Dr. Hazim Dwairi. Example: Continuous beam deflection

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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) + 300 = 2300 mm Or 3000 mm Ł b eff = 2250 mm 537 mm 3f25 9000 mm 2250 mm 300 mm 5f25 2f25 125 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

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 = 486.5 mm < Ł OK 2- Loads and Moments: Self weight = [(3000)(125) + (475)(300)]/10 6 x 25 = 12.94 kn/m w d = (3)(1.0) + 12.94 = 15.94 kn/m w L = (3)(4.8) = 14.40 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 = 92.22 kn.m Pos. M L = w L l n 2 /14 = (14.40)(9) 2 /14 = 83.31 kn.m Pos. M d+l = 92.22 + 83.31 = 175.53 kn.m Pos. M sus = 92.22 + 0.3(83.31) = 117.21 kn.m b. Negative moments Neg. M d = w d l n 2 /10 = (15.94)(9) 2 /10 = 129.11 kn.m Neg. M L = w L l n 2 /10 = (14.40)(9) 2 /10 = 116.64 kn.m Neg. M d+l 129.11 + 116.64 = 245.75 kn.m Neg. M sus = 129.11 + 0.3(116.64) = 164.10 kn.m 3- Modulus of rupture, modulus of elasticity, and modular ratio: f r = 0.7 f c = 0.7 28 = 3.704 MPa E c = 4700 f c = 4700 28 = 24870 MPa n = E s /E c = 200,000/24,870 = 8.0 2

4- Gross and cracked sections moment of inertia: a. Positive moment section: Gross section at mid-span: 537 mm y t = 163.38 mm I g = 1.156 x 10 10 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 + 10.453 y cr 5613.4 = 0 3

Ł y cr = 69.88 mm I cr = 2250(69.88) 3 /3 + 8(1470)(537 69.88) 2 = 2.822 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 + 189.68 y cr - 73164 = 0 y cr = 191.79 mm I cr = 300(191.79) 3 /3 + 7(980)(191.79 63) 2 + 8(2454)(537 191.79) 2 I cr = 3.159 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 = 3.704 x 1.156 x 10 10 / (600 163.38) = 98.07 kn.m M cr /M d = 98.07/92.22 = 1.06 >1 thus, (I e ) d = I g = 1.156 x 10 10 mm 4 M cr /M sus = 98.07/117.21 = 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

= (0.593)(1.156 x 10 10 ) + (1 0.593)( 2.822 x 10 9 ) = 8.00 x 10 9 mm 4 < I g M cr /M d+l = 98.07/175.53 = 0.560 <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 10 10 ) + (1 0.176)( 2.822 x 10 9 ) = 4.360 x 10 9 mm 4 < I g b. Negative moment section M cr = f r I g /y t = 3.704 x 5.40 x 10 9 / (300) = 66.67 kn.m M cr /M d = 66.67/129.11 = 0.516 >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 ) + (1 0.138)( 3.159 x 10 9 ) = 3.468 x 10 9 mm 4 < I g M cr /M sus = 66.67/164.1 = 0.406 >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 ) + (1 0.067)( 3.159 x 10 9 ) = 3.309 x 10 9 mm 4 < I g M cr /M d+l = 66.67/245.75 = 0.271 >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)( 3.159 x 10 9 ) = 3.204 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, 9.5.2.4 suggests using the average I e at the critical positive and negative moment sections. The 1983 commentary on 9.5.2.4 suggested the following approach to obtain improved results: Beams with one end continuous:. =0.85 +0.15. Beams with both ends continuous:. =0.70 +0.15( + ) Where I m refers to I e at midspan, I e1 and I e2 refer to both ends of the beam 5

.( ) =0.85(1.156 10 ) +0.15(3.468 10 ) =1.503 10.( ) =0.85(8.000 10 ) +0.15(3.309 10 ) =7.296 10.( ) =0.85(4.360 10 ) +0.15(3.204 10 ) =4.187 10 6- 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. =1.20 0.20 =1.20 0.20 8 =0.85 14 (Δ ) = 5 48 ( ) =0.85 5 48 Or = 1.769 mm using avrg. (I e ) d = 1.503 x 10 10 mm 4 (92.22 10 )(9000) 24870 1.156 10 =2.301 6

(Δ ) = 5 48 ( ) =0.85 5 48 (117.21 10 )(9000) 24870 8.000 10 =4.225 Or = 4.633 mm using avrg. (I e ) d = 7.296 x 10 9 mm 4 (Δ ) = 5 48 ( ) =0.85 5 48 Or = 12.089 mm using avrg. (I e ) d = 4.187 x 10 9 mm 4 (175.53 10 )(9000) 24870 4.360 10 =11.610 (Δ ) = (Δ ) (Δ ) =11.610 2.301 =9.309 Or = 12.089 1.769 = 10.320 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 >9.309 180 180 - For floors not supporting and not attached to nonstructural elements likely to be damaged by large deflections: (Δ ) = 9000 =25 >9.309 360 360 7- Ultimate long-term deflections: Using ACI method with combined creep and shrinkage effects: = = 2 1+50 1+0 =2.0 Δ = (Δ ) =2.0 4.225 =8.45 Δ +(Δ ) =8.45+9.309 =17.759 Or = 19.586 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

Δ + (Δ ) 480 = 9000 480 =18.75. - For roof or floor construction supporting or attached to nonstructural elements not likely to be damaged by large deflections: Δ + (Δ ) 240 = 9000 240 =37.5 8