### Transcription of Chapter 8. Flexural Analysis of T-Beams

1 **Chapter** 8. **Flexural** **Analysis** of **T-Beams** Reading Assignments Text **Chapter** ; ACI 318, Section Occurrence and Configuration of **T-Beams** Common construction 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. **beam** **beam** **beam** **beam** **beam** **beam** (a) one-way slab (b) two-way slab 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 CIVL 4135 156 T-- **beam** CIVL 4135 157 T-- **beam** Code allows the following maximum effective widths: Symmetrical **beam** ACI318, Section b hf span 1) b . 4. b bw 2) 8h f 2. b bw 1. 3) clear distance between beams 2 2.

2 Bw Flange on one side only (Spandrel **beam** ). ACI318, Section b hf span 1) b b w 12. 2) b b w 6h f 3) b b w 1 clear distance to next web 2. bw Isolated T- **beam** ACI318, Section b hf 1) b 4b w bw 2) hf 2. bw CIVL 4135 158 T-- **beam** **Analysis** of **T-Beams** - ( a > hf). Consider the total section in two parts: 1) Flange overhangs and corresponding steel;. 2) Stem and corresponding steel;. b c . Cc hf a Ts =As fy = +. Asf As - Asf bw Case I Case II. For equilibrium we have: Case I: A sf f y = c h f (b b w) ( ). or c h f (b b w) ( ). A sf =. fy Case II: (A s A sf) f y = c b wa ( ). Solve for a : (A s A sf) f y ( ). a =. c b w and nominal moment capacity will be: hf ( ). M n = A sf f y(d ) + (A s A sf) f y (d a). 2 2. CIVL 4135 159 T-- **beam** Balanced Condition for **T-Beams** See Commentary page 48 of ACI 318-83 (old code). b hf u = c . Cc ab cb h d d-c Ts=Asbfy A bs y bw From geometry: u 87, 000 ( ).

3 Cb = d = d u + y 87, 000 + f y CIVL 4135 160 T-- **beam** **Analysis** of **T-Beams** in Bending: 40 hf =4 u = c . c Cc . d-c Ts=Asfy As = in2 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 < hf As fy 50. a = = = in c b 40. Since in > in, a T- **beam** **Analysis** is required. First find the reinforcement area to balance flanges (Asf = ?). f c . A sf = (b b w)h f = (40 10) 4 = in 2. fy 50. A s A sf = = in 2. Solve for a . c b wa = (A s A sf)f y (A s A sf)f y 50. a = = = in > 4in c b w 10. Assumption is CIVL 4135 161 T-- **beam** c = a = = 1 c = = .279 < Tension-controlled d Find the nominal moment capacity of the **beam** : hf M n = A sf f y (d ) + f y(A s A sf) (d a ). 2 2. M n = (in 2) 50(ksi) ( 4) + 50(ksi) (in 2) ( ). 2 2. M n = 4530 + 1790 = 6, 320 in k Note: This could have been done by statics with T s = A sf y C c = (b b w)(h f) c + ab w( )f c.

4 CIVL 4135 162 T-- **beam** Design of **T-Beams** 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 bw = 11 in. and d = 20 in. What tensile steel area is required at midspan to resist a mo- ment of 6,400 in-kips if fy = 60,000 psi and f'c = 3,000 psi. b hf +. Asf As - Asf bw Case I Case II. Solution First determining the effective flange width from Section ( ) or ACI span 24 12. 1) b = = 72 in 4 4. 2) b 16h f + b w = (16 3) + 11 = 59 in 3) b clear spacing between beams + b w = 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** be- haves like a rectangular shape).

5 Mu 6400 ( ). As = = = in 2. f y(d a 2) 60 (20 3 2). CIVL 4135 163 T-- **beam** Solve for a : A sf y a = = 60 = in > h f = Assumption incorrect c b 3 47. Therefore, the **beam** will act as a T- **beam** and must be designed as a T- **beam** . From Case I given above and Section ( ) we have c h f (b b w) (3ksi) (3in) (47 11) ( ). A sf = = = in 2. fy 60(ksi). hf ( ). M n1 = A sff y d ) = (60ksi) (20 3 2) = 4570 in--kips 2. M n2 = M u M n1 = 6400 4570 = 1830 in--kips ( ). Find a value by iteration. Assume initial a = inches M n2 1830 ( ). A s A sf = = = in 2. f y(d a 2) 60 (20 2). Find an improve a value (A s A sf) f y a = = 60 = in ( ). c b w 3 11. Iterate with the new a = in. M n2 1830 ( ). A s A sf = = = in 2. f y(d a 2) 60 (20 2). Find an improve a value (A s A sf) f y a = = 60 = in ( ). c b w 3 11. M n2 1830 ( ). A s A sf = = = in 2. f y(d a 2) 60 (20 2). CIVL 4135 164 T-- **beam** Since there is no change between equations ( ) and ( ) we have arrived at the answer.

6 There- fore, A s = A sf + (A s A sf) = + = in 2 ( ). Check with ACI requirements for maximum amount of steel (Tension-Controlled). c = a = = ( ). 1 c = = .237 < Tension-controlled d 20. 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** . CIVL 4135 165 T-- **beam**