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

A RCHITECTURAL S TRUCTURES : Concrete Beam Design F ORM, B EHAVIOR, AND D ESIGN • composite of concrete and steel ARCH 331 D R. A NNE N ICHOLS • American Concrete Institute (ACI) S PRING 2018 – design for maximum stresses lecture – limit state design twenty two • 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 Foundations Structures F2008abn Concrete Beams 1 Architectural Structures F2009abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 Concrete Construction Concrete Beams • cast-in-place • types • tilt-up – reinforced – precast • prestressing – prestressed • post-tensioning • shapes – rectangular, I – T, double T’s, bulb T’s – box – spandrel arch.mcgill.ca http:// nisee.berkeley.edu/godden Concrete Beams 3 Foundations Structures F2008abn Concrete Beams 4 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 1

Concrete Beams Concrete • shear • low strength to weight ratio – vertical • relatively inexpensive – horizontal – Portland cement – combination: • types I - V • tensile stresses – aggregate at 45  • course & fine • bearing http://urban.arch.virginia.edu – water – crushing – admixtures • air entraining • superplasticizers Concrete Beams 5 Foundations Structures F2008abn Concrete Beams 6 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 Concrete Concrete • hydration • placement (not pouring!) – chemical reaction • vibrating – workability • screeding – water to cement ratio • floating – mix design • troweling • fire resistant • curing • cover for steel • finishing • creep & shrinkage jci-web.jp Concrete Beams 7 Foundations Structures F2008abn Concrete Beams 8 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 2

Reinforcement Reinforcement • deformed steel bars (rebar) • prestressing strand – Grade 40, F y = 40 ksi • post-tensioning – Grade 60, F y = 60 ksi - most common • stirrups – Grade 75, F y = 75 ksi • detailing – US customary in # of 1/8”  – development length (nominal) • longitudinally placed – anchorage – bottom – splices – top for compression reinforcement http:// nisee.berkeley.edu/godden Concrete Beams 9 Foundations Structures F2008abn Concrete Beams 10 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 Composite Beams Behavior of Composite Members • concrete • plane sections remain plane – in compression • stress distribution changes • steel – in tension • shear studs E y E y         1 2 f E f E 1 1 2 2 R R Concrete Beams 11 Foundations Structures F2008abn Concrete Beams 12 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 3

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 transformed – stresses in the other material need to be Myn   f adjusted by n s I transformed Concrete Beams 13 Foundations Structures F2008abn Concrete Beams 14 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 Reinforced Concrete - stress/strain Reinforced Concrete Analysis • 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 15 Foundations Structures F2008abn Concrete Beams 16 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 4

Location of n.a. T sections • ignore concrete below n.a. • n.a. equation is different if n.a. below flange • transform steel • same area moments, solve for x f f h f h f b w b w   x      x h     h         bx nA ( d x ) 0 f b h x f x h b nA ( d x ) 0 f f  2  f w s s 2 2 Concrete Beams 17 Foundations Structures F2008abn Concrete Beams 18 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 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) C a/2 C c 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 20 Foundations Structures F2008abn Concrete Beams 19 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 5

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 c C a=  1 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 = 0.25            * 0.65 ( ) 0.65 –  1 = factor based on f c f 4000 0 . 85 f b       t y c 0.85   (0.05) 0.65 (0.005 ) c 1 y   1000 – c = location to the n.a. – M u   M n  = 0.9 for flexure* – b = width of stress block –  M n =  T(d-a/2) =  A s f y (d-a/2) – f y = steel yield strength – A s = area of steel reinforcement Concrete Beams 21 Foundations Structures F2008abn Concrete Beams 22 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 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     2 u a 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 ) – *with  steel  0.005,  = 0.9 Concrete Beams 23 Foundations Structures F2008abn Lecture 22 ARCH 331 Concrete Beams 24 Foundations Structures F2008abn Lecture 22 ARCH 331 6

A s for a Given Section (cont) Reinforcement • min for crack control • chart method  • required 3 f – Wang & Salmon Fig. 3.8.1 R n vs.   c A ( bd ) s f M y n  n • not less than 1. calculate R 2 200 bd  A ( bd ) s 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 25 Foundations Structures F2008abn Concrete Beams 26 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 Shells Annunciation Greek Orthodox Church • Wright, 1956 http:// nisee.berkeley.edu/godden http://www.bluffton.edu/~sullivanm/ Concrete Beams 27 Foundations Structures F2008abn Concrete Beams 28 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 7

Annunciation Greek Orthodox Church Cylindrical Shells • Wright, 1956 • can resist tension • shape adds “ depth ” • not vaults • barrel shells Concrete Beams 29 Foundations Structures F2008abn Concrete Beams 30 Foundations Structures F2008abn Lecture 22 ARCH 331 Lecture 22 ARCH 331 Kimball Museum, Kahn 1972 Kimball Museum, Kahn 1972 • outer shell edges Concrete Beams 31 Foundations Structures F2008abn Concrete Beams 32 Foundations Structures F2008abn aasarchitecture.com Lecture 22 ARCH 331 Lecture 22 ARCH 331 8

Kimball Museum, Kahn 1972 Approximate Depths • skylights at peak Concrete Beams 33 Foundations Structures F2008abn Lecture 22 ARCH 331 Concrete Beams 35 Foundations Structures F2008abn Lecture 22 ARCH 331 9