ES2501: Statics/Unit 16-1:
Truss Analysis: the Method of Joints
Truss Structure:
A structure with slender members pin-connected at their
ends, referred as joints, to carry loads at the joints.
P
P
P
Hinged
support
To be a truss:
- Nodal loading only;
- All joints pin-connected
Roller
support
Modeling
Joint/Node
Planar Truss (2D)
Truss
Real physical
Truss
Statically determinate Truss
Truss
Space Truss (3D)
Statically indeterminate Truss
Truss Analysis: the Method of Joints
ES2501: Statics/Unit 16-2:
Significance of Assumptions in Truss Analysis:
Each member in a truss is a two-force member.
 Force of the rest of
P
F1 the truss on member
P
AB through a pin at A
A
P
A
B
Two-force member
in equilibrium

F2
Force of the rest of
the truss on member
AB through a pin at B

F1
Two forces must have
the same amplitude,
opposite direction and
along the same line
B
A
B

F2
- Nodal loading only;
- No moments at node
ES2501: Statics/Unit 16-3:
Truss Analysis: the Method of Joints
Sign Convention:
T
T 0
Tension
T 0
Compressio n
T
In analysis, always starts with the assumed positive direction.
Then, a positive result indicates tension and a negative value
means compression.
Truss Analysis: the Method of Joints
ES2501: Statics/Unit 16-4:
Method of Joints (Nodal Analysis):
Step 1: Find support reactions;
Step 2: Draw a free-body diagram and list equilibrium
equations for each joint;
Step 3: Select independent equations to solve unknowns.
Example 1:
Reactions:
Free-body diagram of the truss, see the lift figure
A
Ax
Ay
45
P
2L
L
0
Ax  Bx  0
Ax   P
iy
0 
Ay  P  0
 Ay  P
L
i
( A)
i
45 o
i
45 o
45 o
B
ix
i
D
Bx
F
F
M
o
E
Equilibrium at C:
L
Equilibrium Equations at Joints:
TCD
TCE
Sign
convention
C
Bx ( 2 L )  P ( 2 L )  0
Bx  P
0
Take moment about a point with the most
unknown forces
TCD sin 45o  P  0
C
P
TCD cos 45o  TCE  0

TCD  2 P
TCE   P
“+” --- tension
“-” --- compression
Truss Analysis: the Method of Joints
ES2501: Statics/Unit 16-5:
Example 1:
Equilibrium Equations at Joints (con’d):
A
Ax
Equilibrium at C:
45 o
Ay
TED
2P
D
0
0
0
Bx
P
TEB
45 o 2 P
45 o
45 o
B P E
Equilibrium at D:
P
Ax   P A
D
TDE
Zero-force
member
D TEC   P
Equilibrium at A:
C
TAD  2 P
TDB
Zero-force
member
 TBE  TEC  0 TBE   P

TCD  0
TCD  0
TDC  2P
0
TDB  0
Ay  P
TAD cos 45o  Ax  0
 TAD sin 45  TAB  Ay  0
o
TAB

TAD  2 P
TAB  0
TAD
Equilibrium at B:
TAB  0
Bx  P
B
TDB  0
TEB  P
Automatically satisfied
ES2501: Statics/Unit 16-6:
Truss Analysis: the Method of Joints
Comments:
-Method of joints uses equilibrium of joints to list
necessary equations for unknowns;
A
Ax
Ay
45 o
2P
0
0
0
Bx
-Method of joints provides complete solution for
P internal forces for all members
D
45
o
B P E
45 o 2 P
45
P
-Identifying zero-force members in a truss may
simplify analysis
o
C
-Sign convention: Use tension as the conventional
direction for the internal force of any member
“+” --- tension; “-” --- compression
T 0
-Presentation of results:
Mark the results on the truss
- If solving problem manually start with finding the reactions and list
equilibrium equations for nodes with least number of unknowns.
Truss Analysis: the Method of Joints
ES2501: Statics/Unit 16-7:
Comments (con’d): -Formulate a set of simultaneous linear equations for
a computer solution
A
Ax
45
Ay
Re-collection of equilibrium equations
o
2P
For truss
D
0
0
0
Bx
45
P
Ay  P  0
Bx ( 2 L )  P ( 2 L )  0
45 o 2 P
o
45
B P E
 1

 0
 2L

 0
 0

 1
 0

 0
 0

 0
Ax  Bx  0
P
o
C
For joint E
 TBE  TEC  0
TCD  0
0 1
0 0 0
0 0 0
1 0
0 0 0
0 0 0
0 0
0 0
0 0 0
0 0 0
0 0 0
0 sin 45o 0
0 0
0 0 0
1 cos 45o
0 0
1 0
0 0 0
1 0 0
0 0
0
0 0
0 0
0 0 0
0 0 1
0 0 0 0  1  cos 45o
1 0
0
0
0
0 0 0
1 0 0
For joint D:
For joint B:
o
o
TAD  TCD  TDE cos 45  0 TBE  TDB cos 45  Bx  0
TDB sin 45o  TAB  0
TDB  TDE sin 45o  0
Select independent 10 equations
 0  for 10 unknown:

 Ax 

A 
 P 
0

y 




0




B
2
LP
x




0
TAB
P 






0
TBE 
0 




T 
 0 
cos 45o 
BD




0 
TEC 
 sin 45o 



TCD 
 0 
0






0 
TDE 
0



1
10 x10 TDA 10 x1  0 10 x1
0
0
For joint C
For joint A:
o
TCD sin 45  P  0
TAD cos 45o  Ax  0
TCD cos 45o  TCE  0  TAD sin 45o  TAB  Ay  0
Ax , Ay , Bx ,TAB,TBE ,TEC ,TED,TCD ,TDA,TDB,
Computer solution
Note: there for more than 10
equations but only 10 of them
are linearly independent