A typical reinforced concrete floor system is shown in the sketches below. Exterior Span Interior Span Exterior Span. Beam Span.

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1 CE 331, Fall 009 Analyi of Reforce Concrete 1 / 6 Typical Reforce Concrete Builg Cat place reforce concrete tructure have monolithic lab to beam an beam to column connection. Monolithic come from the Greek wor mono (one) an litho (tone). Conequence of monolithic contruction clue: beam to column connection tranfer moment makg reforce concrete frame highly etermate, an the beam are "T hape" A typical reforce concrete floor ytem i hown the ketche below. Exterior Span Interior Span Exterior Span A Span lab pan A Plan View of Floor Sytem L n (clear pan) L n L n beam column Elevation View of Floor Sytem

2 CE 331, Fall 009 Analyi of Reforce Concrete / 6 beam pacg / beam pacg / t lab max flange with h b w beam beam pacg Section A A: Slab Elevation an X Section Placement of Reforcement Steel reforcement i place the tenion zone of reforce concrete beam, a icate by the crack pattern hown below. +'ve M teel -'ve M teel Reforcement Tenile Zone Factore Moment ue to Dea + Live Loa (M u ) The moment ue to factore ea an live loa mut be calculate with the ai of a computer program. Deigner oen ue the American Concrete Intitute (ACI) moment coefficient (hown below) which repreent the envelope of moment ue to ea loa plu variou live loa pan loa pattern. w u L n 14 M u w u L n 10 ACI Moment Coefficient

3 CE 331, Fall 009 Analyi of Reforce Concrete 3 / 6 The 3 Stage the Life of a Reforce Concrete Reforce concrete beam are analyze three ifferent way, epeng on whether the concrete ha cracke, whether the teel ha yiele, or whether the concrete i tree to it "non lear" range. 1. Uncracke compreion zone tenion zone reforcement neglecte. Service Loa ` flexure crack compreion zone concrete tenion zone cracke (neglecte) reforcement 3. Ultimate Strength concrete cruhe teel yiel compreion zone concrete tenion zone cracke (neglecte) Reforcement yiele

4 CE 331, Fall 009 Analyi of Reforce Concrete 4 / 6 Example Analyi of Ultimate Strength of a f c = 5000pi fy = 60ki Clear cover =.0 che aume tirrup are #4 bar w u = 1.50 klf b=1 h=0 30 Elevation View of A = 4 #7 bar =.40 Cro-Section of w u = 1.50 klf klf M u = C c T M u =.5 k (.5 k )(15 klf 15 ) (1.50 )(15 )( ) = Neutral Axi.003 c f'c 0.85f'c a=β 1 c a/ C c ε f f T Section A-A Stra Ditribution Actual Stre Ditribution Equivalent (Whitney Stre Block) Stre Ditribution Stre Reultant

5 CE 331, Fall 009 Analyi of Reforce Concrete 5 / 6 1. Calculate M n (nomal moment capacity) a M n = T ( ) T = A f (aume teel yeil, check later) y 1.1 Calculate effective epth () b φ bar / h =0 clear cover φ tirrup φ 4 7 / 8 = effective epth = h clear cover -φ bar tirrup = 0" " = Calculate of epth of tre block (a) from F 0.85 f 0.85(5 ' c H a b = A ki = 0, C ) a (1 c = T f y ) = (.40 )(60 ki ) a =.8 a ki.8 1 M n = A f y ( ) = (.40 )(60 )(17.06 )( ) M n = 188 k 1. Calculate φ (trength reuction factor) φ i a function of the tra the teel. If the tra the teel (ε ) i at leat.5 time the yiel tra (ε y ~= 0.00 = 60 ki / 9,000 ki = f y / E ), then the max. value of φ = 0.90 can be ue. If the tra the teel i le than 0.005, then the equation hown the figure below mut be ue. For beam, ACI require that ε be >= φ ε m for beam = ε

6 CE 331, Fall 009 Analyi of Reforce Concrete 6 / 6 Calculate the tra the teel (ε ) The tra the teel can be calculate ug imilar triangle an the tra itribution hown on pg ε = c an a = β c (from pg 4) ε = a β 1 f'c, pi β 1 <= 4, , , , >=8, For thi example, ε = ε = Therefore, (a) φ = 0.90 (ce ε > 0.005) (b) teel ha yiele (ε > 0.00 = ε y ) a aume at top of pg 5. Check flexure trength We want: φ M n M reuce nomal moment u factore moment φ M n = ( 0.9)(188 ) = 169. > = M u, OK