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I. j.I)QIII I'IIII’III.HM11...I..II‘I'III :I.‘J:I'Q:I.::IM Qi‘rfifia‘ II-1I-I)"A"I-I‘I.,I'I...‘°"_‘ .I_ ml! AM‘\A.“I‘Q.O‘._.i ¢__II_ _I WWII] datu.’ ‘ D . 0' Design and Cost Estimate of a Pedestrian Underpass . Under Saginaw Street, East Lansing, Michigan A Thesis Submitted to The Faculty of MICHIGAN STATE COLLEGE OF AGRICULTURE AND APPLIED SCIENCE by we Me Thatcher Candidate for the Degree 01 Bachelor of Science June 1938 ACKNO‘MEDGWT I am indebted to Professor C. L. Allen and Professor C. A. Miller of the Civil Engineering Department of Michigan State College for their valuable aid and advice, and to the East Lansing Engineering De- partment, the Lansing Engineering Depart- ment, and the Michigan State Highway De- partment for their splendid cooperation. i. 1 ‘5 3 1?} S) INTRODUCTION The School Board of the City of East Lansing, Nuchigan feel that there will be a need for a school in the northwest part of the city in the very near future. There are two possible sites for this school. One site is five acres of land in the southwest corner of the intersection of Saginaw Street and Harrison Avenue, which could be obtained at the price of $1500 an acre. The other site, on the north side of Saginaw Street, is five acres in the southeast corner of the former Inter City Golf Course, Which has been offered to the City of East Lansing gratis for use as a school site. However if the latter site were chosen it would necessitate the con- struction of a pedestrian underpass, because the majority of the school children would be coming from.the south side of Saginaw Street, unchigan trunk line route number seventy-eight, which carries all of the traffic from Flint, Saginaw, Bay City, and other northeastern Michigan cities to Lansing. The problem before the School Board then, is whether it is cheaper to take five acres of free land on the north side of Saginaw Street and build an underpass, or to buy the five acres of land on the south side in which case an underpass is not needed. In this paper, I am going to design and estimate the cost on a underpass under Saginaw Street to serve a school, if built, on the five acres of land in the south- east corner of the former Inter City Golf Course. The logical location for the underpass is at the Touraine Avenue crossing of Saginaw Street. After making a survey of the territory around this inter- section, I fixed the location of the underpass as shown on the map which is in the pocket on the inside of the back cover of this book. The sewer and.water main were located from maps in the East Lansing Engineering Department. They will be cut by the underpass. This will necessitate putting in a sump pump for the sewer. The water main can be lowered. The computations will be given in the first part of the book. The drawings of the completed design are in the pocket on the inside of the back cover of the book. The cost estimate, including cost of mater- ials, construction costs, and engineering costs, will appear in the second part of the book. DESIGN The underpass will slaps from a floor elevation at the south entrance of 52.0 feet at a rate of three per- cent to a floor elevation of 54.7 feet at the north entrance. It will be skewed by an angle of 30 degrees under Saginaw Street, but the entrances will be parallel to the Saginaw Street sidewalks. The inside dimensions will be 7 feet by 7 feet. A 2500 pounds per square inch concrete composed of one part cement, two parts of fine aggregate, and four parts of coarse aggregate will be used. The allowable unit concrete stresses in flexure, shear, and bond will be in accordance with the American Concrete Institution Specifications. These specifications will be followed likewise for the unit stresses in the reinforcement and will, in general, govern the design throughout. The dead load will consist of the weight of the structure, the weight of the pavement slab, and the weight the earth topping. Earth pressure will be acting on the side walls and will be calculated from.Rankine's formula P: c.whz/2 where the constant Ce equals 0.27. Weight of earth will be assumed at 100 pounds per cubic foot and weight of concrete 150 pounds per cubic foot. I For live load, I shall assume the H30 class of high- way loading. Therefore, I shall use a typical truck of 20 tons (as shown in figure la page 5). As only one set of wheels can be over the structure at one time, the live load on the structure will come from.the affect of a coup centrated load of 22,080 pounds, which is the load of one rear wheel of 16,000 pounds plus impact equal to 50/l+125 times 16,000 pounds. The effect of this load on the structure will be in accordance with the empirical formula for distribution of loads set up by the Michigan State Highway Department, which says, that for every foot of depth, from.the pavement surface to the structure, the load will be spread over a square area of depth plus three feet on a side. The effect of two trucks over the underpass at the same trme (figure 1 b) will give the maximum.loading and the effect shown in figure 2 page 0. From.this diagram.it can be seen that the section of maximum.loading'will be the one foot longitudinally of the underpass where the distribution of two wheel loads overlap. The live load on this section will be: azioso : 22 080 3 (area of distribution)2 1 2 -_T37ITS I 2 l700f/ft. for a length of 5.1 feet moving across the width of the underpass. This will be the live load used in design of the structure. The rigid frame type of underpass was selected be- cause it has proved to the most economical type as far as quantity of materials and excavation are concerned. L-v— £53000” LEO-7327 truck H‘EO Loading [—-—-—< raj .; w= Total weight or [r truck: 67 load. ,JL M. 68‘ __J vasé TK‘FV A __.__-+‘$.-_”_ _ ____ .-._-____A___,--_-As,.:_.____a___. u... /4 ’- 0 “a Tap/Cal Truck 0 r4 W . ' ‘0 a ‘1 i n ~° ,__L r F i l a, _,_.:‘_ i if i _ 41“ _ _ O / W O. W F19. I a. Clearances on 20 Pavement a l “In- 1 i : l I I l I L?“ ' W" I “L" ' V wt 1‘. __________ _ll 1 F . 1 l l l I ‘ (2:217? _—‘ — :— 5:4, 1. in ,3.— L§:0;,;_._ 6:0 fl ' a—o a_'-o l ' .' k._ v, /o' -o" ‘ ' L“ "w”,dt' ’ " ! ene h L : fol—OH . r‘ ' ' ‘ “ ’7 FI'C-J I h_ m were ‘W.muhl\ 1.. WNVUM’ WWVQLUBKD *0 10k. 03m; utmEmDpk “ h muted 0‘ w _ _ -t . 0.524 n: h. z w , \ Underpasses of this type have proved very successful in Grand Rapids and Lansing. The slope deflection method of analysis of rigid frames will be used as the basis of design. It consists briefly of a series of simultaneous equations set up and ‘solved, each equation expressing a relation between certain of the bending moments at the ends of the several members of the structure under consideration and giving as a result the value of all these end moments. The relation most commonly used, and the one I will use, is the equilibrium existing between all the and moments at any one of the rigid Joints. The determination of the shears and stresses, completing the solution of the structure, is a simple matter once these moments are- determined. The computations of the end moments are as follows: The general slepe-deflection equation is an.” a 2m: ( sen.” + on, - ea ) 3 on“, the end of the beam for which the moment is written being designated near end. 9 = slope at section designated 3 : modulus of elasticity x -‘.- 1/1. I = moment of inertia H = d/L d = deflection la 3 distance between near and far section on“, 3 the moment caused at the near and of an identical fixedeended beam.csrrying the same load as the given beam. The top slab will be assumed 12 inches thick, the side walls 10 inches thick, and the footing 12 inches thick. The loading will then be as shown in figure 3 page 9, this section being the one foot longitudinal section calculated as the maximum.loaded section. Refer to figure 3 throughout the calculations of the end moments. ab = 12" thick ad 3 10" thick 4 . ion L z 7.44' a : 7w L - 7' rab- .375 and: .375 kd = 5.75" kd = 2.53" .A.t - .0054 x 12 x 10 is, = .0094 x 12 x 7 .Aat : 1.13 age ins Aflt : 0080 3Qe ins I.b = 2597 lad = 1344 - 2397 - 1844 Kai: " “'77:;- Kad " ' 1844 7.44 sex a 2 x m 2: 2,500,000 - 15,520,000 There are no deflections to consider, so R- O in all OQBOIe : 1700 3 - 2 3 Cab 1ms7e44 521e3 fie“ + 4 I 1e3 ) + 2 400 (7ofiéd_.. 5525 - 1845 = 8470'# w imam sexism? in: _ I _ _l coxmeQN nomiQoaxhg 6 0 l (I'- \ . v I. . I . - Ill}... ..0\ I‘ll , .0 _ .5 ad I .v _ _, , , / I. a H . (. -ue, «dfihtmu PM. KOQtfmifi \ . f Km ti... L...\.J\\\ , no! I o a l , wm\.wo.n< 2.. --.1_\ _ a w l l 4 ._ a 4 14 \\ l _ _ n _ a _ ”I _ a H _ 7U if, s. a 5 z i a 1|._ .4* ,LL. 1 _—+ ’ .,. _: NZNQTQQ 661e5 I 7 661e5 I 7 cm1 e 12 + l5 s 555 + 509 - 595% ads , 551.5 x 7 + 551.5 x 7 = 555 + 455 . 549% 7 25156,, - 5470 5" Mad: .‘.’.ll!:K(1..2'566a + 0.686(1)... 595 Md = 25x(1.555,1 + 0.555at ) - 549 Mao I 221: G a. + 11,240 The above expressions contain two unknowns, ea ended . and two equations for finding these unknowns are obtained by the condition of equilibrium existing at each Joint. “no * Mad 3 ° Mda "’ Mdc 3 0 2.55 ea - .555d : T313050?" = .0005751 . - I 10 391 - - - —_——-—l———.-— -C. 5.5595 5, - 1.5045 6,, - .0015572 .4524. e a - 1.5045 e,1 -.0005225 5.1072 5 a - .0015795 9, .0005551 -10- 1.5045 5, + .4524ed . 0003911 1.5045 9. + 5.5595 ed " -.0015157 5.10729d ' -.0022045 6d : -.0004517 'Mad M:b = 15,520,000 (.0005551) . 5470 = ~5495'# 15,520,000 (1.55 x .0005551 4 .55 x 4.0004517) + 595 15,520,000 (.000207) + 595 a 5494'# -------- Check “dd 3 13,520,000 (1.36 x -e0004317 + e68 I e0003681)- 849 Mdo Design of Top Slab: 15,520,000 (-.0005555 ) - 549 s - 5405'} 15,520,000 (-.0004517 ) - 11,240 - 5405*# -----Check 555’ w ' X\\\/700#/fi.\\\\ /// //// //////// > 34.33ma [50"flf-‘t. b 3493”: < , . 3J2 7M547 _ 5623* 5823# Max. M.= 5525 x 5.72 - 400 x 5.72 x 5.72/2 - 1700 x 2.55 x 2.55/2 - 5495 - 21,552 - 2755 - 5525 - 5495 - 9,575'f d - effective depth = Tbti— K a 154 12 x 9 876 - W = - 16! x 12 60.2 7.8" Use 10" d d 8" would be okay, but a check for shear will show that web reinforcement would be necessary in this case. Therefore, I used 10' for d.which made it possible to get away from using web reinforcement as will be shown when I check for shear in the next step. Protective Covering 3 2" Therefore Total Depth 3 12" 511' 5L34' \\\\\/7oo#/ft\ \ . zz////// 250 0#/ft.////// C /50 “/55 > 7m94’ 7/55 # 4.450” Loading for Maximum.Shear v 3 2 + .44 v = 1455 + 5595 = 715555 v = 1315-— : .575 v 3 7186 " 7186 = 680‘ #/.Qe1ne 12 x .875 x 10 - 105 68.4 is less than 75 which is the allowable without web reinforcement, but with special anchorage. Therefore, I shall use special anchorage. -12- A“ (required) = pbd 3 .0094 x 12 x 10 = 1.13 sq. in. 'Bond required - ------ Perfimzier = 1115.55}. bond x J x d stress Total Perimeter : m ' 5.5 inches Try 3 - & inch round bars at 4 inch spacing Total Area = 1.33 sq. inches ------ 0.x. Total Perimeter '-'- 7.06 inches ------ 0.x. 'Shrinkage and Temperature Steel: ‘st 'pbd with p 3 .0025 A.t .0025 x 12 x 10 3 .30 sq. in. Use i inch square bars at 10 inch spacing Therefore for the top slab use: 12'' total thickness 10" effective depth 3 -'£" round bars for main reinforcement at 4' spacing special anchorage of main reinforcement if square bars at 10' spacing for shrinkage and temperature steel. Design of wells: 510/“f 11' 94.5%"1‘ $3495 A ~ H.1 = 5:75" AL’V/ I as “1" ‘a r r d- H4 #045” ’# 263.5%" +5403 59 '75 4* Loading on Wall Assumed a 10" well Calculation of the horisontal forces at the ends ( Hu and ab ). Summation of the moments at either end (a or b) equals zero. 75 I94.5:7x5.5+159x7/2x7/5+5495-5405 2315 d- 1544 + 3493 o 5403 7n n s27af -14- 7 11,1 94.5 x 7 x 5.5 e 159 x 5.5 x 14/5 4- 5405 - 5495 7 11,1 8 2515 + 5055 + 5405 - 5495 104.51E Hd Check: Summation of horizontal forces equal zero.. 94.5 x 7 + 159 x 7/2 I 1045 e 275 661.5 - 661.5 2 1323 1323 ----- _---- Check 1323 Maximum moment 3 5403'} at 12:15:35 = V52.9 = 5.5" Use 3}" protective covering Therefore total depth = 5.5 - 5.5 s 9.5 Use 10" wall effect depth 3 6%“ Check for Shear: Maximum 7 = 1045} 3 1045 _ 1045 ' m - 55:25— ' 15'3””‘1'1n' 15.31 is less than 50 the allowable. Therefore no web reinforcement or special anchorage is required. Alt 8 .0094 x 12 x 6.5 3 .73 sq.in. Use 2-{” round bars - 6“ spacing ”.3 = ;88 UQeine .......... beKe Perimeter 3 4.71 inches Check for Bond: ' - V Unit bond 'tr"' *Perimeter x if! 5* 3 1045 1045 e I e I e Egeg -15- Unit bond stress = 59#/eq.in. --------- 0.x. Shrinkage and Temperature Steel: ‘st 3 pbd = .0025 x 12 x 6.5 = .20 sq. in. Use 1}” square bars at 12" spacing Therefore-for the walls use:- 10' total thickness 6&5 effective depth 2.2" round bars for main reinforcement --- 6" spacing fi’ square bars at 12' spacing for shrinkage and temperature steel. Design of Floor Slab: 6975‘“ 5975” .59’ 7.44’ .59 5403””Q ' .2507”, 9 5403’ ’t // //// v 7‘.” /////7) Loading on Floor Slab Assumed 12“ Thickness -15- .Max. MI 5975 x 5.72 - 1750 x 5.52/2 x 5.52/4 25,951 - 15,551 9,090'# = 12 I 9 090 j _ = fl ‘1 VW - V55" 7‘45 Covering I 1' for drainage groove plus 2'' protective covering. 9 d 3 8' would be okay, but a check for shear will show that web reinforcement would be necessary in this case. Therefore, I used 9“ for d which made it possible to get away from.using web reinforcementas will be shown when I check for shear in the next step. With d = 9" The total depth ' 12' Check for Shear: Max. 7': 7186 (see diagram.for:max. loading of ab for shear --page 12) + 575 - 1750 x .59 = 6867# , 5557 __ 5557 3 ' - 12 x .575 x - ‘92:?“— 72°6fl'q'm' 72.6 is less than 75 which is the allowable without web reinforcement, but with special anchorage. Therefore, I shall use special anchorage. A.t (required) 8 pbd 8 .0094 x 12 x 9 = 1.06 sq. in. Bend required:. = 5557 = 5557 : Mal 14.1-15.5... m 954‘- 5°95" -17- Try 3-%‘ round bars at 4 inch spacing Total Area = 1.33 sq. inches - ------ - O.K. Total Perimeter = 7.06 inches - ------- 0.K. Shrinkage and Temperature Steel: Act 8 pbd with p : .0025 Aat = .0025 x 9 x 12 = .27 sq. in. Use f inch square bars at 10 inch spacing Therefore for the floor slab use: 12" total thickness . 9' effective depth 3-1" round bars for main reinforcement --- 4" spacing special anchorage of main reinforcement f“ square bars at 10" spacing for shrinkage and temperature steel. -13- Other features of the design: The stair slab for the entrance at each end will be eleven inches thick allowing an effective depth of eight inches with a three inch covering. The main reinforce- ment shall consist of three-quarter inch round bars at six inch spacing. The shrinkage and temperature steel shall consist of one-half inch square bars at ten inch spacing. The steps shall consist of safety tread iron grat- ing one to one and.one-quarter inch in thickness, and supported at each end by six inches of concrete to which it is to be bolted. The grates are to be close enough together to prevent a ladies heel from catching and causing damage. The rise shall be seven and one-quarter inches and the tread eleven inches. A railing consisting of two inch round pipe shall be placed on each side of the stairway. For safety purposes, a two-rail guardrail of the same material shall also be placed around the top of the entranceway and in front of the entrance at a distance of five feet. The latter guardrail is to keep children from.running at tap speed into the entrance and possibly falling down the stairs. A.sump compartment with an autmmatic sump pump to pump water frmm it was provided. The sump and.pump were necessary to carry the water from.a short sewer line, which was to be out by the underpass, underneith the underpass. It was also needed to care for drainage of -19- the underpass inside and out. Cast iron floor drains are to be placed at the bottom of the stairs at each entrance to keep the underpass dry inside. These floor drains will drain into the sump, where the water will be pumped into the sewer system of the city. Drain tiles are to be placed on each side of the underpass at the floor slab depth so as to prevent ‘water accumulation in the soil, which would increase the earth pressure considerably. They would also decrease frost action, which would produce cracks in -the walls that would allow'water to seep into the under- pass. These drains also deposit their water into the sump. The fixtures for lighting the underpass shall consist of the standard type for lighting underpasses. Lights are so placed in the concrete of the top slab as to be flush with the ceiling, providing maximum.head room.and still giving ample light. The drawing, showing the completed design, is in the pocket on the inside of the back cover of this book COST ESTIMATE Item. Quantity Concrete (as specified) 150 Reinforcing Steel 22,200 Earth Excavation 680 Gravel Backfill 125 Removing Pavement 37 Concrete Pavement Replaced 37 Removing South Sidewalk 9 Concrete Sidewalk Replaced 9 Relocating Saginaw Street Water Line Automatic Sump Pump 1 6" C.I. Pipe 27 5* 0.1. Pipe 15 5* Clay Tile 270 Cast Iron Floor-Drain 2 Stair Treads & Fittings 20 Floor Drain Grate 2 2' Pipe -- Handrails & 2-Rail ' Guardrails 218 Lighting Equip. Complete Bit..Material, Tar 1,200 . Sub Total Eng. & Contingencies (approx. 10%) Total - 21- Unit o.y. lbs. o.y. o.y. s.y. s.y. s.y. aeye each feet feet feet each each each feet 881. Unit Price $20.00 .05 .60 2.25 .75 4.00 .50 2.25 200.00 2.00 2.15 .60 4.00 11.00 14.00 .65 .12 Amount $3000.00 1110.00 408.00 281.25 27.75 148.00 4.50 20.25 150.00 200.00 54.00 34.40 162.00 8.00 220.00 28.00 141.70 100.00 144.04 6.241089 624.11 $6,866.00 Unit prices given in the eight issues of the hgineering News-Record from March 31, 1938 to May 19, 1938 along with a general knowledge of local prices were used as a basis for the unit prices listed in the estimate. These prices include material costs, trans- portation costs, and installation or construction costs. Therefore, the amount listed for each item is the total cost of that item as an integral part of the completed structure. -22- CONCLUSION The rigid frame type of underpass was used because it was economical. As can be seen from the cost estimate, the total cost of this underpass will be $6,866.00 which for an underpass of such size is a very reasonable figure. The School Board of East Lansing could save approx- imately $634.00 by taking the five acres of free land on the north side of Saginaw Street for a school site and building this underpass in preference to buying five acres of land on the south side where an underpass would not be required. Therefore, the School Board would be wise in accepting the free land and build this underpass. BIBLIOGRAPHY Sutherland, Halc and Clifford, Walter 77., Reinforced Concrete .DOIIE, John Wiley and Sons Inc., New York, 1926. Peabody Jr., Dean, The Desigl of ReinforceeL Concrete . Structures, John Wiley and Sons Inc., New York, 1936. Urquhart, Leonard Church and O'Rourke, Charles Edward, Desig of Concrete Structures, McGraw-Hill Book 00., Inc., New York, 1923. Hool, George, and Pulver, Harry, Concrete Practice, McGraw-Hill Book Co., Inc., New York, 1925. --------, Analysis of Rigid Frame Concrete Brim, Portland Cement Association, 1934. Engineering News-Record, Eight Issues March 31,1938 to May 19, 1938. -24- 5 / TE Qf“ ' EBQEQéfifl QC 557-) 04 crops 5 a p we P7ofiertymne H ‘ CROSS SECT/Ofi THROUQH CENTER OFHPROI-josgp UNDERPflQ-i Scale :— /"= /O'~0" C151g£xrrzlts LOCflTJON OF PROPOSED MgIN/TW 5777515 7‘ UNDE 315755 Efld 7‘ L flNS/Ng, MIC ngflN 5091.15:- /"=50’-o" flPR/L was WM. THflTCHER diam cz_