Simple Design Charts for Helicoidal Stair Slabs with Intermediate LandingsZ Wadud, Non-member Prof S Ahmad, Non-member Helicoidal staircases are becoming popular among architects because of their grand appearances. While design charts for simple helicoidal stair slabs are available in design handbooks, no such design aid is available for fixed ended helicoidal slabs with intermediate landings. This paper attempts to propose such design charts. The theoretical background arises from Solankis (1986) work, which is based on the strain energy method. Because of the symmetric nature of the slab, four of the six stress resultants at mid-span become zero. Thus, if the two redundants at mid-span can be suggested by a design chart, rest of the analysis becomes a statically determinate problem. Based on the analytical work, charts have been proposed to determine these mid-span redundants. Different geometric parameters have been considered and nondimensional parameters have been used to derive the charts. It is expected that use of the design charts will ease the design engineers workload. Keywords: Design charts; Helicoidal stairs NOTATIONS b : width of the stair slab EI F GJ H Ht h Ih K M Mh M sup Mv N Ri Ro : : : : : : : : : : : : : : : flexural rigidity about horizontal axis through midpoint radial horizontal shear force torsional rigidity radial horizontal (redundant) height of the helicoid waist thickness of stair slab moment of inertia about a vertical axis through mid-point ratio of flexural to torsional rigidity vertical moment at mid-span (redundant) lateral moment vertical moment at support vertical moment thrust centreline radius on horizontal projection inner radius on horizontal projection outer radius on horizontal projection shear force at mid-span R1 T V w : : : : radius of centreline of load torsion lateral shear force dead load and live load per unit length of span, measured along the longitudinal centreline of the plan projection slope of the helix contained within the helicoid at radius R total central angle subtended on horizontal projection angular distance from mid-span on a horizontal plane angle measured from bottom support towards top on a horizontal plane angle subtended at the centre by half landing α 2β θ : : : : : θ′ φ R, R2 : INTRODUCTION One of the most important functional elements of a building, be it residential or commercial, high or low rise, is the stair. Depending on the architectural forms, there may be different types of stairs, such as: simple straight stair, dog-legged stair, saw-tooth/slabless stair, free standing stair, and helicoidal stair. Among these, the helicoidal stair has a grand appearance and is increasingly getting popular among architects. However, due to the complex geometrical configuration, the analysis and design of helicoidal stair slabs are more difficult than simple type of stairs. The degree of difficulty is further enhanced for helicoidal stairs with an intermediate landing. While design charts for helicoidal stairs are available in literature1,2 no such aids are found for helicoidal stairs with 269 Z Wadud and Prof S Ahmad are with Department of Civil Engineering, Bangladesh University of Engineering and Technology, Dhaka 1000, Bangladesh. This paper was received on July 7, 2004. Written discussion on this paper will be entertained till April 30, 2005. Vol 85, February 2005 For the helicoidal stair slabs without landings3-6. (e) lateral shear force (V ). (c) torsion (T ). The live load could be uniformly distributed over the surface. also has six stress resultants at any cross section. The slab fixed at both ends is six degree indeterminate. GEOMETRY OF THE HELICOIDAL SLAB WITH INTERMEDIATE LANDING The geometry of a circular helicoidal stair slab with an intermediate landing at mid-span can be defined as (Figure 1). α ′ = tan − 1 Ht R 2β STRESS RESULTANTS Six stress resultants are available at any section of a space structure. However. This necessitates the development of a simple design chart and design procedure for this type of elements. Here. and (f) radial horizontal shear force (F ). partially fixed or hinged. The positive directions of these stress resultants have been illustrated in Figure 2. the helicoid is reduced to its elastic line having the same stiffness as that of the original structure. the helical girder solution fail to take into account the 3-d characteristics of helicoid and its inherent structural efficiency. they treated the structure as a linearly elastic member in space defined by its longitudinal centroidal axis. in this work the live load is considered uniformly distributed over the entire surface on the horizontal projection of the stair. But this simplification neglects the slab action of helicoid and also assumes that the bending stiffness and torsional stiffness of a warped girder are the same as those of a straight beam. Arya and Prakash7 attempted to analyse the case of the helicoidal stairs with intermediate landing. IE (I) JournalCV (1) The coordinates at the mid-surface can be expressed as: x = R cos θ ′ y = R sin θ ′ z = θ ′R tan α ′ z = βR tan α ′ 0 ≤ θ′ ≤ β β ≤ θ′ ≤ β + 2φ (2) (3) (4) (5) z = θ ′ − 2φ R tan α ′ b g β + 2φ ≤ θ′ ≤ 2 β + φ b g (6) LOADING AND BOUNDARY CONDITION The helicoidal stair slab has its self weight. which are (a) vertical moment (Mv ). Helicoidal slab. (b) lateral moment (Mh ). being a space structure. the slab is subjected to live load. 270 . there are six equilibrium equations and twelve unknown reactions. The introduction of an intermediate landing further complicates the situation. (d) thrust (N ). Influence lines were drawn at various cross sections for all the six stress resultants found at such sections for unit vertical load and unit moment about the axis of the structure. Like Scordeilis. They used flexibility approach to analyse internal forces due to dead and live loads in fixed ended circular stairs having an intermediate landing. REVIEW OF PAST WORKS The complex geometry of a regular helicoidal stair slab has made its analysis quite difficult. considered the helicoid as a three dimensional (3-d) helical girder. Evidently. These design charts now stand as helical girder solution for helicoidal stairs. line loads or symmetrical loads about the central axis of the slab. Santathadaporn and Cusens1 presented 36 design charts for helical stairs with a wide range of geometric parameters. point loads.−θ +θ Landing Centre-line of Loads Centre-line of Steps M ν φ β+φ Vertical moment β Radial horizontal shear Torsion Lateral moment θ′ Bottom w R2 R1 Top Thrust Figure 2 Stress resultants in a helicoidal stair slab Lateral shear Figure 1 Plan of a helicoidal stair slab with intermediate landinc intermediate landing. This dead load (self weight) is assumed to be uniformly distributed. In addition. Based on this work. The ends of the slab may be fixed. four design charts were compiled in a modified form in the design handbook by Reynolds and Steedman2. Helical slab with one end fixed and one end hinged is indeterminate to third degree. The stair slab here is considered fixed at its ends in all directions. Other redundants at mid-span have zero values because of symmetry of geometry and loads. are neglected. Stress Resultants in the Helicoidal Stair with Intermediate Landings The analysis method followed here begins from Solankis Approach. the simultaneous solution of which gives the values of the redundants. From this analysis they suggested some generalized behavior of helicoidal star slabs with intermediate landings. four of the six redundants at the mid-span of a helicoidal stair slab with intermediate landing become zero. February 2005 . 271 Vol 85. Because of symmetry in geometry and loading. Solankis findings were similar to those observed by Arya and Prakash.Critical positions of loads were determined to obtain the maximum values of the internal forces. being small in comparison to the deformations caused by twisting and bending moments. Solanki assumed that at the landing level the bending and torsional moments along the upper half of the stair slab to be: l Vertical Moment: M v = M cos θ − 2 wR1 b g (10) l Lateral moment: M h = M sin θ sin α − HR2 θ tan α cos θ sin α 2 − HR2 sin θ cos α + wR1 sin θ − wR1 Rθ sin α e j (11) l Torsion: 2 T = ( M sin θ − HR2 θ tan α cos θ + wR1 sin θ + HR2 sin θ sin α − wR1 R2 θ )cos α + HR2 sin θ sin α (12) l Thrust: N = − H sin θ cos α − wR1θ sin α (13) l Lateral Shear: V = w θ cos α − H sin θ sin α l Radial horizontal shear: F = H cos θ (14) (15) where the radius of centreline of load is b1 − cos θg (7) R1 = 3 2 Ro − Ri3 2 3 Ro − Ri2 F GH I JK (16) l Lateral Moment: M h = − HR2 sin θ (8) l Torsion: 2 T = M sin θ + wR1 sin θ − wR1 R2 θ (9) These equations are valid when the angle subtended by the landing at the centre ( φ ) is small as compared to angle subtended at the centre by the whole stair plus the landing. (iii) The angle subtended at centre by the landing is small compared to the total angle subtended by the stair at the centre. Simply supported or pin jointed helicoids could also be analysed using this method. Only two. l Vertical moment: 2 M v = M cos θ + HR2 θ tan α sin θ − wR1 1 − cos θ (ii) The cross section is symmetric about the two principal axes of the section. Solanki8 analysed the problem of intermediate landing using energy method to find the two unknown redundants at the mid-span section. However. both Arya and Prakash. as well as Solanki did not attempt to develop any simple design methods for helicoidal stair slabs with intermediate landings. ANALYSIS Assumptions To facilitate the analysis procedure. M M C C T H R2 β+φ +θ H Mh o Mv A Bottom Plan B Top Figure 3 Stress resultants and mid-span redundants For the flight the moments and other forces were assumed to follow Morgans5 derivations for slabs without landing. (iv) The moment of inertia of the helicoidal slab section with respect to a horizontal radial axis is negligible as compared to the moment of inertia with respect to the axis perpendicular to it. the assumptions made during the analysis were: (i) Deformation due to shear and direct forces. with the slope of the helix ( α ) put to zero. vertical moment M and radial horizontal shear H remain to be calculated (Figures 1 and 3). He proposed two equations. It should also be noted that the equations for the landing are nothing but the expressions for the flight. 1. k2.05. the first partial derivative of the strain energy with respect to any particular force is equal to the displacement of the point of application of that force in the direction of its line of action.01. However. Because of symmetry in loading and geometry. the derivation of equations (22) and (23) requires tedious mathematical computations. k2. l l l l Total central angle subtended by the stair (range 135° to 360° ). landing angle = 20 ° (20) and ∂U =0 ∂H 2 α = 20 (21) k2 1 k1.The Strain Energy Method The widely used Castiglianos Second theorem states that in any structure. Ratio of radius of the centre line of load to the mean radius of the stair. landing angle = 60 ° IE (I) JournalCV . and δ the deflection in the direction of force. Figures 4 to 6 present a few of the charts. k3 k2 25 30 35 40 The Strain Energy Method Applied to the Helicoidal Stair Slabs The strain energy method has previously been successfully employed by Morgan and Holmes to analyse helicodial stair slabs. This is why. (17) where U is the strain energy. R1 / R2 (1. k3 for R1 / R2 = 1. The strain energy stored by the bending moment is given by: M2 dL U =∑ 2 EI z (18) The vertical moment at the support often becomes the most critical design force. degree Figure 4 Coefficients k1. ∂U =δ ∂P resultants at any section of the helicoidal stair slab. k2. That is. k3 for R1 / R2 = 1. the partial derivatives of the strain energy function with respect to the vertical moment (M ) and radial horizontal force (H ) is equal to zero.01.10). α = 20 And that by the twisting moment is: T2 dL U =∑ 2 GJ z 2 (19) 1 k1. in a helicodial stair slab with a landing at the middle. degree Figure 5 Coefficients k1. according to the Castiglianos second theorem. These charts provide the values of k1 and k2 for a wide range of parameters and are available in Wadud9.05 k3 (22) and H = 2 k2 wR2 1 (23) Landing angle = 60° 150 175 200 225 250 275 300 325 350 Details of the analysis is given in Appendix A. the slope at the mid-span is zero and so is the horizontal deflection. Slope of the tangent helix centre line with respect to the horizontal plane.05 and 1. α ( 20 ° to 40 ° ). Mathematically expressing. ∂U =0 ∂M 0 k1 R1/R2 = 1. P the force. To facilitate the design procedure a series of 21 design charts have been proposed. the material of which is elastic and follows Hookes law and in which the temperature is constant and the supports are unyielding. equations (7) to (15) can be used to determine the six stress 272 Total angle subtended at the centre. k3 25 30 35 40 Solution of these equations yield the values of M and H. k2. SUGGESTION FOR A CHART Once M and H are determined from equations (22) and (23).01 k3 1 Landing angle = 20° 150 175 200 225 250 275 300 325 350 Total angle subtended at the centre. Total angle subtended by the landing (10 ° to 70 ° ). Another factor k3 has been introduced in order to expedite the design process. The strain energy due to shear stress and axial force is neglected. which can be expressed in the form of: M= 2 k1 wR2 0 k1 R1/R2 = 1. because they are small. the basic framework of the analysis is derived from Solankis work. pp 249-256. k2. S Santathadaporn and A R Cusens. Proceedings. pp 127-132. 1986. no 2. A Simple Design Approach for Helicodial Stair Slabs. A C Scordelis. December 1960. E and F N Spon. For a helicoidal stair slab with intermediate landing. (London). there was some discrepancy in his results and therefore the analysis has been carried out independently beginning from the strain energy principle. C E Reynolds and J C Steedman. k3 for R1 / R2 = 1. vol 56. no 6. January 1964. Helicoidal Staircases with Intermediate Landing. Bangladesh University of Engineering and Technology. degree Figure 6 Coefficients k1. REFERENCES 1. February 2005 273 . APPENDIX A: THE ANALYSIS As explained before.10. American Concrete Institute. pp 85-101. Department of Civil Engineering. vol 55. AMC Holmes. A R Cusens and S Trirojna. 8. It is expected that the use of the charts would expedite the design process. A step-by-step procedure for analysis using the charts is given in Appendix. Analysis of Helical Beams Under Symmetrical Loading. no 7. 1988. Dhaka. landing angle = 40 ° 125 100 75 Force values. However. Analysis of Helicoidal Staircases with Intermediate Landing in Analysis of Structural System for Torsion. degree Figure 7 Variation of forces along the span for the design example thrust lateral shear radial horizontal shear 5. 7. k3 0 k1 R1/R2 = 1. ACI Journal. M Sc Engineering Thesis. Part 2. Z Wadud. H T Solanki. Concrete and Construction Engineering. July 1959. but k2 does. (London). CONCLUSION A simple design chart has been proposed for use in design of helicoidal stair slabs with intermediate landings. 9. wR2 z 0 2 φ M2 φ T2 β M2 Mv h ds + v ds ds + ds + 2 EI 0 2 EI h 0 2 GJ 0 2 EI h z z z (35) + β z 0 2 EI 2 Mh ds + β z 0 T2 ds 2 GJ (24) Vol 85. Charts for the Design of Helical Stairs with Fixed Supports. no 3. k2. 2. SP 35. MI. 1973. Concrete and Construction Engineering. Reinforced Concrete Designers Handbook. Design of Helicoidal Staircases-2 Statically Indeterminate Cases. V A Morgan. pp 133-140. no 1. vol 61. ASCE. is given by. London. 10. vol 3. Structural Engineering Practice.10 k3 1 Landing angle = 40° 150 175 200 225 250 275 300 325 350 Total angle subtended at the centre. Helicoidal Staircase Study.2 α = 20 k2 25 30 35 40 The values of k3 are also presented in the proposed design charts. 4. kN-m 100 50 0 50 100 150 200 125 100 75 50 25 0 25 50 vertical moment lateral moment torsion 75 100 125 Angular distance from mid-span. The chart also covers the helicoidal slabs without landing as a central landing angle of 0 ° is equivalent to no landing. J S Cohen. the strain energy function U. Concrete and Construction Engineering. February 1966. pp 1491-1502. Paper No ST 1437. kN 50 25 0 25 50 75 100 125 125 100 75 50 25 0 25 50 75 100 125 Angular distance from mid-span. Proceedings. Tenth Edition. Comparison of Analysis of Helical Staircases. Proceedings. It has been found that k1 and k3 do not vary much on the stair inclination. 6. 200 150 Moment values. A S Arya and A Prakash. vol 54. pp 1437-11437-37. 3. November 1957. degree Figure 8 Variation of moments along the span for the design example U= φ where 2 M sup = k3 . ACI Journal. pp 46-54. Closure to Discussion of Internal Forces in Uniformly Loaded Helicoidal Girder. 1 k1. March 1960. Detroit. Figures 7 and 8 depict the variation of the stress resultants for a prototype stair (Appendix). Having fixed the geometric parameters. the total angle ( θ f ) through which the stair is to rotate to reach its height. B3 . outer radius (or alternatively mean radius and width of stair).where for the landing and (25) 2 2 MR2 sec α G1 + HR2 tan α sec α A − wR1 R2 sec α ( H1 − G1 ) ds = R2 dθ and. A4 . the torsional rigidity can be taken as: A = ( 1 / 8 ) sin 2 β − ( 1 / 4 ) β cos 2 β 2 EII h GJ = I + Ih Then 1 EI 1 I = +1 ≈ 2 GJ 2 I h B = β3 / 6 − ( β2 / 4 − 1 / 8) sin 2 β − ( 1 / 4 ) β cos 2 β (29) C = sin β − β cos β C ′ = sin φ − φ cos φ LM NM OP QP (30) D = ( 1 / 2 ) β − ( 1 / 4 )sin 2 β D ′ = (1 / 2 ) φ − (1 / 4 )sin 2 φ Equation (21) can be rewritten with the help of equations (25). a designer then has to − 1 1 sin α ( D − A) + R2 sin α (C − F ) / R1 ] = 0 2 2 2 ⇒ A1 M + A2 HR2 = A3wR1 2 ⇒ A1 M + A2 HR2 = A4wR2 (33) 274 IE (I) JournalCV . ie. I ≈0 Ih 2 ⇒ B1M + B2 HR2 = B3wR1 2 ⇒ B1M + B2 HR2 = B4wR2 (34) (28) where As per Solanki. A3 . mid-span redundant moment and radial horizontal force: (32) Equations (29) and (30) expands to: 2 3 2 2 MR2 tan α sec α A + HR2 tan2 sec α B − wR1 R2 tan α sec α ( C − A) + 1 2 3 [ MR2 sin α ( D − A) + HR2 sin2 α sec α ( D − 2 A + E ) 2 2 2 3 + wR1 R2 sin α ( D − A) − wR1 R2 sin α ( C − F )] = 0 M= A4 B2 − A2 B4 2 2 wR2 = k1 wR2 A1 B2 − A2 B1 A4 B1 − A1 B4 wR2 = k2 wR2 A2 B1 − A1 B2 (22) ⇒ M [tan α sec α A + + 1 sin α ( D − A)] + HR2 [tan2 α sec α B 2 H= (23) 1 2 sin2 α sec α ( D − 2 A + E )] − wR1 [tan α sec α ( C − A) 2 APPENDIX B: THE DESIGN PROCESS B-1 Analysis The height of the stair. B2 . their values being evident from equations (33) and (34). (26) and (28) as: β 1φ 1β ∂U φ ∂M v ∂T ∂M v ∂T ds + ds + M v ds + ds = 0 T T = Mv 2 0 2 0 ∂H 0 ∂H ∂H ∂H ∂H 0 E = β3 / 6 + ( β2 / 4 − 1 / 8 ) sin 2 β + ( 1 / 4 ) β cos 2 β F = 2 β cos β + ( β2 − 2 )sin β G1 = ( 1 / 2 ) β + ( 1 / 4 )sin 2 β G ′ = ( 1 / 2 ) φ + ( 1 / 4 )sin 2 φ z z z z (31) Similarly equation (20) stands as: β 1φ 1β ∂U φ ∂M v ∂T ∂M v ∂T ds + ds + M v ds + ds = 0 T T = Mv 2 0 2 0 ∂H 0 ∂M ∂M ∂M ∂M 0 z z z z H 1 = sinβ H ′ = sinφ A1 . B4 are constants. The simultaneous solution of equations (33) and (34) yields the values of M and H. are generally suggested by the architect. the length of landing (L) etc. for the flight ds = R2 sec α dθ The partial derivative of the strain energy function with respect to H is. the moment of inertia with respect to a vertical axis Ih is much greater than the moment of inertia about the horizontal axis I. B1 . inner radius. The ratio I/Ih can therefore be neglected. A2 . φ 2M φ 2T ∂T ∂U φ 2 M v ∂M v h ∂M h d s + ds + ds = ∂H 0 2 EI ∂H 0 2 EI h ∂H 0 2 GJ ∂H (26) 1 2 2 2 + [ MR2 cos α D + HR2 sin α ( D − A) + wR1 R2 cos α D − wR1 R2 cos α C ] 2 1 2 2 2 + [ MR2 G ′ − wR1 R2 ( H ′ − G ′ )] + [ MR2 D ′ + wR1 R2 D ′ − wR1 R2 C ′] = 0 2 ⇒ M [sec α G1 + 1 1 1 cos α D + G ′ + D ′ ] + HR2 [tan α sec α A + sin α ( D − A)] 2 2 2 z z z + β 2M v z 0 2 EI β 2M β 2T ∂T ∂M v h ∂M h d s + ds + ds 2 EI h ∂H ∂H 0 0 2 GJ ∂H z z 1 1 2 − wR1 [sec α ( H 1 − G1 ) − cos α D + R2 cos α C / R1 + H ′ − G ′ 2 2 (27) − 1 1 D ′ + R2 C ′ / R1 ] = 0 2 2 Because the stair width is large as compared to its thickness. total dead and live load per unit length along the centreline With the values of R1/R2 and central angle subtended by the landing ( 2φ ) go to the appropriate chart. found using the previously stated equations. The length of landing at the inner edge is 1.477 = 1. inner radius of 1. Determine mean radius R2 from given inner radius Ri and outer radius Ro 8.05 Step 6: Total thickness in the vertical direction is approximately 0.524 × 2 ) = 0. from the mid-span toward the top support using equations (10)-(16). Determine the radius of centre line of loading (R1) from 2 3.406 kN-m (3 250 lb-ft) k2 = 1.433 − 1. The stair is to reach its full height within a 270° turn.73 ⇒ Msup = 0.81 m (12.5 ft). H.6 / ( 1.477 × 3. Determine midspan moment.477 = 81. Concrete unit weight = 23. Find the total angle subtended at the centre by the flights ( 2β ) from θ f and 2φ R2 = ( 1.241 m.35 kN-m ( 65 880 lb-ft) Step 8: The variation of stress resultants along the span. for a 270° stair. The analysis procedure using the charts consists of the following steps: 1. and outer radius of 3.563 kN/m3 (150 pcf).563 + 4.524 2 R1 = 6.665 radian α = tan − 1 3.95 × 2. 7.43) / 2 = 2. refering to Figure 5. find k1 .43 1. Determine other stress resultants at various distance. Determine the angle ( 2φ ) subtended by the landing of length L at a distance R at the centre L = 2φR 3.788 = 10. Msup .88 kN (18400 lb) k3 = 0. θ .467 × (3.036 ⇒ M = 0.95 × 2.6 m (5.524) = 19.241 × 23. is depicted through Figures 7 and 8.432 − 1.152 m (6 inches) high.25 ft). α and β . k1 = 0. February 2005 275 .467 kN/m2 w = 10.788 kN/m2 (100 psf).524 3 = 2.657 × 19. and total angle subtended by flight at the centre ( 2β ) calculate the slope of the tangent to the helix centreline ( α ) as α = tan − 1 Ht R2 2β 2 β = 270 − 2 × 30 = 210° = 3.036 × 19. The stair slab is 0.95 kN/m Step 7: With the calculated values of R1/R2 .152 m (6 inches) thick and the risers are 0. Keep in mind that α = 0 at the landing. Surface UDL = 0.81 = 22. from 2 M = k1 wR2 H = k2 wR2 2 M sup = k3 wR2 Vol 85. 3 2 Ro − Ri3 2 3 Ro − Ri2 R1 / R2 = 2.477m φ = 1.95 × 2. mid-span radial horizontal shear. Find w. Step 1: Step 2: Step 3: Step 4: R2 = ( Ro + Ri ) / 2 2.8° 2. Live load = 4.657 ⇒ H = 1.525 radian = 30° 2β = θ f − 2 φ 4.determine the stress resultants. M. mean radius (R2 ). B-2 Example It is required to analyse a reinforced concrete helicodial stair slab with a height of 3.524 + 3. k2 .6 / 2.73 × 19. and support moment. and k3 for the given value of total angle subtended at the centre ( θ f ) . The introduction of the chart will substantially reduce the tedious computations required to find the design forces and moments.665 Step 5: R1 = 5.6 m 3 3.4772 = 4.524 m (5 ft). From the height of stair (Ht).4772 = 89.43 m (11.25 ft).