Unit - 2

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

2.1 Analysis of redundant trusses by unit load method for external loading

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

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

Step 5) Find and   for each member

Step 6) Find redundant force

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

Problems:

Solution:

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

b.     Consider joint C in equilibrium

c.      Consider joint D in equilibrium

K forces

Remove external load

Apply pair unit load

B.     Consider joint C in equilibrium

Find

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

Key takeaway:

Steps:

Find

and

  for each member

2.2 Lack of fit

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

Problem: 1] 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

Solution:

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

Step 4) Redundant force R

Where,

E= 2x 10 5 N/mm2

A= 100 KN/mm2

2.3 Sinking of support and temperature changes (indeterminacy up to second degree)

1. Sinking of support:

1. Find final moments

Dsi = 1 (Internal)

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

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:

Q =      ∑ PKL

                 AE

              ∑ K2L

                 AE

= - 135.36 x103

          9.66 x103

= 14.01KN

= 14.01 KN (C)

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.

 DSi = (M + R) – 2J

        = (6+4) – 2 (4)

        = 10 – 8

        = 2

DSi (ext) = 1

(DSi) int = 1

2) Q = FAD (Tensile)

3)  Figure

Q =   -        Lt

                ∑ K2L/AE

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

                          22.24 / 200

   = 8.90 KN (T)

Diagram

2.4 Approximate methods of analysis of multi-storied multi-bay 2-D rigid jointed frames by Cantilever method and Portal method

1. 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:

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

Problems:

1)     Analyse the given figure by portal method

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,

2.     Analyse the given figure by portal Method

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=

3.     Analyse the given figure by portal method

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

4.     Analyze the given figure by portal method

P=3.33      Q=3.33     R=10       

5.     Analyze the frame shown in figure below by portal method

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

The cantilever method is based on the following assumption:

      Assumption

Application:

Problem:

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

Key takeaway:

Steps:

References:

1. Structural Analysis: Deodas Menon---Narosa Publishing House.

2. Structural Analysis: Thandavamoorthy---Oxford University Press.

3. Structural Analysis: A Matrix Approach by Pundit and Gupta, McGraw Hills.

4. Structural Analysis by Hibbler, Pearson Education.

5. Structural Analysis: M. M. Das, B. M. Das---PHI Learning Pvt Ltd. Delhi.

6. Fundamentals of Structural Analysis: 2nd ed---West---Wiley.

7. Theory of Structures: Vol. I & II by B. C. Punmia, Laxmi Publication.

8. Theory of Structures: Vol. I & II by Peru mull & Vaidyanathan, Laxmi Publication.

9. Fundamentals of Structural Analysis: K. M. Leet, Vang, Gilbert—McGraw Hills

10. Matrix Methods for structural engineering. by Gere, Weaver.

11. Introduction to the Finite element method, Dr. P.N. Godbole, New Age Publication, Delhi.

12. Finite element Analysis, S.S. Bhavikatti, New Age Publication, Delhi.

13. Basic Structural Analysis: Wilbur and Norris.