Design of Circular Beams
Dr. Mohammed Arafa
Design of Circular Beam Example 1 (ACI 318-99) Design a semi-circular beam supported on three-equally spaced columns. The centers of the columns are on a circular curve of o f diameter 8m. The beam is support supp ort a uniformly distributed factored load of 5.0 t/m, in addition add ition to its own weight. Use f c'
=
350 kg / cm 2 and f y
=
4200 kg / cm 2 B
Solution L
=
hmin
r (π 2 ) = 6.28 m =
L /18. /18.5 = 628 /18. /18.5 = 33.94cm
c.g
use Beam 40×70cm o. w. of the beam =0.4(0.7)(2.5)(1.4)=0.98 t/m’
2R/
total load= 5.14+0.84=5.98 t/m’
C
M max( −ve )
= 0.429wR 2 = 0.429(5.98)(4)2 = 41.05t .m
M max( +ve )
= 0.1514wR 2 = 0.1514(5.98)(4)2 = 14.48t .m
T max
= 0.1042w R 2 = 0.1042(5.98)(4)2 = 9.97t .m
Reactions R A
=
wR
(π − 2 ) =
2 R B = 2wR
5.98(4)
13.65ton (π − 2 ) = 13 2 = 2(5.98)(4) = 47.84ton
Shear at point P:
V θ
=
wR
2
(π − 2 ) −wR θ 23.92
13.65
2.28 13.65 23.92 Shear Force Diagram
Design of Circular Beams
Dr. Mohammed Arafa
41.05
2.28 14.48
14.48
Bending Moment Diagram
Design for Reinforcement
d=70-4.0-0.8-1.25=63.95 2.61⋅ 105 ( 41.05) 0.85⋅ 350 1 − 1 − = 0.0697 > ρ min = 4200 40 ⋅ 63.95 ⋅ 350
ρ −ve
A s ( −ve ) = ( 0.00697 )( 40)( 63.95) = 17.85cm 2
2.61⋅ 105 (14.48) 0.85⋅ 350 1 − 1 − = 0.00238 < ρ min = 4200 40 ⋅ 63.95 ⋅ 350
ρ −ve
A s ( +ve ) = ( 0.0033)( 40 )( 63.95) = 8.44cm
2
Design for Shear V max
φ V c
= =
23.92 ton
(T = 0.0 at middle support)
0.53( 0.85) 350(40)(63.95) = 21.56ton
23.92 − 21.56
= 2.77 ton 0.85 A 2.77 ⋅ 1000 = V 4200(63.95) S AV 2 = 0.0103 cm / cm S
V S
=
Design of Circular Beams
Dr. Mohammed Arafa
Design for Torsion T u
≤
Acp
0.265φ f c' Acp2 pcp
⇒
neglect torsion
= b h, pcp = 2 (b + h ) , A 0 h = x 0 y 0 ,
ph
= 2 ( x 0 + y 0 ) and
A0
= 0.85x 0 y 0
The following equation has to be satisfied to check for ductlity 2
V u Tu p h b d + 1.7A 2 ≤ 0.263φ w oh
f c'
Reinforcement AT S A l
=
T u
2φ f ys A 0
A = T p h and S
A l ,min
=
1.3 f c' Acp f y
−
p h
AT S
At section of maximum torsion is located at θ=59.43 T=9.97 t.m Vu =
wR
2 M = 0.0
(π − 2 ) −wR θ = 13.65 − 5.98(4)(
59.43 360
2π ) = − 11.6ton
φ V c = 0.53( 0.85) 350(40)(63.95) = 21.56ton >V u 0.265φ f c' Acp2 pcp
=
0.265 ( 0.85) f c' ( 40 ⋅ 70) 2 ( 40 + 70 ) 105
i.e torsion must be considered
reinf.required ) ( Noshear
2 =
1.50t .m
<
T u
Design of Circular Beams
Dr. Mohammed Arafa
Ductility Check
The following equation has to be satisfied 2
V u Tu p h ≤ 0.263φ f c' + 2 bw d 1.7Aoh x 0 = 40 − 4 − 4 − 0.8 = 31.2 y 0 = 70 − 4 − 4 − 0.8 = 61.2 p h = 2 ( x 0 + y 0 ) = 2 ( 31.2 + 61.2) = 184.8cm Aoh = 31.2 ⋅ 61.2 = 1909.44 cm
2
2 2 5 V u Tu p h 11.16 ⋅103 ( 9.97 ⋅10 ) (184.8) = 30.04 k g / cm 2 = + + 2 2 40 ⋅ 63.95 1.7 (1909.44 ) bw d 1.7Aoh
0.263φ f c' = 0.263( 0.85) 350 = 41.82 kg /cm 2 > 30.04 kg /cm 2 i.e section dimension are adequate for preventin g brittle failure due to combined shear stresses. AT S
=
T u
2φ f ys A 0
=
9.97 ⋅105 2 ( 0.85) 4200 ( 0.85⋅ 1909.44)
= 0.086 cm 2 / cm
3.5( 40 ) 3.5bw AT = = = 0.0167 cm 2 / cm S 2 ( 4200) min 2f ys
AV AT 2 S + S = 0 + 0.086 = 0.086cm / cm Use φ 12mm @12.5cm (closed stirrups) with
AT S
A l =
A l ,min = A l =
A S
=
1.13 12.5
2 p h = 0.086 (184.8) = 15.98cm
1.3 f c' Acp
15.98
f y
A − T S
1.3 350 ( 40⋅ 70) p = − 15.98 = 0.234cm 2 h 4200
= 5.33cm 2
3 Use 2φ 20 top and 4φ 14skin reinforcement
( A ( A
s , +ve s , −ve
) )
total
total
= 0.0904cm 2 / cm
= 5.33 + 8.44 = 13.77cm 2
use 6φ 18mm
= 17.83 + 8.44 = 23.16cm 2 use8φ 20mm
Design of Circular Beams
Dr. Mohammed Arafa
2φ20mm Stirrups φ
[email protected] 2φ14mm
70 cm
2φ14mm 6φ18mm 40 cm
2φ20mm
6φ20mm
6φ18mm
4φ12mm
