# Application of the Affine Theorem to an Orthotropic Rectangular Reinforced Concrete Slab Having a interior Corner Opening

DOI : 10.17577/IJERTV6IS090052
Text Only Version

#### Application of the Affine Theorem to an Orthotropic Rectangular Reinforced Concrete Slab Having a interior Corner Opening

1. Dr. J. Vikranth,
Associate Professor,
Department of Civil Engineering,
Anil Neerukonda Institute of Technology and Sciences,
(ANITS), Sangivalasa, Bheemunipatnam Mandal,
Visakhapatnam- 531162, India.

2. PVRK Reddy, Assistant Professor,

Department of Civil Engineering,

Anil Neerukonda Institute of Technology and Sciences,(ANITS), Sangivalasa, Bheemunipatnam Mandal, Visakhapatnam- 531162, India.

3. Prof. K. Rambabu, Department of Civil Engineering,

Andhra University College of Engineering, Visakhapatnam – 530003, India.

Abstract – An attempt has been made to applythe affinity theorem to determine collapse load of two-way orthotropic slab withainterior corner openingwith Onelong side continuous and other three sides simply supported slab (OLC) and One long sidesimply supported and other three sides continuous slab (OLD).Keeping in view the basic principles of yield line theory, all possible admissible yield line patterns are considered for the given configuration of the slab subjected to uniformly distributed load (udl). A computer program has been developed to solve the virtual work equations derived in this paper. Illustration of above methodology has been brought out with numerical examples. Relevant tables and charts for given data and the governing admissible failure patterns of the slab for different sizes of openings are presented using the affine theorem. In this paper, authors also present the transformation of orthotropic slab into an equivalent isotropic slab using the affine theorem. The analysis is carried out with aspect ratio of opening quite different to that the slab.

Key Words:aspect ratio, interior corneropening, configuration, affinity theorem, orthotropic slab, uniformly distributed load, ultimate load, ultimatemoment and virtual work equations.

INTRODUCTION

Openings in slabs are usually required for plumbing, fire protection pipes, heat and ventilation ducts and air conditioning. Larger openings that could amount to the elimination of a large area within a slab panel are sometimes required for stairs and elevators shafts. For newly constructed slabs, the locations and sizes of the required openings are usually predetermined in the early stages of design and are accommodated accordingly. Such two way slabs subjected to uniformly distributed load and supported on various edge conditions are being analyzed by using yield-line method as suggested by Johansen, K.W1. Many researchers (Goli,H.B. et al2, Rambabu,K. et al3,Islam,S. et al4, Zaslavsky.Aron5, Siva Rama Prasad et al6, Sudhakar,K.J. et al7, Veerendra Kumaret al8) adopted the yield-line analysis and virtual work method in deriving the virtual work equations of the rectangular reinforced concrete solid slabs subjected to uniformly distributed load and supported on various edge conditions. Johansen, K.W1,also presented the analysis of orthogonal solid slabs implicitly to that of an equivalentisotropic slab by using Affine Theorem provided the ratio of negative to positive moments is same in the orthogonal directions. Various design charts are presented byIslam,S. et al4 for continuous slab(CS) and simply supported(SS) slabs with equalopenings,i.e. ratio of openings and aspect ratio are same. Whereas in this paper similar charts are presented for twoedge conditions with aspect ratio of slab and opening different.

Methodology

The method of determining collapse loads based on principle of virtual work has proved to be a powerful tool for a structural engineer, despite it gives an upper bound value. The work equations are formed by equating the energy absorbed by yield lines and the work done by the external load of the orthogonal rectangular slab with interior corneropenings where a small virtual displacement is given to the slab. The same principle was also used by Islam,S. et al4in their paper. In other words, the work equation is given by

Wult x, ydxdy mult,xx y0 mult, y y x0 ——————– (1)

where Wultis the ultimate load per unit area of slab, (x,y) is the virtual displacement in the direction of the loading at the element of area of dimensions dx,dy, mult,x and mult,yare the yield moments per unit width in the x and y directions, x and y are

the components of the virtual rotation of the slab segments in the x and y directions and x0 and y0are the projected length of the yield lines in x and y directions of slab. The equation (1) contains terms C1, C2, C3and C4 which define the positions of the node points of the yield lines. The values of C1, C2, C3 and C4 to be used in the equation are those which give the minimum load to cause failure. A computer program has been written to find the values of C1, C2, C3 and C4 (in terms of r1, r2, r3 and r4) corresponding to minimum load carrying capacity of the slab.For definitions of various parameters refer notations. Johansen1 has proved that the yield line theory is an upper bound method, so care has been taken to examine all the possible yield line patterns for TAC slab to ensure that the most critical collapse mode is considered otherwise the load carrying capacity of the slab will be overestimated.

Formulation of Virtual Work Equations

There is several possible yield line patterns associated with different edge conditions of the slab. For the OLC edge conditionof slab, the possible admissible failure yield line patterns aretwenty,For the OLD edge conditions of slab, the possible admissible failure yield line patterns aretwenty. These admissible failure yield line patterns are obtained basing on the yield line principleof Johansen.K.W1. For the given configuration of the slab, these twenty failure patterns and corresponding equations have been investigated depending upon the support condition of the slab using a computer program.

The orthogonal reinforced rectangular slab having interior corner opening with the given configuration and the yield criteria are shown in notation Fig.5. The slab is subjected to uniformly distributed load (Wult). Note that the slab is not carrying any load over the area of the opening.

The generalized virtual work equations for continuous edge (CS) condition of slab are derived for the predicted possible admissible failure yield line patterns using the virtual work equation. To get the equations for other edge conditions of the slab, modification should be carried out in the numerators of the equations of each failure patterns.ForOLCslab I1=I2=I3=0 or I1= I3=I4= 0, for OLD slab I2=0 or I4=0.

Virtual Work Equations for Two Adjacent sides Continuous (TAC) Slab

Twentypossible failure patterns are predicted for two edge conditions of the slab.The governing failure pattern for different edge conditions and for different data is presented in Table-1. Let be the virtual displacement at a & b (Fig. 1), for theconsidered failure Pattern-1of aslab. Three unknown dimensions C1, C2, &C3 are necessary to define the yield line propagation completely.All otheradmissible failure patterns are as shown in APPENDIX-A.

y2

D

y1

C5 C7

C6

x2

x3

A y3

I4m

C3=LY/r3

y4

C

I3m

Ly

B

x1

I1m

C8

I2m

C1=Lx/r1 C2=Lx/r2

Lx

FIG 1 (PATTERN 1)

The derived equations below are for the continuously supported rectangular slab with interior corner opening. PATTERN 1:

The external work done by segment A:-

1 C5

C5 1 x1 </p

Wult 2 C5 y1 3C WultC5 y3 2C Wult 2 C8 x1 3C

1

2 r r3 r r2

1

r r

1

r2r3

Wult Ly r 1 5 1 5 1 r8 1 5 3 8

6r 2 r 6r r 12

3

3 1 3

The external work done by segment B:-

1 C8

C8 1

C8

y

y

Wult 2 x1C8 3L

• C3Wult x3C8 2L
• C3Wult 2 x2C8 3L

C3

y

2 r2r3

r2r3

r r 1

1

r2r3

Wult Ly r 3 8 3 8 1 3 8 3 8

6r r 12 2r 1 r 1 r r 6r r 12

1 3

3 3

1 2 2 3

The external work done by segment C:-

1 x2

C7 1 C7

Wult 2 C8 x2 3C Wult y4C7 2C Wult 2 y2C7 3C

2

2 r2r3

2

r r r r2

2

r2r3

Wult Ly r 3 8 1 r8 2 7 2 7 2 7

6r r

12

r 2 6r

2 3 3 3

The external work done by segment D:-

1 y2

C6 1

y1

Wult 2 C7 y2 3C

WultLxC6 2C

Wult 2 C5 y1 3C

2 r2r3

3

r r21 r

• r

3 3

r2r3

Wult Ly r 2 7 3 6 5 7 1 5

6r3 2 6r3

Total work done = work done by segments (A+B+C+D)

r r3

r r2

r r

r2r3

r2r3

r2r3

r r 1

1

r2r3

1 5 1 5 1 r8 1 5 3 8 3 8 3 8 1 3 8 3 8

6r 2 r 6r r 12 6r r 12 2r 1 r 1 r r 6r r 12

W L 2r 3

3 1 3

1 3

3 3

1 2 2 3

ult y

r2r3

r r r r2

r2r3

r2r3

r r2 1 r r

r2r3

3 8 1 r8 2 7 2 7 2 7 2 7 3 6 5 7 1 5

6r r 12

r 2

6r 6r

2 6r

2 3 3 3 3 3

Energy absorbed yield lines: –

m K ‘

y C 1 m

I L 1 m

K ‘ x

x 1 m

I L 1

ult x 1 8 C

ult 1

y C

ult y 1

2 Ly C

ult 2

x Ly C

1 1

3 3

1

1

1 1

mult K ‘x y2 C8 C

• mult I3Ly Cmult K ‘y C5 C7 C
• mult I4 Lx CK1r r r

2

I r

2

K1 rr2r 1 1

I rr

3

K1r

r r

3

I r

x 1 1 5 r 1 1

y 3 8

2 3

x 2

2 7 r

3 2

m r r

8 r

r

12 r r

r 1

r r

8 r

ult 3

3

1 2 3 3

y

3

5

7

4

3

K1 rr r r I rr

Equating total work done by the segments to energy absorbed by yield lines we get

K1r r r

I r

K1 rr2r 1 1

I rr

K1r

r r

I r

x 1 1 5 r 1 1 y 3 8 2 3 x 2 2 7 r 3 2

r r

8 r

r

12 r r

r 1

r r

8 r

3

3

1 2 3 3

W l2

K1 rr r r I rr

ult y

y 3 5 7 4 3

mult

r r3

r r2

r r

r2r3

r2r3

r2r3

r r 1

1

r2r3

1 5 1 5 1 r8 1 5 3 8 3 8 3 8 1 3 8

3 8

6r 2 r 6r r 12 6r r 12 2r 1 r 1 r

r 6r r 12

r 3

3 1 3

1 3

3 3 1 2 2 3

r2r3

r r r r2

r2r3 r2r3

r r2 1 r r r2r3

3 8 1 r8 2 7 2 7 2 7 2 7 3 6 5 7 1 5

6r r

12

r 2 6r 6r

2 6r

2 3 3 3 3 3

Equation 2 for Failure Pattern 2

K1r r2

I r K1rr

I rr K1r2r I r K1 r

I r

x 5 1

1 1 y 3

r r

2 3 x 2 7 3 2 y 3

r r

4 3

W l 2 r r

r 1 5 7

r 1 r

r r 5 7

r

ult y

3

3

mult

r2r3

r2r

• r2r3r1 r2r3r 1 r3r r2r3r 1

8

3

2

7

3

1 5

(1 r1r5 ) 5 1

1 5 3

1 5 3

R * 6r3

2 6r3

6r3

2 r3 1

6r3

r2r3r

1 1 r r r r3

r2r3

r2r3

r2r

r2r3

2 7 3 7 2 2 7

2 7 2 7 6 3 1 5

6r3 2

6r3 6r3 2 6r3

Equation 3 for Failure Pattern 3

K 1 r r r I r K 1 rr r r

I rr

x 1 1 5 r 1 1 y 3

8 3 r 2 3 K 1 rr r r I rr

r r 8

r r

1

r r 1 7

r 1

y 3 7 5

4 3

K 1 r 2 r

3

I r

3

1 3

3

x 2 7 3 2

W l 2

r r

ult y

mult

r 2 r 3

r r r r 2

r 3 r 2

r 3 r 2

r r

r r 2

r 2 r 3 r

1

1 5 1 r

1 5 1 5 8 3 8 3 1 r

8 3 3 8

2 7 3

6r

8 r 2 6r r

12 6r r 12

7 r r 1 2 r 1

6r

R * 3

3 1 3

1 3

1 3 3

3

r 2 r 3 r

1

r r 2

r 2 r 3

r 2 r 3

r r 2

r 2 r 3

2 7 3 1 r r 2 7 2 7 2 7 3 6 1 5

6r3 2 6r 6r 2 6r

2 7

3 3 3

Equation 4 for Failure Pattern 4

K 1 r

r r r

1

I r

K 1 rr

I rr

K 1 r 2 r

I r

x 1 r

1 5 3 2 1 y

3 r

r 2

7

5

3 x 2 7 3 2

6

r

r3

r

r3 1

r3 1 r

r

1 r6 r3

W l 2

K y rr3 r

r7 I3 rr3

ult y

mult

1

r 2 r 2

r r r 1 r r 2 r 2 r 3 r 1 r 2 r 3 r 1

r r 2 r 2 r 3 r 1

6

3 6 1 r

1 5 3 1 5 1 5 3 1 5 3

3 8 2 7 3

R *

6r1

r3 2

6r3

6r3

2 r3 1

6r3

r 2 r 3 r

1

r r 2

r 2 r 3

r 2 r 3

r r r r 2

r 2 r 2

2 7 3

1 r r 2 7 2 7 2 7 1 r

6 3 3 6 3 6

7

6r3

2 7 2

6r3

6r3

r1 2

6r1

Equation 5 for Failure Pattern 5

K 1r

6

8

I r

K 1 rr

y

3

r r

I rr

K 1

I r

x 1 r r

6

1 1

8 3 r

2

3 x

r 2 r

3 2

r r

r 1 r r 1 7

r 1

r 2 7 r

1

r3r6

3 1 3

3

W l 2

K y rr3 r7

r I4 rr3

ult y

mult

r 2 r 3

r r 2

1

r 3r 2

r 3r 2

r r

r 2 r

r 2 r 3 r

1

3 6 1 5 8 3 8 1 r7 3 8 6 3 2 7 3

6r 2 6r r 12 6r r 12 r r 1 2r 1 6r

R * 1

1 3

1 3

1 3 3

3

r 2 r 3 r

1

r r 2

r 2 r 3

r 2 r 3

r r r r 2

r 2 r 3

2 7 3 2 7 1 r2 r7 2 7 2 7 1 r7 3 6 3 6 3 6

6r3 2

6r3

6r3

r1 2

6r1

Equation 6 for Failure Pattern 6

K 1 r

r r r

1

I r

K 1 rr

y

I rr

K 1 r

I r

6

x 1 r

1 5 3 1 1

3 r

r 2

3 x 2 3 2

r r

r r 1 7 5

r 1 r

r

1

r3 r6

3 3

3

W l 2

K y r3 r r

r7 I 4 r3 r

ult y 1

mult r 2 r 3 r 1r r r r 2 r 2 r 3 r 1 r 2 r 3 r 1 r r 2 r 1r r 1 r

1

3 6 1 r

3 5 1 1 5 1 5 3 1 5 3

3 8 3 2 7 3

6r

6 r 2

6r 6r

2 r 1

2r r

6r r

R * 1 3

3 3 3

2 3 2 3

r 1

1 1

r r

1

r r r r 2

r 2 r 3

3

7 2 1 r

6 3 3 6 3 6

6r r

6r r

6r r

2r r 7 r 2 6r

2 3

2 3

2 3

3 2 1 1

Equation 7 for Failure Pattern 7

K1r

r r I r K 1 rr

r r I r r

x 1 r

1 5 1 1 y

3 r

7

8 3 2 3

8

r

r3

r

r3 1

r1 r3 1

r3 1

K1r

I r 1

x 2 3 2 K y rr3 r5 r7 I4 rr3

W l 2

r r

ult y

mult

r 2 r3

r r r r 2

r 2 r3

1 5 1 r8 1 5 1 5 3 8

6r

r 2

6r r

12

3 3 1 3

r 2r3

r r

r r 2

r r 1r 1 r 1

R * 3 8

1 r7

3 8 3 8 2 7

3 3

6r r

12

r r

1 2r

1

2r r

6r r

1 3

1 3 3

3 2 2 3

r 1 1 1 r r 1 r r 2

r 2 r3

3

2 7

3 6 1 5

6r2r3

6r3r2

6r3r2

2r3r2

2 6r3

Equation 8 for Failure Pattern 8

K 1r

I r

K 1 rr r r

I rr

K 1r

I r

x 1 r

r 1 1

y 3 8 3 r

2 3

x 2

3 2

r 6 8

r r

1 r r

1 7

r 1 r

r

1

r3 r6

3 1 3

3

W l 2

K y rr3 r7

r I 4 rr3

ult y

mult

r 2 r 3

r r 2

1

r 3 r 2

r 3r 2

r r

r 2 r

r r

1r

1 r

1

3 6 1 5 8 3 8 3 1 r 3 8 8 3 2 7 3 3

6r

2 6r r

12

6r r

12

7 r r

1 2r

1

2r r

6r r

R * 1

1 3

1 3

1 3 3

3 2 3 2

r

1

1 1

r r

1

r r r r 2

r 2 r 3

3

2 7 1 r

3 6 3 6 3 6

6r r

6r r

6r r

2r r

7 r 2

6r

2 3

2 3 2 3

3 2 1 1

Equation 9 for Failure Pattern 9

K 1 r r r

r r r

1

Ir

K 1 rr

I rr

K 1r

I r

x 1 1 5 1 5 3 1 y 3 r

r 2 3 x 2 3 2 K 1 rr r

• r Irr

W l 2 r r r

r r

1 7 5

r 1 r r

y 3 7 5

4 3

ult y 3

3

3

3

mult

r 2 r 3

r r 2

r 2 r 3 r

1

r 2 r 3 r

1

r 2 r

r r

1r

1 r

1

1 5 1 r r

1 5 1 5 3 1 5 3 8 3 2 7 3 3

R *

6r3

1 5 2

6r3

6r3

2r3

1

2r3r2

6r2 r3

r 1

1 1

r r

1

r 2 r

r 2 r 3

3

2 7 6 3 1 5

6r2 r3

6r2 r3 6r2 r3

2r3r2 2 6r3

Equation 10 for Failure Pattern 10

K1r r r I r 1

x 1 r6

5 1

1 1

K y rr4 r5 r7 I2rr4

r r4 r

K1

r r r

I r

x 1

1 7 r 3 1 K1r(r 2r ) I

rr

W l 2

r r 1

r r 1 6 r r 1

y 3 6

4 3

ult y

1

4 1

1

mult

r 2r3 r

• r rr r r r 2 r 2r3 r 2r3 r 2r
• r 2r33 6

4 5 1

4 6 1 5

1 5

1 5 8 4

1 7

6r 2r 6r

6r 2

6r r

12

1

4 4 4

4 1

r 2r3

r r 2 r r r 2r3r

1

R *

1 7 2

1 7 1

1 7 r6 3 6 1

6r4 r1 1

2r1 1 r4 r1 1

6r1

r 2r3r 1

r r r r 1r r

r 2r3

3 6 1

r3r6 1 3 6 1

3 6 3 6

6r1

2r1

6r1

Equation 11 for Failure Pattern 11

K1r r r r r I r 1 r r r r r 1

x 1 1 5 3 8 1 1 K y rr4 4 8 4 8 1 I2 rr4

r1 r3

r r1

K1r r r I r

x 1 1 7 r 3 1 K1 rr r r I rr

W l 2

rr 1

r r 1 8

rr 1

y 3 5 7 4 3

ult y

1

3 1

1

mult

r 2r3 r

• r r

r r r r 2 r 2r3 r 2r3 r r r

r r (r 1)r r 2 r 2r3 r 1

1 5

3 1 5

3 8 1 5

4 8

4 8

1 4 8

4 8 1

4 8 4 8 1

R * 6r3

2r3

6r1

6r1

2r1

6r1

r 2r3 r 1 r r

r r 2

r 2r3

r 2r3

r 2r

r 2r3

4 8 1 1 1 7 r8 1 7 1 7 1 7 6 3 1 5

6r r r 1 2r r 12 6r r 12 6r r 12 2 6r

1

3 1

3 1

3 3

3 3

3

Equation 12 for Failure Pattern 12

K1r

r r

r r I r 1

x 1 5 1

1 5

1 1

K y rr4 r7 r5

I2rr4

r

r3

r4 r

K1

r 2r

r 2r I r

I r

x 1 7

1 7

1 1

K1 rr r r y 3

W l 2

r r r 12

r r 12

rr 1

y 3 5 7

r

ult y

4 1

3 1

1

mult

r 2r3

r r 2

r r r r

r 2r3

r 2r3

r r 2

r 2r3

1 5 1 5 1 1 5 1 5 1 5 1 5 4 8 1 7

6r 2 r

r 6r

6r 2

6r r 12

3

3 4 4 4

4 1

r 2r3

r r 2 r r

r r

r 2r3

R * 1 7

1 7 1

1 7

1 7

1 7

6r r 12

2r 1 rr 1

r r 1 6r r 12

4 1

1 3 1

4 1 3 1

r 2r3

r r 2

r 2r3

1 7 3 6 1 5

6r r 12 2 6r

3 1 3

Equation 13 for Failure Pattern 13

k1r r r I r 1 r r

x 1 r8 1 5 1 1 K y rr4 4 8 r7 I2rr4

r

r3

r

r1

k1r 2r 1 1 I r

x 1 7

3 1 K1 rr r

r I rr

W l 2

rr 12 r r

rr 1

y 3 5 7

4 3

ult y

mult

r 2r3

1

r r

r r 2

3 4

r 2r3

1

r 2r3

r r r r 2

r 2r3

1 5 1 r5 1 5 1 5 4 8 4 8 1 r7 4 8 4 8 1 7

6r

r 2 6r

6r

r 2

6r r 12

R * 3 3

1 1 1

4 1

r 2r3

r r 1

1

r r 2

r 2r3

r 2r3

r r 2

r 2r3

1 7 1 1 7 1 7 1 7 1 7 3 6 1 5

6r r 12 r 1 r r 2r 1 6r r 12 6r r 12 2 6r

4 1

1 3

4 1

3 1

3 1

3

Equation 14 for Failure Pattern 14

k1r r r I r 1

x 1 r6

1 5

1 1

K y rr4 r5 r7

I2rr4

r

r4 r

k1r 2r 1 1 I r r r

x 1 7

3 1 K1 rr

3 6 r I rr

W l 2

rr 12 r r rr 1

y 3 r 7 4 3

ult y

1 3 4

1 1

mult

r 2r3

r r r r 2

r 2r3

r 2r3

r r 2

r 2r3

3 6 1 r6 1 5 1 5 1 5 1 5 4 8 1 7

6r

r 2 6r

6r 2

6r r 12

1 4

4 4

4 1

r 2r3

r r 1

1

r r 2

r 2r3

R * 1 7

1

1 7

1 7

1 7

6r r 12

r 1 r

r 2r 1

6r r 12

4 1

1 3

4 1

3 1

r 2r3

r r r r 2

r 2r3

1 7 1 r7 3 6 3 6 3 6

6r r 12 r 2 6r

3 1

Equation 15 for Failure Pattern 15

1 1

K1r

I r 1

r r

x 1 r8 r6 1 1 K y rr4 r7 4 8 I2rr4

r

r r1

K1r 2r 1

1 I r r r

x 1 7

3 1 K1 rr r

3 6 I

rr

W l 2

rr

12 r

r rr

1

y 3 7

r 4 3

ult y

mult

1

r 2r3

r r 2

4 3

r 2r3

1

r 2r3

1

r r r r 2

r3r 2

3 6 1 5 4 8 4 8 1 r7 4 8 4 8 7 1

6r 2 6r

6r

r 2

6r r

12

1

1 1 1

4 1

r3r 2

r 2r r r 1 1

r3r 2

R * 7 1

7 1 1

7 1

7 1

6r r 12

2r 1 r 1 r

r 6r r 12

4 1

1 1

4 3 3 1

r3r 2

r r r r 2

r 2r3

7 1 1 r7 3 6 3 6 3 6

6r r 12

r 2 6r

3 1

1 1

Equation 16 for Failure Pattern 16

K 1 r

I r

K 1 r

I r

4

3

x 1

r r

1 1 K 1 rr 2 r

• I rrx 1 r r

4 1 K 1 rr2 r

• I rr4

3

W l 2

r 6 8 r

y 4 8

2 4 rr 1 6 8

rr 1

y 3 6

ult y

mult

r 2 r 3

r r 2

r 2 r 3

r 2 r 3

r 2 r

r 3r 2 r 1

r 3r 2 r 1

r 2 r

r 2 r 3 r 1

3 6 1 5 4 8 4 8 8 4 1 r r 8 4 1 8 4 1 7 1 3 6 1

4 8

R *

6r1

2 6r1

6r1 2

6r1

6r1

2r1 1

6r1

r 2 r 3 r 1

r 2 r

r 2 r 3

3 6 1

6 3 1 r r 3 6

3 6

6r1 2

6r1

Equation 17 for Failure Pattern 17

K 1 r

6

8

I r

K 1 r

I r

4

3

x 1

r r

6

1 1

K 1 rr

• I rrx 1 r r

3 1

K 1 r r 2 r

• I rr4

3

W l 2

r r

y 4 2 4

r r 1 6 8

r r 1

y 3 6

ult y

mult

r 2 r 3

r 2 r

r r

1

1 1

1 1

r 1

3 6 5 1 4 8

1

R * 6r1 2 2r4 r1

6r1r4 6r1r4

6r1r4

r 1 r r 1r 1

r r 2

r 2 r 3 r

1

r 2 r 2 r

1

r 2 r

r 2 r 3

1

4 8

1 1 7 3 6 1 3 6 1 (1 r r ) 6 3 3 6

6r1r4

2r4 r1

2r1

1

6r1

6r1

3 6 2

6r1

Equation 18 for Failure Pattern 18

K 1r

1

I r 1 2

K 1 r

I r 1

2

r6 r8

1 1

K y rr4 r8 I 2 rr4

x 1 r8 r6

3 1

K y rr3 I y rr3

Wultly

r r

r r1 1

r r1 1

mult

1

r r

1

r r 2

r 2 r 3

r 2 r 3

r 2 r 3

r r

1r r 2

r 1r 2r 3

6 3 1 5 4 8 4 8 4 8 8 4 4 8 1 4 8

R * 6r1r3 2r3r1 2

6r1

6r1 6r1 2

6r1

r 1r 2r 3 r r 2 r r 1r 1 r 1 r 1 1

1 4 8 1 7 6 3 1 1 1

6r1

2r1 1

2r3r1

3r3r1

6r3r1

6r3r1

Equation 19 for Failure Pattern 19

K 1 r

I r r r

K 1 r

r r

I r

x 1 r

r 1 1 K 1 rr r

4 8 I

rr x 1 r

2 4

6

7 1

3 1

r 6 8

r

y 4 7

r1

r r1 1

r4 r1 1

r r1 1

W l 2

K 1 rr I

rr

ult y

mult

y 3 4

1

3

r r

1

r r 2

r 2 r 3

r 2 r 3

r r r 2 r

r 2 r 3

3 6 1 5 4 8 4 8 1 r 4 8 8 4 1 7

6r r 2r r 2 6r 6r 7 r 2 6r r 12

R * 3 1 1 3

1 1 1

4 1

r 2 r 3

r r r r 2 r r 1r r 1

r 1 1

1 7 1 r

1 7 1 7 6 3 1 1

1

6r r 12

6 r r 1 2r 1 2r r 1

6r r 6r r

6r r

4 1

4 1 1 3 1

3 1 3 1

3 1

Equation 20 for Failure Pattern 20

K 1 r

I r

x 1

r r

1 1

K 1 rr

• I rr

r 6 8

K 1 r r r

r y 4 2 4

I r

r r

x 1 1 7 r

3 1 K 1 rr 6 3 r

I

rr

W l 2

rr

1

r r

1 8

rr

1

y 3 r 7

4 3

ult y

1

3 1

1 1

mult

r 2 r 3

r 2 r/p>

r r

1 1 1 r 1

3 6 5 1 8 4

1

6r1 2 2r4 r1

6r4 r1 6r4 r1

6r4 r1

r 1 r r 1r 1

r r

r r 2

r 2 r 3

R * 1

8 4

1 1 r

1 7 1 7 1 7

6r r

2r r

8 r r

1 2r

1

6r r

12

4 1

4 1

3 1 1

3 1

r 2 r 3

r r r r 2

r 2 r 3

1 7 1 r

6 3 3 6 3 6

6r r

12

7 r 2

6r

3 1 1 1

The respective equations for corresponding failure patterns can be obtained for OLC and OLD condition by making respective negative yield moment coefficients zero.

Minimization of theVirtual Work Equations

The value

Wult

2

L

y of these equationsconsists of the unknown non dimensional parameters r1, r2,r3 and r4 which define

mult

the positions of the yield lines.A computer program has been developed for various values of the non dimensional parameters r1,

r2,r3 and r4within their allowable ranges in order to find the minimum value of

Wult

2

L

y for the yield line failure patterns

mult

considered. In this computer program, the values of r1, r2,r3 and r4were varied at increments of 0.1. Using the above equations, one can develop useful charts basing on orthogonality which may be used eitherfor design or analysis in general. The governing failure pattern for different edge conditions and for different data is presented in Table-1.

EXAMPLE: One Long sideDiscontinuous Slab (OLD)

(Negative moment to positive moment ratio in both directions is same and unity)

Transform an orthotropic slab to an equivalent isotropic slab in which the ratio of Negative moment to positive moment in both directions is same and unity using the affine theorem.

Since I1

K’x I2

K’y 1.0 , the transformation of the given orthotropic slab(Fig.2 (a)) in X direction is

transformed to an equivalent isotropic slab (Fig.2 (b)) by dividing with . This principle is illustrated in Fig.2. Using the above methodology, few numerical examples are presented in Table 2.

Table 3shows the strength and the failure pattern forOLD based on the principle =r2for different values of coefficient of

orthotropy() and their corresponding orthogonal affine moment coefficients.

Note: In the case of SS, TLC and TSC edge conditions of the slab; the affine theorem cannot be applied because the negative moment is not present along one of the edges of the given slab. Therefore one can design the slab as orthotropic using any available computer program.

Preparation of Charts

Design charts have been prepared for One Long side Discontinuous slab (OLD)slabs.Charts 1-4 are shown in Appendix-

B.Charts1-4 are plotted for

Wult

2

L

y versus Opening (=) for various ratios of K1, K2 and r for different CS conditions.The

mult

L

2

values of rare taken between 1 and 3. The K1 values are plotted varying between 1 and 5 thus ensuring that greater yield moment is in the direction of short span. This is in accordance with elastic theory distribution of bending moments. The values of K2 are

plotted for 1 and 2 for One Long sideDiscontinuous slab (OLD).While preparing the design charts, the least value of

Wult y

mult given by thirteenfailure patterns is considered. Using these charts one can directly design or do analysis of a slab with ainterior corner opening. In addition, a special charts5-6 is presented for various values of openings which can be used for transforming an orthogonal slab to equivalent isotropic slab for the governing failure patterns only.

1. Analysis Problem:(OLC)Determine the safe uniformly distributed load on a rectangular two way slab with interior corneropening all sides continuousslab(Fig.3(a)) for the following data:

A slab 6m X 4m with interior corner opening of size 1.2m X 0.4m at a distance of 1.2m from long edge and 0.8m from short edge is reinforced with 10mm diameter bars @ 200mm c/c perpendicular to long span and 8mm diameter bars @150mm c/c perpendicular to short span. Two meshes are used one at top long side continuous and one at bottom, thickness of the slab is 120mm. Characteristic strength of concrete is 20MPa and yield stress of steel is 415MPa.

Solution:

2

f y Ast

According to IS 456:20009, mult 0.87 f y Astz ,where

Assuming effective depth of slab in short span direction=100.00 mm effective depth of slab in long span direction=90.00 mm

Area of the steel perpendicular to long span=374 mm2

Area of the steel perpendicular to short span=314 mm2

z d 1

f ck bd

———— (2)

The ultimate moments in short and long span directions can be found using the expression (2).

Therefore, multparallel to long span=13.489 kNm/m

mult parallel to short span=10.192kNm/m

6.0

For aspect ratio of slab, r

4.0

1.5 and taking mult = 13.489 kNm/m,

The orthogonal coefficients (Fig.3(b)) will be K`x=0.755, I1=I3=0 and K`y=I2=1.0, I4=0.With these orthogonal coefficients and for

=0.2, =0.1, r =1.5, C5=1.2m, C6=0.8m;

Twenty predicted failure patterns are evaluated by using computer program to find the governing failure pattern and the final results are as follows.

Wult

2

L

y =19.67773, r1=3.46191, r2=3.15185, r3=3.46191 and the failure pattern is 7

mult

Wult= 19.67773 13.489 = 16.5895kN/m2

42

Wult=1.5(Wll+Wdl) =16.5895kN/m2

16.5895

Wll=

1.5

3.5=7.56kN/m2

The intensity of live load on the slab is 7.56kN/m2

2. Design Problem: (OLD)Designall sides continuous slab of 5.0 m X 2.5 m with interior corner openings of size 1.5 m X 0.75 m at a distance of 0.5m from long edge and 0.5m from short edge to carry a uniformly distributed live load of 4.5kN/m2 .Use M20 mix and Fe 415 grade steel.

Given data:Aspect ratio of slab(r) = Lx

Ly

=5.0/2.5=2.0, Lx =1.5 m, Ly=0.75m.

=0.3,=0.3, C5=0.5m, C6=0.5m

Twenty predicted failure patterns are evaluated by using the computer program to find the governing failure pattern,

2

byassumingKx=I1=I3=1.0,Ky=I2=2.0, I4=0.

The value of

Wult Ly

=37.83424andthe failure pattern is 8

mult

Unknown parameters:

r1=

Lx =4.6000, r2=

C1

Lx =3.55185, r3=

C2

Ly

=2.34783

C3

C1 =1.0869 m, C2 =1.4077 m, C3 =1.0648 m

Assuming overall thickness of slab =110 mm Dead load of slab=110 X 25=2.75 kN

Ultimate total load=1.5 X 8.0 =12.0kN/m2

mult=

12.0 2.52

34.83424

=2.15305kNm/m

The orthogonal moments are K1xmult=I1mult= 1.0 X 2.15305=2.15305kNm/m

K1ymult=I2mult=2.0 X 2.15305= 4.3061kNm/m

2.15305106

Effective depth, d

0.138 20 1000 =27.9301 mm

Adopt effective depth as 100 mm and overall depth as 110 mm Area of steel along short span =155.169 mm2

Minimum area of steel required along short span = 204.82 mm2 Use 8 mm diameter bars @ 240 mm c/c

Area of steel along long span=84.3189mm2 Use 6 mm diameter bars @ 300 mm c/c

Details of reinforcement are shown in Fig.4(c).

CONCLUSIONS:

1. The virtual work equations for orthotropic slabs with unequalinterior corner openingwith all sides continuous whose aspect ratio of opening is different from the aspect ratio of slab subjected to udlare presented.
2. Design charts for One LongsideDiscontinuous slabare presented for different aspect ratios of slab.
3. Fe numerical examples are presented based on theorem of VI and VII of affine theorem of Johansen. K.W.1for orthotropic slabs with interior corner openings.
4. Twouseful charts for affine transformation for different sizes of openings arealso presented.
5. In case of One Long sideDiscontinuous, for K2=1, 2 and K1=1, 2,3 and 5 the strength of the slab is decreasing with increase in aspect ratio when compared to solid slab.
6. The charts can be used either for analysis or design for different size of openings in the interior corner of the slab.

REFERENCES:

1. Johansen, K.W., Yield-Line Theory , Cement and Concrete Association, London, 1962, pp. 181.
2. Goli, H B. and Gesund, H., Linearity in Limit State Design of Orthotropic Slabs J. of Structural Division,ASCE,, Oct.1979, Vol.105, No. ST10, pp.1901-1915.
3. Rambabu,K. and Goli, H B. Simplified approach to design orthogonal slabs using affine theorem,Journal of Structural Engineering, Chennai, India, Dec 2006-Jan 2007,Vol.33,No.5,pp.435-442.
4. Islam,S. and Park, R., Yield line analysis of two-way reinforced concrete slabs with openings,The Structural Engineer,June 1971,Vol. 49, No. 6, pp. 269-275.
5. Zaslavsky.Aron,Yield line analysis of rectangular slabs with central opening,Proc. Of ACI, December 1967,Vol. No.64,pp. 838-844.
6. Siva Rama Prasad,Ch., and Goli,H. B., Limit State Coefficients for Orthogonal Slabs. International Journal of Structural Engineering, India, Jan-June 1987, Vol.7, No.1, pp. 93-111.
7. Sudhakar, K.J., and Goli,H B., Limit State Coefficients for Trapezoidal- Shaped Slabs Supported on Three Sides, Journal of Structural Engineering, Chennai, India, June-July 2005, Vol.32, No.2, pp.101-108.
8. Veerendra Kumar and Milan Bandyopadhyay Yield line analysis of two way reinforced concrete slabs having two adjacent edges discontinuous with openings ,Journal of Structural Engineering, Chennai, India, June-July 2009,Vol.36,No.2, pp.82-99.
9. Indian Standard Plain and Reinforced Concrete-Code of Practice, IS 456:2000,BISNewDelhi.

Lx=1.12

Lx=1.584

Ly=0.8

Ly=0.8

Ly=4 Ly=4

Lx=5.6

Lx=7.92

Fig 2 (a): ORTHOTROPIC SLAB Fig 2 (b): EQUIVALENT ISOTROPIC SLAB

Kx=I1=I3=0.5, Ky=I2=1,I4=0. r5=0.2, r6=0.2, =0.2,=0.2,

I3/ Kx= I4/ Ky = 1.0, Âµ = 0.5,K=3

r=Lx/Ly= 5.6/4 =1.4,

In order to check on affine theorem a computer program is used to evaluate the value of WultLy2/multand the value is22.60740

As per affine theorem the transformed Kx =I1=I3=1, Ky=I2=1,I4=0. r5=0.2, r6=0.2, =0.2,=0.2,

Lx=Lx/Âµ= 5.6/0.5 = 7.92m,

r=7.92/4=1.98, Âµ = 1.0,K=4,

The value of WultLy2/multis obtained from chart 5

for r=1.77.Taking this value one can design the given orthotropic slab without using computer program.

Fig 2 – Orthotropic slab and equivalent isotropic slab as per Theorems VI and VII of Johansen

Lx=6.0m

Ly=4.0m

I3mult = 10.192kNm/m

Kxmult = 10.192kNm/m

Kymult = 13.489 kNm/m

I2mult = 0

Fig. 3 (a) Analysis problem Fig. 3 (b) orthogonal moment coefficients

I1 = 1.0 Kx=1.0

Positive steel in

x &y direction

Ly=2.5m

Negative steel

Ky=2.0

Lx= 5.0m

I2 = 2.0

Fig. 4 (a) Design problem Fig. 4 (b) orthogonal moment coefficients

Positive Reinforcement

Ly=2.5 m

Lx=5.0 m

Negative Reinforcement

Fig. 4 (c)

Positive Steel in X&Y Direction

Kxmult

Y

I1mult orI3mult

Negative Steel in X &Y Direction

Kymult

X

I2mult orI4mult

Fig. 5

 Table 1 Governing failure patterns for different data for different edge conditions Failure Patterns OLC (20) OLD (18) Opening Sizes Opening Sizes 1 r5 = 0.3, r6 = 0.2, = 0.2, = 0.1, r = 1.3 * r5 = 0.3, r6 = 0.1, = 0.1, = 0.3, r = 1 2 r5 = 0.4, r6 = 0.1, = 0.1, = 0.2, r = 1.4 r5 = 0.4, r6 = 0.4, = 0.1, = 0.1, r = 1 * 3 r5 = 0.2, r6 = 0.1, = 0.3, = 0.2, r = 1.3 r5 = 0.2, r6 = 0.1, = 0.3, = 0.2, r = 1.5 4 r5 = 0.3, r6 = 0.3, = 0.2, = 0.2, r = 1.1 r5 = 0.4, r6 = 0.4, = 0.1, = 0.1, r = 1.5 5 r5 = 0.2, r6 = 0.2, = 0.3, = 0.1, r = 1 r5 = 0.1, r6 = 0.1, = 0.4, = 0.3, r = 1.5 6 r5 = 0.2, r6 = 0.2, = 0.1, = 0.3, r = 1.8 r5 = 0.3, r6 = 0.3, = 0.1, = 0.2, r = 2 7 r5 = 0.1, r6 = 0.1, = 0.1, = 0.2, r = 1 r5 = 0.1, r6 = 0.1, = 0.1, = 0.2, r = 1 8 r5 = 0.1, r6 = 0.1, = 0.1, = 0.1, r = 1 r5 = 0.1, r6 = 0.1, = 0.1, = 0.1, r = 2 9 r5 = 0.2, r6 = 0.1, = 0.1, = 0.4, r = 1 r5 = 0.4, r6 = 0.1, = 0.1, = 0.1, r = 2 10 r5 = 0.4, r6 = 0.3, = 0.1, = 0.2, r = 1.1 — 11 r5 = 0.2, r6 = 0.1, = 0.2, = 0.2, r = 1 r5 = 0.2, r6 = 0.1, = 0.1, = 0.3, r = 1 12 r5 = 0.4, r6 = 0.1, = 0.1, = 0.4, r = 1.2 — 13 r5 = 0.2, r6 = 0.1, = 0.2, = 0.4, r = 1 r5 = 0.2, r6 = 0.1, = 0.3, = 0.4, r = 1.5 14 r5 = 0.4, r6 = 0.2, = 0.1, = 0.3, r = 1.2 r5 = 0.4, r6 = 0.3, = 0.1, = 0.2, r = 1.8 15 r5 = 0.1, r6 = 0.1, = 0.4, = 0.3, r = 1 r5 = 0.2, r6 = 0.2, = 0.3, = 0.3, r = 1.5 16 r5 = 0.3, r6 = 0.2, = 0.2, = 0.2, r = 1 r5 = 0.2, r6 = 0.3, = 0.2, = 0.2, r = 1 17 r5 = 0.2, r6 = 0.3, = 0.1, = 0.1, r = 1 * r5 = 0.2, r6 = 0.3, = 0.2, = 0.1, r = 1 18 r5 = 0.2, r6 = 0.3, = 0.3, = 0.1, r = 1 r5 = 0.2, r6 = 0.3, = 0.1, = 0.2, r = 1 19 r5 = 0.2, r6 = 0.3, = 0.3, = 0.2, r = 1 r5 = 0.2, r6 = 0.4, = 0.3, = 0.1, r = 1.3 20 r5 = 0.2, r6 = 0.1, = 0.3, = 0.1, r = 1 r5 = 0.2, r6 = 0.2, = 0.2, = 0.3, r = 1 COEFFICIENTS K’x = 1.33, K’y = 0.33, I1 = 0, I2 = 2.33, I3 = 0, I4 = 0, = 0.5, K = 4 K’x = 0.9, K’y = 0.5, I1 = 1.5, I2 = 1.1, I3 = 1.5, I4 = 0, = 1.5, K = 4 *COEFFICIENTS K’x = 2, K’y = 0.5, I1 = 0, I2 = 1.5, I3 = 0, I4 = 0, = 1, K = 4 K’x = 0.67, K’y = 1.33, I1 = 0.67, I2 = 1.33, I3 = 0.67, I4 = 0, = 0.50, K = 4
 Table- 2 Numerical examples based on Theorem VI & VII of Affine Theorem of Johansen. K.W.1 Example (OLD) Orthogonal Moment Co-efficients Aspect Ratio (r) Strength WultLy2/mult Aspect Ratio (r*) S.No. Openingsize 1 =0.2,=0.3 Kx =0.25,K y =1.0, I1=I3=0.25, I2=1.0, Âµ=0.5,,K=2.5 1.0 23.04303 2.0 2 =0.2,=0.2 Kx =0.5,Ky =1.0, I1= I3=0.5, I2=1.0, Âµ=0.5,K=3.0 1.4 22.6074 1.98 3 =0.2,=0.1 Kx =0.667,Ky =1.0, I1= I3=0.667, I2=1.0, Âµ=0.667,K=3.33 1.6 22.87023 1.96 4 =0.1,=0.3 Kx =1,0,Ky =1.0, I1= I3=1.0, I2=1.0, Âµ=1.0,K=4.0 1.3 32.89787 1.3 5 =0.1,=0.2 Kx =1.5,Ky =1.0, I1= I3=1.5, I2=1.0, Âµ=1.5,K=5.0 1.9 27.71505 1.55 6 =0.3,=0.2 Kx =2.0,Ky =1.0, I1= I3=2.0, I2=1.0, Âµ=2.0,K=6.0 1.7 35.05538 1.20 7 =0.3,=0.3 Kx =4.0,Ky =1.0, I1= I3=4.0, I2=1.0, Âµ= I3=4.0,,K=10.0 2.0 44.61118 1.0 NOTE:(1) r*:equivalent isotropic slab aspect ratio, (2)I1/ Kx = I2/ Ky= 1.0, (3)r5=r6=0.2, (4) I4=0.

Table-3

One Long side continuous slab(OLD), based on the principle Âµ=r2

44.61118

Kx =0.5, Ky =1.0

I1= I3=0.5, I2=1.0

0.707

0.5

44.61118

Kx=0.25,Ky=1.0 I1= I3=0.25, I2=1.0

0.5

0.25

Failure pattern for all aspect ratios

mult

L 2

ult y

W

Orthogonal moment coefficients (affine)

Aspect Ratio of

Slab

r

( )

orthotropy

Coefficient of

C2

0.667

Kx =0.667, K

y =1.0

I1=I3=0.667,I2=1.0

44.61118

Kx =1.0, Ky =1.0

I1= I3=1.0, I2=1.0

1.0

1.0

44.61118

0.817

I1m

I3m

C4

I2m

44.61118

Kx =1.5,Ky =1.0

I1= I3=1.5, I2=1.0

1.225

1.5

44.61118

Kx =1.44, Ky =1.0

I1= I3=1.44, I2=1.0

1.2

1.44

C1 Failure pattern –

15

L x -C1

44.61118

Kx =2.0, Ky =1.0

I1= I3=2.0, I2=1.0

1.414

2.0

C1=Lx/r1. r1=2.14783 C3= Ly/r3. r3=2.54783 C4= Ly/r4. r4=1.84607

Interior corner opening coefficient

44.61118

Kx =2.25, Ky =1.0

I1= I3=2.25, I2=1.0

1.5

2.25

I4=0, = 0.3, = 0.3,r5=0.2,r6=0.2.

4.0

Kx =4.0, Ky =1.0

I1= I3=4.0, I2=1.0

44.61118

2.0

Notation:

Continuous edge Simply supported edge Free edge

Negative yield line

CS A slab supported on all sides continuously (restrained)

SS A slab simply supported on all sides

OLC A slab restrained on one long sides and simply supported on other three sides

OLD A slab restrained on three sides and simply supported on one long sides

Kxmult Positive ultimate yield moment per unit length provided by bottom tension

bars placed parallel to X-axis

Kymult Positive ultimate yield moment per unit length provided by bottom tension bars placed parallel to Y-axis

I1mult and I3mult Negative ultimate yield moment perunit length provided by top tension

reinforcement bars placed parallelto x-axis.

I2mult and I4mult Negative ultimate yield moment perunit length provided by top tension

reinforcement bars placed parallelto y-axis.

K1 Ky/Kx

K2 I2/Ky

Lx, Ly Slab dimensions in X and Y directionsrespectively

, coefficients of opening in the slab

mult Ultimate Yield moment per unit lengthof the slab

r Aspect ratio of slab defined by Lx/Ly.

r1, r2, r3, r4 Non dimensional parameters of yield line propagation

r5, r6 Non dimensional parameters of opening distances

Wult Ultimate uniformly distributed loadper unit area of slab.

y

K ‘x I1

Coefficient of orthotropy K ‘

I 2

Appendix-A

I4m

I4m

C3 C3

I1m

I3m

I3m

I1m

LY LY

I2m

I2m

C1 C2

LX

C1 C2

LX

FIG 2 (PATTERN 2)

FIG 4 (PATTERN 4)

I4m

I4m

C3

C3

I3m

I1m

I3m

LY

I1m

LY

I2m

I2m

C1 C2

LX

C1 C2

FIG 5 (PATTERN 5)

LX

FIG – 3 (PATTERN 3)

I4m

I4m

C3 C3

I1m

I3m

I1m

I3m

LY LY

I2m

I2m

C1 C2

LX

C1 C2

LX

FIG 6 (PATTERN 6)

FIG 9 (PATTERN 9)

I4m

I4m

C3

C3

I3m

I1m

I3m

LY

I1m

LY

I2m

C4

I2m

C1 C2

FIG 7 (PATTERN 7)

LX

C1 LX – C1

LX

FIG 10 (PATTERN 10)

I4m

I4m

C3

C3

I3m

I1m

I3m

LY

I1m

LY

I2m

I2m

C4

C1 C2

FIG 8 (PATTERN 8)

LX

C1 LX – C1

LX

FIG 11 (PATTERN 11)

I4m

I4m

C3 C3

I1m

I3m

I3m

I1m

LY LY

C4 C4

I2m

I2m

C1 LX – C1

LX

C1 LX – C1

LX

FIG 12 (PATTERN 12)

FIG 15 (PATTERN 15)

I4m

I4m

C3

C3

I3m

I3m

I1m

LY

I1m

LY

C4

I2m

C4

I2m

C1 LX – C1

FIG 13 (PATTERN 13)

LX

C1 LX – C1

LX

FIG 16 (PATTERN 16)

I4m

I4m

C3 C3

I1m

I3m

I1m

I3m

LY LY

C4 C4

I2m

I2m

C1 LX – C1

LX

C1 LX –

LX

FIG 14 (PATTERN 14)

FIG 17 (PATTERN 17)

I4m

I4m

C3 C3

I1m

I3m

I3m

LY LY

I1m

C4

I2m

I2m

C4

C1 LX – C1

LX

C1 LX – C1

FIG 20 (PATTERN

LX

FIG 18 (PATTERN

I4m

C3

I1m

I3m

LY

C4

I2m

C1 LX – C1

LX

FIG 19 (PATTERN 19)

r=2

r=1

80

r=1.5

90

K1=3 K1=5

100

r=2

r=1

40

r=1.5

55

50

45

Appendix-B

65

r=1

110

60

K1=1 K1=2

r=1

20

r=3

50

r=3

15

40

0

0.1

0.2

0.3

0

0.1

0.2

0.3

=

=

60 r=1.5

r=2

r=3

70

30 r=1.5

25 r=2

r=3

35

2

Strength (WultLy /mult)

2

Strength (WultLy /mult)

Chart-1: OLD, K2=1

Chart-2: OLD, K2=1

<td

 r=1 r=1.5 r=1 r=3 r=1.5 r=2 r=3

80

70

2

Strength (WultLy /mult)

60

50

r=2

40

30

K1=1 K1=2

120

110

r=1.5

100

r=2

90

80

70

r=1.5

60

50

r=2

r=3

r=3

r=1

K1=3

r=1 K1=5

140

130

2

Strength (WultLy /mult)

20

0.3

0.2

0.1

0

0 0.1 0.2 0.3

=

=

35

30

25

40

#### Strength (WultLy2/mult)

Chart-3: OLD, K2=2 Chart-4: OLD, K2=2

45

Isotropic slab with a interior opening basing on Theorem’s VI & VII of affine theorem

=0.2,

=0.3 =0.2,

=0.2 =0.2,

=0.1

=0.1,

=0.3 =0.1,

=0.2

20

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

#### Aspect Ratio(r)

Chart- 5: Strength Vs Aspect Ratio