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# APPLIED ARCH STRUCTURES ARCH 631

Texas A&M

GPA 3.82

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This 3 page Class Notes was uploaded by Eleanora Ernser III on Wednesday October 21, 2015. The Class Notes belongs to ARCH 631 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/225806/arch-631-texas-a-m-university in Architecture at Texas A&M University.

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Date Created: 10/21/15

ARCH 631 Note Set 253 F2008abn Examples Foundations Example 1 PD 150k PL 100k For the 16 in thick 85 fl square reinforced concrete footing canying 150 kips dead load and 100 kips live load on a 24 in square column determine ifthe footing thickness is adequate for 4000 psi A 3 in cover is required with concrete in contact with soil Also determine the moment for reinforced concrete design 16 E 8 6quot 24quot square column SOLUTION 1 Find design soil pressure q i P17 12D 16L 12 150 k 16 100 k 340k 471k1l 2 crilical sacrum 1w moment critical semen in shear q 85M 34 2 Evaluate oneway shear at d away from column face ls V lt We clear d h cc distance to bar area centroid 4 l 139 l presuming two rows of4 bars spaced 1 inch apart to get started 30 c39w 1d magma mm a reinforcmg d 16 in 3 in soil exposure 0 5 in x 15 layers of4 s 1 in betNeen 1125 in V total shear q edge area oneway Shear Ilne V on a 1 ll strip qr edge distance 1 it twoway shear I39me v 471 W 85 11 2 112 1125 in1 1112 in 1 11 1089 k W oneway shear resistance 42 J7 bd for a one foot stn39p b 12in W 0752 4000 psD12 in1125 in 1281 kgt 1089 k 0K 3 Evaluate twoway shear at d2 away from column face ls V lt Wu H 9 b perimeter 424 in 1125 in 43525 in 141 in V total shear on area outside perimeter Pu q punch area 6 c Dian View v 340 k 471 k1123525in2111121n2 2993 kips W twoway shear resistance 1ij bud 075am psi141 in1125 in 301 k gt 2993 k 0K 4 Design for bending at column face M MAM2 for a cantilever L 85 it 2 11y2 325 it and w fora 111 strip qr 1 11 M 471 ksi1 11325 1122 249k11per11 ofwidth To complete the reinforcement design use b 12 in and d 1125 in choose p determine As nd if 41Mw gt M 1 ARCH 631 Note Set 253 F2008abn Example 2 P 100 kips Determine the depth required for the group of 4 friction piles having 12 in diameters if the l column load is 100 kips and the frictional resistance is 400 lbsftz SOLUTION The downward load is resisted by a friction force Friction is determined by multiplying the friction resistance a stress by the area F fASKIN The area of n cylinders is ASKIN n27rL Our solution is to set P g F and solve forlength 100k 400117132 4F 1257r12L1 1quot 12in 100011 L 2199 Example 3 p 300 kips Determine the depth required for the friction and bearing pile having a 36 in diameter if the column load is 300 kips the frictional resistance is 600 lbsft2 and the end bearing pressure allowed is 8000 psf SOLUTION The downward load is resisted by a friction force and a bearing force which can be determined from multiplying the bearing pressure by the area in contact F fASKIN qATIP 4 The area of n cylinders is Am 7rd Our solution is to set P g F and solve forlength 36in 1 1k 36in2 1ft 2 1k 300k600 7 Ir L W warm 7 r z 2 izm 1000m A 4 izm 100011 L226lft ARCH 631 Example 4 Determine the factor of safety for overturning and sliding on the 15 ft retaining wall 16 in wide stem 10 ft base 16 in heigh base when the equivalent uid pressure is 30 lbftj the weight of the stem of the footing is 4 kips the weight of the pad is 5 kips the passive pressure is ignored for this design and the friction coefficient for sliding is 058 The center of the stem is located 3 from the toe SOLUTION Note Set 253 F2008abn iilt gt1 15 fluid pressure 139 3 16 lt 1 gr gt l This is a statics problem there is no design of materials involved Overturning is determined by moments from acting forces and the moment from the resisting force Sliding is determined by the acting horizontal forces and the resisting sliding force which is determined by mulitplying a friction coef cient based on the materials in contact u by a normal force N F uN Find all unknown forces and draw the free body diagram with the weights at the centers of gravity of the stem and base The horizontal lluid equivalent pressure is a triangularly distributed load with the maximum distributed load equal to the density of water multiplied by the height wh 7H wk 30315ft1ft sm39p 450 lbft 4 O k V 5k Wbase N The horizontal force PH wL2 acts at a distance of 13 the height from the fat endquot of the triangle is 1k 1000 PH 4501b 3375k The vertical force from the maximum distributed pressure Pv wL over the right side of the base in the midde of 633 ft is 16in1ft 1k P 450117 1017317 V x f f 2 12171 100011 265k The total downward loads must be resisted by the normal force acting up N4k285k5k1185k F 0581185 k 69k Overturning requirement SF M mm M overturning 21572 The total resisting moment will be from those moments counterclockwise about 0 Mmng 4 k3 ft 5 k5 ft 285 k 10 ft 633 M 565 kft The overturning moment is only from the horizontal lluid force clockwise Moverlurning 3375 k5 ft 16 in1 ft12 in 214 kft 7 565 7 214M 264215 OK SF Sliding requirement SF Fhagantalrresist 2 125 7 2 1 a rig The total resisting force will be from those opposite the hydraulic force to the right Fresisling 69K The sliding force is only from the horizontal uid force to the left FSlldlng 3375 k SF 7 6 7 2042125 OK 3375k

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