Structural analysis 1

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Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry

STRUCTURAL ANALYSIS – 1

UNIT – 1

1. Name any two force methods to analyze the statically indeterminate

structures.

Column analogy method

Flexibility matrix method

Method of consistent deformation

Theorem of least work

2. Write 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 additional

equations.

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.

3. Give example of beams of one degree static indeterminacy.

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In general, 𝐸 = 𝑟 − 𝑒

For this case, 𝑟 = 4 𝑎𝑛𝑑 𝑒 = 3

∴ 𝐸 = 4 – 3 = 1

4. Define 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 DOF

Joint type DOF

Nodal type DOF

5. Differentiate external redundancy and internal redundancy.

In pin jointed frames, redundancy caused by too many members is

called 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).

6. Why to provide redundant members?

To maintain alignment of two members during construction

To increase stability during construction

To maintain stability during loading (Ex: to prevent buckling of

compression members)

To provide support if the applied loading is changed

To act as backup members in case some members fail or require

strengthening

Analysis is difficult but possible

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7. What are the different methods used to analyze indeterminate

structures?

Finite element method

Flexibility matrix method

Stiffness matrix method

8. What 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.

9. Define consistent deformation method.

This method is used for the analysis of indeterminate structure. This

method is suitable when the number of unknown is one or two. When the

number of unknown becomes more, it is a lengthy method.

10. Define primary structure.

A structure formed by the removing the excess or redundant restraints

from 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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11. Write the formulae for degree of indeterminancy.

Two dimensional in jointed truss (2D truss)

𝑖 = (𝑚 + 𝑟) − 2𝑗

Two dimensional rigid frames/plane rigid frames (2D frame)

𝑖 = (3𝑚 + 𝑟) − 3𝑗

Three dimensional space truss (3D truss)

𝑖 = (𝑚 + 𝑟) − 3𝑗

Three dimensional space frame (3D frame)

𝑖 = (6𝑚 + 𝑟) − 6𝑗

Where,

m = number of members

r = number of reactions

j = number of joints

12. What 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 in

displacement of joints. But these would not result in internal stresses or

changes in external reactions.

13. 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.

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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.I

E.I = r – e; I.I = i – EI

Where,

r = no of support reactions and

e = equilibrium conditions

e = 3 (plane frames) and e = 6 (space frames)

14. What are the requirements to be satisfied while analyzing a structure?

Equilibrium condition

Compatibility condition

Force displacement condition

15. Define 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 indeterminancy

E.I =External indeterminancy

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16. Differentiate the statically determinate structures and statically

indeterminate structures.

S. NO

STATICALLY

DETERMINATE

STRUCTURES

STATICALLY

INDETERMINATE

STRUCTURES

1. Conditions of equilibrium are

sufficient to analyze the

structure

Conditions of equilibrium are

insufficient to analyze the

structure

2. Bending moment and shear

force is independent of material

and cross sectional area

Bending moment and shear

force is dependent of material

and independent of cross

sectional area

3. 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.

2. 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 will

considerably reduce the downward deflection that may occur at later stages.

3. Give the procedure for unit load method.

Find the forces P1, P2, ……. in all the members due to external loads

Remove the external loads and apply the unit vertical point load at

the joint if the vertical deflection is required and find the stress

Apply the equation for vertical and horizontal deflection

4. What are the assumptions made in unit load method?

The external & internal forces are in equilibrium

Supports are rigid and no movement is possible

The materials are strained well within the elastic limit

5. 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, the

structure 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.

6. 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.

7. Distinguish between pin jointed and rigidly jointed structure.

S. NO PIN JOINTED STRUCTURE RIGIDLY JOINTED

STRUCTURE

1. The joints permit change of

angle Between connected

members.

The members connected at a

rigid joint will maintain the angle

between them even under

deformation due to loads.

2. The joints are incapable of

transferring 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.

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(i.e.) ∑H = 0; ∑V = 0; ∑M = 0

9. Define trussed beam.

A beam strengthened by providing ties and struts is known as Trussed

Beams.

10. 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 chord

level 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.

11. 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.

12. 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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13. What are the different types of forces acts on a frame?

Dynamic load

Static load

14. What 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 or

inclined fixed ends. Fixed end moments induced in beam ends because of

settlement or rotation of supports.

15. 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. Provision

of rigid joints normally increases the redundancy of the structures.

16. Write down the assumptions made in portal method.

The point of contra-flexure in all the members lies at their middle

points

Horizontal shear taken by each interior column is double the

horizontal shear taken by each of exterior column

17. 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 frame

18. What are the methods used to analyze the beam when it settle at

supports?

Kani’s method

Moment distribution method

Slope deflection method

19. Differentiate symmetry and anti-symmetry frames.

SYMMETRY FRAME ANTI-SYMMETRY FRAME

For symmetric loading, Symmetric

quantities like bending moment,

displacements are symmetrical about

the centroidal vertical axis.

For anti-symmetric loading,

Symmetric quantities like bending

moment, 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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21. Write any two important assumptions made in the analysis of trusses?

The frame is a perfect frame

The frame carries load at the joints

All the members are pin-joined

22. Differentiate perfect and imperfect trusses?

The frame which is composed of such members, which are just

sufficient to keep the frame in equilibrium, when the frame is supporting an

external load, is known as perfect frame. Hence for a perfect frame, the

number of joints and number of members are given by, 𝑛 = 2𝑗 − 3

A 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 number

of members in an imperfect frame will be either more or less than 2𝑗 − 3

23. Write 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

1. 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.

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Between each pair of the supports the beam section is constant.

2. Define fixed end moment.

The moments at the fixed joints of loaded member are called fixed

end moment.

3. Write down the slope deflection equation for a fixed end support.

𝑀𝐴𝐵 = 𝑀𝐹𝐴𝐵 + 2𝐸𝐼

𝑙 [ 2𝜃𝐴 + 𝜃𝐵 +

3𝛿

𝑙 ]

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,

𝑀𝐴𝐵 = 4 𝐸𝐼

𝑙 , 𝑀𝐵𝐴 =

2 𝐸𝐼

𝑙

𝑀𝐴𝐵 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.

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6. 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 members

Unsymmetrical settlement of supports

A combination of the above

7. 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 = 𝑀𝐵𝐴

𝑀𝐶𝐷=

− (

6 𝐸𝐼 𝛿

𝑙2 )

− ( 3 𝐸𝐼 𝛿

𝑙2 )= 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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𝑀𝐴 (𝑑𝑢𝑒 𝑡𝑜 𝑠𝑖𝑛𝑘𝑖𝑛𝑔 𝑜𝑓 𝐵) = 3 𝐸𝐼 𝛿

𝑙2

9. What 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.

10. 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.

11. 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 (∆)

12. 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, 𝜃𝐶

Equilibrium equations used:

𝑀𝐴𝐵 = 0

𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0

𝑀𝐶𝐵 = 0

13. How 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 equation

connecting the joint-moments is used.

14. Write down the equation for sway correction for the portal frame

shown in figure.

𝑆ℎ𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 = 𝑀𝐴𝐵 + 𝑀𝐵𝐴

𝑙1 +

𝑀𝐶𝐷 + 𝑀𝐷𝐶

𝑙2 = 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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16. Write down the equilibrium equations for the frame shown in figure.

Unknowns: 𝜃𝐵, 𝜃𝐶

Equilibrium equations used:

𝑀𝐵𝐴 + 𝑀𝐵𝐶 = 0

𝑀𝐶𝐵 + 𝑀𝐶𝐷 = 0

𝑆ℎ𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 = 𝑀𝐴𝐵 + 𝑀𝐵𝐴 − 𝑃ℎ

𝑙 +


𝑙 + 𝑃 = 0

17. Write down the general slope-deflection equations and state what each

term represents.

General slope deflection equations:

𝑀𝐴𝐵 = 𝑀𝐹𝐴𝐵 + 2𝐸𝐼

𝑙 [ 2𝜃𝐴 + 𝜃𝐵 +

3𝛿

𝑙 ]

𝑀𝐵𝐴 = 𝑀𝐹𝐵𝐴 + 2𝐸𝐼

𝑙 [ 2𝜃𝐵 + 𝜃𝐴 +

3𝛿

𝑙 ]

Where,

MFAB, MFBA = Fixed end moment at A and B respectively due to given

loading

𝜃𝐴, 𝜃𝐵 = 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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19. 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

𝑀𝐹𝐶𝐷 = 𝑀𝐹𝐷𝐶 = − 6 𝐸𝐼 𝛿

𝑙12

𝑀𝐹𝐷𝐸 = 𝑀𝐹𝐸𝐷 = − 6 𝐸𝐼 𝛿

𝑙22

20. 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 𝑀𝐴𝐵 + 𝑀𝐵𝐴 – 𝑃ℎ

𝑙 +


𝑙+ 𝑝 = 0

21. 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 computations

22. Why 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’.

23. 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.

24. Define constant strength beam.

If the flexural Rigidity (EI) is constant over the uniform section, it is

called Constant strength beam.

25. 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 causing

convexity upwards at the supports and sagging (+ ive) moments causing

concavity upwards occur at mid span.

26. 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 – 4

1. Explain 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 an

iterative 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 loads

are obtained.

2. 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 is

distributed 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 = 𝑘1

𝑘1+ 𝑘2+ 𝑘3,

𝑘2

𝑘1+ 𝑘2+ 𝑘3,

𝑘3

𝑘1+ 𝑘2+ 𝑘3

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3. 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 same

rotational sense. C.O =0.5

4. What 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.

5. 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.m

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

𝑆𝑖𝑚𝑝𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 (𝑘) = 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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7. 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 over

moment is the same nature of the applied moment.

8. What is the sum of distribution factors at a joint?

Sum of distribution factors at a joint = 1.

9. What is the moment at a hinged end of a simple beam?

Moment at the hinged end of a simple beam is zero.

10. 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.

11. 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 – 5

1. What is the value of rotation moment at a fixed end considered in

Kani’s method?

𝑀𝐴𝐵 = 2𝐸 𝐾𝐴𝐵 𝜃𝐴

𝑀𝐵𝐴 = 2𝐸 𝐾𝐵𝐴 𝜃𝐵

2. What are the fundamental equations of Kani’s method?

∑𝑀𝑖𝑗 = ∑𝑀𝐹𝑖𝑗 + 2 ∑𝑀𝑖𝑗′ + ∑𝑀𝑗𝑖 = 0

∑𝑀𝑖𝑗′ = −

1

2 ( ∑𝑀𝐹𝑖𝑗 + ∑𝑀𝑗𝑖

′ )

3. 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 distributed

Basic unknown like displacements which are not found directly

4. 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 method

5. State 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 or

bending moment in a structure,

We remove from the structure the resistant to that force quantity

We 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.

6. Define rotation factor.

Rotation factor in Kani’s method is akin to distribution factor in

moment distribution method.

Actually, 𝑢 = − 0.5 × 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

7. Define displacement factor.

∆𝑖𝑗 Is the “displacement factor” for each column, similar to 𝑢𝑖𝑗 we

adopted earlier for rotation factor. Actually, ∆𝑖𝑗 = −1.5 𝐷𝐹 and is

applicable to column only.

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8. 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. Gasper

Kani. 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).

9. 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 the

structure.

10. 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 performed

repeatedly at the joints of a structure, in a chosen sequence.

11. 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 moment

Rotation factor

Resultant restraint moment

Iteration cycle

Final moment

12. What are the methods of analyzing building frame?

Cantilever method

Factor method

Portal method

Engineering

Structural analysis 1