Lec 13-14-15-flexural analysis and design of beams-2007-r

Plain & Reinforced
Concrete-1
CE3601
Lecture # 13, 14 &15
13th
to 20th
April 2012
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)

Plain & Reinforced Concrete-1
Load Carried by the Beam
Beam Supporting One-way Slab
lx
lx
Exterior Beam
Interior Beam
Width of slab supported by interior beam = lx
Width of slab supported by exterior beam = lx/2 + Cantilever width
ly
(ly/lx > 2)

Load Carrie by the Beam
Beam Supporting Two-way Slab (ly/lx≤ 2)
lx
lx
ly ly
Exterior Long Beam
Interior Long Beam
Exterior Short Beam
Interior Short Beam
45o

Load Carrie by the Beam
Beam Supporting Two-way Slab (ly/lx≤ 2) contd…
45olx/2
lx/2
[ ] o
45cos
2/xl
[ ] 2
o
45cos
2/x



= lArea of Square
Shorter Beams
For simplification this
triangular load on both
the sides is to be replaced
by equivalent UDL,
which gives same Mmax as
for the actual triangular
load.
2
2 





=
xl

45o
45o
Equivalent Rectangular
Area
2
2
x
3
2
2
x
3
4
l
l
=×=
Factor of 4/3 convert this VDL into UDL.
Equivalent width supported
by interior short beam
lx
x
x
3
2
l
l 2
= x
3
2
l=
Equivalent width supported
by exterior short beam
+=
3
xl
Cantilever
x
3
2
l

Beam Supporting Two-way Slab (ly/lx≤ 2)contd…
lx
lx
ly
Exterior Long Beam
( )
2
2
x
xy
2
x






+−×=
l
ll
l
Supported Area
lx/2 lx/2ly - lx
4
x
2
x
2
yx 22
llll
+−=
4
x
2
yx 2
lll
−=






−=
2
x
yx
2
1 2
l
ll

Exterior Long Beam
2
R1
3
R1
F
2
−
−
=
y
x
R
l
l
=where
Factor F converts
trapezoidal load into
equivalent UDL for
maximum B.M. at center
of simply supported
beam.For Square panel
R = 1 and F = 4/3

Exterior Long Beam
Equivalent width
lx/2 lx/2ly - lx
Equivalent width
Length...Span
F)Supported..Area( ×
=
ly
y
1
2R1
3R1
2
x
yx
2
1 22
l
l
ll ×





−
−
×





−=






−
−
×





−=
2R1
3R1
2
x
1
2
x 2
lyll
( )



−= 3R1
2
x 2l
+ Cantilever (if present)

Interior Long Beam
Equivalent width
lx/2 lx/2ly - lx
Equivalent width
ly
( )3R1x 2
−l

Wall Load (if present) on Beam
tw (mm)
H
(m)
1000
81.9
1930
1000
tw
Hm1 ××





××=
UDL on beam
Htw019.0 ××= (kN/m)
Htw019.0 ××

Wall Load on the Lintel
Equivalent UDL on lintel if
height of slab above lintel
is greater than 0.866L 0.866L
60o
60o
L
Ltw11.0UDL ××= kN/m
tw = wall thickness in “mm”
L = Opening size in “m”
If the height of slab above lintel is less than 0.866L
Total Wall Load + Load from slab in case of load bearing wall
UDL = (Equivalent width of slab supported) x (Slab load per unit area)
= m x kN/m2
= kN/m

Slab Load per Unit Area
Top Roof
Slab Thickness = 125 mm
Earth Filling = 100 mm
Brick Tiles = 38 mm
Dead Load
2
m/kg3002400
1000
125
=×=Self wt. of R.C. slab
Earth Filling
2
m/kg1801800
1000
100
=×=
Brick Tiles
2
m/kg741930
1000
38
=×=
554 kg/m2
Total Dead Load, Wd =

Slab Load per Unit Area (contd…)
Top Roof
Live Load
WL = 200 kg/m2
Ldu W6.1W2.1W +=
Total Factored Load, Wu
( )
1000
81.9
2006.15542.1Wu ××+×=
2
u m/kN66.9W =

Intermediate Floor
Slab Thickness = 150 mm
Screed (brick ballast + 25% sand) = 75 mm
P.C.C. = 40 mm
Terrazzo Floor = 20 mm
Dead Load
2
m/kg3602400
1000
150
=×=Self wt. of R.C. slab
Screed 2
m/kg1351800
1000
75
=×=
Terrazzo + P.C.C 2
m/kg1382300
1000
)4020(
=×
+
=
633 kg/m2
Total Dead Load, Wd =

Intermediate Floor
Live Load
Occupancy Live Load = 250 kg/m2
Moveable Partition Load = 150 kg/m2
WL= 250 + 150 = 400kg/m2
Ldu W6.1W2.1W +=
Total Factored Load, Wu
( )
1000
81.9
4006.16332.1Wu ××+×=
2
u m/kN73.13W =

Self Weight of Beam
Service Self Wight of Beam = b x h x 1m x 2400
2
L11.112400m1
18
L
12
L
=×××= Kg/m
Factored Self Wight of Beam
22
L131.0
1000
81.9
2.1L11.11 =××= kN/m
Self weight of beam is required to be calculated in at the stage of
analysis, when the beam sizes are not yet decided, so approximate
self weight is computed using above formula.

Bar Bending Schedule
Serial #
Bar
Designation
Number
of Bars
Length
of one
Bar
Dia
of
bar
Weight of Steel Required
Shape of
Bar
#10 #13 #15 #19 #25
1 M-1 #25
2 S-1 #10
3 H-1 #15
Σ
Total weight of steel = 1.05Σ , 5% increase, for wastage during cutting
and bending

Bar Bending Schedule (contd…)
Bent-up Bar
h45o
h
h2
Additional Length = 0.414 h
Total Length = L + 0.414 h
L

90o
-Standard Hooks (ACI)
db
R = 4db
for bar up-to #25
R = 5db
for bar #29 & #36
R = 12db
L
Total Length = L + 18db
For R 4db

180o
-Standard Hooks (ACI)
db
L
Total Length = L + 20db
4db
Same as 90o
hook

Example: Prepare bar bending schedule for the given beam. Clear
cover = 40 mm
2-#20 +1-#15
4000
570
2-#20
2-#10
228
#10 @ 180 c/c
Longitudinal Section
1-# 15

cover = 40 mm
#10 @ 180 c/c
228
375
2-#10
1-# 15
2-#20
(M-2)
(M-1)
(S-1)
(H-1)
Cross Section

Example:
M-1
M-1 = 4000 + 2 x 228 – 2 x 40 + 2 x (18 x 20)
= 5096
M-2h = 375 – 2 x 40 – 2 x 10 – 2 (15/2) = 260
h
M-2 = 4000 + 2 x 228 – 2 x 40 + (0.414 x 260) x 2
= 5091
H-1
H-1 = 4000 + 2 x 228 – 2 x 40
= 4376

cover = 40 mm
Number of Bars = 4000 / 180 +1 = 24 Round-up
Shear stirrups
a
a = 375 – 2 x 40 – 10 = 285 mm
bb = 228– 2 x 40 – 10 =138 mm
Total length of S-1 = 2 (138 + 285 + 18 x 10) = 1206 mm

Bar Bending Schedule
Serial #
Bar
Designation
Number
of Bars
Length
of one
Bar
(m)
Dia
of
bar
Weight of Steel
Bars
Shape of Bar
#10 #15 #20
1 M-1 2 5.096 20 24
2 M-2 1 4.591 15 7.2
3 H-1 2 4.376 10 6.9
4 S-1 24 1.206 10 22.7
1.05 Σ 31.1 7.6 25.2
Total Weight = 64 kg
4376
4376
4376
285138

Design of Singly Reinforced Beam by Strength
Method (for flexure only)
Data:
 Load, Span (SFD, BMD)
 fc’, fy, Es
 Architectural depth, if any
Required:
 Dimensions, b & h
 Area of steel
 Detailing (bar bending schedule)

Design of Singly Reinforced Beam by Strength Method (contd…)
Procedure:
1. Select reasonable steel ratio between ρmin and ρmax.
Then find b, h and As.
2. Select reasonable values of b, h and then
calculate ρ and As.

2. Using Trial Dimensions
I. Calculate loads acting on the beam.
II. Calculate total factored loads and plot SFD and
BMD. Determine Vumax and Mumax.
III. Select suitable value of beam width ‘b’. Usually
between L/20 to L/15. preferably a multiple of
75mm or 114 mm.
IV. Calculate dmin.
b'f205.0
M
d
c
u
min =
hmin = dmin + 60 mm for single layer of steel
hmin = dmin + 75mm for double layer of steel
Round to
upper 75 mm

V. Decide the final depth.
minhh ≥ For strength
minhh ≥ For deflection
ahh ≈ Architectural depth
12
hh ≈
Preferably “h” should be multiple of 75mm.
Recalculate “d” for the new value of “h”

VI. Calculate “ρ” and “As”.








−−=
fc'
2.614R
11wρ
Four methods
y
c
f
'f
0.85w = 2
u
bd
M
R =
Design Table
Design curves
Using trial Method
a)
b)
c)
d)

VII. Check As ≥ Asmin.
Asmin=ρmin bd (ρmin= 1.4/fy to fc’ ≤ 30 MPa)
VIII. Carry out detailing
IX. Prepare detailed sketches/drawings.
X. Prepare bar bending schedule.

1. Using Steel Ratio
I. Step I and II are same as in previous method.
III. Calculate ρmax and ρmin & select some suitable “ρ”.
IV. Calculate bd2
from the formula of moment
V. Select such values of “b” and “d” that “bd2
” value
is satisfied.
VI. Calculate As.
VII. Remaining steps are same as of previous method.
( ) 





−==
'1.7f
ρf
1fρbd0.9MΦM
c
y
y
2
nbu

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