Spherical-dome-20151.pdf

March 26, 2018 | Author: Chandan Narang | Category: Bending, Beam (Structure), Dome, Stress (Mechanics), Chemical Product Engineering


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The Islamic University of GazaDepartment of Civil Engineering ENGC 6353 Design of Spherical Shells (Domes) Shell Structure A thin shell is defined as a shell with a relatively small thickness, compared with its other dimensions. Shell Structure Four commonly occurring Shell Types: Barrel Vault Dome Folded Plate Hyperbolic Paraboloid (Hypar) we have to investigate some important notions of structural design. .What is a shell structure? To answer this question. is some bending also present. the top elements of the beam are compressed and the bottom is extended: The elements through the thickness of the development of internal tension and the arch are being compressed compression is necessary to resist the approximately equally. . Note that there applied vertical loading. Two-dimensional structures: beams and arches A beam responds to loading by bending An arch responds to loading by compressing. bending introduces tension into the convex side of the bent plate. by analogy to the beam. Moreover. Plate Bending A plate responds to transverse loads by bending This is a fundamentally inefficient use of material. . thickness. Plate bending vs. spanning 100 feet. membrane stresses Note: this is an experiment you can try yourself by folding a sheet of paper into a box. This slide shows a concrete plate of 6” If the plate is shaped into a box. The box structure is much stronger and stiffer! . resisting then each of the sides of the box its own weight by plate bending resists bending by the development of membrane stresses. . Note that lines on the shell retain approximately their original shape. Domes A shell is shaped so that it will develop membrane stresses in response to loads The half-dome shell responds to transverse loads by development of membrane forces. ) . Domes The primary response of a dome to loading is development of membrane compressive stresses along the meridians. The dome also develops compressive or tensile membrane stresses along lines of latitude. by analogy to the arch.) Circumferential Hoop Stress (tens. Circumferential Hoop Stress Meridional Compressive Stress (comp. These are known as ‘hoop stresses’ and are tensile at the base and compressive higher up in the dome. A dome that is a segment of a sphere not including latitudes less than 50 does not develop significant hoop tension.’ These are also known as ‘hoop stresses’ . In this figure. intensifying as the colors move towards the red. The colors beyond blue represent circumferential tensile stresses. below about 50 ‘north latitude. the blue color represents zones of compressive stress only. The half-dome shell does develop membrane tensile stresses. developing compressive membrane forces near the crown of the arch. it is an arch developing compressive membrane forces that are transferred to the base of the arch When unsupported along its length. . Barrel Vaults A barrel vault functions two ways compression Arch (compression) tension In the transverse direction. and tensile membrane forces at the base. it is more like a beam. It can be supported on continuous walls along the length. as in this example. with compression on the top and tension on the bottom in the long direction. or at the corners. it functions as an arch across the width. . Barrel Vaults A barrel vault is a simple extension of an arch shape along the width. This form is susceptible to distortion. If supported on the corners. and as a beam. . continued As with any arch. some form of lateral restraint is required-- this figure shows the influence of restraining the base of the arch--the structure is still subject to transverse bending stresses resulting from the distortion of the arch.Barrel Vault. the shell develops membrane tension at the bottom and compression at the top. Folded Plates Folded plate structures were widely favored for their simplicity of forming. in analogy to a beam in bending . Perpendicular to the main span. the shell acts as short span plates in transverse bending In the main span direction. and the variety of forms that were available. and see if you can tell what key element is missing from the folded plate shown. It is missing transverse diaphragms. What’s wrong with this Folded Plate Structure? Compare to the discussion of barrel vaults. . especially at the ends. .This animation shows the effect of adding a diaphragm at the two ends and at midspan. The folded plate shell distorts much less. Meridional stresses along the direction of the meridians 2. Thin Shell Structures Two type of stresses are produced: 1. Hoop stresses along the latitudes Bending stresses are negligible. but become significant when the rise of the dome is very small (if the rise is less than the about1/8 the base diameter the shell is considered as a shallow shell) . 2. Thin Shell Structures Assumption of Analysis 1. Deflection under load are small. Points on the normal to the middle surface deformation will remain on the normal after deformation 3. Shear stresses normal to the middle surface can be neglected . Spherical Shells Internal Forces due to dead load w/m3 Consider the equilibrium of a ring enclosed between two Horizontal section AB and CD  The weight of the ring ABCD itself acting vertically downward  The meridional thrust N per unit length acting tangentially at B  The reaction thrust N +d N per unit unit length at point D E N A F  B C D N+dN  r a H d N a   . E N A r’ F  B C D N+dN  r a H d N Meridional Force N  a   Surface area of shell AEB A  2 a  EF EF  a 1  cos   W  w D  A  D  2  a  EF W  w D  2  a 2 (1  cos  ) N  (2 r ') sin   w D  2  a 2 (1  cos  ) r '  a sin  w  a (1  cos  ) w D  a (1  cos  ) w Da +ve compression N  D   -ve tension sin 2  (1  cos  )(1  cos  ) 1  cos  . N B Hoope force = N  ad  d The horizontal component of N  is N  cos D W N  causes hoop tension N  a cos sin  similarly N+dN The horizontal component of N  +dN  is  N  +dN   cos   d   N  +dN  causes hoop tension N  +dN   cos   d   a cos   d   . E N Spherical Shells A F  B C D N+dN  r a H d N a   Hoop Force N  The difference between the N  and N   dN  which respectively acts at angles  and   with the horizontal give rise to the hoop force. E N Spherical Shells A F B C D N+dN  r a H d N a   When increasein  is small dN  tends to be zero N N  ad   d  N  a cos sin   B d where w  a (1  cos  ) D W N  D sin 2  N+dN  1   cos   cos   1  2 N  w D a  cos .  w D a    1  cos    1  cos   . Spherical Shells HoopForce N   1  N   w D a  cos .   1  cos   wa At crown   0 N   (compression ) 2 At base   90 N   wa (tension ) when N   0   51o 49' for   51o 49' N  will be compressive for   51o 49' N  will be tensile . Summary In order to make the – Negative sign for compression and +Positive sign for tension for Meridional and Hoop forces The previous equations can be rewritten as follows: w Da N   1  cos   1  N  w D a   cos   1  cos   . Spherical Shells . Spherical Shells  Ring Force H  cos  H  N  cos   w D a 1  cos  at   51o 49'  N   0 & H is maximum H max  0.382 w D a . 3535 w L a . Spherical Shells Internal forces due to Live load (wL/m2)horizontal Meridional Force T W  w L   r 2  w L   a 2 sin 2 y  a (1  cos  ) r  a sin  N   2 a sin   sin   w L   a 2 sin 2 w La N   2 Hoop Force N wL a N    cos 2  2 Ring Tension cos  H  N  cos   w L a 2 at   45o  N   0 & H is maximum H max  0. Spherical Shells . 9f y T  H r Vertical Load Vertical Uniform load (w V )  N  sin   o . Ring beam design Design of the Circular Beam Horizontal Load T Ultimate Load As  0.w 2 r Span length  l   # of supports P  2  r wV M max ve  C 1  P r (see the tables of circular beams) M max ve  C 2  P r (see the tables of circular beams) . This change can be done by a transition curve. it is recommended to increase the thickness of the shell in the region of the transition curve. . bending moments will be developed due to the big difference between the high tensile stress in the foot ring and compressive stresses in the adjacent zones It is recommended to use transition curves at the edge and to increase the thickness of the shell at the transition curve. Edge Forces In flat spherical domes. In order to decrease the stress due to the forces at the foot ring. Bending moments can avoided if the shape of meridian is changed in a convenient manner. which when well chosen gives a relief to the stress at the foot ring. Edge Forces In flat spherical domes. bending moments will be developed due to the big difference between the high tensile stress in the foot ring and compressive stresses in the adjacent zones It is recommended to use transition curves at the edge and to increase the thickness of the shell at the transition curve. . In case of hemispherical domes. meridian stresses have a large horizontal component which is taken care of by providing a ring beam there. no ring beam are required since the meridional thrust is vertical at free end .Ring Beam At the free edge of the dome. This ring beam is subjected to hoop tension. If all the meridional lines are led to the crown. To overcome this problem. there will be a lot of congestion of bars and their proper anchorage may be difficult.Reinforcement Steel is generally placed at the center of the thickness of the dome along the meridians and latitudes. Area enclosed by this small circle at the top is reinforced by a separate mesh. . small circle is left at the crown and all the meridional steel bars are stopped at this circle. Assume the following loading: Covering material = 50 kg/m2 and LL= 100 kg/m2  Use ' f c  300 kg / cm 2 and f y  4200 kg / cm 2 y=1. Example: Design of a spherical dome  Design a spherical shell roof for a circular tank 12m in diameter as shown in the figure.4m r=6m a . 42 y=1.896 tan   0.23 cos   0.25 t/m2 Covering materials = 0.5)= 0.4  r=6m 6 sin =  0.493 a Loading on roof Assume shell thickness = 10 cm Own weight = 0.442 13.4m Radius of the Shell a    13.56   26.1 t/m2 Note: the live load is considered as loading per surface area .56m 2y 2 1.05 t/m2 LL= 0. a 2  r 2  a  y  2 a 2  r 2  a 2  y 2  2ay r 2  y 2 62  1.1(2. 2(0.6(0.645 ton/m’ Outward horizontal force =1.35cm 2  f y 0.Design of Ring Beam: Wu= 1.1)=0.4)(0.56)(1.52)=62 ton Vertical Load per meter of cylindrical wall =62/(2*6)=1.52 t/m2 Total load on roof = 2ayWu=2(13.645/tan=3.337 t/m’ Ring tension in beam T  H  r  3.2+0.05)+1.337  5  20 tons T 20*1000 As    5.9  4200  use 8  10 mm . 0018 A s  0. ratio = 0.8 cm 2 use 5  8 mm/m .56  N    3.52*13.0018(10)(100)  1.896 0.56 at   0 N   u    3.52 13. Design of the Shell Meridian Force Meridian force per unit length of circumference Wua N   1  cos  W a 0.896 Use minimum reinf.72 t / m ' (compression) 1  0.52 t / m ' (compression) 2 2 at foot cos   0. 5  8 mm/m .Ring (Hoop) Force  1  N  w u r   cos   1  cos   wr 0.52(13.56) At crown   0 N     3.0018(10)(100)  1.8 cm 2 use minmum reinf.52 t / m ' (compression) 2 2 At foot cos   0.59 t / m ' (compression) A s  0.896 N   2. 6 13.85 cm W u x 2 0.Bending Moment Assume that the thickness at the foot = 15 cm x  0.6 at  0.85  300   1  2.15   0. 5  8 mm/m .188 t / m 2 2 d  15  3  12cm 0.0003   min 100 12   300   2 4200   use minmum reinf.188   1    0.52  0.85  2 Fixing moment M    0.56  0.61106  0. Example: Design of a spherical Dome Reinforcement details . Spherical Shells under General Loading Internal Forces Due to Others Loading .
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