Paper-Design of Circular Base Plate Under Large Bending Moment

Second International Conference on Computer Research and Development

Design of Circular Base Plate under Large Bending Moment Which There Is a Little Gap between Base Plate and Foundation A. A. Gholampour

M. Naghipour

A. Sobati

Department of Civil Engineering University of Tehran Tehran, Iran [email protected]

Department of Civil Engineering Babol University of Technology Babol, Iran [email protected]

Department of Civil Engineering Babol University of Technology Babol, Iran [email protected]

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

code which is equal to 0.7 f c′ ( f c′ is 28 days compressive strength of cylindrical concrete specimen), the transferable maximum compressive force value to foundation will be limited [1]. Dajin Liu presents an approximate method for design of this status [2]. In this paper to resolve this issue, it is considered a little gap between base plate and foundation and because the allowable compressive stress of anchor bolt is equal to 0.6 F y ( Fy is the yielding stress of the anchor

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

Consider the column and base plate with two anchor bolts as the axial load eccentric is large, in Fig. 1. Being R2 enlarge the eccentric means that e > , which e is 4y eccentric value, R is radius of base plate, and y is distance of tensile anchor bolt to column middle axis.

I.

bolt), more compressive force can be transferred to foundation [1]. In other words the proposed method is used to design of columns with axial load by large eccentric (axial load and large bending moment). To do this, we suppose that the utmost compressive anchor bolt reach to the maximum value of compressive force and the compressive zone of plate reach to the critical status simultaneously. II.

INTRODUCTION

Generally steel columns are placed on the foundation with steel plate which is involved on the one hand with column and on the other hand with concrete. Since the steel column undergo the large stresses due to high resistance and concrete has no ability to undergo these stresses, therefore base plate is used for increasing the force transmission level to foundation and reducing it to allowable limit of foundation materials strength. From the types of base plates can be mentioned rectangular or circular base plates, which circular base plates are commonly used for cylindrical columns such as pylons in cabled-stayed bridges, lighting poles, electric power line posts, and even buildings. In this paper from the types of loading states in columns such as; pure axial force, axial force and small bending moment, and axial force and large bending moment, we are going to show a new procedure for designing of base plate in state of loading with axial force and large bending moment. Usually to design base plates in this status, no gap is supposed between plate and foundation. By considering the stress distribution in this status, tensile forces are transferred by anchor bolts and compressive forces are transferred by base plate to foundation. Therefore by considering the maximum tolerable contact stress of concrete foundations according to AISC 978-0-7695-4043-6/10 $26.00 © 2010 IEEE DOI 10.1109/ICCRD.2010.124

PROPOSED METHOD

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 the axial load and large bending moment, some of anchor bolts 588

exposed to tension and others to compression. In Fig. 2 the strain distribution is shown for set of column and base plate R2 that the eccentric is larger than ( y1 is the distance of 4 y1 last tensile anchor bolt to column middle axis).

T1 y1 + 2 A′ = T1 + 2 F1′ y1′ + 2 A= F1′ + 2

2

1

nT

∑T y

i i

i =2 nT

.

∑T

(4)

i

i =2 nF

∑F y j

j = 2′ nF

∑F

j

.

(5)

j

j = 2′

1

in which; nT is half of the total tensile anchor bolts, n F is half of the total compressive anchor bolts, i is a counter for tensile anchor bolts which varies from 2 to nT , j is a counter for compressive anchor bolts which varies from 2′ to n F , T1 is tensile force in utmost tensile anchor bolt, F1′ is compressive force in utmost compressive anchor bolt, yi is the distance of i th tensile anchor bolt to column middle axis, y j is the distance of j th compressive anchor bolt to

2

column middle axis, Fi is the tensile force of i th tensile anchor bolt, and F j is the compressive force of j th compressive anchor bolt. According to the Fig. 2, strain relation for each tensile and compressive anchor bolt using mathematical similarity relation as following [3]: y −B+x ε i = εT . i . (6) x yj + B− x ε j = εF. . (7) 2B − x in which; ε T is the strain of utmost tensile anchor bolt, ε F is the strain of utmost compressive anchor bolt, and B is equal to y1 . On the other hand, tensile force and compressive force relations in each anchor bolt as following [4]: (8) Ti = σ i . Ab = E s .ε i . Ab . F j = σ j . Ab = E s .ε j . Ab . (9)

1=

Figure 2. Strain distribution in base plate section.

Existing equilibrium equations are: A. Force equilibrium equation in direction of vertical on page

∑F

n

= 0 → P +T = F .

(1)

in which; P is column axial force, T is tensile force resultant for tensile anchor bolts, and F is compressive force resultant for compressive anchor bolts. B. Moment equilibrium equation

∑M

o

in which; σ i and σ j are the stress in each anchor bolt, E s

P.e = 0 → M = T . A′ + F . A ⎯M ⎯=⎯ ⎯→ P.e = T . A′ + F . A . (2)

is the elasticity modulus of steel, and Ab is the section area of each anchor bolt. Therefore substituting equation (6) in equation (8) and equation (7) in equation (9), we have: y −B+x y −B+x Ti = E s .ε T . i . Ab = T1 . i . (10) x x yj + B− x yj + B− x F j = E s .ε F . . (11) . Ab = F1′ . 2B − x 2B − x Substituting equation (10) in equation (4) and equation (11) in equation (5), we have:

in which; A′ is the distance of the tensile force resultant to column middle axis and A is the distance of the compressive force resultant to column middle axis. Substituting equation (1) in equation (2), we get: P.e = F( − P . A)′ + F . A . (3) The terms A′ , A , and F are unknowns. We show that these unknowns are the function of distance x (distance between neutral axis and column middle axis). Writing the A′ and A as following:

589

T1 y1 + 2

nT

yi − B + x . yi x

∑T . 1

i =2 nT

A′ =

y −B+x T1 + 2 T1 . i x i =2

∑

F1′ y1′ + 2

nF

yj + B− x

∑F . 1′

2B − x

j = 2′ nF

A=

F1′ + 2

∑F . 1′

T1 = Tmax =

.

≤ FT . Ab . (19) yi − B + x 1+ 2 x i=2 which FT is allowable tensile stress of the anchor bolts. The thickness of the base plate is calculated by determination of x . Now consider Fig. 3.

(12)

.y j

.

yj + B− x

neutral axis

(13)

By simplifying these equations, reduces to: xB + 2 A′ =

∑ (y

i

i =2 nT

x+2

∑ (y

1

− B + x )yi

compressive zone 2 1 2

. i

∑

2B − x

j = 2′ nT

T

nT

(14)

− B + x)

i =2

B(2 B − x ) + 2 A=

nF

∑ (y

j

)

+ B− x yj

j = 2′ nF

(2 B − x ) + 2∑ (y j + B − x )

.

Figure 3. Compressive zone in base plate section.

(15)

The equation for calculate the thickness of base plate is derived from following equation: M (20) f b = cr ≤ Fb . W in which; M cr is critical bending moment of base plate, W is the section modulus of base plate, and Fb is allowable bending stress of base plate. We express the critical bending moment of base plate as following: M cr = sum of the compressive anchor bolts moment around the column Therefore according to the Fig. 4, we have:

j = 2′

Therefore according to the equations (14) and (15), A′ and A are the function of x . Now by calculating tensile and compressive forces resultant: nT nT ⎛ y i − B + x ⎞⎟ ⎜ . (16) T = T1 + 2 Ti = T1 1 + 2 ⎜ ⎟ x i =2 i=2 ⎝ ⎠ nF nF ⎛ y j + B − x ⎞⎟ F = F1′ + 2 F j = F1′ ⎜1 + 2 . (17) ⎜ 2B − x ⎟ j = 2′ j = 2′ ⎝ ⎠ in which; T is the tensile forces resultant and F is the compressive forces resultant. Therefore according to the equations (16) and (17), T and F are the function of x . In equations (16) and (17), T1 and F1′ are also unknowns. Therefore if equations (14), (15), and (17) insert into equation (3), the whole equation (3) will be function of x but be value of F1′ will be unknown. For dispelling this problem, we suppose that the anchor bolt of 1′ reach to allowable compressive stress. Therefore: (18) F1′ = σ 1′ . Ab = 0.6 F y . Ab .

∑

∑

∑

∑

By determination of F1′ , value of T1 is calculated by using of mathematic similarity. Now the whole equation (3) will be function of x which solving this equation is accomplished by using of trial and error iteration for different value of x . Also the allowable tensile stress of utmost anchor bolt can be controlled by the following equation:

Figure 4. Moment arm for calculation of critical moment.

M cr = F1′ .r + 2 F2′ .r + ... = F .r . (21) which r is the vertical distance from center of compressive anchor bolts to column margent. Also the section modulus is calculated based on the relevant strip of compressive part of column margent. If the length of the compressive part of strip is equal to l , according to the Fig. 5, we have:

590

⎛B− x⎞ α = cos −1 ⎜ ⎟ , θ = 2π − 2α . ⎝ a ⎠ ⎛ ⎛ B − x ⎞⎞ l = a.θ = 2a.⎜⎜ π − cos −1 ⎜ ⎟ ⎟⎟ . ⎝ a ⎠⎠ ⎝

(22) (23)

Figure 6. Variation of

F

in (ton) about

e

in (centimeter).

Figure 5. Compressive part of critical strip.

which a is column radius. The section modulus is calculated as following: l.t 2 W= . 6 which t is base plate thickness.

(24)

Therefore by inserting equations (23) into (24), and (24) and (21) into (20), we have: F .r fb = ≤ Fb . (25) 2 ⎛ −1 ⎛ B − x ⎞ ⎞ t 2a⎜⎜ π − cos ⎜ ⎟ ⎟⎟. ⎝ a ⎠⎠ 6 ⎝ By simplifying equation (25): 3F .r

t≥

⎛ ⎛ B − x ⎞⎞ a.Fb .⎜⎜ π − cos −1 ⎜ ⎟ ⎟⎟ ⎝ a ⎠⎠ ⎝ which this equation is also function of x . III.

.

x about e both in (centimeter).

Figure 8. Variation of

t

(26)

PRESENTING RESULTS

After analytical computations and concluding functions of F , T , x and t , more than 130 lines computer program is written in MACRO and some numerical examples is solved to observe how these parameters change when loading or radius of base plate changes. To do so a plate is supposed with 12 bolts of 10 square centimeters in area which placed to plates edge in 5 centimeter. Three statuses are assumed in which the ratio of R is 2: a A.

Figure 7. Variation of

B.

about

e

both in (centimeter).

P = 30 ton

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

P = 15 ton

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

Figure 9. Variation of

591

F

in (ton) about

e

in (centimeter).

C.

Figure 10. Variation of

x about e both in (centimeter).

Figure 13. Variation of

x about e both in (centimeter).

Figure 11. Variation of

t

Figure 14. Variation of

t

about

e

both in (centimeter).

P = 45 ton

F in (ton) about e

e

both in (centimeter).

IV. CONCLUSIONS By observing the graphs, it is seen that also the axial load is very big and its eccentricity is big too, the thickness of base plate is not too much. It means that by combination of plate and bolts for transition of loads to foundation, we can have much more thin plate to do so or on the other hands, with same thickness; we are able to tolerate bigger forces with large eccentricity without failure. By writing a computer program, it is easy to insert axial force, eccentricity, and other dimensions o plate, column and bolts, and have the value of base plate thickness. What's more value of axial force in bolts and other parameters mentioned before can be observed too.

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

Figure 12. Variation of

about

REFERENCES

in (centimeter). [1] [2]

[3] [4]

592

AISC Manual Committee, Manual of Steel Construction: Allowable Stress Design, AISC, 9th ed., Chicago, 1989. 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). E. W. Swokowski, Calculus with Analytic Geometry, 2nd ed., PWS/Kent, Boston, 1988. W. C. Young, Roark’s Formulas for Stress and Strain, 6th ed., McGraw-Hill, New York, 1989.