CHAPTER 6 Space Trusses

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
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.
(Fx = 0; Fy = 0; Fz = 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 = 3j + r  Stable and Internally
Determinate
If m < 3j + r  Unstable
If m > 3j + r  Internally Indeterminate
• Internal stability can sometimes be checked by
careful
inspection
of
the
member
arrangement.
DETERMINACY & STABILITY
Internally
m + r = 3j
m + r > 3j
m + r < 3j
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
3j = 12
m + r = 3j
 Determinate Truss
EXAMPLE 2
m = 15, j = 10, r = 15
Ball &
Socket
m + r = 30
3j = 30
m + r = 3j
 Determinate Truss
EXAMPLE 3
m = 13, j = 8, r = 12
m + r = 25
3j = 24
m + r = 25  3j = 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  Tensile (T)
If the force tends to shorten  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
FBy
Member A
FBx
FAy
Both members A and B has zero force
because both members do not lie in a
straight line.
0
FAx
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:
FAx = -FBx
FAy = -FBy
Member B
Member A
FAy
FBx
FAx
0
0
0
FBy
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
F1= 0 and F2 = 0
Theorem 3: Joint K
F3 = F4 = F5 = 0 and F1= 0
Theorem 3: Joint M
F6 = F7 = F8 = 0 and F2= 0
Theorem 3: Joint J
F9 = F10 = F11= 0 and F5 = F6= 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
Lx = L cos  : Ly = L cos  ; Lz = L cos 
Fx = F cos  : Fy = F cos  ; Fz = F cos 
Therefore:
𝐹𝑥
𝐿𝑥
=
𝐹
𝐿
𝐹𝑦
𝐿𝑦
=
𝐹
𝐿
𝐹𝑧
𝐿𝑧
=
𝐹
𝐿
ANALYSIS METHOD: TENSION
COEFFICIENT
• Tension Coefficient: t = F/L
 Fx = tLx ; Fy = tLy ; Fz = tLz
• If t positive (tension), if t negative (compression)
• The actual force and length are given from:
𝐿=
𝐿𝑥 2 + 𝐿𝑦 2 + 𝐿𝑧 2
𝐹=
• For 3D equilibrium:
Fx = 0 ; Fy = 0 ; Fz = 0
  tLx = 0 ; tLy = 0 ; tLz = 0
𝐹𝑥 2 + 𝐹𝑦 2 + 𝐹𝑧 2
ANALYSIS METHOD: TENSION
COEFFICIENT
y
L, F
Lz , Fz
Ly , Fy
x

A
Lx , Fx
z
The component of length, L, and force, F in the x, y, z direction
ANALYSIS METHOD: TENSION
COEFFICIENT
y
B
Fy
A
F
O
X
D
Fx
Fz
E
z
C
x
𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥
𝐿𝑥
𝐹
𝐹𝑥 = 𝐹
=
𝐿𝑥
𝐿
𝐿
𝐹𝑥 = 𝑡 ∙ 𝐿𝑥
Where t = F/L
t = tension coefficient
ANALYSIS METHOD: TENSION
COEFFICIENT
y
B
Fy
A
Y F
O
D
Fx
Fz
E
z
C
x
𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦
𝐿𝑦
𝐹
𝐹𝑦 = 𝐹
=
𝐿𝑦
𝐿
𝐿
𝐹𝑦 = 𝑡 ∙ 𝐿𝑦
ANALYSIS METHOD: TENSION
COEFFICIENT
y
B
Fy
A
F
O
Z
D
Fx
Fz
E
z
C
x
𝐹𝑧 = 𝐹𝑐𝑜𝑠𝜃𝑧
𝐿𝑧
𝐹
𝐹𝑧 = 𝐹
=
𝐿𝑧
𝐿
𝐿
𝐹𝑧 = 𝑡 ∙ 𝐿𝑧
EXAMPLE 6
The space truss shown in
the figure has roller and
socket support at joint A, B,
C and D
i)
Determine the member
force for all members
at joint F and G
z
y
x
40 kN
10 kN
E
20 kN
40 kN
G
40 kN
2m
F
ii) 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
GF
0
-2
0
Force (kN)
10
0
-40
L (m)
2
4.47
2
t (kN/m) F (kN)
-20
-40
2.5
11.18
-2.5
-5
Fx = 0: -4tGE + 10 = 0
 tGE= 2.5 kN/m
Fz = 0: -2tGC – 40 = 0
 tGC = -20 kN/m
Fy = 0: -2tGE – 2tGF = 0
-2(2.5) – 2tGF = 0
 tGF = -2.5 kN/m
EXAMPLE 6 – Solution
2. Move to Joint F
Member
FC
FD
FE
FG
Force (kN)
Lx (m) Ly (m) Lz (m)
0
2
-2
0
-1
-2
-4
0
0
0
2
0
0
0
-40
L (m)
2.83
2.24
4
t (kN/m)
-5
-15
0
-2.5
F (kN)
-14.14
-33.6
0
Fx = 0: -4tFE = 0
 tFE = 0 kN/m
Fy = 0: 2tFC – tFD + 2tFG = 0
 2tFC – tFD – 5 = 0 ....... (i)
Fz = 0: -2tFC – 2tFD – 40 = 0 .......(ii)
(i) + (ii): -3tFD -45 = 0
 tFD = -15 kN/m
From (i): -2tFC – (-15) – 5 = 0
 tFC = -5 kN/m
EXAMPLE 6 – Solution
3. Calculate the Reaction at Joint C
Member Lx (m) Ly (m) Lz (m)
CF
0
-2
2
CG
0
0
2
Force (kN) RCx
RCy
RCz
L (m)
t (kN/m) F (kN)
-5
-20
Fx = 0: RCx = 0 kN
Fy = 0: -2tCF + RCy = 0
-2(-5) + RCy = 0
RCy = -10 kN (
)
 RCz = 50 kN (
)
Fz = 0: 2tCF + 2tCG + RCz = 0
2(-5) + 2(50) + RCz = 0
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
Ez = 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.
tDC = tDA = tDE = 0
EXAMPLE 7 – Solution
Joint E
Member
EC
EB
ED
EA
Force (kN)
Lx (m) Ly (m) Lz (m)
-2
-4
4
2
-4
4
-2
-4
0
2
-4
0
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
Fz = 0: 4tEB + 4tEC – 4 = 0 .......(i)
Fx = 0: -2tEC + 2tEB – 2tED + 2tEA = 0 ....... (ii)
where tED = 0
Fy = 0: -4tEC – 4tEB – 4tED – 4tEA = 0 .......(iii)
where tED = 0
Solve Eq. (i), (ii) & (iii): tEC = 0 kN/m, tEA = -1 kN/m and tEB = 1 kN/m
EXAMPLE 7 – Solution
Joint C
Joint C has three (3)
unknowns, as tCE = tCD = 0
RCy
z
C
B
Therefore, by Theorem 3:
tCB = tCA = RCy = 0
D
A
Ez = 4 kN
x
y
E
EXAMPLE 7 – Solution
Joint C
Member
CE
CA
CD
CB
Force (kN)
Lx (m) Ly (m) Lz (m)
2
4
-4
4
0
-4
0
0
-4
4
0
0
0
-RCy
0
L (m)
6
5.66
4
4
Fz = 0: 4tCA = 0
 tCA = 0 kN/m
Fx = 0: 4tCA + 4tCB = 0
 tCB = 0 kN/m
Fy = 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
Ez = 4 kN
x
y
E
EXAMPLE 7 – Solution
Joint B
Member
BC
BA
BE
Force (kN)
Lx (m) Ly (m) Lz (m)
-4
0
0
0
0
-4
-2
-4
-4
RBx
-RBy
0
L (m)
4
4
6
t (kN/m)
0
-1
1
Fx = 0: -4tBC – 2tBE + RBx = 0
 RBx = 2 kN
Fy = 0: -4tBE – RBy = 0
 RBy = -4 kN
Fz = 0: -4tBA – 4tBE = 0
 tBA = -1 kN/m
F (kN)
0
-4
6
EXAMPLE 7 – Solution
Joint A
z
C
B
D
RAy
RAx A
x
RAz
Ez = 4 kN
y
E
EXAMPLE 7 – Solution
Joint A
Member
AB
AC
AD
AE
Force (kN)
Lx (m) Ly (m) Lz (m)
0
0
4
-4
0
4
-4
0
0
-2
4
0
-RAx
RAy
RAz
L (m)
4
5.66
4
4.47
t (kN/m)
-1
0
0
-1
Fx = 0: -4tAC – 4tAD – 2tAE – RAx = 0  RAx = -2 kN
Fy = 0: 4tAE + RAy = 0
 RAy = -4 kN
Fz = 0: 4tAB + 4tAC + RAz = 0
 RAz = 4 kN
F (kN)
-4
0
0
-4.47
EXAMPLE 7 – Solution
Method 2
Reactions at A, B and C can
also be obtained from
Method 2
RCy
z
RBy
C
B
RBx
D
RAy
RAx A
Ez = 4 kN
x
RAz
y
E
EXAMPLE 7 – Solution
Method 2
My = 0
Fx = 0
-4(2) + RBx (4) = 0
2 – RAx = 0
 RBx = 2 kN
 RAx = 2 kN
Mz = 0
Fy = 0
 RCy = 0 kN
RAy – 4 = 0
Mx = 0
 RAy = 4 kN
RBy (4) – 4(4) = 0
Fz = 0
 RBy = 4 kN
RAz – 4 = 0
 RAz = 4 kN
CLASS EXERCISE
The space truss shown in the Figure consists of six members and is
supported by a short link at A, two short links at B, and a ball and
socket at D. Analyse the force in each of the members for the given
loading.
y
2m
2m
A
3m
B O
D
z
7m
C
x
1800 N