Design of Circular Beams -...

Design of Circular Beams

The Islamic University of Gaza

Department of Civil Engineering

Design of Special Reinforced Concrete Structures

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 = 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



S.F and B.M at Point P at angle from support C

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



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


Dr. Mohammed Arafa

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



Circular beams loaded uniformly and supported on symmetrically placed columns

A

B


Circular beams at symmetrically placed columns


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)



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)



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



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


2 2P

wRV wR wR

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

16Dr. Mohammed Arafa

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 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


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

RwRT M R R wR R

dTwR

d

M wR

0dT

d

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


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

2

'

''

P rw

M KwR

M K wR

T K wR

O

Mmax(+ve)

Tmax

Tmax

Mmax(-ve)

Vmax

Mmax(-ve)

Vmax

`

/2


20

Example 1

Design a semi-circular beam supported on three-equallyspaced columns. The centers of the columns are on acircular curve of diameter 8m. The beam is support auniformly distributed factored load of 5.14 t/m in additionto 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’

Dr. Mohammed Arafa

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

Dr. Mohammed Arafa

22

Example 1

Bending Moment Diagram

Shear Force Diagram

Dr. Mohammed Arafa


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


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


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


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

27

Example 1

Design for Torsion: Ductility Check

The following equation has to be satisfied

2

'

2

0

0

0 0

2

2 23

2

2.121.7

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

9.9711.16 10

1.7 40 63.95

u u h cc

w oh w

h

oh

u u h

w oh

V T p Vf

b d A b d

x

y

p x y cm

A cm

V T p

b d A

5

2

2

' 2

10 184.830.04 /

1.7 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.Dr. Mohammed Arafa

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

Dr. Mohammed Arafa

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

Dr. Mohammed Arafa

30

Example 1

Design for Torsion

Dr. Mohammed Arafa

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



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



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



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


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


SEMICIRCULAR BEAM FIXED AT END SUPPORTS

If a semicircular beam supports a concrete slab, asshown the ratio of the length to the width of the slab is2r/r = 2, and the slab is considered a one-way slab.

The beam will be subjected to a distributed load, whichcauses torsional moments in addition to the bendingmoments and shearing forces.

The load on the curved beam will be proportional to itsdistance from the support AB



37

2 2

3

3

sin'

2

0.

Load on the curved beam per unit length

Shear Force

Bending Moment at support

Torsional Moment at sup

39

por

0

t

8

3

.11

A B

A B

A

wrw

V V wr wr

wrM M

T wr

Dr. Mohammed Arafa



23

23

3

3

Bending Moment at section N

Torsional Moment at section

1 cossin sin

8 6

1 cossin 0.11sin

8 6

0.116 at =2

1cos 1 sin 2 cos

8 4 24

N

N A

N

Midspan

N A

N

M wr T

M wr

M wr

T wr T

T wr

3 1cos 1 sin 2 0.11cos

8 4 24

0 at =2

MidspanT

39

BMD

TMD

Dr. Mohammed Arafa

Fixed-end Semicircular Beam Under Uniform Loading


Fixed End Semicircular Beam with Uniform Loading

42

2

2


Shear Force

Bending Moment at suppo

1.57

0.

rt

Torsional Moment at support

3

A B

A B

A

w

V V wr

M M wr

T wr

Dr. Mohammed Arafa

43

2

2

2

2

2

sin 1 sin2

4sin 1

0.273 =2

Bending Moment at section N

Torsional Momen

cos cos2 2

4cos

2

0

t at section N

N A

N

Midspan

N A

N

Midspan

M wr T

M wr

M wr

T wr T

T wr

T

=2

Dr. Mohammed Arafa


Fixed End Circular Beam under Uniform Loading



46

2

1 2 3 1 2

4

2


Bending Moment at Centerline

Torsional Moment at Centerline

Bending Moment at Section N

Torsional Moment

0

co

at Se

s

c

1 cos

C

C

N C

w

wrM K K K K K

K

T

M M wr

2si

tion N

n sinN CT M wr

Dr. Mohammed Arafa


47

2

2

Bending Moment at Support A

Torsional Moment at Support A

cos 1 cos

sin sin

A C

A C

M M wr

T M wr

Dr. Mohammed Arafa

Values of and J

48

4

4

3

For circcular section

For square section of side x

For rectangular section of short side x and long side

1

2

0.1 1

y

4

'

The Value of J

J r

J x

J K xy

Dr. Mohammed Arafa

The value of G/E for concrete may be assumed tobe equal to 0.43.

Values of and J

49

For rectangular section of short side x and long side y

Dr. Mohammed Arafa

50

P1- Design a circular beam that supported on six equally spaced

columns, and its centerline lies on a circle 7m in diameter. The

beam carries a uniform dead load of 12 t/m` and live load of

6kg/m’. Use

P2. Find the values of factored shear, moment and torsion at critical

sections for the fixed end semicircular beam, if the diameter of the

circle is 10 m. The beam is apart of floor slab that caries a uniform

dead load (including its own) weight of 800 kg/m2 and live load of

200kg/m2.

P3. Find the values of factored shear, moment and torsion at critical

sections for the 60o V-shape beam, with a=8m. The beam carries a

uniform dead load (including its own) weight of 10 t/m` and live

load of 4t/m`. Assume the ratio of long to short side of the

rectangular section is 2.

Problems

' 2 2300 / and 4200 /c yf kg cm f kg cm

Design of Circular Beams -...

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