تصميم الكمرات Dr.alaa bashandy

‫تتتتت‬ ‫تتتتت‬
‫تتتتتتت‬
20132013
ALB
Dr. Eng. ALaa Ali Bashandy
Menufiya Unv.
Design of Reinforced ConcreteDesign of Reinforced Concrete
StructuresStructures
Design of R. C. Beams
Dr. ALaa Ali Bashandy

RR.. CC.. BBeeaammss
ALB

BeamsBeams
ALB

ALB Dr. Eng. ALaa Ali Bashandy
Members
of
Concrete
Construction
Members
of
Concrete
Construction
Slabs
Beams
Column
Footing
Shallow found.
Deep found.
Solid Slabs
Hollow Block Slabs
Flat Slabs
Paneled Beam system
Isolated Footing
Combined Footing
Strip Footing
Raft/ Mat found.
Friction Pile
Bearing Pile
Axially Loaded Col. Eccentric Loaded Col.or
Simple Beam Continuous Be. Cantilever Beamoror
Dropped Beam Inverted Beam Hidden Beamor or
Long Column Short Columnor
Rectangular Column Circular Columnor
Braced Column Unbraced Columnor
Structural
Sys.
Position ref.
to slab
Loads
Design
Section
Bracing

BeamsBeams
Cantilever BeamSimple Beam
ALB
Continuous Beam
Continuous Beam
- one end -
Continuous Beam
- two ends -
Simple Beam Cantilever Beam

ALB
Cross Sections of BeamsCross Sections of Beams

BeamsBeams
Referring to its position comparing to slabsReferring to its position comparing to slabs
ALB
Hidden BeamDropped Beam Inverted Beam
bb
slabslab

BeamsBeams
BeamsBeams
ALB
Hidden BeamHidden BeamDropped BeamDropped Beam Inverted BeamInverted Beam

21
Sec. 2Sec. 1
1 1 2 2

Sec. 2Sec. 1
T – Sec.((
211 1 2 2
R – Sec.((

211 1 2 2
Sec. 2Sec. 1

Sec. 2Sec. 1
L – Sec.((
211 1 2 2
R – Sec.((

SS44
SS11
SS55
SS66
SS99
SS88
SS77
SS22
SS33
Load DistributionLoad Distribution
ALB

SS44
SS11
SS55
SS66
SS99
SS88
SS77
SS22
SS33
ALB

Different Cases for One-way SlabDifferent Cases for One-way Slab
Cantilever one-way slabCantilever one-way slab
2.0
L
L
S
≥
LLSS
LL
LLSS LLCC
Load DistributionLoad Distribution from one-way slabfrom one-way slab
One-way slabOne-way slab
One directionOne direction
22--sidessides
22--sidessides
11--sideside

One-way S.
Slab
2.00
L
L
s
≥
B1
B2
L
Ls
B1
L









 ++





=
WallB1
S
1B
WO.Wtx1.4
2
L
xWuW S
B2 Ls
L
Ls
Lc
Cantilever Slab
WB2 = 1.4 ( O. Wt B2 + Wwall ) t/m
B1
[ ] 








 ++++





=
WallB1C
S
1B
WO.Wtx4.1Lx
s
Wu
2
L
x
s
WuW
B2
Ls
WB2 = 1.4 ( O. Wt B2 + Wwall ) t/m
B2
B1
L
Load DistributionLoad Distribution from one-way slabfrom one-way slab

Load DistributionLoad Distribution from two-way slabfrom two-way slab
Load from slab
L
B1
B2Ls
Two-way S.
Slab
2.00
L
L
s
<
L
Ls
B1
B2

Two-way S. Slab 2.00
L
L
s
<
L
Ls
B1
B2
L
B1
Wa
L
L
We
Equivalent load forEquivalent load for ShearShear (Wa or Wβ(
Equivalent load forEquivalent load for MomentMoment ( We or Wα(
Momentfor))L/L(
3
1
-1(Cα
Shearfor))L/L(
2
1
-1(Cβ
2
se
sa
==
==
for Trapezoidal Load only
t/m( )[ ]WallS O.WtO.Wtx1.4Cax
2
L
Wu B1
s
XB1Wa ++





=
t/m
Load DistributionLoad Distribution from two-way slabfrom two-way slab
( )[ ]WallS O.WtO.Wtx1.4Cex
2
L
Wu B1
s
xB1We ++





=

L
Ls
B1
B2
Wa
LS
LS
We
Equivalent load forEquivalent load for ShearShear (Wa or Wβ(
Equivalent load f orEquivalent load f or MomentMoment (We or Wα(
for Triangle Load only
t/m
t/m
B2
Ls
Momentfor32Cα
Shearfor21Cβ
e
a
==
==
( )[ ]WallS O.WtO.Wtx1.4Cax
2
L
Wu B2
s
x2BWa ++





=
( )[ ]WallS O.WtO.Wtx4.1Cex
2
L
Wu B2
s
x2BWe ++





=

2
L
L1 >1.1.
L
L
L1
( )













++=
==
wall
W
B
O.Wtx1.4sx Wu
L
Area
ShearforLoadMomentforLoadWB
2
L
L1 <22..
L1 L1
LL
L1
L
Special Cases of LoadingSpecial Cases of Loading
L
L1
( )















++=
==
wall
W
B
O.Wtx1.4sx Wu
L
Area
ShearforLoadMomentforLoadW
1
B
oror
oror
WB
WB

44..
33..
( )





++∑===
wall
W
beam
O.Wtx1.4sx Wu
Span
Area
ShearforLoadMomentforLoadWB
1/2Ca =
1/3Ce =
1/2Ca =
2/3Ce =
1/2Ca =
1/2Ce =
sx Wu
Span
Area
L
WB
WB
WB
WB
WB
L L

ALB
Main and SecondaryMain and Secondary
BeamsBeams

BeamsBeams
Secondary BeamMain Beam
ALB

Main BeamsMain Beams
Main BeamMain Beam

SecondarySecondary
BeamBeam
Secondary BeamsSecondary Beams

Col.Col.Col.Col.
Col.
Beam
Col.
Col.
Col.
Col.
Beam
Beam
Col.
Col.
Col.Col.
Col.
Col.Beam
BeamBeam
B1
B2
B2’
B3
B4
B5
B6
Main and Secondary BeamsMain and Secondary Beams

Col.Col.Col.Col.
Col.
Beam
Col.
Col.
Col.
Col.
Beam
Beam
Col.
Col.
Col.Col.
Col.
Col.Beam
BeamBeam
B1
B2
B2’
B3
B4
B5
B6
Load from Slabs on BeamsLoad from Slabs on Beams

ALB
Design of BeamsDesign of Beams

WBeam (t/m’
) = (Wall Wt. + O. Wt.Beam ) x 1.4 + Wus (From slab)
1.Dead Load (D.L)
22
3
P.C
3
R.C
R.C
ConcreteR.Concrete
t/m0.15kg/m150C.FL.
t/m2.2γ&t/m2.5γ
γ*1.0x1.0xt
x γVW
)F.C(coverfloor-2
(O.Wt)slabofOwn weight-1
s
==
==
=
=
2.Live Load ( L .L)
3.Total Load ( T.L)
Load ValuesLoad Values
1.00m
ts
1.00m
From SlabFrom Slab
From BeamFrom Beam
R.CBConcreteR.Concrete
γxbxtxVW
)W(BeamofOwn weight-1
γWt.O
Wt.O
==B
B
From WallsFrom Walls
( )
cm2512b3t/m15001200
xhxbxtxVW
wWall
Wall
wwWallwallwall
γ
γγ
→=→
==
=
according to code

Live Load ValuesLive Load Values According to Egyptian CodeAccording to Egyptian Code

tB min = 40 cm
Cantilever Beam
Estimation of tEstimation of tBB
t B min = Leff / 16 Leff / 18 Leff / 21 Lc / 5
it is required to check deflection if the span < 10 m
for High Tensile Steel 400/600 (H.T.S)
t B min = previous values ÷ ( 0.4 + fy / 650)
for any other steel type fy / fu
Simple Slab
Continuous Slab
From tow side
Continuous Slab
From one side

but not less thanbut not less than 40 cm40 cm
Cantilever Beam
Value of tValue of tBB
t B = span / 10 span / 12 Lc / 5
GenerallyGenerally
Simple Slab
Continuous Slab
From tow side
Continuous Slab
From one side

LLSS
WWuu t/mt/m
/24WL2
/24WL2
/24LW
2
22
/24LW 2
11
/8WL2
/9WL2
/11LW
2
11 /11LW
2
22
/10WL2
/12LW
2
11 /16LW
2
22
/12WL2
/16LW
2
33
/24LW 2
11
minmin MM + ve+ ve==
8
LxuW 2
Moment ValuesMoment Values
Simply supportedSimply supported continuous two spanscontinuous two spans
continuous more than two spanscontinuous more than two spans
Empirical values for B.MEmpirical values for B.M (Max difference in(Max difference in loadload && spanspan ≤≤ 20%20% and D.L >L.L )and D.L >L.L )

MM supportsupport
)L'L'(8.5
L'WL'W
21
3
22
3
11
supportMLoadUniform
+
+
=
)L'L'(
L'PkL'Pk
21
2
222
2
111
supportMLoadedConcentrat
+
+
=
L1
m
W2
L2
m
W1
L1
m
L2
m
P1 P2
a1 a2
L m
L’ = L
L m
L’ = 0.8L
a1 / L1 & a2 / L2 0.3 0.4 0.5 0.6 0.7
K1 & k2 0.168 0.182 0.176 0.158 0.128
In case of:In case of: difference indifference in loadload oror spanspan ≥≥ 20%20%

Shear ValuesShear Values
Simply supported beamSimply supported beam
Continuous two spansContinuous two spans
Continuous more than two spansContinuous more than two spans
Lm
a b
0.5 W L
S.F.D
+
0.5 W L
-
L1
m
a
0.4 W1 L1
+
0.6 W1 L1
-
L2
m
C
0.6 W2 L2
S.F.D +
0.4 W2 L2
-
b
L1
m
a
0.4 W1 L1
+
0.6 W1 L1
-
L2
m
C
0.5 W2 L2
+
0.5 W2 L2
-
b
-
L3
m
0.5 W3 L3
S.F.D
+

Possible cases of bPossible cases of b
Estimation of bEstimation of b
Rectangular sectionRectangular section R-sec.R-sec.
T - sec. → internal beamT - sec. → internal beam
L - sec. → edge beamL - sec. → edge beam
R - sec.R - sec.
T - sec.T - sec. L - sec.L - sec.
::Values of bValues of b
16 t16 t ss + b+ b
LL22 / 5 + b/ 5 + b
CL. to CL.CL. to CL.
6 t6 t ss + b+ b
LL22 / 10 + b/ 10 + b
½½ CL. to CL.CL. to CL.
bb
LL22 = L= L LL22 = 0.8 L= 0.8 L LL22 = 0.7 L= 0.7 L LL22 = 2 L= 2 L
b.F
M
Cd
cu
u
1=
Least of :Least of :

Given : M u , t s , b =100 cm , Fcu , Fy
Req. : As
d = ts - c (cover) c = 20 - 50 mm(cover) c = 20 - 50 mm
AS min = 0.15 % Ac H.T.S 360/520
= 0.25 % Ac for Mild steel 240/350
.....J.....C
b.F
M
Cd 1
cu
u
1 ===
m′== /cm........
f.d.J
Mu
As 2
y
Ac = b x d
Design of SectionDesign of Section
AS max = 0.4 % Ac recommended

CC1 & J& J

ALB
Shear Behavior of BeamsShear Behavior of Beams

Choice of Shear ValuesChoice of Shear Values
Simply supported beamSimply supported beam
Continuous two spansContinuous two spans
Continuous more than two spansContinuous more than two spans
S.F.D
Chose the larger value of shear along the beam span - axisChose the larger value of shear along the beam span - axis Qmax
Lm
a b
0.5 W L
S.F.D
+
0.5 W L
-
Q1
Q2
L1
m
a
0.4 W1 L1
+
0.6 W1 L1
-
L2
m
C
0.6 W2 L2
S.F.D +
0.4 W2 L2
-
b
Q1
Q2
Q3
Q4
L1
m
a
0.4 W1 L1
+
0.6 W1 L1
-
L2
m
C
0.5 W2 L2
+
0.5 W2 L2
-
b
-
L3
m
0.5 W3 L3
+
Q5
Q4Q2
Q3
Q1

Critical Sections of ShearCritical Sections of Shear

Shear Limitations of BeamsShear Limitations of Beams

Check of ShearCheck of Shear

Cases of qCases of quu

ALB
Reinforcing RebarsReinforcing Rebars

Simple BeamSimple Beam
44++22

22++22++22

11
22
33
44
55
66
77

Cantilever BeamCantilever Beam
11
22
33
55
66
77

ALB
Reinforcement DetailingReinforcement Detailing

UsingUsing StraightStraight RebarsRebars

Using Straight RebarsUsing Straight Rebars
22--spans Beamspans Beam

More than 2-spans BeamMore than 2-spans Beam

Over hanging BeamOver hanging Beam

Using Bent RebarsUsing Bent Rebars

22--spans Beamspans Beam

More than 2-spans BeamMore than 2-spans Beam

ExampleExample::
For the given plan it is required to:
Calculate loads for slabs & Beams
Data:
FL.C = 150 kg/m2
L.L = 300 kg/m2
Steel Grade 360/520
SolutionSolution::
Slabs
D.L = (0.12 m * 2.5 t/m3
)+ 0.15 t/m2
= 0.45 t/m2
L.L = 0.3 t/m2
Wus = 1.4 x 0.45 t/m2
+ 1.6 x 0.3 t/m2
= 1.11 t/m2
To have slab thickness ts
S1 → ts = 400/40 = 10 cm
S2 → ts = 300/45 = 6.67 cm
S3 → ts = 500/45 = 11.1 cm
take ts = 12 cm

1.5
4
6
r ==For S1
→trapezoidal Ca = 0.67 & Ce = 0.85
→triangle Ca = 1/2 & Ce = 2/3
1.33
3
4
r ==For S2
1.2
5
6
r ==For S3
1.67
3
5
r ==For S4

W4 ( Wa & We) t/m
6m
B1 B2
BeamsBeams
O. wt. of beam = 0.25 x 0.6 x 2.5 t/m3
= 0.375 t/m
( ) ( )
t/m'2.951.2x1.4t/m0.375x1.4
2
m1.5
xt/m1.11
x W4.1wt.O.x1.4
2
L
xWuWW
2
wall
s
s4e4a
=++





=
++





==
B4 tB = span / 10 = 600 / 10 = 60 cm
take tB = 60 cm
Wall weight = Wwall = 0.25 (width) x 2.7 (height) x 1.8 (γwall = 1.2 – 1.8 t/m3
) = 1.2 t/m
2.95 t/m
6m
13.275 m.t
8.85 t 8.85 t
B1 B2
B. M. D.

B1
( ) ( )
( ) ( ) t/m'3.2281.2x1.4t/m'0.3125x1.4
2
1
x
2
m4
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
2
walla
s
sa1
=++





=
++





=
( ) ( )
( ) ( ) t/m'3.6051.2x1.4t/m'0.3125x1.4
3
2
x
2
m4
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
2
walle
s
se1
=++





=
++





=
4 m 5 m 1.5 m
8.85 t
a b c
d
Part ab
Part bc
Part cd
t/m'2.11751.2x1.4t/m'0.3125x1.4WW e3a3 =+==
( ) ( )
( ) ( ) t/m'3.5051.2x1.4t/m'0.3125x1.4
2
1
x
2
m5
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
2
walla
s
sa1
=++





=
++





=
( ) ( )
( ) ( ) t/m3.9771.2x1.4t/m'0.3125x1.4
3
2
x
2
m5
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
2
walle
s
se1
=++





=
++





=

3.228 t/m 3.505 t/m
4 m 5 m
Shear
G2G1 G3
B1
1.5 m
3.605 t/m 3.977 t/m
4 m 5 m
Moment
G2G1 G3
1.5 m
2.1175 t/m
8.85 t
8.85 t
Moment Values of B1
W2L2
2
/9
W1L1
2
/9
or
W1L1
2
/11 W2L2
2
/11
WcLc
2
/ 2
Msupp. = 9.514 m.t
6.214 m.t5.244 m.t
P x Lc + WcLc
2
/ 2 = 15.66 m.t
B. M. D.

O. wt. of beam = 0.25 x 0.5 x 2.5 t/m3
= 0.3125 t/m
( ) ( )
t/m'4.2761.2x1.4t/m'0.3125x1.40.63x
2
m3
xt/m1.11
2
1
x
2
m4
xt/m1.11
x W4.1wt.O.x1.4Cx
2
L
xWuW
22
walla
s
sa2
=++





+





=
++





=
( ) ( )
t/m'4.9541.2x1.4t/m'0.3125x1.40.81x
2
m3
xt/m1.11
3
2
x
2
m4
xt/m1.11
x W1.4O.wtx1.4Cx
2
Ls
xWuW
22
wallese2
=++





+





=
++





=
B2
tB = span / 12 = 500 / 12 = 41.67cm
take tB = 50 cm
Part ab
Part bc
t/m'4.4751.2x1.4t/m'0.3125x1.40.582x
2
m3
xt/m1.11
2
1
x
2
m5
xt/m1.11W 22
a2 =++





+





=
t/m'5.2571.2x1.4t/m'0.3125x1.40.769x
2
m3
xt/m1.11
3
2
x
2
m4
xt/m1.11W 22
e2 =++





+





=
4.275 t/m 4.475 t/m
2.1175 t/m
4 m 5 m 1.5 m
Shear
4.954 t/m 5.257 t/m
2.1175 t/m
4 m 5 m 1.5 m
Moment
Part cd
t/m'2.11751.2x1.4t/m0.3125x1.4WW e2a2 =+==
4 m 5 m 1.5 m
R B4
a b c
d

B2
4.954 t/m 5.257 t/m
4 m 5 m 1.5 m
8.85 t
2.1175 t/m
0.4x4.276x4 = 6.84 t
0.4x4.475x5 + 8.85 = 17.8 t
L1
m
a
0.4 W1 L1
+
0.6 W1 L1
-
L2
m
C
0.6 W2 L2
S.F.D +
0.4 W2 L2
-
b
Q1
Q2
Q3
Q4
0.6x4.276x4 = 10.26 t
0.6x4.475x5 = 13.43 t
Wc Lc
3.176 t
8.85 t
+ 8.85 t
4.275 t/m 4.475 t/m 2.1175 t/m
4 m 5 m
1.5 m
Shear
G2G1 G3
8.85 t
W2L2
2
/9
W1L1
2
/9
or
W1L1
2
/11 W2L2
2
/11
WcLc
2
/ 2
Moment Values of B2
10.67 m.t6.656 m.t
P x Lc + WcLc
2
/ 2 = 15.66 m.t
Msupp. = 12.734 m.t
B. M. D.
= 8.81 m.t
= 14.61 m.t

B3
( ) ( )
( ) ( ) ( ) t/m'4.2761.2x1.4t/m'0.3125x1.4m1xt/m1.110.63x
2
m3
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
22
walla
s
sSpanLefta3
=+++





=
++





=
( ) ( )
( ) ( ) ( ) t/m'4.3951.2x1.4t/m'0.3125x1.4m1xt/m1.110.701x
2
m3
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
22
walla
s
sSpanRighta3
=+++





=
++





=
Part ab
Part bc
( ) ( )
( ) ( ) ( ) t/m'4.5761.2x1.4t/m'0.3125x1.4m1xt/m1.110.81x
2
m3
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
22
walle
s
sSpanLefte3
=+++





=
++





=
( ) ( )
( ) ( ) ( ) t/m'4.6941.2x1.4t/m'0.3125x1.4m1xt/m1.110.881x
2
m3
xt/m1.11
x W1.4wtO.x1.4Cx
2
L
xWuW
22
walle
s
sSpanRighte3
=+++





=
++





=
a b c

4.276 t/m 4.395 t/m
4 m 5 m
Shear
G2G1 G3
4.576 t/m 4.694 t/m
4 m 5 m
Moment
G2G1 G3
W2L2
2
/9
W1L1
2
/9
or
W1L1
2
/11 W2L2
2
/11
B3
Moment Values of B3
13.04 m.t
10.67 m.t6.656 m.t
B. M. D.

Main Beams (GirdersMain Beams (Girders))
6.842 t
6.0m 3.0m
G1
RB2 at point a = 0.4 x (4.267 t/m
x 4 m) = 6.842 t
WG1 (wa = we = 3.732 t/m(
( ) ( )
( ) ( ) t/m'3.7321.2x1.4t/m'0.5625x1.4
m9
m1.5xm3x
2
1
m2x
2
m6m2
xt/m1.11
x W1.4wtO.x1.4
Span
Area
WWW
2
wallG1eG1aG1
=++










+










 +
=
++===
∑
RB2
O. wt. of main beam/ girder = 0.25 x 0.9 x 2.5 t/m3
= 0.5625 t/m
tG = span / 10 = 900 / 10 = 90 cm
take tG = 90 cm

6.842 t
6.0m 3.0m
WG1 (wa = we = 3.732 t/m(
RB2
47.29 m.t
21.36 t19.075 t
Mmax = 48.75 m.t
Distance 5.11 m from left supp.
G1
B. M. D.

G2
WG2 (wa = we = 4.997 t/m
(
6.0m 3.0m
RB2
23.69t
RB2 at point b = Wa2 * L = 4.276 t/m
x 4 m + 4.475 t/m
x 5 m = 23.69 t
( ) ( )
( ) ( ) t/m'4.8841.2x1.4t/m'0.5625x1.4
m9
2xm1.5xm3x
2
1
m2.5x
2
m6m1
m2x
2
m6m2
xt/m1.11
x W1.4wtO.x1.4
Span
Area
WWW
2
wallG2eG2aG2a
=++




















+








 +
+








 +
=
++===
∑

WG2 (wa = we = 4.997 t/m(
6.0m 3.0m
RB2
23.69t
91.33 m.t
37.77 t29.87 t
G2
B. M. D.

20.455 t
G3
RB2 at point c = 0.4 x W L + P + M/L = 6.842 t
WG3 (wa = we = 4.38 t/m
(
( ) ( )
( ) ( )
t/m'4.38
1.2x1.4t/m'0.5625x1.4
m9
m6x
2
1.5
m1.5xm3x
2
1
m2.5x
2
m6m1
xt/m1.11
x W1.4wtO.x1.4
Span
Area
WWW
2
wallG3eG3aG3
=
++










+



+










 +
=
++===
∑
RB2 at point c = 0.4 x 4.475 x 5 + 8.85 + (8.85x1.5 + (2.1175 x (Lc
2
/2))/5) = 20.455 t

20.455 t
WG3 (wa = we = 4.38 t/m
(
80.34 m.t
G3
33.35 t26.53 t
B. M. D.

LL
WuWubeambeam t/mt/m
/24WL2
/24WL2
/24WL2
/24WL2
/24WL2
/8WL2
/9WL2
/11WL2
/11WL2
/10WL2
/12WL2 /16WL2
/12WL2
/16WL2
Moment Values of beamsMoment Values of beams

ALB
Dr. ALaa A. Bashandy
E-Mail : Eng_ ALB@yahoo.com
1.1.ECP 203-2007ECP 203-2007
2.2.2032007.
3.3.2032007.
4.4.Design of R. C. StructuresDesign of R. C. Structures– - Vol. 1, 2, 3Vol. 1, 2, 3

تصميم الكمرات Dr.alaa bashandy

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