Structure analysis ii ce331 pdf

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STRUCTURE ANALYSIS II CE331

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Chapter No 01
Introduction to Determinate & indeterminate Structures
Determinate structure: A structure is termed as statically determinate, if it can be analysed
from principles of statics alone, i.e. . Solving a structure
means determining the unknown internal forces, reactions and displacements of the structure.
When a structure can be solved by using the equations of static equilibrium alone, it is known
as determinate structure.
Example of determinate structures are: simply supported beams, cantilever beams, single and
double overhanging beams, three hinged arches, etc.
Indeterminate structure: A structure is termed as statically indeterminate, if it can not be
analysed from principles of statics alone, i.e. .
A structure can be termed as Indeterminate structure if it can not be solved by using the
equations of equilibrium alone. Examples of indeterminate structures are: fixed beams,
continuous beams, fixed arches, two hinged arches, portals, multistoried frames, etc.
Methods for Solving
To solve an indeterminate structure it is necessary to satisfy equilibrium, compatibility and
force-displacement requirements of the structure. The additional equations required to solve
indeterminate structure are obtained by the conditions of compatibility and/or force-
displacement relations. The number of additional equations required to solve an indeterminate
structure is known as degree of indeterminacy. Based on the types of unknown, a structure can
be termed as Statically indeterminate or kinematically indeterminate.
There are following methods to solve indeterminate structures;  Method of consistent
deformation,  Slope-deflection equations,  Moment distribution method,  Stiffness method,
 Flexibility method.
Difference between Determinate and Indeterminate Structures
S.
No.
Determinate Structures Indeterminate Structures
1
Equilibrium conditions are fully adequate to
analyze the structure.
Conditions of equilibrium are not adequate to
fully analyze the structure.
2
Bending moment or shear force at any
section is independent of the material
property of the structure.
Bending moment or shear force at any section
depends upon the material property.

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3
The bending moment or shear force at any
section is independent of the cross-section or
moment of inertia.
The bending moment or shear force at any
section depends upon the cross-section or
moment of inertia.
4
Temperature variations do not cause
stresses.
Temperature variations cause stresses.
5 No stresses are caused due to lack of fit. Stresses are caused due to lack of fit.
6
Extra conditions like compatibility of
displacements are not required to analyze
the structure.
Extra conditions like compatibility of
displacements are required to analyze the
structure along with the equilibrium
equations.
Introduction to force methods & Displacement Method
1 . Force method of analysis:  Primary unknowns are forces and moments,  Deformation
conditions are written depending on pre-selected statically indeterminate reactions.  The
unknown statically indeterminate reactions are evaluated solving these equations.  The
remaining reactions are obtained from the equilibrium equations. OR
Force (Flexibility) Method: • Convert the indeterminate structure to a determinate one by
removing some unknown forces / support reactions and replacing them with (assumed) known
/ unit forces. • Using superposition, calculate the force that would be required to achieve
compatibility with the original structure. • Unknowns to be solved for are usually redundant
forces. • Coefficients of the unknowns in equations to be solved are "flexibility" coefficients.
[A] = b, Where “A” is "flexibility".
2. Displacement method of analysis:  Primary unknowns are displacements.
 Equilibrium equations are written by expressing the unknown joint displacements in terms of
loads by using load-displacement relations.  Unknown joint displacements are calculated by
solving equilibrium equations.  In the next step, the unknown reactions are computed from
compatibility equations using force displacement relations.  This method follows essentially
the same steps for both statically determinate and indeterminate structures.
 Once the structural model is defined, the unknowns (joint rotations and translations) are
automatically chosen unlike the force method of analysis (hence, this method is preferred to
computer implementation).OR
Displacement (Stiffness) Method: • Express local (member) force-displacement
relationships in terms of unknown member displacements. • Using equilibrium of assembled

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members, find unknown displacements. • Unknowns are usually displacements
• Coefficients of the unknowns are "Stiffness" coefficients.
[k]d = f, Where “k” is "Stiffness”.
Chapter No 02
Derivation of Slope Deflection Equation
Procedure
1. For each beam in the frame write the slope deflection equations
2. For each joint in the frame write the moment equilibrium
  0M
3. IF 0 (Frame with sidesway)
(a) For each story in the frame write the equilibrium of shear forces
PVV DCAB 
(b) Relate shear forces to moments
e.g. in the absence of transverse loads,
h
MM
V BAAB
AB


4. Substitute 1 (and 3 if applicable) into 2
5. Solve for unknown rotations and translations
6. Substitute 5 into 1 to compute moments and use 3b to compute shear forces

MAB
MBA
 
AB
AB
BA
AB
AB FEM
LL
EI
M 










 






 322 
BA
AB
BA
AB
BA FEM
LL
EI
M 










 






 322 
D
CB
A
VDCVAB
P
h

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7. Now you have relationships between forces and moments from which you can extract
stiffness coefficients.
Slope-Deflection Method
In this method it is assumed that all deformations are due to bending only. Deformations due to
axial forces are neglected.
In the slope-deflection method, the relationship is established between moments at the ends
of the members and the corresponding rotations and displacements.
The slope-deflection method can be used to analyze statically determinate and indeterminate
beams and frames. In this method it is assumed that all deformations are due to bending only.
In other words deformations due to axial forces are neglected. In the force method of analysis
compatibility equations are written in terms of unknown reactions. It must be noted that all
the unknown reactions appear in each of the compatibility equations making it difficult to
solve resulting equations. The slope-deflection equations are not that lengthy in comparison.
The basic idea of the slope deflection method is to write the equilibrium equations for each
node in terms of the deflections and rotations. Solve for the generalized displacements. Using
moment-displacement relations, moments are then known. The structure is thus reduced to a
determinate structure. The slope-deflection method was originally developed by Heinrich
Manderla and Otto Mohr for computing secondary stresses in trusses. The method as used
today was presented by G.A.Maney in 1915 for analyzing rigid jointed structures.
Assumptions in the slope deflection method
This method is based on the following simplified assumptions:
 All the joints of the frame are rigid,
 Distortion, due to axial and shear stresses, being very small, are neglected.
Applications of Slope deflection method
1. Continuous Beams
2. Frames with out side sway
3. Frames with side sway
Sign Convention
(1) ROTATIONS:– Clockwise joint rotations are considered as (-ve).
(2) END MOMENTS:– clockwise end moments are considered as (+ve).
PROCEDURE
The procedure is as follows:
1. Determine the fixed end moments at the end of each span due to applied loads acting on
span by considering each span as fixed ended. Assign ± Signs w.r.t. above sign convention.
2. Express all end moments in terms of fixed end moments and the joint rotations by using
slope – deflection equations.
3. Establish simultaneous equations with the joint rotations as the unknowns by applying the
condition that sum of the end moments acting on the ends of the two members meeting at a
joint should be equal to zero.
4. Solve for unknown joint rotations.

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5. Substitute back the end rotations in slope – deflection equations and compute the end
moments.
6. Determine all reactions and draw S.F. and B.M. diagrams and also sketch the elastic curve
General Slope Deflection Equation
Equation for Fixed End Moment

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Example for Frame

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Example: Determine support moments using slope deflection method for the frame shown in
figure. Also draw bending moment diagram.

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Slope Deflection for frames: No sidesway
Example 1

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Fixed end moments
Known displacements
Slope deflection equations

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Slope Deflection for frames: Sidesway

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Example:
Fixed end moments
Known displacements
θA = θD = 0
Let horizontal movement (sidesway) = δ
Slope deflection equations

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Joint equilibrium conditions

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Direct Stiffness Method: Deformations due to axial forces are not neglected. The Slope-
deflection method was used for many years before the computer era. After the revolution
occurred in the field of computing direct stiffness method is preferred.
Example 1: Determine the slope at point C of the beam in Fig. a. E = 200 GPa I = 6(106) mm4.
Sol: M/EI Diagram. Fig. b.
Elastic Curve. Since the loading is applied symmetrically to the beam, the elastic curve is
symmetric, as shown in Fig. c. We are required to find θc. This can easily be done, realizing that
the tangent at D is horizontal, and therefore, by the construction, the angle θD/C between tan C
and tan D is equal to θc that is θc = θD/C.
Moment-Area Theorem. Using Theorem 1, θD/C is equal to the shaded area under the M/EI
diagram between points C and D. We have,
Thus,

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Example 2: Determine the slope and deflection at point B of the steel beam
shown in Fig. a. The reactions have been computed. E = 29(103) ksi, I = 800 in4.

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SOLUTION
The negative signs indicate the slope of the beam is measured clockwise and the displacement
is downward, Fig. d.
Example 3: Determine the displacement of point B of the steel beam shown in Fig. a. Take E =
200 GPa, I = 500(106) mm4.
Sol:

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Analysis of frames (No sidesway)

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Analysis of frames (sidesway)

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Chapter No 03
MOMENT DISTRIBUTION METHOD
Concept: This method of analyzing beams and frames was developed by Hardy Cross in 1930.
Moment distribution method is basically a displacement method of analysis. But this method
side steps the calculation of the displacement and instead makes it possible to apply a series of
converging corrections that allow direct calculation of the end moments. This method of
consists of solving slope deflection equations by successive approximation that may be
carried out to any desired degree of accuracy. Essentially, the method begins by assuming
each joint of a structure is fixed. Then by unlocking and locking each joint in succession, the
internal moments at the joints are distributed and balanced until the joints have rotated to
their final or nearly final positions. This method of analysis is both repetitive and easy to
apply. Before explaining the moment distribution method certain definitions and concepts
must be understood.
Sign convention: In the moment distribution table clockwise moments will be treated +ve
and anti clockwise moments will be treated –ve. But for drawing BMD moments causing
concavity upwards (sagging) will be treated +ve and moments causing convexity upwards
(hogging) will be treated –ve.
Any moment acting clockwise is considered to be positive. This differs from the usual engineer’s
sign convention, which employs a Cartesian coordinate system with positive x-axis to the right
and positive y-axis up, resulting in positive moment about the z-axis being counterclockwise.
Fig.1: Moment distribution and common sign convention comparison
Fixed end moments: The moments at the fixed joints of loaded member are called fixed
end moment.
Factors Method
Member stiffness factor:
a) Consider a beam fixed at one end and hinged at other as shown in figure subjected to a
clockwise couple M at end B. The deflected shape is shown by dotted line.
BM at any section xx at a distance x from ‘B’ is given by

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Joint stiffness factor:
If several members are connected to a joint, then by the principle of superposition the total

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stiffness factor at the joint is the sum of the member stiffness factors at the joint i.e.,
KT = K
E.g. For joint ‘0’, KT = K0A + KOB + KOC + KOD.
Distribution factors: If a moment ‘M’ is applied to a rigid joint ‘o’, as shown in figure, the
connecting members will each supply a portion of the resisting moment necessary to satisfy
moment equilibrium at the joint. Distribution factor is that fraction which when multiplied
with applied moment ‘M’ gives resisting moment supplied by the members. To obtain its
value imagine the joint is rigid joint connected to different members. If applied moment M
cause the joint to rotate an amount ‘ ’, Then each member rotates by same amount.
From equilibrium requirement
M = M1 + M2 + M3 + ……
Member relative stiffness factor: In majority of the cases continuous beams and frames will
be made from the same material so that their modulus of electricity E will be same for all
members. It will be easier to determine member stiffness factor by removing term 4E & 3E
from equation (4) and (5) then will be called as relative stiffness factor.

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Unbalanced moments are carried over to the other end of the member when the joint is released.
Added to that, the ratio of the carried-over moment at the other end to the fixed-end moment of
the initial end is the carryover factor.
Lastly, For prismatic members, the carryover moment in each span has the same sign as the
distribution end moment, but is one-half as large.
+ve BM of
𝑀
2
at A indicates clockwise moment of
𝑀
2
at A. In other words the moment ‘M’ at
the pin induces a moment of
𝑀
2
at the fixed end. The carry over factor represents the fraction
of M that is carried over from hinge to fixed end. Hence the carry over factor for the case of
far end fixed is +
1
2
. The plus sign indicates both moments are in the same direction.OR
Carry Over Factor: In MDM, we have to analyze the effects of applying imaginary moments at a
specified point
The beam in Figure, when it receives a moment M at A, will develop at B moment of M/2.
This M/2 is called the carry over moment. If the far end B were hinged, the Carry Over will be
zero.

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Distribution factors: A moment which tends to rotate without translation a joint to which
several members are connected will be divided amongst the connected members in proportion
to their stiffnesses.
i. The rotation of each member at A is obviously 
ii. The moments MAB, MAC, MAD (assuming up to M0 will be in ratio k1;k2:k3.
Mi = Ki , M =  Ki , DFi =
𝑀𝑖
𝑀
=
Ki 
 Ki
, DF =
𝐾
 𝐾
.
Problem 1:
E = Constant, K =
∆𝐄𝐈
𝐋
Solve the beam, draw S.F.D & B.M.D.
As we know, KBA = ∆E
300
15
= ∆E (20) in2/ft. KBC = ∆E
600
20
= ∆E (30) in2/ft.
As, Df = Distribution Factor =
𝐾
∈𝐾
DF(BA) =
∆𝐸 (20)
∆𝐸 (20)+ ∆𝐸 (30)
=
∆𝐸 (20)
∆𝐸 (50)
= DF(BA) = 0.4.
DF(Bc) =
∆𝐸 (30)
∆𝐸 (20)+ ∆𝐸 (30)
=
∆𝐸 (30)
∆𝐸 (50)
= DF(Bc) = 0.6.
DF(AB) =
∆𝐸 (20)
∝ + ∆𝐸 (20)
=
𝑀𝑒𝑚𝑏𝑒𝑟
𝑆𝑢𝑝𝑝𝑜𝑟𝑡 + 𝑚𝑒𝑚𝑏𝑒𝑟
∆𝐸 (20)
∝
= DF(AB) = 0

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(it is zero when both fix ends) now, to solve we consider beam separate.
FEM (BC) = – 8000 Lb – Ft, FEM(CB) = 8000 Lb – Ft.
Example: 2 A continuous beam ABC is shown in figure below. Analyse the beam to for its end
moment and draw the shear force and bending moment diagram. Assume EI is constant.
Sol: i-Distribution Factor (DF)

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ii-Fixed End Moment
iii-Distribution Table

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Example 3 Determine the internal moment at each support of the beam shown in figure
below. The moment of inertial of each span is indicated.
Sol: The moment does not get distributed in the overhanging span AB, so the distributed factor
(DF)AB = 0 and (DF)BA = 1.0.
i-Distribution Factor (DF)

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Chapter No 04
INTRODUCTION TO ROTATIONS & DEFLECTIONS
Deflection
When a structural is loaded may it be Beam or Slab, due the effect of loads acting upon it bends
from its initial position that is before the load was applied. It means the beam is deflected from
its original position it is called as Deflection, and the slope of that deflection is the angle
between the initial position and the deflected position. Consider you are holding a somewhat
flexible stick with your fingers, and tie a little weight on the free end with the help of thread.
that means it has become a Cantilever with point load at its end. you will notice that the flexible
stick will bend down due to the weight to 2.5X to 3X of forces. Now because of this higher
force, beam undergoes in its plastic state and rotates beyond the elastic capacity. Then we
observe this rotation and try to keep it below 3%. If the rotation exceeds then we increase the
capacity of the beam. In general no one talks about end rotation of beam for elastic analysis
and the reason is the beam is fixed on both ends and at elastic level of forces it hardly
experience any rotation. It is earthquake that causes large moments and we evaluate rotation
of beams. So next time of anyone is talking about rotation think of it as LFRS fuses or
system. We do this purposely so that the beam can absorb large amount of forces after
undergoing this cycle of loading and unloading. If we do not provide these fuses and keep
building elastic then it will not dissipate energy so easily.
Deflection of a beam: The deflection at any point on the axis of the beam is the distance
between its position before and after loading. Deflection is vertical displacement of a cross
section of beam from initial position.
Slope of a beam: slope at any section in a deflected beam is defined as the angle in radians
which the tangent at the section makes with the original axis of the beam. Slope is angle
between tangent drawn at a cross section (point) of deflected beam and initial position.
which is hanging at the end. You will also notice that the bending from its initial position is
maximum at the free end and minimum at the point you are holding it. This is Deflection.
Rotation
This term is frequently used in nonlinear analysis. When a beam is subjected to extreme loads
like in case of earthquakes then instead of designing the beam for actual earthquake force and
keep it elastic, we design it for an underestimated force and let the beam crack and rotate
when it reaches its ultimate capacity. For example: You are designing a beam for reduced
earthquake forces say for "X" bending moment. When earthquake occurs and building shakes,
the beam is generally subjected.
Unit Load Methods for Beams
In this method to find rotations and deflections in a determinate beams, first moment M due to
loadings is found. Then a unit load is applied in the direction of required displacement and
resulting moment ‘m’ is found. The required displacement is 
𝑀𝑚𝑑𝑥
𝐸𝐼
, where integration is over
the entire structure. The following example illustrate the method.

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Example: Determine the rotation and deflection at the free end of a cantilever beam of span L
subject to uniformly distributed load over its entire span.
Solution: Figure 2.5(a) shows the beam under consideration. Now due to given load
M =
𝑊𝑥2
2
To find rotation at free end a unit moment is applied (Ref Fig 2.5(b) ). Due to this moment.

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Deflection of beam by unit load Method.

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Introduction to Rotations & Deflections by Conjugate Beam Method
A conjugate beam of a beam is the one which when loaded with
𝑀
𝐸𝐼
diagram of the beam
satisfies the following two conditions.

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 Rotation of the beam with respect to the axis of beam = shear in conjugate beam.
 Deflection of the beam = Moment in conjugate beam.
Hence, from moment area theorem, the conjugate beams for various beams can be obtained.
Table 1 shows conjugate beams for commonly used beams.

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Chapter No 06
Method of Consistent Deformation
Example 1: Analyse propped cantilever beam using Consistent Deformation method.

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Example 2: Determine (approximately) the moment at the joints E and C caused by members
EF and CD of the building bent in Fig. a.

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Sol: For an approximate analysis the frame is modeled as shown in Fig. b. Note that the
cantilevered spans supporting the center portion of the girder have a length of 0.1 L = 0.1(20) =
2 ft. Equilibrium requires the end reactions for the center portion of the girder to be 6400 lb,
Fig. c. The cantilevered spans are then subjected to a reaction moment of
M = 1600(1) + 6400(2) = 14 400 lb ft = 14.4 k ft. Ans.
This approximate moment, with opposite direction, acts on the joints at E and C, Fig. a. Using
the results, the approximate moment diagram for one of the girders is shown in Fig. d.
Castigiano’s theorem for trusses beams and frames
Castigliano’s second theroem or the method of least work applies to only structure that have
constant temperature, unyielding supports & linear elastic material response. The theroem
states that the disp of a point is equal to the first partial derivative of the strain energy in the
structure wrt a force acting at the point & in the direction of the displacement. The slope at a
point in a structure is equal to the first partial derivative of the strain energy in the structure
wrt a couple moment acting at the point & in the direction of the rotation.
Castigiano’s.theorem for trusses

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Example: Determine the vertical disp of joint C of the truss shown in Fig 9.24(a)
The cross-sectional area of each member is A = 400 mm2, Take E = 200Gpa.
Fig. 9.24
Solution:
A vertical force P is applied to the truss at joint C, Fig 9.24(b). The reactions at the truss
supports at A & B are determined & the results shown in Fig 9.24(b), Using method of joints,
the N forces in each member are determined, Fig 9.24(c), These results along with partial
derivatives are listed in the table, Since P does not actually exist as a real load on the truss, we
required P = o in the table above, Applying eqn 9.27.
Castigiano’s.theorem for beams and frames:

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Rather than squaring the expression for internal moment M, integrating, and then taking the
partial derivative, it is generally easier to differentiate prior to integration. Provided E and I are
constant, we have
where
∆ = external displacement of the point caused by the real loads acting on the beam or frame.
P = external force applied to the beam or frame in the direction of
M = internal moment in the beam or frame, expressed as a function of x and caused by both
the force P and the real loads on the beam.
E = modulus of elasticity of beam material.
I = moment of inertia of cross-sectional area computed about the neutral axis.
If the slope θ at a point is to be determined, we must find the partial derivative of the internal
moment M with respect to an external couple moment M’ acting at the point, i.e.,
The above equations are similar to those used for the method of virtual work, Eqs. 9–22 and 9–
23, except 𝜕M / 𝜕P and 𝜕𝑀/𝜕𝑀′replace m and mθ respectively. As in the case for trusses,
slightly more calculation
is generally required to determine the partial derivatives and apply Castigliano’s theorem rather
than use the method of virtual work. Also, recall that this theorem applies only to material
having a linear elastic response. If a more complete accountability of strain energy in the
structure is desired, the strain energy due to shear, axial force, and torsion must be included.
The derivations for shear and torsion follow the same development as Eqs. 9–25 and 9–26. The
strain energies and their derivatives are, respectively,

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These effects, however, will not be included in the analysis of the problems in this text since
beam and frame deflections are caused mainly by bending strain energy. Larger frames, or
those with unusual geometry, can be analyzed by computer, where these effects can readily be
incorporated into the analysis.
Example 1: Determine the displacement of point B of the beam shown in Fig. a. Take E = 200
GPa, I = 500(106) mm4.
Sol: External Force P. A vertical force P is placed on the beam at B as shown in Fig. b.

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Chapter No 06
INFLUENCE LINES
An influence line is a diagram whose ordinates, which are plotted as a function of distance
along the span, give the value of an internal force, a reaction, or a displacement at a particular
point in a structure as a unit load move across the structure.  An influence line is a curve the
ordinate to which at any point equals the value of some particular function due to unit load
acting at that point.  An influence line represents the variation of either the reaction, shear,
moment, or deflection at a specific point in a member as a unit concentrated force moves over
the member.
Influence lines for statically determinate beams and paneled girders
Influence lines for statically determinate beams: Since beams (or girders) often form the main
load-carrying elements of a floor system or bridge deck, it is important to be able to construct
the influence lines for the reactions, shear, or moment at any specified point in a beam.
Loadings. Once the influence line for a function (reaction, shear, or moment) has been
constructed, it will then be possible to position the live loads on the beam which will produce
the maximum value of the function. Two types of loadings will now be considered.
Concentrated Force. Since the numerical values of a function for an influence line are
determined using a dimensionless unit load, then for any concentrated force F acting on the
beam at any position x, the value of the function can be found by multiplying the ordinate of
the influence line at the position x by the magnitude of F. For example, consider the
influence line for the reaction at A for the beam AB, Fig. 6–7. If the unit load is at  =
1
2
the
reaction at A is Ay =
1
2
as indicated from the influence line. Hence, if the force F lb is at this
same point, the reaction is Ay = (
1
2
)(F) lb. Of course, this same value can also be determined by
statics. Obviously, the maximum influence caused by F occurs when it is placed on the beam at
the same location as the peak of the influence line in this case at  = 0 where the reaction
would be is Ay = (1) (F) lb.
Influence Line for Beam Examples
Example 1: Determine the maximum positive shear that can be developed at point C in the
beam shown in Fig. a due to a concentrated moving load of 4000 lb and a uniform moving
load of 2000 lb/ft.
The influence line for the shear at C has been established in Example 1 and is shown in Fig. b.

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Concentrated Force. The maximum positive shear at C will occur when the 4000-lb force is
located at x = 2.5+ ft, since this is the positive peak of the influence line. The ordinate of this
peak is + 0.75; so that, VC = 0.75(4000 lb) = 3000 lb.
Uniform Load. The uniform moving load creates the maximum positive influence for Vc when
the load acts on the beam between x = 2.5+ ft, and x = 10 ft, since within this region the
influence line has a positive area. The magnitude of Vc due to this loading is
Total Maximum Shear at C. (VC)max = 3000 lb + 5625 lb = 8625 lb Ans.
Notice that once the positions of the loads have been established using the influence line,Fig. c,
this value (Vc)max of can also be determined using statics and the method of sections. Show that
this is the case.
Example 2: The frame structure shown in Fig. a is used to support a hoist for transferring
loads for storage at points underneath it. It is anticipated that the load on the dolly is 3 kN
and the beam CB has a mass of 24 kg/m.Assume the dolly has negligible size and can travel
the entire length of the beam. Also, assume A is a pin and B is a roller. Determine the
maximum vertical support reactions at A and B and the maximum moment in the beam at D.
Sol: Maximum Reaction at A. We first draw the influence line for Ay, Fig. b. Specifically, when a
unit load is at A the reaction at A is 1 as shown. The ordinate at C, is 1.33. Here the maximum
value for Ay occurs when the dolly is at C. Since the dead load (beam weight) must be placed
over the entire length of the beam, we have,
Maximum Reaction at B. The influence line (or beam) takes the shape shown in Fig. c. The
values at C and B are determined by statics. Here the dolly must be at B. Thus,

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Maximum Moment at D. The influence line has the shape shown in Fig. d.The values at C and D
are determined from statics. Here,
Example 3: For each beam in Fig. a through c, sketch the influence line for the vertical
reaction at A.
Sol: The support is replaced by a roller guide at A since it will resist, but not Ay. The force Ay is
then applied.
Again, a roller guide is placed at A and the force Ay is applied.
A double-roller guide must be used at A in this case, since this type of support will resist both a
moment MA at the fixed support and axial load Ax but will not resist Ay.

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Example 4: Draw the influence line for the shear in panel CD of the floor girder in Fig. a.
Tabulate Values. The unit load is placed at each floor beam location and the shear in panel CD
is calculated. A table of the results is shown in Fig. b. The details for the calculations when x = 0,
and x = 20 ft, are given in Figs. c and d, respectively. Notice how in each case the reactions of
the floor beams on the girder are calculated first, followed by a determination of the girder
support reaction at F (Gy is not needed), and finally, a segment of the girder is considered and
the internal panel shear VCD is calculated. As an exercise, verify the values for VCD when x = 10
ft, 30 ft, and 40 ft.
Influence Line. When the tabular values are plotted and the points connected with straight line
segments, the resulting influence line for VCD is as shown in Fig. e.

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Example 5: Draw the influence line for the moment at point F for the floor girder in Fig. a.
Sol: Tabulate Values. The unit load is placed at x = 0 and each panel point thereafter. The
corresponding values for MF are calculated and shown in the table, Fig. b. Details of the
calculations for x = 2 m, are shown in Fig. c.As in the previous example, it is first necessary to
determine the reactions of the floor beams on the girder, followed by a determination of the
girder support reaction Gy (Hy is not needed), and finally segment GF of the girder is considered
and the internal moment MF is calculated.As an exercise, determine the other values of MF
listed in Fig. b.
Influence Line. A plot of the tabular values yields the influence line for MF Fig. d.
Example 6: Draw the influence line for the force in member GB of the bridge truss shown in
Fig. a.

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Sol: Tabulate Values. Here each successive joint at the bottom cord is loaded with a unit load
and the force in member GB is calculated using the method of sections, Fig. b. For example,
placing the unit load at x = 6 m (joint B), the support reaction at E is calculated first, Fig. a, then
passing a section through HG, GB, BC and isolating the right segment, the force in GB is
determined, Fig. c. In the same manner, determine the other values listed in the table.
Influence Line. Plotting the tabular data and connecting the points yields the influence line for
member GB, Fig. d. Since the influence line extends over the entire span of the truss, member
GB is referred to as a primary member. This means GB is subjected to a force regardless of
where the bridge deck (roadway) is loaded, except, of course, at x = 8 m. The point of zero force
x = 8 m, is determined by similar triangles between x = 6 m, and x = 12 m, that is,
(0.354 + 0.707)/(12 – 6) = 0.354/x’, x’ = 2 m, so x = 6 + 2 = 8 m.
Example 1: Construct the influence line for the vertical reaction at A of the beam
in Fig. a.

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Sol: Tabulate Values. A unit load is placed on the beam at each selected point x and the value
of Ay is calculated by summing moments about B. For example, when x = 2.5 ft x = 5 ft and see
Figs. b and c, respectively. The results for Ay are entered in the table, Fig. d.A plot of these
values yields the influence line for the reaction at A, Fig. e.
Influence-Line Equation. When the unit load is placed a variable distance x from A, Fig. f, the
reaction Ay as a function of x can be determined from
Example 2: Construct the influence line for the vertical reaction at B of the beam
in Fig. a.
Sol: Tabulate Values. Using statics, verify that the values for the reaction By listed in the table,
Fig. b, are correctly computed for each position x of the unit load. A plot of the values yields the
influence line in Fig. c.

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Influence-Line Equation. Applying the moment equation about A, in Fig. d,
This is plotted in Fig. c
Example: 3 Construct the influence line for the shear at point C of the beam in Fig. a.
Sol: Tabulate Values. At each selected position x of the unit load, the method of sections is
used to calculate the value of Vc Note in particular that the unit load must be placed just to the
left (x = 2.5-) and just to the right (x = 2.5+) of point C since the shear is discontinuous at C, Figs.
b and c. A plot of the values in Fig. d yields the influence line for the shear at C, Fig. e.

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Example 4: Construct the influence line for the shear at point C of the beam in Fig. a.
Sol: Tabulate Values. Using statics and the method of sections, verify that the values of the
shear Vc at point C in Fig. b correspond to each position x of the unit load on the beam. A plot
of the values in Fig. b yields the influence line in Fig. c.

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Influence-Line Equations. From Fig. d, verify that
These equations are plotted in Fig. c.
Example 5: Construct the influence line for the moment at point C of the beam in Fig. a.
Sol: Tabulate Values. At each selected position of the unit load, the value of Mc is calculated
using the method of sections. For example, see Fig. b for x = 2.5 ft. A plot of the values in Fig. c
yields the influence line for the moment at C, Fig. d.

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Influence-Line Equations. The two line segments for the influence line can be determined using
εMc along with the method of sections shown in Fig. e. These equations when plotted yield the
influence line shown in Fig. d.
Example 6: Construct the influence line for the moment at point C of the beam in Fig. a.
Sol: Tabulate Values. Using statics and the method of sections, verify that the values of the
moment Mc at point C in Fig. b correspond to each position x of the unit load. A plot of the
values in Fig. b yields the influence line in Fig. c.
Influence-Line Equations. From Fig. d verify that
These equations are plotted in Fig. c.

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Chapter No 07
Elastic versus Plastic Analysis in Steel Design
The main difference between elastic and plastic design is the assumption of the behaviour of
structure. In elastic design, we assume that structure has failed if it has reached elastic limit but
in plastic design, we consider the structure to have failed when it reaches lower yield point
(Limit state method) or ultimate stress point (Ultimate stress method).
We assume that in elastic design, the structure will regain it’s original shape once the loads
have been removed. In plastic design, we have to consider the deformed shape of the structure
and analyse the secondary loads developed due to that deformation.
Elastic design is comparatively easy. Plastic design is much tougher. Elastic design is generally
opted for concrete, plastic design is generally opted for steel structures and high rise buildings.
Plastic design will give you better results as compared with elastic design. shape factor is a
mathematical term and it does not signify any physical process or phenomenon. OR
The main difference in the two is the point on stress strain curve of steel, that you wish to load
your steel to. In ultimate limit state you may choose to restrict your steel to be only loaded to
elastic limit, but in reality the steel can be loaded more for similar strains i.e. upto the plastic
limit just before rupture. Hence while designing if you make an elastic design, you are
underestimating the strength of steel wildly! The area beyond elastic limit was previously not
that well explored and hence only elastic design was being considered. With advancements in
research and technology, we have successfully explored this earlier locked area of plastic design
and have gained more confidence in executing the same. Plastic design helps design economic
sections and use the full potential of steel while ensuring safety. The shape facotr zp/ze roughly
defines the ratio of plastic to elastic capacity of section. OR
Ze is the Elastic Section Modulus. Zp is the Plastic Section Modulus.
Ze is used to determine the maximum bending moment up to the point where the extreme
fiber has yielded elastically. All other fibers will be below their elastic yield stress.
Zp is used to determine the maximum bending moment at the the point where all fibers have
yielded elastically and the entire section is behaving plastically. This is the point when the so
called plastic hinge has developed. Shape factor is just the ratio of these two parameters, it has
no physical meaning, it is just useful in computation. It is worth noting that Shape factor is very
high for shapes like rectangles (1.5) while for an I beam it is usually around 1.1 to 1.2. The
reason is that the rectangle has a lot more material below the extreme fiber. For the I beam,
once it has reached yielding there isn't much material left, just a little bit of flange and the tiny
strip of web. Of course if we create a new factor that compares the section Modulus (either
one) with the area of the section (The bending strength to area/weight ratio), you'll quickly be
reminded of why we use an I-Beam in the first place.
The Upperbound (Unsafe) Theorem
This can be stated as: If a bending moment diagram is found which satisfies the conditions of
equilibrium and mechanism (but not necessarily yield), then the corresponding load factor is
either greater than or equal to the true load factor at collapse. This is called the unsafe theorem

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because for an arbitrarily assumed mechanism the load factor is either exactly right (when the
yield criterion is met) or is wrong and is too large, leading a designer to think that the frame can
carry more load than is actually possible. Think of it like this: unless it’s exactly right, it’s
dangerous. Since a plastic analysis will generally meet the equilibrium and mechanism criteria,
by this theorem a plastic analysis is either right or dangerous. This is why plastic analyses are
not used as often in practice as one might suppose. The above theorem can be easily seen to
apply to the Illustrative Example. When we varied the position of the hinge we found a collapse
load factor that was either correct ( 1.6875 λ = = λC ) or was too big (λ > λC ).OR
The load factor or Load obtained from the work equation written for any arbitrarily assumed
mechanism is greater or equal to the true collapse load factor or the true collapse load.
The Lowerbound (Safe) Theorem
This can be stated as: If a bending moment diagram is found which satisfies the conditions of
equilibrium and yield (but not necessarily that of mechanism), then the corresponding load
factor is either less than or equal to the true load factor at collapse. This is a safe theorem
because the load factor will be less than (or at best equal to) the collapse load factor once
equilibrium and yield criteria are met leading the designer to think that the structure can carry
less than or equal to its actual capacity. Think of it like this: it’s either wrong and safe or right
and safe. Since an elastic analysis will always meet equilibrium and yield conditions, an elastic
analysis will always be safe. This is the main reason that it is elastic analysis that is used, in spite
of the significant extra capacity that plastic analysis offers. OR
A load factor or Load for which a distribution of bending moments can be found which satisfy
the equilibrium condition and the yield condition is less than or at most equal to the true value
of the collapse load factor or the collapse load
The Uniqueness Theorem
Linking the upper- and lower-bound theorems, we have: If a bending moment distribution can
be found which satisfies the three conditions of equilibrium, mechanism, and yield, then the
corresponding load factor is the true load factor at collapse. So to have identified the correct
load factor (and hence collapse mechanism) for a structure we need to meet all three of the
criteria: 1. Equilibrium; 2. Mechanism; 3. Yield.
The Uniqueness Theorem does not claim that any particular collapse mechanism is unique only
that the collapse load factor is unique. Although rare, it is possible for more than one collapse
mechanism to satisfy the Uniqueness Theorem, but they will have the same load factor. OR
A load factor or a load P for which the distribution of flexural moment is statically acceptable
and for which the value of the flexural moment is equal to the plastic moment in a number of
sections sufficient for the formation of mechanism is the collapse load factor or collapse load.
OR
Lower bound or Static theorem: A load factor (λs ) computed on the basis of an arbitrarily
assumed bending moment diagram which is in equilibrium with the applied loads and where
the fully plastic moment of resistance is nowhere exceeded will always be less than or at best
equal to the load factor at rigid plastic collapse, (λp). In other words, λp is the highest value of

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λs which can be found.
Upper bound or Kinematic theorem: A load factor (λk) computed on the basis of an arbitrarily
assumed mechanism will always be greater than, or at best equal to the load factor at rigid
plastic collapse (λp ). In other words, λp is the lowest value of λk which can be found.
Uniqueness theorem: If both the above criteria are satisfied, then the resulting load factor
corresponds to its value at rigid plastic collapse (λp).
The Equilibrium (or Statical) Method
In this method, free and reactant bending moment diagrams are drawn. These diagrams are
overlaid to identify the likely locations of plastic hinges. This method therefore satisfies the
equilibrium criterion first leaving the two remaining criterion to derived therefrom.
The Kinematic (or Mechanism) Method
In this method, a collapse mechanism is first postulated. Virtual work equations are then
written for this collapse state, allowing the calculations of the collapse bending moment
diagram. This method satisfies the mechanism condition first, leaving the remaining two criteria
to be derived therefrom. We will concentrate mainly on the Kinematic Method, but introduce
now the Incremental Method to illustrate the main concepts.

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Chapter No 08
ARCHES
Mainly three types of arches are used in practice: three-hinged, two-hinged and hingeless
arches.  In 19th century, three-hinged arches were commonly used for the long span
structures.  Then development in structural analysis, for long span structures starting from late
nineteenth century engineers adopted two-hinged and hingeless arches.  Two-hinged arch is
the statically indeterminate structure to degree one.  Usually, the horizontal reaction is
treated as the redundant and is evaluated by the method of least work.
Arch: An arch is a curved beam or structure in vertical plane and subjected to transverse loads
which act on the convex side of the curve and re-sights the external loads by virtue of thrust.
It is subjected to three restraining forces i.e.,  Thrust,  Shear force,  Bending Moment.
Linear arch
If an arch is to take loads, say W1, W2, and W3 (fig) and a Vector diagram and funicular polygon
are plotted as shown, the funicular polygon is known as the linear arch or
theoretical arch. The polar distance „ot‟ represents the horizontal thrust. The links AC, CD, DE,
and EB will be under compression and there will be no bending moment. If an arch of this shape
ACDEB is provided, there will be no bending moment. For a given set of vertical loads W1,
W2…..etc., we can have any number of linear arches depending on where we choose „O‟ or
how much horizontal thrust (or) we choose to introduce.
State Eddy’s theorem.
Eddy‟s theorem states that “The bending moment at any section of an arch is proportional
to the vertical intercept between the linear arch (or theoretical arch) and the centre line of
the actual arch.”
BMx = Ordinate O2O3 x scale factor.

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Three Hinged Arch
An arch hinged at both supports and having third hinge any where in the rib.Three hinged arch is
statically determinate. Third hinge at crown and the other two hinges at each abutments. Mostly
used for long span bridges.
Analysis of 3-hinged arches
It is the process of determining external reactions at the support and internal quantities
such as normal thrust, shear and bending moment at any section in the arch.
Procedure to find reactions at the supports
 Sketch the arch with the loads and reactions at the support.
 Apply equilibrium conditions namely Fx  0, Fy  0 and M 0
 Apply the condition that BM about the hinge at the crown is zero (Moment of all the
forces either to the left or to the right of the crown).
 Solve for unknown quantities.
Analysis of three Hinged Parabolic Arch: Bending moment at the crown hinge is zero. Arch
have two reaction at support (One horizontal & one vertical). Need for four equation to solve and
find the unknown reaction.
The parabolic arch is a curve which has an eccentricity that is equal to one and is obtained by
slicing a cone, with a plane parallel to one side. The point on the axis of symmetry that intersects
the parabola is called the vertex and is the point where the curvature is greatest. In the parabolic
arch, only normal forces and bending moments occurs, but no shear forces. That means when the
arch is subjected to a uniform distributed load and both ends are mounted in fixed bearings, loads
acting upon the inner portion of the arch are primarily compressive forces acting in the direction
of the normal force at every point of the arch.

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We can use three static equilibrium conditions and in addition to that the B.M. at the
crown hinge is equal to zero.
The Circular Arch: Arches with a circular form were commonly employed by the builders of ancient,
heavy masonry arches. Ancient Roman builders relied heavily on the rounded arch to span large, open
areas. Several rounded arches placed in-line (end-to-end) form an arcade, such as the Roman aqueduct.
Shear Force & Bending Moment Diagram

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Shear force diagram Bending moment diagram
Shear Force Diagram
In a statically determinate beam, the numerical value of the shear force can be obtained by
evaluating the algebraic sum of the vertical forces to one side of the section being considered.
The convention adopted in this text to indicate positive and negative shear forces is shown in
Figure 1.
Figure. 1

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The calculation carried out to determine the shear force can be repeated at various locations
along a beam and the values obtained plotted as a graph; this graph is known as the shear force
diagram. The shear force diagram indicates the variation of the shear force along a structural
member.
Bending Moment Diagram
In a statically determinate beam the numerical value of the bending moment (i.e. moments
caused by forces which tend to bend the beam) can be obtained by evaluating the algebraic
sum of the moments of the forces to one side of a section. In the same manner as with shear
forces either the left-hand or the right-hand side of the beam can be considered. The
convention adopted in this text to indicate positive and negative bending moments is shown in
Fig 2(a) and (b). Bending inducing tension on the underside of a beam is considered positive.
Fig. 2(a)
Bending inducing tension on the top of a beam is considered negative.
Fig. 2(b)
Note: Clockwise/anti-clockwise moments do not define +ve or -νe bending moments.
The sign of the bending moment is governed by the location of the tension surface at the
point being considered.
As with shear forces the calculation for bending moments can be carried out at various
locations along a beam and the values plotted on a graph; this graph is known as the ‘bending
moment diagram’. The bending moment diagram indicates the variation in the bending
moment along a structural member.
Influence lines of shear, Reaction and Moment

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