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.
2 4. b bw 2) 8h f 2. b bw 1. 3) clear distance between beams 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 ( ).
3 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 ( ). 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.
4 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 . 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.
5 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.
6 Assumption: Assuming that stress-block depth equals to the flange thickness of 3 inches ( beam be- haves like a rectangular shape). 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 ( ).
7 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. 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.
8 Next steps will be to select the reinforcement and check all the spacing requirements and detail the beam . CIVL 4135 165 T-- beam