a cylinder
Short link
z
C
Determine the force in each member of the space truss shown.
B
D
4m
E z = 4 kN
A x Ball & socket
E
4m 2m
y
2m
EXAMPLE 7 – Solution 1. Start with joints where there are only 3 unknowns force/reaction. Joint B – 5 unknowns Joint C – 5 unknowns Joint A – 7 unknowns Joint E – 4 unknowns
Joint D Theorem 3: Three members at a joint and no external force. Thus , all members have zero forces. t DC = t DA = t DE = 0
EXAMPLE 7 – Solution Joint E Member EC EB ED EA Force (kN)
Lx (m) -2 2 -2 2 0
Ly (m) Lz (m) -4 4 -4 4 -4 0 -4 0 0 -4
L (m) 4.47 6 4.47 4.47
t (kN/m) 0 1 0 -1
F (kN) 0 6 0 -4.47
F z
= 0: 4t EB + 4t EC – 4 = 0 .......(i)
F x
= 0: -2t EC + 2t EB – 2t ED + 2t EA = 0 ....... (ii)
where t ED = 0
F y
= 0: -4t EC – 4t EB – 4t ED – 4t EA = 0 .......(iii)
where t ED = 0
Solve Eq. (i), (ii) & (iii): t EC = 0 kN/m, t EA = -1 kN/m and t EB = 1 kN/m
EXAMPLE 7 – Solution Joint C Joint C has three (3) unknowns, as t CE = t CD = 0
RCy z
C
B
Therefore, by Theorem 3: t CB = t CA = RCy = 0 D
A
E z = 4 kN
x y
E
EXAMPLE 7 – Solution Joint C Member CE CA CD CB Force (kN)
Lx (m) 2 4 0 4 0
Ly (m) Lz (m) 4 -4 0 -4 0 -4 0 0 -RCy 0
L (m) 6 5.66 4 4
F z
= 0: 4t CA = 0
t CA =
0 kN/m
F x
= 0: 4t CA + 4t CB = 0
t CB =
0 kN/m
F y
= 0: -RCy = 0
t (kN/m) 0 0 0 0
F (kN) 0 0 0 0
EXAMPLE 7 – Solution Joint B Joint B has three (3) unknowns.
z
C
RBy B RBx
D
A
E z = 4 kN
x y
E
EXAMPLE 7 – Solution Joint B Member BC BA BE Force (kN)
Lx (m) -4 0 -2 RBx
Ly (m) Lz (m) 0 0 0 -4 -4 -4 -RBy 0
L (m) 4 4 6
t (kN/m) 0 -1 1
F x
= 0: -4t BC – 2t BE + RBx = 0
RBx =
2 kN
F y
= 0: -4t BE – RBy = 0
RBy =
-4 kN
F z
= 0: -4t BA – 4t BE = 0
t BA =
-1 kN/m
F (kN) 0 -4 6
EXAMPLE 7 – Solution Joint A z
C B
D RAy RAx A
E z = 4 kN
x
RAz
y
E
EXAMPLE 7 – Solution Joint A Member AB AC AD AE Force (kN)
Lx (m) 0 -4 -4 -2 -RAx
Ly (m) Lz (m) 0 4 0 4 0 0 4 0 RAy RAz
L (m) 4 5.66 4 4.47
t (kN/m) -1 0 0 -1
F x
= 0: -4t AC – 4t AD – 2t AE – RAx = 0
RAx =
-2 kN
F y
= 0: 4t AE + RAy = 0
RAy =
-4 kN
F z
= 0: 4t AB + 4t AC + RAz = 0
RAz =
4 kN
F (kN) -4 0 0 -4.47
EXAMPLE 7 – Solution Method 2 RCy
Reactions at A, B and C can also be obtained from Method 2
z
C
RBy B RBx
D RAy RAx A
E z = 4 kN
x
RAz
y
E
EXAMPLE 7 – Solution Method 2 F x = 0
My = 0 -4(2) + RBx (4) = 0
2 – RAx = 0
RBx =
RAx =
2 kN
F y = 0
Mz = 0 RCy =
0 kN
RAy – 4 = 0 RAy =
Mx = 0 RBy (4) – 4(4) = 0 RBy =
2 kN
4 kN
4 kN
F z = 0 RAz – 4 = 0 RAz =
4 kN
