FRAME: Interconnected rigid body contains multi-force members
Rigid Non collapsible
Nonrigid collapsible
Structure is not rigid by itself but depends on external support for rigidity
FBD of interconnected bodies:
1. It make necessary to an arbitrary assignment of force components.
2. Force component must be consistently represented in opposite directions on separate FBD.
3. Check the structure whether it is statically determinate or not.
Or
𝒚
y
CE
𝒚
y
x
Analysis of Statically Determinate Truss:
Design of truss involves the determination of the forces (internal forces) in the various
members and selection of appropriate sizes, structural shapes, and material to withstand
the forces.
In order to calculate the forces in individual members, it may be desirable to determine the
reactions at the supports.
To obtain forces in members:
Method of Joints- Useful to find forces in all the member of truss.
Only an idealized truss can be analysed.
Methods of Sections- Quick method to find the forces in selected members of truss.
Method of Joints In order to find the internal forces in members, it may be
desirable to find the reaction at the supports. ,
, and
C
y
a Joint F
T
x
a
We draw FBD of each pins and find the unknows forces by using,
, .
Force away from the pin is indicating tension in the member.
Force towards the pin is indicating compression in the member.
Convention is to indicate the force at the pin on the same side of the member.
Method of Joints:
Joint F
T
We can use symmetry in Geometry as well as in
Loading and Supports.
Zero Force Member:
Third non-collinear member is identified as a zero-force member
Zero Force Member:
If any joint is formed by two non-collinear members without any external force acting on it,
then both the members are identified as zero-force members
Find zero force member:
IE = JI = HI = BE = FG = GH = 0 Zero Force Member
Find zero force member: Q S U W
O
P R T V X
N
A C F M
B D E G H I J K L
F1 F2
If any joint is formed such that only four forces are acting and are collinear in pairs then each
collinear forces are equal.
Cantilever Truss:
Support reactions are not required for the analysis of truss.
Analysis can be started from extreme end of the truss
is satisfied
Problems:
The rectangular frame is composed of four perimeter two-force members and two cables AC
and BD which are incapable of supporting compression.
Find the forces in all members due to load L in position (a) and then in (b).
𝑨𝒚
(a)
𝑨𝒙
𝟒𝑳
𝑨𝒙 = 𝟑
𝟒𝑳
𝑫𝒙 =
𝟑
𝑨𝒚 = 𝑳
𝑫𝒙
Lets assume BD goes slack or loose.
So AC is in Tension So our assumption is valid
𝑨𝒚
(b)
𝑨𝒙
𝟒𝑳
𝑨𝒙 = 𝟑
𝟒𝑳
𝑫𝒙 =
𝟑 𝑫𝒙
𝑨𝒚 = 𝑳
Lets assume BD goes slack or loose.
So AC is in Tension So our assumption is valid
Method of Sections:
Quick method to find the forces in selected members of truss.
Take a cutting section to cut truss into two parts, where we have to find the internal forces on member.
Then use the static equation of equilibrium for the FBD of right/left section part of the truss.
We should select an
Static equation of equilibrium appropriate moment of
∑ 𝐹 = 0, ∑ 𝐹 = 0, ∑ 𝑀 =0 centre through which line of
Moment of centre may or may not lie on the FBD of truss. action of all the cut members
OR ∑ 𝐹 = 0 𝑜𝑟 ∑ 𝐹 = 0, ∑𝑀 = 0, ∑𝑀 =0 excluding one are passing and
answer the required
OR ∑𝑀 = 0, ∑𝑀 = 0, , ∑𝑀 =0 unknown.
𝑩𝑪
FBD of left cut
section of truss 𝑩𝑬
E
𝑨𝑬
In general, we should not cut more than three
members because we have three equations of
equilibrium to find three unknowns. Use 𝑩𝑪
But in exceptional cases we can do this, especially Use 𝑨𝑬
where many members are collinear or concurrent.
Use to find 𝑩𝑬
For the truss loaded as shown in Fig., find the force in members CE and CF by method of sections only.
We take FBD of the right part of truss, section along s-s,
For the pin jointed truss loaded as shown in Fig. . Find
forces in member EC, ED, DF
Note: not all three centre point of moment are in same line.
Find the force in members DJ of
Howe roof truss.
Take first section 1 then find
unknown CJ and CD.
Then take another section find
unknown DJ
𝜶 4 𝜷
𝟐
𝜶=𝐭𝐚𝐧 𝟏 𝟔 = 𝟐𝟔. 𝟓𝟔𝝄 , 𝜷=𝐭𝐚𝐧 𝟏𝟒 = 𝟒𝟓𝝄
𝟏𝟐 𝟒
𝟐𝟏 − 𝟏𝟔
𝑩𝑳 = 𝟏𝟔 + = 𝟐𝟏 𝒇𝒕
𝟐
𝟓
𝜽 = tan 𝟏
𝟏𝟐
𝜷 = tan 𝟏 𝟏𝟐 = 𝟐𝟗. 𝟕𝝄
𝑷𝑪
=
𝟐𝟒
𝑷𝑪 = 𝟑𝟖. 𝟒 𝒇𝒕
𝟐𝟏 𝟏𝟔 𝟐𝟔 𝟏𝟔
Find forces in members AB, CD and EF of the truss shown in
the figure.
Note that there are no joints where members AD, AE, BF and
CF overlap
Consider the section cut by ≈ as shown in figure.
AB = 2/3kN (Compressive)
CD = 0 and
EF = 8/3 kN (Compressive)
Find force in member DE, QE, QJ and
KJ? Section 1-1
. . 𝑴𝑲 = 𝟎
tan 𝛼= , tan 𝜃 =
Zero force member 𝑭𝑫𝑬 = 51685.2 N T
QJ = 0
𝛼 𝑴𝑬 = 𝟎
𝑭𝑲𝑱 = 67500 N T
𝜃
𝑭𝒚 = 𝟎
𝑭𝑸𝑬 = 20581 N T
Find the forces in member in GH, GJ, and CG of the truss
Find the forces in members BF, CF and DF of the arched
trusses shown in the figure.
Truss members are two force members.
