6.1 Simple Trusses
1. Definition of Truss
A structure composed of slender members jointed together at
their end points by bolting, welding or pinning.
F2
F1
F3
Bolting
2. Planar Trusses
planar trusses lie in a single and one often need to support
roofs and bridges.
Roof truss
Bridge truss
3. Assumption for Truss
(1) all loading are applied at the joints
a. members’ weights are neglected (W << external force)
b. members’ weights are included
(2) members are jointed together by smooth pin (No friction forces)
Because of the two assumptions, the truss member is a
two-force member T
T Tensile force
T
C
T
C Compression force
Note: Compression member must be made thicken than tension member
because of the buckling or column effect in compression member.
4. Simple Truss
A simple truss is constructed by starting with a basic triangular
element and connecting two members to form an additional
element.
E
D
A
P
C
B
C
Simple form of rigid or
stable truss
A
D
unstable truss
6.2 the Method of Joints
1. Objective
Evaluate the force acting on each member of a truss structure to
analyze or design a truss
2. Concept
A truss is in equilibrium, so each of its joint also is in equilibrium
Force system acting at pin or joint is coplanar and concurrent
Therefore, we have
(A)member equilibrium is automatically satisfied at each joint
(B)Only force equilibrium ΣFx=0 and ΣFy=0 is necessary
3. Analysis Procedure
Support A is pin constraint.
Support C is roller constraint.
(1) Draw the free-body diagram of a joint having at least
one known force and at most two unknown forces.
For the given truss, only joint B is satisfied.
F.B.D of joint B
(2) Establish the sense of the unknown forces
(a) assume the unknown member forces to be in tension.
(b) Determine correct sense of unknown member forces by
inspection
(3) Apply the force equilibrium equations
ΣFx=0 500+FBC sin45°=0
ΣFy=0 -FAB-FBCcos45°=0
(4) Solve the unknown forces and check their correct sense
FBC=- 500/sin45°=-500√2N (in compression)
FAB=-FBCcos45°=500N (in tension)
(5) Continue to analyze then joints by repeating steps (1) to (4)
choose joint C first, then joint A
F.B.D of joint C
equilibrium equation at joint C
ΣFx=0 500√2 cos45°-FAC=0
ΣFy=0 Cy-500√2 sin45°=0
FAC=500N (in tension); Cy=500N
F.B.D joint A
Ax=Ay=500N
6-3 zero-force members
1. zero-force member
A member supports no loading, which is used to increase
the stability of the truss and to provide support if the applied
loading is changed.
2. Rules for determining zero-force members
(1) If only two members form a truss joint and no
external load or support reaction is applied to the joint.
Ex.
F
D
E
C
A
B
Joint A and joint D are formed by two members AF &AB, ED&CD respectively and
no loading is applied. Hence, members AF, A, ED and CD are zero-force members.
F.B.D of joint A
Equations of equilibrium
FAF
•ΣFx=0 , so FAB=0
A
FAB
• ΣFy=0 , so FAF=0
Equations of equilibrium
F.BD of joint D
D
FED
x
FDC
y
•ΣFx=0, FED+FDC cosθ=0
•ΣFy=0, FDC sinθ=0
So, FDC=0,FED=0
F
The equivalent system
is given as shown.
E
b
B
C
P
(2) If three members form a truss joint for which two of the
members are collinear , the third member is zero-force
members provided no external force or support reaction is
applied to the joint.
EX:
P
E
D
C
A
B
Joint D :3 members AD, ED and CD (ED &DC collinear)
Joint C :3 members AD, DC and CB (DC& CB collinear)
No external force or support reaction is applied to the joint D & C.
∴AD and AC zero-force members
6-4 The method of section
1. Purpose
Determine the loadings acting within a body .
2. Principle
If a body is equilibrium , then any part of the body is also in
equilibrium.
T
T
F  0
F
F
Imaginary
section
In equilibrium
F=T
Internal tension force
F  0
T
T
In equilibrium
In equilibrium
F=T
Apply the equations of equilibrium to the sectioned part to
determine the loading at the section.
3. Analysis procedure
500N
B
2m
A
45
2m
C
(1) Determined the external reactions at the constraints F.B.D
of entire truss.
F.B.D. of entire truss
Equations of equilibrium
B
500N
y 2m
F
+
x
A
Ax
Ay
+
F
+
 MA  0
C
Cy
0
x
y
0
500-Ax=0
Cy-Ax=0
500x2-Cyx2=0
(2) Cut or section the truss through not more than three
members in which the forces are unknown. (Because three
independent equilibrium equations for three unknowns.)
500N
three possible ways
of section for given
truss
(1)
(2)
(3)
C
500N A
500N
500N
(3) Draw the free-body diagram of the part of sectioned
truss which has the least number of forces acting on it.
B
500N
Section(1)
500N
A
C
500N
500N
(4) Establish the sense of the unknown member force
500N
y
45
x
F
AB
FBC
(5) Apply the equations of equilibrium and check the
correct sense of solve forces.
+  Fx  0
+ Fy  0
500N-FBCsin45=0
FBCcos45-FAB=0
FBC=500 2 N
FAB=500N
(in compression)
(in tension)
(a) Moments should be summed about a point lying at the
intersection of the lines of actions of two unknown forces. The
third unknown is determined directly from the moment equation.
(b) Forces may be summed perpendicular to the direction of the two
unknown forces which are parallel.
(6) Continue to analyze other new sections by repeating
steps(1)to(5).
Section (2)
B
C
A
500 2 N
F.B.D. of right part of
section (2)
y
x
FAC
Equations of equilibrium
+ Fy  0
500 2 cos 45  FAC  0
+ Fy  0
 500 2 sin 45  500  0
500N
So, FAC=500N
6.6 Frames and Machines
1. Definition
(1)Frames
A stationary structure composed of pin-connected
multiforce members is used to support loads.
Ex:
bicycle frame
F2
hoist
F1
E
油壓缸
Engine jig.
(2) Machine
A structure composed of pin-connected multiforce
members with moving parts is designed to
transmit and alter the effect of forces.
Ex:crusher
Moving part
F
CAN
2. Assumptions
(1) The structure is properly supported.
(2) The structure contains no more supports or members
than necessary to prevent its collapse.
(3) The joint reactions of the structure can be determined
from the equilibrium equations.
3. Analysis procedure
200N
B
3m
2m
2m
C
60
A
(1) Draw the free body Diagram of entire structure,
a part of structure, or each of its members.
(a) Isolate each part by drawing its outline shape.
Show all forces and/or couple moments act on
the part.
(b) Identify all the two-force member in the
structure.
(c) Forces common to any two contacting
members act with equal magnitudes but
opposite sense on respective members.
Free body diagrams of members AB & BC.
200N
FAB
60
A
60 0
2m
Cx
2
Cy
m
FAB
By inspection, member AB is a two-force member.
(2) Apply the equations of equilibrium
(a) Count the total number of unknowns to make
sure that an equivalent number of equilibrium
equations can be written for solutions.
Unkoown:
FAB , Cx , Cy (3)
(b) Moments should be summed about a point that
lies at the intersection of the lines of action of as
many unknown forces as possible.
Equations of equilibrium for member BC.
 Fx  0 FAB cos 60o  C x  0
  Fy  0 C y  FAB sin 60o  200  0
  M c  0 ( FAB sin 60o )  4  200  2  0
FAB  1154.7 N
C x  577 N
C y  1000 N
(2) Check the correct sense of the unknown forces.
第六章
完