P a g e | 172 STRUCTURAL ANALYSIS - PowerPoint Presentation

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P a g e | 172

STRUCTURAL ANALYSIS

– 1UNIT – 1Name any two force methods to analyze the statically indeterminate structures.Column analogy methodFlexibility matrix methodMethod of consistent deformationTheorem of least workWrite the general steps of the consistent deformation method.By removing the restraint in the direction of redundant forces, released structure (which is a determinate structure) is obtained.In this released structure, displacements are obtained in the direction of the redundant forces.Then the displacement due to each redundant force is obtained and the conditions of displacement compatibility are imposed to get additionalequations.Solution for these equations gives the values of redundant forces.Then the released structure subjected to these known forces gives the forces and moments in the structure.Give example of beams of one degree static indeterminacy.

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In

general, 𝐸 = 𝑟 − 𝑒For this case, 𝑟 = 4 𝑎𝑛𝑑 𝑒 = 3∴ 𝐸 = 4 – 3 = 1Define degree of kinematic indeterminacy (or) Degree Of Freedom.It is defined as the least no of independent displacements required to define the deformed shape of a structure. There are two types of DOFJoint type DOFNodal type DOFDifferentiate external redundancy and internal redundancy.In pin jointed frames, redundancy caused by too many members iscalled internal redundancy. Then there is external redundancy caused by too many supports. When we introduce additional supports/members, we generally ensure more safety and more work (in analysis).Why to provide redundant members?To maintain alignment of two members during constructionTo increase stability during constructionTo maintain stability during loading (Ex: to prevent buckling of compression members)To provide support if the

applied loading is

changedTo act

as backup members in case some

members fail or require strengthening

Analysis is difficult but possible

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What are the

different methods used to analyze indeterminate structures?Finite element methodFlexibility matrix methodStiffness matrix methodWhat are statically indeterminate structures? Give example.If the conditions of statics i.e., ΣH=0, ΣV=0 and ΣM=0 alone are not sufficient to find either external reactions or internal forces in a structure, the structure is called a statically indeterminate structure.Define consistent deformation method.This method is used for the analysis of indeterminate structure. Thismethod is suitable when the number of unknown is one or two. When the number of unknown becomes more, it is a lengthy method.Define primary structure.A structure formed by the removing the excess or redundant restraintsfrom an Indeterminate structure making it statically determinate is called primary structure. This is required for solving indeterminate structures by flexibility matrix method.

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Write

the formulae for degree of indeterminancy.Two dimensional in jointed truss (2D truss)𝑖 = (𝑚 + 𝑟) − 2𝑗Two dimensional rigid frames/plane rigid frames (2D frame)𝑖 = (3𝑚 + 𝑟) − 3𝑗Three dimensional space truss (3D truss)𝑖 = (𝑚 + 𝑟) − 3𝑗Three dimensional space frame (3D frame)𝑖 = (6𝑚 + 𝑟) − 6𝑗Where,m = number of members r = number of reactions j = number of jointsWhat is the effect of temperature on the members of a statically determinate plane truss?In determinate structures temperature changes do not create any internal stresses. The changes in lengths of members may result indisplacement of joints. But these would not result in internal stresses or changes in external reactions.

Define internal

and external indeterminancy.

Internal indeterminacy (I.I) is

the excess no of internal forces present in a

member that make a structure indeterminate.

Prepared by R.Vijayakumar,

B.Tech

(CIVIL), CCET,

Puducherry

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External indeterminacy (E.I) is

the excess no of external reactions in the member that make a structure indeterminate.Indeterminacy (i) = I.I + E.IE.I = r – e; I.I = i – EI Where,r = no of support reactions and e = equilibrium conditionse = 3 (plane frames) and e = 6 (space frames)What are the requirements to be satisfied while analyzing a structure?Equilibrium conditionCompatibility conditionForce displacement conditionDefine degree of indeterminacy.The excess number of reactions take make a structure indeterminate is called degree of indeterminancy. Indeterminancy is also called degree of redundancy.Indeterminancy consists of internal and external indeterminancies. It is denoted by the symbol ‘i’.Degree of redundancy (i) = I.I + E.I Where,I.I = Internal indeterminancyE.I =External indeterminancy

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16.Differentiate the statically determinate structures and statically indeterminate

structures.S. NOSTATICALLYDETERMINATE STRUCTURESSTATICALLYINDETERMINATE STRUCTURES1.Conditions of equilibrium aresufficient to analyze the structureConditions of equilibrium areinsufficient to analyze the structure2.Bending moment and shearforce is independent of material and cross sectional areaBending moment and shearforce is dependent of material and independent of cross sectional area3.

No stresses are caused due to

temperature change and lack of fit

Stresses are caused due

to

temperature change and lack of fit

4.

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.

UNIT

– 2

1.

Distinguish between plane truss and plane

frame.

Plane frames

are

two-dimensional structures constructed with straight elements connected together

by rigid

and/or hinged connections. Frames

are

subjected

to

loads

and

reactions that

lie

in

the

plane

of

the

structure.

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If all the members

of a truss and the applied loads lie in a single plane, the truss is called a plane truss.What is meant by cambering technique in structures?Cambering is a technique applied on site, in which a slight upward curve is made in the structure/beam during construction, so that it will straighten out and attain the straight shape during loading. This willconsiderably reduce the downward deflection that may occur at later stages.Give the procedure for unit load method.Find the forces P1, P2, ……. in all the members due to external loadsRemove the external loads and apply the unit vertical point load at the joint if the vertical deflection is required and find the stressApply the equation for vertical and horizontal deflectionWhat are the assumptions made in unit load method?The external & internal forces are in equilibriumSupports are rigid and no movement is possible

The materials are strained

well within the elastic limit

Why

is it necessary to

compute deflections in structures? Computation of

deflection of structures is necessary for the

following reasons:

If the deflection of a structure is more

than the permissible,

thestructure will not look

aesthetic and will cause psychological upsetting of

the occupants.

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Excessive deflection

may cause cracking in the materials attached to the structure. For example, if the deflection of a floor beam is excessive, the floor finishes and partition walls supported on the beam may get cracked and unserviceable.Define unit load method.The external load is removed and the unit load is applied at the point, where the deflection or rotation is to found.Distinguish between pin jointed and rigidly jointed structure.S. NOPIN JOINTED STRUCTURERIGIDLY JOINTEDSTRUCTURE1.The joints permit change ofangle Between connected members.The members connected at arigid joint will maintain the anglebetween them even under deformation due to loads.2.

The joints are incapable

oftransferring Any moment

to the

connected members and vice- versa.

Members can transmit both

forces and Moments between themselves through the

joint.

3.

The pins transmit

forces

between Connected members by developing

shear.

Provision

of

rigid

joints

normally

increases the

redundancy of

the

structures.

8. What are the

conditions of

equilibrium?

The three conditions

of

equilibrium

are the sum of

horizontal forces, vertical forces and moments

at

any

joint should

be

equal

to

zero.

Prepared by R.Vijayakumar,

B.Tech

(CIVIL), CCET,

Puducherry

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(i.e.) ∑H

= 0; ∑V = 0; ∑M = 0Define trussed beam.A beam strengthened by providing ties and struts is known as Trussed Beams.Define ‘deck’ and ‘through’ type truss.A deck type is truss is one in which the road is at the top chord level of the trusses. We would not see the trusses when we ride on the road way.A through type truss is one in which the road is at the bottom chordlevel of the trusses. When we travel on the road way, we would see the web members of the trusses on our left and right. That gives us the impression that we are going` through’ the bridge.What is meant by lack of fit in a truss?One or more members in a pin jointed statically indeterminate frame may be a little shorter or longer than what is required. Such members will have

to be forced in place during

the assembling. These are called

members having Lack of fit.

Internal forces can develop in

a redundant frame

(without external loads) due to lack of fit.

Give any two situations where sway will

occur in portal

frames.

Eccentric or Unsymmetrical loading

Non-uniform section of the

members

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What are the

different types of forces acts on a frame?Dynamic loadStatic loadWhat is meant by settlement of supports?Support sinks mostly due to soil settlement. Rotation of ‘fixed’ ends can happen either because of soil settlement or upheaval of horizontal orinclined fixed ends. Fixed end moments induced in beam ends because of settlement or rotation of supports.What is a rigid joint?The members connected at a rigid joint will maintain the angle between them even under deformation due to loads. Members can transmit both forces and moments between themselves through the joint. Provisionof rigid joints normally increases the redundancy of the structures.Write down the assumptions made in portal method.The point of contra-flexure in all the members lies at their middle pointsHorizontal shear taken by each interior column is double

thehorizontal shear taken

by each of exterior column

Write

down the assumptions made

in cantilever method.

The point of contra-flexure in all

the members lies at their

middle points

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The direct stress

or axial stress in the columns due to horizontal forces, are directly proportional to their distance from the centroidal vertical axis of the frameWhat are the methods used to analyze the beam when it settle at supports?Kani’s methodMoment distribution methodSlope deflection methodDifferentiate symmetry and anti-symmetry frames.SYMMETRY FRAMEANTI-SYMMETRY FRAMEFor symmetric loading, Symmetricquantities like bending moment,displacements are symmetrical about the centroidal vertical axis.For anti-symmetric loading,Symmetric quantities like bendingmoment, displacements are zero at the point of centroidal vertical axis.Anti-symmetric quantities like slope

and shear force are zero at

the point of centroidal vertical axis.

Anti-symmetric quantities like

slope

and shear force are distributed about the centroidal vertical

axis.

20.What

is meant by

thermal stress?

Thermal stresses are stresses developed in a structure/member due

to change in temperature. Normally, determinate structures do not develop thermal stresses. They can

absorb changes in lengths and consequent displacements without developing

stresses.

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Write

any two important assumptions made in the analysis of trusses?The frame is a perfect frameThe frame carries load at the jointsAll the members are pin-joinedDifferentiate perfect and imperfect trusses?The frame which is composed of such members, which are justsufficient to keep the frame in equilibrium, when the frame is supporting an external load, is known as perfect frame. Hence for a perfect frame, thenumber of joints and number of members are given by, 𝑛 = 2𝑗 − 3A frame in which number of members and number of joints are not given by 𝑛 = 2𝑗 − 3 is known as imperfect frame. This means that numberof members in an imperfect frame will be either more or less than 2𝑗 − 3Write the difference between deficient and redundant frames?

If the

number of members in a frame

are less than (2𝑗 − 3),

then the frame is known as deficient

frame.If the number

of members in a frame

are more than (2𝑗 − 3

), then the frame is known

as redundant frame.

UNIT –

3

What are the

assumptions made

in

slope deflection

method?

This method

is

based

on

the following simplified assumptions.

All the

joints

of

the frame

are

rigid, (i.e.) the angle between

the

members

at

the

joints

does not change, when the members

of

frame

are

loaded.

Prepared by R.Vijayakumar,

B.Tech

(CIVIL), CCET,

Puducherry

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184Between each pair of the supports the beam section is constant.Define fixed end moment.The moments at the fixed joints of loaded member are called fixed end moment.Write down the slope deflection equation for a fixed end support.𝑀𝐴𝐵 =

𝑀𝐹𝐴𝐵

+

2𝐸𝐼

𝑙

[

2𝜃

𝐴 + 𝜃𝐵

+

3𝛿𝑙

]

4. What are the

moments induced

in a

beam member,

when

one

end is

given

a

unit rotation,

the other

end being fixed.

What is

the moment

at the

near

end called?

When

𝜃 =

1,

𝑀

𝐴𝐵

=

4

𝐸𝐼

,

𝑀

𝐵𝐴

=

2

𝐸𝐼

𝑙 𝑙

𝑀

𝐴𝐵

Is

the stiffness

of

AB

at

B

5. Define the

term

sway.

Sway is the

lateral movement

of

joints

in

a

portal

frame

due

to

the

unsymmetrical in dimensions, loads, moments

of

inertia,

end

conditions, etc. Sway

can be

prevented

by

unyielding supports provided

at

the beam

level as

well

as

geometric

or

load symmetry

about

vertical

axis.

Slide14

𝑀

𝐶𝐷

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What are the

situations where in sway will occur in portal frames?

Eccentric or unsymmetrical

loading

Unsymmetrical geometry

Different end conditions of the

column

Non-uniform section of the

membersUnsymmetrical settlement

of supportsA

combination of the

above

What is

the ratio

of

sway moments

at

column

heads when

one

end is

fixed

and

the other

end

hinged? Assume

that the length and M.I of both legs

are

equal.

Assuming the frame

to sway to

the

right

by

δ

Ratio

of sway

moments

=

𝑀

𝐵𝐴

=

𝑙

2

−

(

6

𝐸𝐼

𝛿

)

𝑙

2

−

(

3

𝐸𝐼

𝛿

)

=

2

8. A beam is

fixed at its left

end and

simply supported

at

right.

The

right

end

sinks

to a

lower

level

by

a

distance ‘

∆

’

with

respect

to

the

left

end.

Find

the

magnitude

and

direction

of the

reaction

at

the right

end if

‘l’ is

the beam length and EI, the

flexural

rigidity.

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𝑀

𝐴 (𝑑𝑢𝑒 𝑡𝑜 𝑠𝑖𝑛𝑘𝑖𝑛𝑔 𝑜𝑓 𝐵) =3 𝐸𝐼 𝛿𝑙2What are the

symmetric and anti-symmetric quantities

in structural behavior?When

a symmetrical structure is loaded with symmetrical loading, the

bending moment and deflected shape

will be symmetrical about the same axis. Bending moment

and deflection are symmetrical quantities.

How many

slope deflection equations are available for a two span

continuous beam?

There will be 4 nos. of slope-deflection

equations are available for a two span continuous

beam.

What are the

quantities

in

terms of which

the unknown

moments

are

expressed

in

slope-deflection method?

In

slope-deflection method, unknown moments

are

expressed in

terms

of

Slope

(θ)

Deflection

(∆)

The beam shown in figure

is

to

be

analyzed

by slope-deflection

method.

What are the

unknowns

and

to

determine them.

What are

the conditions

used?

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Unknowns:

𝜃𝐴, 𝜃𝐵, 𝜃𝐶Equilibrium equations used:𝑀𝐴𝐵 = 0𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0𝑀𝐶𝐵 = 0How do your account for sway in slope deflection method for portal frames?Because of sway, there will be rotations in the vertical members of a frame. This causes moments in the vertical members. To account for this, besides the equilibrium, one more equation namely shear equationconnecting the joint-moments is used.Write down the

equation for sway correction for the portal frame

shown in figure.

𝑀𝐴𝐵

𝑆ℎ𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

=

+ 𝑀𝐵𝐴

𝑀

𝐶𝐷

+

+

𝑀𝐷𝐶

𝑙

1

𝑙

2

= 0

15.Who

introduced

slope-deflection method

of

analysis?

Slope-deflection method

was

introduced

by

Prof.

George A.

Maney in

1915.

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|

18816.Write down the equilibrium equations for the frame shown in figure.Unknowns: 𝜃𝐵, 𝜃𝐶Equilibrium equations used:𝑀𝐵𝐴 + 𝑀𝐵𝐶

= 0

𝑀

𝐶𝐵

+ 𝑀

𝐶𝐷 = 0

𝑀

𝐴𝐵𝑆ℎ𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

=

+

𝑀𝐵𝐴 − 𝑃ℎ

+ 𝑀𝐶𝐷 +

𝑀𝐷𝐶

𝑙 𝑙

+ 𝑃 =

0

17.Write

down

the

general slope-deflection equations

and

state what each

term

represents.

General slope deflection

equations:

𝑙

𝑀 =

𝑀 +

2𝐸𝐼

[

2𝜃 +

𝜃

𝐴𝐵

𝐹𝐴𝐵

𝐴

𝐵

𝑙

+

3𝛿

]

𝑀 =

𝑀 +

2𝐸𝐼

[

2𝜃 +

𝜃

𝐵

𝐴

𝐹𝐵𝐴

𝐵

𝐴

𝑙 𝑙

+

3𝛿

]

Where,

M

FAB

,

M

FBA

= Fixed

end moment

at A

and

B

respectively

due

to given loading

𝜃

𝐴

,

𝜃

𝐵

= Slopes at A

and

B

respectively

𝛿

= Sinking of

support

A

with respect to

B

18.How

many

slope-deflection equations are available for each

span?

Two numbers of slope-deflection equations are available for

each

span,

describing the moment

at

each

end of the span.

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18919.In a continuous beam, one of the support sinks. What will happen to the span and support moments associated with the sinking of support.

Let support

D sinks by

𝛿. This will not affect span moments. Fixed end moments (support moments) will get

developed as under

𝑀

𝐹𝐶𝐷 = 𝑀

𝐹𝐷𝐶

= − 6 𝐸𝐼

𝛿

1

𝑙2

𝑀

𝐹𝐷𝐸

=

𝑀

𝐹𝐸𝐷

= −

6

𝐸𝐼

𝛿

2

𝑙

2

What is

the basis on

which the

sway equation

is

formed

for a

structure?

Sway is dealt

with in slope-deflection method

by

considering

the

horizontal equilibrium

of

the whole

frame

taking into account the shears

at

the

base level of

columns

and

external horizontal

forces.

𝑇ℎ𝑒

𝑠ℎ𝑒𝑎𝑟 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑖𝑠

𝑀

𝐴𝐵

+

𝑀

𝐵𝐴

–

𝑃ℎ

+

𝑀

𝐶𝐷

+

𝑀

𝐷𝐶

+ 𝑝 =

0

𝑙 𝑙

State

the limitations

of

slope-deflection

method.

It

is not

easy to

account

for

varying member

sections

It becomes

very

inconvenience when

the unknown displacements

are large

in

number

This method

is

advantageous only

for

the structures with small Kinematic indeterminacy

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The solution

of simultaneous equation makes the method tedious for annual computationsWhy slope-deflection method is called a ‘displacement method’?In slope-deflection method, displacements (like slopes and displacements) are treated as unknowns and hence the method is a ‘displacement method’.Define Flexural rigidity of beams.The product of young’s modulus (E) and moment of inertia (I) is called Flexural Rigidity (EI) of Beams. The unit is Nmm2.Define constant strength beam.If the flexural Rigidity (EI) is constant over the uniform section, it is called Constant strength beam.Define continuous beam.A Continuous beam is one, which is supported on more than two supports. For usual loading on the beam hogging (- ive) moments causingconvexity upwards at the supports and sagging (+ ive) moments causing concavity upwards occur at mid span.What are the

advantages of continuous

beam over simply supported beam?

The maximum bending moment in case of continuous beam is

much less than in case of simply supported beam of same

span carrying same loads.

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In

case of continuous beam, the average bending moment is lesser and hence lighter materials of construction can be used to resist the bending moment.UNIT – 4Explain moment distribution method (Hardy cross method).This method is first introduced by Professor Hardy Cross in 1932. It is widely used for the analysis of indeterminate structures. It uses aniterative technique. The method employs a few basic concepts and a few specialized terms such as fixed end moments, relative stiffness, carry over, distribution factor. In this method, all the members of the structure are first assumed to be fixed in position and fixed end moments due to external loadsare obtained.Define distribution factor.When several members meet at a joint and a moment is applied at the joint to produce rotation without translation of the members, the moment isdistributed among all the members meeting at that joint proportionate to their stiffness.𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 =

𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠

𝑆𝑢𝑚

𝑜𝑓 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑎𝑡 𝑡ℎ𝑒

𝑗𝑜𝑖𝑛𝑡

If there are

three members,

𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟𝑠

=

𝑘

1

𝑘2

, ,

𝑘

3

𝑘

1

+ 𝑘

2

+

𝑘

3

𝑘

1

+ 𝑘

2

+

𝑘

3

𝑘

1

+ 𝑘

2

+

𝑘

3

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Define

carry over factor.A moment applied at the hinged end B “carries over” to the fixed end A, a moment equal to half the amount of applied moment and of the samerotational sense. C.O =0.5What is the difference between absolute and relative stiffness?Absolute stiffness is represented in terms of E, I and l, such as 4EI / l.Relative stiffness is represented in terms of ‘I’ and ‘l’, omitting the constant E. Relative stiffness is the ratio of stiffness to two or more members at a joint.In a member AB, if a moment of -10kN.m is applied at A, what is the moment carried over to B?Carry over moment = Half of the applied moment∴ Carry over moment to B = -10/2 = -5 kN.m𝑆𝑖𝑚𝑝𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 (𝑘) =

6. Define

Stiffness factor.

It is the moment required to

rotate the end while acting on it through a

unit rotation, without translation of the far end being

3 𝐸𝐼

𝑙

𝐹𝑖𝑥𝑒𝑑 𝑖𝑠

𝑔𝑖𝑣𝑒𝑛

𝑏𝑦 (𝑘

) =

4

𝐸𝐼

𝑙

Where,

E =

Young’s modulus

of

the beam material

I =

Moment

of

inertia

of

the

beam

L =

Beam’s

span

length

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Define

carry over moment.It is defined as the moment induced at the fixed end of the beam by the action of a moment applied at the other end, which is hinged. Carry overmoment is the same nature of the applied moment.What is the sum of distribution factors at a joint?Sum of distribution factors at a joint = 1.What is the moment at a hinged end of a simple beam?Moment at the hinged end of a simple beam is zero.A rigid frame is having totally 10 joints including support joints. Out of slope-deflection and moment distribution methods, which method would you prefer for analysis? Why?Moment distribution method is preferable.If we use slope-deflection method, there would be 10 (or more)unknown displacements and an equal number of equilibrium equations. In addition, there would be

2 unknown support moments per span

and the same number of slope-deflection equations. Solving

them is difficult.

What are the

limitations of moment distribution method?

This method is

eminently suited to analyze continuous beams including non-prismatic members but it presents some difficulties when applied to rigid frames, especially when frames are

subjected to side sway

Unsymmetrical frames have to be

analyzed more than once to obtain FM (fixed moments) in the

structures

This method cannot be applied to structures with intermediate hinges

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UNIT –

5What is the value of rotation moment at a fixed end considered in Kani’s method?𝑀𝐴𝐵 = 2𝐸 𝐾𝐴𝐵 𝜃𝐴𝑀𝐵𝐴 = 2𝐸 𝐾𝐵𝐴 𝜃𝐵What are the fundamental equations of Kani’s method?𝑖𝑗∑𝑀𝑖𝑗 = ∑𝑀𝐹𝑖𝑗 + 2 ∑𝑀′ + ∑𝑀𝑗𝑖 = 02𝑖𝑗 𝐹𝑖𝑗𝑗𝑖∑𝑀′ = − 1 ( ∑𝑀 + ∑𝑀′ )What are the limitations of Kani’s method?

Gasper Kani of

Germany gave another distribution procedure in which instead of distributing

entire moment in successive steps, only

the rotation contributions

are distributedBasic unknown like displacements

which are not found

directly

What are the

advantages of Kani’s method?

Hardy Cross method distributed only

the unbalanced moments at joints, whereas Kani’s method distributes the total joint moment

at any stage

of

iteration

The

more

significant

feature of

Kani’s method is that the

process

is

self-

corrective. Any

error at

any

stage of

iteration

is

corrected

in

subsequent steps

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Framed structures

are rarely symmetric and subjected to side sway, hence Kani’s method is best and much simpler than other methods like moment distribution method and slope displacement methodState Miller-Breslau principle.Miller-Breslau principle states that, if we want to sketch the influence line for any force quantity like thrust, shear, reaction, support moment orbending moment in a structure,We remove from the structure the resistant to that force quantityWe apply on the remaining structure a unit displacement corresponding to that force quantity.The resulting displacements in the structure are the influence line ordinates sought.Define rotation factor.Rotation factor in Kani’s method is akin to distribution factor in moment distribution method.Actually, 𝑢 = − 0.5 × 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟Define displacement factor.∆𝑖𝑗 Is the “displacement factor” for each column, similar to 𝑢𝑖𝑗 weadopted earlier for rotation factor. Actually, ∆𝑖𝑗 = −1.5 𝐷𝐹 and is applicable to column only.

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Brief

about Kani’s method of analysis.Kani’s method of analyzing indeterminate structures, particularly, building frames was developed in Germany in the year 1947 by Dr. GasperKani. Like moment distribution it is a method of solving slope deflection equations by an iterative method. Hence, this will fall under the category of stiffness method wherein the level of difficulty would be decided by the degrees of freedom (and not the degree of redundancy).What are the basic principles of compatibility?Compatibility is defined as the continuity condition on the displacements of the structure after external loads are applied to thestructure.Define Kani’s method and how it is better than MDM.Kani’s method is similar to the MDM in that both these methods use Gauss Seidel iteration procedure to solve the slope deflection equations,without explicitly writing them down. However the difference between Kani’s method and the MDM is that Kani’s method iterates the member end moments themselves rather than iterating their increment Kani’s method essentially consists of a single simple numerical operation performedrepeatedly at the joints of a structure, in a chosen sequence.Write the procedure for Kani’s method.While solving structures by this method the following steps may

be kept in

mind.Compute all

fixed end moments

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Compute

and tabulate all rotation factors for all joints that would have rotation.Fixed ends will not have rotation factors. Nor rotation contributions either to the same (fixed end) or to the opposite end.Extreme simply supported ends will initially get a fixed end moment.Iterative process can be formed.(Or)Fixed end momentRotation factorResultant restraint momentIteration cycleFinal momentWhat are the methods of analyzing building frame?Cantilever methodFactor methodPortal method