Method of Joints –Example
Using the method of
joints, determine the
force in each member of
the truss.
Method of Joints –Example
Draw the free body
diagram of the truss and
solve for the equations
F
x
 0  Cx
Cx  0 lb
F
y
 0  2000 lb  1000 lb  E  Cy
 E  Cy  3000 lb
Method of Joints
–Example
Solve the moment about C
M
C
 0  2000 lb  24 ft   1000 lb 12 ft   E  6 ft 
 E  10000 lb
 C y  3000 lb  10000 lb  7000 lb
Method of Joints
–Example
Look at joint A
4
 Fy  0   5 FAD  2000 lb
FAD  2500 lb  FAD  2500 lb  C 
3
3
 Fx  0  5 FAD  FAB  5  2500 lb   FAB
FAB  1500 lb  FAB  1500 lb  T 
Method of Joints
–Example
Look at joint D
4
4
4
4
 Fy  0  5 FAD  5 FDB  5  2500 lb   5 FDB
FDB  2500 lb  FDB  2500 lb  T 
3
3
 Fx  0   5 FAD  5 FDB  FDE
3
3
  2500 lb    2500 lb   FDE
5
5
FDE  3000 lb  FDE  3000 lb  C 
Method of Joints
–Example
Look at joint B
4
4
F

0


F

FBE  1000 lb
 y
BD
5
5
4
4
   2500 lb   FDE  1000 lb
5
5
FDE  3750 lb  FDE  3750 lb  C 
3
3
F

0


F

F

FBE  FBC
 x
BD
BA
5
5
3
3
  2500 lb   1500 lb   3750 lb   FBC
5
5
FBC  5250 lb  FDE  5250 lb  T 
Method of Joints
–Example
Look at joint E
4
4
F

0


F

FEC  10000 lb
 y
EB
5
5
4
4
   3750 lb   FDE  10000 lb
5
5
FEC  8750 lb  FEC  8750 lb  C 
3
3
F

0


F

F

FEC
 x
EB
ED
5
5
3
3
   3750 lb    3000 lb   FEC
5
5
FEC  8750 lb  FEC  8750 lb  C 
Method of Joints
–Example
Look at joint C to check the
solution
4
F
 0   FCE  7000 lb
5
4
   8750 lb   7000 lb  0 OK!
5
3
F

0


FCE  FCB  Cx
 x
5
3
   8750 lb    5250 lb   0  0
5
y
Method of Joints –Class Problem
Determine the forces BC,
DF and GE. Using the
method of Joints.
Method of Sections -Truss
The method of joints is most effective when
the forces in all the members of a truss are to
be determined. If however, the force is only
one or a few members are needed, then the
method of sections is more efficient.
Few simple guidelines of section truss analysis:
• Pass a section through a maximum of 3 members of
the truss, 1 of which is the desired member where it
is dividing the truss into 2 completely separate parts,
• At 1 part of the truss, take moments about the point
(at a joint) where the 2 members intersect and solve
for the member force, using ∑ M = 0,
• Solve the other 2 unknowns by using the equilibrium
equation for forces, using ∑ Fx = 0 and ∑ Fy = 0.
Method of Sections -Truss
If we were interested in
the force of member CE.
We can use a cutting line
or section to breakup the
truss and solve by taking
the moment about B.
Method of Sections – Example
Determine the forces in members FH, GH and GI
of the roof truss.
Method of Sections – Example
Draw a free body diagram and solve for the
reactions.
F
x
 0  RAx
RAx  0 kN
F
y
RAx
0
 L  RAy  20 kN
L
RAy
Method of
Sections –
Example
Solve for the
moment at A.
M
A
RAx
L
RAy
 6 kN  5 m   6 kN 10 m   6 kN 15 m 
1 kN  20 m   1 kN  25 m   L  30 m 
 L  7.5 kN
 RAy  12.5 kN
Method of Sections – Example
Solve for the member GI. Take a cut between the
third and fourth section and draw the free-body
diagram.
lHI
8m
10 m

 lHI 
8 m 
15 m 10 m
15 m
lHI  5.333 m
 8m 
o

28.1

 15 m 
  tan 1 
Method of Sections –
Example
The free-body diagram of
the cut on the right side.
M
H
 1 kN  5 m   7.5 kN 10 m   FGI  5.333 m 
FGI  13.13 kN  FGI  13.13 kN  T 
Method of Sections – Example
Use the line of action of the forces and take the moment
about G it will remove the FGI and FGH and shift FFH to the
perpendicular of G.
Method of
Sections –
Example
Take the moment at G
M
G
 1 kN  5 m   1 kN 10 m   7.5 kN 15 m 
 FFH cos  28.1o   8 m 
FFH  13.82 kN  FFH  13.82 kN  C 
Method of Sections – Example
Use the line of action of the forces and take the moment
about L it will remove the FGI and FFH and shift FGH to point
G.
 5 m 
o


133.2

 5.333 m 
  tan 1 
Method of
Sections –
Example
Take the moment at L
o
M

1
kN
5
m

1
kN
10
m

F
cos
43.2
 

 GH 
 15 m 
 L
FGH  1.372 kN  FGH  1.372 kN  C 
Method of Sections – Class Problem
Determine the forces in members CD and CE using method of sections.
Homework (Due 2/24/03)
Problems:
6-34, 6-37, 6-38, 6-40, 6-45, 6-63
Truss –Bonus Problem
Determine whether the
members are unstable,
determinate or
indeterminate.
Truss –Bonus Problem
Determine the loads in
each of the members.
Truss –Bonus Problem
Determine the loads in
each of the members.
Truss –Bonus Problem
Determine the loads in
each of the members.