Equilibrium and Trusses
ENGR 221
February 17, 2003
Lecture Goals
•
•
•
•
•
6-4 Equilibrium in Three Dimensions
7-1 Introduction to Trusses
7-2 Plane Trusses
7-3 Space Trusses
7-4 Frames and Machines
Equilibrium –Problem
Determine the reactions
at A and the force in bar
CD due to the loading.
Equilibrium –Problem
Draw the free-body
diagram of the main
body.
RAy
RAx
 6 in. 
o
  tan 


26.565

 12 in. 
1

TCD
Equilibrium –Problem
Look at equilibrium
o
F

R

T
cos

26.565
  125 lb  0
 x Ax CD 

RAx   TCD cos  26.565o   125 lb

o
F

R

T
sin

26.565
  40 lb  60 lb  80 lb  0
 y Ay CD 

RAy   TCD sin  26.565o   20 lb

Equilibrium – RAy
RAx
Problem
Take the moment about A

TCD
o
M

0

T
cos

26.565

  6 in.  40 lb  4 in.
 A
CD
 60 lb  8 in.  80 lb 12 in.
TCD cos  26.565o   6 in.  320 lb-in
TCD  59.628 lb
Equilibrium – RAy
RAx
Problem
Take the moment about A

RAx
TCD
   59.628 lb  cos 26.565o  125 lb



 71.667 lb

RAy    59.628 lb  sin  26.565o   20 lb
 6.667 lb


Equilibrium in 3-Dimensions
In two dimensions, the equations are solved
using the summation of forces in the x, y and z
directions and the moment equilibrium includes
moment components in the x, y and z directions.
F
x
0
F
M
x
0
M
y
0
y
0
F
z
0
M
z
0
Trusses -Definition
Trusses are structures
composed entirely of two
force members . They
consists generally of
triangular sub-element and
are constructed and
supported so as to prevent
any motion.
Frames -Definition
Frames are structures that
always contain at least one
member acted on by forces
at three or more points.
Frames are constructed and
supported so as to prevent
any motion. Frame like
structures that are not fully
constrained are called
machines or mechanisms.
Truss
Planar Trusses - lie in a
single plane and all
applied loads must lie in
the same plane.
Truss
Space Trusses - are structures that are not
contained in a single plane and/or are loaded out
of the plane of the structure.
Truss
There are four main assumptions made in the
analysis of truss
1 Truss members are connected together at their
ends only.
2 Truss are connected together by frictionless
pins.
3 The truss structure is loaded only at the joints.
4 The weights of the members may be neglected.
Simple Truss
The basic building block of a
truss is a triangle. Large truss
are constructed by attaching
several triangles together A
new triangle can be added
truss by adding two members
and a joint. A truss
constructed in this fashion is
known as a simple truss.
Simple Truss
It has been observed that the analysis of truss
can be done by counting the number member
and joints on the truss to determine the truss is
determinate, unstable or indeterminate.
Simple Truss
A truss is analysis by using m=2*j-3, where m is
number of members, j represents the number of
joints and 3 represents the external support
reactions.
Simple Truss
If m< 2j-3, then the truss is unstable and will
collapse under load.
If m> 2j-3, then the truss has more unknowns
than know equations and is an indeterminate
structure.
If m= 2j-3, ensures that a simple plane truss is
rigid and solvable, it is neither sufficient nor
necessary to ensure that a non-simple plane truss
is rigid and solvable.
Simple Truss- Identify
Determine type of simple truss is it
determinate, indeterminate or unstable.
Method of Joints -Truss
The truss is made up of single bars, which are
either in compression, tension or no-load. The
means of solving force inside
of the truss use equilibrium
equations at a joint. This
method is known as the
method of joints.
Method of Joints -Truss
The method of joints uses the summation of
forces at a joint to solve the force in the
members. It does not use the
moment equilibrium equation
to solve the problem. In a two
dimensional set of equations,
F
x
0
F
y
In three dimensions,
F
z
0
0
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
 y
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
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.
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
RAx
F
y
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.