
2000, W. E. Haisler Reaction, Truss Structures

Particle vs. Rigid Body

Free Body Diagrams

Reactions

Truss Structures
1


2000, W. E. Haisler Reaction, Truss Structures
Particle vs. Rigid Body
Particle
Rigid Body
Note that for a particle, all forces are directed at the center of the particle. Consequently, there are no net moments on a particle.
This is not the case for a rigid body.
2


2000, W. E. Haisler Reaction, Truss Structures
Basic Equations Available for Statics
COLM: 0
  f ext
COAM: 0

( r f ) ext
or 0
  T ext
Free Body Diagrams
Free Body Diagram is the term frequently used to define the
"system" for a static rigid body. We first draw a free body (sketch) which includes the rigid structure and the external loads applied to it, but we remove all the supports (supports that keep the structure from moving). We replace all the supports by equivalent reaction forces and/or moments.
3


2000, W. E. Haisler Reaction, Truss Structures 4
Free Body Diagrams & Symbols Used for Structure Supports
System
=
System
Rigid Body with Supports & Loads Roller Support Free Body
System
=
System System
=
System
Pined Support Free Body
Fixed Support Free Body


2000, W. E. Haisler Reaction, Truss Structures
Reactions
Reactions are a consequense of Newton's 3rd law:
For a structure which is supported in some way, we can be referring to reactions as those forces and/or moments on either the structure or on the support. But Newtons third law says that these forces/moments must be equal and opposite.
In general, when we use the term reactions , we are referring to the forces produced on the structure by the supports .
5


2000, W. E. Haisler Reaction, Truss Structures 6
100 kN
Consider the rigid frame member below. The structure is pinned at
A and on a roller support at B.
A B
100 kN
This means that the pinned support at A can provide
A x reactions in the horizontal (x) and vertical (y) directions. The roller support at B can provide reactions only in the vertical
A y
A y
A x
B y
B y
Free-body of structure with reactions y direction. The free-body diagram thus shows only the applicable reactions for the type of support which has been removed. Note the equal and opposite forces on the supports themselves. x


2000, W. E. Haisler Reaction, Truss Structures
Note that the reaction forces on the structure have been arbitrarily chosen (assumed) to be positive when they are in the positive x and y directions . This is customary and, in general, we do not worry what the actual direction will be once the reactions are determined. If the solution indicates they a force is negative, this simply means that its direction is opposite to what was assumed .
It is extremely important to put the proper reactions on a free-body consistent with type type of support at that point.
For example, if we had placed a horizontal reaction at point
B, this would mean that point B is pinned, not on a roller support. This would change the problem entirely! structure
Note: a roller support is sometimes shown as:
7


2000, W. E. Haisler Reaction, Truss Structures
3 m
Example: The rigid bar A-B has a mass of 100 kg and supports a mass of 500 kg at point B. The bar is pinned at point A and is supported by a cable between points C and D. Determine the tension in the cable and the reactions at point A.
Step 1. Define and Draw Free Body.
We first draw a free body that includes the rigid bar but not the supporting structure. We replace all support by reaction forces:
C
A rigid bar
4 m
D
1 m
B
A x
A y
A
T
3 rigid bar
2.5 m
4
1.5 m
D
1 m
B
8
980 N 4,900 N


2000, W. E. Haisler Reaction, Truss Structures
Step 2. Apply COLM ( 0
  f ext
) and COAM (
to Solve for Unknown Forces.
0

( r f ) ext
)
Sum moments about point A (CCW +):
0
  M
A

( 3
5
T )( 4 m )

980 N ( 2 .
5 m )

4 , 900 N ( 5 m )

T

11 , 229 N
Sum forces in x direction:
0
  F x

A x
 4
5
( 11 , 299 N )
Sum forces in y direction:
0
  F y

A y
 3
5
( 11 , 299 N )

980 N

4 , 900 N


A
A x y


8

, 983
857
N
N
9


2000, W. E. Haisler Reaction, Truss Structures
Example: The rigid mechanism is pinned at A and has a roller support at B. A couple of 80 Nm is applied at point C.
Determine the reactions.
80 Nm
150 mm
C
10
180 mm
B
120 mm
A
150 mm


2000, W. E. Haisler Reaction, Truss Structures
Step 1. Define and Draw Free Body.
We first draw a free body that includes the rigid bar but not the supporting structure. We replace all support by reaction forces:
80 Nm
150 mm
C
11
180 mm
A
A x
A y
150 mm
B
B y
D
120 mm


2000, W. E. Haisler Reaction, Truss Structures 12
Step 2. Apply COLM ( 0
  f ext
) and COAM (
to Solve for Unknown Forces.
0

( r f ) ext
)
Sum moments about point A (CCW +):
0
  M
A

B y
( 0 .
15 m )

80 Nm

B y

533 .
3 N
Sum forces in y direction:
0
  F y

A y

533 .
3

A y
 
533 .
3 N
Sum forces in x direction:
0
  F x

A x

A x

0
Note that that a moment equation could have been used to obtain
A
 y
A
. Sum moments about point D: y
 
533 .
3 N
0
  M
D
 
A y
( 0 .
15 m )

80 Nm


2000, W. E. Haisler Reaction, Truss Structures
Truss Structures
A truss structure is made up of "two-force" members which are connected by "pinned" joints, and external loads on the truss structure are due to point loads applied only to the pin joints.
A two-force member can carry only a tensile or compression force at each end. These forces must act through and parallel to the centerline of the two-force member.
13
Two-Force,
Pin Joint Member


2000, W. E. Haisler Reaction, Truss Structures 14
Try the following to understand how two-force members work.
Take a stick and drill a hole at each end. Attach a string to each end (through the hole). Now pull on the string. The tension you
T
T exert on the string will straighten out the string such that no matter which way you initially pull on the two strings, the strings will align with the axis of the stick. This is because the joint between the string is a pin joint. The tension in the strings must be equal to the tension in the stick (a two-force member).


2000, W. E. Haisler Reaction, Truss Structures
The analysis of truss structures proceeds in a general way as follows:
Step 1. Define and Draw a Free Body. (notice a, not the)
Step 2. Apply COLM ( 0
  f ext
) and COAM (
to Solve for the Unknown Forces.
0

( r f ) ext
)
Steps 1 and 2 may be done in one of several ways:

Methods of Sections

Method of Joints

Systematic Method of Joints (Solve entire structure at once)
15


2000, W. E. Haisler Reaction, Truss Structures
Method of Sections
Determine the reactions and force in each member. In the method of sections, we choose the free body to be the joints and member(s) connecting two or more joints, or the entire structure.
C
100 lb f
16
20 in.
B
A
30 in.
For example, we may choose the entire structure. Replacing the supports by equivalent reactions provides the free body:


2000, W. E. Haisler Reaction, Truss Structures
C
100 lb f
20 in.
A
B
A x
A y
Sum moments about A:
0
  M
A
 
100 lb f
30 in.
( 20 " )

B y
B y
( 30 " )

B y

66 .
7 lb f
Sum forces in the y direction:
0
  F y

A y

66 .
7 lb f

A y
 
66 .
7 lb f
Sum forces in the x direction:
0
  F x

A x

100 lb f

A x
 
100 lb f
17


2000, W. E. Haisler Reaction, Truss Structures 18
Now cut a section through the structure that includes member AC and joints A and C. Include the values of the reactions just obtained. Assume that the force in the cut members is F
BC
and
F
AB
(tension).
C
C
100 lb f
100 lb f
20 in.
F
BC
A
A
F
AB
B
100 lb f
66.7 lb f 30 in.
B
66.7 lb f
100 lb f
66.7 lb f
66.7 lb f
Notice that at the cut section, we have placed the “reactions” (F
BC and F
AB
) that would occur on the boundary of system (aka free body) due to removing a part of the structure (just like removing a support). Notice that we have assumed that both members are in tension. If we assume member BC is in tension, we must


2000, W. E. Haisler Reaction, Truss Structures show it as tension for both of the free bodies on either side of the cut section. To solve for the member forces, we may consider the free body on either side of the cut. In this case, I will choose the left free body:
C
100 lb f
20
20 in.
30 F
BC
19
100 lb f
A
66.7 lb f
F
AB
Sum moments about A:
0
  M
A
 
100 lb f
( 20 " )

[ 30 /( 1300 )
1 /

F
BC

2
] F
BC
( 20

120 .
2 lb
" ) f
(compression)


2000, W. E. Haisler Reaction, Truss Structures
Sum forces in the x direction:
0
  F x

F
AB

100 lb f

100 lb

120 .
2 [ 30 /( 1300 )
1 / f

F
AB

100 lb f
2
]
(tension)
Now take either joint A or C as a free body. For joint A we have
20
100 lb f
Sum forces in the y direction:
0
  F y

F
AC

66 .
7 lb f
A
100 lb f
66.7 lb f

F
AC

66 .
7 lb f
(tension)


2000, W. E. Haisler Reaction, Truss Structures
Example: Determine the reactions and force in member CD.
Indicate whether the member force is tension or compression.
1,000 lb f
B
D
21
12 ft
A
E
C
10 ft 10 ft
Step 1. Solve for the reactions at the supports with a free body of the entire structure.


2000, W. E. Haisler 22
B
Reaction, Truss Structures
1,000 lb f
D
12 ft
A
E
A x
A y
10 ft
C
10 ft
E y
In the usual way, we find that A x

0 , A y

250 lb f
, E y

750 lb f
.
To determine the force in member CD, consider a section which cuts the structure to the right of joints B and C. We obtain the following two free bodies (with the reactions included):


2000, W. E. Haisler Reaction, Truss Structures 23
1,000 lb f
B
F
BD D
F
CD
12 ft
12
12
A 5
5
250 lb f 10 ft
C
F
CE
Sum forces in the y direction (using left free body):
E
12 ft
750 lb f
0
  F y

250 lb f

F
CD
( 12 / 13 )

F
CD
 
270 .
83 lb f
(compression)
Notice that the solution for the force in member BD is very simple by making this very judicious selection of the free body!!


2000, W. E. Haisler Reaction, Truss Structures 24
Note also that we could have taken the free body to the right of the cut section. We will obviously obtain the same result for F
CD
.
Once we have solved for F
CD
, the remaining forces F
CE
and F
BD may be obtained (in needed) by summing forces and/or moments for either free body above.
If other member forces in the structure are to be determined, we may apply the method of sections repeadedly as above. For example, if we want to determine F
AB
and/or F
AC
, we take a section which isolates joint A by cutting the structure to the left of joints B and C.
F
AB
A
250 lb
F
AC f


2000, W. E. Haisler Reaction, Truss Structures 25
Method of Joints
With the method of joints, we solve for the internal member forces joint by joint.
1) We first do the standard free body of the entire structure to solve for the reactions.
2) We now very carefully select a joint as a free body such that the free body of the joint involves at most two unknown member forces. In each case we assume the unknow member force(s) are tension (+). Use COLM and
COAM to solve for the unknown member forces at this joint.
3) Repeat step 2 for other joints until we have determined all of the member forces (or at least the ones we desire).


2000, W. E. Haisler Reaction, Truss Structures
Example: Consider the previous problem and solve for the force in member BD (F
CD
).
1.
With the method of joints, we first do the standard free body of the entire structure to solve for the reactions. This gives
1,000 lb f
B
D
26
12 ft
A
250 lb f
10 ft
C
10 ft
E
750 lb f


2000, W. E. Haisler Reaction, Truss Structures
2.
To solve for the member forces by the method of joints, we now very carefully select a joint as a free body such that the free body of the joint involves at most two unknown member forces. In each case we assume the unknow member force(s) are tension (+).
To determine the force in member CD, we could analyze joints in the following order:
1) joint F: Solve for F
CE
and F
DE
.
27
E
750 lb f
We obtain F
DE
= -812.5 lb f
(comp) and F
CE
= 312.5 (ten)
F
CE
2) Joint D: Solve for F
CD
. We obtain F
CD
= -270.83 (comp)


2000, W. E. Haisler Reaction, Truss Structures
Example. Solve for all member forces in the truss structure shown below.
1,000 lb f
B
D
12 ft
A E
C
10 ft 10 ft
We first solve for the reactions in the usual manner to obtain:
28


2000, W. E. Haisler
B
Reaction, Truss Structures
1,000 lb f
D
12 ft
A E
C
250 lb f
10 ft 10 ft
750 lb f
Now we assume each member has a tensile force. Drawing the free body diagrams for all five joints gives the following:
29


2000, W. E. Haisler Reaction, Truss Structures
Free body for all joints in structure:
B
1,000 lb f
D
12 ft
A
250 lb f
10 ft
C
F
10 ft
E
750 lb f
We must start with a joint that has only two unknown member forces (since we have only two COLM equations to apply). It is easy to see that one can start at either joint A or E and solve for the two unknown joint forces at that joint. This can be followed by joints B, C and D (in any order). Note: We could not start at
B, C or D since each has 3 unknown member forces!
30


2000, W. E. Haisler Reaction, Truss Structures 31
Systematic Method of Joints
In the method of joints, we can "work our way through a structure" joint-by-joint solving for two unknown member forces at a time. Another approach is to write all the equations defining the member forces and reactions and solve these simultaneously.
Consider the following structure:
1,000 lb f
B
D
12 ft
E A
10 ft
C
10 ft


2000, W. E. Haisler Reaction, Truss Structures 32
Construct a free body diagram of all joints, labeling the unknown member forces and reactions appropriately:
1,000 lb f
B
D
12 ft
A x
A
10 ft
C
F
10 ft
E
A y
E y
Consider a typical joint. It will include 2 or more member forces and the possibility of 2 external forces (either unknown reactions or applied loads). For example, look at joint A or D above.
Assume that the external forces are in x-y coordinates.


2000, W. E. Haisler Reaction, Truss Structures 33
Consider a structure with n joints. For any joint, we can write the
COLM equation as
COLM (x): 0 =

(x component of member force)
+ x component of reaction or applied load
COLM (y): 0 =

(y component of member force)
+ y component of reaction or applied load
If we do this for all joints, this will yield a system of (2n) equations in terms of the unknown member forces and reactions.
Clearly the number of unknows cannot exceed 2n.
Indeterminate structure : A structure is said to be indeterminate if static equilibrium is not sufficient to solve for unknown member forces and/or reactions. I.E., more unknowns than equilibrium equations (we need some other kinds of equations).
Determinate structure : Internal member forces can be determined by using equilibrium alone.


2000, W. E. Haisler Reaction, Truss Structures
Lets work the problem above in this systematic manner. We write the x & y COLM (equilibrium) equations for each joint:
Joint A: x: A x
+ F
AC
+ (5/13) F
AB
= 0 y: A y
+ (12/13) F
AB
= 0
Joint B: x: - (5/13) F
AB
+(5/13) F
BC
+ F
BD
= 0 y: - (12/13) F
AB
- (12/13) F
BC
= 0
Joint C: x: - F
AC
- (5/13) F
BC
+ (5/13) F
CD
+ F
CE
= 0 y: (12/13) F
BC
+ (12/13) F
CD
= 0
Joint D: x: - F
BD
- (5/13) F
CD
+(5/13) F
DE
= 0 y: - (12/13) F
CD
- (12/13) F
DE
- 1,000 = 0
Joint E: x: - F
CE
- (5/13) F
DE
= 0 y: (12/13) F
DE
+ E y
= 0
Note that we have 10 unknows and 10 equations!!!
34


2000, W. E. Haisler Reaction, Truss Structures
1




0


0


0




0





0


0


0




0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0

1
0
0
0
5 / 13
12 / 13

5 / 13

12 / 13
0
0
0
0
0
0
0
0
5 / 13

12 / 13

5 / 13
12 / 13
0
0
0
0
0
0
0
0
5 / 13
12 / 13

5 / 13

12 / 13
0
0
0
0
0
0
0
0
5 / 13

12 / 13

5 / 13
12 / 13
1
0
0
0

1
0
0
0
0
0












F
A
A
E x y y
BD
F
AB













F
F
F
BC
CD
DE
F
AC
F
CE
0
0
0
0
1
0
0
0

1
0



















































0
0
0
0
0
0
0
, 000
0
0


















































1
Solve the above system of 10 equations using EES (or any other equation solver) to obtain
A x
= 0, A y
= 250, E y
= 750, F
BD
= -208.3, F
AB
= -270.8,
F
BC
= 270.8, F
CD
= -270.8, F
DE
= -812.5 F
AC
= 104.2,
F
CE
= 312.5 (lb f
)
35


2000, W. E. Haisler Reaction, Truss Structures 36
Recitation Exercise
Consider the truss structure shown below. Neglect weight of the members. Determine the following:
A
2,000 lb f
2,000 lb f
B a) Reactions.
200 lb f
C
E
6 ft
D
F
6 ft
6 ft i n e m e r c b) F o


2000, W. E. Haisler Reaction, Truss Structures 37
by method of joints or sections. c) Solve for all unknown member mber AD and CF
forces and reactions using the
systematic method to write the
equilibrium equations at all
joints. d) If member CF is removed and
replaced by DE, how does
this change the results? e) If the structure weighs 50 lb, is there a chance that it will try to tip over?
Suppose two members were added to the structure: members BC and DE. Could we solve for all the member forces? The answer is NO because we would now have 14 unknowns (11 member


2000, W. E. Haisler Reaction, Truss Structures forces and 3 reactions) but still only 12 equations. The structure would be statically inderminate since we cannot solve for the member forces by using equilibrium equations alone. You will learn how to solve such a problem later - by adding two additional equations known as kinematic equations.
38


2000, W. E. Haisler Reaction, Truss Structures 39 a) Make a free body of the entire structure. Apply COLM and
0
COAM to obtain:
0
 
F x

200 lbf

E
 x

E x
 

M
E
 
200 lbf
200 lbf (12 )

2, 000(6)

F y ft
A
200 lb f
C
2,000 lb f
2,000 lb f
B
D
6 ft
0


F y


F y
2, 400 lbf
 
2, 000

2, 000 E y
2, 400 lbf

E y

1, 600 lbf
E x
E
F y
6 ft
F
F y
6 ft


2000, W. E. Haisler Reaction, Truss Structures b) To solve for the force in AD, make a cut through AC, AD and
BD, and take the free body above the cut. Assume tension in all members. Apply COLM in the
A
200 lb f
F
AC
2,000 lb f
F
AD horizontal direction:
6 ft
0
 
F x


F
AD
200 lbf

1
2
F
AD
 
282.8
lbf (282.8 lbf compression)
2,000 lb f
B
F
BD
To solve for the force in CF, make a cut through CE, CF and DF, and take the free body about the cut. Apply COLM in the horizontal direction to obtain
0
 
F x

200 lbf

1
2
F
CF

F
CF
 
282.8
lbf (282.8 lbf compression)
40


2000, W. E. Haisler Reaction, Truss Structures c) Write COLM for each joint. Assume tension for all members.
Joint A: x: 0 = 200 lbf + F
AB
+ 1
2
F
AD y: 0 = -2,000 lbf - F
AC
1
2
F
AD
Joint B: x: 0 = -F
AB y: 0 = - 2.000 lbf - F
BD
Joint C: x: 0 = F
CD
+ 1
2
F
CF y: 0 = F
AC
- F
CE
1
2
F
CF
Joint D: x: 0= - F
CD
1
2
F
AD y: 0 = F
BD
+ 1
2
F
AD
- F
DF
Joint E: x: 0 = E x
+ F
EF y: 0 = E y
+ F
CE
A
2,000 lb f
200 lb f
F
AD
F
AB
F
AC 6
6
F
AD
6
F
CD
6
F
D
BD
E
F
DF
F
EC
F
EF
E x
E y
41


2000, W. E. Haisler Reaction, Truss Structures
Joint F: x: 0 = - F
EF
- 1
2
F
CF y: 0 = F y
+ F
DF
+ 1
2
F
CF
Note that we have 12 unknows and 12 equations!!! From EES,
42
Negative values indicate compression.