Design of Circular Base Plate under Large Bending Moment WhichThere Is a Little Gap between Base Plate and Foundation

A. A. GholampourDepartment of Civil Engineering

University of TehranTehran, Iran

aagholampour@ut.ac.ir

M. NaghipourDepartment of Civil EngineeringBabol University of Technology

Babol, Iranm-naghi@nit.ac.ir

A. SobatiDepartment of Civil EngineeringBabol University of Technology

Babol, Irana.sobati@stu.nit.ac.ir

Abstract— Circular base plates are commonly used forcylindrical columns; such as pylons in cabled-stayed bridges,lighting poles, electric power line posts, and even buildings. Insome structures a little gap is considered between base plateand foundation. In this case compressive bolts which transmitcompressive forces to the foundation are also considered indesign of base plate. The purpose of this study is design ofcircular base plates on the effect of applying the axial load andlarge bending moment in the case that tensile and compressiveforces are transmitted to foundation via bolts. An exactmethod has not been presented about design of this status yet,therefore in this study we can offer suitable method for designof circular base plates by expression of mathematic formulas.By considering the graphs in the end of this paper, can beconcluded that, this status is very useful for axial force withlarge eccentric. In other words, this status for axial force withlarge eccentric gives little design thickness than the otherstatus that a gap is not considered.

Keywords: circular base plate; compressive bolt; little gap;large bending moment; mathematic formulas

I. INTRODUCTION

Generally steel columns are placed on the foundationwith steel plate which is involved on the one hand withcolumn and on the other hand with concrete. Since the steelcolumn undergo the large stresses due to high resistance andconcrete has no ability to undergo these stresses, thereforebase plate is used for increasing the force transmission levelto foundation and reducing it to allowable limit of foundationmaterials strength. From the types of base plates can bementioned rectangular or circular base plates, which circularbase plates are commonly used for cylindrical columns suchas pylons in cabled-stayed bridges, lighting poles, electricpower line posts, and even buildings. In this paper from thetypes of loading states in columns such as; pure axial force,axial force and small bending moment, and axial force andlarge bending moment, we are going to show a newprocedure for designing of base plate in state of loading withaxial force and large bending moment. Usually to designbase plates in this status, no gap is supposed between plateand foundation. By considering the stress distribution in thisstatus, tensile forces are transferred by anchor bolts andcompressive forces are transferred by base plate tofoundation. Therefore by considering the maximum tolerablecontact stress of concrete foundations according to AISC

code which is equal to cf ′7.0 ( cf ′ is 28 days compressivestrength of cylindrical concrete specimen), the transferablemaximum compressive force value to foundation will belimited [1]. Dajin Liu presents an approximate method fordesign of this status [2]. In this paper to resolve this issue, itis considered a little gap between base plate and foundationand because the allowable compressive stress of anchor boltis equal to yF6.0 ( yF is the yielding stress of the anchorbolt), more compressive force can be transferred tofoundation [1]. In other words the proposed method is usedto design of columns with axial load by large eccentric (axialload and large bending moment). To do this, we suppose thatthe utmost compressive anchor bolt reach to the maximumvalue of compressive force and the compressive zone ofplate reach to the critical status simultaneously.

II. PROPOSED METHOD

Consider the column and base plate with two anchorbolts as the axial load eccentric is large, in Fig. 1. Being

enlarge the eccentric means thaty

Re4

2> , which e is

eccentric value, R is radius of base plate, and y is distanceof tensile anchor bolt to column middle axis.

Figure 1. Free body diagram of available forces in effect of axial force with large eccentric.

According to the Fig. 1, on the effect of applying theaxial load and large bending moment, some of anchor bolts

Second International Conference on Computer Research and Development

978-0-7695-4043-6/10 $26.00 © 2010 IEEE

DOI 10.1109/ICCRD.2010.124

588

exposed to tension and others to compression. In Fig. 2 thestrain distribution is shown for set of column and base plate

that the eccentric is larger than1

2

4yR ( 1y is the distance of

last tensile anchor bolt to column middle axis).

Figure 2. Strain distribution in base plate section.

Existing equilibrium equations are:

A. Force equilibrium equation in direction of vertical onpage

FTPFn =+→=∑ 0 . (1)in which; P is column axial force, T is tensile forceresultant for tensile anchor bolts, and F is compressiveforce resultant for compressive anchor bolts.

B. Moment equilibrium equation

AFATePAFATMM ePMo .....0 . +′=⎯⎯⎯ →⎯+′=→= =∑ . (2)

in which; A′ is the distance of the tensile force resultant tocolumn middle axis and A is the distance of thecompressive force resultant to column middle axis.Substituting equation (1) in equation (2), we get:

( ) AFAPFeP ... +′−= . (3)The terms A′ , A , and F are unknowns.We show that these unknowns are the function of distancex (distance between neutral axis and column middle axis).Writing the A′ and A as following:

∑

∑

=

=

+

+

=′T

T

n

ii

n

iii

TT

yTyT

A

21

211

2

2. (4)

∑

∑

′=′

′=′′

+

+

=F

F

n

jj

n

jjj

FF

yFyF

A

21

211

2

2

. (5)

in which; Tn is half of the total tensile anchor bolts, Fn ishalf of the total compressive anchor bolts, i is a counter fortensile anchor bolts which varies from 2 to Tn , j is acounter for compressive anchor bolts which varies from 2′to Fn , 1T is tensile force in utmost tensile anchor bolt, 1′Fis compressive force in utmost compressive anchor bolt, iyis the distance of i th tensile anchor bolt to column middleaxis, jy is the distance of j th compressive anchor bolt to

column middle axis, iF is the tensile force of i th tensileanchor bolt, and jF is the compressive force of j thcompressive anchor bolt.

According to the Fig. 2, strain relation for each tensileand compressive anchor bolt using mathematical similarityrelation as following [3]:

xxByi

Ti+−

= .εε . (6)

xBxBy j

Fj −

−+=

2.εε . (7)

in which; Tε is the strain of utmost tensile anchor bolt, Fεis the strain of utmost compressive anchor bolt, and B isequal to 1y .On the other hand, tensile force and compressive forcerelations in each anchor bolt as following [4]:

bisbii AEAT ... εσ == . (8)

bjsbjj AEAF ... εσ == . (9)

in which; iσ and jσ are the stress in each anchor bolt, sE

is the elasticity modulus of steel, and bA is the section areaof each anchor bolt.Therefore substituting equation (6) in equation (8) andequation (7) in equation (9), we have:

xxBy

TAx

xByET i

bi

Tsi+−

=+−

= .... 1ε . (10)

xB

xByFA

xB

xByEF j

bj

Fsj −

−+=

−

−+= ′ 2

..2

.. 1ε . (11)

Substituting equation (10) in equation (4) and equation (11)in equation (5), we have:

=1

12

1

2

589

∑

∑

=

=

+−+

+−+

=′T

T

n

i

i

n

ii

i

xxBy

TT

yx

xByTyT

A

211

2111

.2

..2. (12)

∑

∑

′=′′

′=′′′

−

−++

−

−++

=F

F

n

j

j

n

jj

j

xB

xByFF

yxB

xByFyF

A

211

2111

2.2

.2

.2

. (13)

By simplifying these equations, reduces to:

( )

( )∑

∑

=

=

+−+

+−+

=′T

T

n

ii

n

iii

xByx

yxByxB

A

2

2

2

2. (14)

( ) ( )

( ) ( )∑

∑

′=

′=

−++−

−++−

=F

F

n

jj

n

jjj

xByxB

yxByxBB

A

2

2

22

22

. (15)

Therefore according to the equations (14) and (15), A′ andA are the function of x .

Now by calculating tensile and compressive forces resultant:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +−+=+= ∑∑

==

TT n

i

in

ii x

xByTTTT

21

21 212 . (16)

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−++=+= ∑∑

′=′

′=′

FF n

j

jn

jj xB

xByFFFF

21

21 2

212 . (17)

in which; T is the tensile forces resultant and F is thecompressive forces resultant.Therefore according to the equations (16) and (17), T andF are the function of x .

In equations (16) and (17), 1T and 1′F are also unknowns.Therefore if equations (14), (15), and (17) insert intoequation (3), the whole equation (3) will be function of xbut be value of 1′F will be unknown. For dispelling thisproblem, we suppose that the anchor bolt of 1′ reach toallowable compressive stress. Therefore:

byb AFAF .6.0.11 == ′′ σ . (18)By determination of 1′F , value of 1T is calculated by

using of mathematic similarity. Now the whole equation (3)will be function of x which solving this equation isaccomplished by using of trial and error iteration fordifferent value of x .

Also the allowable tensile stress of utmost anchor boltcan be controlled by the following equation:

bTn

i

i

AF

xxBy

TTTT

.

212

max1 ≤+−

+

==

∑=

. (19)

which TF is allowable tensile stress of the anchor bolts.The thickness of the base plate is calculated by

determination of x . Now consider Fig. 3.

Figure 3. Compressive zone in base plate section.

The equation for calculate the thickness of base plate isderived from following equation:

bcr

b FW

Mf ≤= . (20)

in which; crM is critical bending moment of base plate, Wis the section modulus of base plate, and bF is allowablebending stress of base plate.

We express the critical bending moment of base plate asfollowing:

crM = sum of the compressive anchor bolts moment aroundthe columnTherefore according to the Fig. 4, we have:

Figure 4. Moment arm for calculation of critical moment.

rFrFrFM cr .....2. 21 =++= ′′ . (21)which r is the vertical distance from center of compressiveanchor bolts to column margent.

Also the section modulus is calculated based on therelevant strip of compressive part of column margent. If thelength of the compressive part of strip is equal to l ,according to the Fig. 5, we have:

12

1

2

neutral axis compressivezone

590

απθα 22,cos 1 −=⎟⎠⎞

⎜⎝⎛ −

= −

axB . (22)

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −

−== −

axBaal 1cos.2. πθ . (23)

Figure 5. Compressive part of critical strip.

which a is column radius.The section modulus is calculated as following:

6. 2tlW = . (24)

which t is base plate thickness.

Therefore by inserting equations (23) into (24), and (24) and(21) into (20), we have:

bb Ft

axBa

rFf ≤

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −

−=

−

6.cos2

.2

1π. (25)

By simplifying equation (25):

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −

−≥

−

axBFa

rFt

b1cos..

.3

π. (26)

which this equation is also function of x .

III. PRESENTING RESULTS

After analytical computations and concluding functionsof xTF ,, and t , more than 130 lines computer program iswritten in MACRO and some numerical examples is solvedto observe how these parameters change when loading orradius of base plate changes.

To do so a plate is supposed with 12 bolts of 10 squarecentimeters in area which placed to plates edge in 5centimeter.

Three statuses are assumed in which the ratio of aR is 2:

A. tonP 15=

The variation of xF , and t is as Fig. 6 to 8:

Figure 6. Variation of F in (ton) about e in (centimeter).

Figure 7. Variation of x about e both in (centimeter).

Figure 8. Variation of t about e both in (centimeter).

B. tonP 30=

The variation of xF , and t is as Fig. 9 to 11:

Figure 9. Variation of F in (ton) about e in (centimeter).

591

Figure 10. Variation of x about e both in (centimeter).

Figure 11. Variation of t about e both in (centimeter).

C. tonP 45=

The variation of xF , and t is as Fig. 12 to 14:

Figure 12. Variation of F in (ton) about e in (centimeter).

Figure 13. Variation of x about e both in (centimeter).

Figure 14. Variation of t about e both in (centimeter).

IV. CONCLUSIONS

By observing the graphs, it is seen that also the axial loadis very big and its eccentricity is big too, the thickness ofbase plate is not too much. It means that by combination ofplate and bolts for transition of loads to foundation, we canhave much more thin plate to do so or on the other hands,with same thickness; we are able to tolerate bigger forceswith large eccentricity without failure.

By writing a computer program, it is easy to insert axialforce, eccentricity, and other dimensions o plate, column andbolts, and have the value of base plate thickness. What'smore value of axial force in bolts and other parametersmentioned before can be observed too.

REFERENCES

[1] AISC Manual Committee, Manual of Steel Construction: AllowableStress Design, AISC, 9th ed., Chicago, 1989.

[2] L. Dajin, “Circular base plates with large eccentric loads,” ASCE,Journal of Structural Engineering, vol. 9, Dec. 2004, pp. 142-146, doi:10.1061/(ASCE)1084-0680(2004)9:3(142).

[3] E. W. Swokowski, Calculus with Analytic Geometry, 2nd ed.,PWS/Kent, Boston, 1988.

[4] W. C. Young, Roark’s Formulas for Stress and Strain, 6th ed.,McGraw-Hill, New York, 1989.

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