undefined | unit 1 analysis of indeterminate structures

Unit-1

Analysis Of Indeterminate Structure by force methods

Q1) What is analysis of indeterminate structure by force methods? And explain flexibility method in detail.

A1)

Analysis of indeterminate structure by force methods

The pressure approach of analysis, additionally known because the approach of steady deformation, makes use of equilibrium equations and compatibility situations to decide the unknowns in statically indeterminate systems. In this approach, the unknowns are the redundant forces. For indeterminate systems , Statics (equilibrium) on my own isn't always enough to behavior structural analysis. Compatibility and fabric facts are essential.

Indeterminate structure:

Number of unknown Reactions > Number of equilibrium equations

Advantages:

Smaller deflections for comparable individuals

Redundancy in load sporting capacity (redistribution)

Increased stability

Disadvantages:

Complex connections

Initial / Residual / Settlement Stresses

Methods of analysis:

Equilibrium of forces and moments

Compatibility of deformation amongst individuals and at supports

Material conduct touching on stresses with strains

Strain-displacement relations

Boundary Conditions

Techniques of analysis there are most important techniques:

Flexibility approach ( pressure approach)

Stiffness approach ( Displacement approach)

Flexibility Method or Method of Consistent Deformation:

Flexibility method is also called as method of consistent deformation. Convert the indeterminate shape to a determinate one through disposing of a few unknown forces / assist reactions and changing them with (assumed) known / unit forces

Using superposition, calculate the pressure that might be required to reap compatibility with the unique shape.

This approach is relevant for any type of shape: beam, frame or truss. It is to be cited that beam and frames systems are predominantly bending (flexure) systems whilst trusses are predominantly direct pressure systems (compression) in nature. The truss individuals aren't subjected to bending. In different words, all masses are axial.

Unknowns to be solved for are generally redundant pressure.

The outcomes received through fixing the hassle the use of the pressure approach, are all of the unknown forces (along with reactions on the supports).

Coefficients of the unknowns in equations to be solved are "flexibility" coefficients.

It could be very conceptual in nature. The pressure approach turns into bulky while the Static Indeterminacy of a shape is large.

The pressure approach (that is additionally referred to as the power approach or the approach of steady deformations) makes use of the idea of structural Static Indeterminacy (SI).

If one is inquisitive about locating rotational or translational displacements of an indeterminate shape, they ought to be received one by one the use of any techniques of locating displacements ( Example: unit load approach, second region approach or conjugate beam approach ).

Q2) Explain static indeterminacy and sign conventions in detail.

A2) Static Indeterminacy

The Static Indeterminacy (SI) for beams and frames is described as,

SI = nu — ne

Where,

nu = Number of unknown assist reactions

ne = Number of equations of equilibrium

In fashionable for a -dimensional shape, there are 3 equations of equilibrium (ne = 3) and for a 3-dimensional shape there are six (ne = 6).

For a truss, static indeterminacy entails each outside and inner indeterminacy due to the inner individuals in a truss.

Static indeterminacy (SI) within side the case of a truss is described as,

SI = b + r — 2j

Where,

b = Number of individuals within side the truss

r = Number of reactions on the supports

j = Number of joints within side the truss

Sign Conventions

1). Static Calculations

∑F x = 0, ∑F y = 0 & ∑M = 0

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2). For Structural Actions

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Q3) Explain degree of static indeterminacy with example.

A3) Degree of static indeterminacy (Dsi)

Dsi = No of Unknown Actions –No of Eqn,

Dsi = R – E

External Indeterminacy

When we can't discover assist reactions the use of legal guidelines at statics, it's miles referred to as externally indeterminate structure.

Internal Indeterminacy

When we can't discover inner member movements the use of legal guidelines of statics, it's miles referred to as internally indeterminate structure.

Boundary condition

Displacement Re stained Displacement allow

Δx, Δy, ϴz Nil

Δy,ϴz, Δx

Δx, Δyϴz

Dsi for beams = R – E

Where,

R = No of assist reactions

E = No of Equilibrium equations

Note: For one inner pin, no of extra equations of equilibrium are = m – 1

m = no of individuals linked via way of means of pin.

Sr. No Structure

1. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\11 - Copy.JPG$

R E Dsi

3 3 0

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R E Dsi

3 3 0

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R E Dsi

4 3 1

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R E Dsi

6 3 3

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R E Dsi

6 3 3

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R E Dsi

6 4 4

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R E Dsi

5 4 1

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R E Dsi

9 5 4

Q4) How to calculate Dsi for trusses?

A4) Dsi for Trusses:

(Dsi)Total = (m + R) – (2J)

Where,

M = No of members

R = No of support reactions

J = No of Joints

(Dsi)external = R – 3

(Dsi)internal = (Dsi)total – (Dsi)ext

Dsi = (M + R) – 2j

Sr. No Truss

1. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\17.JPG$

M R J (Dsi)Total

5 3 4 0

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M R J (Dsi)Total

6 3 4 1

(Dsi)int.=0

(Dsi)ext= 1

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M R J (Dsi)Total

5 4 4 1

(Dsi)int.=1

(Dsi)ext= 0

4. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\20.JPG$

M R J (Dsi)Total

5 4 4 1

(Dsi)int.= 1

(Dsi)ext= 0

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M R J (Dsi)Total

19 4 10 3

(Dsi)int.= 1

(Dsi)ext= 2

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M R J (Dsi)Total

6 4 4 2

(Dsi)int. = 1

(Dsi)ext = 1

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M R J (Dsi)Total

11 4 7 1

(Dsi)int. = 1

(Dsi)ext = 0

8. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\24.JPG$

M R J (Dsi)Total

8 4 5 2

(Dsi)int.= 1

(Dsi)ext = 1

9. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\25.JPG$

M R J (Dsi)Total

8 5 5 3

(Dsi)int. = 2

(Dsi)ext = 1

10. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\26.JPG$

M R J (Dsi)Total

9 3 6 0

Q5) How to calculate Dsi for frames?

A5) Dsi for Frames:

(Dsi)Total = 3m + R – 3J

(Dsi)external = R – 3

(Dsi)Internal = (Dsi)Total – (Dsi)ext

Sr. No Structure

1. $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\27.JPG$

M R J (Dsi)Total

2 5 3 2

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M R J (Dsi)Total

3 8 4 3

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M R J (Dsi)Total

6 6 6 6

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M R J (Dsi)Total

4 8 5 5

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M R J (Dsi)Total

3 8 11 10

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M R J (Dsi)Total

5 9 6 6

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M R J (Dsi)Total

15 10 13 16

Q6) How to calculate Dki for frames?

A6) Dki for beams:

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2) $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\40.JPG$

ϴB = 1

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4) $C:\Users\TARUNA\Desktop\internship\february\11 feb2020\42.JPG$

ΔYB&ϴB = 2

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ϴB, ϴC, ϴD = 3

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ϴB, ϴC, ϴDΔYD = 4

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= 0

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ϴBA, ϴBC, ΔBy = 3

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ϴB, ΔBy = 2

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ϴDE, ϴDC, ΔDy = 6

ϴC, ϴBΔBy

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ϴA, ϴB,

ϴD, ϴCB, ϴCD = 10

ΔYC, ϴE

ΔYE, ΔYF1, ϴE

Q7) Explain application of force method to analysis of indeterminate beam and what is flexibility coefficient?

A7) Application of Force Method to Analysis of Indeterminate Beams:

Calculate the Static Indeterminacy (SI) of the shape the use of equation relying upon whether or not the shape is beam, frames, or truss.

Choose one of the response forces (or internal participants of the truss) because the redundant pressure. One at a time if there are more than one redundancy.

Split the statically indeterminate shape right into a determinate shape (acted upon via way of means of implemented hundreds at the shape) and determinate shape(s) acted on via way of means of the redundant forces (one at a time).

Analyze the determinate systems via way of means of the unit load approach to locate the displacement AL, that's the displacement for the implemented loading and redundant removed. Then locate at, that's the displacement for the unit load only, on the factor of redundancy. If a second is taken as redundant, the corresponding displacements may be zero and a

Finally, formulate equation(s) of displacement compatibility on the support(s) (within side the case of beams and frames). In the case of trusses, displacement compatibility of truss bars may be used.

Solve those equation(s) to get the redundant pressure(s).Calculate all of the reactions on the support the use of standards of statics.

Flexibility coefficient:

The flexibility approach is primarily based totally upon the answer of "equilibrium equations and compatibility equations". There will continually be as many compatibility equations as redundant. It is referred to as the flexibility approach due to the fact flexibilities seem within side the equations of compatibility. Another call for the approach is the pressure approach due to the fact forces is the unknown portions in equations of compatibility.

Q= [DQL]+[F][Q]-compatibility equation

DQ=External displacement at redundant position

Displacement at the redundant position because of applied loading

F=flexibility matrix

Q8) Explain Redundant reaction of given example.

A8) Redundant reaction

It is also called as Force Method

In this method, horizontal displacement is neglected

Degree of indeterminacy=R-E

R=Reaction

E=Equilibrium equation

=4-2

The redundant structure is 2.

Use flexibility Matrix Method

Step 1)

=4-2

Redundant structure is 2.

Zone	Limit	Origin	M
CD	0-2	C	0	0	x
DB	0-2	D	-40	0	x+2
BA	0-6	B		x	x+4

= - 11100 /EI

=333.33/

Q9) Explain energy method and strain energy in detail.

A9) Energy methods:

These are methods primarily based totally on linear elastic conduct and conservation of electricity, Energy U = Fx/2 = F2/2k wherein F is the implemented pressure, x is the space moved within side the route of the pressure at its factor of utility and okay is the elastic stiffness of the part, once more within side the route of the pressure at its factor of utility. , i.e. the work done via way of means of outside forces equals the electricity saved within side the structural below

Strain Energy

1) EI = 17250 Kn.m2 find strain energy?

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∑MA = 0

30 x 2 – RB x 4 = 0

RB = 15

VA = VB = 15 KN

BMC = BMD = 15(1) = 15 KN.m

BMC = 15(2) = 30 KN.m

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2) ‘M/EI’ Dia.

3) C.B

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A1 = 1 / 2 x 1 x 15/EI = 7.5/EI

A2 = 2 x 7.5/EI = 15/EI

A3 = 1 /2 x 2 x 7.5/EI = 7.5/EI

A4 = 1 / 2 x 15/EI x 1 = 7.5/EI

Analysis of conjugate beam

∑MA = -RB(4) + 1/EI{7.5(2/3 x 1) + 15(2) + 7.5(2) + 7.5(3 + 1/31)} = 0

= - RB x 4 + 1/EI[ 5 + 30 + 15 + 25 ] = 0

RB x 4 = 75/17250

RB = 18.75/EI

∑fy = 0

-RA + RB + A1 + A2 + A3 + A4 = 0

RA = 18.75/EI

RA = RB = ½{A1 + A2 + A3 + A4} = 18.75/EI

Slope & Deflection

(ϴA)R.B = (SFA)C.B = (-)18.75/EI

= 18.75/EIxC = 1.086 x 10-3radxC

(ϴB)R.B = (SFB)C.B = 18.75/EIxC = 1.086 x 10-3 rad xC

(ΔE)R.B = (BME)CB

(BME)CB= 1/EI{ -18.75 + 7.5(1 + 1/3 x 1) + 7.5(0.5) + 3.75 (1/3 x 1)}

=1/EI{-22.5}

ΔE = 22.5/EI xC = 1.304 mm xC

Q10) Explain principle of minimum potential strength and explain principle of virtual work.

A10) Principle of minimum potential strength:

The minimal overall capability strength precept is a essential idea used in engineering. It dictates that at low temperatures a shape or frame shall deform or displace to a role that (locally) minimizes the overall capability strength, with the misplaced capability strength being transformed into kinetic strength. His general capability strength is the sum of the elastic stress strength, U, saved within side the deformed frame and the capability strength, V, related to the implemented forces

Π = U + V (1)

This strength is at a desk bound function whilst an infinitesimal version from such function entails no alternate in strength

δ Π = δ ( U + V ) = 0 (2)

The precept of minimal general capability strength can be derived as a unique case of the digital work precept for elastic structures issue to conservative forces.

The equality among outside and inner digital work (because of digital displacements) is:

∫ S t δ u T T d S + ∫ V δ u T f d V = ∫ V δ ϵ T σ d V (3)

In the unique case of elastic bodies, the right-hand-aspect of (3) may be taken to be the alternate, of elastic stress strength U because of infinitesimal versions of actual displacements. In addition, whilst the outside forces are conservative forces, the left-hand-aspect of (3) may be visible because the alternate within side the capability strength function V of the forces. The function V is described as

V = − ∫ S t u T T d S − ∫ V u T f d V

Where in: the minus signal implies a lack of capability strength because the force is displaced in its direction. With those subsidiary conditions, (3) becomes:

− δ V = δ U

This leads to (2) as desired. The variation shape of (2) is frequently used as the idea for growing the finite detail approach in structural mechanics.

Principle of virtual work:

Virtual work is the work done with the aid of using a actual pressure appearing via a digital displacement or a digital pressure appearing via a actual displacement.

The Unit Load Method When the precept of digital work is used to calculate the displacement, at a co-ordinate, the device of outside forces,̄ F"> is selected for you to consist best of a unmarried unit pressure at coordinate.

A digital displacement is any displacement constant with the restrictions of the shape, i.e., that fulfill the boundary situations on he supports.

A digital pressure is any device of forces in equilibrium. f(x) and y(x) are actual forces and related displacements. ̄y(x) is a digital displacement constant with the boundary situations.

Q11) Explain the process of Castiglione s theorem in detail.

A11) Castiglione’s 1st theorem:

It states that displacement at a point of application of load in the direction of load is equal to partial derivative of strain energy of structure with respect to load.

1]. ӘU/ӘQi = Δj

U = strain energy

Qj = Force in the direction j

Δj = Displacement in the direction j

2]. Strain energy: Uaxial force = ∫L0P2dx/2EA

UB.M = ∫L0M2dx/2EI

UTorsion = ∫L0T2dx/2GJ

Deflection = translation

3]. Procedure of displacement analysis by Castigliano’s theorem :

Step 1:

Apply imaginary load ‘Q’ at section where displacement is required.

Note:

1>. For horizontal translation apply ‘Q’ in horizontal direction for vertical translation apply force Q in vertical direction

2>. For finding ‘slope’(rotation) apply couple Q.

Step 2:. Find support reactions and draw member F.B.D’s

Step 3: Prepare table:

Zone origin limits EI M ӘM/ӘQ

Step 4: Find displacement using Δj = ∫L0M/EI{ӘM/ӘQj} dx

Find ΔB :

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Zone Origin Limits EI M ӘM/ӘQ

BA B 0 to L EI –Q(x) – W(x2/2)-x

ΔB = ∫L0 M/EI (ӘM/ӘQ)dx = ∫L0 {-W(x2/2)}(-x)dx

= W/2EI{x4/4}L0

ΔB = WL4/8EI ( )

Q12) Find vertical deflection at C.

A12) EI = constant

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M Diagram

$C:\Users\TARUNA\Desktop\internship\february\11 feb2020\last\13.JPG$

From above Fig :

Table

Zone Origin limits EI M ӘM/ӘQj

CB C 0-2 EI -(80 + Q)x - x

BA C 2-5 EI -(80 + Q)x - x

-60(x-2)

ΔC = ∫20-(80)x/EI {-x}dx + ∫52{80x – 60(x-2) (-x)dx/EI}

= ∫2080x2/EIdx + ∫52{80x2 – 60x2 + 2x}dx/EI

= 1/EI{213.33 + 4200}

ΔC = 4413.33/EI( )

Q13) Find vertical deflection at point C.

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A13)

M Diagram

$C:\Users\TARUNA\Desktop\internship\february\11 feb2020\last\11.JPG$

Zone Origin Limit EI M ӘM/ӘQ

CB C 0 – 2 EI - (100 + Q)x -x

BA C 2 – 4 EI -(100 + Q)x -x

ΔC = ∫20 – 100x/EI(-x)dx + ∫42 -100x/EI(-x)dx

= ∫20100x2/EI dx + ∫42100x2/2EIdx

= 266.66 + 1866.6/2 = 266.66 + 933.33

For cantilever always write load from free end. For simply write load from supports.

ΔC = 1200/EI

Q14) Find slope at A and deflection at C.

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A14) M Diagram

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Reactions in M dia.

∑MA = VB(6) – 20 x 6 x 3 – 60(2) – Q = 0

VB = 80 + Q/6

∑Fy = VA + VB – 60 – 20 x 6 = 0

VA = 100 – Q/6

Zone Origin limits EI M ӘM/ӘQ

AC A 0 – 2 EI (100-Q/6)(x) 1 – x/6 + Q – 20(x2/2)

BC B 0 – 4 EI (80+Q/6)(x)-20x2/2 x/6

ϴA = ∫20{100(x) – 10x2}{1-x/6}dx/dEI + ∫40{80x – 10x2}x/6dx/EI

= 1/EI{135.55 + 177.78}

ϴA = 313.32/EI( )

$C:\Users\TARUNA\Desktop\internship\february\11 feb2020\last\8.JPG$

∑MA = VB(6) – 20(6)(3) – (60 + Q)2 = 0

VB = 80 + Q/3

∑Fy = VA + VB – (60 + Q) – 20(6) = 0

VA = -80 – Q/3 + 60 + Q + 120 = 0

VA = 100 + 2/3Q

Q15) Find deflection at point C

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A15)

M Diagram

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Zone Origin Limit EI M ӘM/ӘQ

AC A 0 – 2 EI (25 + Q/2)x x/2

BC B 0 – 2 EI (25 + Q/2)x x/2

ΔC = ∫20(25x)x/2dx/EI + ∫2025x(x/2)dx/2EI

= 33.33 + 16.66/EI

ΔC = 50/EI( )

Q16) Find deflection at D.

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A16)

∑MA = MA – (10+Q)(1) = 0

MA = 10 + Q

∑Fy = VA – (10 + Q) = 0 VA = 10 + Q

∑Fx = HA = 0

$C:\Users\TARUNA\Desktop\internship\february\11 feb2020\last\4.JPG$

Zone origin limits EI M ӘM/ӘQ

AB A 0 – 2 2EI (10+Q)x-(10+Q) x-1

CB C 0 – 1 EI -(10 + Q) -1

DC D 0 – 1 EI -(10 + Q)x - x

ΔDy = ∫20(10x - 10)(x-1)dx/2EI + ∫10 -10(-1)dx/EI + ∫10 -10x(-x)dx/EI

= 3.33 + 10 + 3.33/EI

ΔDy = 16.66/EI ( )

Q17) Explain Castiglione s second theorem in detail.

A17) Castiglione’s second theorem

Basics (1)

Beam fixed at both end is called as Fixed beam

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ii) Fixed beam is statically indeterminate

no. of unknown Reactions = R=6

no. of Equilibrium equation = E=3

Dsi – R-E = 6-3 =3

Note :

For beam subjected to vertical loads lonely-horizontal reactions are zero.

 For Fixed beam subjected to vertical loads only.

no. of unknown Reactions = R=4

no. of Equilibrium equation = E=2

Dsi – R-E = 4-2 =2

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For vertically downward load fixed and moments are begging in nature.

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