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by: Jaylan Rath


Jaylan Rath
Texas A&M
GPA 3.77


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Study Guide
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This 4 page Study Guide was uploaded by Jaylan Rath on Wednesday October 21, 2015. The Study Guide belongs to CVEN 444 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 61 views. For similar materials see /class/226127/cven-444-texas-a-m-university in Civil Engineering at Texas A&M University.

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Date Created: 10/21/15
Twoway Slab Example The floor system consists of solid slab and beams in two directions supported on 20 in square columns Use the ACT Code equations determine minimum slab thickness required for an interior panel Use fc 3 ksi fy 60 ksi 1 To meets the ACT Code minimum h I 800 0005 Minimum h n fy ACI equation 911 360005000 am 051 1i 5 but not less than ln 8000005f h g ACI equat10n 912 36000 5000 01 A The slab thickness need not be more than ln 8000005f h y ACI equat10n 913 36000 Where 1 Clear span in the long direction measured face to face of columns or face to face of beams for slabs with beams 3 The ratio of the long to the short clear spans 35 The ratio of the length of continuous edges to the total perimeter of the slab panel under 39 39 ocm The average value of on for all beams on the sides of a panel a The ratio of the flexural stiffness of a beam section Ed lb to the flexural stiffness of the slab Ecs ls bounded laterally by the centerlines of the panels on each side of the beam E I a cb b ECSIS lb The gross moment of inertia of the beam section about the centroidal axisthe beam sectioOn includes a slab length on each side of the beam equal to the projection of the beam above and below the slab which ever is greater but not more than 4 times the slab thickness Is The moment of inertia of the gross section of the slab Therefore it is required to determine lb Is and on for the beams and slab in the long and short directions 2 The gross moment of inertia of the beam 11 is calculated for the section in the figure 424 quotquot7724 r g 3 A A v A A4 l 2039 E E g 7 wt V 55ij 1 LJ V l 39r 77 NV if 1 A A V iv 887 7 n r 7 v Section BB T The beam and extension of the slab on each side of the beam x y but not more than 4 times the slab thickness Assume h 7 in to checked later then xy22in7in15inlt47in28in therefore be 16 in 2 15 in 46 in and the T section is shown Determine the centroid of the section by taking the moments about the top of the ange The location of the centroid of the ange 7 in 2 35 in and the location of the centroid of the web section is 7 in 15 inZ 145 in Area Width depth Area yi yiAi D d2A Flange 46 7 322 35 1127 13148 47 710544 Web 16 15 240 145 3480 4500 6302 953313 562 4607 58148 166386 Ybar 81975 in I 22453 in EMAi 4607in3 EAi 562in2 7 82in 1 ZIZdZA 58148in 166386inquot 22453in 31 1 The moment of inertia of the slab in the long direction is Is i bh3 Where b 20 ft and h 7 in 12 3 Is 1th 20ft 121n 71n 6860in4 12 Calculating the ocl in the long direction E I 39 4 1 ch b 224531n 2327 E05 6860 in 41 1 The moment of inertia of the slab in the short direction is Is i bh3 Where b 24 ft and h 7 in 12 3 Iszibh324ft 121n 71n 28232in4 12 Calculating the OLE in the short direction E I 39 4 a 2 ch b 224531n 2272 ECSIs 8232 in4 5 The average ocm for the two ocl and ocs alas 327272 am 272 23 2 2 6 Calculate 3 and 35 values for the interior panel Llc 24 20m lftl2m l zll 2 2 2233ft2122 51 Lsc 20 20m lftl2m 1833 2 2 2 Continuous Sides 2 10 Total perimeter 7 Determine hmin using the equation lnl 2233 ft gtllt 39 gtllt 39 2233 ft 121nft800 0005 60000 ps1 538in hmin 36000 5000 1223 05 1 1 1 122 must not be less than h 2233 ft 12 im 800 0005 60000 psi 36000 5000 1221 1 612in nor more than h 2233 ft 12 im 800 0005 60000 psi 36000 819in Minimum h 35 in Therefore h 612 in controls gt 35 in A slab thickness of 65 in or 7 in may be adopted


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