Accordingly in this section, only triangular load is considered, but if the slab had been designed for surpressure, trapezoidal load should be used for the wall design. per ft. No surpressure on the liquid is considered in computing this moment and, therefore, it must also be disregarded in the design of the wall. Balance of 41,000 lb. depth and 12-ft. diameter. Wall with Moment Applied at Base FIG. Wall (a) Fixed end moments (b) Final moments FIG. Use RI = 17.5 ft. and c,2R1 = 0.125 in all subsequent calculations for moments in the slab. Hence, the centre of the prestressing steel (CGS) is . PAGE 19 tank wall. The wall thickness is 14 in. Assume that the temperature is T, in the inner face, T2 in the outer face, and that the temperature decreases uniformly from inner to outer face, T, - Tz being denoted as T. Fig. 43 troweled surface covered with mastic. 35, it will be assumed that a fixed edge with radius R 1* can be substituted for the six intermediate columns. per ft. is applied. in each 5 - of two curtains from the 0.6H point to bottom of wall (see Section 8). per ft. I the outslde 0.411 1 0.5H 0.3H 1 0.6H +.0072 +.0064 +.0040 +.0016 +.0006 +.0002 -.0002 Table IX sign indicates tension in the +.0078 +.0068 +.0054 +.0043 +.0032 +.0026 +.0022 +.0045 +.0039 +.0033 +.0029 Table X Moments in cylindrical wall Rectangular load Fixed base, free top Mom. , per f t . As the concrete tends to expand a prestress is induced in the members. The cross-sectional area of the wall can then be reduced from 1.25 X 20 = 25.0 sq.ft. For illustration: 14,400 X 0.6 = 8,600 ft.lb. For ring tension, multiply coefficients from Table V by VR/H = 2,ooO X 27/20 = 2,700 lb. The presence of the steel bar prevents some of the shortening of the concrete, so the difference in length of the block in Fig. (AS = 1.00) in top of slab and outside of wall at corner. Prestressed concrete circular storage tanks Aman Bagrecha Ppt rcc road design Anjani Shukla FinalReport Taylor Wilson Speedy construction somitra bhardwaj Seismic response of steel beams coupling concrete walls Yahya Ali 357502268-C-clamp-ppt.pptx RajeshJavali2 Pre stress concrete Surendra Gurjar 3.Unbounded post- post-tensioned concrete. 0.811 0.911 1 l.OH - 0 ; T o t a l mom. Now calculate the stresses. due to water A-7 B. Positive +.0094 i.0076 +.0057 Moments in cylindrical wall Shear per ft., V, applied at top Fixed base, free top Mom. 1 0 0 1 / 0 PAGE 9 Section 9. per ft. to, say, zero is much smaller than rotations that may occur when normal settlement takes place in the subgrade. due to water Table A-1 T force due to water pressure B. O.C. Table A-19 Mr due to distributed M Total radial Moment (t.m/m) Radial moment per segment 0.15 R -0.1089 -34.419 -1.594 -9.166 -43.585 -6.54 0.20 R -0.0521 -16.467 -0.93 -5.348 -21.814 -4.36 0.25 R -0.02 -6.321 -0.545 -3.134 -9.455 -2.36 0.30 R 0.0002 0.063 -0.28 -1.610 -1.547 -0.46 0.40 R 0.022 6.953 0.078 0.449 7.402 2.96 0.50 R 0.0293 9.261 0.323 1.857 11.118 5.56 0.60 R 0.0269 8.502 0.51 2.933 11.435 6.86 0.70 R 0.0169 5.341 0.663 3.812 9.154 6.41 0.80 R 0.0006 0.190 0.79 4.543 4.732 3.79 0.90 R -0.0216 -6.827 0.90 5.175 -1.652 -1.49 1.0 R -0.049 -15.487 1.00 5.750 -9.737 -9.74 Tank Base Slab c 1.5 0.15 D 10 Tangential Bending Moment in Base Slab From Table A-17 From Table A-19 T=Coef. When the base is sliding, the displacement is the largest possible, but the reaction is zero. M o m e n t along entire circumference of column capital = -23,200 X r X 4.5 = -328,000 ft.lb. Total moment = 0.0082 X 600 X 352 X 2~ X 26.3 = l,OOO,OOO ft.lb. Design is restricted in principle only by material properties and . 35 shows a layout with seven columns arranged as in a three-way flat slab. 30 AB, which represents the stressless condition due to ;I uniform temperature change throughout, will move to a new position AB. The maximum area of vertical steel in the outside curtain is A,=!$ 5.20 = 0.28 sq.in. F. and f = 4.5T = 4.5 X 75 = 375 p.s.i The stress off = 375 p.s.i. A X 232 X 525 - 298,600 2r X 23 X 12 = 7% 228 in. For all intermediate displacements, the reaction must be between 0 and 3,700 lb. 29. In the rest of the slab, all the way out to the wall, use fourteen x-in. 0 +0.234 = -TI -= __~ _0.2H 0.3H +0.120 to.215 to.101 10.190 0.4H / 0.5H ) 0.6H T to.160 10.130 10.096 +0.254 +0.234 +0.209 +0.180 +0.142 10.268 10.268 +0.266 t0.250 +0.226 to.185 +0.251 +0.273 +0.285 +0.285 10.274 +0.232 to.134 +0.075 +0.02: to.172 10.104 +0.031 to.262 +0.157 10.334 +0.210 +0.203 +0.267 to.322 +0.357 +0.362 10.330 +0.067 10.164 +0.256 +0.339 to.403 +0.429 +0.409 +0.025 +0.137 +0.245 +0.346 +0.428 +0.477 +0.469 +0.018 +0.119 +0.234 +0.344 -to.441 +0.504 +0.514 ~0.011 +0.104 to.219 to.335 to.443 10.534 +0.575 0.0 - 0 . per ft., occurs at rhe edge of the drop panel. is assumed to be distributed uniformly over the subgrade, the upward reaction on the bottom slah will he 2,300,ooO P = n- X 292 = 870 lb. Alternate bars may stop at mid-height, but the other bars extend to top of the wall to serve as support for ring bars during erection. Two types of joints at top of wall are shown in Fig. INFORMATION The final ring Tension are obtained by summing 1 and 2 Wall with Moment Applied at the Top Calculation of Bending moment in the wall 1. However, these tables may be used with good degree of accuracy also when the far end is hinged or fixed. per ft. As = 1.7/1.44 X 7 = 0.17. spans Alternate bars bent 8.9 = 0.77 sq.in. In contrast to this term, "linear prestressing" is used to refer to any other type of prestressing in which a cable is straight or bent but not twisted in a circle around a circular structure. 40 top of the footing is given a trowel finish and then covered with mastic. 16 bt. Inserting As = T/f, in Equation 1 gives CEs -I- fs fc = A,f, + - iT x For illustration, use the data given in Section 4, assume T = 24,100 lb., and compute fc for values off, as given below. t.m/m M=Coef. Prestressing the dome edge keeps the dome in compression, enabling the use of large free . per ft. (M at base). Circular prestressing is also applied in domes and shells. An illustration follows of a typical office procedure in designing a tank similar to that in Fig. 7 and 8 show that assuming the base hinged gives conservative although not wasteful design, and this assumption is therefore recommended. The total moment at this point is M = 14,400 X 27r X 0.6 X 27 = 1,465,OOO ft.lb. per ft. A, = 5.9/1.44 X 7 = 0.59 sq.in. 30. The C/S is more efficiently utilized when compared with a RC section 2. The discussion in Sections 4 through 15 is given in connection with numerical examples most of which apply to a tank having the same dimensions. = 5.96 ft. = 298,600 - 525 X ?r X 5.962 = 7 2 p.s.i. Calculate the ring tension caused by applied moment at the top of the wall using Table A-11 3. The prestressing is done by wires or tendons placed spirally, or over sectors of the circumference of the member. The notch shown in the inside of the joint is optional but appears to be desirable. In this video Modelling and Analysis of Circular Prestressing is shown in midas FEA Software. 31 which is to be placed with a 3-in. Total tang. You might get some help from www.HelpWriting.net Success and best regards! \Vhen the artificial restraint is removed, the rotation of the joint will induce new moments in wall and slab. 23. The induced moments both have positive signs. The economy of PSC is . 28 shows one arrangement with eight radial bars in each quadrant. i 0.5011 Air / to.450 I +0.394 +0.323 +0.236 +0.130 moments, -0.211 -0.233 -0.251 --0.261 -0.259 ___.~ 0.6OR 0.70R to.596 +0.558 +0.510 +0.451 +0.392 ~. A continual spray of water from perforated pipes suspended around the rim of the tank wall makes for excel lent curing, and so does a complete curtain of hurlap kept soaking wet by spraying from a hose. per sq.in. It is difficult to predict the behavior of the subgrade and its effect upon the restraint at the base, but it is more reasonable to assume that the base is hinged than fixed, and the hinged-base assumption gives a safer design. In this case, the taper may be ignored, but under extreme circumstances it may be advisable to take it into account. The sums of the induced moments and the original fixed end moments are the final moments. The design of these structures requires that attention be given not only to strength requirements, but to serviceability requirements as well. It is not considered necessary to use the hinged joint when the wall is supported on an ordinary wall footing since such a footing can transmit little moment to the subgrade. ve moment M u 11.43 t .m / m at 0.6 R d 35 5 0.9 29.1 cm 0.85(300) 2.61(10) 5 (11.43) 1 1 0.00367 min 2 4200 100(29.1) (300) A s (0.00367)(100)(29.1) 10.68 cm 2 / m A s total 2 (0.6 5.0)(10.68) 201.31 cm 2 Tank Base Slab Slab Reinforcement b) Radial Moments At 0.15 R It is reasonable to use a 25% reduction to the theoretical moment at the column capital M u 43.58(0.75) 32.69 t .m / m d 50 5 0.9 44.1 cm 0.85(300) 2.61(10)5 (32.69) 1 1 0.00466 min 2 4200 100(44.1) (300) A s (0.00461)(100)(44.1) 20.29 cm 2 / m A s total 2 (0.15 5.0)(20.29) 95.6 cm 2 Use 32 20 mm @ 12.5 cm Tank Base Slab Slab Reinforcement b) Radial Moments At 0.2 R M u 21.82(0.75) 16.37 t .m / m d 50 5 0.9 44.1 cm 0.85(300) 2.61(10)5 (16.379) 1 1 0.0023 min 2 4200 100(44.1) (300) A s (0.0023)(100)(44.1) 10.14 cm 2 / m A s total 2 (0.2 5.0)(10.14) 63.73 cm 2 Tank Base Slab Slab Reinforcement b) Radial Moments At 0.3 R M u 1.55 t .m / m d 35 5 0.9 29.1 cm 0.85(300) 2.61(10)5 (1.55) 1 1 0.00048 min 2 4200 100(29.1) (300) A s (0.0018)(100)(35) 6.3 cm 2 / m A s total 2 (0.3 5.0)(6.3) 95.4 cm 2 Design of Circular Concrete Tanks Circular Plate Reinforcement Design of Circular Concrete Tanks Radial Reinforcement Design of Circular Concrete Tanks Radial Reinforcement . *(9.725)(5) 2 (1.3) t/m T=Coef. In the top slab where the load is p = 650, the fixed end moment per ft. from Section 12 is -23,200 ft.lb. Design of Circular Concrete Tanks Introduction The report by ACI Committee 350 entitled Environmental Engineering Concrete Structures is essential in understanding the design of tanks. Prestressing compensates for the tensile stresses introduced when the element is loaded. Pre-stressed Concrete By: "The Slackers" Danny Efland Maria Cuellar Joel Irvine. For illustration assuming the shrinkage coefficient, C, of concrete as 0.0003: t _ o.ooo3 x 30 x 106 + 14,ooo - 1 0 x 3 0 0 X T 12 x 300 x 14,000 = 9,ooo + 14,ooo - 3,ooo x T = 0.0004T 50,400,OOO It is felt that an allowable concrete tensile stress for cylindrical tank design of 300 p.s.i. 9 represents merely the loading conditions considered in this section; the effect of a roof slab is treared in subsequent sections. per sq.ft. Allowable steel stress for ring tension is often kept as low as 12,000 or even 10,000 p.s.i. The fixed end moment at base of wall is determined for the triangular loading in Section 4 with coefficients selected from Table VII for H2,Dt = 6. Details Reference has been made in previous sections to sliding, hinged, or continuous joints at base of wall. 28 may be discontinued in accordance with the steel requirements represented by the dash line in Fig. 0 3 1 9 X 6 0 0 X 17.52 = - 5 , 9 0 0 ft.lb. The inside corner of the wall is beveled to minimize the danger of spalling. M Pu 1.007(9.725)(5) 2 9.29 5.75 285.9 ton 1.3 Shear Strength of Base Slab: a) At edge of wall: V u P R 2 column load (9.725)(3.14)(5) 2 285.9 477.9 ton Length of shear section D 3.14(10 100) 3141.59 cm d 35 5 0.9 29.1 cm V c 0.85(0.53)( 300)(3141.59)(29.1)(10) 3 713.34 ton 477.9 O.K. One moment is due to the outward pressure of the liquid, the other due to the upward reaction from the subgrade. 11 2 +0.006 1 + 0 . CORE - Aggregating the world's open access research papers to 25.0 - 0.5 X 0.58 X 10 = 22.1 sq.ft. The tangential moments are computed by selecting coefficients for c/D = 0.15 from Tables XIII and XV, and multiplying them by pR2 = 474,000 ft.lb. mom.. pet seg. 3. Section 2. per ft. V 3,700 = 27 p.s.i. from the center. The column load computed in Section 12 is 508,ooO lb., so it will be assumed here that the column reaction on both top and bottom slab is 508,000 lb. per ft. The wall thickness should be sufficient to keep the concrete from cracking. Within the drop panel, the effective depth is 16.5 in. per ft. Compute ring tension and moment at intervals of 0.1 H (Schedule C). The effect of applying the shear at the top, therefore, is practically the same whether the base is fixed or hinged. Load on area within the section is 650 X P X 6.882 = 96,000 lb. ACI 350 recommends the allowable stress in hoop tension for Grade 60 (4200 Kg/cm2) reinforcing steel as is 1400 Kg/cm2 (fy/3). incl. 18, **See Bibliography, reference No. We've encountered a problem, please try again. Thus the concrete does not crack. Reduction of steel corrosion Increase in . Balance: 508,000 - 382,000 = 126,000 lb. 25 The application of data for design of roof slabs without interior support is illustrated for the tank sketched in Fig. per sq.ft. AI and Machine Learning Demystified by Carol Smith at Midwest UX 2017, Pew Research Center's Internet & American Life Project, Harry Surden - Artificial Intelligence and Law Overview, No public clipboards found for this slide. ( +18,9OO 1+22,100 1+25,100 [+?7,200 J (+27,OOO 1 t22.900 wHS = 62.5 X Multipliers of = 364,500 coefficients from PHJ = 432 X Table VIII = 140,000 1 Compute moments at intervals out height of wall (Schedule B). 0 4 9 0 + 0 . per ft of width \ I 4 E a r s B4 a r s Total : lG-%+~ 18'-3" FIG. 28 and that onlv four types of bent bars are required. * H3 = Coef. In the continuow base, vertical reinforcement extends across the joint which is to be prepared so as to develop maximum bond. The shear on this section is 221,000 - a X 3.252 X 612 = 221,000 - 20,000 = 201,ooO lb. By comparing the ring tension curve in Fig. 11 O.Offmm VIII 0 4 ii 1 O.lH ;- 0.2H , 0.3/f 1 0.4~ ( 0.5H 1 O&H Moments for both hinged and fixed base are plotted in Fig. VR/H -0.4=-8.22V10/5 V=0.02433 ton The change in ring tension is determined by multiplying coefficient taken from Table A-8 by VR/H=0.04866 Cover in Place Example 1 H2 52 5 Dt 20 0.25 Bottom edge hinged Point Ring T Coef. t.m/m Table A-2 Table A-11 Point M. Coef. Then cover the joint and keep it continually wet. 25 which carries a superimposed load of 500 lb. TU fy fs 50 150ton Tu 3.0 50 103 As f y 0.9 4200 39.7 cm 2 / m 19.8 cm 2 / m use 1016 mm at each side provided 20 cm 2 / m Example 1 Bottom edge Sliding Vertical Reinforcement Minimum ratio of vertical reinforcement ACI section (14.3) is taken 0.0012 for deformed bar 16 mm in diameter or less. *(1)(5)(10) t/m M=Coef. By definition, a strand is generally a cable made up of 7 individual steel wires: 6 wires coiled in a long pitch around a central wire. To make certain that the continuous base is watertight, a dam may be placed as shown. 16. 30. per ft. and M = 2,000 ft.lb., respect;vel y (Schedule F). 21,100 + 37,800 Ac + nAs = 1 5 X 1 2 + 1 0 X 2.34 = 289 p.s.i. 3, November, 1933, pages 598-600. The radius of the smallest ring bar may be 1 ft. Use 18-in. Oil Storage Tanks of Unusual Design by Warren Travcll, Cmrme, Vol. Prestressed concrete pressure pipes are covered by BS EN 639 and BS EN 642. You can read the details below. The rotation required to reduce the fixed base moment from 6,700 ft.lb. Data are presented in Tables XIII, XIV and XV for slabs with center support for the following ratios ofcapital to wall diameter: c/D = 0.05,0.10,0.15,0.20, and 0.25. 0.290 Without center Coef. The wires or tendons lay outside the concrete core. 22, June, 1927, pages 386389. per sq.ft. 12. The prestressing is done by wires or tendons placed spirally, or over sectors of the circumference of the member. *(6.37)(5)/(5)2 *(1.65/1.3) t/m = 1.617 *Coef t/m Point Ring T Coef. The ring tension and the moments determined in this section are now added to those in Section 6. F. while the temperature of the outside air To = 30 deg. The inaccuracy may be rectified to some extent by using a relatively low value for EC, such as EC = 1,500,OOO p.s.i., which is used in this section. It reduces the shrinkage stresses, increases the concrete strength and improves the watertightness. 13 with Fig. = 83 = 0.68 sq.in. Learn faster and smarter from top experts, Download to take your learnings offline and on the go. 10. round bars. Tank Base Slab Shear Strength of Base Slab: c) Shear at edge of drop panel: Radius of critical section = 125 + (35 -5 -0.9) = 154.1 cm d 35 5 0.9 29.1 cm Vu (9.725)(3.14)(1.541)2 285.9 213.36 ton Vc 0.85(0.53)( 300)(2 154.1)(29.1)(10)3 219.85 ton 213.36 O.K. per ft. Tang. Table A-17 Mr due to water pressure Mr Coef. Roof Slab with Four Interior Supports M0 ' W, the total load on a panel, equals pL2 = 612 X 192 = 221,000 lb. per ft. (for fixed edge), and by M = 3,200 ft.lb. more than in Sections 4 and 3. 37 shows a segment of a tank wall in two positions, one before and one after a uniform increase in temperature. Looks like youve clipped this slide to already. Reinforcement Minimum concrete cover for reinforcement in the tank wall should be at least 5cm. per Ft. With a 9-in. Table A-19 Mr due to distributed M Total radial Moment (t.m/m) 0.15 R -0.0218 -6.890 -0.319 -1.834 -8.72 0.20 R -0.0284 -8.976 -0.472 -2.714 -11.69 0.25 R -0.0243 -7.680 -0.463 -2.662 -10.34 0.30 R -0.0177 -5.594 -0.404 -2.323 -7.92 0.40 R -0.0051 -1.612 -0.251 -1.443 -3.06 0.50 R 0.0031 0.980 -0.1 -0.575 0.40 0.60 R 0.008 2.529 0.035 0.201 2.73 0.70 R 0.0086 2.718 0.157 0.903 3.62 0.80 R 0.0057 1.802 0.263 1.512 3.31 0.90 R -0.0006 -0.190 0.363 2.087 1.90 1.0 R -0.0098 -3.097 0.451 2.593 -0.50 Tank Base Slab Column Load From Table A-13, load on center support of circular slab is: P coef . per ft. of periphery. Introduction 32. A design based on assumption of fixed base and f, = 12,000 p.s.i. T in tons Check adequacy of wall thickness for resisting moment: fr My I 9.74 t /2 105 (1.7)(1.3) (100)t / 12 t 27.63 cm increase wall thickness at the base to 50 cm using a 25 x 25 cm haunch 2 f c' 2 300 Tank Cylindrical Wall Shear force at the base of the wall, From Table A-12: V u coef . In these types of beams, the prestress beam to act as the formwork to the cast-in-situ concrete. Ring tension and moment are plotted in Fig. 10 and the moment curve with Fig. In the bottom slab, where p = 870, this negative moment will be increased to: -23 2oo x (870 + 650) = -27,100 ft.lb. indicates tension Coefficients at point' 0 . Modification according to ACI 350-06 Permissible Stresses Flexural stress Normal environmental exposures Modification according to ACI 350-06 Permissible Stresses Flexural stress Severe environmental exposures f s ,max 260 s 2 4 2 d b 2 2 17ksi( 120Mpa) for one way members 20ksi( 140Mpa) for two way members. per cu.ft. / A.- ~Reinforcementl I 0.50.1fo 0.2O.lfo O.l5.~Io 0.15.110 = 134 = 54 = 40 = 40 Description 1-1 12.0 8.0 8.0 8.0 7.8 4.7 3.5 3.5 26 16 18 18 >6-in. Fixed end moment including surpressure = -0.049 (400 + 125 - 432) X 232 = -2,400 ft.lb. Design of Circular Concrete Tanks ACI 350R-01 Report This report presents recommendations for structural design, materials, and construction of concrete tanks, reservoirs, and other structures commonly used in water containment, industrial and domestic water, and wastewater treatment works, where dense, impermeable concrete with high resistance to chemical attack is required. Design of Circular Concrete Tanks Load Combination Loading Conditions The tank may also be subjected to uplift forces from hydrostatic pressure at the bottom when empty. 5.Design agencies and regulations. 20 0 3MB Read more . Now customize the name of a clipboard to store your clips. 9.29 X 3,200 Total column load = 508,000 lb. 21, in which the wall is made continuous with a reinforced bottom slab designed for uplift. The assumption of fixed base is not generally in conformity with the actual conditions of restraint, and for other conditions there are little or no published data for designers to use. In this type circular prestressing is directly provided on the concrete core pipe. Shear, bond bearing,camber & deflection in prestressed concrete, Pre stressed concrete- modular construction technology, Prestressed concrete structures and its applications By Mukesh Singh Ghuraiya, DEVELOPMENT OF MODERN PRESTRESSED CONCRETE BRIDGES IN JAPAN, Prestressing Concept, Materilas and Prestressing System, Irresistible content for immovable prospects, How To Build Amazing Products Through Customer Feedback. But surpressure acts on both top and bottom and creates moments in slabs and walls. The amount of reinforcement provided must be sufficient for strength and serviceability including temperature and shrinkage effects. : 13,960 564 Variable thickness (3.5 to 14 in. Inside diameter is used for all calculations here. When the base . per ft. 1g2 ft.lb. per ft. For the whole circumference, the total negative moment is 22,000 X 2~ X 27 = 3,730,OOO ft.lb. The tendency is for the wall load to be distributed to the subgrade near the wall so that the soil reaction is maximum at the wall. Mom. X wItR lb. ring ten. 3. Mom. The slab and the footing are made flush on top in order to keep the wall as low as possible, to avoid waste of concrete in the floor slab and to reduce the length of joint. Laps may be spliced in accordance with code requirements, or the joints may be welded. per ft. When displacement is prevented, the top cannot expand and the ring tension is zero at Point O.OH. The maximum tensile stress in the concrete including effect of shrinkage is: A*=ax.= FIG. 1.44 x 13 per ft. Use x-in. The circumferential prestressing resists the hoop tension generated due to the internal pressure. A continuous concrete sill block is indicated for all intermediate joints. clearance above the surface of the subgrade. 38. In this video results interpretation, od Circular Prestressing is shown in midas FEA Software. Presentation Transcript. The excess amount of ring tension is shown cross-hatched in Fig. Moment applied at an edge is positive when it causes outward rotation at that edge. clarification: the term circular prestressing refers to prestressing in round members, such as tanks and pipes and the members are prestressed in a circular way while the use of bar tendons with threaded anchorages reduces the possibility of pull and are used in post tensioning system, wire tendons are mainly used in post tensioning system, 19. It is seen from the dash line in Fig. The dimension marked t corresponds to the wall thickness. 543 = 35.9 sq.in. But the relatively small values near the base in Fig. Bending Reinforcement: Inside Reinforcement 0.85(300) 2.61(10)5 (9.74) 1 1 0.0074 min 2 4200 100(19.2) (300) A s 0.0074 100 19.2 14.27 cm 2 / m Use 8 16 mm/m on the inside of the wall. Tap here to review the details. F. The area of horizontal steel at the colder face computed as for a cracked section is M 9t2T As = md= sd sq.in. The top slab is as sketched in Fig. As a result, the equation f = YzETe is to be regarded as merely indicative rather than formally correct. is the thickness required for maximum ring tension which occurs at a depth of approximately 0.6H below the top. 1.44 x 10.5 The point of inflection from Table XIII is at a distance from the center of 0.28R1; therefore, with the bars arranged as in Fig. a) Anchorages. = coef. The most convenient way to do this is to add 0.125 to all the coefficients in Table XII, both for radial and tangential moments, and then to multiply the modified coefficients by pR2. As computed in the example, a temperature differential of 75 deg. 284 All of these bars may be discontinued at a distance from inside of wall equal to 0.17 X 27 i- 12 diameters = 4.6 i- 1.0 = 5.6 ft., say, 5 ft. 7 in. Instant access to millions of ebooks, audiobooks, magazines, podcasts and more. By mastic is meant tar, asphalt or synthetic material placed in accordance with manufacturers direction and known to be resistant to the liquid stored. Small Rrinforcrd Conmtc T&J by J. E. Kirkham, Oklahoma Engineering Experiment Station Bulletin 32, Stillwatcr, Okla., 1937, 22 pages. If this is done, the calculations rhat follow will be simplified. Since 300 p.s.i. ActualIy, the condition at the base may be somewhere between hinged and freely sliding, so it is inadvisable to design the ring bars below Point 0.6H Point 1 O.OR 1 O.lH 1 0.211 +32,700! I 11M o m . 1 0.711 / 0.8~ Section 6. Section 11. the coefficient from Table XII at Point l.OOR: -0.125 X pR2 = -0.125 X 625 X 132 = --13,200 ft.lb. With this additional pressure, estimate t = 13 in., which is 5 in. 0 2 4 3 - 0 . 164 +0.138 +0.119 +O.lOO +0.081 +0.067 +0.056 +0.041 +0.086 +0.063 to.047 +0.036 co.021 0.5H - at point* 0.711 1 0.8H 0.9H / 1.OH +0.354 co.220 +0.159 to.125 +0.103 +0.402 +0.224 +0.145 +0.105 +0.080 ;0.535LG& +0.066 co.043 +0.028 co.01 8 +0.007 +0.044 +0.025 +0.013 +0.006 0.000 -0.010 -0.024 -0.010 -0.019 -0.007 -0.011 -0.005 -0.006 -0.002 -0.001 +0.030 +0.006 +0.004 -0.019 to.012 to.007 to.004 +0.001 *When this table is used for shear applied at the base, while the top is fixed, O.OH is the bottom of the wall and l.OH is the top. The prestressing of concrete by using high tensile steel improve the efficiency of the materials. 22, May, 1927, pages 313-318. Table Xl ~/ Moments in cylindrical wall Moment per ft., M, applied at base Hinged base, free top M o m . H E (0.25)3 k 1.010 0.00315625E 5 From Table A-16, the stiffness of the base slab c 1 .5 0.15 D 10 Et 3 k coef . 0 +1.002 L 0.4 0.8 1.2 1.6 2.0 3.0 4.0 5.0 6.0 8.0 - 1.57 - 3.09 - 3.95 - 4.57 - 5.12 I - 6.32 - 7.34 - 8.22 - 9.02 -10.42 Positive sign - 1.32 2.55 3.17 3.54 3.83 - 4.37 4.73 4.99 5.17 5.36 / 0.4H to.311 co.345 10.386 co.433 +0.480 +0.821 +0.154 +0.166 to.198 +0.224 +0.251 +0.310 +0.356 +0.394 +0.427 +0.486 Moment per ft., M, applied at base Hinged base, free top T = coef. unit. at mid-height to 8 in, at top. For this wall, Grays data show that maximum ring tension is approximately 8 per cent greater for triangular than for rectangular wall section, that is, when the sectional area is reduced from 25.0 to 12.5 sq.ft. =0.875bd=0.875X rX2X27X 12X 10.5 Due to increase in moment and shear, it may be advisable to deepen the bottom slab at the wall from 12 in. round bars spaced 11 in. In case the dam should deteriorate to the point where the joint leaks, the notch can easily be talked and the joint made watertight again. A dam shot&~ be specified for all horizontal joints. The base joint is not in equilibrium and when the artificial restraint is removed, it will rotate. AI and Machine Learning Demystified by Carol Smith at Midwest UX 2017, Pew Research Center's Internet & American Life Project, Harry Surden - Artificial Intelligence and Law Overview, No public clipboards found for this slide. 1886 Jackson, P. H., (USA) It is difficult to ascertainwhether the footing is capable of providing a 3,700-lb. The value of sh for plain concrete ranges from 0.0003 to 0.0008. round bars in each band. per Ft. = coef. This rectifies several deficiencies of concrete. The wires or the tendons lay outside the concrete centre. per ft. 1.44 x 10.5 Use seven l-in. per ft. in which t = thickness of wall in in. PR 2 , coef . Circular prestressing is also applied in domes, shells and folded plates. American Standards AWWA C301-92 and C304-92 also cover large diameters. Wall with Moment Applied at Top I + FIG. 0.875 X 2n X 71.5 X 8.5 Shear at edge of wall = = 574,400 lb. in inside and outside curtains. per ft.** A, = 11.0/1.44 X 7 = 1.09 sq.in. Consequently, the investigation made in this section may be omitted in most cases with exception of tanks in which the ring tension is relatively large at the top and the wall is doweled to the roof slab. Sixteen bars, 18 ft. 3 in. Maximum positive radial moment at mid-span. *This moment is not changed by filling the tank since the weight of the liquid is transmitted directly to the subgrade without creating any moments. Concentric-Ring Concrete Tank for Topeka, Enginrrting Rmrd, Vol. -- , placement is zero a .D 8,. Weve updated our privacy policy so that we are compliant with changing global privacy regulations and to provide you with insight into the limited ways in which we use your data. Actual ) 0.6H - 400/- 9001- ) 0.5H tlool- ( 0.6H 1 0.711 ) 0.8H 1 0.9H 1 l.OIf 4001+ 1,100 )+ 4,400 I+ 9,400 / +15,900 I+22,000( Ring tension and moments both for fixed base and for actual base condition are plotted in Fig. 1 0.9H i l.OIf Wall with Hinged Base and Free Top-Trapezoidal load 2.50 = 0.22 sq.in. For each effect, the benefits are listed. It is the temperature differential only, T, which creates stresses. Note that the equation applies to untracked sections only, and that this procedure of stress calculation is to be considered merely as a method by which the problem can be approached. As discussed in Section 12, reduce this moment 28 per cent which gives 0.72 X (-328,ooO) = -236,000 ft.lb. 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