Design of T and L Beams in Flexure - Dr.Qaisar Ali

18-May-21

1

Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Prof. Dr. Qaisar Ali CE 320 Reinforced Concrete Design

Lecture 04

Design of T and L Beams in

Flexure

By: Prof. Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

[email protected]



Topics Addressed

Introduction to T and L Beams

ACI Code provisions for T and L Beams

Design Cases

Design of Rectangular T-beam

Design of True T-beam

References

2

18-May-21

2



Objectives

At the end of this lecture, students will be able to;

Differentiate between T-beam and L-beam

Explain Mechanics of Rectangular T-beam and true T-beam

Design T-beam for flexure

3


Prof. Dr. Qaisar Ali CE 320 Reinforced Concrete Design 4

The T or L Beam gets its name when the slab and beam produce

the cross sections having the typical T and L shapes in a monolithic

reinforced concrete construction.

Introduction to T and L Beam

18-May-21

3



In casting of reinforced concrete floors/roofs, forms are built for

beam sides, the underside of slabs, and the entire concrete is

mostly poured at once, from the bottom of the deepest beam to the

top of the slab.





18-May-21

4







18-May-21

5




a

a

a

a

b

b

Tension

Section a-a

Compression

Compression

Tension

Section b-b

a

a

b

b

a

a



Positive Bending Moment

In the analysis and design of floor and roof systems, it is common practice

to assume that the monolithically placed slab and supporting beam

interact as a unit in resisting the positive bending moment.

As shown, the slab becomes the compression flange, while the supporting

beam becomes the web or stem.


Tension

Section a-a

Compression

Flange

Web

18-May-21

6



Negative Bending Moment

In the case of negative bending moment, the slab at the top of the stem

(web) will be in tension, while the bottom of the stem will be in

compression. This usually occurs at interior support of continuous beam.


Tension

Section b-b

Compression



For T and L beams supporting monolithic or composite slabs, the effective

flange width bf shall include the beam web width bw plus an effective

overhanging flange width in accordance with ACI 318-19 Table 6.3.2.1.

ACI Code Provisions for T and L Beams

sw

hf

Effective Flange Width

bfbf

bw

Web or Stem

Slab

Flange

18-May-21

7



Calculation of Effective Flange Width (bf) (ACI 318-19, 6.3.2.1)


T - Beam

1 bw + 16hf

2 bw + sw

3 bw + ℓn/4Least of the above values is selected as bf

swsw

Where,

bw = Width of the beam;

hf = Slab thickness;

sw = Clear distance to the adjacent beam;

ℓn = Clear length of beam.

Note: In ACI 318-19 (6.3.2.1), h is used for

slab thickness, while in ACI 318-19

(9.3.1.1), h is used for beam depth.

However, in this lecture, for differentiation

between beam depth and slab thickness, hf

will be used to denote slab thickness.



14

Calculation of Effective Flange Width (bf) (ACI 6.3.2.1)


L - Beam

1 bw + 6hf

2 bw + sw/2

3 bw + ℓn/12

Least of the above values is selected as bf

s

w

hf

Effective Flange Width

bfbf

b

w

Web or Stem

Slab

Flange

swbw

14

18-May-21

8



Design Cases

In designing a T-Beam for positive bending moment, there exists two

conditions:

Case 1. The depth of the compression block may be less than or equal to

the slab depth i.e. flange thickness (a ≤ hf)

In such a condition the T-Beam is designed as rectangular beam for positive

bending with the width of compression block equal to bf.

bf

hf

bw

N.A

As

d

a



Design Cases

Case 2. The compression block cover the flange and extend into the web (a ˃ hf)

In such condition the T-Beam is designed as true T-beam.

bf

(bf - bw)/2

hf

bw

N.A

Ast

d

a

18-May-21

9




Flexural Capacity

Case 1. When a ≤ hf

bf

hf

bw

N.A

As

d

a



Flexural Capacity

(∑Fx = 0)

0.85fc′ abf = Asfy

a = Asfy/ 0.85fc′ bf

(∑M = 0)

Mn = T*la = Asfy (d – a/2)

As ΦMn = Mu ; ΦAsfy (d – a/2) = Mu

Therefore, As = Mu/Φfy(d – a/2)

The other checks remain same as that of rectangular beam design.

Note: In calculating Asmax and Asmin, use bw , not bf.


bf

hf

bw

As

d

a

N.A

T

Ca/2

la = (d - a/2)

0.85fc ′εc

εs

a

18-May-21

10


Prof. Dr. Qaisar Ali CE 320 Reinforced Concrete Design 19`

Example 01

The roof of a hall has a 5″ thick slab with beams having 30 feet. c/c and

28.5 feet clear length. The beams are having 9 feet clear spacing and

have been cast monolithically with slab. Overall depth of beam (including

slab thickness) being 24 in. and width of beam web being 14 in. Calculate

the steel reinforcement area for the simply supported beam against a

total factored load (including self weight of beam) of 3 k/ft. Use fc′ = 3 ksi

and fy = 60 ksi.





5″

14″

As

21.5″

19″

Solution:

Span length (lc/c) = 30′ ; clear length (ln) = 28.5′

Wu = 3 k/ft

d = 24-2.5 = 21.5″, bw = 14″; hf = 5″

Effective flange width (bf) is minimum of,

bw + 16hf = 14 + 16 × 5 = 94″

bw + sw = 14 + 9 × 12 = 122″

bw + ln/4 + = 14 + 28.5 × 12/4 = 99.5″

Therefore, bf = 94″

18-May-21

11



Solution:

Check if the beam behaviour is T or rectangular.

Mu = wulc/c2/8 = 3 x 302 x12 / 8 = 4050 in-kips

Trial # 01

Let a = hf = 5″

As = Mu/Φfy(d – a/2) = 4050/{0.90 × 60 × (21.5 – 5/2)} = 3.94 in2

Trial # 02

a = Asfy/(0.85fc′bf) = 3.94 × 60/ (0.85 × 3 × 94) = 0.98″ ˂ hf = 5″

Therefore, design as Rectangular beam.

As = Mu/Φfy(d – a/2) = 4050/{0.90 × 60 × (21.5 – 0.98/2)} = 3.56 in2





Solution:

Trial # 03

a = Asfy/(0.85fc′bf) = 3.56 × 60/ (0.85 × 3 × 94) = 0.89″

As = Mu/Φfy(d – a/2) = 4050/{0.90 × 60 × (21.5 – 0.89/2)} = 3.56 in2

Therefore As = 3.56 in2

Try #8 bars, No of Bars = 3.56 / 0.8 = 4.45, say 5 bars

18-May-21

12




Solution:

Check for maximum and minimum reinforcement allowed by ACI:

Asmin = 3 ( fc′/ fy) bwd ≥ (200/fy) bwd ; (Greater of these two for fy ≤ 80 ksi)

3 ( fc′ /fy) bwd = 3 × ( 3000 /60000) bwd = 0.82 in2

(200/fy) bwd = (200/60000) x 14 × 21.5 = 1.0 in2

Therefore, Asmin = 1.0 in2

Asmax = 0.27 (fc′ / fy) bwd = 0.27 x (3/60) x 14 × 21.5 = 4.06 in2

Asmin (1.0) < As (4.0) < Asmax (4.06) O.K!




(3+2),#8 bars

24″

=14″

5″

bw

= 94″bf

Drafting

18-May-21

13




Class Activity: Design the Beam B1 for the following moments

fc′ = 4 ksi, fy = 60 ksi, beam web width = 15′′, Slab thickness = 6′′

Overall beam depth (including slab thickness) = 24′′,

3122 ′′ K

1764 ′′ K

24′ clear length 24′ clear length

B1-Moment Diagram

1764 ′′ K




bfbf

bf

(bf -bw)/2

d d= +

hf

bw

N.A

Ast AsAsf

bw bw

d

a

ΦMn ΦMn2ΦMn1

ΦMn = ΦMn1 + ΦMn2

(bf -bw)/2

hf a

Flexural Capacity

Case 2. When a > hf

18-May-21

14




Flexural Capacity

ΦMn1 Calculation:

From stress diagram

T1 = C1

C1 = 0.85 fc′ (bf - bw)hf

T1 = Asf fy

Asf fy = 0.85fc′ (bf - bw)hf

Everything in the equation is known except Asf

Therefore, Asf = 0.85fc′ (bf - bw)hf / fy

ΦMn1 = T1 x l1 = ΦAsf fy (d – hf/2)

bf

d

N.A

Asf

bw

hf

T1

C1

hf/2

l1 = d - hf/2

(bf -bw)/2

ΦMn1




Flexural Capacity

ΦMn2 Calculation:

From stress diagram

T2 = C2

C2 = 0.85 fc′ abw

T2 = As fy

As fy = 0.85fc′ abw

a = As fy / (0.85 fc′ bw)

ΦMn2 = T2 x l2 = Φ As fy (d – a/2)

bf

(bf -bw)/2

d

As

bw

hf

T2

a/2a C2

l2 = d - a/2

ΦMn2

N.A

18-May-21

15




As Calculation

We know that ΦMn = Mu

ΦMn1 + ΦMn2 = Mu

ΦMn1 is already known to us,

Therefore ΦMn2 = Mu – ΦMn1

And as, ΦMn2 = T2 x l2 = Φ As fy (d – a/2)

Also ΦMn2 = Mu – ΦMn1

Therefore, As = (Mu – ΦMn1)/ Φfy (d – a/2);

and a = As fy / (0.85 fc′ bw)

Calculate As by trial and success method.

bf

(bf -

bw)/2

d

As

bw

hf

T2

a/2a C2

l2 = d - a/2

ΦMn2

N.A




Ductility Requirements

T = C1 + C2 [ ∑Fx = 0 ]

Astfy =0.85fc′(bf – bw)hf + 0.85fc′abw

Astfy = Asffy + 0.85fc′abw

For ductility εs = εt = εty + 0.003 (ACI 318-19 table 21.2.2)

For a = β1c, Ast will become Astmax, Therefore,

Astmax fy= 0.85fc′β1cbw + Asffy (For fc′ ≤ 4000 psi, β1 = 0.85)

Astmax = 0.27 (fc′/ fy) bwd + Asf (fy =60 ksi), Astmax = 0.3 (fc′/ fy) bwd + Asf (fy =40 ksi)

Alternatively Astmax (True T beam) = Asmax (singly) + Asf

So, for T-beam to behave in a ductile manner Ast, provided ≤ Astmax (True T beam)

• For fy = 40 ksi

c = 0.41d and, a = β1c = β10.41d

• For fy = 60 ksi

c = 0.37d and, a = β1c = β10.37d

18-May-21

16



Example 02

Design a simply supported T beam to resist a factored positive moment

equal to 6700 in-kip. The beam is 12″ wide and is having 24″ total depth

including a slab thickness of 3 inches. The centre to centre and clear

lengths of the beam are 25.5′ and 24′ respectively. The clear spacing

between the adjacent beams is 3 ft. Take fc′ = 3 ksi and fy = 60 ksi.




Solution:

Span length (lc/c) = 25.5′ ; clear length (ln) = 24′

d = 21.5″; bw = 12″; hf = 3″

Effective flange width (bf) is minimum of,

bw + 16hf = 12 + 16 × 3 = 60″

bw + sw = 12 + 3 × 12 = 48″

bw + ln/4 = 12 + 24 × 12/4 = 84″

Therefore, bf = 48″


18-May-21

17



Solution:

Check if the beam behaviour is T or rectangular.

Let a = hf = 3″

As = Mu/Φfy(d – a/2) = 6700/{0.90 × 60 × (21.5 – 3/2)} = 6.203 in2

a = Asfy/(0.85fc′bf) = 6.203 × 60/ (0.85 × 3 × 48) = 3.041″ > hf

Therefore, design as True T-beam.




Solution:

Design:

Calculate Asf

Asf = 0.85fc′ (bf – bw) hf/fy

= 0.85 × 3 × (48 – 12) × 3/60 = 4.59 in2

The nominal moment resistance (ФMn1), provided by Asf is,

ФMn1 = ФAsf fy {d – hf/2} = 0.9 × 4.59 × 60 × {21.5 – 3/2} = 4957.2 in-kip


18-May-21

18




Solution:

Design:

The nominal moment resistance (ФMn2), provided by remaining steel As is,

ФMn2 = Mu – ФMn1 = 6700 – 4957.2 = 1742.8 in-kip

Let a = 0.2d = 0.2 × 21.5 = 4.3″

As = ФMn2/ {Фfy(d – a/2)} = 1742.8 / {0.9 × 60 × (21.5 – 4.3/2)}= 1.667 in2

a = Asfy/(0.85fc′bw) = 1.667 × 60 / (0.85 × 3 ×12) = 3.27″

By trial and success method, finally As= 1.62 in2

Ast = Asf + As = 4.59 + 1.62 = 6.21 in2 (Use 9, #8 Bars)

Ast(Provided) = 9 x 0.8 = 7.2 in2



Solution:

Ductility requirements, Asmin ≤ (Ast = As + Asf) ≤ Astmax (True T-Beam)

Asmin = 3 ( fc′/ fy) bwd ≥ (200/fy) bwd (for fy ≤ 80 ksi)

3 ( fc′/ fy) bwd = 3 × (√(3000)/60000) x 12 x 21.5 = 0.706 in2

200/fy bw d = (200/60000) x 12 x 21.5 = 0.86 in2

Asmax (singly) =0.27(fc′/fy)bd = 0.27x(3/60)x12x21.5 = 3.48 in2

Astmax (True T- Beam) = Asmax (singly) + Asf = 3.48 + 4.59 = 8.07 in2

Asmin (0.8) < Ast(provided) (7.2) < Astmax (True T- Beam) (8.07) , O.K.


18-May-21

19



Solution: (Class Activity)

Check design capacity your self.





Drafting:

(3+3+3),#8 bars

d=19.125″

=12″

3″

bw

=48″bf

18-May-21

20



References

Design of Concrete Structures 14th Ed. by Nilson, Darwin and Dolan.

Building Code Requirements for Structural Concrete (ACI 318-19)

Design of T and L Beams in Flexure - Dr.Qaisar Ali

Documents

Transcript of Design of T and L Beams in Flexure - Dr.Qaisar Ali