design of circular beams - الصفحات الشخصية | الجامعة...
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Design of Circular Beams
The Islamic University of Gaza
Department of Civil Engineering
Design of Special Reinforced Concrete Structures
ENGC 6353 Dr. Mohammed Arafa 1
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Circular Beams
They are most frequently used in circular reservoirs,
spherical dome, curved balconies … etc.
Circular beams loaded and supported loads normal
to their plans.
The center of gravity of loads does not coincide
with the centerline axis of the member.
ENGC 6353 Dr. Mohammed Arafa 2
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Circular Beams
Circular beam are subjected to torsional moments
in addition to shear and bending moment
The torsional moment causes overturning of the
beams unless the end supported of the beams are
properly restrained.
ENGC 6353 Dr. Mohammed Arafa 3
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Semi-circular beam simply supported on three
equally spaced supports
ENGC 6353 Dr. Mohammed Arafa 4
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Supports Reactions
Let
RA = RC = V1
RB = V2 Taking the moment of the reaction about the line
passing through B and parallel to AC
1
1
0
22
22
M
RV R wR R
WRV
2
2
0
2 22
2
yF
WRV wR
V Rw
Semi-circular beam supported on three supports
ENGC 6353 Dr. Mohammed Arafa 5
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ENGC 6353 Dr. Mohammed Arafa 6
S.F and B.M at Point P at angle from support C
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Taking the bending moment at point P at angle
from the support say, C equal:
1
2 2
2 2
sin / 2sin sin / 2
/ 2
2 sin 2 sin / 22
2sin 2sin / 2
2
RM V R wR
wRM R wR
M wR
ENGC 6353 Dr. Mohammed Arafa 7
Semi-circular beam supported on three supports
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Maximum negative moment occurs at point B
max.
2
( )2
0.429
ve B
B
M M at
M wR
Maximum Positive moment
2
2
max
. ( 0)
2cos 2sin / 2 cos / 2 0
2
2tan 29 43'
2
0.1514
ve
dMMax M where
d
dMwR
d
M wR
8
Semi-circular beam supported on three supports
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Torsional Moment at point P
1
2
2
max.
sin / 2cos cos / 2
/ 2
2 2cos sin
2 2
dTThe section of maximum torsion 0
d
dT0 59 26 '
d
0.1042
RT V R R wR R
T wR
at
T wR
ENGC 6353 Dr. Mohammed Arafa 9
Semi-circular beam supported on three supports
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Circular beams loaded uniformly and supported on
symmetrically placed columns
A
B
ENGC 6353 Dr. Mohammed Arafa 10
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Circular beams at symmetrically placed columns
ENGC 6353 Dr. Mohammed Arafa 11
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Consider portion AB of beam between two
consecutive columns A and B be with angle
The load on portion AB of beam =wR
The center of gravity of the load will lie at a
distance from the center
Let M0 be the bending moment and V0 be the
shear at the supports
R sin( /2) ( /2)
ENGC 6353 Dr. Mohammed Arafa 12
Circular beams at symmetrically placed columns
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From geometry
The moment M0 at the support can be resolved
along the line AB and at right angle to it. The
component along line AB is and
at right angle of it.
Taking the moment of all forces about line AB
0 / 2V wR
0M sin( /2) 0M cos( /2)
ENGC 6353 Dr. Mohammed Arafa 13
Circular beams at symmetrically placed columns
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Taking the moment of all forces about line AB
0
2
0
2
0
sin / 22 sin / 2 cos / 2
/ 2
sin / 2 cos / 2
2sin / 2 / 2 2sin / 2
1 cot / 22
RM wR R
M wR
M wR
ENGC 6353 Dr. Mohammed Arafa 14
Circular beams at symmetrically placed columns
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S.F and B.M. at point P, at angle from the support
• Load between points A and P is wR • Shearing force at point P
ENGC 6353 Dr. Mohammed Arafa 15
2 2P
wRV wR wR
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The direction of vector representing bending moment at point P will act along line PO
Shear force and bending moment at point P
0 0
22 2 2
2 2
2
2
sin / 2sin cos sin / 2
/ 2
sin 1 cot / 2 cos 2 sin / 22 2
sin cos cot / 2 cos 2sin / 22 2
since cos 1 2sin / 2
1 sin cot / 2 cos2 2
RM V R M wR
wRM wR wR
M wR
M wR
16
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The direction of vector representing torsion at point P will act at
right angle to line PO
Shear force and bending moment at point P
0
2
2 2 2 2
2 2 2 2
sin / 2sin cos cos / 2
2 / 2
since cos 1 cos 2 sin / 2
21 cot / 2 sin sin / 2 1 sin / 2 cos / 2
2
2 11 cot / 2 sin sin / 2 1 sin
2 2
RwRT M R R wR R
R R R R
T wR wR wR
T wR wR wR
2 2
2 2
2
2
sin cot / 2 sin sin / 2 sin2
cot / 2 sin sin / 22
since 2sin / 2 1 cos
cos sin cot / 2 sin2 2 2
T wR
T wR
T wR
17
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To obtain maximum value of torsional moment
Shear force and bending moment at point P
0
2
2
sin / 2sin cos cos / 2
2 / 2
1 sin sin cot / 2 cos 02 2
1 sin sin cot / 2 cos 02 2
RwRT M R R wR R
dTwR
d
M wR
0dT
d
i.e. point of maximum torsion will be point of zero moment
ENGC 6353 Dr. Mohammed Arafa 18
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Circular beams loaded uniformly and supported on
symmetrically n placed columns
# of Supports R V K K` K``
Angle
for Max
Torsion
4 P/4 P/8 90 0.137 0.07 0.021 19 21
5 P/5 P/10 72 0.108 0.054 0.014 15 45
6 P/6 P/12 60 0.089 0.045 0.009 12 44
7 P/7 P/14 51.4 3 0.077 0.037 0.007 10 45
8 P/8 P/16 45 0.066 0.030 0.005 9 33
9 P/9 P/18 40 0.060 0.027 0.004 8 30
10 P/10 P/20 36 0.054 0.023 0.003 7 30
12 P/12 P/24 30 0.045 0.017 0.002 6 21
2
support
` 2
Midspan
`` 2
max
2P rw
M KwR
M K wR
T K wR
O
Mmax(+ve)
Tmax
Tmax
Mmax(-ve)
Vmax
Mmax(-ve)
Vmax
`
/2
19
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20
Example 1
Design a semi-circular beam supported on three-equally spaced columns. The centers of the columns are on a
circular curve of diameter 8m. The beam is support a uniformly distributed factored load of 5.14 t/m in addition to its own weight.
' 2 2350 / and 4200 /c yf kg cm f kg cm
min
2 6.28
/18.5 628 /18.5 33.94
use Beam 40×70cm
L r m
h L cm
o. w. of the beam =0.4(0.7)(2.5)(1.2)
= 0.84 t/m’
Total load= 5.14+0.84=5.98 t/m’
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21
Example 1
2 2
max( )
2 2
max( )
2 2
max
0.429 0.429(5.98)(4) 41.05 .
0.1514 0.1514(5.98)(4) 14.48 .
0.1042 0.1042(5.98)(4
Re
) 9.97 .
5.98(4)2 2 13.65
2 2
2 2(5.98)(4) 47.84
actions
Shear at
ve
ve
A
B
M wR t m
M wR t m
T wR t m
wRR ton
R wR ton
point P: 22
wRV wR
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ENGC 6353 22
Example 1
Bending Moment Diagram
Shear Force Diagram
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ENGC 6353 Dr. Mohammed Arafa 23
Example 1
Design for Reinforcement
5
min
2
( )
5
min
( )
d=70-4.0-0.8-1.25=63.95
2.61 10 41.050.85 3501 1 0.00697
4200 40 63.95 350
0.00697 40 63.95 17.85
2.61 10 14.480.85 3501 1 0.00238
4200 40 63.95 350
0.00
ve
s ve
ve
s ve
A cm
A
233 40 63.95 8.44cm
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ENGC 6353 Dr. Mohammed Arafa 24
Example 1
Design for Shear
max
2
23.92 ( 0.0at middle support)
0.53 0.75 350(40)(63.95) 19.02
23.92 19.026.53
0.75
6.53 1000 4200(63.95)
0.0243 /
c
S
V
V
V ton T
V ton
V ton
A
S
Acm cm
S
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ENGC 6353 Dr. Mohammed Arafa 25
Example 1
Design for Torsion
' 2
0 0 0 0 0 0 0 0
2 2
'
2
0.265neglect torsion
, 2 , , 2 and 0.85
The following equation has to be satisfied to check for ductlity
2.121.7
c cp
u
cp
cp cp h h
u u h cc
w oh w
f Aif T
p
A bh p b h A x y p x y A x y
V T p Vf
b d A b d
0
'
,min
Reinforcement
2
1.3and
uT
ys
c cpT Tl h l h
y
TA
S f A
f AA AA p A p
S f S
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ENGC 6353 Dr. Mohammed Arafa 26
Example 1
Design for Torsion
At section of maximum torsion is located at =59.43
T=9.97 t.m
59.432 13.65 5.98(4)( 2 ) 11.6
2 360
0.0
0.53 0.75 350(40)(63.95) 19.02 Noshear reinf.required
u
c u
wRV wR ton
M
V ton V
2' 2 '
5
0.265 0.265 0.75 40 701.32 .
2 40 70 10
c cp c
u
cp
f A ft m T
p
i.e torsion must be considered
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27
Example 1
Design for Torsion: Ductility Check The following equation has to be satisfied
2
'
2
0
0
0 0
2
2 2 53
2
0.2631.7
40 4 4 0.8 31.2
70 4 4 0.8 61.2
2 2 31.2 61.2 184.8
31.2 61.2 1909.44
9.97 10 184.811.16 10
1.7 40 63.95 1.7
u u hc
w oh
h
oh
u u h
w oh
V T pf
b d A
x
y
p x y cm
A cm
V T p
b d A
2
2
' 2
30.04 /1909.44
19.022.12 0.75 2.12 350 49.6 / 30.04
40 63.95
cc
w
kg cm
Vf kg cm
b d
i.e section dimension are adequate for preventing brittle failure due to combined shear stresses.
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28
Example 1
Design for Torsion
52
0
2
min
2
9.97 100.097 /
2 2 0.75 4200 0.85 1909.44
3.5 403.50.0167 /
2 2 4200
0 0.097 0.086 /
A 1.13Use 12mm @11.5cm (closed stirrups) with 0.0983
S 11.5
uT
ys
wT
ys
V T
TAcm cm
S f A
bAcm cm
S f
A Acm cm
S S
c
2 /m cm
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29
Example 1
Design for Torsion
2
'
2
,min
2
2
,
,
0.086 184.8 15.98
1.3 1.3 350 40 7015.98 0.234
4200
15.985.33
3
Use 2 20 top and 4 14skin reinforcement
8.44 5.33 13.77 use6 18mm
Tl h
c cp Tl h
y
l
s ve total
s ve total
AA p cm
S
f A AA p cm
f S
A cm
A cm
A
217.83 5.33 23.16 use8 20mmcm
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30
Example 1
Design for Torsion
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Internal Forces in Circular beams Using polar coordinate
A
B
P
as = constant
dVw ds rd
ds
dVwr
d
dM dTV r
ds dr
dT T
dr r
dM TV
ds r
dMT Vr
d
ENGC 6353 Dr. Mohammed Arafa 31
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Internal Forces in Circular beams Using polar coordinate
The component of the moment M along ds in the radial direction must be equal to the difference of the torsional
moments along the same element
MdsdT Md
r
dTM
d
This means that the torsional moment is maximum at point of zero bending moment
ENGC 6353 Dr. Mohammed Arafa 32
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Internal Forces in Circular beams Using polar coordinate
Differentiating the equation
The solution of this deferential equation gives
22
2
22
2
d Vrd M dT dVr r w
d d d d
d MM wr
d
dMT Vr
d
2sin cosM A B wr
ENGC 6353 Dr. Mohammed Arafa 33
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Internal Forces in Circular beams Using polar coordinate
The constants can be determined from conditions at supports
The internal forces are in this manner statically indeterminate. In many cases, they can be determined from the conditions of equilibrium alone because the
statically indeterminate values are either known or equal to zero.
2sin cosM A B wr
ENGC 6353 Dr. Mohammed Arafa 34
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Circular beams loaded uniformly and supported on
symmetrically n placed columns
The bending moment and shearing force V at any section at an angle from the centerline between two
successive supports are given by:
0 0
22
2
n n
rw rwR V
n n
2
0
2
0
cos1
sin
sin
sin
M r wn
T r wn
V rw
ENGC 6353 Dr. Mohammed Arafa 35