Circular Beam

Design of Circular Beams

The Islamic University of GazaDepartment of Civil EngineeringDesign of Special Reinforced Concrete Structures

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 of loads does not coincide with the centerline axis of the member.


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.


Semi-circular beam simply supported on three equally spaced supports


Supports Reactions LetRA = RB = V1RB = V2Taking 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

ππ

π

=

= −

= −

∑

( )2

2

0

2 22

2

yFWRV wR

V Rw

π π

=

= − −

=

∑



Taking the bending moment at point P at angle θfrom 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

θ

θ

θ

θθ θ θ

θ

π θ θ

πθ θ

= −

= − −

− = −



Maximum negative moment occurs at point B

max.

2

( )2

0.429

ve B

B

M M at

M wR

πθ− = =

= −

Maximum Positive moment

( ) ( ) ( )2

2max

. ( 0)

2cos 2sin / 2 cos / 2 0

22tan 29 43'

20.1514

vedMMax M whered

dM wRd

M wR

θ

θ

θπ

θ θ θθ

πθ θ

+ = =

− = − =

−

= =

=



Torsional Moment at point P


( ) ( ) ( )1

2

2max.

sin / 2cos cos / 2

/ 2

2 2 cos sin2 2

dTThesection of max. torsion 0d

dT 0 59 26 'd

0.1042

RT V R R wR R

T wR

at

T wR

θ

θ

θ

θ

θθ θ θ

θ

π π θ θ θ

θ

θθ

= ⋅ − − −

− − = − − +

=

= =

=


Circular beams loaded uniformly and supported on symmetrically placed columns

A

B

θ


Consider portion AB of beam between two consecutive columns A and B be θ

The load on portion AB of beam =wRθ

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) θ θ


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 wRθ=

0M sin( /2)θ 0M cos( /2)θ


Taking the moment of all forces about line AB


( ) ( ) ( )

( )( ) ( )

( )

( )

0

20

20

sin / 22 sin / 2 co s / 2

/ 2

sin / 2 co s / 22sin / 2 / 2 2sin / 2

1 cot / 22

RM wR R

M wR

M wR

θθ θ θ

θ

θ θθ θθ θ θ

θ θ

= −

= ⋅ − = −


Shear force and bending moment at point P, at angle φ from the support

2 2PwRV wR wRθ θφ φ = − = −

• Load between points A and P is wRφ• Shearing force at point P


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 2sin / 2 1

1 sin cot / 2 cos2 2

RM V R M wR

wRM wR wR

M wR

M wR

φ

φ

φ

φ

φφ φ φ φ

φθ θφ θ φ φ

θ θφ φ θ φ φ

φ φ

θ θφ θ φ

= − −

= − − − = − + − + =

= − + +


The direction of vector representing torsion at point P will act at right angle to line PO


( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

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 co t / 2 sin sin / 2 1 sin / 2 co s / 22

2 11 cot / 2 sin sin / 2 1 sin2 2

RwRT M R R wR R

R R R R

T wR wR wR

T wR wR wR

φ

φ

φ

φθφ φ φ φφ

φ φ φ

θ θ φ θ φ φ φ φφ

θ θ φ θ φ φ φφ

= − − + −

− = − =

= − − + −

= − − + −

( ) ( )

( ) ( )

( )

( )

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

φ

φ

φ

θφ θ φ θ φ φ φ

θφ θ φ θ φ

φ φ

θ θ θφ φ φ θ φ

= − − + − = − −

= −

= − + − − ENGC 6353 Dr. Mohammed Arafa 17

To obtain maximum value of torsional moment


( ) ( ) ( )

( )

( )

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

RwRM R R wR R

dTwR

d

M wR

φ

φ

φθφ φ φ φφ

θ θφ φ θ φφ

θ θφ φ θ φ

= − − + −

= − − − = = − − − =

0dTd

φ

φ=

i.e. point of maximum torsion will be point of zero moment


Internal Forces in Circular beams Using polar coordinate

A

B

ϕP

as r= constant

dV w ds rddsdV wrdsdM dT Vds dr

dT Tdr rdM T Vds r

dM T Vrd

ϕ

ϕ

= − =

= −

+ =

=

+ =

+ =



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 Mdr

dT Md

ϕ

ϕ

= =

=

This means that the torsional moment is maximum at point of zero bending moment



Differentiating the equation

The solution of this deferential equation gives

( )22

2

22

2

d Vrd M dT dVr r wd d d dd M M wrd

ϕ ϕ ϕ ϕ

ϕ

+ = = = −

+ = −

dM T Vrdϕ

+ =

2sin cosM A B wrϕ ϕ= + −



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 wrϕ ϕ= + −


Circular beams loaded uniformly and supported on symmetrically n placed columns

The bending moment and shearing force V at any section at an angle ϕ from the centerline between two successive supports are given by:

0 022

2n nrw rwR Vn n

π πϕ ϕ

π π

= =

= = ±

2

0

2

0

cos 1sin

sinsin

M r wn

T r wn

V rw

π ϕϕ

π ϕ ϕϕ

ϕ

= −

= − −

= −ENGC 6353 Dr. Mohammed Arafa 23

Circular beams loaded uniformly and supported on symmetrically n placed columns

# of Supports R V θ K K` K``Angle

for Max Torsion

4 P/4 P/8 90 0.137 0.07 0.021 19 215 P/5 P/10 72 0.108 0.054 0.014 15 456 P/6 P/12 60 0.089 0.045 0.009 12 447 P/7 P/14 51 3/7 0.077 0.037 0.007 10 45 8 P/8 P/16 45 0.066 0.030 0.005 9 339 P/9 P/18 40 0.060 0.027 0.004 8 3010 P/10 P/20 36 0.054 0.023 0.003 7 3012 P/12 P/24 30 0.045 0.017 0.002 6 21

2support

` 2Midspan

`` 2max

2P rwM KwR

M K wR

T K wR

π

θ

θ

θ

=

=

=

=O

Mmax(+ve)

Tmax

Tmax

Mmax(-ve)

V

Mmax(-ve)

Vmax

φ`

θ/2


Circular Beam

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