Comparative Study of Strengths of Two-Way Rectangular Continuous Slabs with Two Openings Parallel to Long Side and Short Side

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Volume 6, No. 2, February 17 9 Comparative Study of Strengths of Two-Way Rectangular Continuous Slabs with Two Openings Parallel to Long Side and Short Side D.

1 Volume 6, No. 2, February 17 9 Comparative Study of Strengths of Two-Way Rectangular Continuous Slabs with Two Openings Parallel to Long Side and Short Side D. Manasa, Assistant Professor, Gayatri Vidya Parishad Technical Campus, Rushikonda, Visakhapatnam Dr. K. Rambabu, Department of Civil Engg., AU College of Engineering, Andhra University, Visakhapatnam ABSTRACT Keeping in view the basic principles of yield line theory, collapse load of two-way orthotropic rectangular slabs with openings parallel to long side and openings parallel to short side are determined when the slab is subjected to uniformly distributed load (udl).two computer program have been developed to solve the virtual work equations for the admissible yield line failure patterns of the slab. Relevant tables and charts are presented. Four sides continuous edges with various sizes of openings are considered in order to plot the design charts for isotropic reinforcement coefficients. Comparative study of strengths of two-way rectangular continuous slabs with two openings parallel to long side and short side is made for same type of openings. Affine transformation is also performed for slab with openings. Some continuous slabs are cast and tested by applying uniformly distributed load using pneumatic pressure. Some photographs of Testing are discussed. Keywords Aspect ratio, openings parallel to long side, affine theorem, Orthotropic slab, uniformly distributed load, Ultimate load and Moment, Affine transformation, Yield line theory orthogonal solid slabs implicitly to that of equivalent Isotropic slab by using Affine Theorem provided the ratio of negative to positive reinforcement areas are same in orthogonal directions. However, this is a limitation on affine theorem and same is clearly brought out in this paper. Design charts are presented for Two-way slabs with two openings parallel to long side and short side subjected to uniformly distributed load for four edges continuous slab (CS). The concept of yield-line analysis was frst presented by A. Ingerslev in , K. W. Johansen 1 developed modern yield-line theory. The method involves postulating a yield-line pattern (failure mechanism) which is compatible with the boundary conditions and then using the principle of virtual work to compute the ultimate load carrying capacity. However, due to the upper-bound nature of the yield-line method, a range of yield-line patterns will often need to be explored. Yield line theory is particularly suitable for obtaining an ultimate limit state solution for an irregularly shaped slab, slabs with openings and slabs of various shapes 1. Hillerborg s strip method 1,3,6,14 which is a lower bound method can also be used to compute the ultimate load carrying capacity. INTRODUCTION Two way reinforced slabs often carry pipes, ducts and other services of considerable sizes through its openings. Such two way slabs when subjected to uniformly distributed load supported on four edges are being analyzed by using yield-line method as suggested by Johansen, K.W 1. Many researchers (Rambabu 9, Goli 11, Zaslavsky.Aron 8 ) 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 supported on various edge conditions. Johansen, K.W 2,also presented the analysis of Fig. 1 Orthotropic Slab with Openings Parallel to Long Side

Volume 6, No. 2, February 17 10 any load over the area of the opening.

2 Volume 6, No. 2, February any load over the area of the opening. The ultimate load equations are derived for the assumed possible admissible failure yield patterns using the virtual work method for continuous edge (CS) condition of slab. TRANSFORMATION OF ORTHOTROPIC SLAB TO ISOTROPIC SLAB (Using theorems VI & VII of Johansen, K.W 1 ) Fig. 2 Orthotropic Slab with Openings Parallel to Short Side VIRTUAL WORK EQUATIONS: FORMULATION OF VIRTUAL WORK EQUATIONS FOR SLAB WITH DIFFERENT OPENINGS There are several possible yield line patterns associated with different openings of the slab (Appendix A). For any opening of slab, all the possible admissible failure yield line patterns are considered keeping in view the basic principles of yield line theory. These admissible failure yield line patterns are obtained basing on the yield line principles 1,6,14 for the given configuration of the slab. These failure patterns and corresponding equations can be investigated using a computer program. In order to solve the complicated virtual work equations, a computer program was developed in FORTRAN for all the cases separately which gives the least value of W ult L 2 y /m ult for the given input data (K' x,k' y,i 1, I 2, I 3, I 4, α,β, r 5 and r 6). It is difficult to solve virtual work equations to analyse and design orthogonal slabs for every input data. The design charts presented in this paper simplifies the analysis and design of orthogonal slabs. Many researchers [6 13] have produced work equations for uniformly loaded two-way rectangular slabs with and without openings for different edge conditions. The slab is subjected to ultimate uniform distributed load (W ult ) and supported on four edges. Note that the slab is not carrying L x = 4.8m Affine transformation based on theorems VI and VII of Johansen, K.W 1 is a technique that transforms an orthogonal slab into that of an equivalent isotropic slab, whose strength is one and the same. The transformation of orthotropic slab to that of equivalent isotropic slab is presented with some examples. In order to convert the slabs, charts are required and these charts are prepared using isotropic reinforcement, which means that in each direction the reinforcement is same and unity(rambabu 4 et.al). These charts show the minimum value of W ult L 2 y /m ult versus aspect ratio of slab. According to theorems VI and VII of Johansen, K.W any orthogonal slab can be transformed into an equivalent isotropic slab provided that the ratio of negative to positive moments in the slab is same. When the ratio of negative to positive moment in the orthogonal slab is same, then such a slab can be transformed into an equivalent isotropic slab directly, which means that the given orthogonal slab solution can be obtained by analyzing the isotropic slab with modified dimensions in the X direction. Problem 4.6.1: Transform an orthotropic continuous slab into an equivalent slab in which the ratio of Negative moment to positive moment in both directions is same and unity using modified (Rambabuet al 10 ) affine theorem. Since I 1 /K' x = I 2 /K' y = 1.0, (refer Fig.4.16), the transformation of the given orthotropic slab (Fig.3(a)) to an equivalent isotropic slab(fig.3(b)) is done by simply by dividing the span in X direction with μ. This principle is illustrated in Fig 3 using affine theorem. L' x = m I 2 =1.0 K' x = 0.5 I 1 =0.5 K' y = 1.0 I 3 = 0.5 I 4 =1.0 L y = 4m I 2 =1.0 K' x = 1.0 I 1 =1.0K' y = 1.0 I 3 =1.0 I 4 =1.0 L y =4m Fig. 3(a): Orthotropic slab Fig. 3(b): Equivalent Isotropic slab

3 Volume 6, No. 2, February K' x =I 1 = I 3 =0.5, K' y =I 2 = I 4 =1.0, α=0.1, β=0.2,r 5 =0.2, r 6 =0.2 I 3 / K' x = I 4 / K' y = 1.0, µ = 0.5, K=3 r=l x /L y = 4.8/4 = 1.2 In order to check on affine theorem a computer program is used to evaluate the value of 2 /m ult and the value is As per affine theorem the transformed factors are: K' x =I 1 = I 3 =1.0, K' y =I 2 = I 4 =1.0, α=0.1, β=0.2,r 5 =0.2, r 6 =0.2 L' x = L x / µ = 4.8/ 0.5 = m, r=6.7882/4=1.697, µ = 1.0, K=4, The value of 2 /m ult is obtained from Chart -5 For corresponding value of r= Taking this value one can design the given orthotropic slab without using computer program. Fig: 3 Affine transformation For Continuous Slab With Openings Parallel To Long Side DEVELOPMENT OF ANALYSIS AND DESIGN CURVES Design charts have been prepared for continuous slab (CS), For openings parallel to long side and Short Side for all sizes of openings and are shown in Appendix A. Design charts (Chart 1 26 of Appendix A) are plotted for 2 /m ult against aspect ratio, r for different size of openings. These charts are prepared based on affine coefficients (K' x = K' y = I 1 = I 2 = I 3 = I 4 = 1.0), designers choice coefficients (K' x =0.5, K' y =0.6, I 1 = I 3 =1.5, I 2 = I 4 = 1.4), Reverse of designers choice coefficients (K' x =1.5, K' y =1.4, I 1 = I 3 =0.5, I 2 = I 4 = 0.6). The values of r are taken between 1.0 and 2.0 and size of openings up to 0.2 of the length of the slab sides are considered. Since two-way slabs are considered, the aspect ratio of slab, r is limited to 2.0. With increase in value of r, the value of 2 /m ult of a slab decreases. And size of opening is limited to 0.2 so that the slab behaviour doesn t change to cantilever action from two-way action particularly near openings. If the sizes of opening are changed the value of 2 /m ult of a slab may increase or decrease depending upon the type of opening. While preparing the design charts, the least value of 2 /m ult given by all the failure patterns is considered for the corresponding opening. For example in case of a CS slab with opening α=0.1, β=0.2,r 5 =0.2, r 6 =0.2 (Chart 5 of Appendix A), it consists of three lines. Each line represents the least value of 2 /m ult of a slab with opening s (two openings parallel to long side) Using these charts one can directly design/analyze a slab with an opening at different locations. ANALYSIS AND DESIGN PROBLEMS Analysis Problem: Determine the safe uniformly distributed load on a rectangular two way slab with longer side corner openings supported on four sides (continuous slab, Fig. 9.a) for the following data: A slab 6 m x 5 m with openings of size 0.9 m x 1.0 m towards the shorter edge are placed at a distance of 1.0 m from long edge and 1.2 m from short edge is reinforced with 10mm φ 0 mm c/c perpendicular to long span and 10 mm φ mm c/c perpendicular to short span is Considered. Two meshes are used one at top and one at bottom. Thickness of the slab is 1 mm. Characteristic strength of concrete is MPa and yield stress of steel is 415 MPa. According to IS 6:00 7, z d 1 m f ult y A st 0.87f /f ck bd y A st z, where Assuming effective depth of slab in short span direction = mm Effective depth of slab in long span direction = mm Area of the steel perpendicular to long span=392.7 mm 2 Area of the steel perpendicular to short span= mm 2 The ultimate moments in short and long span directions can be found using the expression (1) Therefore, M u parallel to long span = K' y m ult =14.144kNm/m M u parallel to short span = K xm ult = knm/m Assume K' y =1.0 then K' x =14.144/14.273=0.991 ~1.0 For aspect ratio of slab, r=6.0/5.0=1.2 and taking m ult = kNm/m, the orthogonal moment Coefficients will be K' x= K' y =I 1 =I 2 = I 3 =I 4 =1.0 with these orthogonal coefficients From Chart 6 of Appendix A for α = 0.15, β = 0.2, r 5 = 0.2, r 6 = 0.2, we get 2 /m ult 41.0 W ult = x /5 2 = kN/m 2

4 Volume 6, No. 2, February W dl = (Dead load including finishing) = 0.12 x +0.5 = 3.5 kn/m 2 W = 1.5(w ll +w dl ) = kn/m 2 W ll = / = kN/m 2 The intensity of live load on the slab is kN/m 2 Design Problem: Design a continuous slab 5m X 4m with openings parallel to short side at two corners of size each 0.75mx0.4m located at 1.0m from long edge and 1.0m from short side for a uniformly distributed live load of 4.5kN/m 2.use M mix and Fe 415 grade steel. Given: Aspect ratio of slab =L x /L y =5/4=1., αl x =0.75m, βly=0.4m, α=0.15, β=0.1, r 5 =0.2, r 6 =0. Assume K' x =I 1 =I 3 =0.5,K' y =I 2 =I 4 =1.0 The value of 2 /m ult is obtained by performing affine transformation. 2 /m ult 27. (refer SI.No.3 of Table 1) Assuming Overall thickness of slab = 1 mm Dead load of slab = 1 x = 3.0 kn/m 2 Dead loads including finishing s = 4.5 kn/m 2 Total load = 9.0 kn/m 2 Ultimate total load = 1.5 x 9.0 = 13.5 kn/m 2 m ult =13.5x4 2 /27.= 7.239kNm/m The orthogonal moments are K' x m ult = I 1 m ult = 0.5 x 7.239= 3.62kNm/m K' y m ult = I 1 m ult = 1.0 x 7.239= knm/m As per IS 6:00 5 Mu lim = 0.36 X umax d X umax d bd 2 f ck --- (2) x 10 6 =0.36(0.48) (1-.042(0.48)) (1000) (d 2 ) d=51.22 mm Effective depth required, d = mm Adopt cover 15 mm and effective depth as 100 mm and overall depth as 1 mm Positive/Negative short span moment = knm/m, Ast (required) = 9.72 mm 2 Provided 10 mm diameter mm c/c = mm 2 Positive/negative long span moment = 3.62kNm/m, Ast (required) = mm 2 Provided 8 mm diameter mm c/c = mm 2 Table: 1 Affine Transformation Examples for Continuous Slab Examples S. No Edge conditions 1 2 Openings Parallel To Long Side Size of openings & Edge distances α=0.15, β=0.1, r 5 =0.2, r 6 =0. α=0.1, β=0.15, r 5 =0.2, r 6 =0.2 Orthogonal Moment Co-efficients K' x =3.0, K' y =1.0, I 1= I 3 =3.0, I 2 =I 4 =1.0, µ=3.0, K=8.0 K' x =0.667, K' y =1.0, I 1 =0.667, I 2 =I 4 =1.0, µ=0.667, K=3.33 Aspect Ratio (r) Strength 2 /m ult Aspect Ratio (r*) Graph No Openings Parallel To Short Side α=0.15, β=0.1, r 5 =0.2, r 6 =0. α=0.15, β=0.1, r 5 =0.2, r 6 =0.2 K' x =0.75, K' y =1.0, I 1 = I 3 =0.75, I 2 =1.0, µ=0.75, K=3.5 K' x =2.0, K' y =1.0, I 1= I 3 =2.0, I 2 =I 4 =1.0, µ=2.0, K=6.0 NOTE: (1) r* : Equivalent isotropic slab aspect ratio, I 1 / K' x = I 2 / K' y = 1.0,

5 Volume 6, No. 2, February TESTING OF CONTINUOUS SLABS (CS): The experimental work on orthogonal slabs with openings parallel to long side and openings parallel to short side has not been reported so far in literature. The Slab 1.88m 1.28m is lifted up onto the testing platform and adjusted to have a continuous edges condition on all sides of the slab. Now the Slab with openings parallel to short side of size 0.288m 0.24m located at a distance 0.4m from the short edge and 0.22m from long edge, having the reinforcement details µ = 0.5, K = 4.0, K' x =0.67, K' y =1.33, I 2 = I 4 = 1.33,I 1 = I 3 =0.67 whose aspect ratio r = 1.5 and other relevant parameters were α= 0.16,β = 0.2, r 5 = 0.2, r 6 = 0.15, was loaded uniformly using pneumatic air pressure. The first crack was observed in dial gauge at a pressure of 0.2 Kg/cm 2 which was equivalent to a load of kn. The testing was continued and crack pattern was observed at a pressure of 0.37 Kg/cm 2 which was equivalent to 78.1kN and the failure pattern was marked with permanent marker (black) and was photographed and presented in Appendix B. The computer program for the above parameters gave the value of 2 /m ult =38.75 and. The theoretical predicted pattern and the pattern obtained after testing seems to be coinciding and hence W ult = = kn The predicted load was kN, whereas the load recorded was 78.1kN.These tests were planned to get the conformation of the failure pattern and ultimate load. The ratio of ultimate load to the predicted ultimate load is 1.. CONCLUSIONS 1. Design charts for two-way slabs with openings parallel to long side and short side for Continuous slabs with different size of openings at different locations for different aspect ratios 2. The charts developed can help any design engineer or architect to pick up once the choice of location of opening depending upon the plan or required strength criteria. 3. The charts can be used either for analysis or design of two-way slabs with different size of openings and different openings. 4. The charts can be used either for analysis or design of two-way slabs with different size of openings and at different locations. 5. A comparative study of ultimate strength of these cases is presented with the help of charts. (Appendix-B) 6. Few numerical examples are presented based on theorem of VI and VII of affine theorem of Johansen 1 for orthotropic slabs with two openings parallel to long side and short side. 7. The strength of slabs with openings parallel to short side is less when compared to other slabs with openings parallel to Long side 8. Casting and testing of one third scale models is done and observed that the strength obtained theoreticallyand experimentally is same 9. The details of testing and failure mechanisms of slab after test are photographed and given in Appendix-B. APPENDIX - A Charts Legend: Affine Coefficients Designers Choice Coefficients Reverse of Designers Choice Coefficients For Slabs with Openings Parallel to Long Side α=0.15,β=0.1,r 5 =0.2,r 6 =0.2 Chart-1 : α=0.15,β=0.1,r 5 =0.2,r 6 =0.2 α=0.15,β=0.1, r 5 =0.2, r 6 =0.2 Chart-2 : α=0.15,β=0.1,r 5 =0.2,r 6 =0.2

6 Volume 6, No. 2, February α=0.1,β=0.15, r 5 =0.2, r 6 =0.2 Aspect ratio,r α=0.1, β=0.1, r 5 =0., r 6 =0.2 Chart-3 : α=0.1,β=0.15,r 5 =0.2,r 6 =0.2 Chart-7 : α=0.1,β=0.1,r 5 =0.,r 6 =0.2 α=0.15,β=0.15, r 5 =0.2, r 6 =0.2 Aspect ratio,r α=0.1, β=0.15, r 5 =0., r 6 =0.2 Chart-4 : α=0.15,β=0.15,r 5 =0.2,r 6 =0.2 Chart-8 : α=0.1,β=0.15,r 5 =0.,r 6 =0.2 α=0.1, β=0.2, r 5 =0.2, r 6 =0.2 α=0.1, β=0.2,r 5 =0.,r 6 =0.2 Chart-5 : α=0.1,β=0.2,r 5 =0.2,r 6 =0.2 Chart-9 : α=0.1,β=0.2,r 5 =0.,r 6 =0.2 α=0.15, β=0.2, r 5 =0.2, r 6 =0.2 α=0.1, β=0.1, r 5 =0.2, r 6 =0. Chart-6 : α=0.15,β=0.2,r 5 =0.2,r 6 =0.2 Chart-10 : α=0.1,β=0.1,r 5 =0.2,r 6 =0.

7 Volume 6, No. 2, February α=0.15,β=0.1,r 5 =0.2,r 6 =0. Chart-11 : α=0.15,β=0.1,r 5 =0.2,r 6 =0. α=0.1,β=0.15,r 5 =0.2,r 6 =0. Chart-12 : α=0.1,β=0.15,r 5 =0.2,r 6 =0. α=0.15,β=0.15,r 5 =0.2,r 6 =0. Chart-13 : α=0.15,β=0.15,r 5 =0.2,r 6 =0. α=0.1,β=0.1,r 5 =0.,r 6 =0. Chart-14 : α=0.1,β=0.1,r 5 =0.,r 6 =0. For Slabs with Openings Parallel to Short Side α=0.1,β=0.15,r 5 =0.2,r 6 =0.2 Chart-15 : α=0.1,β=0.15,r 5 =0.2,r 6 =0.2 α=0.15,β=0.15,r 5 =0.2,r 6 =0.2 Chart-16 : α=0.15,β=0.15,r 5 =0.2,r 6 =0.2 α=0.2,β=0.15,r 5 =0.2,r 6 =0.2 Chart-17 : α=0.2,β=0.15,r 5 =0.2,r 6 =0.2 α=0.1,β=0.1,r 5 =0.2,r 6 =0.2 Chart-18 : α=0.1,β=0.1,r 5 =0.2,r 6 =0.2

8 Volume 6, No. 2, February α=0.15,β=0.1,r 5 =0.2,r 6 =0.2 Aspect ratio,r α=0.1,β=0.1,r 5 =0.,r 6 =0.2 Chart-19 : α=0.15,β=0.1,r 5 =0.2,r 6 =0.2 Chart-23 : α=0.1,β=0.1,r 5 =0.,r 6 =0.2 α=0.2,β=0.1,r 5 =0.2,r 6 =0.2 α=0.15,β=0.1,r 5 =0.,r 6 =0.2 Aspect ratio,r Chart- : α=0.2,β=0.1,r 5 =0.2,r 6 =0.2 Chart-24 : α=0.15,β=0.1,r 5 =0.,r 6 =0.2 α=0.1,β=0.15,r 5 =0.,r 6 =0.2 α=0.1,β=0.1,r 5 =0.2,r 6 =0. Chart-21 : α=0.1,β=0.15,r 5 =0.,r 6 =0.2 Chart- : α=0.1,β=0.1,r 5 =0.2,r 6 =0. α=0.15,β=0.15,r 5 =0.,r 6 =0.2 α=0.15,β=0.1,r 5 =0.2, r 6 =0. Chart-22: α=0.15,β=0.15,r 5 =0.,r 6 =0.2 Chart-26 : α=0.15,β=0.1,r 5 =0.2,r 6 =0.

Volume 6, No. 2, February 17 17 α=0.1,β=0.1,r 5 =0.2,r 6 =0.2, K=4.0 µ=1,ls µ=1,ss 1 1.5 2 Chart -27 : α=0.1,β=0.1,r 5 =0.2,r 6 =0.2 APPENDIX-B α=0.

5 2 µ=1,ls µ=1,ss Chart -29 : α=0.15,β=0.1,r 5 =0.2,r 6 =0. α=0.1,β=0.15,r 5 =0.,r 6 =0.2, K=4 µ=1,ls µ=1,ss 1 1.5 2 Chart - : α=0.1,β=0.15,r 5 =0.,r 6 =0.2 PLATE 2: Inflatable Balloon which will Apply Pressure on Slab

9 Volume 6, No. 2, February α=0.1,β=0.1,r 5 =0.2,r 6 =0.2, K=4.0 µ=1,ls µ=1,ss Chart -27 : α=0.1,β=0.1,r 5 =0.2,r 6 =0.2 APPENDIX-B α=0.15,β=0.1,r 5 =0.2,r 6 =0.2, K=4.0 µ=1,ls µ=1,ss Chart -28 : α=0.15,β=0.1,r 5 =0.2,r 6 =0.2 α=0.15,β=0.1,r 5 =0.2,r 6 =0., K=4 PLATE 1: Box Setup made by Welding Mild Steel Channels and Angles and Bottom Part Covered with Wooden Plank µ=1,ls µ=1,ss Chart -29 : α=0.15,β=0.1,r 5 =0.2,r 6 =0. α=0.1,β=0.15,r 5 =0.,r 6 =0.2, K=4 µ=1,ls µ=1,ss Chart - : α=0.1,β=0.15,r 5 =0.,r 6 =0.2 PLATE 2: Inflatable Balloon which will Apply Pressure on Slab

Volume 6, No. 2, February 17 18 PLATE 5: Failure Mechanism at a recorded load of 78.

Per Edge Condition) List of Symbols CS A slab supported on all sides continuously,(restrained) D Overall depth of the slab DC Designer s Choice Coefficients I 1, I 2, I 3 Negative moment Coefficients

Negative ultimate yield moment per unit length provided by top tension reinforcement bars placed parallel to x- axis.

10 Volume 6, No. 2, February PLATE 5: Failure Mechanism at a recorded load of 78.1 kn of CS slab with Long side openings PLATE 3 Inflatable Balloon which Will Apply Pressure on Slab is arranged inside the box setup PLATE 4 Slab Mounted on Setup and Fixed with Bolt and Nuts (As Per Edge Condition) List of Symbols CS A slab supported on all sides continuously,(restrained) D Overall depth of the slab DC Designer s Choice Coefficients I 1, I 2, I 3 Negative moment Coefficients in their and I 4, I 1 m ult, I 3 m ult I 2 m ult, I 4 m ult K ' xm ult K ' ym ult corresponding directions. Negative ultimate yield moment per unit length provided by top tension reinforcement bars placed parallel to x- axis. Negative ultimate yield moment per unit length provided by top tension reinforcement bars placed parallel to y- axis. Positive ultimate yield moment per unit length provided by bottom tension bars placed parallel to x-axis Positive ultimate yield moment per unit length provided by bottom tension bars placed parallel to y-axis L x, L y Slab dimensions in X and Y directions respectively RDC Reversal of Designer s Choice VWM W ult d f ck Coefficients Virtual Work Method. Ultimate uniformly distributed load per unit area of slab. Effective depth of slab in mm. Characteristic Compressive Strength of concrete, N/mm 2 f y Yield Strength of steel, N/mm 2 m ult Ultimate Yield moment per unit length of the slab r Aspect ratio of slab defined by L x /L y. r 5, r 6 Coefficients of Distances of openings from supports

11 Volume 6, No. 2, February UDL 2 m ult α, β K REFERENCES uniformly distributed load Often called as strength of slab at ultimate stage Coefficients of openings in the slab Sum of orthogonal moment coefficients in x and y directions i.e K ' x+ K ' y+i 1 (or I 3 )+ I 2 (or I 4 ) µ Coefficient of orthotropy= [K ' x+i 1 (or I 3 )]/[ K ' y + I 2 (or I 4 )] Negative yield line Positive yield line Continuous edge [1] K.W. Johansen, Yield-Line Theory (Cement and Concrete Association, London, 1962), p. 181 [2] K.W. Johansen, Yield-Line Formulae for Slabs (Cement and Concrete Association, London, 1972) [3] ACI 421.3R-15, Guide to design of reinforced twoway slab systems. American Concrete Institute, p. 11 [4] British Standards Institution, BS : The structural use of concrete part 1, Code of practice for design and construction. BSI (1997) [5] IS 6:00, Indian standard plain and reinforced concrete-code of practice, BIS New Delhi, India [6] R. Park, W.L. Gamble, Reinforced Concrete Slabs (Wiley, New York, 00) [7] S. Islam, R. Park, Yield line analysis of two-way reinforced concrete slabs with openings. J. Struct. Eng. 49(6), (1971) [8] A. Zaslavasky, Yield line analysis of rectangular slabs with central opening. Proc. ACI 64, (1967) [9] K. Rambabu, Yield line analysis of orthotropic rectangular two- way slab supported on four sides with openings. Ph.D. Thesis in Struct. Engg., Vol. 2, Andhra University, India (01) [10] K. Rambabu, H.B. Goli, Y.D. Shaik, Application of affine theorem to orthotropic rectangular reinforced concrete slab with unequal corner opening. J. Struct. Eng. 36(3), (09). (India, 13) [11] Sudhakar,K.J., and andgoli, H.B., Detailing of regular and irregular shape of RC slabs by yield line method, National Seminar on Detailing of RC and Steel Structures, Madras, rd Dec [12] M. Ravindra, K. Rambabu, H.B. Goli, Application of the affine theorem to an orthotropic rectangular reinforced concrete slab continuous over two adjacent sides and simply supported on other two sides having a short side opening. J. Struct. Eng. 41(2), (09). (India, 14) [13] Manasa, K. Rambabu and K. DurgaRani, Application of modified affine theorem to orthotropic rectangular reinforced concrete slab with interior corner opening parallel to shorter side, Journal of Structural Engineer, SERC,Vol. No., Issue No.4. pp [14] 14. C.E. Reynolds, J.C. Steedman, Reinforced Concrete Designers Hand Book (Taylor and Francis, London, 08). (pp , 137 1)

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