Three-Hinged Arches PDF

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Three Hinged Arches...

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Chapter 4

Three-Hinged Arches

The arches are widely used in modern engineering. Arches permit to cover a larger span. The greater is the span than an arch becomes more economical than a truss. From esthetic point of view the arches are more attractive than trusses. Materials of the modern arches are concrete, steel, and wood. The body of the arch may be solid or consist of separate members. The arches are classified as three-hinged, two-hinged, and arch with fixed supports. A three-hinged arch is geometrically unchangeable statically determinate structure which consists of two curvilinear members, connected together by means of a hinge, with two-hinged supports resting on the abutment. This chapter is devoted to analyze only three-hinged arches. Two-hinged arch and arch with fixed supports will be considered in Part 2.

4.1 Preliminary Remarks 4.1.1 Design Diagram of Three-Hinged Arch The arch with overarched members is shown in Fig. 4.1a. The arch contains three hinges. Two of them are located at the supports and third is placed at the crown. These hinges are distinguishing features of the three-hinged arch. Design diagram also contains information about the shape of the neutral line of the arch. Usually this shape is given by the expression, y D f .x/. a

b C

y

y = f (x )

j

f

f

A l

B

HA RA

l

x HB

RB

Fig. 4.1 (a) Design diagrams for deck-arch bridge; (b) Design diagram of three-hinged arch

I.A. Karnovsky and O. Lebed, Advanced Methods of Structural Analysis, DOI 10.1007/978-1-4419-1047-9 4,  c Springer Science+Business Media, LLC 2010

77

78

4 Three-Hinged Arches

The each post which connects the beams of overarched structure with arch itself has the hinges at the ends. It means that in the poles only axial force arises. Idealized design diagram of the three-hinged arch without overarched members is shown in Fig. 4.1b. Degrees of freedom of three-hinged arch according to Chebushev formula (1.1) is W D 0, so this structure is geometrically unchangeable. Indeed, two rigid discs AC and BC are connected with the ground by two hinges A and B and line AB does not pass through the intermediate hinge C . This structure has four unknown reactions, i.e., two vertical reactions RA , RB and two horizontal reactions HA , HB . For their determination, three equilibrium equations can be formulated considering the structure in whole. Since bending moment at the hinge C is zero, this provides additional equation for equilibrium of the part of the system. It means that the sum of the moments of all external forces, which are located on the right (or on the left) part of the structure with respect to hinge C is zero X X MC D 0: (4.1) MC D 0 or left

right

These four equations of equilibrium determine all four reactions at the supports. Therefore, three-hinged arch is a geometrically unchangeable and statically determinate structure.

4.1.2 Peculiarities of the Arches The fundamental feature of arched structure is that horizontal reactions appear even if the structure is subjected to vertical load only. These horizontal reactions HA D HB D H are called a thrust; such types of structures are often called as thrusted structures. One type of trusted structures (thrusted frame) was considered in Chap. 3. It will be shown later that at any cross section of the arch the bending moments, shear, and axial forces arise. However, the bending moments and shear forces are considerably smaller than corresponding internal forces in a simply supported beam covering the same span and subjected to the same load. This is the fundamental property of the arch thanks to thrust. Thrusts in both supports are oriented toward each other and reduce the bending moments that would arise in beams of the same span and load. Therefore, the height of the cross section of the arch can be much less than the height of a beam to resist the same loading. So the three-hinged arch is more economical than simply supported beam, especially for large-span structures. Both parts of the arch may be connected by a tie. In this case in order for the structure to remain statically determinate, one of the supports of the arch should be rolled (Fig. 4.2a, b). Prestressed tie allows controlling the internal forces in arch itself. Tie is an element connected by its ends to the arch by mean of hinges, therefore the tie is subjected only to an axial internal force. So even if horizontal reactions of supports equal zero, an extended force (thrust) arises in the tie.

4.1 Preliminary Remarks

a

y

A

79

b

C

Tie

RA

B

RB

x

y

A

C Elevated tie

B

RA

x

RB

Fig. 4.2 (a, b) Design diagram of three-hinged arch with tie on the level supports and elevated tie

Thus, the arch is characterized by two fundamental markers such as a curvilinear axis and appearance of the thrust. Therefore the structure in Fig. 4.3 presents the curvilinear trustless simply supported element, i.e., this is just members with curvilinear axis, but no arch. It is obvious that, unlike the beam, in this structure the axial compressed forces arise; however, the distribution of bending moments for this structure and for beam of the same span and load will not differ, while the shear forces are less in this structure than that in beam. Thus, the fundamental feature of the arch (decreasing of the bending moments due to the appearance of the thrust) for structure in Fig. 4.3 is not observed. P

Fig. 4.3 Simply supported thrustless curvilinear member

A

B

RA

RB

4.1.3 Geometric Parameters of Circular and Parabolic Arches Distribution of internal forces in arches depends on a shape of the central line of an arch. Equation of the central line and some necessary formulae for circular and parabolic arches are presented below. For both cases, origin of coordinate axis is located at point A (Figs. 4.1b and 4.2).

Circular Arch Ordinate y of any point of the central line of the arch is defined by formula s  2 l f l2 y D R2  x RCfI R D ; (4.2) C 2 2 8f where x is the abscissa of the same point of the central line of the arch; R the radius of curvature of the arch; f and l are the rise and span of the arch.

80

4 Three-Hinged Arches

The angle ' between the tangent to the center line of the arch at point .x; y/ and horizontal axis is shown in Fig. 4.1b. Trigonometric functions for this angle are as follows: 1 1 sin' D .l  2x/ I cos' D .y C R  f / : (4.3) 2R R Parabolic Arch Ordinate y of any point of the central line of the arch y D 4f x .l  x/

1 : l2

(4.4)

Trigonometric functions of the angle between the tangent to the center line of the arch at point .x; y/ and a horizontal axis are as follows tan' D

dy 4f 1 I sin' D cos'  tan' D 2 .l  2x/ I cos' D p dx l 1 C tan2 

(4.5)

For the left-hand half-arch, the functions sin' > 0; cos' > 0, and for the right-hand half-arch the functions sin' < 0 and cos' > 0.

4.2 Internal Forces Design diagram of a three-hinged symmetrical arch with intermediate hinge C at the highest point of the arch and with supports A and B on one elevation is presented in Fig. 4.4. The span and rise of the arch are labeled as l and f , respectively; equation of central line of the arch is y D y .x/. Reactions of Supports Determination of internal forces, and especially, construction of influence lines for internal forces of the three-hinged arch may be easily and attractively performed using the conception of the “reference (or substitute) beam.” The reference beam is a simply supported beam of the same span as the given arch and subjected to the same loads, which act on the arch (Fig. 4.4a). The following reactions arise in the arch: RA ; RB ; H A ; HB . The vertical reactions of three-hinged arches carrying the vertical loads have same values as the reactions of the reference beam 0 RA D RA I RB D R0B :

(4.6)

4.2 Internal Forces

a P1

y

81

P2

f

yk

A HA

P1

Pn

P2

jk Tangent

yk

y(x)

B

Nk

k Qk

H

x

HB

xk l

RA P1

Mk

b

Tangent C k jk

RB Pn

P2

RA

x1 x2

xk

C 0

RA

x1 x2

Reference beam R B0

l

Fig. 4.4 Three-hinged arch. (a) Design diagram and reference beam. (b) Positive internal forces at any section k

The horizontal reactions (thrust) at both supports of three-hinged arches subjected to the vertical loads are equal in magnitude and opposite in direction HA D HB D H:

(4.7)

Bending moment at the hinge C of the arch is zero. Therefore, by definition of the bending moment     l l l MC D RA  P1  x1  P2  x2 HA  f D 0: 2 2 2 „ ƒ‚ … 0 MC

Underlined set of terms is the bending moment acting over section C of the reference beam (this section is located under the hinge of the arch). Therefore last equation may be rewritten in the form M C0  HA f D 0; which allows immediately to calculate the thrust HD

MC0 : f

(4.8)

Thus, the thrust of the arch equals to bending moment at section C of the reference beam divided by the rise of the arch.

82

4 Three-Hinged Arches

Internal Forces In any section k of the arch, the following internal forces arise: the bending moment Mk , shear Qk , and axial force Nk . The positive directions of internal forces are shown in Fig. 4.4b. Internal forces acting over a cross section k may be obtained considering the equilibrium of free-body diagram of the left or right part of the arch. It is convenient to use the left part of the arch. By definition X Pi .xk  xi /  Hyk ; Mk D RA xk  left

X Q k D RA  P left

Nk D  RA 

X left

!

cos'k  H sin'k ;

P

!

sin'k  H cos'k ;

(4.9)

where Pi are forces which are located at the left side of the section k; xi are corresponding abscises of the points of application; xk , yk are coordinates of the point k; and 'k is angle between the tangent to the center line of the arch at point k and a horizontal. These equations may be represented in the following convenient form Mk D Mk0  Hyk ;

Qk D Q0k cos'k  H sin'k ;

Nk D Q 0k sin'k  H cos'k ;

(4.10)

where expressions Mk0 and Qk0 represent the bending moment and shear force at the section k for the reference beam (beam’s bending moment and beam’s shear).

Analysis of Formulae [(4.8), (4.10)] 1. Thrust of the arch is inversely proportional to the rise of the arch. 2. In order to calculate the bending moment in any cross section of the three-hinged arch, the bending moment at the same section of the reference beam should be decreased by value Hyk . Therefore, the bending moment in the arch is less than in the reference beam. This is the reason why the three-hinged arch is more economical than simply supported beam, especially for large-span structures. In order to calculate shear force in any cross section of the three-hinged arch, the shear force at the same section of the reference beam should be multiplied by cos'k and this value should be decreased by H sin'k . 3. Unlike beams loaded by vertical loads, there are axial forces, which arise in arches loaded by vertical loads only. These axial forces are always compressed.

4.2 Internal Forces

83

Analysis of three-hinged arch subjected to fixed loads is presented below. This analysis implies determination of reactions of supports and construction of internal force diagrams. Design diagram of the three-hinged circular arch subjected to fixed loads is presented in Fig. 4.5. The forces P1 D 10 kN; P2 D 8 kN; q D 2 kN=m. It is necessary to construct the internal force diagrams M , Q, N . P1=10kN Design diagram

q =2kN/m k

y

3

2

C 4

n

5

Circle

6

f=8m

1

A

P2=8kN

7

B

x

H

H 4m

4m

l=32m

RA P1

RB P2

q

Reference A beam

B C

RA0 14.5

RB0 4.5

+

Q 0 (kN) − 11.5

19.5 M0 (kNm)

58

78 116

124

125 152

154

Fig. 4.5 Three-hinged circular arch. Design diagram, reference beam, and corresponding internal force diagrams

Solution. Reference beam The reactions are determined from the equilibrium equations of all the external forces acting on the reference beam X MB D 0 W RA0 ! RA0  32 C P1  24 C q  8  12 C P2  4 D 0 ! R0A D 14:5 kN; X MA D 0 W RB0 ! R0B  32  P1  8  q  8  20  P2  28 D 0 ! RB0 D 19:5 kN:

The values reactions just found should be checked using the equilibrium Pof the two 0 C R 0  P  q  8  P D 14:5 C 19:5  10  2  8  8 D equation Y D RA 1 2 B 34  34 D 0.

84

4 Three-Hinged Arches

The bending moment M 0 and shear Q 0 diagrams for reference beam are presented in Fig. 4.5. At point C .x D 16 m/ the bending moment is MC0 D 152 kN m. Three-hinged arch The vertical reactions and thrust of the arch are 0 D 14:5 kN; RB D R0B D 19:5 kN; H D RA D RA

M C0 152 D D 19 kN: f 8

For construction of internal forces diagrams of the arch, a set of sections has to be considered and for each section the internal forces should be calculated. All computations concerning geometrical parameters and internal forces of the arch are presented in Table 4.1. Table 4.1 Internal forces in three-hinged circular arch (Fig. 4.5); (RA D 14:5 kN I RB D 19:5 kNI H D 19 kN) Section

x.m/ y.m/

0 A 1

1 0.0 4

2

M x0 Hy .kNm/ .kNm/

sin '

cos '

2 0:0 4

3 0:8 0:6

4 0:6 0:8

8

6:330

0:4

0:9165

116

k 3 4 (C ) 5 6 n

10 12 16 20 24 26

7:0788 0:3 7:596 0:2 8:0 0:0 7:596 0:20 6:330 0:40 5:3205 0:50

0:9539 0:9798 1:0 0:9798 0:9165 0:8660

125 134 152 154 124 101

7

28

4

0:6

0:8

78

B

32

0:0

0:8

0:6

0

5 0 58

Q0x Nx Qx Mx .kN / .kN / .kNm/ .kN/ 6 7 8 9 50 0:0 0 14:5 6:5 23 76 18 14:5 0:2 23:9 14:5 5:6892 23:213 120:27 4:27 4:5 3:4757 19:213 134:497 9:497 4:5 1:4074 19:474 144:324 10:324 4:5 0:6091 19:516 152 0:0 4:5 4:5 19:00 144:324 9:676 3:5 0:3707 19:316 120:27 3:73 11:5 2:9397 22:013 101:089 0:089 11:5 0:459 22:204 11:5 2:2 22:1 76 2 19:5 4:2 26:9 19:5 4:2 26:9 0:0 0 19:5 3:5 27

Notes: Values in nominator and denominator (columns 8 and 9) mean value of the force to the left and to the right of corresponding section. Values of discontinuity due to concentrated load equal P cos' and P sin' in shear and normal force diagrams, respectively

The column 0 contains the numbers of sections. For specified sections A, 1–7, and B the abscissa x and corresponding ordinate y (in meters) are presented in columns 1 and 2, respectively. Radius of curvature of the arch is RD

f 8 322 l2 D C D 20 m: C 8f 88 2 2

Coordinates y are calculated using the following expression

y D y .x/ D

s

R2 



l x 2

2

RCf D

q 400  .16  x/2  12:

4.2 Internal Forces

85

Column 3 and 4 contain values of sin' and cos', which are calculated by formulae l  2x 32  2x yCRf y C 12 : D ; cos' D D 20 2R 40 R

sin' D

Values of bending moment and shear for reference beam, which are presented in columns 5 and 7, are taken directly from corresponding diagrams in Fig. 4.5. Values for Hy are contained in column 50. Columns containing separate terms for Q 0cos'; Q0sin'; H cos'; H sin' are not presented. Values of bending moment, shear and normal forces for three-hinged arch are tabulated in columns 6, 8, and 9. They have been computed using (4.10). For example, for section A we have 0 cos'A  H sin'A D 14:5  0:6  19  0:8 D 6:5 kN; QA D QA

NA D Q 0A sin'A  H cos'A D 14:5  0:8  19  0:6 D 23 kN:

The final internal force diagrams for arch are presented in Fig. 4.6. q=2kN/m

P1=10kN C 3 4

k y

2

n

5

6

f=8m

1

A

P2=8kN

7

x

B

H

H 4m RA

4m

l=32m

RB

18 4.27

9.49 10.32 0.089 M (kNm) 3.73

2.0

9.676 P1cos j2

5.69

4.5 2.2 +

0.2

3.5 Q (kN)

− 6.5

1.4 2.94

3.47

4.2 N (kN)

−

23 23.21 19.21

19.47 P1sin j2

22.1 26.9

27

Fig. 4.6 Design diagram of three-hinged circular arch. Internal forces diagrams

Bending moment diagram is shown on the side of the extended fibers, thus the signs of bending moments are omitted. As for beam, the bending moment and shear

86

4 Three-Hinged Arches

diagrams satisfy to Schwedler’s differential relationships. In particularly, if at any point a shear changes sign, then a slope of the bending moment diagram equals zero, i.e., at this point the bending moment has local extreme (for example, points 2, 7, etc.). It can be seen that the bending moments which arise in cross sections of the arch are much less than in a reference beam.

4.3 Influence Lines for Reactions and Internal Forces Equations (4.6), (4.8), and (4.10) can be used for deriving of equations for influence lines. Vertical reactions The equations for influence lines for vertical reactions of the arch are derived from (4.6). Therefore the equations for influence lines become   IL .RA / D IL RA0 I

  IL .RB / D IL R0B :

(4.11)

Thus, influence lines for vertical reactions of the arch do not differ from influence lines for reactions of the reference simply supported beam. Thrust The equation of influence lines for thrust is derived from (4.8). Since for given arch a rise f is a fixed number, then the equation for influence lines become IL .H / D

  1  IL M C0 : f

(4.12)

Thus, influence line for trust H may be obtained from the influence line for bending moment at section C of the reference beam, if all ordinates of the latter will be divided by parameter f . Internal forces The equations for influence lines for internal forces at any section k may be derived from (4.10). Since for given section k, the parameters yk , sin'k , and cos'k are fixed numbers, then the equations for influence lines become   IL .Mk / D IL ...