A Study of the Sufficiency of Materials In a One Story Residence Fuilt to Withstand Hurricane Winds A Thesis Submitted to The Faculty of MICHIGAN STaTE COLLEGE of AGRICULTURE AND APPLIED SCIENCE BY JOhn A. McCall Candidate for the Degree of Bachelor of Science June 1947 TH ESIS (L./ Acknowledgments The writer wishes to acknowledge his appreciation to members of the Civil Engineering Department, particularly Professor C. A. Miller, for their kind assistance in this study. he). CS w‘. «z a} Cf Dedication To my wife. Table of Contents Introduction.............................. .Roof Overhang............................. Floor Thickness........................... Hind Velocity and Pressure................ Distribution of Hind Forces............... Transverse Strength of fialls.............. Strength of Hortar Joints................. Roof Strength............................. Design of Li tels......................... Doors................................ Easter fiedroom fiindow................ Utility Room Jindow.................. Laundry and Treakfast Alcove Windows. Kitchex sundown...”................ Dining Room Window................... Gallery Windows...................... Living Room jindow................... Investigation of Windows.................. Easter Bedroom....................... Utility Room......................... Laundry and treakfast Alcove......... Kitchen.............................. Dining Room.......................... Living doom.......................... Gallery.............................. Chilflney FOOtIAIE’; DeSignOOOOOOO0.0.0.0000... ..17 .038 [I 00/ .u‘ ..58 Table of Contents ‘I‘Iall Footiftlg‘JeSiEnOOOOOOOOOOOOOO0....0.00....0.00.061 J‘Orms for. :‘\OOf SlabOO...OOOOOOOOOOOOOCOOOOOO...0.00.62 Heat Transmission Losses............................64 d! HOBOC. in. Symbols and Notation coefficient used in A3=E_ ad cross sectional area American Concrete Institute : area of tensile reinforcement or of column bars ares of compressive reinforcement in flexual members area of web reinforcement : width of rectangular beam bending moment coefficient used in A;::M-KF; also distance from . cd neutral axis to extreme fiber of a section : resultant of compresSive stresses effective depth of flexual members distance from extreme fiber to compressive reinforce- ment declination (degrees) compressive stress in extreme fiber stress in tensile reinforcement - foot : stress in web reinforcement ° 12,'o'o"o" allowable working stress bd ; used in determination of resisting moment of concrete sections Building Code, City of Houston, Texas moment of inertia : inch : ratio of distance (jd) between resultants of com— K.C.M.A. psf .Q pressive and’tensile stresses to effective depth ratio of distance (kt) between extreme fiber and neutral axis to effective depth %fc3k : length of Span pound : external moment (ft. kips) O np+(n-l)p‘; used in determination of k ° ratio of modulus of elasticity of steel (3;) to that of concrete (EC) : National Concrete Masonry Association ° number of stirrups latitude (degrees) - ratio of compressive reinforcement in beams . pounds per square foot np+(n-l)p'd'; used in determination of k d sum of perimeters of bars spacing of stirrups (in.) ° base length of shear diagram (ft.) resultant of tensile stresses bond stress : shearing stress : shearing stress taken by web reinforcement total shear ° the total external vertical shear in excess of that allocated to the unreinforced web unit load per lineaI ft. ratio of distance (zkd) between extreme fiber and resultant of compressive stresses to distance kd meridian zenith distance--angular distance from a point directly overhead (zenith) to the sun and measured along a meridian. (1) Introduction In these days of the ever present housing shortage everyone has a dream house. 'One day that dream will come true. But long before this, tie wise builder did a lot of thinking and planning. it is with this idea in mind that the writer approaches the subject. The first consideration is a floor plan which will give the physical measurements necessary to carry on an investigat- ion. Features that were to be included were an entry, break- fast alcove, laundry room, sewing room, heater room, cold storage room, dressing room off master bedroom, and all rooms Opening on a gallery. Since no such plan was available it was necessary to act as my own architect. The reader is reminded at this point that it is bad business to try to make your own working drawings in case you are planning to build your own home. All construction should be supervised by a competent builder. For instance, to leave out one small detail such as roof flashings around a chimney would be a very costly detail indeed. To satisfy the requirement that all rooms Open on a gallery suggested a U-shaned floor plan with the gallery Open- ing on an interior courtyard. Start with the kitchen since most of the fixtures are standardized as to size and deciding on an arrangement of the fixtures will determine the size of the room. It was decided to include a combination sin; and dishwasher, gas stove, gas refrigerator separated by a base cabinet 18 inches wide, desk (2) to hold a telephone and file recipes, corner base cabinets, and other base cabinets such as are necessary to fill out the circumference of the room. Base cabinets are 19 inches deep and 30 inches high. Wall cabinets extend along the outside wall over the sink-dishwasher combination and along one interior wall over the stove and refrigerator. The north wall is Open over the base cabinets. The base cabinets act as a serving counter for the breakfast alcove in the next room. A door on the east wall Opens off the gallery. The breakfast alcove has built-in seats along the outside wall and along the back of the kitchen base cabinets. There are three chairs on the north side of the table. The laundry and sewing room has a Bendix washer and dryer, clothes hampers for the three classifications of soiled clothing, a desk to hold the sewing machine and store buttons, paterns, etc., wall cabinet storage space for other sewing aids and necessities, laundry sink, rotary ironer, ironing board that folds up to fit into a wall cabinet. There is also storage space in wall cabinets for pottery and this room can also act as flower arranging center. A door Opens onto the gallery. The utility room contains the hot water boiler and auto- matic hot water heater. Both use natural gas. The valves for the adjustment and control of the radiant heating panels are also in this room. The east wall is taken up by a refriger- ated cold storage room. It is not intended to heat the utility room. Wall.cabinets could be built to provide additional ’ w<.deU-k, TI o. . so. a x. x :k :N u...“ ..._... -3me .| lllll . . H jflni‘.‘ t 61 1 i .‘ II... ‘II . .‘FI. t1 .I‘l‘ :l; |¢-I.|n|l ‘. .ll - l...» ’- V’mvf'tr . s .u‘. o a .‘ .7 Intro, Os- uni. {Jib ‘0‘ .4 .WLNMZNDKI‘U UFQHM, : sz‘mfiu‘; Gituwm D _\ _ J I‘ll '1. II) ..o.. it..* .vo'l c all I. \1 VIII; 1. t l .i- | FEE- 3- I T In". I'II‘ 'I! .‘III ' emu. some Khmo _ i .l WLMZ~Dow4< . xi. W¢mX. ..‘ J. i O‘ 5‘ Lav???” viii-7 .' .‘f :ai v: 6V5?” "3‘3.“ w . \ o‘ H :14 . ~“ ... . . .\3-Mr. .z‘ . .. ' . .T \ 3..“ A. $1 .v- _ v . i an!— .- ‘ .- Jy‘. u r "i‘ ’v .1..- Aflfi - 4.33%? 71». - o . 1...; .51 v. . v t; . . , . o. . . , .. ... L . . , . o u a. . . o . o .I .v I at .4 , . 4 x. . ,.. y . . w .. 4 1 4 I I . i , . Sr}. 4 u... .. ,2 . .. ... I, n o .. .3. . ., ..-:A , l I a . . . . . .a . u‘ :_ .'.. a“ -- 'Qi' 3.00m »H_I_::D . 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I -7: :2: ::.—.:"_::' ."'"....“‘.’. :2. :: _ _ _ ._ 1‘; I ALCOVE I , I ————-—II I 1 i . I I I I I I I I __I VXZIIZ 1:“ I' I l I ,KITCH EN II I DINING , ROOM r_‘_h::::_“.::: 2::444... ***““I “““““ W‘fiflr“ **“ I . . II | W ' ' LIhG II . :‘ CONSTRUCTFNJ I ROWA I FLOOR 'PLAN SCALE? ICM.=_5' FI6. 9 (13) until it reaches its lower culmination on December 21. We know that the slant rays of a December'sun have very little warmth and so we want all Of the sun to enter the windows. These sun angles are best shown by the diagram of Tig. 14. The clear distance from floor to ceiling is 7 ft. 6 in. Figure the overhang by using the tan Of the angle of the suns rays. This angle varies with different localities. Since Houston, Texas is in the hurricane area, calculations will be for that locality. Houston is located at 29 north latit- ude. From Astronomy is given the formula z,_=¢-6 for a heavenly body south of the zenith 5‘: 23°26.5' (1946 Ephemeris) L-fow zm=¢-5=29123° 26 . 5 ' =5°33 . 5 '. Overhang=7.5 tan 5°33.5E 7.5x0.09731=O.73 ft. Use a one foot overhang as was used on all other sides of the house. Floor Thickness Next check to see if the floor is thick enough. Standard thickness for floors with pipes embeded for radiant heating is six inches. Assuming one inch of cover coat over the tension reinforcement would give an effective depth of 5 inches. Table 2 of the American Concrete Institute Teinforced Concrete Design Handbook gives a resisting moment of a rectangular section one foot wide with an effective depth Of 5 inches as 6.2 ft. kips. (7) storage Space. The master bedroom has a dressing room with built—in wardrobe closet. Future prOposed additions to this wing are three smaller bedrooms for the children. A bathroom would serve two bedrooms. Access to the dining room is by means of a swing door. off the kitchen. The west side of the room was left entirely Open. The living room has a mirror that runs the full width of the room above the fireplace mantel. Below the mirror and flanking the fireplace are two bookcases. Future prOposed additions to this wing are a study room, a large recreation room, and a guest room with bath. Both the front and the side entrances have storage Space - for clothing as you come from the outside. The two halls running east and west will have doors at their far ends when the proposed additions are completed. It will only be necessary to investigate the original construction since conditions are representative in this part of the house. Roof Overhang The first step is to calculate the roof overhang on the south side of the house. This overhang acts as a sun shade for the windows on this side of the house. This sun shade is needed when the sun reaches its upper culmination or its high point in the sky. Upper culmination occurs on June 21. Then the sun gradually travels southward a small amount each day (14) Sec. 2612, H.B.C. gives other values as 5” H 3,000 psi. f} =‘20,OOO psi. for billet-steel bars 5i II 0.40 1;": 1,200 psi. n = 10 The bending moment is calculated to see that it does not exceed the resisting moment. Span= 18 ft. 11 in. use 19 ft. Live load: 40 psf. Sec. 2304, H.B.C. Dead load: §_ x 150: 75 psf. 2 s.M.= % wl‘ where w: 115 lb. per lineal ft. H 13.1%.: %W11= % (119(19):: 5,200 ft. lb. 5,200<:6,200, therefore the floor meets the code specificat- ions. Figure the tension steel required for the floor slab. M: TJd= Asfs 3d From Table l, A.C.I. Design Handbook 3: 0.875 A a M = 6,200xl2 = 0.851 sq. in. f;3d 20,000x0.875x5 Table 3, “.C.I. Design Handbook gives 0.88 sq. in. for E in. round bars spaced 6 in. center to center. Use % in. round bars Spaced 1 ft. center to center as Spacers. Wind Velocity and Pressure Before designing for stresses set up in the building an equivalent wind loading must be decided upon. Hurricane winds frequently attain velocities of 100- miles per hour. Records indicate that these strongest winds often reach and, for 5 minutes or more, maintain velocities (15) of 100 miles per hour, and occasionally even 125 to 150 miles p") h per hour with gusts of still greater velocity. No state along the South Atlantic and Gulf Coasts has an average of one trOpical storm of full hurricane force in a year. According to engineers, Coral Gables, Florida was cent- rally located in the path of the hurricane of September 18, 1926. This storm, in the Opinion Of R. W. Gray, weather observer at Miami, exceeded all prior hurricanes in the United States from the standpoint of severity. The wind velocity was 125 miles per hour. A deluge of rain amounting to between 8 and 15 inches in 16 hours fell during the storm. At the time of the hurricane Coral Gables contained some 2,500 residences, apartments and other buildings, all of which had concrete masonry walls. These buildings erected in ac- cordance with provisions of the local building code, came thru the storm without a single case of destruction and with but slight damage. George E. Merrick, president, Coral Gables Corporation, in commenting on the storm, wrote: "Coral Gables, wisely restricted to concrete construction, withstood the force of the gale.......The total damage to the entire city of Coral Gables will not exceed $1,500,000 or about 1 per cent of the amount of money that has been spent in Coral Gables in construction and develOpment." Mr. Merrick was speaking of concrete masonry when he said concrete construction because at the time concrete masonry had been used in 93% of the buildings in Coral Gables. (16) The building code that was adOpted Was in practical conformity with the recommendations of the 2uilding Code Committee, U.S. Department of Commerce. The quality of mortar was definitely regulated, the Specification for portland cement mortar requiring 1 part portland cement and not to exceed 3 parts sand, by volume; for portland cement-lime' mortar, 1 part portland cement, 1 part slaged lime, and not more than 6 parts sand, by volume. Under the provisions of the code, anchoring of roofs and tying-in of florrs are required. Pressure exerted on a building by wind is governed by many factors, such as wind velocity, atmospheric pressure and temperature, height above ground, shape of building, and its orrientation towards the air stream. In these investigations use velocity pressure P, in pounds per square foot equals 0.0033 V , where V: velocity of wind in miles per hour. Based on the foregoing formula with V: 125, the approximate velocity pressure is 50 pounds per square foot. H. L. Whittemore, according to unpublished data on stren- gth of low-cost houses of the National Bureau of Standards, worked out velocity pressures that could be anticipated at 30 feet a ove ground in any loc lity in the United States. Pressures at 20 feet would be approximately 10 per cent less. His pressure curves are based on the maxi um average wind velocity for a 5-minute interval as reported by the U. S. Weather Bureau plus 50 per cent gust. The formula (l7) (P-0.0033V ) used above was the Weather Bureau formula. 0n the strength of Whittemore's data, it seems reason- able tO assume for design purposes a velocity pressure of 45 lb. per sq. ft. after a 10 per cent reduction. Distribution of Wind Forces Table 1 shows the average distribution in terms of vel- ocity pressure P on walls, roof, and eaves. Table 1 Average Distribution in Terms of Velocity Pressure P, on Malls, Roofs, and Eaves Building Element Orientation Velocity Pressure Walls Windward . 0.80 P pressure Walls Leeward 0.50 P suction Walls Parallel 0.58 P suction Roof Windward (0 to 20 deg. slope) 0.77 P suction Roof Leeward, flat 0.77 P suction Roof Parallel, any slope - 0.77 P suction Eaves Maximum pressure up 1.57 P Eaves Maximum pressure down 1.16 P The wind forces shown above are based on a windtight building. Loads from internal wind forces must be added for buildings with Openings or potential Openings such as windows or doors. For buildings having 30 per cent or more wall Openings in windward side add 0.77 P pressure outward on roof and walls, and for Openings in leeward or parallel sides add 0.58 P suction in. For Openings between 0 and 30 per cent of wall area, assume prOportional additional loads. The percentage of Openings on each side are as follows: (18) East side - 18.8 + 48.2 + 10.6 + 20.9 + 20.9 = 119.4 = 25.7% 62 x 7.5 05 South side 2.6 1 116.6 3 119.2 ; 48% 33 x 7.5 248 West side 7.5 x 62 565 North side 13.9 + 18.8 + 31.4 g 64.1 = 22.1% 7.5 x 58.5 289 Wind from east For Openings in windward side add 25.] x 0.77 x 45 = 29.7 lb. pressure out on roof & walls 30 For Openings in leeward side and parallel walls add ”119 + 42 + 64 x 1 x 0.58 x 45 g 19.6 lb. suction in on 248 + 465 , 289 30 re 0 f and walls ThiJB gives a net result due to wall Openings of 10.1 lb. pressure out on walls and roof. If the roof slab were poured using 2 x 6 in. edge foruns its thickness would be 5 g in. and would weigh 55625 x l x 1 x 150 a 70.3 lb. per sq. ft. 22 Pressure on east wall (1.8 x 45 - 10.1 g 25.9 lb. pressure in Pressure on west wall (3.55 x 45 x 1 - 10.1 a 32.6 lb. suction out (19) Pressure on north and south walls 0.58 x 45 + 10.1 ; 32.6 lb. outward suction Pressure on roof 70.3 - 0.77 X.45 - 10.1 = 25.5 lb. pressure down Max. pressure up on eaves - 1.57 x 45 - 70.3 = 0.3 lb. pressure up _ Max. pressure down on eaves = 1.16 x 45 + 70.3 ; 122.5 lb. pressure down Wind from west For Openings in windward side add 2_ x 0.77 x 45 ; 10.4 lb. pressure out on roof and walls 0 \N For Openings in leeward side and parallel walls add 119 + 119 + 64 x 1 x 0.58 x 45 = 26.1 lb. suction in on 465 + 248 f 289 3 . walls and roof This gives a net result due to wall Openings of 15.7 lb pressure in on walls and roof Pressure on west wall 0.8 x 45 + 26.1 g 62.1 lb. pressure in Pressure on east wall 26.1 - 0.5 x 45 g 3.6 lb. pressure in Pressure on north and south walls 26.1 - 0.58 x 45 ; 0 lb. pressure Pressure on roof 70.3 + 15.7 - 0.77 x 45 = 51.3 lb pressure down (20) Wind from north For Openings in windward side add ‘22.1 x 0.77 x 45 g 25.5 lb. pressure out on roof & walls For Openings in leeward side and parallel walls add‘ 119 + 119 + 42 x 1_ x 0.58 x 45 = 10.2 lb. suction in on 248 + 465 1 465 30 walls and roof This gives a net result due to wall Openings of 15.3 lb. pressure out on roof and walls. Pressure on north wall 0.8 x 45 - 15.3 = 20.7 lb. pressure in Pressure on south wall 0.5 X 45 i 15.3 = 37.8 lb. suction out Pressure on east and west walls 0.58 x 45 + 15.3 ; 41.4 lb. suction out Pressure on roof 70.3 - 15.3 - 0.77 x 45 g 20.4 lb. pressure down Wind from south For Openings in windward side add 0.77 x 45 = 34.7 lb. pressure out on walls and roof For openings in leeward side and parallel walls add 64 + 119 + 42 x 1_ x 0.58 x 45 g 16 lb. suction in on 289'+ 465 + 465 30 roof and walls This gives a net result due to wall Openings of 18.7 lb. pressure out on walls and roof Pressure on south wall 0.8 x 45 - 18.7 = 17.3 lb. pressure in ‘ (21) Pressure on north wall 7 0.5 x 45 + 18.7 = 41.2 lb. suction out Pressure on east and west walls 0.58 x 45 + 18.7 : 44.8 lb. suction out Pressure on roof 73.3 - 18.7 - 0.77 x 45 ; 16.9 lb. pressure down Wind loadings are sumarized in Table 2. From Table 2 the greatest velocity pressure exerted was 62.1 psf. pressure on the west wall with wind from the west. Modulus of rupture as given by the Portland Cement Association is 130 psi. for course construction of dry tamped concrete as in a double hollow wall. The Underwriters' Laboratories' Standard for Concrete Masonry Units defines hollow units as having average core areas in excess of 25 per cent of gross volume. The unit used in these walls was a 3-Oval core 8 x 8 x 16 in. cinder block. This unit falls within the Underwriters' Standard for a hollow unit. The units alone will not rupture. Transverse Strength of Walls The University of Illinois conducted tests for determ- ining the modulus of rupture of walls. Results in psi. for 3-Ova1 core 8 x 8 x 16 in. cinder blocks were 34, 42, 18, 22, and 23. The flexure test was made on large walls. These walls, which failed by cracking along a horizontal joint near mid— height, were later tested in cOmpression , and developed as good compressive strength as the uncracked walls. (22) Tmfle2 Velocity Pressures on Building Elements According TO 0rrientation With Respect to Wind Wind from Building Element Velocity Pressure East E. wall 25.9 psf. pressure W. wall 32.6 psf. suction N. & S. walls 32.6 psf. suction Roof 25.5 psf. pressure Eaves 0.3 psf. pressure up " 122.5 psf. pressure down West W. wall 62.1 psf. pressure E. wall 3.6 psf. pressure N. & S. walls 0.0 psf. Roof 51.3 psf. pressure Eaves 0.3 psf. pressure up " 122.5 psf. pressure down North N. wall 20.7 psf. pressure S. wall 37.8 psf. suction E. & W. walls 41.4 psf. suction Roof 20.4 psf. pressure Eaves 0.3 psf. pressure up " 122.5 psf. pressure down South S. wall 17.3 psf. pressure N. wall 41.2 psf. suction E. & W. wall 44.8 psf. suction Roof 16.9 psf. pressure Eaves , 0.3 psf. pressure up " 122.5 psf. pressure down (23) The flexure test was made by placing the wall against a vertical structural steel framework and applying a lateral concentrated load along the horizontal center line with the slab supported along the tOp and bottom edges. The Span was 9 feet. The load was applied by means of a 10-ton screw Jack, a Spring dynamometer and a steel I-beam. Values of modulus of rupture were computed on the basis of a beam of solid rectangular section, with a width equal to the length of the wall and a depth equal to the wall thickness. This assumption is used to facillitate computation and gives values that err on the side of conservatism. The failure of the walls was in all cases a failure in adhesion of mortar to unit. In all of the oval core units, molded on a metal pallet, the failure was between the smooth, molded lower surface of the block and the mortar. The walls with face—shell bedding gave about as good results, considering the mortar strengths, as do corresponding walls with full bedding. This is lOgical, since the area occupied by cross webs should have little effect on flexual resistance. The low values for modulus of rupture were for low strength mortars. Therefore we are safe in throw- ing out the low value which gives an average for the modulus of rupture as 30 pounds per square inch. The Portland Cement Association, in their pullication, Reinforced Concrete Houses, calculate the resistance of a dwelling wall to wind pressure by assuming the minimum hor- izontal section with Openings deducted and the wind load carided.on the vertical span. (24) From Table 2 the wihd load on the west wall is 62.1 psf. The clear Span between the roof and floor is 7.5 ft. x 1; 1 wl‘ = 62.1 x 7.5 = 436 ft. lb. 8 8 For combined bending with axial load f ; P 1 fig where c = 4 in. A I P ; 51.3 x 21.583 x 16 f 32.5 x 6 z 934 lb. 2 12 A ; 0.60 x 15.75 x 8 a 75.7 Sq in. I = 1_ x 15.75 x 8' - 4 x 0.0491 x 3.21 x 5' = 592.4 in. 12 f ; 934 1 436 x 12 x 4 = 12.3 1 35.3 ; 23 psi.tension 75.7 592.4 Even though the tests indicate that a wall could be expected to rupture at an average of 30 psi., the wall will be desigmed so there is no resulting tension in the mortar joint. If the cinder units above the midpoint of the vertical Span are filled with concrete, it will give additional compressive forces. All cores above the midpoint are filled in sections of the west wall that are unbroken by Openings. The cores of the first course can be filled on the ground and allowed to set before the blocks are placed in the wall. Cores of the other four courses above this one can be filled in place in the wall. Core volumes in each block when filled with concrete give additional weight as figured below 4 x 0.7854 x 3.21 x 5 x 7.75 x 150 = 34 1b. 1728 This added weight gives stresses in the wall due to combined \ compressive and bending_forces as follows: (25) f 1104 .t 436 x 12 x 4 .-_ 81's 1 31.2 ,-_ 22.4 psi. tension 7 126 671 It is apparant that there is no advantage in filling the cores with concrete. In fact it is a definite disadvantage as far as the insulating prOperties of the wall are concerned and makes a damp wall. The wall can be designed with tension steel spaced 6 ft. center to center and anchored in mortar placed in the cores. Use an 8 x 8 in. bond beam along the top of the wall and use hooked bars. Embed hooked dowels in the footing and cut them off two feet above the tOp of the footing. These dowels are of the same size as the vertical steel.‘ The vertical reinfor- cement is placed 1% in. in from both the outside and the inside wall surfaces on 6 ft. centers. With a stress intensity of 23 psi. 8 the extreme fiber, the stress intensity 1% in. in (thickness of face shell), by similar triangles is 2.5 ; _g_ x ; 14.4 psi. 23 Then the tension in the mortar joint along the length of one block is 14.4 + 23 x 1.5 x 15.75 = 442 1b. 2 Tension over 4.5 blocks or 6 ft. is 442 x 4.5 g 1,985 lb. tension A: 3-1_ =,l;2§§_ : 0.0994 sq. in. 1:, 20 ,000 A for % in. round rods is 0.20 sq. in. This method of reinforcing is used in the west wall only. ' (26) Strength'Of Mortar Joints NO wall is stronger than its mortar joints. The amount of bedding area is important because the greater the area provided for mortar bedding, the stronger the wall. Full mortar bedding results in a stronger wall than face shell bedding, as the load is distributed over a greater area. The design Of the unit also is an important factor, as it governs the bedding area, the shell and web thickness of the units, and their alignment in the wall. With face shell bedding of the 3 - oval core, 8 x 8 x 16 in. unit, the mortar bedding area is 45 per cent of the gross Varea as given in Table 8, Facts About Concrete Masonry. The unit stress under a loading of 80 psi. on the wall is 178 psi. To provide a factor of safety of 4, the mortar should have a strength of 4 x 178 or 712 psi. A 1:1:6 portland cement-lime or stronger mortar is recommended for all masonry wall construction. The 1:1:6 portland cement-lime mortar has a strength of 1,000 psi. The National Concrete Masonry Association figures that for face shell bedding of mortar the wall strength is 42 per cent of the strength of the units. The Houston Building Code requires a hollow concrete unit to test in compression to 1,000 psi. over the gross area. This would give a wall strength of 420 psi. gross area and a factor Of safety of 5.25 for allowable loads of 80 psi. A factor of safety of 4 is usually required. (27) Roof Strength The greatest velocity pressure exerted on the roof during the hurricane was 51.3 psf. pressure. This pressure was a combination of the live and dead loads. The dead load alone was 70.3 psf. Part of this weight was Canceled by suction from the wind. Therefore the design will be for the deal load and a live load of 25 psf. as required by the Houston Build- ing Code. Effective depth is 4.5 in. with a 1 in. cover coat over the tension reinforcement. Table 2 of the A.C.I. Design Handbook gives a resisting moment of a rectangular section 1 ft. wide with an effective depth of 4.5 in. as 5 ft. kips. Other data used was f3 = 20,000 psi. for reinforcement bars of structural grade fc 3 3,000 p81. f... ,-_ 0.40 f; = 1,200 psi. n = 10 The bending moment is now calculated Span : 19 ft. 7 in. = 19.58 ft. w = 95.3 psf. 3.1/1. :: %w12=%x 95.3 x 19.58’ = 4,570 ft. 1b. This moment neglects the overhang and gives results that are conservative; 4,570<:5,000, therefore the roof meets the code specifications. Tension steel required for the roof slab is now figured (28) From Table 1, A.3.I. Design Handbook J = 0.875 53 =_ M g 5,000 x 12 ; 0.761 sq. in. fi;3§ 20,000 x 0.875 x E15 From Table 3, A.C.I. Design Handbook, 0.75 sq. in. call for II I _... 3 in. round bars Spaced 7 inches center to center. Use 5 in. round bars spaced 1 ft. center to center as spacers. Steel is also required for stresses deve10ped due to negative moment where the roof slab passes over the outside wall. A; a M _ 84.3 x 12 = 0.0129 sq. in. fsjd 20,000 x 0.875 x 4.5 Use a 2 in. round bar (A5 = 0.11 sq. in.) two feet long and 03 spaced 21 in. center to center around the outside walls. When the soffit forms for pouring the roof slab are placed, quarter round moulding is placed around the circum- ference of the roof and three inches in from the edge, to be later removed and form a drip. The roof does not need anchor- ing as it has sufficient weight. Design of Lintels Door lintel The National Concrete Masonry Association calls for a lintel 5 g in. high and eight inches wide for clear Spans up to 7 ft. It uses two g in. round deformed bars placed 1% in. above the bottom and—g in. in from the sides. >w d IL DOW .mnz ..m..w.i w memo z ‘ 1 . . I. ‘0 .0 (1“ " III I l o!!! ‘BII. 1.! l i t i. 'I I'll l l I ‘7 I i 9 I l I I HI‘.,IIIIIIIII I‘lluantit: .I‘ .Olull. O'IIIII III‘ 'llrlll. Iv! ..I :lrll...'|ll 1......‘II .. Il'vll. I .pl - .n'l' - ~ . r . c. I I. on rul- “""'LJ _- ——— .4 but (33) i” """“‘"‘"DEC E Mb‘s R 2‘ I 5 T .._.__..M;.,;.::,;,., §SEPTEMBER 2|” L ’JUNE 2&5?” \ \ 0, i x. \\ \T '\\ \“Q \ M V .5 4 . Z M- ii._...._-.,,:..-.. __._ .. '/ \E M \ _-.--o--..-. . -. -—... U. ' ---._.._H-.._-...-—.-L4_.- i E ‘ . --- .. - \ \o‘ .“\‘\ .\.\\ \'\ \‘.§\\‘V\- E DIAGRAM CF SUN ANGLES FIG. l4 I \ J ‘ 1‘ ’ . a): VP 000R LHWTEL H6. \5 NO SCALE (34) With a clear Opening of 33 in. and a bearing area extending 4 in. on each side use a design Span of 37 in. and design as a simple beam. Roof load is 70.3 psf. Suggested design by N.C.M.A. is shown in Fig. 15. The design will be investigated as shown on page 17, A.C.I. Design Handbook. Unit load per running foot of lintel is 70.3 x 21é583 + g: x §_ x 1 x 150 = 827 lb. per ft. 12 M = 1 wl‘: 827 x 3.08‘ .-_ 983 ft. lb. 5 —8_ Given: 6b = 8 in.; d = 6: in.; A:.; 0.22 sq. in.; n = 10 M 983 ft. lb. III : q z 11%: 10 X 0.22 : 0.044 bd 8 x 6.25 From Table 11 k :im 'f 2Q - II]. : Y0.044 f 0.088 - 0.088 = 0.212 From Table 13 for z - 0.33 and k g 0.212: 3 z 0.93 13. ,-_ 12M 3 12 x 983 ; 9,240 psi. jdA 0.93 x 6.25 x 0.22 Sec. 2613, H.B.C.: allowable f = 20,000 psi. 5 f; -»§3x: k = 9,240 x 0.212 = 249 psi. n 1-k 10 1 - 0.212 Sec. 2613, H.D.C.: allowable f; =,1,200 psi. The design is acceptable. Master Fedroom Window Lintel Suggested design by N.C.M.A. is shown in Fig. 16. The design will be investigated as shown on page 17, A.C.I. Design (35) .3qu 02 C .0; .5223 Zoom 1,, 23:5 JUPZZ >>OQ>2>> EOOXQMQ IMHMQE fimvngm _..m _..m_..n _..n wnvfim... . ..__.N 4. .. NJ . . ~ II" ‘.|.I‘ .1le\ II. \n‘l... Ill‘II'I. ll}: 7.130.}!!! I I... 11.1311; .1. 11+..._I4 .TlisrlT _ . . . . (36) Handbook. The Opening is 6 ft. 10 ini’and the design Span is 7 ft. 6 in. Given: b = 8 in.; d ; 6; in.; d' - 18 in.; A: = 1 sq. in. A} ; 1 sq. in.; n = 10; w 827 lb. per ft. M = 1 wl ; 827 x'7.5z a 5,820 ft. lb. 8 8 m : nA; f (n - 1) Ai : 10 x 1 + 9 x l = 0.20 + 0.18 bd bd 8 x 6.25 E X 5.25 .g 0.38 : nA + gn-IZAg x g'; 10 x 1 + 9 x 1 x 1.5 ; 0.20 + bd bd d 8 x 5.5 8 x 6.5 5.25 0.043 = 0.243 From Table 11, for m = 0.38 and q = 0.243: k = 0.414 For entering Table 12 determine x 9 x 1 - 0.434 X n-lA..' z 1 bd 0.414 8 x 6.25 1 k 1x9': 1 x 15.-.0579 k d 0.414 6.25 From Table 12: + gn-lz Ag x d; x (1 — dl) Z:l 6 kbd kd kd : 0.167 + 0.434 x 0.579 1 + fln-liaE x (l-dI) 0.5 1 0.434 x 0.421 2 kbd kd x 0.421 = 0.167 + 0.0768 : 0.359 0.5 + 0.18 From Table 13, for z = 0.359 and k : 0.414: 3 g 0.853 f: :_ 12M : 12 X 5320 : 13,100 p81. ‘ JdA; 0.853 x‘6125 x 1 Sec. 2613, H.£.C.: allowable f5 - 20,000 psi. féalways less than nfi; or less than 9,280 psi. when f. : it x k = 13,100 x 0.414 ; 928 psi. n 1-k 10 1-O.Elz - (37) Shear ; wl ; 827 x 7.5- 3,100 lb“ 2 2 Sec. 2618, H.B.C. v :. V 31100 -72.7 psi. bjd :8 x 0. 853 x 6.25 Sec. 2613, H.B.C. gives allowable v z 0.02 fl: : 0.02 x 3,000 ; 60 psi. with no web rein:orcement. Requires web reinforcement. Allowable shear with web reinforcement ; 0.06 x 3,000 = 180 psi. Stirrups are made of N0. 6 gage cold drawn steel wire with a tensile strength of 16,000 psi. Sec. 3613, H.B.C. Formerly stirrups were allowed to carry 20,000 psi. Design will be by the method on page 30, A.C.I. Design Handbook. The shear diagram is triangular. Given: v' = 12.7 psi.; b : 8 in.; d z 6.25 in.; fr: 16,000 psi. f2 = 3,000 psi.; 10. 6 wire stirrups; Av.. 0.0648 sq. in. _§_ - 12. ___7_ s ; 0.656 ft. 3.75 72. 7 Avag 1,6 000 x 0.0648 -.- 1,040 l = 11b : 12.7 X 8 ; 0.098 S Avf'v' l , 0Z0 N : 55(l) : 6 x 0.656 x 0.098 g l stirrup S , Index 2 1.5 S : 1.5 X 0.656 : 10 0.098 mu: From diagram 17, l stirrup is required. Stirrups can be securely fastened to compression steel and hence will not require any minimum depth of embedment. The suggested design by the N.C.M.A. is shown in Fig. 16 and will be adopted. _ (38) Utility Room Lintel ’ 01ear Opening is 3 ft. 8 in. and the design span is 4 ft. 4 in. Suggested design by the N.C.M.A. is shown in Fig. 17. The design will be investigated by the method shown on page 17, A.C.I. Design Handbook. Given: b : 8 in.; d 6.25 in. ; d' = 1.5 in.; A 0.40 sq. in. I; 6’: A;: 0.22 sq. in.; n 2 10; w = 827 lb. per ft. 2 M : %'x 827 x 4.25 ; 1,865 ft. lb. . m : g%§.+ (ng§2As ; 1% i 2.22 + g i 0.3: = 0.08 + 0.0397 = 0.120 2 = pg; + gn-le; x g'= 0.08 + 0.0379 x 1.5 2 0.08 + 0.0091 bd bd d 3:25 ; 0.0891 From Table 11, for m = 0.12 and q = 0.0891: R ; 0.318 For entering Table 12 determine 1 x (n-lZAi : 1 x 9 x 0.22 = 0.125 R bd 0.318 8 x 6.25 1 a x g' ; 1 X 1.5 : 0.756 d 0.318 6.25 From Table 12: z ; 0.377 determined as follows l + (n-leg x g; x (1- d' d) z ; 6 kbd kd - 0.167+ 0125x 0.756 x 0.244 1 + [n-iz A; x (1 dd ) .5 + 0.0125 x 0.244 2 kbd = 0.167 + 0.023 2 0.190 x 0.377 0.5 + 0.003 0.503 From Table 13, for z = 0.377 and k = 0.318: J : 0.87 fg : 12M 2, 12 x 1,865 2,10,300 psi.- jdA: 0.87 x 6.25 x 0.40 (39) f.-.-_ & x k -_-_ 0,300 x 0.218=482 psi. 10 1-k 10 0.0 2 fs' is always less than n11 or less than 4,820 psi. Sec. 2613, H.B.C. : allowable f, - 20,000 psi. allowable fig - 1,200 psi. Shear ; wl ; 827 x 4.25 ; 1,760 lb. 2 2 Sec. 2618, H.B.C. V = V = 11760 = 40.4 1381. bjd 87x 0.87 x 6.25 Sec. 2613, H.B.C. gives allowable v z 0.02 f; = 60 psi. without web reinforcement. Stirrups shown in Fig. 17 are not necessary. Laundry and Breakfast Alcove Lintels Clear Opening is 5 ft. 7 in. Design span is 6 ft. 3 in. Suggested design by N.C.M.A. is shown in Fig. 18. The design will be investigated by the method sh wn on page 17, A.C.I. Design Handbook. Given: b g 8 in.; d g 6.25 in.; d' = 1.5 in.; AJ = 0.88 sq. in. A = 0.88 sq. in.; I1 : 10; W .1: 827 lb. p81“ 11). s' 11,-, 822,1 x 6.25" _-_ 4,030 ft. lb. m = nAs f (n-1)A§ = 10 x 0.88 + 9 x 0.88 ; 0.176 + 0.159 'bd bd 8 x 6.25 8 x 0.25 :.0.335 q ; pg; + (n-lZAi x d” ; 0.176 1 0.159 x 0.24 = 0.214 bd bd d Froszable 11, for m = 0.335 and q 3 0.214: k n O :- O For:entering Table 12 determine d (40) M... .IM... 02 vb. _ .m w .m .1. m .irz 3 22,832 _ Z. Z m: U.:: r - (E I 1.. 7m__mm.Tm_N_ . wifitw 4. : G 1 .ll. .1 )I. I'll II) ..... (c. i... )4 1+!) _ '4 4|! 4 . .11) ”Nu/1m ..(.mk\1N _ . _ . _ _ _ _ _ . «In I... (I II: II. (I I Ir i.| L LI _|L.. Ll. mm}... W a .. (iii)!!! (lit- 1) 1.. 1 ..® ii .I‘ Il‘l-rbllalll.|.|-. . .l v 4 S.r.‘l \ (I! (41) x (n-lZA; : l x 9 x 0.88 0.396 1 E bd 0.4 8 x 6.25 1 k X 0.24 z; 0.60 + (n-lZAE x g; x (l-gl) kbd .kd kd 1 (n—lZAE x (l-di) kbd kd 0.5 + 0.159 From Table 13, for z : 0.397 and k = 0.40: J ; l-zk z .60 x 0.40 O NIH (NH H CDC 0 o 1-O.4O x 0.397 = 0.841 81, 0.84 x 6.25 x 0:88 10,500 psi. L.» f. - is x k = 10,500 x 0.4 701 psi. n 1-k 10 1:573 f} is always less than nfc or less than 7,010 psi. Sec. 2613, H.B.C.: allowable f5 ; 20,000 psi allowable f< =.l,200 psi. Shear : El : 827 X 6.25 : 2,580 lb. 2 2 Sec. 2618, H.B.C. V :.;Z_ 2.580 = 61.4 psi. bjd 8 x 0.841 x 6.25 n Sec. 2613, H.B.C. gives allowable v z 60 psi. without web reinforcement. Since our span is 5 in. less than the span of the suggested design the stirrups will be considered as unnecessary. Kitchen Lintel Clear Opening is 5 ft. Design span is 5 ft. 8 in. Suggested design by N.C.M.A. is shown in Fig 19. The design will be investigated by the method shown on page 17, A.C.I. ’ (42) Design Handbook. Given: A; =_0.22 sq. in.; n =_10 M = 827 x 5.67‘=_3,320 ft. lb. _8_ m ; £51 + gn—lQAE = 10 x 0.62 + bd bd 8 x 6.25 g 0.144 q c EA; + gn-lZAé x g'; 0.124 1 bd bd d = 0.124 1 0.005 a 0.129 From Table 11, for m = 0.144 and q = 0.129: k For entering Table 12 determine From Table 12 + iszlléi X EL X (l-QL) kbd kd kd . 91-139,; x (riff kbd Rd 0.5 + 0.039 From Table 13, for z 0.354 if: = 12314 JdAJ 12 x 3,320 0.868 x 6.25 x 0.62 ; 11.860 x 0.384 10 0.016 f‘-§.Xk n l-k Sec. 2613, H.B.C.: 'allowable g. allowable f. Shear ; l: 8 E 2 2 0.354 and k i d A5 1.5 in.; = 0.62 sq. in. 9 X 0.22 = 0.124 + 0.0198 8 x .25 0.0198 x 0.24 0.384 1 x gn-lzgg = ‘1 x 9 x 0.22 = 0.103 k bd 0.58E ’“56““ l X g': 1 X 0.24 = 0.625 k d 07384 x 0.625 X 0.375 0.375 0.384: j = 0.868 11,860 psi. u 20,000 psi. 1,200 psi. 37 X 5067 = 23340 lb- (43) Sec. 2618, H.B.C. -‘ V = _X_ z 2,540 = 53.8 psi. bjd 50 x 0.868 Sec. 2613, H.B.C. gives allowable v = 0.02 fl; ; 60 psi. without web reinforcement. Stirrups shown in Fig. 19 are not necessary. Dining Room Lintel Clear opening is 8 ft. 6 in. Design span is 9 ft. 2 in. The N.C.M.A. gives no suggested designs for Spans over 7 ft. Design will be by the method on page 9, A.C.I. Design Handbook. Given: £3: 20,000 psi.; n g 10; f; = 1,200 psi.; b = 8 in. g 6.25 in.; d' z 1.5 in. d M g 827 x 9.167‘ ; 8,680 ft. lb. _8_ From Table l, for 20,000/lO/l,200: k z 197 From Table 4, for b x d = 8 x 6.25: F a 0.026 then M ; 8.68 KF a 197 x 0.026 = 2212 M - KF = 3.56 Compressive reinforcement is required when (M - RF) is positive, since this is the residual moment not taken by the concrete. From Table 7, for 20,000/10/1,200 and g' g 0.24 d 0 II is (n-l)(l-gm)(k—gf) d d = 20,000 x 9(1-0.24)§0.375-0.24) 12,000 n (l-k) _ 12,000 x 10(1-0.375I 180,000 x 0.76 x 0.155 3 0.246 120,000 x 0.625 (44> mZoW 02 mi; 83.7.: 2,002.2, kiwi/10 mm OJ Juli/i: 9.83232 5.00m 762.25 1%---.17-1.-- I. 1. ;..._ ¥_¥_zmfim_ mrj - wvmmo .0 1&4 . Ci-.. rHMIr-M- (45) M-KF = 3.56 - = 2.32 sq. in. cd 0.246 x 6.25 A; From Table 1, for is = 20,000: a z 1.44, therefore #5 z’é_ _ 8.68 g 0.967 Sq. in. ad 1.44 x 6.25 With u ; allowable bond stress, computeZi-z 8 000 V select 7ud bars from Table 5, and check width of web required to accommodate bars from Table 6. Sec. 2613, H.B.C., u = 0.04 f; = 0.04 X 3,000 3 120 p81. v -.-. 11.1. = 5.3.81 x 9.167 -.- 3.780 lb. 2 2 = 8,000 x 5,78 = 5.78 in. 7 x 120 x 6.25 From Table 5, use two 1 l in. Square bars at the tOp of the lintel and two 1 in. round bars at the bottom of the lintel 8 to give 3.74 sq. in. andZi’g 14.5 in. From Table 6, minimum web widths for these two combinations are 8.5 in. and 8 in. respectively. Jith Special anchorage minimum web width is 8 in. In accordance with Sec. 2619, H.B.C., the two 1 l in. bars will be bent down to form a semi-circular hook of min- imum radius of 4 bar diameters. Sec. 2618, H.B.C. V = ;Z_ = 3,780 = 86.5 psi. bjd 50 x 0.875 Sec. 2613, H.s.c. gives allowable v = 0.02 f; = 60 psi. without web reinforcement. Stirrup design will be as shown on page 30, A.C.I. Design Handbook. given 1 v' = 26.5 psi.;. b = 8 in.; d = 6.25 in.; iv = 16,000 psi. f2 ; 3,000 psi.; ho. 6 wire U-stirrups; 40-: 0.0648 sq. in. - 26.5 S = 1.41 4758-8675 From diagram 17, for fv'= 16,000 and No. 6 wire U-stirrups: A,f,.; 1,040 l:V'b=2.65X6.25:O.16 s Ava. 1,040 N ; 6S 1 z 6 x 1.41 x 0.16 z 2 stirrups s S - 1. 5 X 1. 41 Index - 1.5 13.2 T 0.16 S From diagram 17, 2 stirrups are needed. The report of the A.C.I. Joint Committee on Sta dard Specifications for Concrete and Reinforced Concrete gives the spacing of vertical stirrups as s = Avflwad, where V' = v'bjd : 26.5 x 0.875 x 50 ; 1,160 lb. V S = 1,040 x 0.875 x 6.25 = 4.9 in. 1,160 3 : 1.41 X 12 = 15.9 in. Use 7 Stirrups, Spaced 2,3,3,3,3,3,3 in. as Shown in Fig. 20. Stirrups are to be securely fastened to compression steel. Gallery Window Lintels Lintels exterd over two double window units and have a 8 in. bearing width at each end. Clear Opening of each double unit is 10 ft. 8 in. Two identical designs will be used to eliminate the necessity of additional steel at the top of the (47) lintel to take care of the negative mdment whichvvould result with a lintel continuous over two spans. Design Span is 11 ft. 4 in. Design will be by the method on page 9, A.C.I. Design Handbook. Given: f5 = 20,000 psi.; n ; 10; f; g 3,000 psi.; b = 8 in. f§=.1,200 psi.; w = 821 x 11.335= 13,260 ft. lb. From Table l, for 20,000/10/1,200; k = 197 From Table 4, for b x d z 8 x 6.25: F = 0.026 KF : 197 X 0.026 = 5.12 M-KF : 8.14 Compressive reinforcement is required since (m-KF) is positive. From Table 7, for 20,000/10/1,200 and g' = 0.24: c = 0.246 d A; = ‘,’:_pr = 8014 = 503 Sq. in. 0.246 x 6.25 cé‘ This would require 4-1: in. square bars and a minimum web width of 9 in. without Special anchorage or 8.5 in. with Spec- ial anchorage. Redesign with an 8 in. width and 11 2 in. depth. w a 827 + 8_ x 4_ x l x 150 ; 860 lb. per ft. 12 12 From Table 4, for b x d = 8 x 10.25: 4F - 0.0705 M .-_., 860 x 11332": 13.80 -8_ KF = 197 x 0.0705 15.90 (I 1"IE-KF = - 0.10 (48) Compressive reinforcement is not required since (A-KF) is negative. From Table l, for f, = 20,000: a = 1.44, therefore A: = fl- : 1308 2 00935 Sq. in. ad 1.44fix 10.25 See. 2613, H.B.C., u z 0.04 f; g 120 p81. V W1 : C80 X 11. 33- 4 ,870 lb. :w— 2 2F.- 8 000 v 8,000 x 4.87 z 4.51 in. 7ud 7 x 120 x 10.25 From Table 5, forzig 4.51 in. and-As = 0.935 sq. in. Two 8 in. Square bars give A5 = 0.94 sq. in. andggg 6.4 in. From Table 6, minimum web width is 6.5 in. Sec. 2618, H.B.C. 03 V = _E_ g 4 1870 :6 bjd 8 x 0. 875 x 10. 25 psi. Sec. 2613, H.B.C. gives allowable v = 0.02 f; = 60 psi. without web reinforcement. Stirrups are needed. Stirrup design will be as Shown on page 30, A.C.I. Design Handbook. Given: v' 2 8 psi.; b a 8 in.; tv. 16, 000 psi.; f2: 3,000 psi. No. 6 wire U-Stirrups; A”: 0.0648 sq. in. s 2 5 s = 0.666 ft. 5.66 08 A,£,= 16,000 x 0.0648 =,1,040 1 =_ V'b : 8 X 8 ; 0.0615 S Ayfv. 1,040 N H 68(1) =6x 0.666 x 0.0615 g 1 stirrup 8 Index = 1.5 S = 1.5 x 0.666 ; 16.3 1 0.0615 8 (49) From diagram 17; 1 stirrup is required. The A.C.I. Joint Committee gives the spacing of vertical sitrrups as s = A,f¥fld.where V' = v'bjd ; 8 x 0.875 x 8 X 10.25 g 573 lb. V 11040 x 0.875 x 10.25 : 16.3 in. 8 x 0.875 x 8 x 10.25 m N U) N Use 5 U-stirrups spaced 2,3,3,3,3 in. It will be necessary to use two 2 in. round bars at the t0p of the lintel to anchor 8 the stirrups. Design is shown in Fig. 21. Living Room Lintel Clear Opening is 20 ft. Design Span is 20 ft. 8 in. Design will be by methods of pages 8 and 9, A.C.I. Design Handbook. Given: 13,; 20,000 psi.; n _ 10; I“: 1,200 psi.; b = 8 in. fl; = 3,000 psi.; w 827 lb. per ft. The value of w is only approximate since it is the loading due to comained dead and live loads on a lintel 7 2 in. deep. t M = 827 X 20.667 = 44,100 ft. lb. _8_ From Table 1, for 20,000/10/1,200, K = 197 K 197 From Table 4, select b x d = 8 x 18.5 (F = 0.228) An overall depth of 20 in. would be required with no com- pressive reinforcement. (50) .fl .4. aw 0 Z J I _ 9 ti.» 2.2: mm .010 {/1 ml. 0 I 0.2;: 7zflflmim an... -m (51) Redesign with a depth of 15 % in. - w - 827 +‘§_ x 8_ x l x 150 = 894 lb. per ft. 12 12 47,750 ft. lb. M .-_ 894 x 20.667z “8— From Table 4, for b x d = 8 x 14.25: F ; 0.1355 KF : 197 x 0.1355 = 26.7 M-KF = 21-05 Compressive reinforcement is required since (M-KF) is positive. From Table 7, for 20,000/10/l,200 and g'- 0.105: c - 0.59 d A' - M-KF - 21.05 - 2.5 Sq. in. Cd 0059 X 14025 From Table 1, for f - 20,000: a - 1.44 A " :_'I__ - 47.75 " 2033 Sq. in. ad 1.44 x 14.25 Sec. 2613, H.B.C., u - 0.04 f‘ - 120 psi. v - g; - 894 x 20.667 - 9,230 lb. 2 2 g - 8,000 v - 8,000 x 9.23 - 6.18 in. ' 7ud 7 x 120 x 14.25 From Table 5, use two 1 l in square bars at both the t0p and bottom of the lintel to give A - 5.08 sq. in. andzig 18 in. From Table 6, minimum web width is 8 in. with Special anchorage. Sec. 2619, H.B.C. requires that the two 1 1 in. t0p bars be 8 bent down to form a semi-circular hook of minimum radius of 4 bar diameters. The bottom bars will be bent up to the same radius. (52) Sec. 2618, HQBQC. - V = _y_ = 9,230 = 92.5 $381. bjd 8“x 0.675 x 14.25 Sec. 2613, 5.5.0. gives allowable v ; 0.03 fig = 90 psi. with no web reinforcement cut with anchorage of longitudinal reinforcement. Use 5 U-stirrups made of No. 6 wire and Spaced at 2.3.3,},3 in. as shown in Fig. 22. Investigation of Window Areas ‘he next step is an investigation to see if the windows will withstand the force of the wind. Don Graf's Technical, Sheets give a formula to examine a glass pane on the pressure side for strength as PA : 3.48 Mt‘F where M - modulus of rupture which is taken as 6,000 pounds per sq. in. The formula now becomes approximately 21,000 t‘F in which AS P the pressure in pounds per sq. ft. the thickness in inches ('9' [I z;- u the area in Square feet the factor for ratio of width to height of the pane t ‘I U) II Safety factor. Safety factor of either 5 to 10 is recommended, depending on glass application. For example, for glass subjected to pressures less than 40 pounds per square inch, a safety factor of 5 is used. When the pressure exceeds 40 psi., a safety factor of 10 arbitrarily is used. (53) Ratio Factor Width-Height (F) 10:10 (Square) 1.000 9:10 1.005 8:10 1.02 7:10 1.07 6:10 1.14 5:10 1.25 4:10 1°45 3:10 1.8 2:10 2.6 1:10 5.0 The Libbey-Owens-Ford Glass Also makes a tempered plate glass called Tuf-flex which has a modulus of rupture of 30,000 psi. Master Bedroom windows This window is shown in Fig. 13. The center unit is a Thermopane unit 36 in. or 3 ft. wide and 48 E in. or 4 ft. high. Thermopane is a factory-built transparent insulating glass unit for windows composed of two or more lights of glass separated by l in. or % in. of dehydrated air space and hermetically sealed around the edges at the factory with a patented metal-to-glass bond. The bond between the metal seal and the glass will withstand a shearing force greater than 1,000 psi. From Table 2 with wind from the south there is 41.2 psf. ‘ (54) suction on the north side. If we assume that glazing methods will anchor the glass sufficiently to keep the panes from popping out, we can use this pressure to investigate. The thickness necessary to withstand the pressure is eXpressed by 21,000 F V21.000 x 1.07 32 t =J PAS = E1.2 x 12 x 10 — 0.468 =71; in. The thickest glass made for ThermOpare units'is : in. The Z_ 2 DJ in. sheet glass is double strength. With double strength glass the thickness required would be 15 in. This glass can be considered as safe since the factor of safety might have been taken as 5 with the consideration that the pressure was so near the 40 psf. recammended for a safety factor of 5. The swinging easements on each side of the picture window have lights 16 57 in. or 1.36 ft. wide and 12 in. or 10 _ 1 ft. high.‘ A = 1.36 sq. ft. 1 = 7.4 F g 1.07 with width and height dimensions 1.35 10 reversed in. t ; PAS : $1.2 X 1.36 X 10 ; 0.157 ; £1,000 F V21.000 x 1.07 3 in. sheet glass satisfy the requirements for area and vTfi N thickness. The swinging easements are double glazed. The double glazing is essentially an inside storm window that can be replaced by screens in the summer. This is a very desirable feature from the standpoint of reducing heat losses thru the glass areas. (55) Utility Room Window ~ This window has 6 lights 16 in. or 1.36 ft. wide and 2, lo 1 ft. high as shown in Fig. 13. Again we can use 26 in. thick 1 sheet glass. It is a double glazed horizontal gliding unit. Laundry and Breakfast Alcove Windows Fig. 11 shows these double glazed horizontal gliding units. Both windows have 6 lights 27 g in. or 2.3 ft. wide and 1 ft. high. Area of 1 light is 2.3 Sq. ft. 1 = 4.34 F ; 1.45 with width and height dimensions reversed 2.3 10 From Table 2, with wind fram the south, the maximum velocity pressure on the east wall is 44.8 psf. suction. Again assum- ing that the glazing methods will prevent the panes from p0pping out, the 44.8 psf. can be used as a design factor. t: : PAS :‘ 4.8x 2.} X 1’3 2 0.184 : jg in. 21,000 F 21,000 x 1.45 1 Use 33 in. thick sheet glass. 1 Kitchen Window This double glazed horizontal gliding window is pictured in Fig. 11. There are 4 lights 24 g in. or 2.05 ft. wide and 10 in. or 0.833 ft. high. The area of one light is 1.71 sq.ft. 0.8%: ; 4.06 F ; 1.45 with width and height dimensions 2. 5 10 reversed t -.-..‘ PAS =F+4.sx 1.7 x10 _._ 0.1595 = 1 21,000 F V21,000 x 1.45 [__.l in. 31 'Use 26 in. thick sheet glass. 1 (56) Dining Room Window - The unit as shown in Fig.11 consists of a picture window of ThermOpane with swinging casement windows on each side with double glazing. The swinging casement lights are 162 in. or 13.5 ft. wide and 1 ft. high. : 4.8 x 15.5 x 10 = 0.164 = 11 in. 21,000 F 21,000 x 1.07 64 t - PAS q‘ Use 26 in thick sheet glass 1 The picture window is 55 in. or 4.56 ft. wide and 60 % in. or 5.07 ft. high. Area is 5.07 x 4.56 = 23.1 Sq. ft. 4.56 z _9__ F = 1.005 5.07 10 t = PAS = 44T8 x 23.1 x 10 = 0.686 in. = 11 in. 21,000 F 21,000 x 1.005 16 The thickest glass made for ThermOpane units is i in. Neither the double strength sheet glass nor the Tuf-flex heat treated glass are fabricated in this width. The i in. plate glass will stand P a 21,000 t‘F = 21,000 x 1.005 = 11.44 psf. AS 162x 23.1 x 5 The wind velocity in miles per hour (V) which is eqivalent to given pressures in pounds per square foot may be found from the formula Pressure :3: 0.004'V‘ V g P ; 11.53 ; 53.4 miles per hour .004 0.004 This window can be expected to rupture. From Table 2, it can be seen that the window will pop out with wind from the south with 44.8 psf. suction and also pop out with wind from the (57)- north with 41.4 psf. suction. With wind from the east, the window will rupture inward under a pressure of 25.9 psf. Living Room Jindow This window is shown in Fig. 10. The center unit is 75 9 in. or 6.29 ft. high and 48 11 in. or 4.05 ft. wide. 16 16 The unit is a ThermOpane unit and has an area of 25.5 Sq ft. With the wind from the north, the maximum velocity press- ure is 57.8 psf. suction. Assuming that glazing methods are adequate to keep the glass from pOpping out the 37.8 psf. can be used as a design criterion. 4.05 = 6.43 F = 1.14 with the width and height dimensions 5.29 10 reversed t :4 PAS : 37.8 X 25.5 X 5 :- O.458 :, 1.2 111. [21,000 F 21,000 x 1.14 32 Again the thickest glass made is 4 in. and double strength sheet glass and heat treated Tuf-flex are not fabricated in this width. With 5 in. plate glass, the window will stand p = 21,000 t‘F ; 21,000 x 1.14 = 11.73 AS 16 x 25.5 X 5 The approximate equivalent wind velocity is V :1 P ; 11.2% ; 54.1 miles per hour 0.00 0.00 From Table 2, it is apearant that with wind from the east the window would pop out with a velocity pressure of 32.6 psf. suction. With wind from the n rth, the window w uld again pop out with a velocity pressure of 37.8 psf. suction. With wind from the south, the window would rupture inward under a ‘pressure of 17.3 psf. ’ The companion units on each side of the picture window are dou le glazed horizontal gliding units. These are units having panes 30 g in. or 2.55 ft wide and 1 ft. high. Area is 2.55 Sq. ft. 1 g §.92 F - 1.45 with width and height dimensions 2.55 10 ' reversed t =.[ PAS =JEZ.8 x 2.55 x7: = 0.1255 = 1 in. [21,000 F [21,000 x 1:45 6 Use 23 in. sheet glass 1 Gallerv Windows As shown in Fig. 12, panes are 24 5 in. or 2.05 ft. wide 8 and 10 in. or 0.834 ft. high. These units are double glazed horizontal gliding units. The area of one pane is 1.7 sq. ft. F = 1.45 with width and height dimensions reversed With wind from the west a pressure of 62.1 psf. is exerted. t a" PAS :J62.1 X 1.71 X 10 = 0.188 g, 26 in. ’21,000 F 121,000 x 1.45 l Use 26 in. sheet glass 1 All small panes will withstand a 125 mph. wind. Picture windows on the south and west sides will rupture and have an equal chance of pOpping out. These windows should be insured with Windstorm insurance. Chimney Footing The footing will have its base five feet below the floor slab to duplicate the worst conditions that might be encount- ered such as soft soil bearing areas or a lepe that would require a fill to the level of the floor slab. Heights of the (59) concrete blocks will average approximately 40 pounds with mortar joints where concrete blocks are used below grade and cinder blocks above grade. Fireplace dimensions approximate 32 in. by 56 in. The footing is 8 in. thick and 8 in. wider than the width and length of the fireplace making its dimensions 40 in. x 64 in. Weight of the footing is 40 x 64 x 8 x 150 g 1,545 lb. 1,728 Above the footing are 4 tiers with 9 blocks around the circumference of each tier. The weight of these blocks are 40 x 9 x 4 g 1,440 lb. On these blocks is a 4 in. concrete slab which is the floor of the ash pit. This slab weight is :2 x_§6 x 4 x 150 z 622 1b. 1,728 Above the slab are 2 tiers of blocks with 9 blocks ar und the circumference of each tier. These blocks form the ash pit. One block will be left out of the outside wall to remove the ashes. fiei;ht of these blocks is 17 x 40 ; 680 pounds Assume that the weight of one half the flgor Span across the width of the f replace is Carried down to the footing. Then this weight is 6 x 56 x 9.5 x 150 ; 3,320 lb. 144 The tOp of the fireplace mantel is 6 blocks high. There are 5; blocks on the back side and the ends in each tier. They weigh (60) 5.5 X 6 X 40 = 1,320 lb. '- The smoke shelf weighs approximately 8 x 8 x 40 1 8 x 16 x 40 (150) = 444 1b. 1,72%?» 2 x 1,728 The precast concrete mantel and concrete forming the smoke chamber weigh 8 x 8 x 56 1 8 x 4 x 56 (150) = 560 1b. 2 x 1,728 1,728 There are ten more tiers to the tOp of the chinney. There are two 8 x 8 in. flue linings. One tier consists of 9 masonry units and 8 concrete bricks with 4 hollow units 4 in. thick. The bricks weigh 5.5 lb. The weights of these units are 10 x 11 x 43 1 10 x :.5 x 8 = 4,840 lb. For details of fireplace and chimney construction re:er to Fin. 38, Facts About Concrete Kasonry by the National Concrete Kasonry Association. The total weight bearing on the soil is 1,545 + 1,440 + 622 + 3,320 f 444 + 360 1 4,840 :12,571 lb. Sec. 2802, H.B.C. gives the allowable bearing capacity of soft clay, sandy loam or silt as 1 ten per sq. ft. This section also says that the footing shall be designed so that the allowable bearing capacity Specified for the particular soil shall not be exceeded. Pressure exerted on the soil is 12,571 ; 708 lb. per sq. ft. 3.35 X 5.55 The allowable soil pressure is not exceeded. Sec. 2622, H.€.C. gives the allowasle conpressive uuit stresses in footings as 0.25 f; without reinforcement. (61) This is 0.25 x 3,000 = 750 psi4 Pressure exerted on the foot- ing with gross area of each block equal to 128 sq. in. is 112026 = 906 p81. 9 x 128 This does not exceed the allowacle. Wall Footing The heaviest load that could be brought on the footing would ;e the case where the core areas of all clocks above the floor slab were filled with concrete. Roof slab has a dead weight of 70.3 psf. and a span of 21.58 ft. including the overhang. The floor slag has a scan of 19 ft. and a dead load of 75 psf. Load along a 16 in. length of wall is 1.33 x 70.3 x 21.58 + 8 x 16 x 7.5 x 150 + 75 x 1.33 x 12 2 144 2 + 6 x 32 = 3,152 lb. The footing extends out 4 in. on each side of the wall and is 8 in. thick. Its weight per 16 in. length is 16 x 16 x 8 x 150 = 178 1b. 1,725 The com ression in the footin: is P M 2:152 = 2496 p810 128 Sec. 2622, H.S.C. gives the allowable compressive unit stress as 0.25 fig without reinforcement. This is 0.25 x 3,000 = 750 psi. Pressure exerted on the soil is 13.152 + 17:1 x 144 = 1,875 psf. 16 x 16 ,. Sec. 2802, L.B.C. gives allowacle bearing capacity of soft clay, sandy loam or silt as 1 ton per sq. ft. This allowable bearing yu- capacity is not exceeded. (62) Forms for Roof Slab Forms for the roof slap will be designed as suggested on pages 50 to 60, ieinforced Joncrete Design. For soffit f?PmS x Z in. 3 4 0 use 1 x 8 in. square edge timber. Finished size is —\'] U1 *6 f 1') Live load of pouring crew «3 o d U) - w Pb 0 fit. of concrete slab \ T1 U1 rd (D Fab 0 Formwork, etc. 1 fw = :2 and l g S I 0 figs where S for a rectangle z 1 bh n .‘L n ml 010‘ adj z»: .-. fwbh 3‘ fiwis co monly taken as 1,800 psi. 93 :3 Cu .2 351‘; fwbd 12 '6" As the stringers will probably require closer Spacing than the sheathing, check their capacity first. Try 4 x 10 in. stringers T . 1 l , and compute the maxisum Spacing from wsl 12 = fun with a 19 ft. Span. 1553 x 19 x 19 x 1.5 = 1,800 x 56.41 s ; 1,800 x 56.41 = 1.21 ft. or 14.5 in. 155 x 561 x 1.5 The flexual stress of the decking, using F - wl for full continuity is 2- 155 x 1.21 ; 12 x g x g x % x fw f”: 155 x 1.46 x 96 = 202 psi. 2 x 9 (63) I‘ This is within the allowable so a 14 in. sgwaitg will is used. Ine decking can be left with Spaces between the boards since corkboard is placed on the decking before the concrete is poured. The ceiling forms are constructed 1 in. deeper than the thickness of the concrete slab. Before placing the reinforcing steel and the pouring f the concrete, one course of corkboard 1 in. thick shall be laid down in the Jorms. All transverse joints shall be broken and all joints made tight. Galvanized wire nails shall be driven obliquely into the corkboard-two to the square foot-the heads to be left protrud- ing approximately 1% in. The reinforcing steel shall then be put in and the concrete poured. With 4 x 4 in. shoring spaced 6 ft. apart, the load on a shore with the slab weighing 70 psf. and forms, etc. weigh- ing 25 psf., is 95 x 6 x 14 ; 665 lb. 12 Cross sectional area of the shoring is 13.14 sq. in. and the shoring is stressed to 665 ; 51 p81. $3.14 Allowable stress is 1,200 l- L g 1,200 l- 76 = 916 psi. 80 D 80 x 4 All lumber must be graded Ho. 2 or better to use these stresses. Shoring has 4 x 4 in. tee-head braced with 1 x 6 in. boards. In erecting provide a sill piece under the shores to distribute the weight over the Slab. 1 x 4 in. ribbon bracing is placed high enough above the floor to permit walking below. ‘ (64) Heat Transmission Losses Heat transmission of the various materials will be given as a matter of comparative efficiency as an insulating material. Heating and ventilating engineers measure heat losses in terms of ‘ritish thermal units (B.t.u.)-units used to measure heat in much the same manner as we use pounds to measure weight. U) tahdard wood frane construction with g in. plaster on metal H ath has a heat loss of 0.26 £.t.u. per sq. ft. per hour per degree F'difference in temperature. This figure, 0.26 B.t.u. loss, is considered by heating and ventilating engineers as not excessive as far as efficiency is concerned. In fact, it is quite genera ly accepted as the basing point on which efficiency or inefficiency of heat losses may be measured. The University of Kinnesota in tests conducted , found that filling the cores of a standard 3-oval core 8 x 8 x 16 in. unit with regranulated COFJ, rock wool, or similar granular or loose fill material with equivalent insulation values would reduce the heat losses through the plain concrete masonry wall substantially 50 per cent. The coefficient of heat transmission of U of a plain wall, either with or without plaster aid with loose or granular fill in cores is 0.20 B.t.u. per sq. ft. per hour per degree F‘ difference in temperature. Two coats of portland cement saint reduce the value of U on plain concrete masonry walls 0.03 B.t.u. Without cores fill- ed U is O.4O B.t.u. While a certain amount of wall insulation is desirable in all types of modern buildings, in general the importance of additional wall insulation in reducing fuel costs has been (65) over-emphasized. The increcsing use of larger window areas in buildings of all types is placing greater emphasis on window insulation than on wall insulation. If the window area of a building is increased only 3 per cent, the relative importance of wall insulation in reducing fuel costs is reduc- ed 6 per cent. This is true because ordinarily a greater amount of heat is 10‘: through the windows than through the wall of each buildirg. The coefficient U for d uble Thermcpane is 0.57 and U for double glazing is 0.61 As a comparison a single glass pane has a U of 1.07 Q.t.u. Heat losses of a 27 x 34 ft. uninsulated 1% story house built with light-weight concrete masonry exterior walls, furred metal lath and plastered interior finish, with 25 per cent of the wall area in windows and doors, will be approximately 35 per cent through windows and doors 30 " " " the roof 2O " " " the walls 15 " " around all wall Openings Heating engineers generally agree that little iS'b be gained by reducing the heat loss beyond 0.15 B.t.u. In other words, the cost of insulation to provide a wall with a heat loss below this point may be out of preportion to the fuel saved. To secure this condition in E in. concrete masonry or exterior frame walls requires the addition of l in. of rigid insulation or its equiValent. This amount of insulation in the above typiCal house would reduce the actual fuel bill only I 8 per cent. (66) Providing storm windowstand doors for this house would reduce the fuel cost aoout 18 per cent and providing insulat- ion in the roof equivalent to 3% in. of rock wool or similar material would reduce the heatin; cost 15 per cent. The floor slab is poured on a layer of Foamglas 4 in. thick which has a thermal coefficient of 0.40 B.t.u. at 50 F. Foamglas is glass made in cellular form and cut into accurately sized blocks. Each cell is closed and impervious to air or water. Foamglas is permanent and fireproof and has a crushing strength of 150 psi. The roof slab with l in. corkboard and standard roofing has a heat transmission coefficient U of 0.20. The refrigerated cold storage room will require 4 inches of corkboard to keep the room at the recommended 30 F. FINIS _Bibliobraphy Molander, E.G., "Engineering Analysis of Wind Loads on Farm ?uildings," Agricultural Engineering, Jan. 1947. "Reinforced Concrete Houses," Portland Cement Association, 33 West Grand Avenue Chicago Illinois. ’ i. 3 "Building Code,” City of Houston, Texas. ' American Concrete "Reinforced Concrete Design Tandbook,‘ Institute, Detroit, hichigan. Sutherland, Hale and heese, Raymond C., "Reinforced Concrete Design," John Wiley & Sons, Inc. "Report of Joint Committee on Standard Specifications for Concrete and Reinforced Joncrete Submitting Recommended Practice and Standard Specifications for Concrete and Reingorced Concrete," third printing, Aug. 1944, American Concrete Institute, fetroit, Tichigan. "Don Graf‘s Technical Sheets on ThermOpane," Libbey-Gwen- ?ord Glass Company, Toledo 3, Chio. RiChELrt , F 0E Q , .JOOQVJor’th ’ :3 0.": o , ‘.P)0r11189n ’ AL. I? Q T: Q , "TeSt S of the Stability of Concrete masonry Walls," A.S.T.I. Proceedings, Vol. 51, Part 2. u ‘l I J:'.‘,~;v q < n J ‘.‘ \.\ Axanuayf -‘ra‘qP'T-V ff 4- . f?“ "'i-fcv. - McCall S E L. AH rh. B 4.. l 7 KATE UNIVERSITY I 1 ' ‘1 v?"