INTRODUCTION TO COLUMN ANALOGY METHOD.docx

COLUMN ANALOGY METHOD

315

CHAPTER = 7 INTRODUCTION TO COLUMN ANALOGY METHOD The column analogy method was proposed by Prof. Hardy Cross and is a powerful technique to analyze the beams with fixed supports, fixed ended gable frames, closed frames & fixed arches etc., These members may be of uniform or variable moment of inertia throughout their lengths but the method is ideally suited to the calculation of the stiffness factor and the carryover factor for the members having variable moment of inertia. The method is strictly applicable to a maximum of 3rd degree of indeterminacy. This method is essentially an indirect application of the consistent deformation method. This method is based on a mathematical similarity (i.e. analogy) between the stresses developed on a column section subjected to eccentric load and the moments imposed on a member due to fixity of its supports .*(We have already used an analogy in the form of method of moment and shear in which it was assumed that parallel chord trusses behave as a deep beam). In the analysis of actual engineering structures of modern times, so many analogies are used like slab analogy, shell analogy & girder analogy etc. In all these methods, calculations are not made directly on the actual structure but, infact it is always assumed that the actual structure has been replaced by its mathematical model and the calculations are made on the model. The final results are related to the actual structure through same logical engineering interpretation. In the method of column analogy, the actual structure is considered under the action of applied loads and the redundants acting simultaneously. The load on the top of the analogous column is usually the B.M.D. due to applied loads on simple spans and therefore the reaction to this applied load is the B.M.D. due to redundants on simple spans. P1

P2

WKN/m

Ma A

B MB L E1=Constt. (a) Given beam

0

(d) Loading on top of analogous column, Ms diagram, same as(b). 1 (Unity)

0 (b) B.M.D. due to applied loads, plotted on the compressin side.

0

L

0

(e) X-section of analogous column.

MB MA (c) B.M.D. due to redundant, plotted on the compression side.

(f) Pressure on bottom of analogous column, Mi diagram.

The resultant of B.M.D’s due to applied loads does not fall on the mid point of analogous column section which is eccentrically loaded.

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THEORY OF INDETERMINATE STRUCTURES

Msdiagram = BDS moment diagram due to applied loads. Mi diagram = Indeterminate moment diagram due to redundants. If we plot (+ve) B.M.D. above the zero line and (-ve) B.M.D below the zero line then we can say that these diagrams have been plotted on the compression side. (The conditions from which MA & MB can be determined, when the method of consistent deformation is used, are). From the Geometry requirements, we know that (1) the change of slope between points A & B = 0; or sum of area of moment diagrams between A & B = 0 (note that EI = Constt:), or area of moment diagrams of fig.b = area of moment diagram of fig..c. (2)

the deviation of point B from tangent at A = 0; or sum of moment of moment diagrams between A & B about B = 0, or Moment of moment diagram of fig.(b) about B = moment of moment diagram of fig.(c) about B. above two requirements can be stated as follows. (1)

total load on the top is equal to the total pressure at the bottom and;

(2)

moment of load about B is equal to the moment of pressure about B), indicate that the analogous column is on equilibrium under the action of applied loads and the redundants.

SIGN CONVENTIONS:It is necessary to establish a sign convention regarding the nature of the applied load (Ms diagram) and the pressures acting at the base of the analogous column (Mi-diagram.) FIRST SIGN CONVENTION:Load ( P) on top of the analogous column is downward if Ms diagram is (+ve) which means that it causes compression on the outside or vice- versa.. Outside Inside Outside

C T

2ND SIGN CONVENTION :Upward pressure on bottom of the analogous column ( Mi - diagram) is considered as (+ve). 3RD SIGN CONVENTION :Moment (M) at any point of the given indeterminate structure ( maximum to 3 rd degree) is given by the formula.

COLUMN ANALOGY METHOD

317

M = Ms - Mi, which is (+ve) if it causes compression on the outside. PROBLEM :-

Determine the fixed-ended moments for the beam shown below by the method of column

analogy. SOLUTION: W/Unit length. A

B

EI=Constt. L

2 WL 8 + 3 WL 12 3 WL 12

0

0

L 2 WL 0 12 2 WL 12 0

+

WL2 /24

Ms-diagram (B.M.D. due to applied loads on B.D.S.) Loading on top of analogous column.

X-section of 1 analogous column Mi-diagam Pressure on bottom of analogous column.(uniform as resultant falls on the mid point of 2 analogous column section WL 12 2 WL 12

Pressure at the base of the column = P A = WL3 12(Lx1) Mi = WL2 12 (MS)a = 0 Ma = (Ms - Mi)a = 0 - WL2 12 Ma = - WL2 12 Mb = (Ms-Mi)b = (0 - WL2) = - WL2 12 12 Mc = (Ms - Mi)c = WL2 - WL2 8 12 Mc = 3 WL2 - 2 WL2 = WL2 24 24

318

THEORY OF INDETERMINATE STRUCTURES

IF B.D.S. IS A CANTILEVER SUPPORTED AT ‘A’. W/unit-length A

EI=Constt. L

B

3

L/4

WL 6

3/4L

0

0 Ms-diagram

2

WL 2

L/4

3 WL 6

L/2 yo M 1

L

yo

X-section of analogous column. 3 Eccentric load wl /6 acts on centre of analogous column x-section with an associated moment as well

A = bh = L x WL2 = WL3 (n+1) 2 (2+1) 6 X = b = L = L (n+2) (2+2) 4 X =  Mxdx  Mdx  Mdx = oL (- WX2 ) dx = - W X3 oL = - WL3 2 2 3 6 L 2 L 3  Mxdx = o (- WX ) dx = o - WX dx 2 2

( Same as above)

= - W  X4 oL = - WL4 2 4 8 X=  Mxdx  Mdx X = - WL4 . 6 = 3 L. (from the origin of moment 8 (-WL3) 4 expression or moment area formula can be used) NOTE : Moment expression is independent of the variation of inertia.

Properties of Analogous Column X-section :A =L1=L

COLUMN ANALOGY METHOD

319

I yo yo = L 3 12 C = L 2 A e’=M = ( WL3) ( L ) = WL4 6 4 24

( L/4 is distance between axis yo- yo and the centroid of Ms diagram)

(Mi)a = P  Mc ( P is the area of Ms diagram ) A I = -WL3 - WL4 . L . 12 (Load P on analogous column is negative) 6.L 24 . 2 . L3 = - WL2 - WL2 ( Reaction due to MC/I would be having the same 6 4 direction at A as that due to P while at B these = -2WL2 - 3 WL2 two would be opposite) 12 = - 5 WL2 12 (Ms)a = - WL2 2 Ma = (Ms - Mi)a = - WL2 + 5 WL2 2 12 = - 6 WL2 + 5 WL2 12 Ma = - WL2 12 Mb = (Ms - Mi)b (Mi)b = P  Mc A I = - WL3 + WL4  L  12 6L 24  2  L3 = -WL2 + WL2 6 4 = - 2WL2 + 3 WL2 12 = WL2 12 (Ms)b = 0 Mb = (Ms - Mi)b = 0 - WL2 = - WL2 12 12

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THEORY OF INDETERMINATE STRUCTURES

PROBLEM :-

Determine the F.E.Ms. by the method of column analogy.

SOLUTION:- CASE 1 ( WHEN BDS IS A SIMPLE BEAM ) P b

a L Pab L +

L+a 3

Ms-diagram

L+b 3 1 (P ab) P ab xL= 2 2 L a

P ab 2

M

1 x-section of analogous column L

e = L - (L + a) = 3 L - 2 L - 2a = ( L - 2 a) 2 3 6 6

( The eccentricity of load w.r.t mid point of analogous column)

M = (Pab) (L - 2a ) = Pab (L - 2a) 2 6 12

Properties of Analogous Column X - section . A =L1=L I = L2 12 C = L 2 (Mi)a = P  Mc A I = Pab + Pab (L - 2a)  L  12 2 L 12 2  L3 = Pab + Pab (L - 2a) 2 L 2 L2 = PabL + PabL - 2 Pa2b 2 L2 = 2 PabL - 2 Pa2b 2 L2

COLUMN ANALOGY METHOD

(Mi)a

321

= PabL - Pa2b L2 = Pab (L - a) L2

a+b=L b=L-a

= Pab . b L2 (Mi)

= Pab2 L2

(Ms) a = 0 Ma = (Ms - Mi) a = 0 - Pab2 L2 Ma =- Pab2 L2 The (-ve) sign means that it gives us tension at the top when applied at A. (Mi)b = P  MC A I = Pab - Pab (L - 2a)  L  12 2L 12L2 2  L3 = Pab - Pab (L - 2a) 2L 2L2 = PabL - PabL + 2Pa2b 2L2 = 2Pa2b 2L2 (Mi)b = Pa2b L2 (Ms)b = 0 Mb = (Ms - Mi)a = 0 - Pa2b L2 2 Mb = - Pa b L2 The minus sign means that it gives us tension at the top.

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THEORY OF INDETERMINATE STRUCTURES

CASE (2) If B.D.S. is a cantilever supported at A:P a

A

0

b

B

EI=Constt L

1 Pa(a) 2 Pa 2 2

0 Ms-diagram

Pa 2 Pa 2

e M

1 x-section of analogous column L L/2

e=L - a 2 3 Pe

= ( 3L - 2a ) 6

=M = Pa2 ( 3L - 2a ) = Pa2 (3L - 2a) 2 6 12

Properties of Analogous Column section :- A = L , I = L3 , C = L 12 (Mi)a = P  MC A I = - Pa2 - Pa2 (3L - 2a) . L . 12 2L 12 . 2 . L3

2

(Due to upward P= Pa2/2, reaction at A and B is downwards while due to moment

= - Pa2 - Pa2 (3L - 2a) 2L 2L2

, reaction at B is upwards while at A it is downwards. Similar directions will have

= -Pa2L - 3Pa2L + 2Pa3 2L2

the same sign to be additive or vice-versa)

= -4 Pa2L + 2Pa3 2L2 = -2Pa2L + Pa3 L2 = Pa2 (a - 2L) L2

COLUMN ANALOGY METHOD

323

= -Pa2 (2L - a) L2 (Mi)a = -Pa2 (L + b) L2 (Ms)a = - Pa Ma

= (Ms - Mi)a = -Pa + Pa2(L + b) L2 = - PaL2 + Pa2 L + Pa2b L2 = - PaL (L - a) + Pa2 b L2 = - PabL + Pa2 b L2 = - Pab (L - a) L2

Ma

(Mi)b

(Mi)b (Ms)b Mb

Mb

= - Pab . b L2 = - Pab2 L2

( Same result as was obtained with a different BDS)

= P  MC A I = - Pa2 + Pa2 (3L - 2a) 2L 2L2 2 2 = - Pa L + 3Pa L - 2Pa3 2L2 2 = 2 Pa L - 2Pa3 2L2 2 = Pa L - Pa3 L2 2 = +Pa (L - a ) L2 2 = Pa b L2 = 0 = (Ms - Mi)b = 0 - Pa2 b L2 2 = - Pa b ( Same result as obtained with a different BDS) L2

324

THEORY OF INDETERMINATE STRUCTURES

PROBLEM:-

Determine the F.F.Ms. by the method of column analogy.

SOLUTION:w/unit length B

A L/2

L/2 EI = Constt

If B.D.S. is a cantilever supported 0 at b.

WL3 48 L 8

0

Ms-diagram

= WL x L 2 4 2 WL 8

WL3 48 e=3/8 L M

1 L

Analogous col. section.

e = L - L = 4L - L = 3L 2 8 8 8 Pe=M = WL3  3L = WL4 48 8 128

Where P= Area of Ms diagram=WL3/48

Properties of Analogous column section. A = L, I = L3 and C = L 12 2 Step !: Apply P= Area Of BMD(Ms diagram ) due to applied loads in a BDS at the center of analogous column section i.e. at L/2 from either side. Step 2: The accompanying moment Pe, where e is the eccentricity between mid point of analogous column section and the point of application of area of Ms diagram, is also applied to at the same point alongwith P. Step 3: Imagine reactions due to P and M=Pe. Use appropriate signs. (Mi)a = P  MC ( Subtractive reaction at A due to P) A I = - WL3 + WL x L x 12 ( P is upwards, so negative. Reactions due to this P 48L 128x 2 x L3 at A and B will be downwards and those due to moment term will be upward at A and downward = - WL2 + 3WL2 at B. Use opposite signs now) 48 64 = - 4WL2 + 9WL2 192

COLUMN ANALOGY METHOD

325

= + 5 WL2 192 (Ms)a = 0

( Inspect BMD drawn on simple determinate span)

Ma = (Ms - Mi)a = 0 - 5WL2 192 Ma = - 5WL2 192 (Mi)b = P  MC A I = -4 WL2 - 9WL2 192 = - 13 WL2 192 (Ms)b = - WL2 8 Mb = (Ms - Mi)b

( Additive reactions at B)

= - WL2 + 13 WL2 = -24 WL2 + 13 WL2 8 192 192 Mb = -11 WL2 192 Determine the F.E M’s by the method of column analogy for the following loaded beam..

PROBLEM:SOLUTION:-

W/Unit length B

A L/2

(L)WL 192 b X= n+2

X=

L 10 0

0 Ms-diagram

L

3 WL 192

2(3+2) L 10

WL 192

3

=

=

L/2 EI=Constt:

bh A= n+1

e

M

2 (1xWxL) x L (L) = WL 2 2 3 2 24 1

L

Analogous column section

326

THEORY OF INDETERMINATE STRUCTURES

e = L - L = 5L - L = 4L =2 L 2 10 10 10 5 M = ( WL3 )  ( 2 L ) = WL4 192 5 480

Properties of Analogous column section. A= L , I= L3/12, C=L/2 (Mi)a

= P  MC A I

(Mi)a

= - WL3 + WL4  L  12 ( Subtractive reaction at A) 192L 480  2  L3 = - WL2 + WL2 192 80 = - 80WL + 192 WL2 15360 = 112 WL2 15360

( Divide by 16)

(Mi)a = 7 WL2 960 (Ms)a = 0 Ma

= (Ms - Mi)a

Ma

= 0 - 7 WL2 = - 7 WL2 960 960

(Mi)b = P  MC A I = - WL2 - WL2 192 80 = - 80 WL2 - 192 WL2 15360 = - 272 WL2 15360 = - 17 WL2 960

COLUMN ANALOGY METHOD

327

(Ms)b = - WL2 24 Mb = (Ms - Mi) b = - WL2 + 17 WL2 24 960 = - 40 WL2 + 17 WL2 960 Mb = - 23 WL2 960

Note : After these redundant end moments have been determined, the beam is statically determinate and reactions , S.F, B.M, rotations and deflections anywhere can be found.

328

THEORY OF INDETERMINATE STRUCTURES

STRGIGHT MEMBERS WITH VERIABLE CROSS - SECTION. PROBLEM:- Determine the fixed - end moments for the beam shown by the method of column analogy

SOLUTION:-

90kn 3kn/m A

4m B

6m I=2

10m I=1 2

3.83m Ms dia. due EI to U.D.L. 0a only.

Ms dia due EI to point load only.

3x16 = 96 8

P2 b90

3kn/m

45 0

8m P4 135 4m

24kn 6m 10m 24kn M=24x6-3 x (6)2 2 90kn =90kn-m 4m 12m

8m

P1

90x12x4 16

= 270

67.5 yo

P3 x

1/2

16+4 3 =6.67m

67.5kn

M=22.5x6 =135kn-m 1

M 9.15m

90x4 =22.5kn 16

6.85m

Analogous column x-section.

In this solution, two basic determinate structures are possible. (1) a simply supported beam. (2) a cantilever beam. This problem is different from the previous one in the following respects. (a) Ms - diagram has to be divided by a given value of I for various portions of span. (b) The thickness of the analogous column x - section will also vary with the variation of inertia. Normally, the width 1/EI can be set equal to unity as was the case in previous problem. (c) As the dimension of the analogous column x - section also varies in this case, we will have to locate the centroidal axis of the column and determine the moment of inertia about it.

COLUMN ANALOGY METHOD

(1) By choosing a simple beam as a B.D.S. P1 = 2  16  96 = 1024 KN ( Corresponding to entire BMD due to UDL) 3 Mdx = o6 (24x - 1.5 x2) dx ( Simply supported beam moment due to UDL) = 12x2 - 0.5x3o6 = 12  36 - 0.5  216=432-108=324 area abc = 324 Mxdx = o6 (24x - 1.5x2) x dx = o6 (24x2 - 1.l5x3) dx =  24 x3 - 1.5 x4 o6 = 8  63 - 1.5  64 3 4 4 = 1242 x =  Mxdx = 1242 = 3.83 m from A.  Mdx 342 P2 = 1 ( area abc) = 324 = 162 KN( To be subtracted from Ms diagram ) 2 2 P3 = 1  16  270 = 2160 KN ( Area of BMD due to concentrated Load) 2 P4 = 1  6  67.5 = 202.5 KN ( To be subtracted from Ms diagram ) 2

Properties of Analogous column x - section. Area = A = 1  10 + 1  6 = 13 m2 2

x = xdA = (1  10) 5 + (1/2x 6  13) A 13 = 6.85 m ( From point B) . It is the location of centroidal axis Yo-Yo. Iy0 y0

= 1  103 + 10(1.85)2 + 0.5  63 + (0.5  6)  (6.15)2 = 240 m4 12 12

Total load to be applied at the centroid of analogous column x - section. = P1 + P3 - P2 - P4 = 1024 + 2160 - 162 - 202.5 = 2819.5 KN

329

330

THEORY OF INDETERMINATE STRUCTURES

Applied Moment about centroidal axis = M = + 1024 (1.15) + 2160 (0.18) - 162 (5.32) - 202.5 (5.15) = - 1116 KN-m The (-ve) sign indicates that the net applied moment is clockwise. (Mi)a

= P  MC A I

( subtractive reactions at A)

= 2819.5 - 1116  9.15 13 240 = + 174.34 KN-m (Ms)a = 0 Ma = (Ms - Mi)a = 0 - 174.34 = - 174.34 KN-m (Mi)b = 2819.5 + 1116  6.85 ( Note the difference in the values of C for points A and B.) 13 240 = + 248.74 KN-m

(Ms)b = 0

Mb = (Ms - Mi)b

= 0 - 248.74

= - 248.74 KN-m

The -ve sign with Ma & Mb indicate that these cause compression on the inside.

COLUMN ANALOGY METHOD

331

PROBLEM:- Determine the F.E.Ms. by the method of column analogy. SOLUTION:90kn 3kn/m A

1 Chosing a simple beam as a.B.D.S.

4m B

IC 6m

2IC 3m P2 1.95 m

3x13 8 =63.4

58.5 b

d

54

c = 29.25 e

a

P3 2.58 m

Ms EI

dia. due to U.D.L..

f 6.5 m 90x9x4 =249.23 13 P6 Ms dia. due to pt load. 2.67 EI 124.62 m

P1 P5 83 41.5

2m

(13+4) yo P 3 =5.67m 175.9kn 4 ½

x 877.6kn-m

1

1/2 Analogous column x-section

6.66m

6.34m yo

3KN/m 19.5

13m

19.5

(M3)L

= 19.5 x 3 - 1.5(3)2 = 45 KN-m ( 3m from A )

(M4)R

= 19.5 x 4 - 3 (4)2 = 54 KN-m ( 4m from B) 2 9 0

9 m 2 7 .6 9

4 m 6 2 .3 0 7

(M3)1 = 27.69  3 = 83 KN-m ( 3m from A) ( M4)R = 62.307x4=249.22 Mdx

= area abc = o3 (19.5 x - 1.5 x2)dx = 19.5 x2 - 1.5 x3 o3 = 74.25 2 3

332

THEORY OF INDETERMINATE STRUCTURES

Mxdx = o3 (19.5 x2 - 1.5 x3) dx = 19.5 x3 - 1.5 x4o3 3 4 = 145.12 x

= 145.12 74.25

= 1.95.m ( From point A as shown )

Area def = Mdx = o4 (19.5x - 1.5 x2)dx = 124 Mxdx = o4 (19.5 x2 - 1.5 x3)dx = 19.5x3 - 1.5 x4o4 3 4 = 320 x

P1 P2

= 320 = 2.58 m ( From point B ) 124 = 2  63.4  13 = 549.5 KN( Due to entire BMD due to UDL ) 3 = 1 (area abc) = 1 (74.25) = 37.125 KN ( To be subtracted ) 2 2

P3

= 1 (area def) = 1 (124) = 62 KN ( To be subtracted ) 2 2

P4

= 1  249.23  13 = 1620 KN ( Entire area of BMD due to point load) 2

P5

= 1  41.5  3 = 62.25 KN ( To be subtracted ) 2

P6

= 1  4  124.62 = 249.23 KN ( To be subtracted ) 2

Properties of Analogous column x - section. A = 1  4 + 1  6 + 1  3 = 9.5m2 2 2 x x Iyoyo

= (0.5  6)  2 + (1  6)  7 + (0.5  3)  (11.5) 9.5 = 6.66 ( From point B) = 0.5  43 + (0.5  4)(4.68)2 + 1  63 + (1  6)(0.34)2 12 12 + 0.5  32 + (1.5)(4.84)2 12 = 101.05

COLUMN ANALOGY METHOD

333

Total concentric load on analogous column x – section to be applied at centroidal column axis ) = P1 - P2 - P3 + P4 - P5 - P6 = 1759 KN Total applied moment at centroid of analogous column. = 549.5 (0.16) + 37.125 (4.39) - 62(4.08) + 1624 (0.99) + 62.25 (4.34) - 249.2 (3.99) = + 877.6 clockwise. (Mi)a = P + MC A I

( Reactions due to P and M are subtractive at A)

= 1759 - 877.6  6.34 9.5 101.05 = + 130 KN-m (Ms)a = 0 Ma = (Ms - Mi)a = 0 - 130 = - 130 KN-m (Mi)b = P  MC A I = 1759 + 877.6  6  6.66 9.5 101.05 = + 243 KN-m (Ms)b = 0 Mb = (Ms - Mi)b = 0 - 243 Mb = - 243 KN-m

( Reactions due to P and M are additive at B)

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THEORY OF INDETERMINATE STRUCTURES

Now the beam has become determinate. Let us now solve the same question by choosing another BDS (2) Choosing cantilever supported at B as a B.D.S:3KN/m 90KN A

B 3m

6m

3m P2 bh A = n+1 13x253.5 = 3 =1098.5 X' =

X' =

2.25m

b = n+2

0f

60.79 C

P2= 6.75

13.5

126.75 3x 13x13 P4=720 g =253.5

e121.5

Ms/EI diagram due to u.d.l

4x360 =720 2 4 3

3.25m

d

b

0a

b 13 = 4 =3.25 n+2

A= bh = n+1

P1=1098.5

P3= 367

=1.33

1.37m

180 90x4=360

Ms/EI diagram due to point load

yo

1/2

1.33m

1089.75Kn

Ps=360KN

3894KN-m

1

1/2 6.66m

6.34m yo

Analogous column section

3KN/m

4m

A

B 3m

10m

39

253.5

Area abc = Mdx = 3 (- 3 x2)dx o 2 = - 1.5 x3 3 = 0.5  33 = - 13.5 3 o 3 Mxdx =  (1.5x3)dx = - 1.5x43 o 4 o = - 30.375

( Upwards to be subtracted)

COLUMN ANALOGY METHOD

335

Location of centroidal axis from B: ( 1/2x3 + 1x6+1/2x4)X =(1/2x4x2+1x6x7+1/2x3x11.5) 9.5X’= 63.25 0r X’ = 6.66m from B or 6.34 m from A. x

= - 30.375 = 2.25 m - 13.5

( From A)

Area defg = Mdx = o4 (39x - 253.5 - 1.5x2)dx = 39 x2 - 253.5 x - 1.5 x3 o4 2 3 = - 734 Mxdx = o4 (39x2 - 253.5x - 1.5x3)dx = 39x3 - 253.5x2 - 1.5x4 o4 3 2 4 = - 1292 x = - 1292 - 734 x = + 1.76 m From B P1 = 1098.5 KN ( Area of entire BMD due to UDL ) P2 = 1 (area abc) = 1 (13.5) = 6.75 K( To be subtracted) 2 2 P3 = 1 ( area defg) = 1 (734) = 367 KN( To be subtracted ) 2 2 P4 = 720 KN( Area of entire BMD due to point Load ) P5 = 1  180  4 = 360 KN 2 Total concentric load on analogous column x - section = P1 + P2 + P3 - P4 + P 5 = - 1098.5 + 6.75 + 367 - 720 + 360 = - 1084.75 KN( It is upward so reaction due to this will be downward) Total applied moment at centroid of column = 6.75 (6.34 - 2.25) + 1098.5 (6.66 - 3.25) - 367 (6.66 - 1.76) + 720 (6.66 -1.33) - 360 (6.66 - 1.33) = 3894 KN-m (anticlockwise)

336

THEORY OF INDETERMINATE STRUCTURES

Properties of Analogous column x - section. A = 1  4 + 1  6 + 1  3 = 9.5 2 2 x = 6.66 From B as in previous problem. Iyoyo = 101.05 as in previous problem. (Mi)a = P  MC ( Reactions are subtractive at A) A I = - 1084.75 + 3894  6.34 9.5 101.05 (Mi)a = + 130 KN-m( Same answer as in previous problem ) (Ms)a = 0 Ma = (Ms - Mi)a Ma = ( 0 - 130) = - 130 KN-m (Mi)b = P  MC ( Reactions are additive at B ) A I = - 1084.75 - 3894  6.66 9.5 101.05 = - 370.83 KN-m (Ms)b = - 253.5 - 360 = - 613.5 KN-m Mb = (Ms - Mi)b = - 613.5 + 370.83 Mb = - 243 KN-m Now beam is determinate.

COLUMN ANALOGY METHOD

337

STIFFNESS AND CARRYOVER FACTORS FOR STRAIGHT MEMBERS WITH CONSTANT SECTION:Ma=K a A

Mb=(COF)Ma

a b

Ma EI

B L

EI=Constt:

+ 0

L/3

2/3L 1 MaL xbLx Ma = 2EI 2 EI MbL 2EI 2/3L

0

0 Loading on the conjugate beam L/3

__

Reaction on the 0 Mb conjugate EI beam.

aL/2 a

a L

1 EI

Analogous column section.

By choosing a B.D.S. as simple beam under the action of Ma and Mb, we can verify by the use of conjugate beam method that b = 0. In this case, we are required to find that how much rotation at end A is required to produce the required moment Ma. In other words, a (which is in terms of Ma and Mb can be considered as an applied load on the analogous column section). The moments computed by using the formula P/A  MC/I will give us the end moments directly because in this case Ms diagram will be

zero. Properties of analogous column section:A= L EI I = 1 L3 = L3 EI 12 12EI Downward load on analogous column = a at A. Accompanying moment = a  L ( About centroidal column axis ) 2. C= L 2 Ma = P + MC A I = a EI + L

a  L  L  12EI ( Reactions are additive at A and are upwards) 2  2  L3

338

THEORY OF INDETERMINATE STRUCTURES

= a EI + 3a EI L L Ma = 4 EI a L Where 4EI = Ka L Where Ka = stiffness factor at A. Mb = P  MC ( Reactions are subtractive at B) A I = a EI - 3a EI L L = - 2a EI L = - 2EI . a L The (-ve) sign with Mb indicates that it is a (-ve) moment which gives us tension at the top or compression at the bottom.

(COF) a  b Carry-over factor from A to B =Mb = 2 = + 1 Ma 4 2 “BY PUTING A EQUAL TO UNITY , MA & MB WILL BE THE STIFFNESS FACTORS AT THE CORRESPONDING JOINTS”. In the onward problems of members having variable x-section, we will consider a = b = 1 radians and will apply them on points A & B on the top of the analogous column section. The resulting moments by using the above set formula will give us stiffness factor and COF directly. EXAMPLE:Determine the stiffness factors at A & at B and the carry-over factors from A to B and from B to A for the straight members with variable x-sections shown in the figure below..

COLUMN ANALOGY METHOD

339

SOLUTION:-

A

B 2I 4m

1 1 1 x6+ x6+ x4 2EI EI 2EI 1 rad 3 6 2 = + + EI EI EI 1 A 11 = EI 2EI

I 6m

1 rad

A=

7.73

x I

I

B

1 EI

7.73m

x

2I 6m

1 Analogous column section 2EI

8.27m

= (0.5  6)  3 + (6  1)  9 + (4  0.5)  14 11 = 8.27 = 0.5  63 + (0.5  6)  (5.27)2 + 1  63 + (1  6)  12 12 (0.73)2 + 0.5  43 + (0.5  4)  (5.73)2 12 = 181.85 EI

Ma = P  MC A I = P + MC = 1  EI + 7.73  7.73  EI A I 11 181.85 Ma = 0.419 EI = 0.419  16 EI L Ma = 6.71 EI L Ka = 6.71 Mb = EI - 7.73  8.27  EI x 16 11 181.85 L = - 4.17 EI L (COF)AB = Mb = 4.17 = 0.62 Ma 6.71 (COF)AB = 0.62

Now applying unit radian load at B.

340

THEORY OF INDETERMINATE STRUCTURES

1ra d 1ra d 8 .2 7

7 .7 3

8 .2 7

Ma = [EI/11- ( 8.27x7.73xEI)/181.85] 16/L Ma = - 4.17 EI L Mb = [ EI + 8.27  8.27  EI] 16/L 11 181.85 Mb = 7.47 EI L Kb = 7.47

(COF)ba Carry-over factor from B to A = Ma = 4.17 Mb 7.47

(COF)ba = 0.56

APPLICATION TO FRAMES WITH ONE AXIS OF SYMMETRY:EXAMPLE:

Analyze the quadrangular frame shown by the method of column analogy. Check the

solution by using a different B.D.S. SOLUTION:B

6m

C 5I

12KN

2I

2I

6m

D

A 10m

COLUMN ANALOGY METHOD

341

The term “axis of symmetry” implies that the shown frame is geometrically symmetrical (M.O.I. is symmetrical) w.r.t. one axis as shown in the diagram. The term does not include the loading symmetry ( the loading can be and is unsymmetrical). Choosing the B.D.S. as a cantilever supported. at A. B

C

12KN

6m

6m

D

A 72 kN-m 10m Ms-diagram

B

C

5I

12 kN-m

6m 2I 6m 2I Force= 108 EI 36 EI

-

2 A

D

Ms - Diagram EI

According to our sign convention for column analogy, the loading arising out of negative Ms /EI will act upwards on the analogous column section. y B y=2.27m

5m Mxx

x 3.73m

C 1 5

108 EI

5m x

Myy

2m A

D 1 2

y

1/2

342

THEORY OF INDETERMINATE STRUCTURES

 Properties of Analogous Column Section:A = (1  6)  2 + 1  10 = 8 2 5 EI y = (1/5  10)  1/10 + 2 [ 1/2  6]  3 = 2.27 about line BC. 8 Ixx = 2[ 0.5  63 + (1/2  6) x (0.73)2 ] + 10  (1/5)3+ (0.2  10)  (2.27)2 12 12 = 31.51 EI Iyy = 0.2  103 + 2 [ 6  0.53 + (6  0.5)  (5)2] 12 12 = 167 EI Mxx = 108  1.73 = 187 clockwise. Myy = 108  5 = 540 clockwise. Applying the formulae in a tabular form. Ma = ( Ms- Mi)a ( Mi)a= P/A  Mx y/Ix  My x/Iy)

POINT

Ms

P/A

A B C D

- 72 0 0 0

- 13.5 - 13.5 - 13.5 - 13.5

Mx y Ix - 22.14 + 13.47 + 13.47 - 22.14

My x Iy - 16.17 - 16.17 + 16.17 + 16.17

Mi

M

- 51.81 - 16.20 + 16.14 - 19.47

- 20.19 + 16.20 - 16.14 + 19.47

Note : Imagine the direction of reaction due to P, Mx and My at various points. Use appropriate signs. PROBLEM:-

Analyze the quadrangular frame shown by the method of column analogy. 3KN/m

B

C

5I

6m

2I

2I

A

6m

D 10m

COLUMN ANALOGY METHOD

343

Choosing B.D.S. as a cantilever supported at A. 3KN/m C

B

MA+3x 10x 5=0 MA=-3x10x5 = 150KN-m 150K n-m D

A

15

B.D.S under loads Draw Ms-diagram by parts and then superimpose

Free Body Diagrams 15

150 B

3 KN/m C

150 15 B B

150

C

C

150 A

150

30 150

D

15 3KN/m

150 B -

C

150 I

A

150 Ms-Diagram

D

344

THEORY OF INDETERMINATE STRUCTURES

100 2.5 75

B

C

-

30

450 Ms - Diagram EI

6m I

3m

75 A

D

10m

2.25m

100

y C

B 450

2.275m

1/5 Mx

X

X

6m

0.725m My

3.725

3m 1/2

10m

1/2

D

Analogus colmun section y

Properties of analogous column section:A = 2 [ 1/2  6] + 1  10 = 8 5 EI y = (1/5  10)  1/10 + 2[(6  1/2)  3] = 2.275 about line BC 8 Ix = 2 [ 1/2  63 + (1/2  6)  (0.725)2] + [ 10  (1/5)3 + (10  1 )  (2.275)2] 12 12 5 = 31.51 EI Iy = 2 [ 6  0.53 + (6  0.5)  52] + 0.2  103 12 12 = 166.79 / EI Mx = 450  0.725 - 100  2.275 = 95.75 KN-m Clockwise My = 450  5 + 100  2.75 = 2525 KN-m clockwise. P = 100 + 450 = 550 KN

COLUMN ANALOGY METHOD

345

Applying the formulae in a tabular form. Ma = ( Ms- Mi)a ( Mi)a= P/A  Mx y/Ix  My x/Iy) POINT

Ms

P/A

A B C D

- 150 - 150 0 0

- 68.75 -68.75 -68.75 -68.75

Mx y Ix - 11.32 + 6.91 + 6.91 -11.32

My x Iy - 75.69 - 75.69 + 75.69 + 75.69

Mi

M

- 155.76 - 137.53 13.85 -4.38

5.76 -12.47 -13.85 4.38

Example.4 :Determine stiffness factors corresponding to each end and carry-over factors in both direction of the following beam. SOLUTION:-

2 I 5 I 4 I 3 I I 2 m 2 m 1 . 5 m m 2 m1

yo

1/5

½

1/EI

¼

4.74m

1/3EI

3.76m yo

Properties of Analogous Column Section :A = 1  2 + 1  1.5 + 1  2 + 1  1 + 1  2 5 2 4 3 A = 3.32 EI X = (1  2) (1)+(1  1)(2.5)+  2) 4+(1  1.5)(5.75) + (1  2)(7.5) 3.32 X = 3.76 Iyoyo = 1/3  23 + ( 1/3  2) + (2.76)2 + 1  13 + (1  1)(2.26)2 12 12 + (1/4)  (2)3 + (1/4  2)(0.24)2 + (1/2)  (1.5)3 12 12

346

THEORY OF INDETERMINATE STRUCTURES

+ (1/2  1.5)(1.99)2 + (1/5)  (2)3 + (1/5  2)(3.74)2 12 = 19.53 EI (1) ka & (COF)a  b :1 rad

A

B

8.5m 1 4.74

A

4.74

B

3.76

Ma= P/A  MC/I = 1xEI/3.32 + 4.74x4.74 EI/19.53 = 1.45 EI Ma = 1.45  8.5  EI = 12.33 EI L L Ka = 12.33

Mb = EI - 4.74  3.26  EI 3.32 19.53 = - 0.61 EI = - 0.61  8.5  EI = - 5.19EI L L Mb = - 5.19 EI L (COF)a  b = Mb = 5.19 = 0.42 Ma 12.33 (COF)a  b = 0.42

COLUMN ANALOGY METHOD

347

(2) Kb & (COF) b  a :Ma = P  Mc A I 1 rad

A

B

8.5m 1 3.76

A

4.74

Ma

Mb

Mb

B

3.76

EI 3.76 x 4.74 x EI 3.32 19.53 = -0.61EI EI = - 0.61 x 8.5 x = -5.19 EI/L L = P  Mc A I = EI + 3.76  3.76  EI 3.32 19.53 =

=1..03 EI = 1.03  EI x 8.5 L = 8.76 EI L Kb = 8.76

(COF)b  a = Ma = 5.19 = 0.6 Mb 8.76 (COF) b  a = 0.6

348

THEORY OF INDETERMINATE STRUCTURES

Example.5:-

Analyze the following gable frame by column analogy method.

SOLUTION :-

3 kN/m

B

3I

C

7m

I

I

A

3m

D

3I

14 m

E

3kN/m

7.62

C

B 7

D

7 3 .5

7 3 .5

B

7 .6 2

C D

E

A

21

21 B.D.S under loads

A

E M s -d iag ra m

COLUMN ANALOGY METHOD

2. 4.7

2 .8 6

D

M s diagram EI

E

6

6 4 .7

a =

x B

86

24.5 24.5 C

349

4.375

A

Taking the B.D.S. as a simply supported beam. Mx Mx

= 21X – 1.5X2 = Mc at x = 7m

Mc

= 21 x 7 – 1.5 x 72 = 73.5 KN-m

Sin 

=

Cos 

3 = 0.394 7.62 = 7 = 0.919 7.62 P1 = P2 = 2/3  24.5  7.62 = 124.46 P = P1 + P2 = 248.92  Mx dx = o7 (21 X - 1.5X2) dx = 21X2 - 1.5 X3 o7 = 343 2 3  (Mx)X dx = o7 (21 X2 - 1.5X3)dx = 21X3 - 1,5 X4 o7 3 4 7 x 73 - 1.5 x 74 = 1500.625 4 X =  (Mx) X dx = 1500.625  Mx dx 343 X = 4.375 Horizontally from D or A. Shift it on the inclined surface. Cos  = 4.375 a a = 4.375 = 4.375 Cos  0.919 a = 4.76

350

THEORY OF INDETERMINATE STRUCTURES

124.46 124.46 2.8 6

3m B

4.7 6

C 2.17

D

1/3

x

x Mx

4.83

A

E 1

1

PROPERTIES OF ANALOGOUS COLUMN SECTION A = 2 (1  7) + 2 (1/3  7.62) = 19.08 Y = 2[ (1  7)  3.5 ] + 2 [ (1/3  7.62)  8.5)  8.5 ] 19.08 Y = 4.83 from A or E Ix = 2 [ 1  73 + (1  7) (4.83 - 3.5)2 ] 12 + 2 [ (1/3)  (7.62)3  ( 0.394 )2+ ( 1/3(7.62) ( 1.5 + 2.17)2 ] 12 = 154.17 Ix  154 Iy = 2 [ 7  13 + 7(7  1)  72 ] 12 + 2 [(1/3)  (7.62)3  12 =770.16 Iy  770

(0.919 )2 + (1/3  7.62)  (3.5)2 ]

bL3 Sin2 12

bL3 Cos2 12

COLUMN ANALOGY METHOD

351

Total load on centroid of analogous column P = P1 + P2 = 124.46 + 124.46 = 248.92 Mx = 2  [124.46  4.05 ] Mx = 1007 (clockwise). My = 0 Applying the general formulae in a tabular form. Ma = ( Ms- Mi)a ( Mi)a= P/A  Mx y/Ix  My x/Iy) Point

Ms (A)

P/A (1)

A B C D E

0 0 + 73.5 0 0

+ 13.05 + 13.05 + 13.05 + 13.05 + 13.05

PROBLEM:

Mx .Y Ix (2) - 31.58 + 14.19 + 33.81 + 14.19 - 31.58

My .X Iy (3) 0 0 0 0 0

(B)=Mi (1)+(2) +(3) - 18.53 + 27.24 + 46.86 + 27.24 - 18.53

Analyze the frame shown in fig by Column Analogy Method.

10kN B

C 3I

2kN/m

2I

2I

A

4m

D 3m

Choosing the B.D.S. as a cantilever supported at A. 4 MA + 10 x 1.5 + 2 x 4 x = 0 2 MA = - 31 KN-m

M (A)-(B) + 18.53 - 27.24 +26.64 - 27.24 + 18.53

352

THEORY OF INDETERMINATE STRUCTURES

10 kN

2kN/m

8 31

B.D.S

10

Free Body Diagrams :10

15 1.5

10

B

C 10

15

C

15

B

1.5

15 2kN/m no B.M.D

A 8

31

31

D

10

10 15 15

31

M s-diagram

COLUMN ANALOGY METHOD

353

Properties Of Analogous Column Section :A = (1  4 )  2 + ( 1  3) = 5 2 3 (3  1 )  ( 1 ) + 2 [ ( 1  4 )  2 ] Y = 3 6 2 . 5 Y = 1.63 From line BC Ix = 3  (1/3)3 + (1/3  3)  (1.63)2 +2[0.5  43  (0.5  4)  (0.37)2] 12 12 = 8.55 Iy = (1/3)  (3)3 + 2 [ 4  0.53 ( 4  0.5)  (1.5)2] 12 = 9.83 Total load on top of analogous column section acting at the centroid. P = 3.75 + 30 + 10.67 = 44.42 upward. Mx = - 3.75 x 1.63 + 30 x 0.37 + 10.67 x 1.37 = 19.61 clockwise. My = 10.67  1.5 + 30  1.5 + 3.75  1 = 64.76 clockwise. Applying the general formulae in a tabular form. Ma = ( Ms- Mi)a ( Mi)a= P/A  Mx y/Ix  My x/Iy) Point Ms P/A Mx . y My . x Mi Ix Iy A - 31 -8.88 - 5.44 - 9.88 - 24.2 B - 15 - 8.88 + 3.74 - 9.88 - 15.02 C 0 - 8.88 + 3.74 + 9.88 + 4.74 D 0 - 8.88 - 5.44 + 9.88 - 4.44

M - 6.8 + 0.02 - 4.74 + 4.44

Example: Analyze the following beam by column analogy method. SOLUTION :3 k N /m

3 I

1 0 k N

I

1 .5 I

Choosing B.D.S as cantilever supported at A

M s d i a g r a m d u e t o u . d . l o n l y

6 2 4

7 2

354

THEORY OF INDETERMINATE STRUCTURES

4 0

a

b

2c 1.5m 6 P2=1.33

P1=24x4 + 48x4 24x4 = 224 + 3 2 2.14 d 16 e 24

24 72

3.21m P3=18.67

Ms-diagram EI due to u.d.l P 4=80 1.33

40

yo Analogous 1 column section

1/1.5

1/3 4.78

yo

3.22

area (abc)=  Mxdx = o2-1.5X2 dx =  - 1.5 X3 o2 = -4 3 2 3 4 2 (Mx) x dx = o -1.5X dx =  - 1.5 X o = - 6 4 X = - 6 = 1.5m from A -4 area (bcde) = (Mx) dx = o4 - 1.5X2dx - o2 - 1.5 X2 dx =  - 1.5 X3 o4 -1.5 X3 o2 = - 28 3 3  (Mx)x dx = o4- 1.5 X3dx - o2 - 1.5 X3dx = - 90 P3 = 1 (area bcde) = 1 (28) = 18.67 KN 1.5 1.5 P4 = 80 KN

COLUMN ANALOGY METHOD

Total concentric load on analogous column section. P = - P1 + P2 + P3 - P4 = - 224 + 1.33 + 18.67 - 80 = - 284 KN (upward)

Properties of Analogous Column Section. A= 1 2+ 1 2+14=6 3 1.5 X = (1  4)  2 + (2  1/1.5)  5 + (1/3  2 )  7 6 = 3.22 Iyoyo = 1  43 + (1  4)(1.22)2 + (1/1.5)  23 + (1/1.52)(1.78)2 12 + (1/3)  23 + (1/3  2) (3.78)2 12 = 25.70 Total applied moment at centroid of column. = - 224  1.68 - 80  1.89 - 18.67  1.57 - 1.33  3.28 = - 426.79 (counterclockwise). (Mi)a = P  Mc A I = - 284 + 426.79  4.78 6 25.7 = + 32.05 (Ms)a = 0 Ma = (Ms - Mi)a = 0 - 32.05 Ma = - 32.05 KN-m

355

356

THEORY OF INDETERMINATE STRUCTURES

(Mi)b = P - Mc A I = - 284 - 426.79  3.22 6 25.7 = - 100.81 (Ms)b = - 72 - 40 = - 112 Mb = (Ms - Mi)b = - 112 + 100.81

Mb = - 11.19 KN-m