DISEÑO DE VÍAS - Curva Circular Espiral CircularDescripción completa
gives the design details.
The article discusses about the "Circular Economy. A circular economy is a contrasting option to a linear economy influence, use, to arrange in which we keep assets being used for whatever length of time that conceivable, remove the most extreme ince
Descripción: EJERCIOSIOS DE CANALES
partes de la carta circular uso y esquemasDescripción completa
Metode Analisa Struktur dengan Menggunakan Balok Konjugasi, hak cipta kepada penulis terkait, pengupload cuma berusaha berbagi summary yang telah dibuat oleh penulis terkait, hak cipta meng…Full description
Design spreadsheet for the beamsFull description
mektur 2Full description
JKR MALAYSIA BEAMFull description
Full description
The Isl The Isla amic Un Uniiversi sitty of Ga Gaza za Department of Civil Engineering Design of Special Reinforced Concrete Structures
Design of Circular Beams
ENGC 6353
Dr Mohammed Arafa
1
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 gravity of loads does not coincide with the centerline axis of the member.
ENGC 6353
Dr Mohammed Arafa
2
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
Semi-circular beam simply supported on three equally spaced supports
ENGC 6353
Dr Mohammed Arafa
4
Semi-circular beam simply supported on three equally spaced supports Supports Reactions Let R A = R B = V1 R B = V2 Taking the moment of the reaction about the line passing through B and parallel to AC
∑ M = 0 2V 1R V 1
=
2 R = wR π R − π
WR
2
(π − 2 )
ENGC 6353
∑ F
y
=0
WR V 2 = wR π − 2 (π − 2 ) 2 V 2 = 2Rw Dr Mohammed Arafa
5
Semi-circular beam simply supported on three equally spaced supports
ENGC 6353
Dr Mohammed Arafa
6
Semi-circular beam simply supported on three equally spaced supports Taking the bending moment at point P at angle θ from the support say, C equal: R sin (θ / 2 ) M θ =V 1R sin θ − wR θ sin (θ / 2 ) θ / 2 M θ =
Semi-circular beam simply supported on three equally spaced supports Maximum negative moment occurs at point B M
π
max. −ve
M B
= M B (at θ = ) 2
= −0.429wR 2
Maximum Positive moment Max . M +ve dM θ d θ
= wR
tan θ =
= where ( 2
d θ
= 0)
(π − 2 ) cos θ − 2 sin (θ / 2) cos (θ / 2) = 0 2
π − 2 2
= 0.1514wR 2
M 6353 ENGC
dM θ
θ = 29 43'
Dr Mohammed Arafa
8
Semi-circular beam simply supported on three equally spaced supports Torsional Moment at point P R sin (θ / 2 ) T θ =V 1 ⋅ ( R − R cos θ ) −wR θ R − cos (θ / 2 ) θ / 2 π − 2 2 π − 2 T θ = wR cos θ − θ + sin θ − 2 2
The section of max. torsion dTθ d θ
=0
dTθ d θ
=0
at θ = 59 26 '
T max. = 0.1042wR 2 ENGC 6353
Dr Mohammed Arafa
9
Circular beams loaded uniformly and supported on symmetrically placed columns
A
θ
B
ENGC 6353
Dr Mohammed Arafa
10
Circular beams loaded uniformly and supported on symmetrically placed columns
ENGC 6353
Dr Mohammed Arafa
11
Circular beams loaded uniformly and supported on symmetrically placed columns Consider
portion AB of beam between two consecutive columns A and B be θ
The
load on portion AB of beam =wR θ
The
center of gravity of the load will lie at a distance R sin(θ /2) (θ /2) from the center
Let
M0 be the bending moment and V0 be the shear at the supports
ENGC 6353
Dr Mohammed Arafa
12
Circular beams loaded uniformly and supported on symmetrically placed columns From
geometry
V0
= wR θ / 2
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 M 0 sin(θ /2) and M 0 cos(θ /2) at right angle of it.
Taking
the moment of all forces about line AB
ENGC 6353
Dr Mohammed Arafa
13
Circular beams loaded uniformly and supported on symmetrically placed columns Taking
the moment of all forces about line AB
R sin (θ / 2 ) − R co s(θ / 2 ) 2 M 0 sin (θ / 2 ) = wR θ θ / 2 sin (θ / 2 ) co s(θ / 2 ) θ 2 ⋅ − θ M 0 = wR θ /2 2sin (θ / 2 ) 2sin (θ / 2 ) θ 2 M 0 = wR 1 − cot (θ / 2 ) 2 ENGC 6353
Dr Mohammed Arafa
14
Shear force and bending moment at point P, at angle φ from the support
• •
Load between points A and P is Shearing force at point P wR θ θ VP = −wR φ = wR − φ 2 2 ENGC 6353
wR φ
Dr Mohammed Arafa
15
Shear force and bending moment at point P The direction of vector representing bending moment at point P will act along line PO M φ =V 0 R sin φ − M 0 cos φ −wR φ
R sin (φ / 2 )
φ / 2
sin (φ / 2 )
2
θ sin φ −wR 1 − cot (θ / 2 ) cos φ − 2wR 2 sin 2 (φ / 2) M φ = 2 2 θ θ M φ = wR 2 sin φ − cos φ + cot (θ / 2 ) cos φ − 2 sin 2 (φ / 2) 2 2 since cos φ + 2sin 2 (φ / 2 ) = 1 wR θ
2
θ θ M φ = wR −1 + sin φ + cot (θ / 2 ) cos φ 2 2 2
ENGC 6353
Dr Mohammed Arafa
16
Shear force and bending moment at point P The direction of vector representing torsion at point P will act at right angle to line PO T φ
= M 0 sin φ −
wR θ
2
( R − R cos φ ) +wR φ R −
R
sin (φ / 2 ) φ / 2
cos (φ / 2 )
since R − R cos φ = R (1 − cos φ ) = 2 R sin 2 (φ / 2 ) T φ
Shear force and bending moment at point P To obtain maximum value of torsional moment dT φ d φ
=0
R sin (φ / 2 ) cos (φ / 2 ) = M 0 sin φ − ( R − R cos φ ) + wR φ R − 2 φ / 2 dT φ θ θ = wR 2 1 − sin φ − sin φ − cot (θ / 2 ) cos φ = 0 2 d φ 2 θ θ 2 M φ = wR 1 − sin φ − sin φ − cot (θ / 2 ) cos φ = 0 2 2 wR θ
i.e. point of maximum torsion will be point of zero moment
ENGC 6353
Dr Mohammed Arafa
18
Internal Forces in Circular beams Using polar coordinate dV ds dV ds dM ds dT dr dM ds dM d ϕ
= −w
ds = rd ϕ
A
= −wr + =
dT
dr T
+
r T r
ϕ
=V
as r= constant
P
B
=V
+T =V r ENGC 6353
Dr Mohammed Arafa
19
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
dT dT d ϕ
= Md ϕ =
Mds r
= M
This means that the torsional moment is maximum at point of zero bending moment
ENGC 6353
Dr Mohammed Arafa
20
Internal Forces in Circular beams Using polar coordinate Differentiating the equation dM
+T =V r
d ϕ 2
d M dϕ
2
+
dT dϕ
=
d (V r ) dϕ
=r
dV
= − r 2w
dϕ
2
d M 2 d ϕ
+ M = −wr 2
The solution of this deferential equation gives
M = A sin ϕ + B cos ϕ −wr
2
ENGC 6353
Dr Mohammed Arafa
21
Internal Forces in Circular beams Using polar coordinate M = A sin ϕ + B cos ϕ −wr
2
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.
ENGC 6353
Dr Mohammed Arafa
22
Circular beams loaded uniformly and supported on symmetrically n placed columns 2ϕ0 = R
=
2π n
2π rw n
ϕ 0 = V
π n
=±
π rw n
The
bending moment and shearing force V at any section at an angle ϕ from the centerline between two successive supports are given by: π cos ϕ M = r w − 1 n sin ϕ 0 π sin ϕ 2 T = −r w − ϕ n sin ϕ 0 VENGC = −6353 rw ϕ 2
Dr Mohammed Arafa
23
Circular beams loaded uniformly and supported on symmetrically n placed columns # of Supports
R
V
θ
K
K`
K``
4 5 6 7 8 9 10 12
P/4 P/5 P/6 P/7 P/8 P/9 P/10 P/12
P/8 P/10 P/12 P/14 P/16 P/18 P/20 P/24
90 72 60 51 3/7 45 40 36 30
0.137 0.108 0.089 0.077 0.066 0.060 0.054 0.045
0.07 0.054 0.045 0.037 0.030 0.027 0.023 0.017
0.021 0.014 0.009 0.007 0.005 0.004 0.003 0.002
Angle for Max Torsion 19 21 15 45 12 44 10 45 9 33 8 30 7 30 6 21
