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SA

Unit - 2

Analysis of redundant pin jointed frames and multi-storied multi-bay 2-D rigid jointed frames

Q1) Explain the analysis of redundant trusses by unit load method for external loading?

A1) Consider the truss which has hinged supports at both ends.

External indeterminacy Number of unknown reaction (R) -equilibrium equation (E) 04-03= 01 Internal indeterminacy=m-2j+3-9-2x6+3=0

Degree of static indeterminacy = 01

Assuming vertical component at D as a redundant force

Then, restraint in that direction is to be removed and vertical force R, is treated as an additional unknown force acting on that structure. The condition for consistency is that turns of should have zero vertical displacement to represent the given truss.

As the truss is a determinate truss, its displacement can be evaluated and the consistency condition can be imposed on it.

 For finding the displacements, unit load method may be used because this method is ideally suited for finding the displacement in trusses.

Now, the total displacement in the truss may be split into two parts:

One due to the given loadings (i.e., P- Analysis) and the ether due to the redundant force Rp Let the forces developed in the members of the truss, due to given loadings be P Analysis and that due to unit load at D in the direction of Ro be K.

Then According to unit load method, the vertical displacement of B due to the given loading is given by

Vertical displacement due to RD

The displacement of B

But according to consistency condition the vertical displacement is equal to zero

 

Q2) Write down the procedure of analysis the indeterminate truss of one degree redundancy?

A2) Procedure to analyze the indeterminate truss of one degree redundancy

Step 1) Find degree of indeterminacy

Step 2) Select the redundant member

Step 3) Remove the redundant member and find the force P is members of the truss for the given load system

Step 4) Remove the load given system

  • Apply a unit load pair in place of redundant
  • Find the force K in the members of the structure
  • Step 5) Find and   for each member

    Step 6) Find redundant force

    Step 7) Find the force in any member by the reaction

     

    Q3) Analyze the frame shown in fig. for the force in various members. All the members may be assumed to have the same (l/AE)

    A3) Step 1) Degree of redundancy

    R-E= 3-3 = 0

    Degree of internal redundancy= m-(2j-3) = 6-(2x4-3) =1

    Total degree of redundancy= 0+1 = 1

    Step 2) Let AD be the redundant

    Remove member AD

    Step 3) P forces

  • Consider joint A in equilibrium
  • b.     Consider joint C in equilibrium

    c.      Consider joint D in equilibrium

    K forces

    Remove external load

    Apply pair unit load

  • Consider the joint A in equilibrium
  • B.     Consider joint C in equilibrium

    Find

     

    Member

    P

    K

    PK

    K^2

    Final forces (F)

    =P+KR

    Remark

    AB

    0

    +0.71

    0

    0.504

    -0.62

     

    +

    Tension

    -Compression

    AC

    0

    +0.71

    0

    0.504

    -0.62

    AD

    0

    -1

    0

    1

    0.88

    BD

    5

    +0.71

    +3.55

    0.504

    4.37

    BC

    0

    -1

    0

    1

    0.88

    CD

    0

    -0.71

    0

    0.504

    -0.62

     

    Step 6) Find redundant force R

    Step 7) Final forces

    Final force in all member is given by F= P+KR

    Final force is shown in table

     

    Q4) What do you mean by lack of fit?

    A4) Lack of fit is defined as little longer or little shorter length of any member in truss.

    If any one member of a redundant frame had lack of fit stress will be induced an all members of the redundant frame when that member is forced in position. The redundant member (R) due to lack of fit can be determine by using the following formula

    Where

     λ= Shorting or excess of member in mm

    + ve = if member is short in length

    -Ve if members excess in length

    Final force in various are determined by the formula F= P+KR

     

    Q5) Find the force of all member of the frame shown in fig. If the member BC short in length by 10 mm and is force into position Take E= 2x10^5 N/mm2 All member have same area of cross section of 10 mm2

    A5)

  • Degree of indeterminacy
  • Dsi = R-E= 0

    2.     P force

    m- (2j-3) = 1

    There is no external force acting on truss

    p force due to external loading in various member will be zero

     

    3.     K force

    Remove member BC

    Apply unit load at joint c

    Find the K force

     

    Member

    L

    P

    K(KN)

    K^2 L

    Final force F= P+KR

    Remark

    AB

    4000

    0

    -0.80

    2560

    9.26

     

    += Tension

    - = Compression

    AC

    3000

    0

    -0.60

    1080

    -6.94

    AD

    5000

    0

    +1.00

    5000

    +11.57

    BD

    3000

    0

    -0.60

    1080

    -6.74

    BC

    5000

    0

    +1.00

    5000

    +11.57

    CD

    4000

    0

    -0.80

    2560

    -9.26

     

     

     

     

    17280

     

     

    Step 4) Redundant force R

    Where,

    E= 2x 10 5 N/mm2

    A= 100 KN/mm2

     

    Q6) Explain the concept dinking of support and temperature changes (indeterminacy up to second degree)?

    A6) 1. Sinking of support:

  • Find final moments
  • Dsi = 1 (Internal)

    24

    DSi = (m+r) – 2 J

            = (8 +3) – 2 (5)

            = 11 – 10

    DSi = 1 (total)

    (DSi) external = R – 3 = 0

    (DSi) internal = 1

     

    2) Selection of ‘Q’

     Let Q = FBE (Tensile)

     

    3) P – Dia & K-Dia

    25

    26

     

    15 diagonals in tension then all sides are in compression + DA = - 1 2

     Sin 270 = 1 sin45

                    = -1 x -1

                     = 1

    P-analysis

    ∑MA=0

    -50 x 2 + VC x 4 = 0

    VC = 25KN

     

     

    ∑FY=0

    HA .VA + VC = 0

    VA = - 25 KN

     

    ∑FX=0

        HA = - 50KN

     

    K-analysis:

     

    Sr. No.

    Member

    L (mm)

    P (KN)

    K

    PKL

    (x103)

    K2L

    (x103)

    Q

    F=P + KQ

    1.

    AB

    2000

    25

    -1/2

    -35.35

    1

     

    34.9

    2.

    BC

    2000

    25

    0

    0

    0

     

    25

    3.

    CD

    20002

    -252

    0

    0

    0

     

    -25

    4.

    DE

    2000

    -50

    -1/2

    70.71

    1

    -14

    -40

    5.

    EA

    20002

    0

    -1/2

    0

    1

     

    9.9

    6.

    DB

    2000

    0

    -1/2

    0

    1

     

    9.9

    7.

    AD

    20002

    252

    1

    100

    22

     

    21.35

    8.

    BE

    20002

    0

    1

    0

    22

     

    -/4

     

     

     

     

     

    135.36 x103

    9.66 x103

     

     

     

    Q =      ∑ PKL

                     AE

                  ∑ K2L

                     AE

     

    = - 135.36 x103

    9.66 x103

    = 14.01KN

    = 14.01 KN (C)

    27

     

     

    2. Temperature changes (indeterminacy up to second degree)

    Change in temperature cause change in length of a member. In redundant trusses the change in length of any member gives rise is force all other members.

    Let temperature of member AB decreases by T0c then the contraction of the member AB is given by

    But the free contraction is not possible in the truss. Hence tensile force of magnitude R develops in the member AB this cause movement of joint A and B in the truss. The compatibility condition demand elongation of member AB and movement of joint A and B the value of R is given by

    Where,

    K= force in various member due to unit load applied for the member under consideration.

  • Find final member force & reaction
  • 28

     DSi = (M + R) – 2J

            = (6+4) – 2 (4)

            = 10 – 8

            = 2

    DSi (ext) = 1

    (DSi) int = 1

     

    2) Q = FAD (Tensile)

    3)  Figure

    30

     

     

    Sr. No.

    Member

    Length

    Area

    K

    K2L/A

     

    Q

    P = QK

    1.

    AB

    4000

    2000

    4/3

    3.56

     

    11.86

    2.

    BC

    3000

    1500

    1

    2

     

    8.90

    3.

    CD

    4000

    2000

    4/3

    3.56

    8.90

    11.86

    4.

    DA

    3000

    1500

    1

    2

     

    - 8.90

    5.

    AC

    5000

    2500

    -5/3

    5.56

     

    -14.84

    6.

    BD

    5000

    2500

    -5/3

    5.56

     

    -14.84

     

     

     

     

     

    22.24

     

     

     

    Q =   -        Lt

                    ∑ K2L/AE

     

         =   -     - 1.1. x 10-5 x 30 x 3000

                              22.24 / 200

       = 8.90 KN (T)

     

    Diagram

    29

     

    Q7) Explain the approximate methods of analysis of multi-storied multi-bay 2-D rigid jointed frames by Cantilever method?

    A7) Portal method

    This method is applicable for low rise structures. In low rise structures shear deformations are dominant, therefore this method simplifying assumptions regarding horizontal shear in column. Each bay of a structure as a portal frame and horizontal force is distributed equally among them. This means each interior column takes twice as much as the exterior column

    Assumptions:

  • The points of inflection or point of contra flexure are located at the mid height of each column. However, for hinged column base it assumed at hinged column base
  • The point of inflection is located at mid span of beam
  • The horizontal shear at any floor is divided among all the basis that each interior column takes twice as much as the exterior column
  • Steps

    1)     Assume first assumption -> mark point of contra flexure at centre of each column & beam

    2)     Apply 2nd assumption -> Horizontal shear is double at interior column.

    3)     Find P& Q force.

    4)     Find moment at beam = always same as moment of column.

    5)     Find shear force at beam=moment at

     

    Q8) Analyze the given figure by portal method

     

    A8) Step 1) 1st Assumption-> Mark point of contra flexure of each member

    Step 2) 2nd Assumption -> Interior Horizontal shear is double than (extreme) lost member.

    P+2P+P=12

    P=3KN

    Q+2Q+Q=12+24

    Q=9KN

    Step 3) Moment at the end of column

    Step 4) Moment at column

    Step5) Moment at the end of roof Beam

    Step 6) Moment at the end of floor Beam

    Step 7) Shear force in Beam

    Similarly,

     

    Q9) Analyze the given figure by portal Method

    A9) Mark point of contra flexure at centre of each beam& column.

    Horizontal share is double at interior column.

    Find p & q

    P+2P+P=20

    P=5kN

    Q +2Q +Q=50

    Q=12.5KN

    Moment at column                          AD= =

    MBE=

    MCF=

    MDG=MGD=

    MEH=MHE=

    MFI=MIF=

     

    Q10) Analyze the given figure by portal method

     

    A10)                                                         P+2P+2P+P=25

    4P=25

    P=4.16

    Q+2Q+2Q+Q=60

    6Q=60

    Q=10KN

    MAE=MEA==4.16

    MBF=MFB=

    MCG=MGC=

    MDH=MHD=

    MEI=MIE==10

    MFI=MIF=

    MGK=MKG=

    MHL=MLH=

    Shear force

    SFAB=20.8/2=10.4

    SFBC=20.8/3=6.93

    SFCD=20.8/4=5.2

    SFEF=60.8/2=30.4

    SFFG=60.8/3=20.26

    SFGH=60.8/4=15.3

    BMD

    Q11) Analyze the given figure by portal method

     

    A11)                                            P=3.33      Q=3.33     R=10       

     

    Q12) Analyze the frame shown in figure below by portal method

    A12) 1) Apply 1st assumption

    2) Apply 2nd Assumption-> Horizontal shear is double at interior member

    P+2p+p=25

    P=6.25KN

    Q+2Q+Q=25+50

    Q=18.75KN

    3) Moment at the end of column

    4) Moment at the end of floor column

    5) Moment at beam

    6) Moment at end of beam

    7) Shear force at (beam) floor

     

    Q13) What are the assumption and application of cantilever method?

    A13) The cantilever method is based on the following assumption:

          Assumption

  • There is a point of contra flexure at the centre of each member.
  • The intensity of axial stress in each column storey is proportional to the horizontal distribution of that column from the centre of gravity of all columns of the storey under consideration.
  • Application:

  • The method is more applicable to high rise structure since bending action is more predominant in these cases
  • The column having different area of cross section can be taken into account.
  •  

    Q14) Determine the approximate value of bending moment shearing force and axial force in a two bay two storey portal frame shown in fig. using cantilever method assume equal cross-sectional areas for all columns?

    A14) Step-1 Find centroidal distance

    Take moment about A

    Step-2> As per 2nd assumption

    Vertical stresses in column are proportional to their absence from the C.G. of the columns in that storey.

    1>  Consider top storey

    Assume is downward & is upward.

    Put values

    Taken about point of contra flexure i.e., at 0 point

    Clockwise +ve               Anti clockwise  -ve

    Upward      +ve                Downward       -ve

    Consider lowest storey

    Assume are downward

    are upward

    As per 2nd assumption

    Take a moment about 0 point.

    Step-3) To find S.F. in proof Beam

    S.F. in AB=1KN

    SF in BC=1+0.42=1.42

    SF in CD=1+0.42-0.28=1.14

    S.F. in EF=6-1=5KN

    S.F. in FG=6+2.57+1-0.48=7.15KN

    SF in GH=6+2.37-1.71-1-1-0.42+0.28=5.72KN

    Step-4) To find Moment in roof beam

    Formula => Moment in roof beam=S.F. in that Beam*Balt of Beam span

    ii> To find moment in floor beam

    Step -5) To find Moment in column

    Moment in column of floor beam

    Step-6) S.F. in column

    S.F. AE=2/1.5=1.33KNM

    SF BF=5.55/1.5=3.7KN

    S.F. CG=6.9/1.5=4.6KN

    S.F. DH=3.42/1.5=2.28KN

    S.F. in column EI=1.33*1.5+x+2=10

    X=4KN

    S.F. in column FJ=3.725*1.5+x+2= (17.85+10)

    X=11.128KN

    S.F. in column GK=4.667*1.5+x*2= (17.14+17.85)

    X=13.99KN

    S.F. in column HL=2.28*1.5+x+2= (17.142)

    X=6.86KN