Reinforced Concrete
Topic: T-sections
Presented by: CSMantawil
Introduction
Floors and roofs in reinforced concrete buildings may be
composed of slabs that are supported so that loads are
carried to columns and then to the building foundation.
One such system, called a beam and girder system, is
composed of a slab on supporting reinforced concrete
beams and girders. The beam and girder framework, in
turn, is supported by columns. In such a system, the beams
and girders are commonly placed monolithically with the
slab. See figure 7.1
2
Fig 7.1 Beams and Girder Floor System
3
The beams are commonly spaced so that they
intersect the girders at the midpoint, third points, or
quarter points, as shown in Figure 7-2.
Fig 7.2 Common beam and girder layout 4
Cont’n of Fig 7.2
Fig 7.2 Common beam and girder layout
Source: RC Design, 8th Ed by Aghayere & Limbrunner 5
Except precast systems, most floors, roofs,roof decks,
and beams are often monolithically constructed. When
beams become an integral part of the slab they form
flanged section that resembles to a tee.
Actual and equivalent stress distribution over flange width
Source: Reinforced Concrete Design by Wang & Salmon 6
Basic Elements
bw = web width
Sw = clear spacing
between adjacent web
t or hf = flange thickness
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ACI Code 6.3.2.1
The effective width of the flange shall be taken as the smallest of:
(Based on ACI318- 14)
1 ) be = bw + 16hf
2 ) be = bw + ln/4
where ln = clear span of the beam
3 ) be = bw + Sw
where sw = average clear spacing between webs
Note: in some books, the number 3 criterion
may be taken as the center to center spacing of beams
Note that hf and t are often used interchangeably
8
For exterior T section (slab on only one side of web)
For isolated T-sections: the following relations must
both be satisfied:
a) b e ≤4 b w
1
b) t≥ b w
2
Note: There are slight changes in ACI 319-19 chapter 6.3.2.1 9
From Wight & McGregor,7th Ed. p. 175
10
Case 1: a ≤ t (treat as rectangular
section)
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Case 2: a > t (treat as real t-beam)
Let C1 = Cweb and C2 = Cflange
C 1=0.85 f ' c b w a
C 2=0.85 f ' c (b e −b w )t 12
Since A s= A s 1 + A sf hence A s 1= A s − A sf
( A s− A sf ) fy
a=
0.85 f ' c b w
13
It may be necessary to check:
1) The total tension steel ratio relative to the web effective area
is equal to or greater than ρ min
0.25 √ (f ' c ) 1.4
ρ min = ≥
fy fy
2) The net tensile strain is equal to greater than ety + 0.003
for tension-controlled sections.
3) See other ACI provisions as described by authors
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Example 1
Determine the nominal and design moment
strength within the span of a floor beam as shown
in Fig 1 whose projection below a 115-mm slab is
330 mm x 610 mm (effective depth is 635 mm for
two layers of steel). Tension reinforcement is 8-
#25 mm f bars. The span length of the beam,
measured to the center of the supporting columns,
is 7.30 m, and the beams are centered 4 m apart.
Supporting columns have a square cross section
with 715 mm side dimension. Use f’c = 31 MPa and
fy = 415 MPa
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It is assumed that clear spacing
between layers of bars = 25 mm
Figure 1
Part of solution:
1) be = bw + 16t = 330+16(115) = 2170 mm
2) be = bw + ln/4 = 330 +[7300-715]/4 = 1976.25 mm
3) be = c to c spacing of beams = 4000 m
use be = 1976.25 mm
0.05(31−28)
As = 8 (0.7854)(25)2 = 3927 mm2 β 1=0.85− ≈0.83
7
25 25
d t =d +s /2+db /2=635+ + =660 mm 16
2 2
Example 2
Determine the Design moment strength of
the isolated T-section shown in the figure
when it is provided by 8-#36 mm f bars.
Use f’c = 28 MPa and fy = 420 MPa.
Soln:
2
(36)
A s =π x 8=8143 mm2
4
25 36
d t =915+ + =945.5 mm
2 2
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Check requirements for isolated section
a) b e ≤4 bw ;
760 mm<4 (350)=1400 mm ok
bw
b) hf ≥
2
350
175 mm>( )=175 mm ok
2
Assuming rectangular section behavior, i.e. a ≤ hf
As f y Alternately, we can test it by the comparing
a= Ac and Af
0.85 f ' c b e
As f y 8143(420)
8143(420) A c= = =143,700 mm2
a= =189 mm 0.85 f ' c 0.85(28)
0.85 f ' c 760 A f =760(175)=133,000 mm2
Since a > hf = 175 mm, Since Ac > Af, a > hf
assumption is incorrect. 18
1st solution. Using the entire section,
A c ( ȳ)= A 1 y 1 + A 2 y 2
175 30.6
143,700( ȳ)=760(175)( )+10,700(175+ )
2 2
ȳ=95.15≈95 mm
The lever arm z=d− ȳ=915−95=820 mm
10,700
y 2= =30.6 mm checking for ductility ,
350 c / d t =242/ 945.5=0.256<0.375 ,
then from the figure , hence ϕ=0.90
a=175+30.6=205.6 mm and ϕ M n =0.90 A s f y ( z)
205.6
c= ≈242 mm −6
0.85 ϕ M n =0.90(8143)(420)(820) x 10
ϕ=2524 kN −m 19
2nd Sol’n: By two-couple method
T = A s f y =C=C w +C f
(8143)(420)=0.85 f ' c b w (a)+0.85 f ' c (b e −b w )
a=205.6 mm as before and c =242 mm
The moment In terms of compression forces,
[ a hf
ϕ M n=ϕ 0.85 f ' c b w a(d− )+0.85 f ' c (b e −b w )h f (d− )
2 2 ] 20
205.6
M n=(0.85)(28)(350)(205.6)(915− )+
2
175 −6
(0.85)(28)(760−350)(175)(915− ) x 10
2
M n=2804.093 kN −m
With f = 0.90 as earlier computed,
ϕ M n =0.90(2804.093)=2523.7 kN −m
Alternately,, by use of formula as presented in the earlier slides,
in terms of tensile forces
A sf f y =0.85 f ' c (b e −b w )h f
0.85(f ' c )(h f )(b e −b w )
a) A sf =
fy
0.85(28)(175)(760−350)
A sf = =4065.83 mm2
420 21
b) A sw = A s − A sf =8143−4065.83=4077.17 mm2
( A s − A sf ) f y (4077.17)(420)
a= = =205.6 mm
0.85 f ' c b w 0.85(28)(350)
c = 242 mm and f = 0.90 as computed before
M n=M n 1 + M n 2
a hf
M n=( A s − A sf )(f y )(d− )+ A sf f y (d− )
2 2
205.6 175
M n =(4077.17)(420)(915− )+(4065.83)(420)(915− ) x 10−6
2 2
M n=2803.9 kN −m
ϕ M n =0.90(2803.9)=2523.5 kN −m
Note the slight difference between the results of methods used
22
The floor system shown consists of 100 mm
slabs supported by 2.85 m clear span beams
spaced 3 m on center. The beams have a
web width bw of 350 mm. Determine the
necessary reinforcement for a typical interior
beam if the factored applied moment is 574
kN-m. Use d = 470 mm, f’c = 21 MPa, and fy
= 420 Mpa.
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Next topic: Design of T-sections
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