NALYSIS AND DESIGN OF T A NALYSIS
BEAMS AND DOUBLY REINFORCED BEAMS
TOPICS Analysis and Design of T-Beams
T-Beams Analysis of T-Beams
Design of T-Beams
Design of T-Beams for Negative Moments
L-Shaped Beams
Analysis and Design of Doubly Reinforced Beams
Compression Steel Analysis of Doubly Reinforced Beams Design of Doubly Reinforced Beams
OBJECTIVES Analyze T-beams.
Design T-beams
Distinguish singly reinforced concrete beams from doubly reinforced beams
Analyze doubly reinforced beams
Design doubly reinforced beams
T-BEAMS
Reinforced concrete floor systems normally consist of slabs and beams that are placed monolithically. As a result, the two parts act together to resist loads. In effect, the beams have extra widths at their tops, called flanges called flanges,, and the resulting T-shaped beams are called Tbeams. The part of the T-beam below the slab is referred to as the web or stem.
T-BEAMS
Effective size of T-Beams
ACI Specs for effective flange width, width, b, of T-beams For Symmetrical T-Beams
b ≤ ¼ (beam span) ½ (b – (b – b bw) ≤ 8 (h f ) ↔ b ≤ bw +16 (hf ) b ≤ (center to center distance to the next web)
T-BEAMS
ACI Specs for effective flange width, width, b, of T-beams
For Isolated T-Beams
hf ≥ ½bw
b ≤ 4bw
A NALYSIS OF T-BEAMS
specifications relating to the strains in the reinforcing of T-Beams are identical to rectangular beams
t ≥ 0.005
t ≤ 0.004 unless the member is subjected to an axial loaad ≥ 0.10 f’c Ag
A NALYSIS OF T-BEAMS
The neutral axis for T-beams can fall either in the flange or in the stem, depending on the proportions of the slabs and stems.
A NALYSIS OF T-BEAMS
If the neutral axis falls in the flange, (a ≤ h f ) the section above the neutral axis is rectangular ass umed to be the concrete below the neutral axis is assumed cracked , and its shape has no effect effect on the flexure flexure calculation other than weight the rectangular formulas apply
A NALYSIS OF T-BEAMS
If the neutral axis falls in the stem, (a > h f ) the compression concrete section above the neutral axis is no longer a rectangle the rectangular formulas do not apply
A NALYSIS OF T-BEAMS
various cross sections of T-beams (a beam does not really have to look like a T beam to be one)
the concrete on the tension side is assumed to be cracked therefore the shape or size of the concrete has no effect on the theoretical resisting moments.
however, the shape, size, and weight of the tensile concrete affect deflections that occur and the dead weights affect the magnitudes magnitudes of the moments to be resisted
A NALYSIS OF T-BEAMS
Steps: 1.
Check Asmin
2.
Compute T = As fy
3.
Determine the area of the concrete in compression (Ac) stressed to 0.85 f’c C = T = 0.85 f’c Ac Ac = T/(0.85 f’c)
4.
Calculate a, c, and t
5.
Calculate Mn
A NALYSIS OF T-BEAMS
Example: Determine the design strength of the T-beam shown with f’c = 4000 psi and fy = 60000 psi.
A NALYSIS OF T-BEAMS
Solution:
Check Asmin: 3 f ' c
Asmin
Asmin
Asmin
fy
bw d
3 4000 60000
(10in)(24in)
0.76in 2 Asmin 0.80in
2
As As
Asmin
Asmin
Asmin
200bw d fy 200(10in)(24in)
6.00in 2 ok
60000 0.80in 2
A NALYSIS OF T-BEAMS
Solution:
Compute T: T Asfy Asfy
T
6.00in 2 (60ksi ksi )
T 360 360 k
A NALYSIS OF T-BEAMS
Solution:
Determine Ac:
T C 0.85 f ' cAc T 0.85 f ' cAc Ac Ac
T 0.85 f ' c 360k 0.85(4ksi)
Ac 105.88in 2
required concrete area on the compression side
A NALYSIS OF T-BEAMS
Solution:
Calculate the flange area: A flang e b h f A flang e 60in(4in) A flang e 240in 2
required concrete area on the compression side N.A. N.A. is in the flange
A NALYSIS OF T-BEAMS
0.85f’c
Solution:
c a
N.A.
C
T
Calculate a, c, and t: Ac b a a a
Ac
a 1 c c
b 105.88in 2
60in a 1.76in
c
a 1 1.76in
0.85 c 2.07in
t
0.003 003
d c c 0.003 003(d c) t c 0.003 003(24in 2.07in) t 2.07in 005 t 0.0318 0.005 section is ductile and and 0 90
A NALYSIS OF T-BEAMS
0.85f’c
Solution:
c a
N.A.
T
Calculate Mn: a Mn T (d ) 2 360 k )(24in Mn 0.90(360
Mn 7490 .9in.k 624 .2 ft .k Mn 624
an s
.
1.76in 2
)
C
A NALYSIS OF T-BEAMS
Example: Compute the design strenght for the T-beam shown in which f’c = 4000 psi and fy = 60000 psi.
A NALYSIS OF T-BEAMS
Solution:
Check Asmin: 3 f ' c
Asmin
Asmin
Asmin
fy
bw d
3 4000 60000
(14in)(30in)
1.33in 2 Asmin 1.40in
2
As As
Asmin
Asmin
Asmin
200bw d fy 200(14in)(30in) 60000 1.40in 2
10.12in2 ok
A NALYSIS OF T-BEAMS
Solution:
Compute T: T Asfy Asfy
T 10.12in 2 (60ksi ksi )
T
607 607.2k
A NALYSIS OF T-BEAMS
Solution:
Determine Ac:
T C 0.85 f ' cAc T 0.85 f ' cAc Ac Ac
T 0.85 f ' c 607.2k 0.85(4ksi)
Ac 178.59in 2
required concrete area on the compression side
A NALYSIS OF T-BEAMS
Solution:
Calculate the flange area: A flang e b h f A flang e 30in(4in) A flang e 120in 2
required concrete area on the compression side N.A. N.A. is in the stem
A NALYSIS OF T-BEAMS
Solution:
Locating the center of gravity of Ac: AT y Ay Ay y AT
2
2
120 120in (2in) 58.6in (4in y y 3 34in
2
120 120in 58.6in
2
4.19in 2
)
A NALYSIS OF T-BEAMS
0.85f’c
Solution:
a
c
C
y
N.A.
d-y
T
Calculate a, c, and t:
t
a 4in 4.19in
a
a 8.19in
c
c c
1
c
a
1 8.19in
0.85 9.64in
0.003 003
d c c 0.003 003(d c) t c 0.003 003(30in 9.64in) t 9.64in 005 t 0.00634 0.005 section is ductile and and 0 90
A NOTHER NOTHER METHOD IN A NALYZING NALYZING T-BEAMS
The beam is divided into a set of rectangular parts consisting of the overhanging parts of the flange and the compression part of the web
A NOTHER NOTHER METHOD IN A NALYZING NALYZING T-BEAMS
The total compression Cw in the web rectangle:
C w
0.85 f ' c
a bw
The total compression in the overhanging flange Cf :
C f 0.85 f ' c(b bw )(h f ) but if a < h f , replace h f with a
A NOTHER NOTHER METHOD IN A NALYZING NALYZING T-BEAMS
The nominal moment is determined by multiplying Cw and Cf by their respective lever arms from their centroids to the centroid of the steel:
a
h f
Mn C w (d ) C f (d ) 2 2
A NALYSIS OF T-BEAMS
Example: Compute the design strenght for the T-beam shown in which f’c = 4000 psi and fy = 60000 psi.
A NALYSIS OF T-BEAMS
Solution:
Check Asmin: 3 f ' c
Asmin
Asmin
Asmin
fy
bw d
3 4000 60000
(14in)(30in)
1.33in 2 Asmin 1.40in
2
As As
Asmin
Asmin
Asmin
200bw d fy 200(14in)(30in) 60000 1.40in 2
10.12in2 ok
A NALYSIS OF T-BEAMS
Solution:
Compute T: T Asfy Asfy
T 10.12in 2 (60ksi ksi )
T
607 607.2k
A NALYSIS OF T-BEAMS
Solution:
Determine Ac:
T C 0.85 f ' cAc T 0.85 f ' cAc Ac Ac
T 0.85 f ' c 607.2k 0.85(4ksi)
Ac 178.59in 2
required concrete area on the compression side
A NALYSIS OF T-BEAMS
Solution:
Calculate the flange area: A flang e b h f A flang e 30in(4in) A flang e 120in 2
required concrete area on the compression side N.A. N.A. is in the stem
A NALYSIS OF T-BEAMS
Solution:
Calculate a: A stem A stem A stem
A stem
Ac A flang e
178 178 .59in 58.6in
2
120 120in
2
h stem
h stem
2
bw ( h stem )
a h f h stem
58.6in 2
a 4in 4.19in
14in 4.19in
a 8.19in
a h f
A NALYSIS OF T-BEAMS
Solution:
Compute Cw : C w
0.85 f ' c a bw
C w
0.85(4ksi ksi )(8.19in)(14in)
C w
389 389.8k
Compute Cf : C f
0.85 f ' c(b bw )(h f )
C f
0.85(4ksi ksi )(30in 14in)(4in)
C f
217 217 .6k
A NALYSIS OF T-BEAMS
Solution:
Compute c :
a c
c c
1
c
a
1 8.19in
0.85 9.64in
Compute t: t
0.003 003
d c c 0.003 003(d c) t c 0.003 003(30in 9.64in) t 9.64in t 0.00634 0.005 section is ductile and 0.90
A NALYSIS OF T-BEAMS
Solution:
Calculate Mn: h f a Mn C w (d ) C f (d ) 2 2 8.19in 4in 389 .8k (30in ) 217 217 .6k (30in ) Mn 389 2 2 Mn 16190 in.k Mn 1349 ft .k
Calculate Mn: Mn 1349 ft .k
Mn (1349 ft .k ) Mn 0.90(1349 ft .k ) Mn 1214 ft .k
A NALYSIS OF T-BEAMS
Seatwork: Determine the design strength of the T-beam shown if f’c = 35MPa and fy = 420MPa and Es = 200000 MPa
PLATE NO. 20 A NALYSIS OF T-BEAMS
#5.46/146
PLATE NO. 21 A NALYSIS OF T-BEAMS
#5.47/146
DESIGN OF T-BEAMS
Steps
Determine the effective flange width
Determine the required reinforcement by trial and error
Compute Mn assuming = 0.90
Assume lever arm z equal equal to the larger larger of: z = 0.90d z = d – d – h hf /2
Compute trial steel area
Check a and z
Calculate As and compare with trial steel area (As trial steel area or previous area)
Check Asmin, c, t, and
DESIGN OF T-BEAMS
Example: Design a T-beam for the floor floo r system shown. MD = 80 ft.k, ML = 100 ft.k, f’c = 4000 psi, fy = 60000 psi, and simple span = 20 ft.
DESIGN OF T-BEAMS
Solution:
Effective flange width is the least of: (a) b
1
( beam span) 4 1 (20 ft ) 5 ft b 4 b 60in
(b) b bw 16h f
(c) b
(c to c dist to the next web)
b 12in 16( 4in)
b 10 ft
b 76in
120 in b 120
and A flange 240in b 60in and flang e b h f 60in( 4in) 240
2
DESIGN OF T-BEAMS
Solution: Compute Mu: Mu 1.2 M D 1.6 M L Mu 1.2(80 ft .k ) 1.6(100 100 ft .k ) Mu 256 256 ft .k
Design reinforcement by trial and error:
Compute Mn assuming = 0.90: Mn
Mu
Mu
Mn
Mn
Mn
256 256 ft .k 0.90 284 284 4 ft k
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Assume lever arm z equal to the larger of: (a) z 0.90d
(b) z d
z 0.90(18in)
z 18in
z 16.20in
h f 2 4in
z 16in
z 16.20in
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
As
12in 284 284 .4 ft .k ( ) 1 ft 60ksi(16.2in)
As 3 51in 2
2
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Compute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(4ksi) Ac 3.51in 2 (60ksi) Ac 61.9in 2
A flang e 240in 2
N.A. N.A. in flange
Ac ab a a
Ac b 61.9in 2 60in
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z with N.A. in the flange: z d
a
2
z 18in
1.03in 2
z 17.48in
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
As
12in 284 284 .4 ft .k ( ) 1 ft 60ksi(17.48in)
As 3.25in 2 not close to
i
As
try again
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(4ksi) Ac 3.25in 2 (60ksi) Ac 57.4in 2
A flang e 240in 2
N.A. N.A. in flange
Ac a
a
ab Ac
b 57.4in
60in
2
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z with N.A. in the flange: z
z z
d
a
2 0.96in
18in
17.52in
2
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
As
12in 284 284 .4 ft .k ( ) 1 ft 60ksi(17.52in)
As 3.25in 2
close to previ previous As ok
DESIGN OF T-BEAMS
Solution: Check Asmin: 3 f ' c
Asmin
Asmin
Asmin
fy
bw d
3 4000 60000
(12in)(18in)
0.68in 2
Asmin
0.72in
2
As As
Asmin
Asmin
Asmin
3.25in2
200 200bw d fy 200 200 (12in)(18in) 60000 0.72in 2
ok
DESIGN OF T-BEAMS
Solution: Compute c : a c
c
1
t
c
a
1 0.96in
0.85 c
Compute t:
1.13in
0.003
d c c 0.003(d c) t c 0.003(18in 1.13in) t 1.13in 005 t 0.045 0.005
section is ductile and
0.90 as assumed
Required steel reinforcement :
As 3.25in As
2
DESIGN OF T-BEAMS
Example: Design a T-beam for the floor floo r system shown. MD = 200 ft.k, ML = 425 ft.k, f’c = 3000 psi, fy = 60000 psi, and simple span = 18 ft.
DESIGN OF T-BEAMS
Solution:
Effective flange width is the least of: (a) b
1
( beam span) 4 1 (18 ft ) 4.5 ft b 4 b 54in
(c) b
(b) b bw 16h f
(c to c dist to the next web)
b 15in 16(3in)
b
6 ft
b 63in
b
72in
and A flange b 54in and flang e
b h f
54in(3in) 162 162in 2
DESIGN OF T-BEAMS
Solution: Compute Mu: Mu 1.2 M D 1.6 M L Mu 1.2(200 200 ft .k ) 1.6( 425 425 ft .k ) Mu 920 920 ft .k
Design reinforcement by trial and error:
Compute Mn assuming = 0.90: Mn
Mu
Mu
Mn
Mn
Mn
920 920 ft .k 0.90 1022 ft k
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Assume lever arm z equal to the larger of: (a)
z
0.90d
z
0.90(24in)
(b) z d
h f
z 24in
z
21.6in
2 3in
z 22.5in
z
22.5in
Compute trial steel area: Mn Tz Mn
12in 1022 ft .k ( ) 1 ft As 60ksi (22.5in)
fy z
As 9 08in 2
Mn Asfyz Asfyz As
2
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Compute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(3ksi ksi ) Ac 9.08in 2 (60ksi ksi ) 213 .6in 2 Ac 213
162in 2 A flang e 162
N.A N.A.. in stem or web
A stem Ac A flang e 2 2 A stem 213 213 .6in 162 162in 2 A stem 51.6in
A stem bw h stem 51.6in 2 15in(h stem ) h stem 3.44in
: DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Locating the neutral axis of Ac : AT y Ay Ay y AT
162 162in 2 (1.5in) 51.6in 2 (3in y y 2.28in
162 162in 2 51.6in 2
3.44in 2
)
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z : z d y
z
24in
z
21.72in
2.28in
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
12in 1022 ft .k ( ) 1 ft As 60ksi (21.72in) As 9.41in 2 not close to
i
As
try again
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(3ksi ksi ) Ac 9.41in 2 (60ksi ksi ) 221 .4in 2 Ac 221
162in 2 A flang e 162
N.A N.A.. in
stem
A stem Ac A flang e
A stem bw h stem
A stem 221 221 .4in 2 162 162in 2
59.4in 2 15in(h stem )
A stem 59.4in 2
h stem 3.96in
: DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Locating the neutral axis of Ac : AT y Ay Ay y AT
2
2
162 162in (1.5in) 59.4in (3in y y 2.43in
2
162 162in 59.4in
2
3.96in 2
)
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z : z d y
z
24in
z
21.57in
2.43in
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
12in 1022 ft .k ( ) 1 ft As 60ksi (21.57in) As 9.48in 2 not close to
i
As
try again
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(3ksi) Ac 9.48in 2 (60ksi) Ac 223.1in 2
A flang e 162 162in 2
N.A N.A.. in
A stem
A stem
223 223 .1in
A stem
61.1in
A stem bw h stem
Ac A flang e
2
2
162 162in
stem
2
61.1in 2 15in(h stem ) h stem 4.1in
: DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Locating the neutral axis of Ac : AT y Ay Ay y AT
162 162in 2 (1.5in) 61.1in 2 (3in y y 2.47in
162 162in 2 61.1in 2
4.1in 2
)
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z : z d y
z
24in
z
21.53in
2.47in
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
12in 1022 ft .k ( ) 1 ft As 60ksi (21.53in) As 9.49in 2 not close to
i
As
try again
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(3ksi ksi ) Ac 9.49in 2 (60ksi ksi ) 223 .3in 2 Ac 223
A flang e 162in 2
N.A N.A.. in
stem
A stem Ac A flang e
A stem bw h stem
A stem 223 223 .3in 2 162 162in 2
61.3in 2 15in(h stem )
A stem 61.3in 2
h stem 4.09in
: DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Locating the neutral axis of Ac : AT y Ay Ay y AT
162 162in 2 (1.5in) 61.3in 2 (3in y y 2.5in
162 162in 2 61.3in 2
4.09in 2
)
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z : z d y
z
24in
z
21.5in
2.5in
Compute trial steel area: Mn Tz Mn Asfyz Asfyz As
Mn fy z
12in 1022 ft .k ( ) 1 ft As 60ksi (21.5in) As 9.51in 2 not close to
i
As
try again
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute a : C T 0.85 f ' c Ac Asfy Asfy 0.85(3ksi ksi ) Ac 9.51in 2 (60ksi ksi ) 223 .8in 2 Ac 223
162in 2 A flang e 162
N.A N.A.. in
stem
A stem Ac A flang e
A stem bw h stem
A stem 223 223 .8in 2 162 162in 2
61.8in 2 15in(h stem )
A stem 61.8in 2
h stem 4.12in
: DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Locating the neutral axis of Ac : AT y Ay Ay y AT
162 162in 2 (1.5in) 61.8in 2 (3in y y 2.5in
162 162in 2 61.8in 2
4.12in 2
)
DESIGN OF T-BEAMS
Solution: Design reinforcement by trial and error:
Recompute z : z d y
z
24in
z
21.5in
2.5in
Compute trial steel area: Mn Tz Mn
12in 1022 ft .k ( ) 1 ft As 60ksi (21.5in)
fy z
As 9.51in 2
Mn Asfyz Asfyz As
close to
i
As
k
DESIGN OF T-BEAMS
Solution: Check Asmin: Asmin
Asmin
Asmin
3 f ' c fy
bw d
3 4000 60000
(15in)(24in)
1.14in 2
Asmin 1.2in
2
As
Asmin
Asmin
Asmin
9.51in 2
200 200bw d fy 200 200(15in)(24in) 60000 1.2in 2
ok
DESIGN OF T-BEAMS
Solution: Compute c : a c
c c
Compute t: d
a
0.003 003
t
1 c 1 3in 4.12 0.85 8.38in
c
c
0.003 003(d
t
t
t
c)
c
0.003 003(24in 8.38in)
8.38in 0.0056
0.005 005 section is ductile and
0.90 as assumed
Required steel reinforcement :
As As 9.51in 2
A LTERNATE LTERNATE METHOD IN DESIGNING T-BEAMS
Steps:
Assume a value of z
Compute a trial steel area
Determine a for that steel area assuming a rectangular section
If a < hf , then design as rectangular beam with b = effective flange width
If a > hf , then the section is a real T-beam
Divide the beam into its rectangular components The compressive force provided by the overhanging flange rectangles will be balanced by the tensile force A sf
A LTERNATE LTERNATE METHOD IN DESIGNING T-BEAMS
Steps:
The compression force in the web is balanced by the remaining tensile force A sw
For the overhanging flange 0.85 f ' c(b bw )h f As f fy As f
0.85 f ' c(b bw )h f fy
Mu f As f fy(d
h f 2
)
For the web Muw Mu Mu f
Asw
bw d
Required steel reinforcement reinforcement
Ass A A Ass f As w
DESIGN OF T-BEAMS
Example: Design a T-beam for the floor floo r system shown. MD = 200 ft.k, ML = 425 ft.k, f’c = 3000 psi, fy = 60000 psi, and simple span = 18 ft.
DESIGN OF T-BEAMS
Solution:
Effective flange width is the least of: (a) b
1
( beam span) 4 1 (18 ft ) 4.5 ft b 4 b 54in
(b) b bw 16h f
(c) b
(c to c dist to the next web)
b 15in 16(3in)
b
6 ft
b 63in
b
72in
b 54in
DESIGN OF T-BEAMS
Solution: Compute Mu: Mu 1.2 M D 1.6 M L Mu 1.2(200 200 ft .k ) 1.6( 425 425 ft .k ) Mu 920 920 ft .k
Design reinforcement:
Solve Asf assuming = 0.90: 0.85 f ' c(b bw )h f As f fy As f
0.85(3ksi)(54in 15in)(3in)
As f 4.97in 2
60ksi ksi
DESIGN OF T-BEAMS
Solution: Design reinforcement:
Solve Muf : Mu f
As f fy (d
Mu f
Mu f
6039 in.k
Mu f
503 ft .k
h f 2
)
0.90(4.97in 2 )(60ksi)(24in
Solve Muw : Muw Mu Mu f Muw 920 920 ft .k 503 503 ft .k Mu
417 417 ft k
3in
2
)
DESIGN OF T-BEAMS
Solution: Design reinforcement:
Solve Asw :
Muw
12in 1000lb 417 ft .k ( )( ) 1 ft 1k
fybw d 2 (1
0.59 fy
f ' c
)
0.90 (60000 psi)(15in)(24in) 2 (1
0.0126
Required As :
Asw
bw d
Asw
0.0126 (15in)(24in)
Asw
4.54in 2
As As f Asw As 4.97in 2 4.54in 2 2
0.59 (60)
4
)
DESIGN OF T-BEAMS FOR NEGATIVE MOMENTS
When T-beams are resisting negative moments, their flanges will be in tension and the bottom of their stems will be in compression.
in tension
in compression (a rectangular beam)
DESIGN OF T-BEAMS FOR NEGATIVE MOMENTS ACI Code requires that part of of the flexural steel in the top of the beam in the negative moment region be distributed over the effective width of the flange or over a width equal to one-tenth of the beam span, whichever is smaller.
in tension
in compression (a rectangular beam)
DESIGN OF T-BEAMS FOR NEGATIVE MOMENTS
Should the effective width be greater than oneone tenth of the span length, the Code requires that some additional longitudinal steel be placed in the outer portions of the flange.
in tension
in compression (a rectangular beam)
DESIGN OF T-BEAMS FOR NEGATIVE MOMENTS
For statically determinate members with their flanges in tension, bw in the equations is to be replaced with either 2bw or the width of the flange, whichever is smaller.
in tension
in compression (a rectangular beam)
L-SHAPED BEAMS
edge T-beams with a flange on one side only
Effective width of the overhanging flange: b 1/12 (span length of beam) b bw + 6 hf b ½ (center to center dist. to the next web)
PLATE NO. 22 DESIGN OF T-BEAMS
#5.23/142
PLATE NO. 23 DESIGN OF T-BEAMS
#5.49/147
DOUBLY REINFORCED CONCRETE BEAM
concrete beam that has compression steel as well as tensile steel
COMPRESSION STEEL
steel used on the compression side of the beam
needed in addition to tensile steel when space or aesthetic requirements limit beams to small sizes
increases not only the resisting moments of concrete sections, but also the amount of curvature that a member can take before flexural failure
COMPRESSION STEEL
makes beam tough and ductile, enabling them to withstand large moments, deformations,and stress reversals that might occur during earthquakes
effective in reducing long term deflections due to shrinkage and plastic flow
helpful for positioning stirrups by tying them to the compression bars and keeping them in place during concrete placement and vibration
DOUBLY REINFORCED BEAMS
when compression steel is used, the nominal resisting moment of the beam consist of two part:
the part due to the resistance of the compression concrete and the balancing tensile reinforcement
the part due to the nominal moment capacity of compression steel and the balancing amount of the additional tensile steel.
DOUBLY REINFORCED BEAMS
doubly reinforced beam
compression in concrete is balanced by the tensile reinforcement
compression steel is balanced by the additional tensile reinforcement
DOUBLY REINFORCED BEAMS
Nominal moment moment of beam, assuming assuming the tensile tensil e steel has yielded yielded
Mn Mn1 Mn2 a
Mn As As1 fy(d ) As As2 fy(d d ' ) 2 a As1 fy(d ) As As2 fy(d d ' )] Mn [ As 2
DOUBLY REINFORCED BEAMS
Nominal moment moment of beam, assuming assuming the compression steel has yielded
Mn Mn1 Mn2 a
Mn As As1 fy(d ) As As ' fy(d d ' ) 2 a As1 fy(d ) As As ' fy(d d ' )] Mn [ As 2
DOUBLY REINFORCED BEAMS
Nominal moment moment of beam, assuming assuming the compression steel has not yielded yielded
Mn Mn1 Mn2 a Mn As As1 fy(d ) As As ' fs' (d d ' ) 2 a As1 fy(d ) As As ' fs' (d d ' )] Mn [ As 2
A NALYSIS OF DOUBLY REINFORCED BEAMS Steps:
Check whether the compression steel has yielded or not. if s ' s
s '
s E s
then, compression steel has yielded and fs' fy
A NALYSIS OF DOUBLY REINFORCED BEAMS Steps:
Check the ductility of the section a)
005 if t 0.005 then, the section is ductile and 0.90
b)
004 t 0.005 005 if 0.004 then, the section is in transition range and
0.48 83
t
A NALYSIS OF DOUBLY REINFORCED BEAMS Steps:
Check the ductility of the section c)
004 if t 0.004 then, the section is not ductile and the section may not be used
Calculate the moment capacity of the section
a
Mn [ As As1 fy(d ) As As ' fy(d d ' )] 2 or a As1 fy(d ) As As ' fs' (d d ' )] Mn [ As 2
A NALYSIS OF DOUBLY REINFORCED BEAMS
Maximum permissible As to ensure tensile failure
As Asmax
maxbd
As As ' fs' fy
EXAMPLE
Determine the design moment capacity of the beam shown for which fy = 60000 psi and fc’ = 3000 psi.
SOLUTION
check if compression steel has yielded Solving for c: Asf Asfyy
0.85 fc' 1cb As' ( As
6.25in
2
fy
60000 psi
fc'
3000 psi
1
b
14in
As' d '
Es
0.85
2in
2
2.5in
29000 ksi
by quadratic formula 8 34in
c d ' c
)(0.003 003) Es
SOLUTION
check if compression steel has yielded Solving for a:
1c
a
a
0.85(8.34in)
a
7.09in
SOLUTION
check if compression steel has yielded Solving for s’:
s ' c
0.003
d '
s '
s
'
c
(c
d ' )0.003 c
(8.34in 2.5in)0.003
8.34in
s ' 0.00210
SOLUTION
check if compression steel has yielded Solving for s:
s
E s
s
fy E s 60000 psi p si
s
s
29000000 psi 0.00207
SOLUTION
check if compression steel has yielded Compare s‘ and s: ' s
0.00210
s
0.00207
the compression steel has yielded and fs’ = fy
SOLUTION
check ductility of the section
t d
0.003
c
t t
c
(d
c)0.003 c
(24in 8.34in)0.003
8.34in
t 0.00563 > 0.005 section is ductile and = 0.90
SOLUTION
calculate the design moment strength a Mn [ As As1 fy(d ) As As ' fy(d d ' )] 2 Solving for As1 : T 2 C '
As1 As As2 but but As As
As2 fy As' fs ' since
fs' fy
Ass2 fy A A Ass' fy
As As2
As '
As As 2 2in 2
As1 As As As As2 As1
As1
6.25in
2
4.25in
2
2in
2
SOLUTION
calculate the design moment strength
a
Mn [ As As1 fy(d ) As As ' fy(d d ' )] 2
0.90
fy
60000 psi
d
24in
d '
2.5in
a
As ' As1
7.09in
2in
2
4.25in
Mn 7016 in.k 584 .7 ft .k Mn 584
2
EXAMPLE
Compute the design moment strength of the section se ction shown. fy = 60000 psi and fc’ = 4000 psi.
SOLUTION
check if compression steel has yielded Solving for c: Asf Asfyy
0.85 fc' 1cb As' ( As
5.06in
2
fy
60000 psi
fc'
4000 psi
1
b
14in
As' d '
Es
0.85
1.2in
2
2.5in
29000 ksi
by quadratic formula 6in
c d ' c
)(0.003 003) Es
SOLUTION
check if compression steel has yielded Solving for a: a
1c
a
0.85(6in)
a
5.10in
SOLUTION
check if compression steel has yielded Solving for s’:
s ' c
0.003
d '
s '
s
'
c
(c
d ' )0.003 c
(6in 2.5in)0.003
6in
s ' 0.00175
SOLUTION
check if compression steel has yielded Solving for s:
s s
s E s
fy E s 60000 psi p si
s
s
29000000 psi 0.00207
SOLUTION
check if compression steel has yielded Compare s‘ and s: ' s
0.00175
s
0.00207
the compression steel has not yielded
fs ' s ' E s
fs' 0.00175 (29000000 psi) fs' 50750 psi
SOLUTION
check ductility of the section
t d
0.003
c
c
(d
t
c)0.003 c
(24in 6in)0.003 003
t
t
6in
0.009 009
> 0.005
section is ductile and = 0.90
SOLUTION
calculate the design moment strength a Mn [ As As1 fy(d ) As As ' fy(d d ' )] 2 Solving for As1 : T 2 C '
As1 As As2 but but As As
As2 fy As' fs ' As As ' fs' As As2 fy As2
As1
(1.2in 2 )(50750 psi)
As2
60000 psi
1 01 015 5in
As1 As As As As2
2
As1
5.06in
2
1.015 015in
4.045 045in
2
2
SOLUTION
calculate the design moment strength
a
Mn [ As As1 fy(d ) As As ' fs' (d d ' )] 2
0.90
fy
60000 psi
fs '
50750 psi
d
24in
d '
2.5in
a
As ' As1
5.10in
1.2in
2
4.045 045in
Mn 5863in.k Mn 488 488 6 ft k
2
PLATE NO. 24 #5.31/144 A NALYSIS OF DOUBLY REINFORCED BEAMS PLATE NO. 25 #5.33/144 A NALYSIS OF DOUBLY REINFORCED BEAMS PLATE NO. 26 #5.51/148 A NALYSIS OF DOUBLY REINFORCED BEAMS
DESIGN OF DOUBLY REINFORCED BEAM Steps:
Compute Mu
Calculate Mn assuming = 0.90
Compute beam’s nominal moment strength assuming maximum possible tensile steel with no compression steel
EXAMPLE
Design a rectangular beam for MD = 325ft.k and ML = 400ft.k if fc’ = 4000psi 4000ps i and fy = 60000psi. The maximum permissible beam dimension are shown.
SOLUTION
Compute Mu
Mu 1.2 M D 1.6M L Mu 1.2(325 ft .k ) 1.6(400 ft .k ) Mu 1030 ft .k
Compute Mn (assume = 0.90) Mu
Mn Mn
Mn Mu
1030 ft .k
0.90
1144 .4 ft .k
total nominal moment the section should resist
SOLUTION
Compute maximum possible tensile steel ratio, max: 0.85 1 f ' c fy
375 max 0.375
0.85(0.85)(4000 ) 60000
max 0.375 375
max 0.0181
SOLUTION
Compute beam’s nominal moment strength, Mn 1, assuming maximum possible tensile steel with no compression steel Mn1 fybd fybd 2 (1
Mn1
0.59 fy
f ' c
0.0181(60000 psi)(15in)(28in) 2 [1
) 0.59(0.0181)(60ksi ksi )
4ksi ksi
]
Mn1 10725579 .7in.lb Mn1 893.8 ft .k
since Mn1 < Mn, design section as doubly reinforced beam
SOLUTION
Compute As1, the amount of tensile reinforcement that will balance the compression concrete section As1
maxbd
0.85 1 f ' c fy
375 max 0.375
0.85(0.85)(4000 ) 60000
375 max 0.375
max 0.0181
As As1
0.181(15in)(28in)
As As1 7.6in 2
SOLUTION
Solve for Mn2, the nominal moment that the compression steel must resist
Mn Mn1 Mn2
Mn2 Mn Mn1
Mn2
1144.4 ft .k 893.8 ft .k
Mn2
250.6 ft .k
SOLUTION
Design compression reinforcement, As’
Mn2 As'
As As' fy(d d ' ) Mn2 fy(d d ' )
12in 250 250.6 ft .k ( ) 1 ft As' 60ksi ksi (28in 3in)
try #9 rebars n
'
Ad
b
n n
As' 2in 2
As
2in 2 1in 2 2
use 2#9 rebars As As 'actual
2in
2
SOLUTION
check if compression steel has yielded Solving for a: T 1
Ass1 fy A
a a
C
0.85 fc' ab
As1 fy 0.85 fc' b
7.6in 2 (60ksi ksi )
0.85(4ksi ksi )(15in) a
8.94in
SOLUTION
check if compression steel has yielded Solving for c: a
1c a
c
c
1 8.94in
0.85 c
10.52in
SOLUTION
check if compression steel has yielded Solving for s’:
s ' c
0.003
d '
s '
s
c
(c
'
d ' )0.003 c
(10.52in 3in)0.003
10.52in
s ' 0.00214
SOLUTION
check if compression steel has yielded Solving for s:
s s
s E s
fy E s 60000 psi p si
s
s
29000000 psi 0.00207
SOLUTION
check if compression steel has yielded Compare s‘ and s: ' s
0.00214
s
0.00207
the compression steel has yielded
fs
'
fy
fs' 60000 psi
SOLUTION
Compute As2, the required tensile reinforcement that will balance the compression steel T 2 C '
As As2 fy As As ' fs' Ass2 fy A A Ass ' fy
As As2
As '
As As 2
2in
2
SOLUTION
Design tensile reinforcement, As As As As As1 As2
try #10 rebars
As As 7.60in 2 2in 2 As
9.60in
n
2
As
'
Ad
b
n
n
9.60in 1.27in
7.56
2
2
say 8
use 8#10 rebars
As Asactual 10.16in 2
SOLUTION
re-check if compression steel has yielded Solving for c: Asf Asfyy
0.85 fc' 1cb As' ( As
10.16in
2
fy
60000 psi
fc '
4000 psi
1
b
15in
As' d '
Es
0.85
2in
2
3in
29000 ksi ksi
by quadratic formula 10 43in
c d ' c
)(0.003 003) Es
SOLUTION
check if compression steel has yielded Solving for a:
1c
a
a
0.85(10.43in)
a
8.86in
SOLUTION
check if compression steel has yielded Solving for s’:
s ' c
0.003
d '
s '
s
'
c
(c
d ' )0.003 c
(11.08in 3in)0.003
11.08in
s ' 0.00219
SOLUTION
check if compression steel has yielded Compare s‘ and s: ' s
0.00219
s
0.00207
the compression steel has yielded ok
SOLUTION
check ductility of the section
t d
0.003
c
t t
c
(d
c)0.003
c
(28in 11.08in)0.003
t
11.08in
0.0046 < 0.005
section is in transition range
SOLUTION
check ductility of the section determine :
0.48 83 t
0.48 83(0.0046 )
0.86
SOLUTION
check the adequacy of the designed section
a
Mn [ As As1 fy(d ) As As ' fs' (d d ' )] 2 0.86 As1( actual )
Asactual As2 ( actual )
As1( actual )
10.16in 2 2in 2
fy 60000 psi fc ' 4000 psi 1
0.85
b 15in As'actual 2in 2 d ' 3in Es 29000 ksi
8.16in 2
SOLUTION
check the adequacy of the designed section
a
Mn [ As As1 fy(d ) As As ' fs' (d d ' )] 2 Mn 12910in.k Mn
1076 ft .k
Mu
1030 ft .k
ok, beam is adequate
EXAMPLE A beam is limited to the dimensions shown. If MD = 170ft.k, ML = 225ft.k, fc’ = 4000psi and fy = 60000psi, select the reinforcing required .
SOLUTION
Compute Mu
Mu 1.2 M D 1.6M L Mu 1.2(170 ft .k ) 1.6(225 ft .k ) Mu
564 56 4 ft .k
Compute Mn (assume = 0.90) Mu
Mn Mn
Mn Mu
564 ft .k 0.90
626 62 6.7 ft .k
total nominal moment the section should resist
SOLUTION
Compute maximum possible tensile steel ratio, max: 0.85 1 f ' c fy
max 0.375
0.85(0.85)(4000 ) 60000
375 max 0.375
max 0.0181
SOLUTION
Compute beam’s nominal moment strength, Mn 1, assuming maximum possible tensile steel with no compression steel Mn1 fybd fybd 2 (1
Mn1
0.59 fy
f ' c
0.0181(60000 psi)(15in)(20in) 2 [1
) 0.59(0.0181)(60ksi ksi )
4ksi ksi
]
Mn1 5472000 in.lb Mn1
456 ft .k
since Mn1 < Mn design section as doubly reinforced beam
SOLUTION
Compute As1, the amount of tensile reinforcement that will balance the compression concrete section As1
maxbd
0.85 1 f ' c fy
max 0.375
0.85(0.85)(4000 ) 60000
max 0.375
max 0.0181
As As1 As As1
0.181(15in)(20in)
5.43in 2
SOLUTION
Solve for Mn2, the nominal moment that the compression steel must resist
Mn Mn1 Mn2
Mn2 Mn Mn1 626.7 ft .k 456 ft .k
Mn2
Mn M n2
170.7
ft .k
SOLUTION
Design compression reinforcement, As’
Mn2 As'
As As' fy(d d ' ) try #8 rebars
Mn2 fy(d d ' )
As'
n
12in 170 170.7 ft .k ( ) 1 ft 60ksi ksi (20in 4in)
As
'
Ad
b
n
3.15in 2 0.79in 2
n 3.98 say 4 As As '
3.15in
2
use 4#8 rebars As As 'actual
3.16in
2
SOLUTION
check if compression steel has yielded Solving for a: T 1
C
As As1 fy 0.85 fc' ab a a
As1 fy 0.85 fc' b
5.43in 2 (60ksi ksi )
0.85(4ksi ksi )(15in)
a 6.39in
SOLUTION
check if compression steel has yielded Solving for c: a
1c a
c
c
1 6.39in
0.85 c
7.52in
SOLUTION
check if compression steel has yielded Solving for s’:
s ' c
0.003
d '
s '
s
c
(c
'
d ' )0.003 c
(7.52in 4in)0.003
7.52in
s ' 0.00140
SOLUTION
check if compression steel has yielded Solving for s:
s s
s E s
fy E s 60000 psi p si
s
s
29000000 psi 0.00207
SOLUTION
check if compression steel has yielded Compare s‘ and s: ' s
0.00140
s
0.00207
the compression steel has not yielded
fs ' s ' E s
fs' 0.00140 (29000000 psi)
fs' 40600 psi
SOLUTION
Compute As2, the required tensile reinforcement that will balance the compression steel T 2 C '
As As2 fy As As ' fs' As As 2 (60 ksi ksi ) As2
(3.16in 2 )(40.6ksi ksi )
2.14in
2
SOLUTION
Design tensile reinforcement, As As As As As1 As2 As A s As As
5.43in
2
7.57in
try #10 rebars 2.14in
2
n
As
2
'
Ad
b
n
n
7.57in 1.27in
5.95
2
2
say 6
use 6#10 rebars As Asactual
7.59in
2
SOLUTION
re-check if compression steel has yielded Solving for c: Asf Asfyy
0.85 fc' 1cb As' ( As
7.59in
2
fy
60000 psi
fc '
4000 psi
1
b
15in
As' d '
Es
0.85
3.16in
2
4in
29000 ksi ksi
by quadratic formula 7 53in
c d ' c
)(0.003 003) Es
SOLUTION
check if compression steel has yielded Solving for a:
1c
a
a
0.85(7.56in)
a
6.4in
SOLUTION
check ductility of the section
t d
0.003
c
t t
c
(d
c)0.003 c
(20in 7.53in)0.003
7.53in
t 0.00497
< 0.005
section is in transition range
SOLUTION
check ductility of the section determine :
0.48 83 t 0.48 83(0.00497 )
0.89
SOLUTION
check the adequacy of the designed section
a
Mn [ As As1 fy(d ) As As ' fs' (d d ' )] 2 0.89
As1( actual )
As1( actual )
fy
Asactual As2 ( actual )
7.59in 2
60000 psi
fc' 4000 psi
1
0.85
b 15in
As 'actual 3.16in 2
d ' 4in
Es
29000 ksi
2.13in 2
5.46in 2
SOLUTION
check the adequacy of the designed section
a
Mn [ As As1 fy(d ) As As ' fs' (d d ' )] 2 Mn
Mn
6800 .8in.k
566 56 6.7 ft .k
Mu
564 56 4 ft .k
ok, beam is adequate
SEATWORK (1WHOLE)
Solve Prob. #538/145
PLATE NO. 27 #5.35/145 DESIGN OF DOUBLY REINFORCED BEAMS PLATE NO. 28 #5.53/148 DESIGN OF DOUBLY REINFORCED BEAMS PLATE NO. 29 DESIGN OF DOUBLY REINFORCED BEAMS FLOWCHART
F05_19
p113
p125
p126
pr05_05
pr05_08
pr05_11
pr05_12
pr05_13
pr05_14
pr05_15
pr05_16
pr05_18
pr05_21
pr05_22
pr05_23
pr05_26
pr05_27
pr05_28
pr05_29
pr05_30
pr05_31
pr05_32
pr05_34
pr05_35
pr05_36
pr05_37
pr05_37
pr05_46
pr05_47
pr05_48
pr05_49
pr05_50
pr05_51
pr05_52
pr05_53
pr05_54
pr05_59
