Dr. Hazim Dwairi. Example: Continuous beam deflection

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Leslie Gilmore
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1 Example: Continuous beam deflection Analyze the short-term and ultimate long-term deflections of end-span of multi-span beam shown below. Ignore comp steel Beam spacing = 3000 mm b eff = 9000/4 = 2250 mm Or 16(125) = 2300 mm Or 3000 mm Ł b eff = 2250 mm 537 mm 3f mm 2250 mm 300 mm 5f25 2f mm Data: f c = 28 MPa f y = 420 MPa γ c = 25 kn/m 3 Beam spacing 3000 mm Superimposed dead load (not including beam self weight) = 1.0 kn/m 2 Live load = 4.80 kn/m 2 (30% sustained) A s is not required for strength. 1

2 1- Minimum Slab Thinkness: Minimum thickness, for members not supporting or attached to partitions or other construction likely to be damaged by large deflections: h min = l/18.5 = 9000/18.5 = mm < Ł OK 2- Loads and Moments: Self weight = [(3000)(125) + (475)(300)]/10 6 x 25 = kn/m w d = (3)(1.0) = kn/m w L = (3)(4.8) = kn/m In lieu of a moment analysis, the ACI approximate moment coefficients (ACI 8.3.3) may be used as follows: Pos. M = wl n 2 /14 for positive I e and maximum deflection, Neg. M = wl n 2 /10 for negative I e. a. Positive moments Pos. M d = w d l n 2 /14 = (15.94)(9) 2 /14 = kn.m Pos. M L = w L l n 2 /14 = (14.40)(9) 2 /14 = kn.m Pos. M d+l = = kn.m Pos. M sus = (83.31) = kn.m b. Negative moments Neg. M d = w d l n 2 /10 = (15.94)(9) 2 /10 = kn.m Neg. M L = w L l n 2 /10 = (14.40)(9) 2 /10 = kn.m Neg. M d+l = kn.m Neg. M sus = (116.64) = kn.m 3- Modulus of rupture, modulus of elasticity, and modular ratio: f r = 0.7 f c = = MPa E c = 4700 f c = = MPa n = E s /E c = 200,000/24,870 = 8.0 2

3 4- Gross and cracked sections moment of inertia: a. Positive moment section: Gross section at mid-span: 537 mm y t = mm I g = x mm 4 Cracked section at mid-span: 2250 mm 125 mm 300 mm 2250 mm y cr 537 mm na s 300 mm 125 mm (2250)( y cr ) 2 /2 = 8(1470)(537 - y cr ) y 2 cr y cr = 0 3

4 Ł y cr = mm I cr = 2250(69.88) 3 /3 + 8(1470)( ) 2 = x 10 9 mm 4 b. Negative moment section: Gross section at support: I g = (300)(600) 3 /12 = 5.40 x 10 9 mm 4 Cracked section at support: 537 mm For A s = 2454 mm 2, A s = 980 mm 2, d = 537 mm and d = 63 mm, then (300)( y cr ) 2 /2 + (8-1)(980)( y cr - 63) = 8(2454)(537 - y cr ) y 2 cr y cr = 0 y cr = mm I cr = 300(191.79) 3 /3 + 7(980)( ) 2 + 8(2454)( ) 2 I cr = x 10 9 mm 4 (n-1)a s na s 300 mm 5- Effective moments of inertia: a. Positive moment section M cr = f r I g /y t = x x / ( ) = kn.m M cr /M d = 98.07/92.22 = 1.06 >1 thus, (I e ) d = I g = x mm 4 M cr /M sus = 98.07/ = 0.84 <1 thus, (I e ) sus = (M cr / M sus ) 3 I g + [1 - (M cr / M sus ) 3 ]I cr y cr 4

5 = (0.593)(1.156 x ) + ( )( x 10 9 ) = 8.00 x 10 9 mm 4 < I g M cr /M d+l = 98.07/ = <1 thus, (I e ) d+l = (M cr / M d+l ) 3 I g + [1 - (M cr / M d+l ) 3 ]I cr = (0.176)(1.156 x ) + ( )( x 10 9 ) = x 10 9 mm 4 < I g b. Negative moment section M cr = f r I g /y t = x 5.40 x 10 9 / (300) = kn.m M cr /M d = 66.67/ = >1 thus, (I e ) d = (M cr / M d ) 3 I g + [1 - (M cr / M d ) 3 ]I cr = (0.138)(5.40 x 10 9 ) + ( )( x 10 9 ) = x 10 9 mm 4 < I g M cr /M sus = 66.67/164.1 = >1 thus, (I e ) sus = (M cr / M sus ) 3 I g + [1 - (M cr / M sus ) 3 ]I cr = (0.067)(5.40 x 10 9 ) + ( )( x 10 9 ) = x 10 9 mm 4 < I g M cr /M d+l = 66.67/ = >1 thus, (I e ) d+l = (M cr / M d+l ) 3 I g + [1 - (M cr / M d+l ) 3 ]I cr = (0.02)(5.40 x 10 9 ) + (1 0.02)( x 10 9 ) = x 10 9 mm 4 < I g c. Average inertia values For prismatic members (including T-beams with different cracked sections in positive and negative moment regions), I e may be determined at the support section for cantilevers and at the midspan section for simple and continuous spans. The use of the midspan section properties for continuous prismatic members is considered satisfactory in approximate calculations primarily because the midspan rigidity has the dominant effect on deflections. Alternatively, for continuous prismatic and nonprismatic members, suggests using the average I e at the critical positive and negative moment sections. The 1983 commentary on suggested the following approach to obtain improved results: Beams with one end continuous:. = Beams with both ends continuous:. = ( + ) Where I m refers to I e at midspan, I e1 and I e2 refer to both ends of the beam 5

6 .( ) =0.85( ) +0.15( ) = ( ) =0.85( ) +0.15( ) = ( ) =0.85( ) +0.15( ) = Initial of short-term deflections: Δ = 5 48 M a is the support moment for cantilevers and the midspan moment (when K is so defined) for simple and continuous beams. = = = (Δ ) = 5 48 ( ) = Or = mm using avrg. (I e ) d = x mm 4 ( )(9000) =

7 (Δ ) = 5 48 ( ) = ( )(9000) =4.225 Or = mm using avrg. (I e ) d = x 10 9 mm 4 (Δ ) = 5 48 ( ) = Or = mm using avrg. (I e ) d = x 10 9 mm 4 ( )(9000) = (Δ ) = (Δ ) (Δ ) = =9.309 Or = = mm a. Allowable deflections: - For flat roofs not supporting and not attached to nonstructural elements likely to be damaged by large deflections: (Δ ) = 9000 =50 > For floors not supporting and not attached to nonstructural elements likely to be damaged by large deflections: (Δ ) = 9000 =25 > Ultimate long-term deflections: Using ACI method with combined creep and shrinkage effects: = = =2.0 Δ = (Δ ) = =8.45 Δ +(Δ ) = = Or = mm using avrg. I e a. Allowable deflections: - For roof or floor construction supporting or attached to nonstructural elements likely to be damaged by large deflections (very stringent limitation): 7

8 Δ + (Δ ) 480 = = For roof or floor construction supporting or attached to nonstructural elements not likely to be damaged by large deflections: Δ + (Δ ) 240 = =37.5 8

Serviceability Deflection calculation

Chp-6:Lecture Goals Serviceability Deflection calculation Deflection example Structural Design Profession is concerned with: Limit States Philosophy: Strength Limit State (safety-fracture, fatigue, overturning