234 STEPS IN FINDING OM, OF A BEAM WITH KNOWN TENSION STEEL AREA 4,
AND OTHER BEAM PROPERTIES.
Given: b, d, ff, fy, and Ag
I. Solve for p:
Is
°* ba
Pmin shall be the greater of (a) or (b):
o.25/f
@) Pmin = i
) es
) Pmin =
Il. Check if steel yields by computing py
sp, £000
Pr = 0.8! BF 600+ hy
ifp py steel does not yield, proceed to Step IV
NOTE: if p < Pmin, the given A, is not adequate for the beam dimension.
— 7
UL. p < py
' 0.85f!ab = Asfy
a
iff, > 1000 MPa, tension controlled,
iff, < 1000 MPa, transition region
fe-fy
1000 -
= a a
OM, = 00.85fab (d — a
9 =0.65+ 0.25 3: ANALYSIS AND DESIGN OF BEAMS FOR FLE
LEXURE
CHAPTER
(STRENGTH DESIGN METHOD)
IV. p > Pb
compression controlled, @ = 0.65
CC. =T
0.85flab = Asfs
a= fie &f, = 6005—=
d-c C
0.85f{Bycb = As600——
ce
a=fyc = —
d
fe = 600—— = —
Thus,
om, or (6-$)=04s (4-5)
or
OM, = 06. (a 5) = 90.85/eab (a-$) ‘HAPTER 2: ANALYSIS AND DI OF BEAMS FOR FLEXURE
mae C
(STRENGTH DESIGN METHOD)
ILLUSTRATIVE PROBLEMS:
Solution:
3 fz (__600
Pn = ose ( )
600+ f
» By = 0.85 for f = 21 MPa
21/600
=08 (= __) = 0.02792
/ Pb Te DBALO. 85) 335 (goo rl coe
b.
: ag
Pb ra
bd
Asp = 0.02792(300)(500)
Asp = 4188 mm?
Solution:
, 600d
% S600 = are ili. caine
b.
a is 600
nae i (eorng)
0.05
B, = 0.85 — |
— 28) = 0.85 8 30 — 28) = 0.836
bo CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLE vs
(STRENGTH DESIGN METHOD) LEXURE
6) 32 ( 600)
Pr = 0.85(0.836) 773 (G99 4 =z) = 0.03047
As,
Po =a
bd
‘Agy = 0.03047(300)(500)
Agy = 4570.5 mm?
Solution:
a
a (se
“lag
By = 0.85 for f = 21 MPa
2
Pmax = 5(——Se
345 0.01885
b.
Asmasx ;
Agmax = 0.01885(300)(500)
Agmax = 2827.5 mm?
c ]
51 3
= O0—A#, fibd? (1-—,
OMamax = OF Ps k'ba? (1-= 61)
800-f, 800 - 345
S A ae = BZA
Ones = 0165 + 0.25 75555 = 065 + 0.25 7959-545 = °
OM pmax = 08244 (0 2 (1-2 08s)
inmax = 0.824 (0,85)(21)(300)(500)? ( 1 - + (0.
0 328.660 kNm
CONC AEIED REINFORGED e a
CONCRETE DESIGN
te
MJBCASTRO 1D) DESIGN OF BEAMS FOR FLEXURE |
28 CHAPTER 2: ANALYSIS
(STRENGTH DESIGN METHOD)
Solution:
7 3
ee 3a = 5 (500) = 214.286 mm
b.
a ( Bee
Pmax = 7\—
By = 0.85 for f/ = 28 MPa
85)
pn =a
es
Pmax =a
Asmax = 0.02094(300)(500)
Asmax = 3141 mm?
Solution:
Solve for OMpmax:
oe SL , 3
Mynax = OF ap Pifibd? (1 -ah)
| penance ‘HAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTA DESIGN METHOD)
ae 800 ~ 276
max = 0.65 + 0.25 ——__ = 0.65 4.0.25 0 276 _
1000 = fy 787000 = 276 = 0.831
By = 0.85 for
27.6 MPa
Then,
51 4
OMpmax = (0.831) 755 (0.85)(27.6)(300)(490)2 (1 a 740089)
OMpmax = 418.371 kNm
~~
a. My =20kNm <@Mymax — Singly Reinforced
Determine if the section is tension-controlled or transition:
459, 3
OMae = egg Prlibd? (1- hr)
459 3
OMne = FeG9 (0-85)(27.6)(300)(490)* (1 aa (085))
OMe = 407.508 kNm
OMnr > M, — Tension — controlled, 0 = 0.90
Solve for Ry:
M, = OR,,bd®
20x 106 = 0.90R,,(300)(490)?
Ry, = 0.30851
a.
Pt BSF:
08576), |, _ 2(0.30851)
276 rR 0.85(27.6)
p= 0.00113
Solve for p:
Check min?
4
Pmin =” = 976 = 0.00507 > 0.00113
+ Use p= min = 0.00507
Thus, see
Az = pbd = 0.00507(300)(490) = 745.29 mm?
Fe ee
REINFORCED
CONCRETE DESIGN MJBCASTRO CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR URI
(STRENGTH DESIGN METHOD)
b. My = 140 KNm < OMymax > Singly Reinforced
Determine if the section is tension-controlled or transition:
OMe > My > Tension ~ controlled, @ = 0.90
Solve for R,
OR bd®
140 x 10° = 0.90R,,(300)(490)?
Ry, = 2.15959
Solve for p:
O85 f, 2Rn
Pah (.- =
oas(a7.6)/,_ |, _2(2:15959)
276 i 0.85(27.6)
p= 0.00822
J Check Prin
14
14
=== = 0005 00:
Pmin = = 776 = 9° 07 < 0.00822
= Use p = 0.00822
Thus,
‘A, = phd = 0.00822(300)(490) = 1208.34 mm?
cc. My =410kNm < OMpmax > Singly Reinforced
Determine if the section is tension-controlled or transition:
My, < My > Transition re
Solve for c and As:
M,, = 00.85ffab(a-<)
a=fe
d
@ = 0.65 + 0.25: where f, = 600.
1000 jf,
4) (0.85! 8,cb) (a a Ae
ee
SIMPLIFIED REINFORCED
“acm aa ‘CONCRETE DESIGN
— § «=—sli(isd ——"“™~S—=——s—s—~S~—i—“i—O—s—sSO
CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTH DESIGN METHOD)
iG 49
410x108 = (os +0254 —a59
276
(0.88)(27.6)(0.85)(e)(300) (490 ~ 9255)
276
c= 189.534 mm
a = 0,85(189.534) = 161.104 mm
490 — 189.534
189.534
Pieced aeiranz7e
Bree (-22" 1000276
= 951.173 MPa > fy, fy = fy
a0"
f, = 600—
= 0.883
Then,
ea
0.85fab = Asf,
0.85(27.6)(161.104)(300) = A.(276)
A, = 4108.152 mm?
@-My=500KNm > OMymax —* Doubly Reinforced
‘Compression reinforcement is necessary (see Chapter 3)
Solution:
I Solve for p:
2
Ag _ 3525
= 5a = 300500) ~ °°
Pmin Shall be the greater of (a) or (b):
_ 0.25) fe _ 0.25V3:
Pas ee = 0.00353
©) Pmin = Rae 0.00338
* Pin = 0.00353
if pin (34 — 28) = 0.807
34
Pp = 0.85(0.807) —
( 600
m4
oneal = 0.03333
ifp > pp steel does not yield, f, # f,
IIL p > pp
‘compression controlled, 0 = 0.65
C.=T
O.85f/ab = Ayf,
fs = 600:
a=Byec
0.85(34)(,c)(300) = 97 (28? (600
orem SY SAE
SIMPLIFIED REINFORCED
CONCRETE DESIGN MJBCASTRO-
d-c
c wy ee eS
‘STRENGTH DESIGN METHOD ETON OF BEAMS FOR FLEXURE '
9.85(34)(0.807)(c)(300)
a 500 —
= 97 (28)? (600 “)
= 304.673 mm
@ = 0.807 (304.673) = 245.871 mm
as 500 — 304.673
f, = 600 600?" = 384.662 MP.
‘ 0 s0sG 7am ean
Thus,
OM, = 00.85 fab (a ~ 5)
OM,, = (0.65)(0.85)(34)(245.871) (300) (500 os
OM, = 522.463 kNm
cae)
Z
Solution:
Wy = 12Wp, + 1.6,
Wy = 1.2(20) + 1.6(40) = 88 kN/m
For a simply supported beam:
wl? _ 88(6)?
Noe cae
= 396kNm
Solve for DMymax?
$1 3
= 02g pnd? (1-2,
OMnmax = O55 if’ ba? (1-61)
800-f,
i000
Dmax = 0.65 + 0.25
fy, = 0.85 for f! = 21MPa
a CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLE:
(STRENGTH DESIGN METHOD) ao
Then,
OMymax = (04 927) 5 (0 85)(21)(300)(550)?(1 (:
OMpmax = 399-126 kN
40 85)
My * 600+ f, 600+ 345
Gp = B,C = 0.85(460.317) = 391.269 mm
% = Gy — 150 = 241.269 mm
+ SIMPLIFIED REINFORCED.
a a as CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTH DESIGN METHOD)
500mm
From the figure:
C, +20 =T
0.85/cA; + 2(0.85fA2) = Acfy
(0.85) (28)(150)(500) + 2(0.85)(28)(125)(241.269) = Aq,(345)
Asy = 9334,929 mm?
b. Balanced moment capacity of beam
Balanced condition, 0 = 0.65
OMyy = 06; (a -) +200,(a- 150-2)
=
2
+2(0.65)(0.85)(28)(4125)(241.269) (725 — 150 ~
OM» = 1178.135 kNm
OMyy = (0:65)(0.85)(28)(150)(500) (725 —
241.269
2 )
©. Maximum steel area
max = 54 = 5 (725) = 310.714mm
Gmax = BiCmax = 0.85(310.714) = 264.107 mm
2max = Omax ~ 150 = 114.107 mm
SIMPLIFIED REINFO!
CONCRETE DESIGN MIBCASTRO
rt .
semmmage CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
~ "(STRENGTH DESIGN METHOD)
From the figure:
+20, =T
0.85 A, + 2(0.85f' 2) = Ashy
(0.85)(28)(150)(500) + 2(0.85)(28)!
Agy = 7141845 mm?
(125)(114.107) = Asmax (345)
d. Maximum moment capacity
OMpmax = 06, (d- =) +206, (d - 150-4)
800-fy 800
eA BOOED = 0,65 + 0.25
Omax = 0:65 + 0257599 = 5 To00
150
OMynay = (0:824)(0.85)(28)(150)(500) (725 -+)
114.107
+2(0.824)(0.85)(28)(125)(114.107) (72 —150- =“)
OMymax = 1245.808 kNm
e. Required tension steel area when M,, = 800 kNm
M, = 800 kNm <@Mymax — Singly Reinforced
Note: Some formulas derived may not be applicable for a non-rectangular sections
Assume a = 150 mm;
ee 176471
Oey aa
76.471
= 1864.994 MPa > 1000 MPa,
Tension — controlled @ = 0.90
a
OM, = 06, (a ~$) = 0.90(0.85)(28)(150)(500) (725 s =)
OM, = 1044.25 kNm E
Since the required M, = 800 kNm < 1044.225 kNm, a < 150mm
eerie ee Be rd, yea
CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTH DESIGN METHOD)
Assuming @ = 0.90
o
My = 06e(d-$)
£800 x 10° = 0.90(0.85)(28)(a)(500) (725 ~ 5)
a= 111.623 mm
a _ 111.623
BR 085
c= = 131.321 mm
Check Assumption:
ie 725 - 131.321
me = 2712:
f= 600—— = 600 12.494 MPa > 1000 MPa,
Tension ~ controlled $ = 0.90
Assumption is ok!
Since f, > fy, therefore steel yields, f; = fy
co
O.85fJab = Acf,
0.85(28)(111.623)(500) = 4,(345)
A, = 3850.185 mm?
f, Required tension steel area when M,, = 1200 kNm
M, = 1200 Nm < OMymax — Singly Reinforced
‘Note: Some formulas derived may not be applicable for a non-rectangular sections
Assume a = 150 mm;
nao 176.471
OAS aa
71
= 1864.994 MPa > 1000 MPa,
90
Tension — controlled 6 =
a
OM, = 0C-(a— $=
OM, = 1044.225 kNm
.90(0.85)(28)(150)(500) (72s = =)
Since the required M,, = 1200 kNm > 1044.225 kNm, a > 150 mm
———E—EE
sIMPLIFIED
CONCRETE DESIGN MIECAEERY:
Es ie . |
CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTH DESIGN METHOD)
M, = 06, (a- =) + 200, (d - 150 - 3)
6c, = o.ssta, = 0:85(28)(150)(600) = 1798000
Cy = 0.85f2A2 = 0.85(28)(z)(125) = 2975(0.85¢ — 150)
z=a—150 = 0.85¢ — 150
a=fc
-e
fy where f; = 600
fs =
0 = 0.65 + 0.255599 — f
(ott) (2)
= (oasoas aT
id
(600:
+2 ( 65 + 0.25 oe) [2975(0-85¢ — 150) G —150—
0.85¢ — =n
1000 = fy
725 -¢)
1200 x 10° = | 0.65 + pp es ees [1zaso00 (725 -4)|
PE 1000 — 345 2
0.85¢ — 150))
(600 25—*| — 345
42[ 0.65 +0.25- 5355 — [es7scosse ~ 150) (725 - 150-———
1000 — 345
98.733 mm.
a = 0.85(398.733) = 338.923 mm
z = 338.923 — 150 = 188.923 mm
725 — 398.733
= 600: = =
h = = 600-So5733 490.956 MPa > fy
Since f, > fy, therefore steel yields, f; = fy
490.956 — 345
0 = 0.65 + 0.2: ‘
Soop =a45)( sine
Thus,
G+, =T
0.85f'A; + 2(0.85f/ Az) = Asfs
(0.85)(28)(150)(500) + 2(0.85)(28)(188.923)(125) = A,(345)
A, = 8432.150 mm?
ernest CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE.
(STRENGTH DESIGN METHOD)
700mm,
Solution:
450mm,
d=700-75 =625mm
"
Asproviaea = 4 (32)? = 3216.991 mm?
Note: Some formulas derived may not be applicable for a non-rectangular sections
Solve for Asp
600d _ 600(625)
~ 600+f, 600+345
dy = By cy = 0.85(396.825) = 337.301 mm
& = 396.825 mm
2p = dy — 150 = 187.301 mm
From the figure:
GQ+Q=T
(0.85/24,) + 0.85/0A2 = Asf,
(0.85)(21)(150)(150) + (0.85)(21)(187.301)(450) = A,,(345)
Asp = 5524.986 mm?
EERE ENS ASR
SIMPLIFIED REINFO!
CONCRETE DESIGN MJBCASTRO CHAPT
d
R 2: ANALYSIS AND DESIG
” NALYSIS ESIGN OF BE, f E
IGTH DESIGN METHOD) a
150.150 150
150
a
:
|
Since As provided < Asp, Steel will yield, fz = fy
G+
(O.85f/A;) + 0.85/42 = Asfs
(0.85)(21)(150)(150) + (0.85)(21)(a — 150)(450) = a (32)?(345,
a = 238.171 mm
= 280.201 mm
8.171 mm
$25 — 260201 738.325 MPa < 1000 MP.
280201 fy as -
+ Transition region
738.325 — 345
Tora = Pe
® = 0.65 + 0.25
M, = OC, (a - 1) +00 (a < re)
My = 0(0.85f¢A,) (a -) + 00.85f4, (d ~ 150-5)
M,, = (0:800)(0.85)(21)(150)(150) (625 ~ 5°)
93.17!
+ (0,800) (0.85)(21)(88.171)(450) (625 ~ 150-~ >
M, = 420.865 KNm CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTH DESIGN METHOD)
125_ 100, 125_
Solution:
= 700 ~ 75 = 625 mm
=
Asproviaea = 43 (32)? = 3216.991 mm?
Note: Some formulas derived may not be applicable for a non-rectangular sections
Solve for Asp:
600d 600(625)
00+ f 600+ 345
y= Bicy = 0.85(396.825) = 337.301 mm
is = 396.825 mm
2p = dy — 125 = 212301 mm
125_ 100.125,
oe MJBCASTRO- CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXUR)
(STRENGTH DESIGN METHOD)
E
From the figure:
24,+G=T
2(0.85 fy) + 0.85f'A2 = Ashy
2(0. 85)(21)(125)(125) + (0. 1 85)(21)(212.301)(350) = Asy(345)
Agy = 5461342 mm?
Since As provided < Asby Steel will yield, f = fy
20, + =T
2(0.85 FAs) + 0.85f'A2 = Asfs
2(0.85)(21)(125)(125) + (0.85)(21) (a —
a = 213.363 mm
125)(350) = 47 (32)"(345)
a _ 213.363
=! = 251.015
ag = 251.015 mm
88.363 mm
625 — 251.015
jae eS: Fe
600-5 893.935 MPa < 1000 MPa
=: Transition region
893,935 — 345,
Beet 025 — pergge = 0860
My = 200, (4-4) + 00, (a = 125-2
My = 20(0.85f24,) (a -*5°) + 00.85//43 (a — 125 - 5)
M,, = 2(0.860)(0.85)(21)(125)(125) (62s es =)
+ (0.860)(0.85)(21)(350)(88.363 ) (62s = 125- ae
= 486.247 KNm CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS F1
(STRENGTH DESIGN METHOD) ‘OR FLEXURE
SUPPLEMENTARY PROBLEMS:
Mos ue
= Pa, f, = 345;
Oh
— _ CHAPTI ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE
(STRENGTH D HOD) ==——_— & °° -. °° °&«»}aar i
CHAPTER
3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
CHAPTER 3:
ANALYSIS AND DESIGN OF T-BEAMS AND
DOUBLY REINFORCED BEAMS
3.0 NOTATIONS & SYMBOLS:
depth of equivalent stress block, mm
area of tension reinforcement, mm?
As
A, area of compression reinforcement, mm?
> width of compression face of member, mm
c distance from extreme compression fiber to neutral axis, mm.
re balanced c, mm
Ge compressive force of concrete, N
d distance from extreme compression fiber to centroid of tension
reinforcement, mm
@ distance from extreme compression fiber to centroid of compression
reinforcement, mm
E. Modulus of Elasticity of concrete, MPa
Ey Modulus of Elasticity of steel, = 200,000MPa
ft specified compressive stress of concrete, MPa
fe calculated stress in reinforcement at service loads, MPa
im specified yield strength of steel, MPa
h overall thickness of member, mm.
My nominal moment, Nmm_
My ultimate moment at section, Nmm_
oD tensile force of steel reinforcement, N
A factor depending on the value of f!
& strain in steel below yield point = f,/E,
e strain in concrete, maximum = 0.003
y strain in steel at yield point =f /E,
o strength reduction factor
Po balanced steel ratio
CONCRETE DESTN See as CHAPTER 3:
)
ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
3.1 T-BEAMS !
otslabs and beams that are placed monolithically,
ss have extra widths at their tops, ca
‘of a T-beam below the slab is rete,
Reinforced conerete floor systems normally consist
a result, the two parts act together to resist loads, In effect, the beam:
flanges, and resulting T-shaped beams are called T-beums. The part
to as the web ot stem. ;
> 3.1.1 T-BEAM GEOMETRY (NSCP 2015 SECTION 40632)
+ 406.3.2.1 For non-prestressed T-beams supporting monolithic Or composite slabs, the effec
flange width by shall include the beam web jvidth By plus an effective overhanging flange wic
in accordance with Table 406.3.2.1, where itis the slab thickness and Sw is the clear distance
the adjacent web.
Table 406.3.2.1
Dimensional Limits for Effective Overhanging Flange
Width for T-Beams
i 8h
Each side of web Least of: Sw/2
1/8
; 6h
One side of web Least of: Sw/2
L,/12
SIMPLIFIED REINFORCE!
CONCRETE DESIG CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS j wR
> 3.2 MINIMUM FLEXURAL REINFORCEMENT IN NON-PRESTR BEAMS (NSCP
2015 SECTION 409.6.1)
except as provided in Section 409.6.1,3. For
II be the greater of (a) and (b).
the value of by, shall be the lesser of by
© 409,6.1.2 Asgnin Sha
team with a flange in tension,
a statically determinate
and 2B
o.25Vfe, A
ae eae
ty
)
Note:
vn is in tension, the amount of tension
nge of a seetior
the unreinforced
ACI Commentary R9.6.1.2. If the fla
reinforcement needed to ntake the strength of the reinforced section equal that of
ection is approximately twice that for a rectangular section OF that of a flanged section with
the flange in compression. A greater amount ‘of minimum tension reinforcement is particularly
necessary in cantilevers and other statically determinate beams where there is no possibility for
redistribution of moments
Thus,
sith their flange in tension, By in the above formula is to be
For statically determinate members W
of the flange by, whichever is smaller.
replaced with either 2by or the width o
3,2 ANALYSIS AND DESIGN OF T-BEAM
CASE 1: when a <¢
as those for Rectangular Beam
Note: Analysis and Design is the same
a
SIMPLIFIED REINFORCED
CONCRETE DESIGN ee MS AND DOUBI
4
CHAPTER 3: ANALYSIS AND DESIGN OF T-BEA‘
REINFORCED BEAMS
CASE 2: whena >¢
Nominal Moment:
b-h
1000-f,
@ = 0.65 for balanced condition & compression — controlled
0 = 0.90 for tension — controlled
0 = 0.65 + 0.25. for transition region
> 32.1 STEPS IN FINDING THE REQUIRED TENSION STEEL AREA A, OF A SINGLY
REINFORCED T-BEAMS WITH KNOWN REQUIRED MOMENT M,, AND OTHER BE \\
PROPERTIES.
Given: bd, fé, fy and M,
1. Solve for Mamax
3
Cmax = 54 =
= BiCmax = —_
z=a-t
=Gides z
Mynax = Gi (a geclane-)
‘s he
® 0.65 + 0.25 705
Mraz
MJBCASTRO SIMPLIFIED REINFORC!Y
CONCRETE DI CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
SIMPLIFIED REINFORCED
if OMpmax 2 My design as Singly Reinforced
if OMpmax < My design as Doubly Reinforced
Il. Check moment capacity of flange OMny
Assume
c
By
fe = 600
iff, 1000 MPa, tension controlled, 0 = 0.90
iff, < 1000 MPa, transition region
es fi-fy
9 = 065 + 0257905
ifM, OMpp, a>¢ analyze as T—beam
IML Analysis as T-beam
Assume @ = 0.90
M, = 06, (4-5) +00, (a-¢
* iff, = 1000 MPa, tension controlled, @ = 0.90, assumption is ok!
G+G=T
O.85f/A; + 0.85f!A2 = Asfy
As=
Cheek Asmin:
Asmin Shall be the greater of (a) or (b):
nee
B
(©) Asmin =
CONCRETE DESIGN
ie 7 BEAMS AND DOUBLY
jot ok!
CHAPTER 3: ANALYSIS AND DES:
Hl 2 SIGN OF T-
REINFORCED BEAMS
© if f, < 1000 MPa, transition region ,assumption ™
= tora
0 = 065 + 0253559- f
G+G=T
O.85flAr + 0.85/cA2 = Ashy
As
Check Asmin:
‘Aemin Shall be the greater of (a) oF (®):
@ amin = aE ed
by
14
=— bd
©) Asmin =F
> 3.2.2 STEPS IN FINDING OM, OF A T-BEAM WITH KNOWN TENSION STEEL AREA 4,
AND OTHER BEAM PROPERTIES.
Given: b, df, fy,and As
1. Solve for cp;
, - hod
> = 600+ fy
a= Bil»
ifae; proceed to step II — Sas og
IL. Solve for: CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
C+, =T
0.85/24; + 0.85f!A2 =
Aa = —
fy
+ ifA,> As; tension steel does not yield, @ = 0.65
7 2 = by (at) = by (he ~t)
f= 6002
GQ+G=
0.85 fA; + O.85f!A2 = Ach,
O85 fdby t+ 0.85 fiby (Bye — t) = A, 600
ee
c
d-c
¢
OM, = OC, iG -
"ifs < Asp; tension steel yields
c=T
0.85/24, = Asfy
Ac = __
ifAc Ay; thena>t analyze asT— beam
My =G (4-3) +0 (¢-e-4)
My = O85 f'by (a -5) +085 {iby 2(d—e
Solve for 0:
c
iff, = 1000 MPa, tension controlled, = 0.90
iff, <1000MPa, transition region
h—fy
= 0.65 +0.25—5
o +028 000% >»
AND DESIGN OF T-BEAMS AND DOUBLY
Rea CH APT 3: ANALY!
REINFORCED BEAMS
"ORCED BEAMS
3.3 ANALYSIS AND DESIGN OF DOUBLY-R)
ce o aesthetic) and other consideration
quired to resist the given bend,
pression. These by
use of architectural (spa
compression force re
forced with steel to carry com
Tesion steel reinforcement, resulting in & dup,
If a beam cross section is limited be
may happen that the conerete could not develop the e'
moment and the compression conerete should be rein
sections are designed to have both tension and compr
reinforced beam
cement is used for reasons other than resistin
tions where reversal of moments might occ
reinforcement reduces the long-ter
However, there are situations in which compression reinfo
moments, Doubly reinforced beams are also used in situal
specially during earthquakes. In addition, inclusion of compression
deflections of beams and increase the ductility of the sections.
ed beam is analyzed by dividing the beam into two couples Myx and Myz as shown.
ind the part of the tension steel As
ff the tension steel area A..
Doubly reinforce
jon conerete ai
sssion steel A’ and the other part of
= Myz is the couple due to compres:
= Mya is the couple due to compre
L
i
eee
|
=
Fy/Es =~“ i#2+;« ii.
CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
Ultimate Moment Capacity:
My = Mus + Mu
Mus = Mu — Muz = ©Mnmax
a= 0 (t-3)
Balanced condition @ = 0.65
600d
600+h,
&
Maximum Condition
800 - fy
1000 - fy
DOUBLY-REINFORCED RECTANGULAR
AND OTHER BEAM PROPERTIES.
0 = 0.65 + 0.25
> 33.1 STEPS TO DETERMINE A, AND A; OF A
BEAM WITH KNOWN REQUIRED MOMENT M,
Given: b,d, fe, fy,and My
1. Solve for @Mpmax
OR
if @Mpmax 2 My design as Singly Reinforced
if Mymax fyi _ tension steel yields f, = fy
= iff, fy; compression steel yields ff = fy
= iff! fyi__ tension steel yields f; = fy
+ iff, f,; compression steel yields f = fy
= if ff < fy; compression steel does not yield if
I. Ifall steel yields
iff, = 1000 MPa, tension controlled, @ = 0.90
iff, < 1000 MPa, _ transition region = 0.65 + 0.25
My = Mus + Mug
My = Mus + Muz
a
M, = 07, (d-5) + 072(d - 4) M, = 06. (4 -3) +00,(d-d')
or
My = OAgify (4-5) + Aczfeld— 2) M, = 0085{fab(d-$) + 04 f4(d - 4)
My= My =
III. If compression steel does not yield
T=Ce+Cy
Achy = 0.85 fcab + Asfe
Asfy = 0.85 f¢Bycb + As (00! 7
‘SIMPLIFIED REINFORCED
CONCRETE DESIGN D DOUBLY |
oR? CHAPTER 3: ANALYSIS AND DESIGN OF T-| AMS AN’
REINFORCED BEAMS : ft
tension controlled, = 0.90
if f, = 1000 MPa,
nsition region 0 = 0.65 + 0.25
iff, <1000MPa, tra
My = Mus + Muz My = Mus + Mug
= 07, (a- 5)+ ord - 4’) a‘ My = Ce (4-3) + 6(d-a’)
or , i ie
=0Anf, (4-5) + OAgf(d - 4) M, = 00.85 fab (d —5) + 94; f1(4 ~ 4
he —
ae
IV. If tension steel does not yield @ = 0.65
T=Cco+
Asfs = bbe gab+ aiff
As (600==*) = = 0.85f/Bicb + Achy
c=
a= 2
ff = 6005
y d
f= 6005 = __
My = Mus + Muz M= Mus + Mua
Z " @
= 0% (d-5) + 07(d - a) My = 06,(d-$) + 06,(4 4)
or
My, = OAgsfy (4-5) + OAsafe(d - 4) M, = 90.85 fab (a — 5) + 0Arfi(d-¢
Me i
M= = °° ° & .:
CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
ILLUSTRATIVE PROBLEMS:
Solution:
From NSCP 2015 Table 406.3.2.1
Effective Flange width by is the smallest of
a. by + 16h = 250 + 16(120) = 2170 mm
be By + (Sws/2) + (Sw2/2) = 250 + (3000/2) + (3000/2) = 3250 mm
cc. by + (Ly/4) = 250 + (5000/4) = 1500 mm .
Therefore, by = 1500 mm
Solution:
a, Balanced steel area
600d __ 600(460) _ 57> 199
% = S004 f oo0t414 mm
fy = 0.85 for fe! = 20.7 MPa
ay = BiCp = 0.85(272,189-) = 231.361 mm
nl
SIMPLIFIED REINFORCED
CONCRETE DESIGN CHAPTER 3: ANALYSIS AND DESIGN OF
f LYSIS ESIG! BEAMS AND
REINFORCED BEAMS. NOM DOUBL
Since ay >t
231.361 — 110 = 121.361mm
Thus,
Gt
on5f (Ay + 0.85f2A2 = Asly
0.85 f: (thy) + 0.85/¢ @nbw) = Ashy
0.85(20.7)(110)(900) + 04 {85(20.7)(121.361) (310) = Asp (414)
Agp = 5806.431 mm?
b. Nominal and ultimate balanced moment capacity
Myy = Gs + Co¥2
Myy = 0.85f¢ (thy) iG -5) +0.85f! @ebw)(d-¢
Myy = 0.85(20.7)(110)(900) (460 — “)
+0.85(20.7)(121.361)(310) (460 -110-F
Mp = 896.989 KN
For Balanced condition, 6 = 0.65
OMyy = 0.65(896.989)
OM yy = 583.043 kNm
MJBCASTRO SIMPLIFIED REINFORC!?
‘CONCRETE DESIG*
o Oe en ee ee ee
CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY meee
REINFORCED BEAMS
c. Maximum steel area
ae aha = $460) = 197.143 mm
max = BrCmax = 0.85(197.143) = 167.572 mm
Since dmax > ti
2max = Amax — ¢ = 167.572 — 110 = 57.572 mm
Thus,
Q+C,=T
O.85f/A; + 0.85f/A, = Asfy
0.85,¢ (thy) + 0.85f¢ Zmaxby) =
0.85(20.7)(110)(900) + a 85(20.' Bn 572)(310) = Ay (414)
Asmax = 4966.011 mm?
4. Nominal and ultimate maximum moment capacity
Mnmax = CY: + Co¥2 i
Mama = 0.85fe(tby) (d ~ §) + 0.854! @nazbw) (a-1-")
Mnmax = 0.85(20.7)(110)(900) (460 — *)
+0.85(20.7(57.572)(310) (460 ~ 110 ~ 7572)
Momax = 806.340 kNm
For Maximum condition,
9=065+025 = b - o65 4025 S005 458
000 f, Too0— 414 ~ 9815
OMnmax = 0.815(806.340)
OMnmax = 657.167 kNm
CONCRETE DESK eeecee MJBCASTRO Solution:
IL Solve for OMamax
Cmax (470) = 201.429 mm
B, = 0.85 for fo = 20.7 MPa
max = B1Cmax = 0.85(201.429) = 171.215 mm
Since amax >t}
2max = Imax ~ ¢ = 171.215 — 100 = 71.215 mm
OMnmax = OC1Y1 + OC2y2
OMymax = 00.85 ¢(tby) (d ~ 3) + 00.85/£ émaxby) (d—¢-
Reo 5 tg.95 800-4
tooo f, ~ °° + °?° 7900-414
0 = 0.65 + 0.25.
ly a (ek TP ee) | |
CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY ‘eos.
REINFORCED BEAMS
100)
OMmax = 01815(0.85)(20.7)(100)(820) (470 - 400)
+0.815(0.85)(20.7)(71.215)(250) (470 ~ 100 -)
OMymax = 579.239 kNm
Il. Check moment capacity of flange @Mny
00
=f =< 2 = 117647
aes on
470 - 1174
ee ee 797 0
600 797.001 MPa
f,>1000MPa, tension controlled, = 0.90
a 100.
Olay = 86, (d ~ 3) = 0.90(0.85)(207)(100)(820) (470 - 02)
OMyy = 545.375 kNm
Mp = 150kNm & M, = 120kNm
My = 1.2Mp + 1.6M, = 1.2(150) + 1.6(120) = 372 kNm
> OMnmax >My design as Singly Reinforced
>My 1000 MPa, tension controlled, @ = 0.90 Assumption is correct!
SIMPLIFIED
REINFORCED
CONCRETE DESIGN
MJBCASTRO
et F CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS A)
REINFORCED BEAMS we goueLy
G=T
O.85flaby = Ash,
(0.85) (20.7)(65.521)(820) = As(414)
Ag = 2283.407 mm?
Flange is in compression,
shall be the greater of (a) or (b):
Asmin §
(a) Asmin = a d= 925V207 (250)(470) = 322.822 mm?
(b) Asmin = ite d= 24 (2501470) = = 397.343 mm?
Therefore, Ay = 2283.407 mm?
Mp = 175kNm & M, = 190kNm
b.
= 1.2Mp + 1.6M; = 1.2(175) + 1.6(190) = 514 kNm
5 OMpmax > My design as Singly Reinforced
My, 1000 MPa, tension controlled, 0 = 0.90 Assumption is cor
fe > fy therefore f, =
aT
0.85faby = Asfe
(0.85)(20.7)(93: 526)(820) = 4,(414)
Ag = 3259.381 mm? Coo ee
CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS ora
Flange is in compression,
Asin Shall be the greater of (a) ot (b):
o2s/ fe, , _ 0.25V207
(a) Asmin = Ee = Fig (250)(470) = 322.822 mm?
14 14 ae
(0) Asmin = yet = Gia (250)(470) = 397.343 mm
Therefore, Ay = 3259. 381 mm?
Mp = 195kNm &M, = 210kNm
My, = 1.2Mp + 1.6M, = 1.2(195) + 1.6(210) = 570 kNm
+ OMpma > My design as Singly Reinforced
>My > OMye, a>t analyze as T— beam
Assume @ = 0.90
t z
M, =0¢,(¢-5) +06, (a-«-3)
570x 108 = 0,90(0.85)(20.7)(100(820) (470 — Ss)
+ 0.90(0.85)(20.7)(z)(250) (470 - 100 - 3)
17.212 mm
t+z=100+17.212= 117.212 mm
a _ 117212 _ 147996
Ani 085 Vignes
d-c__ 470-137.896
acne Tagewasrase
1445.019 MPa
Je > 1000 MPa, tension controlled, © = 0.90 Assumption is correct!
fe > fy therefore f, = fy
,+G=
O.85f'A1 + 0.85/A2 = Acf,
0.85 fi tby + 0.85f/zby = Acfy
(0.85)(20.7)(100)(820) + (0.85)(20.7)(17.212)(250) = As(414)
As = 3667.878 mm?
EERE EARN A
SIMPLIFIED REINFORCED
CONCRETE DESIGN er CHAPTER 3: ANALYSIS AND DESIGN OF T-BEAMS AND DOUBLY
REINFORCED BEAMS
Flange is in compression,
Asmin Shall be the greater of (a) or (b):
(0) Agmin = ooVEE pg = 225N207 (950)(470) = 322.822 mm?
fy 414
14 14 2
eS pais = 397.343 mm
Cb) Apia he byd Fig (25000470)
Therefore, A, = 3667.878 mm?
Solution:
L Solve for cp;
600d 600(600)
= =———_ = 1.952
> = Coors mao aa uaa
dy = Bycy = 0.85(380.952) = 323.809 mm > t
IL. Solve for:
323.809 — 100 = 223.809 mm
2p = ay
G+G=T
O.B5f'A; + O.85f'Az = Acfy
O.B5fthy + 0.85f!2pby = Asp fy
(0.85) (20.7)(100)(1500) + (0.85)(20.7)(223.809)(250) = Asy(345)
Asp = 10503.564 mm?
"
Ay = 63 (28)? = 3694.513 mm?
As A, 1000 MPa
fs > 1000 MPa, tension controlled, @ = 0.90
fe > fy therefore f, = fy
Thus,
a
OM, = 6C.(d-5)
OM, = 00.85f’aby (4-5)
48.
OM,, = 0.90(0.85)(20.7)(48.294)(1500) (600 ne,
OM, = 660.584 kKNm
