J50I ELEMENTARY Reinforced Concrete Building Design by LEONARD C. URQUHART Instructor of Bridge Engineering, Cornell University ELEMENTARY Reinforced Concrete Building Design by LEONARD C. URQUHART Instructor of Bridge Engineering, Cornell University CARPENTER & COMPANY ITHACA. NEW YORK 1915 v) COPYRIGHT 1915 CARPENTER & Co. INTRODUCTION The following pages are intended to furnish the computing side of an elementary course in reinforced concrete construction and to supplement a text book used in the class room. The references made to Diagrams and Tables are to those of Reinforced Concrete Construction, Volume I, by George A. Hool. It has been the intention to give a complete design for the roofs, floors, columns, and footings of a reinforced concrete building of the beam-and-girder type. Other details such as walls, windows, stairways, elevator shafts, etc., do not as a rule affect the main design, involve more practical considerations, and will not be taken up here. For an example of these details the student is referred to Volume II of the text book mentioned above. The design of the floors and columns is given in a very precise form, and although so much exactness is not required in ordinary building design, it has been the intention to familiarize the student with methods applicable to large and important members, and also to show methods of checking each portion of the design. There are many methods of form construction, but the one briefly given in the following design seems to be more general than many others. Here again more exactness will be found than is ordinarily required, but in buildings having many similar units such exactness would result in final economy. The design of a combined footing is given in order to furnish an outline for a problem of this type in the computing room. CORNELL UNIVERSITY, October 31, 1915. SPECIFICATIONS 1 . The concrete in this structure shall consist of a wet mixture of one part Portland cement to six parts of aggregate, fine and coarse, and shall be capable of developing an average compressive stress of 2000 pounds per square inch at twenty-eight days when tested in cylinders eight inches in diameter and sixteen inches long under laboratory conditions of manufacture. 2 . All reinforcement shall consist of medium open hearth steel having an elastic limit of not less than 50000 pounds per square inch, and shall be of plain round section. 3. All flexure formulae shall be based on the "straight line theory." 4. The span lengths used in computations shall be taken as the distance center to center of supports. 5. Slabs, beams, and girders shall be considered as fully con- tinuous, and bending moments shall be computed accordingly both for uniform and concentrated loads. 6. The negative moment which exists at the supports shall be taken equal to the maximum positive moment sustained by the slab, beam, or girder. Reinforcement shall be provided to resist this negative moment and all such reinforcement shall be continued beyond the support to the quarter point of the adjacent span, but in no case shall this distance be less than that required to develop the full strength of the rods at the allowable bond stress. 7. The lateral spacing of parallel rods shall not be less than two and one-half diameters center to center, nor shall the distance from the side of the beam or girder to the center of the nearest rod be less than two diameters. Layers of rods, in beams and girders ? shall have at least one-half inch vertically in the clear. In girders and columns the metal shall be protected by a mininium of two inches of concrete ; (i. e. from the surface of the rod to the surface of the concrete) in beams, by one and one-half inches; in floor slabs, by one-half inch of concrete. Main reinforcement in slabs, beams, and girders shall not be less than three-eighths of an inch nor more than one inch in diameter. Stirrups shall not be less than one quarter inch in diameter. 8 . In columns the concrete to a depth of one and one-half inches shall be considered as protective covering and not included in the effective section. 9. The minimum allowable thickness for slabs is four inches, and the main reinforcement when consisting of rods shall not be placed farther apart than two-and-one-half times the slab thick- ness. Rods placed perpendicular to the main reinforcement for the prevention of cracks, due to shrinkage, etc., shall be provided to an amount not less than one-fourth of one percent of the slab section. 10. Beams whose axes are perpendicular to the main reinforce- ment of the slab may be designed as Tee beams, provided the slab is poured with the beams and properly bonded to the same. Under these conditions the effective width of flange shall be taken so as not to exceed (a) one-fourth of the span length of the beam,(b) eight times the thickness of the slab plus the width of the stem. 11. When the maximum shearing stresses exceed the value allowed for the concrete alone, web reinforcement must be provided to aid in carrying the diagonal tension stresses. The horizontal reinforcement is to be bent up where possible to assist in taking the diagonal tension stresses, but this is not to be considered in determining the required amount of web reinforce- ment. In the calculations of web reinforcement the concrete may be counted upon as carrying one-third of the shear, and the remainder is to be provided for by means of metal reinforcement. The maximum longitudinal spacing of vertical stirrups shall not exceed three-fourths of the working depth of the beam. The effective length of stirrups in bond shall be taken as six-tenths of the working depth of the beam. 12. The ratio of length to diameter of concrete columns shall not exceed fifteen. Columns shall not have less than one per cent, nor more than four per cent, of longitudinal reinforcement, which shall be held together by bands or hoops, not less than one-fourth inch in diameter, spaced center to center not more than one-half of the diameter of the column. Not less than four rods shall be used for the reinforcement of any column, and rods shall not be less than one-half inch nor more than one inch in diameter. Bending stresses due to eccentric loads must be provided for by increasing the section until the maximum stress does not exceed the allowable value. 13 . The footings shall be so designed that they can be poured at one operation. The bottom of the footing shall not be nearer to the lowest layer of steel than four inches. 14. The allowable working stresses shall not exceed the follow- ing: Axial compression on columns reinforced with longitudinal steel and spirals or bands 500 pounds per square inch Concrete in compression in slabs, beams, or girders 650 pounds per square inch Concrete in compression over the support of continuous slabs, beams and girders 750 pounds per square inch Shearing strength of plain concrete as a measure of diagonal ten- sion 40 pounds per square inch Shearing strength of concrete with proper metal reinforcement 120 pounds per square inch Bond stress between concrete and plain reinforcing rods v 80 pounds per square inch Tensile stress in reinforcing steel .... 16000 pounds per square inch Compressive stress in reinforcing steel not more than fifteen times the working compressive stress in the concrete. Ratio of the modulus of elasticity of steel to that of concrete .... 15 The building whose partial design is given in the following pages has three stories above ground and a basement, and has the follow- ing dimensions: Width, center to center of wall columns 54 / o" Length, center to center of wall columns loo'o* Span of slabs 9' o" Span of beams 20' o* Span of girders 18' o* Distance floor to floor 12' o" It is designed for a live load on the floors of 100 pounds per square foot, a live load on the roof of 35 pounds per square foot, and the allowable soil pressure is taken as 4000 pounds per square foot. DESIGN OF THE FLOOR SLAB Assumed thickness from Table 6 4 inches Total load per square foot 150 pounds Bending moment for unit slab ^ x 1 50 x 9 2 x 1 2 = 1 2 1 50 in-lbs. From Table 2p= 0077 k = 368 j = : . .874 107.4 12150 . = 3.07 inches 107.4 x 12 Taking the value of d to the nearest quarter of an inch greater than the theoretical value determined, and adding either three quarters of one inch or one inch for insulation, so that the total thickness of the slab is determined to the nearest one-half inch, the actual value of d is 3>< inches and the total thickness of the slab 4 inches as assumed. 12150 . , - 267 square inches i6ooox .874x3.25 150x4.5 . e = o - Q - = .................... 2.97 inches Using | inch round rods s = x 12 = .... 4. 96 inches 1.178 or s = x 12 = ..... 4. 76 inches 2.97 Rods will be spaced ............ 4 J inches center to center. To provide for the negative bending moment near the support each alternate rod is bent upward near the quarter point of the span and continued over the support to the quarter point of the adjacent span. This distance beyond the support must not be less than that required to develop the full strength of the rod in bond which is asf s -?- perimeter of rod x u' for a f inch rod the distance is i8f inches which is less than one-quarter of the span (27 inches). The reinforcement required in the other direction is .0025 x 4 x 12 = .12 square inches per foot section of slab, requiring that f inch round rods be placed '- - xi2 = 11.04 or n inches center to .120 center. IO REVIEW OF THE FLOOR SLAB .1104 p = - = ................................. 0075 4.5X3-25 From Table 3 k= ................................ 375 j = ......... - ...................... 875 a s for a unit slab = - : x .1104 .......... 294 square inches 4-5 t 12150 L = - - - = .......................... 145500 = 2 x 14500 x. 0075 = 8o 375 T ^ 12150 K= - ^=2 = .............................. 95.9 12x3.25 From Diagram i, with the value of K = 95.9 and p = .0075, f s = 14500 and fc = 580 which check the above computations. 150x4.5 v = - ^r- ^ - = ............................ 19.8 i2X.875X3.25 1 50x4.5 u = - - - - = 76.1; 12 x i.i78x.875X3-25 * '-REINFORCEMENT $"4-f'c,TOC. -8-0" I EACH ALTERNATE ROP LIKE a ROP a I. rrA.'of,&7- rr T& .- J. rr .1 ROPb r FlGURE I DESIGN AND REVIEW OF THE FLOOR BEAM The weight of slab carried by one floor beam is 50 x 20 x 9 = .................................. 9000 pounds The live load carried by one floor beam is 100x20x9= 1 8000 pounds The weight of the beam below the slab may be assumed as 10% of the above weight = ................. 2700 pounds 29700 pounds II The bending moment is T V x 29700 x 20 x 12 = 594000 inch pounds In order that the allowable unit shearing^stf^SF shall not be exceeded, the beam must have a cross-section b'd equal to -* - IO S = 141.4 square inches. With b' = 7! inches, d = 18.25 inches. For economy with b' = 7 p and the ratio of cost between the steel and concrete taken as 60, the depth d should be 60 x 594000 4 16000 X 7f 2 = 18.95 inches. Therefore d must be equal to or greater than 18.25 inches and as nearly equal to 18.95 inches as the forms will permit. (See actual sizes of lumber under Design of Forms) . The I I 4x4" ROUGH Li CAP FIGURE 2 distance to be covered by the side lagging (see Figure 2) must then be at least 18.25 + 2 J + if - 4 = 18.75 inches. Two pieces J i' x 5!" and one piece if* x 7!* furnish 19} inches; d is taken as i8f inches, and the total depth of the beam i8j * + 2p = i\\ inches. The distance of 2\ inches allows for two rows of rods, whose centers are i \ inches apart vertically, and assuming that the rods are not larger than one inch in diameter, this allows a clear distance below the lower rods of i \ inches according to the specifi- cations. The actual weight of the beam below the slab is 17^x7! x i5ox 20 = 144 The revised total load on the beam 29830 pounds The revised bending moment 596600 inch pounds The revised maximum shear , 14915 pounds Taking the value of jd equal to d - ft, 596600 14915 2830 pounds i6jx 16000 = 2.23 sq. in. and s = i6|x8o = 11.13 m - 12 Six | inch round rods furnish an area of 2.65 square inches and a total perimeter of 14.14 inches, but over the support the stresses in the steel and in the concrete cannot be brought within the allow- able value unless all of the lower rods are carried through to the quarter point of the adjacent span. Five J inch round rods, furnishing very little more steel do not require these long lower rods and are adopted. The latter furnish an area of 3.01 square inches and a total perimeter of 13.75 inches. The width of the flange (6) is taken as large as the specifications will permit, i. e. either J x 20' -o" = 60 inches, or 8 x 4 -f yf = 39.75 inches; the latter value (40 inches) governing in this case. p = - = .0040 and -5 = -33 = .21. On Diagram 9 18.75 x 40 d 18} the point located by the above values of p and -. falls above the dotted line showing that the neutral axis is in the web. From Table 9, with- = .21 andp = .0040 k = .304 / = .914 From Diagram 9, k = .30 and / = .91, thus checking the above values from Table 9. f s = - - = ........................ 11600 3.01 x .9i4x 18.75 .314 x 11600 fc = 15 d - .304) = ............................ Three rods from each beam are bent up and carried over the support to the quarter point of the span of the adjacent beam. One of these rods from each beam reaches its upper position directly over the center of the support so that these two rods together only have the effective strength of one rod. Thus with three rods from each beam bent up the effective steel area over the support is equal to five f inch round rods or 3.01 square inches. The two rods from each beam which are not bent up are carried beyond the. ends of the rods from the adjacent beam far enough to develop the full strength of the rods in bond, (43.75 inches for a J inch rod) making the effective steel area at the bottom of the beam over the support equal to two j inch round rods or 1.20 square inches. In the case of the six f inch round rods mentioned 13 above it would have been necessary to carry three rods from each beam through along the bottom to the quarter point of the adjacent span in order to obtain a large enough steel area to keep the unit stresses within the allowable limits. (For beams having an even number of rods, one-half of the total number of rods from each beam can be bent up and carried through to the quarter point of the adjacent beam, while the remaining half is carried past the rods from the adjacent beam far enough to develop the full strength of the rods in bond. This gives an effective area at the top of the beam equal to the total number of FULL LINES GIVE LENGTHS OF ROPS FOR FLOOR BEAM; POTTEP LINES FOR ROOF BEAM. FIGURE 3 rods in one beam, and an effective area at the bottom equal to one- half of this amount.) The points at which the three rods may be bent up are deter- mined as follows : One rod at I i - / f 1 = 5- 53 feet from the support. One rod at ^ f i - / _ J = 3 68 feet from the support. One rod at 2 - fi - / 3 1 = 2 . 26 feet from the support. These points are also checked graphically by means of Figure 3. The upper layer of rods over the support is placed as high as the slab rods will permit so that their centers may be considered to be i| inches from the upper surface of the slab. b = 7! inches d = 21 J- ij-f 19! inches d' = 2f inches a s = 3.01 sq. in. i. 20 sq. in. 3.01 a'. P = P' = P' = P d' = d = .0204 .0082 4 143 (See Figure 4) From Table 1 1, interpolating f or p = .0204 p' = -25P P' = -5op p' = -4P L K j L K j L K d' d d' d d' d 10 .262 .0173 .848 .305 .0177 .864 Interpolating 15 .255 .0169 .836 .288 .0172 .845 143 .256 .0170 .838 .291 .0173 .848 .277 .0172 .844 596600 -= 749 I 2 100 ** "* ^T! I i ! 1 J STIRR I I itthemidspanis.9i6x i8f = 17.18 The value of yd at the support is. 844 x 19 J = 16.25 FIGURE 4 Ji The value of jd would of course change at the I point of inflection but the design will be on the _[ side of safety if the smaller value of jd computed above is used in all of the computations involving shear and bond. 15 The maximum shear at the support is 149*5 pounds The maximum shear at midspan is J x i 8000 = . . 2250 pounds The unit shear at the support is 3 149 * 5 - = .. 118.5 pounds ?4 X l0 - 2 5 The unit shear at midspan is g - = 17.9 pounds 74 x 10.25 The distance from the support beyond which stirrups are not required (i. e. the point where the unit shear is equal to 40 pounds WlX 2 per square inch) may be determined by the equation ^ + w d x = 4objd -f- , where x = the distance from the section to the opposite support, Wd = the dead load per linear foot and wi = the live load per linear foot. wj = 100 x 9 = 900 pounds, wa = 50 x 9+ 142 = 592 pounds 9oox 2 . 3 . 592 x 20 , - -f 592X = 40 x 7! x 16.25 + x + 26 -3 2 x = 487-0. 2 X 2O 2 and x = 12.53 f eet > so that stirrups are required over a distance of 20 - 12.53 = 7.47 feet from the support. Selecting a stirrup that requires a spacing of about 6 inches at the support, a s = - X 149 1 5 x _ square inches; a f inch round U stirrup 16000 x 16.25 furnishes .2208 square inches and is adopted. WlX 2 1 The maximum shear at any section is =- + wa (x - - ). W, Wdl -> = 22 . 5 = 5920 Dis. from Support x x 2 WlX 2 ~2T W dX V 3 Spacing inches 20 400 9000 11840 14920 86100 5-8 i 19 36i 8120 11250 13450 86100 6-4 2 18 324 7290 10660 12030 86100 7.2 3 17 289 6500 10060 10640 86100 8.1 4 16 256 5760 9470 9310 86100 9-2 5 15 225 5060 8880 8020 86100 10.7 6 H 196 4410 8290 6780 86100 12.7 7 13 169 3800 7700 5580 86100 154 7-47 12.53 157 3530 7420 5030 86100 17.1 i6 The maximum spacing allowed by the specifications is f x i8f = 14.06 inches so that in drawing a curve for the stirrup spacings similar to Figure 46 of the text, it becomes a horizontal line when its ordinate reaches a value of 14 inches. In drawing this curve it is necessary that the horizontal and vertical scales be the same in order that the spacings plotted as ordinates may be transferred to the horizontal axis by means of 45 lines. These lines cannot be drawn until the location of the first stirrup is determined, which will be placed two or three inches from the edge of the support, and the width of the column or girder which supports the floor beam is not yet known. , , , j i ^u -L- .2208 x 16000 The bond stress developed on the stirrups is 2 x 1.178 x .6 x 18.75 = 133 pounds per square inch. This high value is safe, however, since the upper ends of the stirrups are bent into the form of hooks. The bond stress on the upper horizontal rods at the support is 14915 5 x 2.74QX 16.25 allowable value = 66.8 pounds per square inch and is within the FIGURE 5 DESIGN AND REVIEW OF GIRDER. The maximum moment in the girder has the same value when all of the load in one bay is considered as uniformly distributed on the girder, as it does when the load from one floor beam is considered as concentrated at the center of the girder. The total load sup- ported by two floor beams is 59660 pounds, and the weight of the girder below the slab may be assumed as i itLS, where L is the span of the beam and 5 the span of the slab. This assumed weight is 11x4^x20x9 = 8910 pounds, and the total load is 68570 pounds. 1234260 inch pounds M = x 68570 xi8xi2 = 12 MI = 1,234,260 = 107.4 With 6=13^ inches, d must be at least 29.2 inches. The distance to be covered by the side lagging must be at least 29.2 -f- 2\ + if - 4 = 29.45 inches (see Figure 6). Four pieces 7f x 1 1 furnish 31 inches; d is taken as 30.75 inches and the total depth of the beam (h) as 33.25 inches. '^~ ^Tl . 1 . ft, JHT* 4x4" CAP i* I.I FIGURE 6 The actual weight of the beam below the slab is 144 x 150 x 18 = 7410 pounds, the revised total load 67070 pounds, and the revised bending moment 1207260 inch pounds. 1207260 as = - r - ^ = 2. 8 1 square inches. 16000 x .874x30.75 Five | inch round rods furnish an area of 3.01 square inches and can be placed in one row as assumed. P = 7TT 3.01 13^x30.75 Table 3, k= j = 1,207,260 K = 3.01 x. 877x30.75 2 X .OO72 X I49OO 369 I,2O7,26o I3-5X 30-75 .0072 369 .877 14900 582 94.6 i8 From Diagram i with K = 94.6 and p = .0072, f, = 15000 and f = 580, which check the above computations. In order more nearly to approach the actual conditions of load distribution in the girder, it is assumed that the girder supports the slab and its live load for a distance on each side of the girder equal to its own width, and the load carried by one floor beam not in- cluded in the above is assumed as concentrated at the center of the girder. Accordingly, (i) The uniform load = ^JLIll lg _|_ = ^520 pounds (2) The concentrated load = 29830 - 9110 -r- 2 = 25275 pounds M u = /j x 16520 x i8x 12 = ......... 297360 inch pounds M c = 1x25275x18x12 ............ 909900 inch pounds FIGURE 7 Using these two moments Figure 7 is constructed and the points at which the rods may be bent up are determined. Both Figures 3 and 7 neglect the continuity of the construction and therefore give very conservative results. As both the bending moment and the cross-section of the beam have the same value at the support as at midspan, five rods will be required at the top of the beam over the support, while none is required at the bottom. However, it is better design to carry the lower rods through in a manner similar to that of the floor beam. 25275 + 16520 Maximum shear at the support = 20900 pounds 19 2 r 2 7 C Maximum shear at midspan - - = 12640 pounds Unit shear at the support = . . 57.4 pounds I3-5X30-75X.877 Unit shear at midspan = . 14.7 pounds 13-5 x 30.75 x. 877 The true maximum shear at midspan would occur when a strip equal to three times the width of the girder times the span of the slab is unloaded, which would give a value of 13020 pounds, but the difference is so small that it may be neglected and the shear may be assumed to vary uniformly throughout the span. The point beyond which stirrups are not required is = 2.10 feet from midspan. Selecting a stirrup which requires a 4 x 20000 x 6 spacing of about 6 inches at the support a s = - 16000 x 30.75 x .877 = .194 square inches; a finch round U stirrup furnishes .2208 square inches and is adopted. . . . f x .2208 x 16000 Ine spacing required at the support is ; = 6.8 57-4 x 13} inches and that at a point 2.10 feet from midspan 9.8 inches. A curve similar to that used for the floor beam may be constructed by connecting the points located by the two spacings computed above by a straight line, and the actual spacing of the stirrups determined when the width of the column is known. The actual bond stress on the horizontal rods is 20900 . , = 56.4 pounds per square inch. 5 X2.749X .877 x 30.75 A rod schedule similar to Figure 5 can now be constructed for the girder. DESIGN OF THE ROOF. The design of the roof is similar to that of the floor, but as the loads are smaller, the construction will be somewhat lighter and only the design will be considered, the construction not being heavy enough to call for a review. Roof Slab. Assuming the minimum allowable thickness of slab which is four inches, the total load carried is 50 + 35 = 85 pounds 20 per square foot. The bending moment for a unit slab is 6885 inch pounds and the required depth to the steel 2.31 inches. Using a depth as large as the thickness of the slab will permit (i. e. 3.25 inches), the steel area required per unit slab is .152 square inches, and f inch round rods spaced 8 J inches center to center furnish this area. The reinforcement required in the other direction is .12 square inches per foot section which is furnished by f inch round rods spaced n inches center to center. The main reinforcing rods are bent upward at the same points as those used in the floor slab. Roof Beam. The total weight of slab and live load sustained by the beam is85X2ox9 = 15300, and assuming the weight of the beam as 10% of this load the total assumed load is 16830 pounds, the end shear 8415 pounds, and the bending moment 336600 inch pounds. The required value of b'd for shear is 80.2 square inches and with b' = 5! inches, d must be at least 13.9 inches. The economical value of d is 16.8 inches, and the distance to be covered by the side lagging should be somewhat more than 13.9 + 2f + if -4 = 14.4 inches. Two pieces if" x yf " furnish 15.5 inches mak- ing the value of d 15.0 inches and the actual weight of the beam below the slab -^ x 150x20= 1650 pounds. The revised 144 values of the total load, end shear, and bending moment are 16950 pounds, 8475 pounds, and 339,000 inch pounds respectively. The area of steel required is 1.63 square inches and the total perimeter 8.15 inches, which can be furnished by three f inch round rods with an area of 1.80 square inches and a total perimeter of 8.25 inches. The points at which two of the rods may be bent up can be determined by the use of Figure 3 , the dotted lines representing the length of rods for the roof beam. It will be assumed that an effective area of steel at the top of the beam over the support of three J inch round rods and one J inch round rod at the bottom of the beam will be sufficient to keep the unit stresses within the allowable values. Figures similar to 2, 4, and 5 can now be constructed for the roof beam. The maximum shear at the center of the beam is f x 6300 = 788 pounds, and taking jd = d - \t 13 inches, the unit shear at the 21 support is 113.5 pounds and at midspan 10.5 pounds. Since the live load is small it may be assumed that the shear varies uniformly. Then the point beyond which stirrups are not required is = 2.86 feet from midspan or 7.14 feet from the sup- 113-5-10.5 port. By using the same method of finding the location of this point as was used in the design of the floor beam, this distance is found to be 7.03 feet from the support so that the error in the above computation is small and on the side of safety. Selecting a stirrup that requires a spacing of about six inches at the support a^ = fx . = .163 square inches; a A inch 16000 x 13 round U stirrup furnishes . 1 534 square inches and is adopted. The spacing required at the support is 5.6 inches and at the point 7.14 feet from the support 16.0 inches. According to the specifications 11.25 inches is the maximum allowable distance between stirrups. With these spacings as ordinates a curve of stirrup spacings can be constructed similar to that of the floor girder. Roof Girder. The total load sustained by two roof beams is 33900 pounds and assuming that the weight of the girder below the slab is 3.2LS 2 pounds = 5180 pounds, the total assumed load is 39080 pounds, and the bending moment 703440 inch pounds. Then bd 2 = 6550, and with 6 = i if inches, d must be at least 23.6 inches, and the distance to be covered by the side lagging must be at least 23.6 + 2\ -f- if - 4 = 24.1 inches. Three pieces if* x 5!" and one piece if* x 7}" furnish 25 inches, making d = 24.5 inches and the actual weight of the girder below the slab x 1 50 x 18 = 144 5070 pounds, and the revised total load 38970 pounds. The revised bending moment is 701,460 inch pounds, and the steel area required is 2.05 square inches, which is furnished by four | inch round rods, which can all be placed in one row, as assumed. Assuming the same distribution of loads as in the floor girder, the uniform load is - x 85 x 18 + 5070 = 9570 pounds, and 22 the concentrated load 16950 - = 14700 pounds, the bending moment due to the uniform load 172,260 inch pounds, and that due to the concentrated load 529,200 inch pounds. With these moments a diagram similar to Figure 7 is constructed and it is found that two of the rods may be bent up at distances of 6.45 feet and 4.05 feet respectively from the support. The maximum shear at the support is 12135 pounds and the unit shear 48.2 pounds, and the corresponding values at midspan are 7350 pounds and 29.2 pounds respectively. The point beyond which stirrups are not required is 3.36 feet from the support. A T 5 ^ inch round U stirrup requires a spacing of 6.5 inches at the sup- port and a spacing of 7.8 inches at a point 3.36 feet from the sup- port. The difference between these spacings is so slight that no diagram need be constructed and the stirrups may be spaced 7 inches center to center over the distance they are required. Fig- ures similar to 5 and 6 can now be constructed for the roof girder. COLUMNS. Two columns are to be designed: one, the typical interior column supporting the main floor, and being in turn supported by the basement footing; two, an exterior column or wall column supporting the main floor and having a girder framing into it whose axis is perpendicular to the wall. In low buildings the columns are often made the same size throughout, but in high buildings large quantities of material may be saved by decreasing the sizes of the columns toward the top of the building. The same general method of design applies. The eccentric moment in the column due to a possible eccentric load on the floor is to be taken equal to one-fourth of the moment produced by the eccentric load in the girder framing into the column. This reduction is allowable on account of the stiffness and the continuity of the construction. REQUIREP WORKING SIDE OF A SQUARE COLUMN FOR KNOWN LOAPS ANP ECCENTRICITIES 60C ! I C 5 ^ I 3 ( HO A 55( 5^ "^ . \ ^x,^ , ^, s, ^--^ \, * *^^ ' ^.^ ^ ^^^ *-. > ""f*.,^ \ f --~ s ^ - ~~ : _^ ^= 10 4= 5C *" 1 .^^^ >-^. 12345 ECCENTRICITIES IN INCHES FIGURE 8 DESIGN OF INTERIOR COLUMN Weight of roof supported by one column 38970 pounds Weight of floor supported by one column 67070 pounds Weight of three floors and roof supported by one column 240180 pounds Live load on one floor beam 18000 pounds The moment produced in the girder due to this live load on the floor beam is 648000 inch pounds, and one-fourth of this moment M is 162000 inch pounds. The approximate value of x = , = 24 - = .67 inches, and referring to Figure 8 the probable size of 240100 column is found to be 23 " x 23 ". With one and one-half inches of fireproofing on all sides the outside dimensions of the column are 2 f) X 26 26" x 26". The weight of one column is ' x 150 x 12 = 8450 144 pounds, and of three columns 25350 pounds. The total load to be supported by the basement column with the main floor loaded on one side of the column only is 240180 + 25350 18000 = 247530 pounds. x = .65 and referring again to Figure 8, the column selected is of sufficient size. As it is uneconomical to use steel reinforcement in columns, the smallest amount allowable by the specifications is used. One per cent. of23X23 = 5.29 square inches, which can be furnished by twelve f inch round rods, the combined area of which is 5.30 square inches. a + na s = 23 x 23 + 15 x 5.30 = ................... 609 s = 5 .3ox(8.i 7 ) 2 = ............................. 354 = 23320+15x354= .......................... 28630 247530 162000x11.5 - ' And checking this result by another method 12 x .4418 Po = - = ................................ oioo 23x23 r 8.17 ' l = ^ = ....................................... 35S f 2 = ............................................. 126 z = .i26x 180 = ................................. 22.7 K= - - + '^X r - - - =.870 + .138=1.008 i + 15 x. oioo 23 i +22. 7 x. oioo 23x23 4/ 25 Using Diagram 13, K = i.oo and f c = 468. Under full load, with no eccentricity to consider, the stress in the concrete only reaches 436 pounds per square inch. According to the specifications the rods in the column are held together by one- quarter inch bands placed about nj inches center to center. DESIGN OF EXTERIOR COLUMN. The weight of roof supported by one column is: Live load 180 square feet at 35 pounds 6300 pounds Slab load 180 square feet at 50 pounds 9000 pounds Weight of i \ beams below the slab i J x 1650 2470 pounds Weight of \ girder below the slab \ x 5070 2540 pounds 20310 pounds In a similar manner the weight from one floor is 34950 pounds. Assuming that a six inch wall is to fill the space between columns and beams, the weight on each beam due to the wall above it is (12 - f |) (20 - f |) ( T 6 f ) 150 = 13610 pounds. As this is less than one-half of the live and slab loads carried by one floor beam (28080 pounds) if the wall beam is made equal in section to the inter- mediate beam it will sustain the wall in addition to its proportion of floor load. The weight of walls, floors, and roof is then 165990 pounds, and the total load on one floor beam being 29830 pounds, the eccentric moment is i ,073 ,880 inch pounds and the approximate value of x is 1.62 inches. From Figure 8 the probable size of the column is 22" x 22" and with ij inches of fireproofing on all sides the outside dimensions of the column are 2 5 * x 2 5 *, and the weight of one column 7810 pounds. The total load is 189420 pounds, and the actual value of x is 1.42 inches, and referring again to Figure 8 the size of the column selected is sufficient. One per cent, of 2 2 X2 2 = 4.84 square inches, and as eight f inch round rods furnish only 4.81 square inches, twleve J inch round rods furnish the smallest section allowed by the specifications. a c -f na s = 564, I c = 19520, r = 7.78, I s = 321, I = 24330, and f c = 458- r r 2 p = .0109, - = .353,- = .125, z = 22.5, K = 1.170, andfc = 458- 26 Considering the slight difference in the size of the interior and exterior column it would be advisable to make them of the same size for the sake of the uniformity of construction. COLUMN FOOTINGS. In the design under consideration it is assumed that no property line interefers with the construction of the footings and a simple square footing is to be designed for each column. Recent tests have shown that it makes little or no difference whether rods in a footing are placed in two or in four directions. The design of the two way type follows. For an example of the four way footing see Concrete Plain and Reinforced by Taylor and Thompson or Reinforced Concrete Construction Vol II by George A. Hool. Tests have also shown that a diagonal tension failure will not occur nearer the edge of the column than at a distance equal to the depth of the footing at the edge of the column. In most footings where this condition is considered no web reinforcement is required and in case the unit shear in the concrete does slightly exceed the allowable value at the section mentioned above, a slight increase in the depth of the footing will bring the unit shear within the limiting value. For a design of a footing with stirrups see Concrete Plain and Reinforced. DESIGN OF INTERIOR COLUMN FOOTING Load from Roof '. 38970 pounds Load from three floors 201210 pounds Load from four columns 33800 pounds Bearing area required 2 68.50 sq. ft. 4000 Allowing about 10% for the weight of the footing and taking the length of the side of the footing to the nearest one-half foot, a foot- ing 9' - o" x 9' - o" furnishing an area of 81 square feet is adopted. 8l 2 3 X2 3 Referring to Figure 9 the area of the trapezoid abed is 144 = 1 9 -3 3 square feet. The upward pressure of the soil on this area is 19.33 x 4000 = 77300 pounds. The distance of the center of gravity of the trapezoid from the edge of the column (ab) is BANDS i TC.TO c i "PIPE: PLATES w>^ r -2-9" . i\ ' i i/i 1 1 3-6" 2-9"- ,-,'f> 6z"c.TOC. IN THE CLEAR LJJj ' l_Hl.L Sl-Ui+if I-I--I- 42^ /23 + 2 x io8\ T V 2 3 + 108 ) 3 \^ 23 + 108 the edge of the column is 1998000 inch pounds. FIGURE 9 25.85 inches, and the bending moment at Depth required for moment . / ^ = 28.4 inches ^107.4x23 28 Depth required for punching shear =32.0 inches 1008000 a s = ^-r = 4.46 square inches 16000 x. 875x32 2 o = 773 = 34 .5inches Sox. 875x32 The effective reinforcement is to be spaced over a distance equal to the width of the column plus twice the depth of the footing at the edge of the column plus one-half of the remaining distance = 23 + 2x32 H =97^ inches; J inch round rods spaced 6| inches center to center furnish in a width of 97 J inches an area of 6.63 square inches and a total perimeter of 35.3 inches. The top of the footing is made 42 inches square, thus providing a 6 inch ledge all around the outside of the column so that the column forms may be erected before the basement floor is poured. The depth T is taken equal to about one-half of d or 16 inches. Then the depth of the footing at a distance of 32 inches from the working side of the column (along ef) is 16 H -x 16 = 21.10 oo inches, and the distance ef is 108 - 2 x loj = 87 inches. The area of the trapezoid cdef is 7.11 square feet, making the upward pres- sure from the soil on this area 28440 pounds and the unit shear along ef equal to = 17.7 pounds. 87 X2i.i x .875 The total maximum depth of the footing at the column edge is made 37 inches and at the outside edge 21 inches, thus allowing 4 inches clear insulation below the lower layer of rods. The weight of this footing is about 31000 pounds, making the total load sustained about 305000 pounds, requiring an area of 76.13 square feet, or a footing 8.72 feet square. Therefore no revision is necessary. DESIGN OF EXTERIOR COLUMN FOOTING Load from roof .' 203 10 pounds Load from three floors 104850 pounds Load from three walls 40830 pounds Load from four columns 31240 pounds 197230 pounds Bearing area required 49.3 1 square feet 29 Proceeding as in the design of the interior column footing, a footing 7'-6*x7'-6* is selected. The area of the trapezoid abed is 13.22 square feet, the upward pressure of the soil on this area 52900 pounds, and the bending moment at the edge of the column 934,000 inch pounds. The depth required for moment is 19.9 inches, and that for punching shear 22.9 inches. Adopting a depth of 23 inches the steel area required is 2.90 square inches and the total perimeter 32.8 inches. The effective reinforcement will be spaced over a distance of 79 inches, and J inch round rods spaced 5^ inches center to center will furnish in a width of 79 inches an area of 6.35 square inches and a total perimeter of 33.8 inches. Making the top of the footing 42 inches square and the depth T, 12 inches, the depth of the footing at a distance of 23 inches from the working side of the column is 17.04 inches and. the distance ef 68 inches. The area of the trapezoid cdef is 6.03 square feet, the upward pressure from the soil on this area 24120 pounds and the unit shear along 6/23.8 pounds. The total maximum depth of the footing at the column edge is made 28 inches and at the outside edge 17 inches. FORMS. i . The forms are to be like those shown on posted sheet, and in accordance with the following specifications : Material to be of spruce, and the allowable unit stresses as follows Modulus of elasticity 1300000 pounds per square inch Extreme fibre stress in flexure . 1 200 pounds per square inch Compression on sides of fibres 300 pounds per square inch Compression on ends of fibres . 1200 pounds per square inch Shear parallel to the grain .... 80 pounds per square inch Compression on struts of length under 1 5 times least diameter of side 900 pounds per square inch Compression on struts over 1 5 times least dimension d. 1 200 (i - 1 /6od) pounds per square inch Deflection not to exceed span / 360. Basing the deflection on the values specified above with E = 1,300,000 and D = Span / 360, the following equations result and are to be used : 30 Simple beams. Equation i , W = 277 ,000 I /I 2 Equation 2,1 = Partially continuous beams. Equation 3, W = 462,000 I /I 2 Equation 4, 1 = / 462,000 I/w S/ On the basis of strength Simple beams. Equation 5, W = . ' : i7 4.- * 1 /g6oo.I Equation 6, 1 = Partially continuous beams Equation 7, W = 12000.! ~1 e / 277,000 I/w T^ o 1 / I2OOO.I Equation 8, 1 = In the above formulae W = total load in pounds on the span; w = the load per linear inch of span; / = the length of the unsup- ported span in inches ; I = the moment of inertia of the board or plank in biquadratic inches. The forms shall be designed for a construction live load of 75 pounds per square foot, in addition to the weight of the concrete they sustain. The weight of the forms may be neglected. In the design of the column forms, the yokes are to be spaced on the assumption that the wet concrete exerts a fluid pressure of 80 pounds per square foot on the lagging, but the yokes in no case are to be spaced further apart than 24 inches center to center. The thickness of lumber for the forms depends upon the number of times it is to be used and the manner in which it is erected, but the general practice is i inch stock dressed to J inch for floors, i J or i \ inch stock dressed to i J or if for columns and the sides of beams and girders, and 2 inch stock dressed to if inches for beam and girder bottoms. The actual dimensions of the material are to be used in designing and the various parts of the forms are to be made from the following stock : Actual Size Nominal Size Lagging for i|*by 3 f, 5! "or 7! " I'or if by 4', 6', or 8" Planks if" by 5 f , 7} or gf 2 " by 6", 8" or 10" Battens ij" by 3!" rough 2 "by 4* Joists if'byaKsl.or?! 2 "by4*,6",or8" Yokes 1 1 "by 3! "rough 2 "by 4" Posts 3 1 "by 3!" rough 4" by 4" Caps 3! " by 3!" rough 4" by 4" Braces J" by 35* rough i"by4" Stringers 3 J'by sf, sJ'by sf' 4' by 6", 6*by6',or6'by 8', or sf "by 7 J" rough DESIGN OF THE FORMS Span of floor beams 2o'-o" Center to center floor beams p'-o* Span of girder iS'-o* Thickness of slab 4* Cross-section of beam below slab i7i*x 7!" Cross-section of girder below slab 29!' x 13?" Distance floor to floor i2 / -o* Cross-section of column 26" x 26* Spacing of Joists. The weight of slab and construction live load is 125 pounds per square foot or 10.4 pounds per linear inch of 12 inch strip. The Iofai2"xJ" section is .67 biquadratic inches and the safe span of the lagging by Equation 4 is 3 1 .o inches. When the span of the joist is more than about 5 feet, 2" x 8" joists are requirecTin order that they may be spaced far enough apart to develop a reasonable percentage of the strength of the lagging. It then becomes more economical to support the joists midway between floor beams, allowing the use of much smaller joists. The distance between the centers of the joist bearers attached to adjacent floor beams is 108 - 7! - 2 (i| + if ) - 2 = 92 J inches. Then with a center support the span of the joist is about 46 inches. - of if* x 5!" joist 10.35 cubic inches | of if x 3 f* joist 4.40 " By Equation 7 W = 2700 pounds for i| r x 5 J" joist and 1 150 pounds for i| * x 3! * joist 32 The required spacing of the larger joist is = 67.6 inches 12 and of the smaller 28.8 inches. The latter is very close to the spacing allowed by the lagging and the smaller joist is adopted. The actual spacing of the joists may be determined as follows : Center to center of girders 20' - o" Width of girder 13 J* Side lagging 2 x if = 2 J" Battens 2 x if $\" Joist if" 21 inches Center to center of joists attached to girders 18' - 3 * Approximate number of spaces r-f- 7.61 2o.o 18'- V Therefore, joists are spaced . . . .27!" center to center. o Supports for the beam forms: Construction live load per lineal foot of beam x 7 5 = 48 pounds Weight of beam per linear foot = 174 pounds 144 222 pounds Total load per linear inch 18.5 pounds I of 7 f "x if" plank 3.46 pounds Safe span by Equation 4 44.2 inches Posts supporting the beam form: Construction live load per linear foor I - -^ I 7 5 = 388 pounds Weight of slab per linear foot / - - J 50 = 258 pounds Weight of beam below slab per linear foot I7 ^ X7JXI5 = . .141 pounds 144 787 pounds Total load per linear inch 65.6 pounds Safe bearing of post on fibres of cap 3jx3jx3oo= 4500 pounds 33 % Safe spacing of posts j~-r . ; ;;....: 68.6 inches The posts are spaced 54! inches center to center, thus bringing a post under each alternate joist. Under the remaining joists a plank support for the beam form is placed. Center support for joists (Stringer): The load sustained by each joist over a span of 46 inches is : Construction live load -^- - x 75 = 656 pounds 144 Slab load -" - x 50 = 437 pounds 1093 pounds One post under the center of the stringer would sustain the load from the joist without injuring the fibres of the stringer, but it would require a 6* x 8" stringer. By placing a post under each alternate joist the bending moment for the span of 54! inches is I x 27! I T 8 ^ = i igyoinch pounds. Then bd 2 = 59.9 and with b = 3! inches, d = 3.93 or 5! inches. By using a 6" x 4' stringer with three posts the amount of lumber is less than with a 6" x 8* stringer and only one post and the use of the heavier timber is avoided. Supports for the girder forms. In a manner similar to that used in determining the spacing of supports for the beam forms the safe spacing is found to be 39.3 inches. Posts supporting the girder forms. The joist adjacent to the girder is supported by the girder forms which are in turn supported by the posts. Area of slab supported per linear foot 12' (21" + 27!*) Load from slab per linear foot xi25= .. 504 pounds 144 Weight of girder below slab per linear foot x 150 = 412 pounds 916 pounds Total load per linear inch 76.3 pounds 34 Safe spacing of posts ........................... 59.0 inches Reaction from stringer not included in the above, 547 pounds. The safe distance between the center and adjacent posts is 2 =55.4 inches 76-3 The actual spacing of the posts may be determined as follows : Center to center of column ....................... 18' - o r Width of column .............................. 26" Side lagging 2 x i| .............................. 2 J Clear distance between column forms .............. 15' 7! ' With one post in the center and one on each side 47 inches (46 T f ) center to center the full distance will be covered. A plank support may be placed midway between the posts, but as the clear span between the posts is only 43 inches this is hardly necessary. Column Forms: lof i2"x i|"section .......................... 1.42 Pressure at bottom on strip 1 2 " x i " = - -.... 80 pounds Pressure at J point on strip 1 2 " x i " = ........... 60 pounds Pressure at \ point on strip 1 2 " x i " ............. 40 pounds Pressure at f point on strip 1 2 " x i " ............. 20 pounds SAFE SPAN OF LAGGING Point By Equation 4 By Equation 8 B ott om 20.2 inches 19.4 inches 22.2 inches 22.4 inches 25.4 inches 27.5 inches 32.0 inches 38.9 inches The span of the 2 x 4 is about 26 + 2 (if + i J -f i) = 34 inches lof 2 X4 .................................... 9.1 W by Equation 5 ............................ 1325 pounds SAFE SPACING OF YOKES Bottom Qgo 7. 6 inches oO X y^ J 10.2 inches \ 1 5. 3 inches 30.5 inches 35 As the strength of the yokes governs their spacing, the bolts 1^2^ must take a stress of J pounds requiring a section of .0414 square inches. A f inch bolt has a net area of .068 square inches and is adopted. A washer having a net area of =2.21 square 2 x 300 inches and a gross area of 2.40 square inches must be used. A if inch round washer furnishes this area and a thicknes of | inch is sufficient. 36 DESIGN OF A COMBINED FOOTING When, on account of limited ground area it is impossible to construct simple spread footings to support the exterior columns, . K^fcfer-r-K [ i 200000 F?= 71 r i"$ STIRRUPS- KIN EACH ROW r.~^JTJ, I K x --r l"_ i"$ STIRRUPS I I..L 7J U ABOUT eO"C.TOC FIGURE 10 either a cantilever construction must be used or the footing must be combined with one or more of those for the interior columns. The simplest type of a combined footing is a trapezoidal slab the center of gravity of which coincides with the center of gravity 37 of the loads that the footing sustains. The following design shows the general manner of determining the shape of this type of footing, its depth, and the necessary amount of reinforcement. Two columns i2 / -o* center to center, one having a cross-section of 1 8" x 1 8" and the other a cross-section of 24" x 24* sustain loads of 200000 pounds and 300000 pounds respectively. The allowable pressure on the soil is 5000 pounds per square foot, and the allow- able bond stress on the reinforcing rods is 100 pounds per square inch. Otherwise the specifications heretofore used are to be fol- lowed. (See Figure i o). The bearing area required for these loads is 100 square feet. Allowing the footing to project six inches beyond the edge of the larger column the total length of the footing is 14.25 feet. The center of gravity of the footing must be at a distance of -xi2 + 1.5 = 6.30 feet from b\. Using the equations involv- 500000 ing the area and center of gravity of a trapezoid , - 2 X 100 bi + t>2 = - = 14-04 14-25 and bl + 2b2 ^3X 14.04x6.30 = 14-25 whence bi = 9.44 feet and b 2 = 4.60 feet. The width of the foot- ing at a point midway between the two columns is 9.44 - 14.25 (9.44 - 4.60) = 6.94 feet and the center of gravity of the larger of the two trapezoids on either side of this section is I -~ ; ^^ 1 3 y 6.94+9-44 / = 3.94 feet from this mid-point. The bending moment at this section is 300000 (6.0-3.94)12 = 7,310,000 inch pounds and the approximate depth of footing required is 28.6 inches. Allowing for insulation, the total depth of the footing is assumed as 3 2 inches, which adds to the weight to be sustained 400 pounds per square foot. The above computation was only for the purpose of determining the approximate weight of the footing. Final dimensions may now be computed in the same manner. The area required is 108.7 square feet, and the revised values of bi and 02 are 10.28 feet and 4.98 feet respectively. The section of true maximum moment is of course at the point where the shear is equal to zero. Let the distance of this point from bi be called y, the load on the larger column PI, and the difference between the allowable soil pressure and the weight of the footing w. Then PI = w I l " r-^- 1 , from which the distance y is determined V / as 7.33 feet. The width of the footing at this point (6 4 ) is 7.55 feet and the distance of the center of gravity of the trapezoid bounded by 61 and 6 4 is 3 .85 feet from 6 4 . The maximum bending moment is 300000 (7.33 - 1.5) - 3.85 I 1 7.33x4600 = 590000 foot pounds or 7080000 inch pounds and the depth of footing required is 27.0 inches. The width of the transverse distributing beam under the larger column is taken as 36 inches and that under the smaller column as 1 8 inches. The moment at the edge of the column due to the upward pressure of the soil is f- (10.28- 2.o) 2 x 12 = 3000000 oXIO.2o inch pounds for the longer beam, and - - (4.98-1.5)2 x 12 = o x 4-9^ 732000 inch pounds for the shorter beam. The maximum shear at the edge of the column for the former is - -- 1 - 1 = 2 y 10.28 j 1 2 1000 pounds, and for the latter 70000 pounds. The depths required are 27.9 inches and 19.5 inches respectively. A depth of 28 inches is adopted for the entire footing and with four inches of insulation the total depth is 32 inches as assumed. The distance from the point of zero shear to the point where the unit shear reaches the allowable value for plain concrete is deter- mined by solving the following equation for z. z 2 -+- zwb 4 - - - -v'jdb 4 = o (In using this equation inches I . (bi-b2) I must be used with pounds per square inch, and feet with pounds per square foot). From the above equation the value of z is 2.70 feet, and the width of the footing at this point (6 6 ) is 8.56 feet and that at the edge of the larger column (6 6 ) is 9.3 5 feet. The distance over which web reinforcement is required (q) is 2.13 feet. The 39 maximum shear along 6 6 is^-^ - ^^ (4-83) (4600) = 187700 2 pounds, and the unit shear 68.3 pounds. The average unit shear over the distance q is 54. 1 5 pounds, the average width of the footing 8.95 feet and the total area of web reinforcement required 2 x ^4.iqx8.o';x2.i^x 144 , . . A , . L - 6.20 square inches, requiring thirty- 3 x 16000 two \ inch round single stirrups. The spacing of the stirrups is determined graphically on Figure 10. In a similar manner forty-two \ inch round single stirrups spaced over a distance of 3.01 feet are required at the other end of the footing. Computing the horizontal reinforcement for the main slab 7080000 16000 x. 875x28 . , , = 1 8. i square inches and 2 = - - = 76.6 inches; which will be furnished by IQOX .875x28 twenty-five i inch round rods. Similarly for the transverse beams sixteen rods and nine rods respectively are required. The same condition with regard to diagonal tension failures may be assumed for the distributing beams as was assumed in the square footing. Computing the unit shear 28 inches from the edge of the column it is found to be 59.7 pounds so that stirrups are required from tjiis point to the point where the shear is equal to 40 pounds per square inch which is at a distance of 33 inches from the edge of the column. 2 x 40. o x 36 x 5 The area of web reinforcement required is - f^ = .374 3 x 16000 square inches so that one double \ inch round stirrup will be suffi- cient. Two double stirrups are placed 30 inches from the edge of the column in order to distribute the reinforcement over the full width of the beam. Gaylord Bros. Maker* Syracuse N. Y. FAT. JM. 2 1,1 Ml YC 13071 f&i UNIVERSITY OF CALIFORNIA LIBRARY