CiteSeerX — A New Theoretical Model

Abstract— The paper reports theoretical analysis of the beam (concrete, steel and composite) with any condition of support stiffness under concentrated loading acting at any point on the beam. Equations are available only for simply supported and fixed condition. But for the intermediate support condition (in between pin and rigid) the existing equations are not valid. Using area-moment theorem and assuming elastic behavior; equations have been derived for end moment, maximum deflection and location of maximum deflection for different position of loads and different support conditions. Then with the help of EXCEL workbook the design chart have been prepared. Finally the results obtained by the derived equations are checked with standard equations.

Index Terms— Moment, Deflection, Support stiffness, Area moment.

I. INTRODUCTION

ONCRETE structure should be designed in such a way so that it is economical and safe for the users. Generally, a beam to column connection of a frame is considered either fully rigid or pin. But from numerous tests (Davison et.

al., 1990 and Li et. al., 1996) and numerical studies (Ahmed, B. and Nethercot, D. A., 1995, 1996, 1997, 1998) it is seen that the actual behavior at the connection between beam and column is neither pin nor rigid but rather partially rigid.

Practically all the beam and column are connected to each other between rigid and pin connection. Support condition has a great influence on moment and deflection e.g. the end moment of simply supported full span uniformly loaded beam is zero where as the fixed supported identical beam is wl²/12 and the first one deflects five times more than the second condition. Thus the moment and deflection are changed from

Manuscript received March 7, 2010.

1Md. Wahid Ferdous, Lecturer, Department of Civil Engineering, Rajshahi University of Engineering & Technology, Rajshahi – 6204, Bangladesh. (e-mail: [email protected]).

2Kamrun Nahar, Lecturer, Department of Civil Engineering, Rajshahi University of Engineering & Technology, Rajshahi – 6204, Bangladesh. (e-mail: [email protected]).

3Dr. Shaikh Md. Nizamud-doulah, Professor, Department of Civil Engineering, Rajshahi University of Engineering & Technology, Rajshahi – 6204, Bangladesh. (e-mail: [email protected]).

4Khaleda Ferdous, Department of Biotechnology and Genetic Engineering, Islamic University, Kustia, Bangladesh. (e-mail:

[email protected]).

the actual moment and deflection when the support condition is considered as pin or rigid not considering the actual one. As a result the design is affected. To overcome this problem the stiffness of support should be considered for calculating moment and deflection and this paper provides an opportunity to analysis the beam from a new angle.

II. GENERALEXPRESSIONFORENDMOMENT

Let us assume a concentrated load

P

acts on the beam of length

L

at a distance

X

from the support point

A

where

K

_A and

K

_B are the support stiffness of point

A

and

B

respectively.

From the definition of stiffness we get,

A A

A

K

= M

θ

and

B B

B

K

= M θ

Where

M

_A and

M

_B are the end moment of corresponding point respectively. By using area-moment theorem (Pytel, A.

and Singer, F. L.), in Fig. 1(c).

L t

_{B A}

A

tan θ =

/

Since

θ

_A is very small, hence

tan θ

_A

= θ

_A

]

) 1 (

1 [

BA B A

A

Area X

EI L K

M = ×

Similarly,

] )

1 ( 1 [

AB A B

B

Area X

EI L K

M = ×

1

)

( f

L X L

M

_A

= PX −

....………..…………. (1)

2

)

( f

L X L

M

_B

= PX −

... (2)

A New Theoretical Model for Moment and Deflection

Considering the Effect of Support Stiffness

Md. Wahid Ferdous, Kamrun Nahar, Dr. Shaikh Md. Nizamud-doulah and Khaleda Ferdous.

C

Fig. 1. Semi rigidly connected beams with one concentrated load.

Where,

] 3 ) 3 )(

( 1 6 6

3 ) )(

2 ( ) 1 6 (

1

[

m n n m

m n L X L

X n f

+ +

−

+

−

− +

=

……… (3)

] 3 ) 3 )(

( 1 6 6

3 ) )( 1 1 ( ) 2 6 (

2

[

m n n m

L m n X L X n f

+ +

−

+ +

−

−

=

………... (4)

A B

K

n = K

= Support stiffness ratio and

K

A

L EI m

) (

=

= Member to support stiffness ratio.

Equation (1) and (2) gives the general expression for end moments of beam having different support stiffness when a concentrated load

P

acts at a distance

X

from the support point A.

The values of f and ₁

f

₂ are obtained from equations (3) and (4) using different values of

n

,

m

and different

X L

_ratios.

These f and ₁

f

₂ values are furnished in Fig.2 to Fig.11.

0.0 0.2 0.4 0.6 0.8 1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 1

n

Fig. 2. Variation of

f

₁ with

n

for different values of

m

and for

X L = 0

0.0 0.2 0.4 0.6 0.8 1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f1

n

Fig. 3. Variation of

f

₁ with

n

for different values of

m

and for

X L = 0 . 25

X

P

A B

K

_A

K

_B

L

(a) Concentrated load acting on beam

L X L PX ( − )

M

_A

M

_B (b) Moment diagram by parts

X

₀

A C B tA/C Dmax

tB/A

(c) Deflected shape showing tangent

0.0 0.2 0.4 0.6 0.8 1.0 0.0

0.1 0.2 0.3 0.4 0.5 0.6 0.7

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f1

n

Fig. 4. Variation of

f

₁ with

n

for different values of

m

and

for

X L = 0 . 5

0.0 0.2 0.4 0.6 0.8 1.0 0.0

0.1 0.2 0.3 0.4 0.5 0.6

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 1

n

Fig. 5. Variation of

f

₁ with

n

for different values of

m

and

for

X L = 0 . 75

0.0 0.2 0.4 0.6 0.8 1.0 0.0

0.1 0.2 0.3 0.4 0.5 0.6

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 1

n

Fig. 6. Variation of

f

₁ with

n

for different values of

m

and for

X L = 1

0.0 0.2 0.4 0.6 0.8 1.0 0.00

0.02 0.04 0.06 0.08 0.10 0.12 0.14

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 2

n

Fig. 7. Variation of

f

₂ with

n

for different values of

m

and for

X L = 0

0.0 0.2 0.4 0.6 0.8 1.0 0.00

0.05 0.10 0.15 0.20 0.25 0.30

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 2

n

Fig. 8. Variation of

f

₂ with

n

for different values of

m

and for

X L = 0 . 25

0.0 0.2 0.4 0.6 0.8 1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 2

n

Fig. 9. Variation of

f

₂ with

n

for different values of

m

and for

X L = 0 . 5

0.0 0.2 0.4 0.6 0.8 1.0 0.0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 2

n

Fig. 10. Variation of

f

₂ with

n

for different values of

m

and for

X L = 0 . 75

0.0 0.2 0.4 0.6 0.8 1.0 0.0

0.2 0.4 0.6 0.8 1.0

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

f 2

n

Fig. 11. Variation of

f

₂ with

n

for different values of

m

and for

X L = 1

III. GENERALEXPRESSIONFORTHELOCATIONAND

MAGNITUDEOFMAXIMUMDEFLECTION

Let us assume

X

₀ is the distance of maximum deflection measured from support point

A

and

C

is the point of maximum deflection that is shown in Fig. 1.

To find out the location and magnitude of maximum deflection two cases may arises-

Case I:

Distance of maximum deflection measured from support point

A

is less than or equal to the distance of the concentrated load measured from support point

A

i.e.

X

₀

≤ X

Case II:

Distance of maximum deflection measured from support point

A

is more than or equal to the distance of the concentrated load measured from support point

A

i.e.

X

₀

≥ X

From area moment theorem, _AC ^{B A}

Area

_AC

EI

L

t

_/

= 1 ( ) θ =

1) For Case I: (

X

₀

≤ X

)

1

1 1 2 1 0 1

2

) 4 (

a

c a b b L

X − ± −

=

……… (5)

Where,

₁

1 ( f

₁

f

₂

) L

a = + X −

₁

2 f

₁

L b = − X

{ 2 ( 2 )}

3

^{1 2}

1

f f

L X L

c = − X − − +

2) For Case II: (

X

₀

≥ X

)

2

2 2 2 2 0 2

2

) 4 (

a

c a b b

L

X − ± −

=

……… (6)

Where,

₂

= ( 1 − )( f

₁

− f

₂

) − 1 L

a X

₂

2 [ 1 ( 1 ) f

₁

] L b = − − X

L f X L f

X L

c = − ( 1 − X )[( 2 − ) − ( 2 + )] − 3

1

2 1

2

IV. VARIATION OF

X

₀

L

FOR DIFFERENT VALUES OF

n

,

m

AND DIFFERENT POSITION OF LOADS

We got two equations (equation 5 and equation 6) for calculating

X

₀

L

. Equation (5) should be used only when the values of

X

₀

L

obtained from equation (5) becomes less than the values of

X L

which is almost known to us.

Equation (6) is valid only when the values of

X

₀

L

obtained from equation (6) becomes more than the values of

X L

. Fig. 12 to Fig

. 14 are drawn by considering this case.

0.0 0.2 0.4 0.6 0.8 1.0 0.39

0.40 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

X 0/L

n

Fig. 12. Variation of

X

₀

L

with

n

for different values of

m

and for

X L = 0 . 25

0.0 0.2 0.4 0.6 0.8 1.0 0.50

0.51 0.52 0.53 0.54 0.55 0.56

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

X0/L

n

Fig. 13. Variation of

X

₀

L

with

n

for different values of

m

and for

X L = 0 . 5

0.0 0.2 0.4 0.6 0.8 1.0 0.56

0.57 0.58 0.59 0.60 0.61 0.62 0.63 0.64

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

X0/L

n

Fig. 14. Variation of

X

₀

L

with

n

for different values of

m

and for

X L = 0 . 75

V. GENERAL EXPRESSION FOR MAXIMUM DEFLECTION

In this case, _{A C}

Area

_AC

X

_A

t EI

D = = 1 ( ) ×

/ max

1) For Case I: (

X

₀

≤ X

)

)]

( 2 3 )[

1 )(

( ) 6 (

) 1 ( ) 3 (

2 1 0 1

0 2 3

0 3 3

max

f L f

f X L X L

X L X EI PL

L X L

X EI D PL

−

−

−

−

−

=

……… (7)

r s s

s

D

D

D =

₋

−

₋

⇒

_max

Where,

) 1 ( ) 3 (

3 0 3

L X L

X EI

D

_s−s

= PL −

= Deflection due to simply supported action.

)]

( 2 3 )[

1 )(

( )

6 (

^{1 2}

0 1

0 2 3

f L f

f X L X L

X L X EI

D

_s−r

= PL − − −

= Reduction of deflection due to partially rigid action.

2) For Case II: (

X

₀

≥ X

)

] ) )(

2 ( ) )(

)(

1 ( 3

) 1 ( ) ( 2 )[

6 (

0 2 0 0

0 0

2 3

max

L X L X L X L X L

X L X L X L X L X

L X L

X L X EI D PL

− +

+ +

−

− +

−

=

)]

2 ( )

1 ( 3 )[

1 ( ) )(

6 ( ^{1 2}

0 1 2 0

0 3

f L f

f X L X L

X L

X L X EI

PL − − + +

−

………. (8)

r s s

s

D

D

D =

₋

−

₋

⇒

_max

Where,

] ) )(

2 ( ) )(

)(

1 ( 3

) 1 ( ) ( 2 )[

6 (

0 2 0 0

0 0

2 3

L X L X L X L X L

X L X L X L X L X

L X L

X L X EI D_{s s} PL

− +

+ +

−

− +

−

− =

= Deflection due to simply supported action.

)]

2 ( )

1 ( 3 )[

1 ( ) )(

6 ( ^{1 2}

0 1 0 2

0 3

f L f

f X L X L

X L

X L X EI

D_s−r =PL − − + +

= Reduction of deflection due to partially rigid action.

From equation (7) we get,

)]

( 2 3 )[

1 )(

( ) 6 ( 1

) 1 ( ) 3 ( 1

2 1 0 1

0 2

0 3 3

max

f L f

f X L X L

X L X

L X L

X PL

EID

−

−

−

−

−

=

………..…. (9) From equation (8) we get,

] ) )(

2 ( ) )(

)(

1 ( 3

) 1 ( ) ( 2 )[

6( 1

2 0 0 0

0 0

2 3

max

L X L X L X L X L

X L X L X L X L X

L X L

X L X PL

EID

− +

+ +

−

− +

−

=

)]

2 ( )

1 ( 3 )[

1 ( ) )(

6 ( 1

2 1 0 1 0 2

0

f f

L f X L X L

X L

X L

X − − + +

−

……….. (10)

VI. VARIATION OF ^max₃

PL EID

FOR DIFFERENT VALUES OF

n

,

m

AND DIFFERENT POSITION OF LOADS

Again we got two equations (equation 9 and equation 10) for calculating ^max₃

PL

EID

. Equation (9) can be used only

when the values of

X

₀

L

is obtained from equation (5) and equation (10) is valid only when the values of

X

₀

L

is obtained from equation (6). Fig. 15 to Fig. 17 are drawn by considering this case.

It should be noted that knowing the support stiffness the end moments can be obtained; using these end moments location of maximum deflection can be computed. The magnitude of maximum deflection can be obtained using the end moments and the location of maximum deflection.

0.0 0.2 0.4 0.6 0.8 1.0

0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

EID max/PL3

n

Fig. 15. Variation of

EID

_max

PL

³ with

n

for different values of

m

and for

X L = 0 . 25

0.0 0.2 0.4 0.6 0.8 1.0

0.005 0.010 0.015 0.020 0.025 0.030

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

EIDmax/PL3

n

Fig. 16. Variation of

EID

_max

PL

³ with

n

for different values of

m

and for

X L = 0 . 5

0.0 0.2 0.4 0.6 0.8 1.0

0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 0.020 0.022

m = 0.01 m = 0.05 m = 0.1 m = 0.2 m = 1 m = 100

EIDmax/PL3

n

Fig. 17. Variation of

EID

_max

PL

³ with

n

for different values of

m

and for

X L = 0 . 75

VII. DESIGN CHART

Fig. 2 to Fig. 11 provides the coefficients (with respect to connection stiffness ratio

n

and ratio of beam to connection stiffness

m

) required for computing the end moments. Fig. 12 to Fig. 14 provides the location of maximum deflection with respect to

n

and

m

. Fig. 15 to Fig. 17 shows the generalized deflection calculation chart in terms of

3 max

PL

EID

with respect to

n

and

m

. For known span, load, allowable deflection and preferred ratio of

n

it is possible to obtain the required beam section.

VIII. RESULT AND DISCUSSION

A. For rigid condition and center point loading

Fig. 18. Beam with rigid condition and center point loading.

When beam and column are connected to each other by identical support with completely rigid condition and if the concentrated load acts at the mid point of the beam (which is shown in Fig. 18) then

K

_A

= K

_B and both tends to infinity.

So,

= = 1

A B

K

n K

and

m

indicates very small value. From

equation (3) and (4) we get,

2 1

2

1

= f =

f

.

8 ) ( )

* ( ) (

2

2

PL

L X L PX L

X L L

X L

M

_A

= PX − − = − =

8 )

* ( ) (

2

2

PL

L X L PX L X L

X L

M

_B

= PX − = − =

2

0

= 1 L

X

and

EI D PL

192

3

max

=

.

B. For simply supported condition and center point loading

Fig. 19. Beam with simply supported condition and center point loading.

In simply supported structure (which is shown in Fig. 19), with identical support condition

K

_A

= K

_B and both tends to zero. So,

n = 1

and

m

tends to infinity i.e. indicates very large value. Putting these values in equation (3) and (4) we get,

f

₁

= f

₂

= 0

So,

M

_A

= M

_B

= 0

,

2

0

= 1 L

X

and

EI D PL

max

= 48

.

C. For propped cantilever beam with right end pined and center point loading

Fig. 20. Propped cantilever beam with right end pined and center point loading.

In propped cantilever beam (which is shown in Fig. 20),

K

A tends to infinity and

K

_B tends to zero. So,

m

indicates very small value and

n = 0

. Putting this value in equation (3) and (4) we get,

4 3

1

=

f

and

f

₂

= 0

.

So,

16

M

_A

= 3PL

and

M

_B

= 0

,

5 1 1

0

= −

L

X

and

EI D PL

5 48

3

max

=

Numerous codes attempt to check the deflection of beams (concrete, steel and composite) by imposing various depth-to- span ratios, but the contribution of beam-to-column connection stiffness is neglected in all the cases. Also from the codes the actual deflection is not known. This paper provides a systematic approach to compute the deflection and beam-to- column connection stiffness to ensure that the deflection can be kept within the allowable limit.

IX. CONCLUSION

From this study the following conclusions can be drawn, 1) A new theoretical model for moment and deflection has been developed.

2) This model can explain the variation of moment and deflection of a beam for varying support condition.

3) From this model, the actual moment developed at the end of the beam under concentrated load acting at any point of the beam with any support stiffness can easily be calculate.

4) The actual magnitude of maximum deflection and its location of a beam under concentrated load acting at any point of the beam with any support stiffness can easily be computed.

P

X

A B

L

P

X

A B

L

P X

A B L

Finally, it can be stated that the support stiffness of beam- to-column connection has an important role for the actual analysis. Every engineer engaged in construction engineering must pay prime attention to it. For economic design and analysis, the traditional concept of frame analysis should be change.

X. REFERENCES

[1] Davison, J. B., Lam, D. and Nethercot, D. A. (1990).

“Semi-rigid action of composite joints”. The Structural Engineer, Vol. 68, No. 24, 489-499.

[2] Li, T. Q., Nethercot, D. A. and Choo, B. S. (1996).

“Behaviour of flush end plate composite connections with unbalanced moment and variable shear/moment ratios: part 1:

experimental behaviour”. Journal of Construction Steel Research, Vol. 38, No. 2, 125-164.

[3] Ahmed, B. and Nethercot, D. A. (1998). “Effect of column axial load on composite connection behaviour”. Engineering Structures. Vol. 20, No. 1-2, 113-128.

[4] Ahmed, B. and Nethercot, D. A. (1997). “Prediction of initial stiffness and available rotation capacity of major axis composite flush end plate connections”. Journal of Construction Steel Research, Vol. 41, No. 1, 30-60.

[5] Ahmed, B. and Nethercot, D. A. (1996). “Effect of high shear on the moment capacity of composite cruciform end plate connections”. Journal of Construction Steel Research, Vol. 40, No. 2, 129-163.

[6] Ahmed, B. and Nethercot, D. A. (1995). “Numerical modeling of composite flush end plate connections”. Journal of Singapore Structural Steel Society, Vol. 6, No. 1, 87-102.

[7] Ahmed, B. “Deflection of semi-rigidly connected beams”.

Journal of Civil Engineering, The Institution of Engineers, Bangladesh. Vol. CE 29, No. 2, pp. 133-150. December, 2001.

[8] Pytel, A. and Singer, F. L. Strength of materials, 4^th edition, McGraw-Hill Book Company.

Md. Wahid Ferdous

Mr. Md. Wahid Ferdous was born in Harinakundu, in the district of Jhenidah in Bangladesh on 2 December 1985. He obtained the degree of Bachelor of Science in Civil Engineering from Rajshahi Univerisity of Engineering &

Technology in the year 2008 and secured First Class Honors with First Position out of One Hundred Students. He is selected for the University Gold Medal for the best overall performance in the final year of Bachelor of Science in Civil Engineering Degree. He also got best student awards in the year of 2004 and 2007. His major field is Civil Engineering.

Just after getting his B.Sc. Degree, he joined as a Lecturer in the Department of Civil Engineering, Rajshahi Univerisity of Engineering & Technology, Rajshahi, Bangladesh.

Kamrun Nahar

Kamrun Nahar was born in Rajshahi District in Bangladesh on 6 July 1986.

She obtained the degree of Bachelor of Science in Civil Engineering from Rajshahi Univerisity of Engineering &

Technology in the year 2008 and secured First Class Honors with Second Position out of One Hundred Students. She was awarded the Government Board Scholarship for four years due to her outstanding performance in the Higher Secondary Examination. She also got the best student awards in the year of 2005 and 2006. Her major field is Civil Engineering. Just after getting his B.Sc. Degree, she joined as a Lecturer in the Department of Civil Engineering, Rajshahi Univerisity of Engineering & Technology, Rajshahi, Bangladesh.

Dr. Shaikh Md. Nizamud-doulah

Dr. Shaikh Md. Nizamud-doulah is presently the Professor and Dean of the Faculty of Civil Engineering, Rajshahi Univerisity of Engineering &

Technology, Rajshahi, Bangladesh. He got his B.Sc. Engineering degree from Rajshahi University; M.Tech. from IIT Kharagpur, India; & Ph.D. from BUET.

He is an expert in Structural Engineering.

Khaleda Ferdous

Khaleda Ferdous was born in Harinakundu, in the district of Jhenidah in Bangladesh on 10 April 1984. She obtained the degree of Bachelor of Science (Honors) from Islamic University with First Class in 2008. Now she is doing her Master program from the same university. She is interested to give her knowledge among the people for the betterment of them as well as for her country.