Chapter 8.

Flexural Analysis of T-Beams

8.1. Reading Assignments

Assignments
Text Chapter 3.7; ACI 318, Section 8.10.

8.2. Occurrence and Configuration of T-Beams

• Common construction type.- used in conjunction
conjunction with either on-way or two-way
two-way slabs.
• Sections consists
consists of the flange
flange and web or stem;
stem; the slab forms the beam
beam flange, while
while
the part of the beam projecting below the slab forms is what is called web or stem.

Beam

m m m
a a m a
e e a e
B e
B B B

Beam

(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

Longitudinal
compression
compression sress Simplified equivalent
width, stress

CIVL 4135 156 T-- Beam

CIVL 4135 157 T-- Beam
Code allows the following maximum effective widths:
8.3.1. Symmetrical Beam

ACI318, Section 8.10.2.

hf span
1) b ≤
4
b − bw
2)
2
≤ 8h f

b − bw 1 cle
3)
2
≤ 2
clear
ar distan
distance
ce betwee
between
n beams
beams
bw

8.3.2. Flange on one side only (Spandrel Beam)

ACI318, Section 8.10.3.

hf span
1) b − bw ≤
12

2) b − b w ≤ 6h f

3) b − b w ≤ 1 clea
clearr dist
distan
ance
ce to nex
nextt web
web
2
bw

8.3.3. Isolated T-Beam

ACI318, Section 8.10.4.

hf

1) b ≤ 4b w

bw
2)
2
≤ hf

bw

CIVL 4135 158 T-- Beam

8.4. Analysis of T-Beams - ( a > hf)

Consider the total section in two parts:

1) Flang
Flangee ov
overh
erhang
angss and
and corre
corresp
spond
onding
ing steel;
steel;
2) St
Stem
em an
and
d co
corr
rres
espo
pond
ndin
ing
g stee
steel;
l;

b 0.85fc′
Cc
hf
a

Ts =As fy
= +
Asf As - Asf
bw

Case I Case II

For equilibrium we have:

8.4.1. Case I:

A sf f y = 0.85f c′h f ( b − b w) (8.1)

or
0.85f c′ h f (b − b w)
(8.2)
A sf = fy

8.4.2. Case II:

(A s − A sf) f y = 0.85f c′ b wa (8.3)

Solve for “a”:

(A s − A sf) f y (8.4)
a =
0.85f c′ b w
and nominal moment capacity will be:
h
A sf f y(d f
Mn = − 2) + (A s − A sf) f y (d − 2)
a (8.5)

CIVL 4135 159 T-- Beam

8.5. Balanced Condition for T-Beams
See Commentary page 48 of ACI 318-83 (old code).

b hf Áu = 0.003 0.85f c′
Cc
ab
cb

h d
d-c Ts=Asbfy
A bs Á
y

bw

From geometry:
Áu 87, 000
87, (8.6)
cb = d = d
Áu + Áy 87
87,, 000 + fy
000

CIVL 4135 160 T-- Beam

8.6. Example.- Analysis of T-Beams
T-Beams in Bending:
Bending:
40” hf =4” Áu = 0.003 0.85f c′

c Cc

”
20.5
d-c Ts=Asfy
As = 6.88 in2 Á
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 f y 6.88 × 50
a = = 0.85 × 2.4 × 40
= 4.22 in
0.85f c′ b

Since 4.22 in > 4.00 in , a T-beam analysis is required.

First find the reinforcement area to balance flanges (A sf = ?)

f c′
A sf = 0.85 (b − b w)h f = 0.85 × 2.4 × (40 − 10) × 4 = 4.90 in 2
fy 50

As − A sf = 6.88 − 4.90 = 1.98 in 2

Solve for “a”

0.85f c′b wa = (A s − A sf)f y

(A s − A sf)f y 1.98 × 50
a = = 0.85 × 2.4 × 10
= 4.86 in > 4in o.k.
0.85f c′b w
Assumption
Assumption is o.k.

CIVL 4135 161 T-- Beam

 c = a = 4.86 = 5.72
β1 0.85

c = 5.72 = .279 < 0.375 Tension-controlled

d 20.5

Find the nominal moment capacity of the beam:

hf a)
M n = A sf f y (d − ) + f y(A s − A sf) ( d −
2 2

M n = 4.9(in 2) × 50(ksi) × (20.5 − 4) + 50(ksi) × 1.98(in 2) × (20.5 − 4.86)

2 2

M 4530 1790 6,320 in k

n = + = −

Note:
This could have been done by statics with
T s = A sf y

Cc = (b − b w)(h f) × 0.85f c′ + ab w(0.85)fc′

CIVL 4135 162 T-- Beam

8.7. Example.-
Example.- Design of T-Beams
T-Beams in Bending-
Bending- Determination
Determination of
of Steel Area for
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
negative-moment requirements
requirements at the
supports
supports,, are b = 11
11 in
in.. an
and
d d = 20 in. What tensile
tensile steel area is required
required at midspan to
to resist a mo-
mo-
w
ment of 6,400 in-kips if fy = 60,000 psi and f ’
c = 3,000 psi.

hf

+
Asf As - Asf
bw

Case I Case II

Solution

First determining the effective flange width from Section (8.3.1.) or ACI 8.10.2

span 24 × 12
1) b ≤ = = 72 in
4 4
2) b 16h bw (16 3) 11 59 in
≤ + = f × + =
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.

ion: Assuming that stress-block

Assumption:
Assumpt stress-block depth equals to the flange thickness of 3 inches (beam be-
haves like a rectangular shape).

Mu 6400 2
s
A = φf y(d − a∕2) = 0.9 × 60 × (20 − 3∕2) = 6.40 in (8.7)

CIVL 4135 163 T-- Beam

Solve for “a”:

A sf y 6.40 × 60
a = = 0.85 × 3 × 47 = 3.2 in > h f = 3.0 Assumpti
Assumption
on incorrec
incorrectt
0.85f c′b

There
Therefo
fore,
re,thebeam
thebeam will
will actas a T-beam
-beam andmust be desig
designedas
nedas a T-beam
-beam.. From
From Case
Case I given
given above
above
and Section (8.4.1.) we have

0.85f c′ h f (b − b w) 0.85 × (3ksi) × (3in) × (47 − 11) (8.8)

A sf = fy
=
60(ksi)
= 4.58 in 2

hf (8.9)
φM n1 = φA sff yd − ) = 0.9 × 4.58 × (60ksi) × (20 − 3∕2) = 4570 in-
in--kips
-kips
2

φM n2 = M u − φM n1 = 6400 − 4570 = 1830 in-

i n--k
-kips
ips (8.10)

Find “a” value by iteration.

iteration. Assume initial
initial a = 3.5 inches

φM n2 1830
A s − A sf = = = 1.86 in 2 (8.11)
φf y(d − a∕2) 0.9 × 60 × (20 − 3.5∕2)

Find an improve “a” value

(A s − A sf) f y 1.86 × 60 = 3.97 (8.12)

a = = 0.85 × 3 × 11 in
0.85f c′ b w

Iterate with the new a = 3.97 in.

φM n2 1830
A s − A sf = = = 1.88 in 2 (8.13)
φf y(d − a∕2) 0.9 × 60 × (20 − 3.97∕2)

Find an improve “a” value

(A s − A sf) f y 1.88 × 60 = 4.02 (8.14)

a = = 0.85 × 3 × 11 in
0.85f c′ b w

φM n2 1830 2
s sf
A − A = φf y(d − a∕2) = 0.9 × 60 × (20 − 4.02∕2) = 1.88 in (8.15)

CIVL 4135 164 T-- Beam

Si
Sinc
ncee th
ther
eree is no ch
chan
ange
ge betw
betwee
een
n eq
equa
uati
tion
onss (8
(8.1
.13)
3) an
and
d (8.1
(8.15)
5) we ha
have
ve arri
arrive
ved
d at th
thee an
answ
swer
er.. Ther
There-
e-
fore,

A s = A sf + (A s − A sf) = 4.58 + 1.88 = 6.46 in 2 (8.16)

Check with ACI requirements for maximum amount of steel (Tension-Contro

(Tension-Controlled)
lled)

c = a = 4.02 = 4.73 (8.17)

β1 0.85

c
= 4.73 = .237 < 0.375 Tension-controlled
d 20

Theref
Therefore
ore,, theT-bea
theT-beam
m satisf
satisfies
iestheACI
theACI provis
provision
ionss forten
for tensio
sion
n failur
failure.
e. Next
Next steps
steps will
will be to select
selectthe
the
reinforcement and check all the spacing requirements and detail the beam.

CIVL 4135 165 T-- Beam