twenty one service loads x load factors concrete holds no tension - PowerPoint PPT Presentation

E LEMENTS OF A RCHITECTURAL S TRUCTURES : Concrete Beam Design F ORM, B EHAVIOR, AND D ESIGN • composite of concrete and steel ARCH 614 D R. A NNE N ICHOLS • American Concrete Institute (ACI) S PRING 2019 – design for maximum stresses lecture – limit state design twenty one • service loads x load factors • concrete holds no tension • failure criteria is yield of reinforcement • failure capacity x reduction factor http:// nisee.berkeley.edu/godden concrete construction: • factored loads < reduced capacity – concrete strength = f’ c materials & beams Concrete Beams 2 Elements of Architectural Structures S2007abn Concrete Beams 1 Elements of Architectural Structures S2009abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 Concrete Construction Concrete • cast-in-place • low strength to weight ratio • tilt-up • relatively inexpensive • prestressing – Portland cement – aggregate • post-tensioning – water • hydration • fire resistant • creep & shrink http://nisee.berkeley.edu/godden Concrete Beams 3 Elements of Architectural Structures S2007abn Concrete Beams 4 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 1

Behavior of Composite Members Reinforcement • plane sections remain plane • deformed steel bars (rebar) • stress distribution changes – Grade 40, F y = 40 ksi – Grade 60, F y = 60 ksi - most common – Grade 75, F y = 75 ksi – US customary in # of 1/8 ”  • longitudinally placed – bottom E y E y – top for compression reinforcement         1 2 f E f E 1 1 2 2 – spliced, hooked, terminated... R R Concrete Beams 6 Elements of Architectural Structures S2007abn Concrete Beams 5 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 Transformation of Material Stresses in Composite Section • n is the ratio of E ’ s E • with a section n  2 E E   2 steel transformed to one n E 1 E E material, new I • effectively widens a material to get 1 concrete – stresses in that same stress distribution My material are   f c determined as usual I transforme d – stresses in the other material need to be Myn   f adjusted by n s I transforme d Concrete Beams 7 Elements of Architectural Structures S2007abn Concrete Beams 8 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 2

Reinforced Concrete Analysis Reinforced Concrete - stress/strain • for stress calculations – steel is transformed to concrete – concrete is in compression above n.a. and represented by an equivalent stress block – concrete takes no tension – steel takes tension – force ductile failure Concrete Beams 10 Elements of Architectural Structures S2007abn Concrete Beams 9 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 Location of n.a. T sections • ignore concrete below n.a. • n.a. equation is different if n.a. below • transform steel flange • same area moments, solve for x f f h f h f b w b w x            x h h ( ) 0 bx nA d x         f b h x x h b nA ( d x ) 0 f s  2  f f f w s 2 2 Concrete Beams 11 Elements of Architectural Structures S2007abn Concrete Beams 12 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 3

ACI Load Combinations* Reinforced Concrete Design • 1.4D • stress distribution in bending • 1.2D + 1.6L + 0.5(L r or S or R) 0.85 f’ c b • 1.2D + 1.6(L r or S or R) + (1.0L or 0.5W) a/2 C C x a=  1 c • 1.2D + 1.0W + 1.0L + 0.5(L r or S or R) d h NA A s • 1.2D + 1.0E + 1.0L + 0.2S T T • 0.9D + 1.0W actual stress Whitney stress block • 0.9D + 1.0E *can also use old ACI factors Wang & Salmon, Chapter 3 Concrete Beams 13 Elements of Architectural Structures S2007abn Concrete Beams 14 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 Force Equations Equilibrium • C = 0.85 f c ba • T = C 0.85 f’ c 0.85 f’ c • T = A s f y a/2 a/2 • M n = T(d-a/2) a=  1 x a=  1 c C C • where d – d = depth to the steel n.a. – f c = concrete compressive • with A s T T strength A f – a = height of stress block s y – a =  –  1 = factor based on f c 0 . 85 f b c – x = location to the n.a. – M u   M n  = 0.9 for flexure – b = width of stress block – M u =  T(d-a/2) =  A s f y (d-a/2) – f y = steel yield strength – A s = area of steel reinforcement Concrete Beams 16 Elements of Architectural Structures S2007abn Concrete Beams 15 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 4

Over and Under-reinforcement A s for a Given Section • over-reinforced • several methods – steel won’t yield – guess a and iterate • under-reinforced 1. guess a (less than n.a.)  0 . 85 f ba 2.  – steel will yield c A s f • reinforcement ratio y 3. solve for a from M u =  A s f y (d-a/2) http://people.bath.ac.uk/abstji/concrete_video/virtual_lab.htm A    ρ – s M     u a 2 d   bd  A f   – use as a design estimate to find A s ,b,d s y 4. repeat from 2. until a from 3. matches a in 2. – max  is found with  steel  0.004 (not  bal ) Concrete Beams 17 Elements of Architectural Structures S2009abn Lecture 21 ARCH 614 Concrete Beams 18 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 A s for a Given Section (cont) Reinforcement • chart method • min for crack control  • required 3 f – Wang & Salmon Fig. 3.8.1 R n vs.   c A ( bd ) s f M y n  n • not less than R 1. calculate 200 bd 2 bd s  A ( ) f 2. find curve for f ’ c and f y to get  y • A s-max :   a ( 0 . 375 d ) 3. calculate A s and a 1 cover • simplify by setting h = 1.1d • typical cover – 1.5 in, 3 in with soil • bar spacing spacing Concrete Beams 20 Elements of Architectural Structures S2007abn Concrete Beams 19 Elements of Architectural Structures S2007abn Lecture 21 ARCH 614 Lecture 21 ARCH 614 5

Approximate Depths Concrete Beams 21 Elements of Architectural Structures S2009abn Lecture 21 ARCH 614 6