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by: Mrs. Estell Kuhic


Marketplace > Texas A&M University > Architecture > ARCH 614 > ELEMENTS OF ARCH STRUC
Mrs. Estell Kuhic
Texas A&M
GPA 3.72

Anne Nichols

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About this Document

Anne Nichols
Class Notes
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This 6 page Class Notes was uploaded by Mrs. Estell Kuhic on Wednesday October 21, 2015. The Class Notes belongs to ARCH 614 at Texas A&M University taught by Anne Nichols in Fall. Since its upload, it has received 27 views. For similar materials see /class/225803/arch-614-texas-a-m-university in Architecture at Texas A&M University.

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Date Created: 10/21/15
Concrete Beam Design composite of concrete and steel American Concrete Institute ACI design for maximum stresses lecwre limit state design service loads X load factors concrete holds no tension L 7 7 V failure criteria is yield of reinforcement 39 39 failure capacityx reduction factor concrete constructlon 39 factored loads lt reduced capacity concrete strength f C materials amp beams Concrete Construction Concrete castinplace low strength to weight ratio tiltup 1 relatively inexpensive prestressing TE 1 Portland cement J aggregate posttensioning 633 Sp rai l m water 1 hydration re resistant creep amp shrink Concrete Beams 3 Elements of Architectural Structures 82007abn Concrete Beams 4 Elements of Architectural Structures 8200 7abn Lecture 21 ARCH614 Lecture 21 ARCH 614 Reinforcement deformed steel bars rebar Grade 40 Fy 40 ksi Grade 60 Fy 60 ksi most common Grade 75 Fy 75 ksi n n US customary in of 18 0 longitudinaly placed bottom top for compression reinforcement spliced hooked terminated Comets Beams 5 Elements olArcvllectural Structures SQUWam Lecture21 ARCH 614 Transformation of Material n is the ratio of E s n E E1 effectivey Widens a material to get same stress distribution I39 quoti t 1 Concrete Beams 7 Elements aArcnrtecturastructures SQDWam Lecture 21 ARCH 614 Behavior of Composite Members plane sections remain plane stress distribution changes Concrete Beams 6 Elerrems orArcrntecturastructures 52007abn Lecture 21 ARCH 614 Stresses in Composite Section With a section E E transformed to one n 2 quoteel material neW E1 Econcrete stresses in that material are f 2 My determined as usual 5 I f d stresses in the other mm mm material need to be Myn adjusted by n ft 2 I transformed Concrete Beams 5 Elerrems orArcrntecturastructures 52007abn Lecture 21 ARCH 614 Reinforced Concrete stressstrain s4 245mmquot Hem 4m W4 F Location of na ignore concrete below na transform steel same area moments solve forX 47H quotA1 bxg nAsd x0 Concrete Sam 44 5m mmmmmm Structure 520mmquot Lecture 24 ANCH a4 Reinforced Concrete Analysis 52mmquot for stress calculations steel is transformed to concrete concrete is in compression above na and represented by an equivalent stress block concrete takes no tension steel takes tension force ductile failure Concrete Seam 4o ammommmw Srucmre 52mmquot mm 24 21mm T sections n a equation is different if na below ange bfhf xe hjxehfbw X hanAsm 7X 0 Concrete Seam 42 ammommmw Srucmre 52mmquot mm 24 21mm AC Load Combinations Reinforced Concrete Design 14D F stress distribution in bending 12D F T 16L H b 085f c 05L or S or R XI 0 a al2 0 12D16L orSorR10L or08W h d V Rx r I I 12D16W10L05L0rSOrR ifs T T o 13920 1390E 1390L 03928 actual stress Whitney stress 09D 16W16H blOCk 090 105 16H can also use old AC fem Force Equations Equilibrium C 085 f cba 085m T C 085r C al2 quotquotquotquotquot quot al2 T Asfy a 1x C M Td a2 d a 1x C Where d depth to the steel na f C concrete compressive T With As T strength a s y a height of stress block 085fc39b b 39dth f t bl k f gee CDimitri if M lt Mn 09 for flexure y 39 y g M Tda2 Asfyda2 AS area of steel relnforcement concrete Beams 75 Elements ofArchltecturai structures 52007am concrete Beams 76 Elements of Archltectural structures 52007am Lecture 27 ARCH 674 Lecture 27 ARCH 674 Over and Underreinforcement overreinforced steel won t yield underreinforced steel will yield reinforcement ratio bd use as a design estimate to find Asbd max 3 is found with 3599 20004 not p13 AS for a Given Section cont chart method Wang amp Salmon Fig 381 Rn vs 3 My 1 calculate Iquot 2 bd 2 find curve forf cand fy to get7 3 calculate As and a simplify by setting h 11d AS for a Given Section several methods guess a and iterate 1 guessa less than n a 2 A 39 a J J 3 solve for a from Mu nAsfylda2 M a2 d7 x r 4 repeatfrom 2 unila from 3 matches a in 2 Reinforcement min for crack control required 3 39 A Z bd Fi not less than 39 A bd ti typical cover 15 in 3 in with soil bar spacing co ver SPECquot94 Approximate Depths Dams mm m Concrers Lecture


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