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# 03a Pin jointed trusses

```Fundamental Mechanics
ENGG 1300
Pin-Jointed Trusses
- Method of Joints
Ir Dr. Ray Su
Faculty of Engineering
THE UNIVERSITY OF HONG KONG
1
Structural Analysis of Pin-Jointed Trusses
2
Pin-Jointed Truss
A truss is a structure composed of
a series of slender members
jointed together at their ends.
Roof truss
Planar trusses lie in a single plane.
They are often used for roof
structures or bridges.
They are typically constructed
from triangular (but not
quadrilateral) elements to achieve a
stable structural form.
Truss bridge
footbridge
3
Connections
The connections of a truss are
usually formed by bolting or
welding the ends to a common
plate, called a gusset plate.
bolts
Gusset plate
Some of them are formed by
simply passing a large bolt or pin
through each of the members.
A pin
4
Assumptions for analysing pin-jointed trusses
Analytical
model of
a bridge truss

All loadings are applied at the joints






In most situations, such as for bridge and roof trusses, this
assumption is true. The external loads are applied at the joints.
The self-weight of a member is usually evenly distributed at the
end of the member.
For simplicity, the self-weight of trusses is neglected in this
course, unless otherwise specified.
The members are jointed together by smooth (frictionless) pins.
The truss members take axial load only, either in tension or in
compression.
Deformations of the truss are small.
5
Sign convention of pin-jointed truss members
The reaction forces at the ends of the truss member must be
Why?
directed along the axis of the member.
If the force tends to elongate the member, it is considered to
be in tension.
T
T
If the force tends to shorten the member, it is considered to
be in compression.
C
C
Often, compression members must be made thicker to resist
buckling effects.
7
For information
Buckling of a compression member
Buckling is an instability
phenomenon in
structural engineering,
where a small increase
in the compression load
can lead to a sudden
large deformation and
catastrophic failure of a
member under
compression.
Before buckling
After buckling
8
Method of Joints
In previous lecture, we have already learnt how to calculate
external reaction forces for beams and frames.
In order to fully analyse or solve a truss structure, we need to
calculate all the internal forces within every member of the
truss.
The method of joints simply consists of:
1.
Selecting a solvable joint within the truss and isolating it.
2.
Drawing a free body of that solvable joint – replacing the
truss members with unknown axial forces.
3.
Applying the two equilibrium equations to solve for the
unknown forces. Why?
F
H
0
and
F
V
0
9
Example
Determine all the internal forces of the truss by the Method
of Joints.
Solution
The free body diagram of Joint B
is shown below.
Applying ΣFH = 0, then
A pinned
support
A roller
support
+
+ 500 – FBC sin 45&ordm; = 0
 FBC = 500 / sin 45&ordm; = 707.1 N
10
After updating the free body
diagram of Joint B, we can now
solve for member BA.
FBC = 707.1N
Applying ΣFV = 0, then
+
+ 707.1 cos 45&ordm; – FBA = 0
 FBA = 707.1 cos 45&ordm; = 500 N
11
Sign convention for tension and compression
When using the
method of joints, we
are considering
internal forces of
members (in black).
Which one is
in tension?
The forces shown (in red) are the
external forces.
Remember that when using the
method of joints – the joint must be
in equilibrium (i.e. external loads
balance by internal forces).
The best way to remember the
sign convention is:
All tension forces are drawn
AWAY from the joint.
12
Continue from the previous Example, the next joint to analyse is
Joint C.
FBD of Joint C
(C)
(T)
RC
F
H
0 
 + 707.1 cos 45&ordm; – FCA = 0
 FCA = 500 N
F
V
0 
– 707.1 sin 45&ordm; + RC = 0  RC = 500 N
You can now transfer your answers to the main FBD.
13
The last joint to analyse is Joint A.
FBD of Joint A
RAH
RAV
FAC = 500N (T)
RC = 500N
F
F
H
0 
 RAH = 500 N
V
0 
 RAV = 500 N
You can now transfer your answers to the main FBD.
14
RAH = 500N
FAC = 500N (T)
RAV = 500N
RC = 500N
Free body diagrams of joints and
members
Note: You can check your answers by working out the reactions
separately and seeing if they match.
Example: Take moment at A, -500 N2+Rc2 = 0, Rc = 500 N
15
Procedure of method of joints
1. Draw the FBD for the complete truss.
2. Label all the joints in the truss (preferably A to Z, usually
omitting I).
3. Draw and label the external support reactions.
4. Decide whether you need to calculate the external support
reactions in order to progress or complete the truss analysis.
5. Determine the forces in each truss member by method of joints:
i.
Choose an appropriate solvable joint (i.e. a joint with at most
two unknown member forces).
ii. Draw the FBD of the chosen joint – assuming the unknown
member forces to be in tension or compression.
iii. Apply the equilibrium equations to solve for the unknown
forces.
iv. Transfer each member force to the main FBD (of the complete
truss), ensuring correct tension or compression signs are
expressed.
v. Repeat steps i to iv until the entire truss has been solved. 16
Example
Determine all the internal forces of the truss by the Method of Joints.
pinned
support
Roller
support
Here, we need to
calculate the
reactions first
because Joints B
and D both have
more than 2
unknown member
forces.
17
Solution
M
RCV
FBD
B
RCH
 400 x 3 + 600 x 4 - RA x 6 = 0
 RA = 600 N
F
V
D
0 
 -RCV + RA – 400 = 0, Rcv = 200 N
F
H
RA
0 
0 
 RCH = 600 N
Now it is possible to analyse either Joint A or C.
18
Choosing Joint A:
200 N
FBD
B
FBD of Joint A
600 N
(C)
(T)
D
600 N
F
F
V
 0    + 600 – (FAB x 4/5) = 0  FAB = 600 x 5/4 = 750 N
H
 0    – (750 x 3/5) + FAD = 0  FAD = 450 N
You can now transfer your answers to the main FBD.
19
Choosing Joint C next:
200 N
FBD
B
FCB
600 N
FCD
You should be able to see
via inspection that:
FCB = 600 N (C)
FCD = 200 N (C)
450 N (T)
D
600 N
You can now transfer your answers to the main FBD.
20
Choosing Joint D next:
B 600 N (C)
200 N (C)
600 N
450 N (T)
FBD of Joint D
FDB (C)
200 N
200 N
FBD
D
600 N
F
V
 0    – 200 – (FDB x 4/5) = 0  FDB = – 200 x 5/4 = – 250 N
As FDB is negative, this means that the initial assumption of force sense
was incorrect.
You can check the results by summing the horizontal forces.
FH  0    – 450 + 600 +(–250 x 3/5) = 0
21
You can now transfer your answers to the main FBD.

200 N
FBD
B 600 N (C)
200 N (C)
600 N
450 N (T)
D
As FDB is negative, this
means that the initial
assumption of force
sense (C) was incorrect.
So we can simply change
the sense of the force
when transferring it to
the main free body
diagram as shown.
600 N
22
Method of Joints
To sum up, the method of joints consists of:
1.
Selecting a solvable joint within the truss and isolating it.
2.
Drawing a free body diagram of that solvable joint –
replacing the truss members with unknown axial forces.
3.
Applying the equilibrium equations to solve for the unknown
forces.
23
Ended
24
For information
Force equilibrium
A joint
F
H
0
F1
Unknown force
F3
Applied forces
+
The usual sign convention
Then
F2
-
F1 - F2 - F3 = 0  F1 = + F2 + F3
The second equation can actually be interpreted like that the
unknown force (F1) is balanced by the applied loads (F2 &amp; F3).
The corresponding sign convention is
+
Unknown
force
=
+
Applied