6.4 Space Trusses
6.4 Space Trusses Example 1, page 1 of 6
1. The truss is supported by short links at B and D and by a
ball and socket at C. Determine the force in each member,
and state whether the force is tension or compression.
y
4m
4m
D
C
2m
5m
B
x
A
z
200 N
7m
6.4 Space Trusses Example 1, page 2 of 6
1
y
At joint A, only three unknown forces are
present, so begin the analysis there.
4m
2
4m
Free-body diagram of joint A.
FAD
D
FAB
FAC
C
2m
A
5m
B
x
200 N
Equilibrium equation
FAD + FAC + FAB
{200j}N= 0
(1)
A
z
200 N
7m
6.4 Space Trusses Example 1, page 3 of 6
y
3
Express the forces in component form.
4m
FAD = FAD
= FAD
unit vector pointing from A to D
4i + 5j
7k
D
42 + 52 + 72
= FAD( 0.4216i + 0.5270j
0.7379k)
C
2m
(2)
FAC = FAC(0.4216i + 0.5270j
3j
0.7379k)
(3)
7k
A
32 + 72
z
= FAB(0.3939j
0.9191k)
5m
B
x
FAC and FAD differ only in sign of x component
FAB = FAB
4m
(4)
200 N
7m
6.4 Space Trusses Example 1, page 4 of 6
4
Substitute the component forms of the force vectors into Eq.1:
FAD + FAC + FAB
200j = 0
0.4216FADi + 0.5270FADj
(Eq. 1 repeated)
0.7379FADk
+ 0.4216FACi + 0.5270FACj
0.7379FACk)
+ 0.3939FABj
200j = 0
0.9191FABk
Or, in scalar form,
Fx = 0:
0.4216FAD + 0.4216FAC = 0
(5)
Fy = 0: 0.5270FAD + 0.5270FAC + 0.3939 FAB 200 = 0
(6)
Fz = 0:
(7)
0.7379FAD
0.7379FAC 0.9191FAB = 0
Solving gives
FAD = 474.3616 N (T)
Ans.
(8)
FAC = 474.3616 N (T)
Ans.
(9)
FAB = 761.5773 N = 761.5773 N (C)
Ans.
(10)
6.4 Space Trusses Example 1, page 5 of 6
y
4m
5 Free-body diagram of joint B. Now that the force
in member AB is known, the member forces FBD
and FBC can be found.
4m
Bzk (Reaction force)
D
C
2m
FBC = FBC
FBD
5m
42 + 22
= FBC (0.8944i + 0.4472j)
B
B
4i + 2j
x
FBD = FBD
FBA = FAB (Force of member AB on joint B is equal
and opposite to force on joint A.)
4i + 2j
42 + 22
= FAB(0.9191j
= FBD( 0.8944i + 0.4472j)
A
7m
761.5773 N, by Eq. 10
= (300j
z
0.3939k), by Eq. 4
700k) N
200 N
6
Summing the forces in the x and y directions gives
+
+
Fx = 0: FBC(0.8944)
FBD(0.8944) = 0
Fy = 0: FBC(0.4472) + FBD(0.4472) + 300 = 0
7 Solving gives
FBC = 335.4102 N = 335.4102 N (C)
Ans.
FBD = 335.4102 N = 335.4102 N (C)
Ans.
(11)
6.4 Space Trusses Example 1, page 6 of 6
y
4m
8 Free-body diagram of joint D. Now that the force
in member BD is known, the remaining unknown
member force FDC can be found.
4m
Reaction forces
Dyj
Dzk
D
C
2m
5m
D
B
FDC = FDCi
FDB = FBD
x
= ( 335.4102)( 0.8944i + 0.4472j)
= 300i +
FDA = FAD
= FAD( 0.4216i + 0.5270j
A
z
200 N
by Eq. 11
j
0.7379k), by Eq. 2
474.3616 N, by Eq. 8
7m
= (200i
250j + 350k) N
9 Summing the x components of the forces acting on
joint D gives
+
Fx = 0: 200
300 + FDC = 0
Solving gives
FDC = 100 N (T)
Ans.
6.4 Space Trusses Example 2, page 1 of 4
2. Determine the force in each member of the space truss, and
state whether the force is tension or compression. The truss is
supported by short links at C and D and by ball-and-socket
supports at A and E.
y
1
Only three unknowns two member forces and a
reaction force are present at D, so a free-body
diagram of D is a good place to start.
4m
Free-body diagram of joint D
B
A
C
x
4m
FDB
Dzk
(Reaction force
from link at D)
3m
3m
FDE
D
E
700 N
D
700 N
z
5m
Equilibrium equation
FDB + FDE {700j}N+ Dzk = 0
(1)
6.4 Space Trusses Example 2, page 2 of 4
2 Express the forces FDB and FDEin component form.
FDB = FDB
y
unit vector pointing from D to B
4m
5i + 4j
= FDB
3k
B
52 + 42 + 32
A
= FDB( 0.7071i + 0.5657j
0.4243k)
FDE = FDEi
(2)
C
x
4m
3m
(3)
3m
Substitute the component form of the force vectors in Eqs. 2
and 3 into Eq. 1:
E
D
FDB + FDE 700j + Dzk = 0
(Eq. 1 repeated)
700 N
z
or
0.7071FDBi + 0.5657 FDBj 0.4243FDBk
FDEi 700j + Dzk = 0
In scalar form,
Fx = 0:
Solving gives
FDB = 1237.4369 N (T)
0.7071FDB
Fy = 0: 0.5657FDB
Fz = 0:
3
5m
FDE = 0
700 = 0
0.4243FDB + Dz = 0
Ans.
(4)
Ans.
(5)
FDE = 875 N
= 875 N (C)
Dz = 525 N
6.4 Space Trusses Example 2, page 3 of 4
y
Now that the force in member DB is known,
there are only three unknowns at joint B.
4
4m
Free-body diagram of joint B
B
B
FBA
A
x
4m
FBC
3m
FBD
FBE
3m
Equilibrium equation
E
FBE + FBC + FBA + FBD = 0
5
C
Express each force in component form
FBE = FBE
700 N
z
5m
4 j + 3k
42 + 32
= FBE ( 0.8j + 0.6k)
FBC = FBC
D
(6)
4j
3k
2
2
(7)
(9)
FBD = FDB
1237.4369 N, by Eq. 4
= FDB( 0.7071i + 0.5657j 0.4243k), by Eq. 2
4 +3
= FBC ( 0.8j
FBA = FBA i
Force of member
BD on joint B is
equal and opposite
force of member
BD on joint D.
0.6k)
(8)
= {875i
700j + 525k} N
(10)
6.4 Space Trusses Example 2, page 4 of 4
6
Substitute the component form of the force vectors in Eq. 7-10
into Eq. 6:
FBE + FBC + FBA + FBD = 0
7
Free-body diagram of joint C.
y
B
(Eq. 6 repeated)
or,
0.8FBEj + 0.6FBEk 0.8FBCj 0.6FBCk
FBAi 875i 700j + 525k = 0
4m
FBC = 0
In scalar form,
Fx = 0:
FBA + 875 = 0
Fy = 0:
0.8FBE
x, C
z
FEC
Cyj
0.8FBC 700 = 0
Fz = 0: 0.6FBE 0.6FBC
3m
525 = 0
Fz = 0: FEC + 0 = 0
Solving gives
FBE = 875 N = 875 N (C)
Ans.
FBA = 875 N (T)
Ans.
FBC = 0
Ans.
Thus
FEC = 0
Ans.
6.4 Space Trusses Example 3, page 1 of 6
3. The truss is supported by ball-and-socket
joints at A, C, D, and F. Determine the force
in each member and state whether the force
is tension or compression.
y
10 ft
8 ft
D
F
C
x
A
E
6 ft
B
5 kip
2 kip
z
4 kip
6.4 Space Trusses Example 3, page 2 of 6
1
y
At joint B, only three unknown forces are
present, so begin the analysis there.
10 ft
2
Free-body diagram of joint B
FBA
FBE
8 ft
FBC
B
D
F
5 kip
2 kip
C
4 kip
x
A
3
E
Equilibrium equation:
FBA + FBE + FBC + {5i} kip {4j} kip
+ 2k} kip = 0
(1)
6 ft
4 Express the forces in component form
B
FBA = FBA j
(2)
FBE = FBE k
(3)
FBC = FBC
10i + 6j
2
2
(4)
10 + 6
= FBC(0.8575i + 0.5145j)
(5)
5 kip
z
2 kip 4 kip
6.4 Space Trusses Example 3, page 3 of 6
5
Substitute the component form of the force vectors in Eqs.
2-5 into Eq. 1:
FBA + FBE + FBC + 5i
+ 2k = 0
FBAj
4j
(Eq. 1 repeated)
FBEk + 0.8575FBCi + 0.5145FBCj
+ 5i 4j + 2k = 0
In scalar form,
Fx = 0: 0.8575FBC + 5 = 0
Fy = 0: FBA + 0.5145FBC
Fz = 0:
(6)
4=0
(7)
FBE + 2 = 0
(8)
Solving gives
FBA = 7 kips (T)
Ans.
(9)
Ans.
(10)
Ans.
(11)
FBC = 5.8310 kip (T)
= 5.8310 kip (C)
FBE = 2 kip (T)
6.4 Space Trusses Example 3, page 4 of 6
y
6 Free-body diagram of joint E. Now that the force
in member BE is known, the forces in members
ED, EC, and EF can be found.
10 ft
FED
8 ft
FEF
E
D
FEC
C
FEB
x
A
E
Equilibrium equation
FEB + FED + FEF + FEC = 0
F
(12)
6 ft
B
5 kip
2 kip
z
4 kip
6.4 Space Trusses Example 3, page 5 of 6
y
7 Express each force in component form.
10 ft
FEB = force in direction opposite to FBE
= FBE
by Eqs. 3 and 11
= { 2}k kip
FED = FEDj
8 ft
D
F
(13)
(14)
C
FEF = FEF
10i + 6j
x
A
E
102 + 62
= FEF(0.8575i + 0.5145j)
(15)
6 ft
FEC = FEC
10i + 6j + 8k
B
102 + 62 + 82
5 kip
2 kip
= FEC(0.7071i + 0.4243j
+ 0.5657k)
(16)
z
4 kip
6.4 Space Trusses Example 3, page 6 of 6
8
Substitute the component form of the force vector in Eqs. 13-16 into Eq. 12:
FEB + FED + FEF + FEC = 0
(Eq. 12 repeated)
2k + FEDj + 0.8575FEFi + 0.5145FEFj + 0.7071FECi
+ 0.4243FECj + 0.5657FECk = 0
In scalar form,
Fx = 0: 0.8575FEF + 0.7071FEC = 0
Fy = 0: FED + 0.5145FEF + 0.4243FEC = 0
Fz = 0: 2 + 0.5657FEC = 0
Solving gives
FEC = 3.54 kip = 3.54 kip (C)
Ans.
FED = 0
Ans.
FEF = 2.92 kip (T)
Ans.
6.4 Space Trusses Example 4, page 1 of 7
4. Determine the force in each member of the space
truss. State whether the force is tension or
compression. The supports at C and D are short
links; at A a ball-and-socket support is present.
y
400 N
2m
800 N
B
5m
C
x
A
4m
D
8m
z
6.4 Space Trusses Example 4, page 2 of 7
1 Joint B is the only joint with no more
than three unknown forces, so begin
with a free-body diagram of joint B.
y
400 N
2m
800 N
B
2 Free-body diagram of joint B
800 N
5m
B
C
400 N
FBC
FBA
A
FBD
4m
D
Equilibrium equation
8m
FBA + FBD + FBC
{800j}N + (400k}N= 0
x
(1)
z
6.4 Space Trusses Example 4, page 3 of 7
3
Express the forces in component form.
FBA = FBA
= FBA
2m
800 N
B
2i
5j
22 + 52
6i
0.9285j)
5m
(2)
5j
C
62 + 52
= FBC(0.7682i
FBD = FBD
400 N
unit vector pointing from B to A
= FBA( 0.3714i
FBC = FBC
y
6i
A
0.6402j)
(3)
5j + 4k
62 + 52 + 42
= FBD(0.6838i
0.5698j + 0.4558k)
(4)
4m
D
z
8m
x
6.4 Space Trusses Example 4, page 4 of 7
4
Substituting the component form of the forces in Eqs. 2, 3, and 4 into
Eq. 1 gives
FBA + FBD + FBC
800j + 400k = 0
(Eq. 1 repeated)
or, in scalar form,
Fx = 0:
0.3714FBA + 0.7682FBC + 0.6838FBD = 0
(5)
Fy = 0:
0.9285FBA
(6)
0.6402FBC
0.5698FBD
800 = 0
Fz = 0: 0.4558 FBD + 400 = 0
5
(7)
Solving Eqs. 5, 6, and 7 simultaneously gives
FBA = 646.2198 N = 646.2198 N (C)
Ans.
(8).
FBC = 468.6150 N (T)
Ans.
(9)
FBD = 877.4964 N = 877.4964 N (C)
Ans.
(10)
6.4 Space Trusses Example 4, page 5 of 7
6 Now that the force in member BC is known, there are
only three unknowns at joint C.
y
400 N
2m
Free-body diagram of joint C
800 N
B
FCB = FBC
= (468.6150 N)(0.7682i
FCAi
0.6402j), by Eq. 3
C
5m
= 360i + 300j
FCDk
C
Cyj (Reaction from link C)
A
Equilibrium equations:
Fx = 0:
FCA
4m
D
360 = 0
8m
Fy = 0: Cy + 300 = 0
z
Fz = 0: FCD = 0
7 Solving simultaneously gives
FCA = 360 N
= 360 N (C)
FCD = 0
Cy = 300 N
x
Ans.
(11)
Ans.
(12)
(13)
6.4 Space Trusses Example 4, page 6 of 7
8
Now that the forces in members CD and BD are known,
a free-body diagram of joint D will give an equation for
the remaining unknown member force, FDA.
y
400 N
2m
800 N
B
9
Free-body diagram of joint D
FDB
FCD = 0
FDA
5m
Dxi
D
C
Dyj
x
Reactions from links
A
4m
D
10 Equilibrium equation:
FDA + FDB + Dxi + Dyj = 0
(14)
8m
z
11 Express the forces in component form.
FDB = FBD
= FBD(0.6838i
0.5698j + 0.4558k)
877.4964 N, by Eq. 10
= {600i
500j + 400k} N
(15)
(Eq. 4 repeated)
12 FDA = FDA
8i
4k
82 + 42
= FDA( 0.8944i
0.4472k )
(16)
6.4 Space Trusses Example 4, page 7 of 7
13 Substitute the component form of the force vectors in Eqs. 15 and
16 into Eq. 14:
FDA + FDB + Dxi + Dyj = 0
(Eq. 14 repeated)
or
0.8944FDAi 0.4472FDAk + 600i
+ 400k + Dxi + Dyj = 0
500j
Then the summation of the z components gives
Fz = 0:
0.4472FDA + 400 = 0
Solving gives
FDA = 894 N (T)
Ans.
6.4 Space Trusses Example 5, page 1 of 5
5. The truss is supported by ball-and-socket
joints at A, B, C, and D. Determine the forces
in members FE and EC. State whether the
forces are tension or compression.
y
800 N
800 N
800 N
800 N
G
H
E
F
8m
B
C
x
4m
2m
4m
A
D
3m
z
5m
3m
6.4 Space Trusses Example 5, page 2 of 5
y
800 N
800 N
800 N
800 N
Section
G
H
E
F
8m
B
C
x
4m
2m
4m
A
D
3m
z
5m
3m
1 Because we have been asked to find only two
member forces, the method of sections will be
used. Pass a section through the truss that cuts
members DF, CF, FG, and FE.
6.4 Space Trusses Example 5, page 3 of 5
2
Free-body diagram of portion of truss on
side of section that includes joint F.
y
800 N
FFG (parallel to the z axis)
F
FFE
FBD
3
+
C
8m
x
FEF can be computed by summing moments about the z axis:
Mz = 0:
(800 N)(3 m)
FFE(8 m) = 0
Solving gives
FFE = 300 N = 300 N (C)
3m
D
z
Ans.
6.4 Space Trusses Example 5, page 4 of 5
y
800 N
800 N
800 N
800 N
G
H
Section
E
F
8m
B
C
x
4m
2m
4m
A
D
3m
z
5m
3m
4 To calculate the force in member EC, pass a
section through the truss that cuts members EC,
EF, EA, and EH.
6.4 Space Trusses Example 5, page 5 of 5
5
Free-body diagram of portion of truss on
side of section that includes joint E
FEC can be found by summing moments about line AB. To
sum moments, first we have to express FEC in component
form.
8i 8j 6k
FEC = FEC
82 + 82 + 62
6
800 N
FEC (Parallel to line AB)
E
FFE = 300 N (C)
y
Position
vector
from B
to E
FEC
FEA
8m
C
= FEC( 0.6247i
0.6247j
0.4685k)
The moment of FEC about the line AB is given by
MAB = u rBE FEC
B
x
6m
Unit vector
parallel to AB
= k { 3i + 8j + 6k} m
FEC( 0.6247i
0.6247j
0.4685k)
= 6.8716 FEC
8m
z
7
A
3m
+
Sum moments about AB, calling a moment positive if it
produces a counterclockwise moment about AB, as
viewed from A looking back at B.
MAB = (800 N)(3 m)
300 N)(8 m) + 6.8716 FEC = 0
Solving gives
FEC = 0
Ans.