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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645

Contribution of Axial Deformation in the Analysis of Rigidly Fixed Portal Frames Okonkwo V. O.1, Aginam C. H.2, and Chidolue C. A.3 123

Department of Civil Engineering Nnamdi Azikiwe University, Awka Anambra State-Nigeria

ABSTRACT: In this work the stiffness equations for evaluating the internal stress of rigidly fixed portal frames by the displacement method were generated. But obtaining the equations for the internal stresses required non-scalar or parametric inversion of the structure stiffness matrix. To circumvent this problem, the flexibility method was used taking advantage of the symmetrical nature of the portal frame and the method of virtual work. These were used to obtain the internal stresses on rigidly fixed portal frames for different cases of external loads when axial deformation is considered. A dimensionless constant s was used to capture the effect of axial deformation in the equations. When it is set to zero, the effect of axial deformation is ignored and the equations become the same as what can be obtained in any structural engineering textbook. These equations were used to investigate the contribution of axial deformation to the calculated internal stresses and how they vary with the ratio of the flexural rigidity of the beam and columns and the height to length ratio of the loaded portal frames.

Keywords: Axial deformation, flexural rigidity, flexibility method, Portal frames, stiffness matrix,

I. INTRODUCTION Structural frames are primarily responsible for strength and rigidity of buildings. For simpler single storey structures like warehouses, garages etc portal frames are usually adequate. It is estimated that about 50% of the hotrolled constructional steel used in the UK is fabricated into single-storey buildings (Graham and Alan, 2007). This shows the increasing importance of this fundamental structural assemblage. The analysis of portal frames are usually done with predetermined equations obtained from structural engineering textbooks or design manuals. The equations in these texts were derived with an underlying assumption that deformation of structures due to axial forces is negligible giving rise to the need to undertake this study. The twenty first century has seen an astronomical use of computers in the analysis of structures (Samuelsson and Zienkiewi, 2006) but this has not completely eliminated the use of manual calculations form simple structures and for easy cross-checking of computer output (Hibbeler, 2006). Hence the need for the development of equations that capture the contribution of axial deformation in portal frames for different loading conditions.

II. DEVELOPMENT OF THE MODEL

Figure 1: A simple portal frame showing its dimensions and the 6 degrees of freedom The analysis of portal frames by the stiffness method requires the determination of the structure’s degrees of freedom and the development of the structure’s stiffness matrix. For the structure shown in Figure 1a, the degrees of freedom are as shown in Figure 1b. I1 and A1 are respectively the second moment of inertia and crosssectional area of the columns while I2 and A2 are the second moment of inertia and cross-sectional area of the beam respectively. The stiffness coefficients for the various degrees of freedom considering shear deformation can be obtained from equations developed in Ghali et al (1985) and Okonkwo (2012) and are presented below The structure’s stiffness matrix can be written as: đ?‘˜11 đ?‘˜21 đ?‘˜ đ??ž = 31 đ?‘˜41 đ?‘˜51 đ?‘˜61

đ?‘˜12 đ?‘˜22 đ?‘˜32 đ?‘˜42 đ?‘˜52 đ?‘˜62

đ?‘˜13 đ?‘˜23 đ?‘˜33 đ?‘˜43 đ?‘˜53 đ?‘˜63

đ?‘˜14 đ?‘˜24 đ?‘˜34 đ?‘˜44 đ?‘˜54 đ?‘˜64

đ?‘˜15 đ?‘˜25 đ?‘˜35 đ?‘˜45 đ?‘˜55 đ?‘˜65

đ?‘˜16 đ?‘˜26 đ?‘˜36 . đ?‘˜46 đ?‘˜56 đ?‘˜66

(1)

Where kij is the force in coordinate (degree of freedom) i when there is a unit displacement in coordinate (degree of freedom) j. They are as follows: đ?‘˜11 =

đ??¸đ??´1 â„Ž

đ?‘˜12 = 0 www.ijmer.com

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 đ?‘˜13 =

6đ??¸đ??ź2

đ?‘‘1 đ?‘‘2 đ?‘‘3 {đ??ˇ} = đ?‘‘4 đ?‘‘5 đ?‘‘6

đ?‘™2

đ?‘˜14 = −

12đ??¸đ??ź1 â„Ž3

đ?‘˜15 = 0 đ?‘˜16 = đ?‘˜22 = đ?‘˜23 =

.

.

.

.

.

(5)

6đ??¸đ??ź2

Where di is the displacement in coordinate i.

đ?‘™2 12đ??¸đ??ź1 â„Ž3

+

đ??¸đ??´2

đ??š1 đ??š2 đ??š {đ??š} = 3 đ??š4 đ??š5 đ??š6

đ?‘™

6đ??¸đ??ź1 â„Ž2

đ?‘˜24 = 0 đ?‘˜25 = −

đ??¸đ??´2

.

đ?‘™

.

(2)

.

.

.

.

.

.

(6)

Where Fi is the external load with a direction coinciding with the coordinate i (McGuire et al, 2000).

đ?‘˜26 = 0 đ?‘˜33 =

4đ??¸đ??ź2 đ?‘™

đ?‘˜34 = −

+

By making {D} the subject of the formula in equation (3)

4đ??¸đ??ź1 â„Ž

đ??ˇ = đ??ž

6đ??¸đ??ź2

đ?‘˜44 =

2đ??¸đ??ź2

đ?‘™3

+

đ??¸đ??´1 â„Ž

đ?‘˜45 = 0 đ?‘˜46 = − đ?‘˜55 =

6đ??¸đ??ź2 đ?‘™2

12đ??¸đ??ź1 â„Ž3

.

.

đ?‘€ = đ?‘€đ?‘&#x; + đ?‘€1 đ?‘‘1 + đ?‘€2 đ?‘‘2 + đ?‘€3 đ?‘‘3 .

đ?‘™ 12đ??¸đ??ź2

{đ??š − đ??šđ?‘œ }

+

đ??¸đ??´2 đ?‘™

From Maxwell’s Reciprocal theorem and Betti’s Law k ij = kji ( Leet and Uang, 2002). When there are external loads on the structure on the structure there is need to calculate the forces in the restrained structure Fo as a result of the external load. The structure’s equilibrium equations are then written as {đ??š} = {đ??šđ?‘œ } + đ??ž {đ??ˇ} . . . đ?‘˜10 đ?‘˜20 đ?‘˜ {đ??šđ?‘œ } = 30 đ?‘˜40 đ?‘˜50 đ?‘˜60

.

. (7)

Once {D} is obtained the internal stresses in the frame can be easily obtained by writing the structure’s compatibility equations given as

đ?‘™2

đ?‘˜35 = 0 đ?‘˜36 =

−1

.

.

.

.

.

(3)

.

.

(8)

Where M is the internal stress (bending moment) at any point on the frame, Mr is the internal stress at the point under consideration in the restrained structure while Mi is the internal stress at the point when there is a unit displacement in coordinate i. To solve equation (8) there is need to obtain {D}. {D} can be obtained from the inversion of [K] in equation (7). Finding the inverse of K parametrically (i.e. without substituting the numerical values of E, h, l etc) is a difficult task. This problem is circumvented by using the flexibility method to solve the same problem, taking advantage of the symmetrical nature of the structure and the principle of virtual work.

III. APPLICATION OF THE FLEXIBILITY MODEL The basic system or primary structure for the structure in Figure 1a is given in Figure 2. The removed redundant force is depicted with X1.

(4)

Where kio is the force due to external load in coordinate i when the other degrees of freedom are restrained. Figure 2: The Basic System showing the removed redundant forces

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 The flexibility matrix of the structure can be determined using the principle of virtual work. By applying the unit load theorem the deflection in beams or frames can be determined for the combined action of the internal stresses, bending moment and axial forces with đ??ˇ=

đ?‘€đ?‘€ đ??¸đ??ź

đ?‘ đ?‘

đ?‘‘đ?‘ +

đ??¸đ??´

đ?‘‘đ?‘ .

.

.

1 đ?‘‘ 11

12đ??¸đ??ź1 đ??ź2 đ??´1

=

đ??ź1 đ??´1 đ?‘™ 3 +6đ??ź2 đ??´1 đ?‘™ 2 â„Ž +24đ??ź1 đ??ź2 â„Ž

đ?‘‘ 33 2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23

(9)

đ?‘‘ 32

Where đ?‘€ and đ?‘ are the virtual internal stresses while M and N are the real/actual internal stresses.

2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23

đ?‘‘ 22

.

. .

= 2đ??´

3đ??¸đ??ź1 đ??´2 (đ?‘™đ??ź1 +2â„Žđ??ź2 ) 2 2 3 4 2 â„Ž đ?‘™đ??ź1 +3đ??ź1 đ?‘™ +đ??´2 â„Ž đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

= 2đ??´

−3đ??¸đ??ź1 đ??ź2 đ??şđ??´2 â„Ž 2 2 2 3 4 2 â„Ž đ?‘™đ??ź1 +3đ??ź1 đ?‘™ +đ??´2 â„Ž đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

(15a)

.

(15c)

đ??¸đ??ź1 đ??ź2 2đ??´2 â„Ž 3 +3đ??ź1 đ?‘™ 3 2đ??´2 â„Ž đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +đ??´2 â„Ž 4 đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

=

(15b)

.

(15d)

E is the modulus of elasticity of the structural material

2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23

A is the cross-sectional area of the element (McGuire et al, 2000; Nash, 1998)

Equation (12) is evaluated to get the redundant forces and these are substituted into the structures force equilibrium (superposition) equations to obtain the internal stresses at any point.

If dij is the deformation in the direction of i due to a unit load at j then by evaluating equation (9) the following are obtained: đ?‘‘11 =

đ??ź1 đ??şđ??´1 đ?‘™ 3 +6đ??ź2 đ??´1 đ?‘™ 2 â„Ž +12đ??ź1 đ??ź2 â„Ž 12đ??¸đ??ź1 đ??ź2 đ??´1

đ?‘‘12 = 0 .

.

đ?‘‘13 = 0 . đ?‘‘22 = đ?‘‘23 = đ?‘‘33 =

−ℎ 2 đ??¸đ??ź1

.

.

đ??¸đ??ź1 đ??ź2

.

. .

.

đ?‘™đ??ź1 +2â„Ž đ??ź2

(10a)

. .

.

3đ??¸đ??ź1 đ??´2

.

. .

.

2đ??´2 â„Ž 3 +3đ??ź1 đ?‘™

.

(10b)

.

(10c) (10d)

.

.

.

(10e)

.

.

(10f)

From Maxwell’s Reciprocal theorem and Betti’s Law dij = dji. The structure’s compatibility equations can be written thus �11 �21 �31

đ?‘‘11 đ?‘‘21 đ?‘‘31

đ?‘‘11 đ?‘‘21 đ?‘‘31

đ?‘‘10 đ?‘‹1 đ?‘‹2 + đ?‘‘20 = 0 . đ?‘‹3 đ?‘‘30

i.e. đ?‘“đ??š + đ?‘‘đ?‘œ = 0 .

đ?‘“ −1 =

đ??´đ?‘‘đ?‘— đ?‘“

.

(12)

.

1

đ?‘“ −1 = 0 0

0

8đ??¸đ??ź1

.

.

.

.

.

.

.

.

.

(17b)

đ?‘‘30 =

−đ?‘¤ đ?‘™ 2 (đ?‘™đ??ź1 +6â„Žđ??ź2 ) 24đ??¸đ??ź1 đ??ź2

.

.

(17c)

By substituting the values of equations (17a) – (17c) into equations (12)

đ?‘‹3 =

.

.

.

.

(18a)

−đ??´2 đ??ź1 đ?‘¤â„Ž 2 đ?‘™ 3

.

2đ??´2 â„Ž 3 đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +đ??´2 â„Ž 4 đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

đ?‘¤ đ?‘™ 2 2đ??´2 â„Ž 3 đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +3đ??´2 â„Ž 4 đ??ź2 +18â„Žđ?‘™đ??ź1 đ??ź2 24 2đ??´2 â„Ž 3 đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +đ??´2 â„Ž 4 đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

.

(18b) (18c)

Evaluating equation (16) for different points on the structure using the force factors obtained in equations (18a) – (18c) 2đ?‘¤đ?‘™ 3 đ??ź1 â„Ž 3 đ??´2 −3đ?‘™đ??ź1

đ?‘€đ??´ = 12

2đ??´2 â„Ž 3 đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +đ??´2 â„Ž 4 đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

đ?‘€đ??ľ = 24

2đ??´2 â„Ž 3 đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +đ??´2 â„Ž 4 đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2

. .

(19)

−2đ?‘¤đ?‘™ 3 đ??ź1 2â„Ž 3 đ??´2 +3đ?‘™đ??ź1

. .

(20)

.

(21)

0

đ?‘‘ 33

−đ?‘‘ 32

2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23 −đ?‘‘ 23

2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23 đ?‘‘ 22

2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23

2 đ?‘‘ 22 đ?‘‘ 33 −đ?‘‘ 23

From equations 10a – 10f

(16)

(13)

(Stroud, 1995) đ?‘‘ 11

� ℎ 2�2

.

(11b)

.

.

đ?‘‘đ?‘’đ?‘Ą đ?‘“

đ?‘‘20 =

đ?‘‹2 = 4 .

.

đ?‘‘10 = 0 . . (17a)

đ?‘‹1 = 0 .

.

.

Where M is the required stress at a point, Mo is the stress at that point for the reduced structure, Mi is stress at that point when the only the redundant force Xi =1 acts on the reduced structure. For the loaded portal frame of Figure 3, the deformations of the reduced structure due to external loads are

(11a)

Where F is the vector of redundant forces X1, X2, X3 and do is the vector deformation d10, d20, d30 due to external load on the basic system (Jenkins, 1990). đ??š = đ?‘“ −1 (−đ?‘‘đ?‘œ ) .

đ?‘€ = đ?‘€đ?‘œ + đ?‘€1 đ?‘‹1 + đ?‘€2 đ?‘‹2 + đ?‘€3 đ?‘‹3 .

.

(14)

đ?‘€đ??ś = đ?‘€đ??ś =

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−2đ?‘¤ đ?‘™ 3 đ??ź1 2â„Ž 3 đ??´2 +3đ?‘™đ??ź1 24 2đ??´2 â„Ž 3 đ?‘™đ??ź1 +3đ??ź12 đ?‘™ 2 +đ??´2 â„Ž 4 đ??ź2 +6â„Žđ?‘™đ??ź1 đ??ź2 −đ?‘¤ đ?‘™ 3 đ??ź1 8đ??ź2 â„Ž 3 +đ?‘ 2 đ?‘™ 3 đ??ź1 12 4â„Ž 3 đ??ź2 2đ?‘™đ??ź1 +â„Ž đ??ź2 +đ?‘ 2 đ?‘™ 3 đ??ź1 đ?‘™đ??ź1 +2â„Ž đ??ź2

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 2𝑤𝑙 3 𝐼1 ℎ 3 𝐴2 −3𝑙𝐼1

𝑀𝐷 =

. .

12 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

(22)

−𝑤 ℎ 3

𝑑30 =

.

6𝐸𝐼1

.

. .

.

(30c)

By substituting the values of equations (30a) – (30c) into equations (12) For the loaded portal frame of Figure 4, the deformations of the reduced structure due to external loads are −𝑤𝑙 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ 𝐼1 𝐼2

𝑑10 =

.

128 𝐸𝐼1 𝐼2 𝐴1 𝑤 ℎ 2𝑙2

𝑑20 =

.

16𝐸𝐼1

.

−𝑤 𝑙 2 (𝑙𝐼1 +6ℎ𝐼2 )

𝑑30 =

.

.

48𝐸𝐼1 𝐼2

(23a) (23b)

.

(23c)

By substituting the values of equations (23a) – (23c) into equations (12) 𝑋1 =

𝑋3 =

.

32 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2

.

(24a)

−2𝑤 ℎ 2 𝑙 3 𝐴2 𝐼1

.

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑤 ℎ 3 𝑙 2 𝐴2 3ℎ 𝐼2 +2𝑙𝐼1 +3𝑤𝑙 3 𝐼1 𝑙𝐼1 +6ℎ 𝐼2

.

48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑠2

Let 𝛽 = 𝑠 .

.

1

.

.

.

𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +8𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2

− 64

48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

.

− 64

𝑤 ℎ 2 𝐼1 𝐴1 𝑙 3 +5𝐴1 𝑙 2 ℎ 𝐼2 +24ℎ𝐸𝐼1 𝐼2

(24c)

𝑀𝐵 =

. (25)

𝑤 𝑙 4 𝐴1 5𝑙𝐼1 +24ℎ 𝐼2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2

. .

48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

.

𝑤 ℎ 3 𝑙 2 𝐼2 𝐴1 3 2 𝐼1 𝐴1 𝑙 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2

.

64 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +2𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2

𝑀𝐶 = −2

. .

.

𝑑20 =

12𝐸𝐼1

𝑤ℎ4 8𝐸𝐼1

.

.

. .

.

+ 24

.

.

.

.

(32)

.

.

.

𝑤ℎ 3 𝐼2 −ℎ 3 𝐴2 +12𝑙𝐼1 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

(34) 𝑤ℎ 3 9ℎ 2 𝑙𝐼1 𝐴2 +5𝐴2 ℎ 3 𝐼2 +12𝑙𝐼1 𝐼2

𝑤 ℎ 3 𝑙 2 𝐼2 𝐴1 3 𝐼1 𝐴1 𝑙 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2

𝑋1 = 𝐴

+ . .

. .

𝑤 ℎ 3 𝑙 2 𝐼2 𝐴1

𝑑30 = 0

(28)

.

.

+ 24

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

(35)

𝑃𝑎 2 𝑙 2𝐸𝐼1

.

. .

.

.

.

(36a)

.

.

(36b)

.

. .

6𝑃𝑎 2 𝑙𝐼2 𝐴1 3 2 1 𝐼1 𝑙 +6𝐴1 ℎ 𝑙 𝐼2 +24ℎ 𝐼1 𝐼2

𝑋2 = 0

(29)

For the loaded portal frame of Figure 5, the deformations of the reduced structure due to external loads are 𝑤𝑙 ℎ 3

. .

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑀𝐷 = −2

+

(36c)

By substituting the values of equations (36a) – (36c) into equations (12)

64 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +8𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2 48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑑10 = −

(31c)

𝑤ℎ 3 𝐼2 −ℎ 3 𝐴2 +12𝑙𝐼1

+ 24

𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2

𝑑20 = 0 .

+

𝑀𝐷 = −

.

(33)

𝑑10 = −

48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

3𝑤 𝑙 2 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ𝐼1 𝐼2

.

(27)

𝑀𝐶 =

3𝑤 𝑙 2 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ𝐼1 𝐼2

.

(31b)

For the loaded portal frame of Figure 6, the deformations of the reduced structure due to external loads are

+

𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +2𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2

.

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

(24b)

(26)

𝑤 𝑙 4 𝐴1 5𝑙𝐼1 +24ℎ 𝐼2

(31a)

Evaluating equation (16) for different points on the structure using the force factors obtained in equations (31a) – (31c)

. 𝑀𝐵 =

.

𝑤ℎ 3 𝐼2 −ℎ 3 𝐴2 +12𝑙𝐼1

𝑋3 = 24

. 𝑀𝐴 =

. .

.

−𝑤ℎ 4 𝐴2 3𝑙𝐼1 +2ℎ 𝐼2 3 2𝐴2 ℎ 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑋2 = 8

2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑤 ℎ 3 9ℎ 2 𝑙𝐼1 𝐴2 +5𝐴2 ℎ 3 𝐼2 +12𝑙𝐼1 𝐼2 24 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

Evaluating equation (16) for different points on the structure using the force factors obtained in equations (24a) – (24c)

.

.

𝑀𝐴 =

3𝑤𝑙 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ𝐼1 𝐼2

𝑋2 = 16

𝑤 ℎ 3 𝑙𝐼2 𝐴1 3 2 1 𝐴1 𝑙 +6𝐼2 𝐴1 𝑙 ℎ +24ℎ 𝐼1 𝐼2

𝑋1 = 𝐼

(30a)

𝑋3 = 0 .

.

. . .

.

.

. .

. .

.

(37a) (37b)

.

(37c)

Evaluating equation (16) for different points on the structure using the force factors obtained in equations (37a) – (37c)

(30b) www.ijmer.com

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝑀𝐴 = −𝑃𝑎 1 − 𝐴 𝑀𝐵 = 𝐴

3𝑎𝑙 2 𝐼2 𝐴1

.

3 2 1 𝐼1 𝑙 +6𝐴1 ℎ 𝑙 𝐼2 +24ℎ 𝐼1 𝐼2

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 3 2 1 𝐼1 𝑙 +6𝐴1 ℎ 𝑙 𝐼2 +24ℎ 𝐼1 𝐼2

.

.

(38)

For the loaded portal frame of Figure 8, the deformations of the reduced structure due to external loads are

(39)

𝑑10 = −𝑃 𝑑20 =

𝑀𝐶 = − 𝐴

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1

.

3 2 1 𝐼1 𝑙 +6𝐴1 ℎ 𝑙 𝐼2 +24ℎ𝐼1 𝐼2

𝑀𝐷 = 𝑃𝑎 1 − 𝐴

. .

3 2 1 𝐼1 𝑙 +6𝐴1 ℎ 𝑙 𝐼2 +24ℎ 𝐼1 𝐼2

.

(41)

𝑑10 = − 4𝐸𝐼

.

1

.

𝑃𝑎 2 3ℎ −𝑎

𝑑20 =

.

6𝐸𝐼1 𝑃𝑎 2

𝑑30 = − 2𝐸𝐼

.

1

.

.

.

(42b)

.

(42c)

𝑋3 =

3𝑃𝑎 2 𝑙𝐼2 𝐴1 3 2 1 𝐴1 𝑙 +6𝐼2 𝐴1 𝑙 ℎ +24ℎ 𝐼1 𝐼2

𝑋2 = 2

.

.

.

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

.

2𝐸𝐼1 𝐼2

.

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1

(43b)

2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑀𝐵 =

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1

2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2

.

.

+2

.

(49a)

−3𝑃𝑎 ℎ 2 𝐼1 𝐴2 𝑙−𝑎

(49b)

𝑃𝑎 ℎ 3 𝐴2 2𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2

(49c)

2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑃𝑙 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1

+ (50)

𝑃 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1

+

2(𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 ) 𝑃𝑎 ℎ 3 𝐴2 2𝑎𝐼1 +ℎ𝐼2 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2

(43c)

.

.

.

(51)

𝑀𝐶 = −

𝑃𝑙 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1

2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑃𝑎 ℎ 3 𝐴2 2𝑎𝐼1 +ℎ𝐼2 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

+ . .

.

.

(52)

𝑀𝐷 = .

𝑃𝑎 2 𝐼2 −ℎ 3 𝐴2 +𝑎ℎ 2 𝐴2 +3𝑙𝐼1

𝑃𝑙 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1

+

2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑃𝑎 𝐴2 ℎ 3 −𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

. (44)

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

.

.

(53)

.

(45)

IV. DISCUSSION OF RESULTS

𝑀𝐶 = −2

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 3 𝐼1 𝐴1 𝑙 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2

.

.

.

+2

𝑃𝑎 2 𝐼2 −ℎ 3 𝐴2 +𝑎ℎ 2 𝐴2 +3𝑙𝐼1 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

(46)

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2 𝑃𝑎 2 𝐴2 𝐼1 𝑙 3ℎ 2 −𝑎ℎ +𝐼2 2𝐴2 ℎ 3 −𝑎𝐴2 ℎ 2 +3𝐼1 𝑙 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

The internal stress on the loaded frames is summarized in table 1. The effect of axial deformation is captured by the dimensionless constant s taken as the ratio of the end translational stiffness to the shear stiffness of a member. 𝑠1 =

𝑀𝐷 = −

.

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑀𝐵 = −𝑃𝑎 +

+

2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑃𝑎 2 𝐴2 𝐼1 𝑙 3ℎ 2 −𝑎ℎ +𝐼2 2𝐴2 ℎ 3 −𝑎𝐴2 ℎ 2 +3𝐼1 𝑙

(48c)

2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑃𝑎 𝐴2 ℎ 3 −𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

Evaluating equation (16) for different points on the structure using the force factors obtained in equations (43a) – (43c)

−𝑃𝑎 +

.

𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2

2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑀𝐴 =

(48b)

𝑃 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ𝐼2 𝑎𝑙 𝐴1 +2𝐼1

𝑀𝐴 = −𝑃𝑎 +

(43a) .

2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼12 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2

𝑋3 = 2

𝑃𝑎 𝑎𝐼1 +2ℎ 𝐼2

.

. .

−𝑃𝑎 2 𝐴2 𝑙𝐼1 3ℎ−𝑎 +ℎ 𝐼2 3ℎ −2𝑎

−𝑃𝑎 2 𝐼2 −ℎ 3 𝐴2 +𝑎ℎ 2 𝐴2 +3𝑙𝐼1

.

Evaluating equation (16) for different points on the structure using the force factors obtained in equations (49a) – (49c)

By substituting the values of equations (42a) – (42c) into equations (12) 𝑋1 = 𝐼

𝑋1 =

𝑋2 = 2

(42a)

.

.

(48a)

By substituting the values of equations (48a) – (48c) into equations (12)

For the loaded portal frame of Figure 7, the deformations of the reduced structure due to external loads are 𝑃𝑎 2 𝑙

2𝐸𝐼1

(40)

.

12𝐸𝐼1 𝐼2 𝐴1

𝑃𝑎ℎ 2

𝑑30 = −

3𝑎𝑙 2 𝐼2 𝐴1

𝑎 2 𝐼1 𝐴1 3𝑙 −2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1

+

.

.

(47)

𝑠2 =

12𝐸𝐼1 ℎ3 12𝐸𝐼2 𝑙3

12𝐼

∙ 𝐸𝐴 = ℎ 2 𝐴1 . 1

𝑙 𝐸𝐴2

1

=

12𝐼2 𝑙 2 𝐴2

.

.

.

(54)

.

(55)

When the axial deformation in the columns is ignored 𝑠1 = 0 and likewise when axial deformation in the beam is www.ijmer.com

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 ignored đ?‘ 2 = 0 . If axial deformation is ignored in the whole structure, đ?‘ 1 = đ?‘ 2 = 0. The internal stress equations enable an easy investigation into the contribution of axial deformation to the internal stresses of statically loaded frames for different kinds of external loads. For frame 1 (figure 3), the moment at A, MA is given by equation (19). The contribution of axial deformation in the column ∆đ?‘€đ??´ is given by ∆đ?‘€đ??´ = đ?‘€đ??´(đ?‘“đ?‘&#x;đ?‘œđ?‘š = .

đ?‘¤ đ?‘™ 3 đ??ź1 12

đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› 19)

− đ?‘€đ??´(đ?‘“đ?‘&#x;đ?‘œđ?‘š

đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› 19 đ?‘¤â„Žđ?‘’đ?‘› đ?‘ 2 =0)

4â„Ž 3 đ??ź2 +đ?‘ 2 đ?‘™ 3 đ??ź1 3 4â„Ž đ??ź2 2đ?‘™đ??ź1 +â„Ž đ??ź1 +đ?‘ 2 đ?‘™ 3 đ??ź1 đ?‘™đ??ź1 +2â„Ž đ??ź2

.

−

1 2đ?‘™đ??ź1 +â„Ž đ??ź2

.

(56)

Equation (56) gives the contribution of axial deformation to MA as a function of h, l, I1, I2 and s By considering the case of a portal frame of length l = 5m, h = 4m. Equation 56 was evaluated to show how the contribution of axial deformation varied with đ??ź1 đ??ź2 . The result is shown in Table 2. When plotted on a uniform scale (Figure 9) the relationship between ∆đ?‘€đ??ľ and đ??ź1 đ??ź2 is seen to be linear. This was further justified by a linear regression analysis of the results in Table 2 which was fitted into the đ??ź model ∆đ?‘€đ??´ = đ?‘ƒ1 đ??ź1 + đ?‘ƒ2 to obtain đ?‘ƒ1 = −0.004815 and 2

đ?‘ƒ2 = 0.0001109 and the fitness parameters sum square of errors (SSE), coefficient of multiple determination (R2 ) and the root mean squared error (RMSE) gave 1.099 x 10 -7, 0.999952 and 0.0001105 respectively. From Table 2 when đ??ź1 đ??ź2 = 0 , ∆đ?‘€đ??´ ≅ 0 and when đ??ź1 đ??ź2 = 10; ∆đ?‘€đ??´ = −0.0479 which represent only a 5% reduction in the calculated bending moment. In like manner by evaluating the axial contribution in the beam, ∆đ?‘€đ??ľ for varying đ??ź1 đ??ź2 of the portal frame, Table 3 was produced. This was plotted on a uniform scale in Figure 10. From Figure 10 it would be observed that there is also a linear relationship between ∆đ?‘€đ??ľ and đ??ź1 đ??ź2 . When đ??ź fitted into the model ∆đ?‘€đ??ľ = đ?‘ƒ3 đ??ź1 + đ?‘ƒ4 it gave đ?‘ƒ3 = 2

−0.006502 and đ?‘ƒ4 = −0.0003969 for the fitness parameters sum square of errors (SSE), coefficient of multiple determination (R2 ) and the root mean squared error (RMSE) of 6.51 x 10-7, 0.9999 and 0.000269 respectively. By pegging đ??ź1 đ??ź2 to a constant value of 0.296 and the variation of ∆đ?‘€đ??´ with respect to â„Ž đ??ż investigated, Table 4 was generated. A detailed plot of Table 4 was presented in Figure 11. From Table 4 when â„Ž đ??ż = 0; ∆đ?‘€đ??´ = −3.125 (about 30% drop in calculated bending moment value) while at â„Ž đ??ż > 0, ∆đ?‘€đ??´ dropped in magnitude exponentially to values below 1. In like manner, when the variation of ∆đ?‘€đ??ľ with respect to â„Ž đ??ż was investigated, Table 5 was generated. A detailed plot of Table 5 was presented in Figure 12. From Table 5 when â„Ž đ??ż = 0; ∆đ?‘€đ??´ = −4.1667 (about 40% drop in calculated bending moment value) while at â„Ž đ??ż > 0, ∆đ?‘€đ??´ dropped in magnitude exponentially to values below 1.

V. CONCLUSION The flexibility method was used to simplify the analysis and a summary of the results are presented in table 1. The equations in table 1 would enable an easy evaluation of the internal stresses in loaded rigidly fixed portal frames considering the effect of axial deformation. From a detailed analysis of frame 1 (Figure 3), it was observed that the contribution of axial deformation is generally very small and can be neglected for reasonable values of đ??ź1 đ??ź2 . However, its contribution skyrockets at very low values of â„Ž đ?‘™ i.e. as â„Ž đ?‘™ => 0. This depicts the case of an encased single span beam and a complete departure from portal frames under study. This analysis can be extended to the other loaded frames (Figures 4 – 8) using the equations in Table 1. These would enable the determination of safe conditions for ignoring axial deformation under different kinds of loading.

REFERENCES Ghali A, Neville A. M. (1996) Structural Analysis: A Unified Classical and Matrix Approach (3rd Edition) Chapman & Hall London Graham R, Alan P.(2007). Single Storey Buildings: Steel Designer’s Manual Sixth Edition, Blackwell Science Ltd, United Kingdom Hibbeler, R. C.(2006). Structural Analysis. Sixth Edition, Pearson Prentice Hall, New Jersey Leet, K. M., Uang, C.,(2002). Fundamental of Structural Analysis. McGraw-Hill ,New York McGuire, W., Gallagher R. H., Ziemian, R. D.(2000). Matrix Structural Analysis,Second Edition, John Wiley & Sons, Inc. New York Nash, W.,(1998). Schaum’s Outline of Theory and Problems of Strength of Materials. Fourth Edition, McGraw-Hill Companies, New York Okonkwo V. O (2012). Computer-aided Analysis of Multistorey Steel Frames. M. Eng. Thesis, Nnamdi Azikiwe University, Awka, Nigeria. Reynolds, C. E.,and Steedman J. C. (2001). Reinforced Concrete Designer’s Handbook, 10th Edition) E&FN Spon, Taylor & Francis Group, London Samuelsson A., and Zienkiewicz O. C.(2006), Review: History of the Stiffness Method. International Journal for Numerical Methods in Engineering. Vol. 67: 149 – 157 Stroud K. A.,(1995). Engineering Mathematics. Fourth Edition, Macmillan Press Ltd, London.

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645

Table 1: Internal stresses for a loaded rigid frame

S/No

LOADED FRAME

REMARKS

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝑤 𝑙 3 𝐼1 4ℎ 3 𝐼2 −𝑠2 𝑙 3 𝐼1

𝑀𝐴 = 𝑀𝐷 = 12 4ℎ 3 𝐼

3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝑀𝐵 = 𝑀𝐶 = 12 4ℎ 3 𝐼

−𝑤𝑙 3 𝐼1 8𝐼2 ℎ 3 +𝑠2 𝑙 3 𝐼1

3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2

𝑉𝐴 = 𝑉𝐷 =

𝑤𝑙

𝑤 ℎ 2 𝑙 3 𝐼1 𝐼2

𝐻𝐴 = 𝐻𝐷 = 4ℎ 3 𝐼

3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2

2

See equations (19) – (22)

𝑤𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +8𝑙𝐼1 +𝑙 3 𝐼1 𝑠1 𝑙𝐼1 +6𝑠2 ℎ 𝐼2 48 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝑙𝐼1 +2ℎ𝐼2

𝑀𝐴 =

𝑀𝐵 = − 64 𝑀𝐶 = − 𝑀𝐷 = 𝑉𝐷 =

𝑤𝑙 4 5ℎ𝐼1 +24ℎ𝐼2

+

𝑙 2 𝐼1 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑤𝑙 3𝐼1 𝑙 3 +24𝐼2 𝑙 2 ℎ +16𝑠1 ℎ 3 𝐼2 64 𝐼1 𝑙 3 +6𝐼2 𝑙 2 ℎ+2𝑠1 ℎ 3 𝐼2

𝑤 𝑙 4 5ℎ 𝐼1 +24ℎ 𝐼2 64 𝑙 2 𝐼1 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑤 𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +2𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +6ℎ 𝐼2 48 4ℎ 3 𝐼2 2𝑙ℎ 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2

+

𝑤 𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +2𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +6ℎ𝐼2 48 4ℎ 3 𝐼2 2𝑙ℎ 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝑤𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +8𝑙𝐼1 +𝑙 3 𝐼1 𝑠1 𝑙𝐼1 +6𝑠2 ℎ 𝐼2 48 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝑙𝐼1 +2ℎ𝐼2

𝑤𝑙 3𝑙 3 𝐼1 +24ℎ𝑙 2 𝐼2 +16𝑠1 ℎ 3 𝐼2 32 𝑙 3 𝐼1 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝐻𝐴 = 𝐻𝐷 = 2 4ℎ 3 𝐼

𝑉𝐴 =

𝑤𝑙 3𝐼1 𝑙 3 +24𝐼2 𝑙 2 ℎ +16𝑠1 ℎ 3 𝐼2 64 𝐼1 𝑙 3 +6𝐼2 𝑙 2 ℎ +2𝑠1 ℎ 3 𝐼2 𝑤𝑙 2

− 𝑉𝐷

𝑤 ℎ 2 𝑙 3 𝐼1 𝐼2

3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝑙𝐼1 +2ℎ𝐼2

See equations (26) – (29)

𝑤ℎ 2 𝐼1 𝑙 3 +5ℎ 𝑙 2 𝐼2 +2𝑠2 ℎ 3 𝐼2

𝑀𝐴 = − 𝑀𝐵 = 2

2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 𝑤ℎ 3 𝑙 2 𝐼2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑀𝐶 = − 2 𝑀𝐷 = −

3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝑤 ℎ 3 𝐼2 −4ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

+ 6 4ℎ 3 𝐼

𝑤 ℎ 3 𝑙 2 𝐼2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 𝑤ℎ 3 𝑙 2 𝐼2

2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑉𝐴 = 𝑉𝐵 = 𝐼

𝑤 ℎ 3 𝐼2 9ℎ 2 𝑙𝐼1 +5ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1

+ 6 4ℎ 3 𝐼

+ 6 4ℎ 3 𝐼

𝑤 ℎ 3 𝐼2 −4ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1

3 2 2𝑙 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2

+

𝑤 ℎ 3 𝐼2 9ℎ 2 𝑙𝐼1 +5ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 6 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝑤 ℎ 3 𝑙𝐼2

3 2 3 1 𝑙 +6𝐼2 𝑙 ℎ +2𝑠1 𝐼2 ℎ

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝐻𝐷 =

𝑤 ℎ 4 𝐼2 3𝑙𝐼1 +2ℎ𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝐻𝐴 = 𝑤ℎ − 𝐻𝐷

See equations (32) – (35)

3𝑎 𝑙 2 𝐼2 3 2 3 1 𝑙 +6ℎ 𝑙 𝐼2 +2𝑠1 ℎ 𝐼2

3𝑃𝑎 2 𝑙 2 𝐼2 3 2 3 1 𝑙 +6ℎ 𝑙 𝐼2 +2𝑠1 ℎ 𝐼2

𝑀𝐴 = −𝑃𝑎 1 − 𝐼 𝑀𝐶 = − 𝐴

𝑀𝐵 = 𝐼

3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1

𝑀𝐷 = 𝑃𝑎 1 − 𝐼

3 2 1 𝐼1 𝑙 +6𝐴1 ℎ 𝑙 𝐼2 +24ℎ𝐼1 𝐼2

3 2 3 1 𝑙 +6ℎ𝑙 𝐼2 +2𝑠1 ℎ 𝐼2

6𝑃𝑎 2 𝑙𝐼2

𝑉𝐴 = 𝑉𝐵 = 𝑙 3 𝐼

HD

3𝑎𝑙 2 𝐼2

𝐻𝐴 = 𝐻𝐵 = 𝑃

2 3 1 +6𝑙 ℎ 𝐼2 +2𝑠1 ℎ 𝐼2

See equations (38) – (41)

𝑀𝐴 = −𝑃𝑎 + 2 𝑀𝐵 = 2

3𝑃𝑎 2 𝑙 2 𝐼2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑀𝐶 = − 2 𝑀𝐷 = −

+

𝑃𝑎 2 4𝐼1 𝐼2 ℎ𝑙 3ℎ −𝑎 +𝐼2 8ℎ 3 𝐼2 −4𝑎ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2

𝑃𝑎 2 𝐼2 −4ℎ 3 𝐼2 +4𝑎ℎ 2 𝐼2 +𝑠2 𝑙 3 𝐼1 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

+ 2 4ℎ 3 𝐼

3𝑃𝑎 2 𝑙 2 𝐼2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 3𝑃𝑎 2 𝑙 2 𝐼2

2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑉𝐴 = 𝑉𝐵 = 𝐻𝐷 =

3𝑃𝑎 2 𝑙 2 𝐼2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑃𝑎 2 𝐼2 −4ℎ 3 𝐼2 +4𝑎ℎ 2 𝐼2 +𝑠2 𝑙 3 𝐼1 3 2 2𝑙 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2

+ 2 4ℎ 3 𝐼 +

𝑃𝑎 2 4𝐼1 𝐼2 ℎ𝑙 3ℎ −𝑎 +𝐼2 8ℎ 3 𝐼2 −4𝑎ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

3𝑃𝑎 2 𝑙𝐼2 𝑙 3 𝐼1 +6𝑙 2 ℎ 𝐼2 +2𝑠1 ℎ 3 𝐼2

2𝑃𝑎 2 𝐼2 𝑙𝐼1 3ℎ −𝑎 +ℎ𝐼2 3ℎ −2𝑎 4ℎ 3 𝐼2 2𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2

𝐻𝐴 = 𝑃 − 𝐻𝐷

See equations (44) – (47)

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4252 | Page


International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝑀𝐴 = −𝑃𝑎 + 𝑀𝐵 = −𝑃𝑎 + 𝑀𝐶 = −

𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ 𝐼2 6𝑎𝑙 +𝑠1 ℎ 2 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ 𝐼2 6𝑎𝑙 +𝑠1 ℎ 2 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑃 𝑎 2 𝐼1 3𝑙−2𝑎 +6𝑎ℎ𝑙𝐼2 +𝑠1 ℎ 3 𝐼2 𝑙 3 𝐼1 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝐻𝐴 = 𝐻𝐵 =

+

2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2

𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ 𝐼2 6𝑎𝑙 +𝑠1 ℎ 2

𝑀𝐷 = − 𝑉𝐷 =

𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ𝐼2 6𝑎𝑙 +𝑠1 ℎ 2

+ +

+

𝑃𝑎 4ℎ 3 𝐼2 −𝑎𝐼1 +ℎ𝐼2 +3𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2 𝑃𝑎 4ℎ 3 𝐼2 2𝑎𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝑃𝑎 4ℎ 3 𝐼2 2𝑎𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑃𝑎 4ℎ 3 𝐼2 −𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

𝑉𝐴 = 𝑃 − 𝑉𝐷

6𝑃𝑎 ℎ 2 𝐼1 𝐼2 𝑙−𝑎 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2

See equations (50) – (53) Table 2: Axial deformation contribution ∆𝑴𝑨 for different values of

𝑰𝟏

𝑰𝟐

w =1kN/m L = 5m H = 0.4m h = 4m 𝑰𝟏

𝑰𝟐

0

1

2

3

4

5

∆𝑴𝑨

0

-0.0045

-0.0095

-0.0144

-0.0192

-0.0241

𝑰𝟏

6

7

8

9

10

-0.0289

-0.0337

-0.0384

-0.0432

-0.0479

𝑰𝟐

∆𝑴𝑨

Table 3: Axial deformation contribution ∆𝑴𝑩 for different values of

𝑰𝟏

𝑰𝟐

w =1kN/m L = 5m H = 0.4m h = 4m 𝑰𝟏

𝑰𝟐

0

1

2

3

4

5

∆𝑴𝑩

0

-0.0066

-0.0135

-0.0201

-0.0267

-0.0332

𝑰𝟏

6

7

8

9

10

-0.0396

-0.0460

-0.0524

-0.0587

-0.0651

𝑰𝟐

∆𝑴𝑩

Table 4: Axial deformation contribution ∆𝑴𝑨 for different values of 𝒉 𝒍 w =1kN/m L = 5m 𝒉

𝒍

∆𝑴𝑨 𝒉

𝒍

∆𝑴𝑨

𝐼1

𝐼2 = 0.296

0

0.1

0.2

0.3

0.4

0.5

-3.1250

-0.5409

-0.0823

-0.0239

-0.0096

-0.0047

0.6

0.7

0.8

0.9

1.0

-0.0026

-0.0015

-0.0010

-0.0006

-0.0004

Table 5: Axial deformation contribution ∆𝑴𝑩 for different values of 𝒉 𝒍 www.ijmer.com

4253 | Page


International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 w =1kN/m L = 5m đ?’‰

đ??ź1

đ??ź2 = 0.296

0

0.1

0.2

0.3

0.4

0.5

∆đ?‘´đ?‘¨

-4.1667

-0.7668

-0.1208

-0.0359

-0.0146

-0.0072

đ?’‰

0.6

0.7

0.8

0.9

1.0

-0.0040

-0.0024

-0.0015

-0.0010

-0.0007

đ?’?

đ?’?

∆đ?‘´đ?‘¨

Figure 9: A graph of ∆đ?‘´đ?‘¨ against

đ?‘°đ?&#x;?

Figure 10: A graph of ∆đ?‘´đ?‘Š against

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đ?‘°đ?&#x;?

đ?‘°đ?&#x;?

đ?‘°đ?&#x;?

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International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645

Figure 11: A graph of ∆đ?‘´đ?‘¨ against đ?’‰ đ?‘ł

Figure 12: A graph of ∆đ?‘´đ?‘Š against đ?’‰ đ?‘ł

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