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Lecture 5 - Page 1 of 9
Lecture 5 – T- Beams Concrete beams are often poured integrally with the slab, forming a much stronger “T” – shaped beam. These beams are very efficient because the slab portion carries the compressive loads and the reinforcing bars placed at the bottom of the stem carry the tension. A T-beam typically has a narrower stem than an ordinary rectangular beam. These stems are typically spaced from 4’-0” apart to more than 12’-0”. The slab portion above the stem is designed as a one-way slab spanning between stems (see Lecture 6).
A typical T-beam has the following dimensions and notations:
bw bw
d Overhang width
Clear distance
hf = Slab thickness
b = Effective flange width
NOTE: Stirrups in T-beam are required (not shown in this sketch)
Lecture 5 - Page 2 of 9
Assuming T-beams are symmetrical, the following design dimensions are used:
Overhang width = smaller
b = smaller T-Beam Analysis
T-beams are analyzed similarly to rectangular beams, except the compression area is a narrow “strip” usually located in the slab.
hf
a = Effective conc. compressive thickness
8hf or ½(Clear distance)
¼(Beam span) or (2 x overhang width) + bw
bw
Z = (d - 2a )
b = Effective flange width
d
Ac = Shaded area = Effective concrete compression area = (a)(b)
As = Total area of main tension bars
Lecture 5 - Page 3 of 9
Mu = Usable moment capacity of T-beam
= φTZ
where: φ = 0.9 T = Tension force developed in main bars = Asfy Ac = Effective concrete compression area
= cf
T'85.0
a = Effective concrete compressive thickness
= bAc
Z = Moment arm distance between center of compression to center of tension
= d - 2a
Lecture 5 - Page 4 of 9
Example 1 GIVEN: A commercial building has T-beams spaced 6’-6” (center-to-center) with a 4” thick concrete slab as shown in the framing plan and cross-section views below. Use the following information:
• Superimposed service floor dead load (NOT including conc. wt.) = 40 PSF • Superimposed service floor live load = 100 PSF • Concrete f’c = 3000 PSI • ASTM A615 Grade 60 bars
REQUIRED:
1) Determine the maximum factored moment, Mmax, on the T-beam. 2) Determine the usable moment capacity, Mu, for the T-beam.
A A
Typ.
6’-6”
T-be
am s
pan
= 20
’-0”
T-beam
Perimeter girder Column
Framing Plan
Lecture 5 - Page 5 of 9
Step 1 – Determine maximum factored moment, Mmax, on T-beam:
Determine area of T-beam = Slab area + Stem area = (6.5’)(0.333’) + (1.1667’)(0.666’) = 2.94 ft2 Determine service weight of T-beam = Area of T-beam x Conc. unit wt. = 2.94 ft2(150 lb/ft3) = 441 PLF Det. factored uniform load on T-beam wu = 1.2D + 1.6L
= 1.2[(6.5’)(40 PSF) + 441 PLF] + 1.6[(6.5’)(100 PSF)] = 841 PLF + 1040 PLF = 1881 PLF → Use wu = 1.9 KLF
Det. Maximum factored moment, Mmax = 8
2Lwu
= 8
)"0'20)(9.1( 2−KLF
Mmax = 95 KIP-FT
18”
Service Dead Load Service Live Load
8”
16”
6’-6” hf = 4”
2 - #9 bars
Section “A-A” Thru T-Beams (NOTE: stirrups and top bars not shown)
Lecture 5 - Page 6 of 9
Step 2 – Determine effective concrete slab width “b”:
Overhang width = smaller
b = smaller
Step 3 – Determine effective conc. compression area Ac:
T = Tension force developed in main bars = Asfy = 2 bars(1.00 in2 per #9 bar)(60 KSI) = 120 KIPS Ac = Effective concrete compression area
= cf
T'85.0
= )3(85.0
120KSI
KIPS
= 47.1 in2
¼(Beam span) = ¼(20’-0” x 12”/ft) = 60” ← USE or (2 x overhang width) + bw = (2 x 32” + 8”) = 72”
8hf = 8(4”) = 32” ← USE or ½(Clear distance) = ½(78” – 8”) = 35”
Lecture 5 - Page 7 of 9
Step 4 – Determine usable moment capacity, Mu for the T-beam:
a = Effective concrete compressive thickness
= bAc
= "60
1.47 2in
a = 0.79”
Z = Moment arm distance between center of compression to center of tension
= d - 2a
= 16” - 2
"79.0
Z = 15.6”
Mu = φTZ = 0.9(120 KIPS)(15.6”) = 1685 KIP-IN Mu = 140.4 KIP-FT NOTE: Since Mu = 140.4 KIP-FT > Mmax = 95 KIP-FT, T-beam is ACCEPTABLE.
Lecture 5 - Page 8 of 9
Heavily-Reinforced T-Beams
T-beams with a lot of tension reinforcement may have a portion of the effective concrete area located within the stem as shown below: The location of the centroid of the effective concrete compression area is found by methods discussed in AECT 210 – Structural Theory (see Lecture 5). After the location is found, analysis is exactly the same as ordinary T-beams. Similar to ordinary rectangular reinforced concrete beams, the ACI 318 limits the amount of tension steel in T-beams so that the steel will yield prior to concrete compression failure. The maximum area of steel, As is shown in the table below.
Maximum Tensile Steel Permitted in T-Beams Concrete and Steel Properties: Formula (As = in2) Concrete f’c = 3000 PSI Steel fy = 40 KSI
As max = 0.0478[bhf + bw(0.582d – hf)]
Concrete f’c = 3000 PSI Steel fy = 60 KSI
As max = 0.0319[bhf + bw(0.503d – hf)]
Concrete f’c = 4000 PSI Steel fy = 40 KSI
As max = 0.0638[bhf + bw(0.582d – hf)]
Concrete f’c = 4000 PSI Steel fy = 60 KSI
As max = 0.0425[bhf + bw(0.503d – hf)]
bw
hf
b
Z d
As
Ac = Shaded area = Effective concrete compression area
Lecture 5 - Page 9 of 9
Example 2 GIVEN: The T-beam from Example 1. REQUIRED: Determine the maximum area of tension steel permitted, As max:
Step 1 – Determine As max:
From Example 1:
Concrete f’c = 3000 PSI Steel fy = 60 KSI b = 60” hf = 4” bw = 8”
As max = 0.0319[bhf + bw(0.503d – hf)] = 0.0319[(60”)(4”) + 8”(0.503(16”) – 4”)] As max = 8.7 in2
NOTE: This area of tension steel As = 8.7 in2 is a LOT!! In order to supply this much steel the beam would require 9 - #9 bars, 15 - #7 bars or 20 - #6 bars! It would be far better to change the beam dimensions than to try to squeeze this many bars into the beam.
