Trusses Method Of Sections

1. Erie Canal Lift Bridge – Main Street, Brockport, NY
©2007 Coon All Rights Reserved

2. Supports
A pinned support can support a
structure in two dimensions.
A roller support can support a
structure in only one dimension.

3. Static Determinacy Formula
RMJ +=2
ReactionsofNumber R
MembersofNumber
Joints ofNumber
=
=
=
M
J

4. Statically Indeterminate
A truss is considered statically indeterminate when the static equilibrium
equations are not sufficient to find the reactions on that structure.
There are simply too many unknowns.
RMJ +=2Try It
Did you notice
the two pinned
connections?
B
A CD
lbDF500=

5. Statically Determinate
A truss is considered statically determinate when the static equilibrium
equations can be used to find the reactions on that structure.
RMJ +=2Try It
Is the truss
statically
determinate
now?
B
A CD
lbDF500=

6. Static Determinacy Example
()
3838
335192
2
=
+=
+= RMJ
Each side of the main street bridge in Brockport, NY has 19 joints, 35 members
and three reaction forces (pin and roller) making it a statically determinate truss.
What if these
numbers were
different?
ReactionsofNumber R
MembersofNumber
Joints ofNumber
:Remember
=
=
=
M
J

7. Equilibrium Equations
0=SXF
The sum of all forces in the X- direction is zero.
0=SYF
The sum of all forces in the Y- direction is zero.
0=SM
The sum of the moments about a given point is zero.

8. Momentary Review
RESISTANCE ARM
L
r
EFFORT
ARM
L
e
F
e
EFFORT FORCE
F
r
RESISTANCE FORCE
A moment is a twisting or turning force
sometime referred to as torque.
A moment arm is nothing more than a
lever. The wheelbarrow pictured to the
right is a third class lever.
Given the following information, you
could calculate how much force would be
needed to lift the handles of the
wheelbarrow.
•Distance from the fulcrum (A) to
the effort (C)
•Distance from the fulcrum (A) to
the resistance (B).
•Resistance Load (B)
A
D C

9. RESISTANCE ARM
L
r
EFFORT
ARM
L
e
F
e
EFFORT FORCE
F
r
RESISTANCE FORCE
Now lets replace the wheelbarrow
with a truss. Likewise, joint A would
be the fulcrum, the load is applied at
joint D, and the reaction at joint C is
counteracting force F
D
.
Remember the truss is in static
equilibrium, therefore, all forces
must sum to zero.
If we sum the moments about point
A, we can find the reaction force R
CY

at point C.
Momentary Review
lbDF500=
B
A C
AXR
AYR CYR
D
3’ 7’

10. Finding the Reaction Forces
“For every action, there is an
equal and opposite reaction”
- Sir Isaac Newton

11. Using Moments to Find R
CY
A force that would cause a
clockwise moment is
negative.
A force that causes a
counterclockwise moment is
positive.
lb
lbftft
ftlbft
ftftlb
ftft
150
500,1)10(
0)10(500,1
0)10()3(500
0)10()3(
0
=
=
=+-
=+-
=+-
=S
×
×
CY
CY
CY
CY
CYD
A
R
R
R
R
RF
M
F
D
is negative because it
causes a clockwise
moment.
R
CY
is positive because it
causes a
counterclockwise moment.
B
A C
AXR
AYR CYR
D
3’ 7’
lbDF500=

12. Sum the Y Forces to Find R
AY
We know two out of the three forces
acting in the Y-direction. By simply
summing those forces together we
can find the unknown reaction at point A.
Please note that F
B
is a shown as a
negative because of its direction.
See Cartesian coordinate system.
lbAY
AYlb
AYlblb
AYCYD
Y
R
R
R
RRF
F
350
0350
0150500
0
0
=
=+-
=++-
=++-
=S
B
A C
AXR
AYR
D
lbDF500=

13. Sum the X Forces to Find R
AX
Because joint A is pinned, it could react
to a force applied in the X-direction.
However, Since the only load applied to
this truss (F
B
) has no X-component, R
AX

must be zero.
0
0
=
=S
AXR
Fx
B
A C
AXR
AYR
D
lbDF500=

14. A
B
C
D
E
F
If you can solve a truss using the
Method of Joints, you can solve a
truss using the Method of Sections.

15. A
B
C
D
E
F
R
AY
R
AX
R
FY
Calculate Reaction Forces R
AX
, R
AY
& R
FY
0=SXF 0=SYF 0=SM

16. A
B
C
D
E
F
R
AY
R
AX
R
FY
Let’s find the force in member CD.
F
CD
known
known
known

17. A
B
C
D
E
F
R
AY
R
AX
R
FY
Cut across two or three members, but no
more than three.
known
known
known

18. A
B
C
R
AY
R
AX
Treat this cut section as a RIDGID BODY.
known
known

19. A
B
C
R
AY
R
AX
Assume the forces on cut members act as
external forces on the cut
F
BD
Assumed Compression
F
CD
Assumed Tension
F
CE
Assumed Tension
known
known

20. A
B
C
R
AY
R
AX
Treat left section as a RIDGID BODY.
F
BD
F
CD
F
CE
0=SXF
0=SYF
0=SCM
3 unknowns
BD, CD
X
& BC
1 unknown
CD
y
can be found
1 unknown
FE can be found
known
known

21. 0=SXF
The sum of all forces in the X- direction is zero.
0=SYF
The sum of all forces in the Y- direction is zero.
0=SM
The sum of the moments about a given point is zero.
Why can we only cut three members?

22. D
E
F
R
FY
You could use the right side of the truss as well.
Start by cutting through two or three members.
F
BD
F
CD
F
CE
Assumed Compression
Assumed Tension
Assumed Compression
known

23. D
E
F
R
FY
You could use the right side of the truss as well.
Start by cutting through two or three members.
F
BD
F
CD
F
CE
0=SXF
0=SYF
0=SCM
3 unknowns
BD, CD
X
& BC
1 unknown
CD
y
can be found
1 unknown
FE can be found
known

24. Erie Canal Lift Bridge – Main Street, Brockport, NY
©2007 Coon All Rights Reserved