Chapter 8. Flexural Analysis of T-Beams

Transcription

1 Chapter 8. Flexural Analysis of T-s 8.1. Reading Assignments Text Chapter 3.7; ACI 318, Section Occurrence and Configuration of T-s Common construction type.- used in conjunction with either on-way or two-way slabs. Sections consists of the flange and web or stem; the slab forms the beam flange, while the part of the beam projecting below the slab forms is what is called web or stem. (a) one-way slab (b) two-way slab 8.3. Concepts of the effective width, Code allowable values In reality the maximum compression stress in T-section varies with distance from section Web. Real max, Longitudinal compression sress Simplified equivalent width, stress 156

2 157

3 Code allows the following maximum effective widths: Symmetrical ACI318, Section b 1) b span 4 2) b Flange on one side only (Spandrel ) ACI318, Section ) b clear distance between beams b 1) b span 12 2) b 6 3) b 1 2 clear distance to next web Isolated T- ACI318, Section b 1) b 4 2) 2 158

4 8.4. Analysis of T-s - ( a > ) Consider the total section in two parts: 1) Flange overhangs and corresponding steel; 2) Stem and corresponding steel; b 0.85f c C c a T s =A s f y = + A sf Case I A s- A sf Case II For equilibrium we have: Case I: A sf f y = 0.85f c (b ) (8.1) or A sf = 0.85f c (b ) f y (8.2) Case II: Solve for a : (A s A sf ) f y = 0.85f c a (8.3) a = (A s A sf ) f y 0.85f c (8.4) and nominal moment capacity will be: M n = A sf f y (d 2 ) + (A s A sf ) f y (d a 2 ) (8.5) 159

5 8.5. Balanced Condition for T-s See Commentary page 48 of ACI (old code). b Áu = c b a b 0.85f c C c h d A b s Á y d-c T s =A s b f y From geometry: c b = Á u Á u + Á y d = 87, , f y d (8.6) 160

6 8.6. Example.- Analysis of T-s in Bending: 40 =4 Á u = f c c C c 20.5 A s = 6.88 in 2 Á y d-c T s =A s f y 10 Find the nominal moment capacity of the beam given above: f c = 2, 400 psi f y = 50, 000 psi Solution: Check to see if a T-beam analysis is required: Assume a< a = A s f y 0.85f c b = = 4.22 in Since 4.22 in > 4.00 in, a T-beam analysis is required. First find the reinforcement area to balance flanges (A sf =?) A sf = 0.85 f c (b b f w ) = (40 10) 4 = 4.90 in2 y 50 A s A sf = = 1.98 in 2 Solve for a 0.85f c a = (A s A sf )f y a = (A s A sf )f y = f c = 4.86 in > 4in o.k. Assumption is o.k. 161

7 c = a β 1 = = 5.72 c d = =.279 < Tension-controlled Find the nominal moment capacity of the beam: M n = A sf f y (d 2 ) + f y(a s A sf )(d a 2 ) M n = 4.9(in 2 ) 50(ksi) ( ) + 50(ksi) 1.98(in2) ( ) M n = = 6, 320 in k Note: This could have been done by statics with T s = A s f y C c = (b )( ) 0.85f c + a (0.85)f c 162

8 8.7. Example.- Design of T-s in Bending- Determination of Steel Area for a given Moment: A floor system consists of a 3 in. concrete slab supported by continuous T beams of 24 ft span, 47 in. on centers. Web dimensions, as determined by negative-moment requirements at the supports, are = 11 in. and d = 20 in. What tensile steel area is required at midspan to resist a moment of 6,400 in-kips if f y = 60,000 psi and f c = 3,000 psi. b + A sf Case I A s- A sf Case II Solution First determining the effective flange widtrom Section (8.3.1.) or ACI ) b span 4 = = 72 in 2) b 16 + = (16 3) + 11 = 59 in 3) b clear spacing between beams + = center to center spacing between beams = 47 in The centerline T beam spacing controls in this case, and b = 47 inches. Assumption: Assuming that stress-block depth equals to the flange thickness of 3 inches (beam behaves like a rectangular shape). A s = M u φf y (d a 2) = (20 3 2) = 6.40 in2 (8.7) 163

9 Solve for a : a = A sf y = f c b = 3.2 in > = 3.0 Assumption incorrect Therefore, the beam will act as a T-beam and must be designed as a T-beam. From Case I given above and Section (8.4.1.) we have A sf = 0.85f c (b ) f y = 0.85 (3ksi) (3in) (47 11) 60(ksi) = 4.58 in 2 (8.8) φm n1 = φa sf f y d ) = (60ksi) (20 3 2) = 4570 in -kips 2 (8.9) φm n2 = M u φm n1 = = 1830 in -kips (8.10) Find a value by iteration. Assume initial a = 3.5 inches A s A sf = Find an improve a value φm n2 φf y (d a 2) = ( ) = 1.86 in2 (8.11) a = (A s A sf ) f y = f c = 3.97 in (8.12) Iterate with the new a = 3.97 in. A s A sf = Find an improve a value φm n2 φf y (d a 2) = ( ) = 1.88 in2 (8.13) a = (A s A sf ) f y = f c = 4.02 in (8.14) A s A sf = φm n2 φf y (d a 2) = ( ) = 1.88 in2 (8.15) 164

10 Since there is no change between equations (8.13) and (8.15) we have arrived at the answer. Therefore, A s = A sf + (A s A sf ) = = 6.46 in 2 (8.16) Check with ACI requirements for maximum amount of steel (Tension-Controlled) c = a β 1 = = 4.73 (8.17) c d = =.237 < Tension-controlled Therefore, the T-beam satisfies the ACI provisions for tension failure. Next steps will be to select the reinforcement and check all the spacing requirements and detail the beam. 165