CET301: STRUCTURAL ANALYSIS-I

Introduction, module 1

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 Text books
1. Wang C. K, Intermediate Structural Analysis, McGraw Hill

2.DevdasMenon, structural Analysis, Narosa Publications

3.Hibbeler, Structural Analysis , Pearson Education

4.Vazirani and Ratwani, Analysis of structures, Khanna Publishers

5.R.Vaidyanathan and P.Perumal, Comprohensive Structural Analysis, Vol

1, Laxmi Publications

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 Module 1

• Statically determinate trusses: Method of joints and method of sections

(simple illustrative numerical problems only) –
• Deflection of statically determinate structures: Introduction and simple
illustrative examples of simple beams and cantilever beams only on: a)
Method of successive integrations, b) Moment area method and c)
Castigliano’s theorem Part I

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Example 1:

Q. Determine the forces in each

member of the truss using method
of joints?

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 Structural Analysis???

It is the application of solid mechanics to predict the response (forces and

displacements) of a given structure (existing or proposed) subjected to
specified loads

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What is a structure?

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 structure

A structure is an elastic system which upon loading will deform and can
come back to its original position completely on removal of loading because
of internal resistance
A structure will have a non-linear deformation.

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Structural Analysis I MANU P RAJ Govt. College of
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 Classification of structures
1. Based on the predominance of dimensions
a)skeletal structures
b)surface structures
c)solid structures
Note: the analysis of surface and solid structures and require higher level
mathematics

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2. Based on type of joints
a) rigid jointed structures
1. rigid jointed plane structures
2. rigid jointed space structures
b) pin jointed structures
1. Pin Jointed plane structure
2. Pin jointed space structure

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 Rigid jointed structures
• At a rigid joint, joint rotate as a whole ,each member connected to rigid
joint will have same rotation or slope.
• At a rigid joint relative slope is zero ( included angle between different
members remains constant)
• At rigid joint absolute slope is non zero
• At a rigid joint the design forces may be
1. axial force
2. shear force
3. bending moment

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 Pin jointed structures
These are the skeletal structures in which different members are connected
to frictionless hinges

These hinges cannot resist moment. So moments are not transferred to the
members

Members of pin-jointed structures designed for axial force only. No bending

moment, no shear force

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 Equilibrium equations for structures
• space structure( no. of equilibrium equations=6)
rigid jointed or pin jointed

σ𝐹𝑥 = σ𝐹𝑦 = σ𝐹𝑧 = σ𝑀𝑥 = σ𝑀𝑦 = σ𝑀𝑧 = 0

• plane structure (no. of equilibrium equations =3)

rigid jointed or pin jointed

σ 𝐹 𝑥 = σ 𝐹 𝑦=σ 𝑀 𝑧 = 0

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 Equilibrium equations of joints
1. Pin joint of a plane frame( 2)
σ𝐹𝑥 = σ𝐹𝑦 = 0
Note: individual pin joint cannot resist moment, but structure as a whole
will resist moment.

2. Pin joint of a space frame (3)

σ 𝐹 𝑥 = σ 𝐹 𝑦 = σ 𝐹 z =0

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3. Rigid joint of a plane frame ( 3)
σ𝐹x = σ𝐹𝑦 = σ𝑀𝑧 = 0

Note: pin joint cannot resist moment, where as rigid joint can resist
moment

4. Rigid joint of a space frame (6)

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 Compatibility equations
• Equilibrium equations are related to forces , where as compatibility
equations are related to boundary conditions or deformations
• Number of compatibility equations at a support are equal to number of
reactions

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 Statically determinate structure
• If the number of unknown forces ( external & internal ) of a given
structure is equal to the equilibrium equations.
Examples: simply supported beam, cantilever beam , three hinged arch etc.

Note: to analyse the statically determinate structure equilibrium equations

are sufficient

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 Statically indeterminate structure
• If the number of unknown forces ( external & internal ) of a given
structure is more than the equilibrium equations
Examples: propped cantilever beam, fixed beam, two hinged arch etc.

Note: to analyse the statically indeterminate structure equilibrium equations

are not sufficient. We have to consider compatibility conditions also.

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 Degree of static indeterminacy( Ds)
• Static indeterminacy is related to number of unknown forces

Ds= Dse + Dsi - number of force releases in the structure

Where,
Dse= external static indeterminacy
Dsi= internal static indeterminacy

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Notes
1. Dse is related reactions at support
2. Dsi is related internal configuration of the member
3. Forces are released due to moment hinge or vertical shear hinge or
horizontal shear hinge

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 External static indeterminacy (Dse)

Dse = r- equilibrium equations of a structure

r= total number of reaction components of the given structure

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Ds = 0, statically determinate structure
Ds> 0, statically indeterminate structure
Ds<0, unstable or mechanism

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 External static indeterminacy( Dse)
Plane frame( rigid jointed or pin jointed)

Dse= r- 3
Space frame (rigid jointed or pin jointed)

Dse= r- 6
Ex: ???

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 Internal static indeterminacy ( Dsi)
• Rigid jointed plane frame

Dsi= 3 C

Where,
C= number of cuts required to convert a closed structure to an
open structure
= number of closed boxes
3 represent number of force released by making a cut in a rigid
jointed plane frame
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 Rigid jointed plane frame

Total static indeterminacy = Ds= Dse+ Dsi= ( r-3) + 3C

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• Rigid jointed space frame

Dsi= 6 C
Where,
C= number of cuts required to convert a closed structure to an open
structure
= number of closed boxes
6 represent number of force released by making a cut in a rigid jointed space
frame
Ds= Dse+ Dsi = ( r-6) + 6C

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 Rigid jointed space frame

Total static indeterminacy = Ds= Dse+ Dsi= ( r-6)+ 6c

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 Force releases
1. moment hinge
number of force released due to a moment hinge is = n-1
where,
‘ n’ is the number of members passing through the stomach of the
moment hinge
2. Horizontal shear hinge
number of force released due to a horizontal hinge is = 1
3. Vertical shear hinge
number of force released due to a Vertical shear hinge is = 1

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 Internal static indeterminacy
Pin jointed plane frame

Dsi = m - ( 2j-3)

where,
m= number of members
j= number of joints

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When Dsi= 0, m = 2j – 3 and the structure is statically determinate internally
and is stable. Such a truss is called a perfect truss

When Dsi< 0, the number of members provided is less than that required for
stability. Such a truss is called unstable or deficient truss

When Dsi> 0, number of members is more than the number required. Such
a truss is called statically indeterminate or redundant truss

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 Pin jointed plane frame

Total static indeterminacy=Ds= Dse+ Dsi

Ds= (r-3)+ ( m-2j+3)= m+r-2j

m= number of members
j= number of joints
r= number of reaction components

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Static indeterminacy of a pin jointed frame= m+r-2j

ex:??

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 Kinematic indeterminacy( Dk)

• Kinematic indeterminacy (degrees of freedom)is the number of unknown

joint displacements in the structure.
• The degree of kinematic indeterminacy (Dk )is defined as the total number
of degrees of freedom ( independent displacement coordinates ) at the
various joints in the skeletal structure.
• It is related to displacement of a structure
• joints: where members meet; supports; free ends; where moment of inertia
changes etc
• joints undergo translations or rotations
• some of the joint displacements are known from the support conditions.
Others are unknown, to be determined
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Types of support Degree of freedom

Free end 3( 2 translations, and 1 rotation)

Roller support 2( 1 translation & 1 rotation)

Hinged support 1( 1 rotation)

Fixed support 0( completely restrained, 0 rotation, 0 translation)

Pin joint of a plane truss 2( 2 translation, moments are not considered in

the design of truss, hence rotation is not
considered )

Pin joint of space frame 3( 3 translation)

Rigid joint of plane frame 3( 2 translation, 1 rotation)

Rigid joint of a space frame 6 (3 translation, 3 rotation)

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Dk= 6 ( rotation at B& C, one sway, 3 axial deformations)

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Dk= 3 ( neglecting axial deformations)

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In rigid joint structural analysis the displacement due to flexure only
considered . The displacement due to shear force and axial forces are
neglected

In the plane frame shown in Figure the joints B and C have 3 degrees of
freedom as shown in the figure. However if axial deformations of the
members are neglected then u2 and u5 can be neglected and u1=u4. Hence,
we have 3 independent joint displacements. rotations at B and C and one
translation.( u3, u6, u1)

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Truss analysis

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Truss : A truss is a structure composed of slender members joined together
at their end points

Types of trusses? homework

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 Analysis of determinate truss
Assumptions
1. Members are subjected to axial force only ( axial compression &
tension).shear force and bending moment negligible
2. Self weight of the members negligible. Compared to the external loads
self weight of the member is negligible.
3. All connections are smooth , frictionless and hinges
4. Members ( prismatic or non-prismatic) are linear
5. Loads and reactions act at the joints only( to eliminate bending moment)
6. All the loads are applied in the plane of the truss

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 Sign conventions
• If the force is towards the joint or pushing the joint, is compression

bolt

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• If the force is pulling the joint or moving away from the joint, is tension

bolt

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• Note:
If the bolt is subjected to compression , members are also subjected to
compression

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 Methods of truss analysis

1. The method 2. The method

of joints of sections

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 The method of joints
• Equilibrium of joint is considered in method of joints
• For pin jointed plane frame the number of equilibrium equations are two
.σ𝐹𝑥 = σ𝐹𝑦 = 0
• Using the two equilibrium equations we can find two unknown forces at a
time
• At the pin joint, if the number of unknowns are more than two this
method is not suitable
• To find the forces in a internal member, we have to find the forces in the
prior member , hence this method is a time taking method
• To find force in all members method of joints is preferred

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 Find Find the reactions at the supports

Draw the FBD of all the joints. Consider one

of the joints where the number of unknown
Consider
Steps in forces are not more than two (number of
equilibrium equations are two)

method of
joints Apply Apply the equilibrium equation and
work out the unknown

Move Move to another joint in similar way

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Example 1:

Q. Determine the forces in each

member of the truss using method
of joints?

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 solution

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Example 2: Determine the force in each member of the truss
shown below

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Determine the reaction forces on the entire truss

400 N C
RCy
B
RCx

A D

RAy 600 N

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RCx = 600 N ←
RCy = 200 N ↓

RAy = 600 N ↑

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Draw a FBD of each joint, assume all members in tension, write and solve the joint
equilibrium equations
FAB
400 N
A B FBC
FAD
600 N FAB FBD

FBD of joint A FBD of joint B

FBD FCD
200 N
FBC
FAD D 600 N C 600 N

FCD
FBD of joint D FBD of joint C

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Solution 2:

• FAB = 750 N (C)

• FAD = 450 N (T)
• FBC = 600 N (C)
• FBD = 250 N (T)
• FCD = 200 N (C)

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Example 3:

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Solution 3:

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Example 4:

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Solution 4:
• FGD = 32.5 kN (C)
• FCD = 30 kN (T)
• FCG = 0
• FFG = 32.5 kN
• FCF = 19.53 kN (C)
• FBC = 45.0 kN (T)
• FBE = 24.0 kN (C)
• FAB = 60.0 kN (T)
• FEF = 48.8 kN (C)
• FBF = 6.25 kN (T)
• FAE = 37.5 kN (T)
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 The method of sections
• In this method we can directly determine the force in any member without
proceeding to that member by a joint by joint analysis
• Equilibrium of a structure is considered in method of sections
• For pin jointed plane frame number of equilibrium equations are

• Using the 3 equilibrium equations we can find 3 unknowns at a time

• To find the force in the selected member, pass a section( imaginary cut)
through selected member and the section should divides the entire
structure to two separate parts

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• The cutting line should be chosen in such a way that it cuts a maximum
of three members in which forces are unknown

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 Principle
• If a body is in equilibrium, then any part of the body is also in
equilibrium.
• Forces in few member can be directly found out quickly without
solving each joint of the truss sequentially
• Method of Sections and Method of Joints can be conveniently
combined
• A section need not be straight
• More than one section can be used to solve a given problem

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 Steps in method of sections
Subcase (a): Applying σ 𝑀 = 0 concept
FED ????

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1. Calculate reactions at support

2. Pass a section through the chosen member and two other members

so that the other members pass through a common joint( other

members cut cannot have moment about the above joint)

3. Consider any one side of the section

4. Apply σ 𝑀 = 0 at common joint B

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 Steps in method of sections
Subcase (b): Applying σ 𝐹𝑥 = 0 concept
FCF ????

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1. Calculate reactions

2. Pass a section through the chosen member and other vertical

members so that vertical members cut cannot have horizontal
component

3. Consider one side of the section

4. Apply σ 𝐹𝑥 = 0

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 Steps in method of sections
Subcase (c): Applying σ 𝐹𝑦 = 0 concept

FBE ????

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1. Calculate the reactions

2. Pass a section through the chosen member and other horizontal

member so that the horizontal members cut cannot have vertical
component

3. Consider one side of the section

4. Apply σ 𝐹𝑦 = 0

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 Example 1
Q. Find out the internal forces in members FH, GH, and GI

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1.Find out the reactions
RA = 12.5 kN
RL = 7.5 kN
2. Pass a section through members FH, GH, and GI and take the right-
hand section as a free body

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3. Apply the conditions for static equilibrium to determine the desired
member forces.

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Solution:
FGI = 13.13 kN (T)
FGH = 13.82 kN (C)
FGI = 1.371 kN ( C)

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Example 2
Determine the force in the member CF of the given truss. Indicate whether the
member is in tension or compression. Assume each member is pin connected.

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Notes: Zero Force Members in truss
1. If only two non-collinear members form a truss joint and no external
load or support reaction is applied to the joint, the two members must
be zero force members

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2. If three members form a truss joint for which two of the members
are collinear, the third member is a zero-force member provided no
external force or support reaction is applied to the joint

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3. When two pairs of collinear members are joined as shown in figure,
the forces in each pair must be equal and opposite

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 Deflections of beams
• Deflections of structures can come from loads, temperature, fabrication
errors or settlement
• In designs, deflections must be limited in order to prevent cracking of
attached brittle materials
• A structure must not vibrate or deflect severely for the comfort of
occupants
• Deflections at specified points must be determined if one is to analyze
statically indeterminate structures

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Slope and deflection of beams by
• method of successive integration
• Macaulay’s method
• Moment area method
• Conjugate beam method
• Castigliano’s theorem

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• R= radius of curvature
1
• = 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒
𝑅
• M= Bending moment
• E= Modulus of Elasticity
• I= Moment of Inertia
• y=deflection

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Sign convention for BM, SF and load

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Sign convention for curvature

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 Method of successive integrations

Deflection of beam at any cross section is obtained if the beam differential

equation is integrated twice.
ⅆ2 𝑦
E𝐼 2 = M …………….bending moment
ⅆ𝑥

ⅆ𝒚 ⅆ𝑦
EI. = ‫ 𝑀 ׬‬ⅆ𝑥 + 𝑪𝟏, from which slope ( ) can be calculated
ⅆ𝒙 ⅆ𝑥

EI. y= ධ ‫ 𝑀(׬‬ⅆ𝑥) + 𝑪𝟏𝒙 + C2 , from which deflection (y) can be calculated

The constants of integration are found by applying the end conditions

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Example 1

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Example2

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 The Moment-Area Method
• The moment-area method, developed by Otto Mohr in 1868
• Semi graphical method
• Beams with sudden change in EI can be analyzed
• For deflections of beams especially cantilever beams For deflections of
beams, especially cantilever beams
• Suitable when slopes and deflections at some points are required not the
complete equation of the deflection curve

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Moment area theorem 1
Angle between tangent drawn at any two points on elastic curve is equal to
𝑴
area of diagram between the same points
𝑬𝐈

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Moment area theorem 2
The vertical ordinate taken from a point on elastic curve to the tangent
𝑴
drawn from other point on elastic curve is equal to moment of area of
𝑬𝐈
diagram between the two point about which the ordinate is taken

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