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 diameter 8m. The beam is support a uniformly distributed factored load of 5.0 t/m, in addition to its own weight. Use f c' 350 = = kg / cm 2 and f y 4200 kg / cm 2 B
Solution
= L r= (π 2 ) 6.28 m = hmin L= /18.5 628= /18.5 33.94cm
c.g
P
use Beam 40×70 cm 2R/
o. w. of the beam =0.4(0.7)(2.5)(1.4)=0.98 t/m’ total load= 5.14+0.84=5.98 t/m’
A
= M max( −ve ) 0.429 = wR 0.429(5.98)(4) = 41.05 t .m 2
2
θ O
2 = M max( +ve ) 0.1514 = wR 2 0.1514(5.98)(4) = 14.48 t .m 2 T max 0.1042 wR 2 0.1042(5.98)(4) = = = 9.97 t .m
Reactions wR 5.98(4) R= 2) 2 ) 13.65 ton (π − = (π − = A 2 2 R B 2= wR 2(5.98)(4) = = 47.84 ton Shear at point P:
V = θ
wR (π − 2 ) −wR θ 2 23.92
13.65
2.28 13.65 23.92 Shear Force Diagram
C
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 ρ= = −ve 4200 40 ⋅ 63.95 ⋅ 350 0.00697 )( 40 )( 63.95 ) 17.85 cm 2 = A s ( −ve ) (= = ρ −ve A s ( +ve )
2.61 ⋅105 (14.48 ) 0.85 ⋅ 350 1 − 1 − 0.00238 < ρ min = 4200 40 ⋅ 63.95 ⋅ 350 2 = ( 0.0033)( 40 )( 63.95 ) = 8.44cm
Design for Shear
(T 0.0 at middle support) V max 23.92 ton = =
φV c
0.53 = ( 0.85) 350(40)(63.95) 21.56 ton
23.92 − 21.56 = 2.77 ton 0.85 AV 2.77 ⋅1000 = 4200(63.95) S AV = 0.0103 cm 2 / cm S
VS =
Design of Circular Beams
Dr. Mohammed Arafa
Design for Torsion Tu ≤
0.265φ f c' Acp2 pcp
⇒
neglect torsion
2 (b + h ) , A 0 h = 2 ( x 0 + y 0 ) and A 0 = 0.85x 0 y 0 Acp = b h, pcp = x 0 y 0 , ph = The following equation has to be satisfied to check for ductlity 2
V u Tu p h ≤ 0.263φ f c' + 2 bw d 1.7Aoh Reinforcement Tu AT = 2φ f ys A 0 S 1.3 f c' Acp AT AT Al A l ,min = = − p h and ph fy S S At section of maximum torsion is located at θ=59.43 T=9.97 t.m 59.43 wR 13.65 − 5.98(4)( 2π ) = V u = (π − 2 ) −wR θ = −11.6 ton 2 360 M = 0.0
φV c 0.53 ( 0.85 ) 350(40)(63.95) = = 21.56 ton >V u
( No shear reinf.required )
0.265φ f c' Acp2 0.265 ( 0.85 ) f c' ( 40 ⋅ 70 ) = = 1.50 t .m < T u pcp 2 ( 40 + 70 )105 2
i.e torsion must be considered
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 ) = 184.8 cm
Aoh = 31.2 ⋅ 61.2 = 1909.44 cm 2 2
V u Tu p h = + 2 bw d 1.7 Aoh
2 5 11.16 ⋅103 ( 9.97 ⋅10 ) (184.8 ) = 30.04 kg / cm 2 + 2 40 63.95 ⋅ 1.7 (1909.44 )
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 preventing brittle failure due to combined shear stresses. Tu AT 9.97 ⋅105 = = = 0.086 cm 2 / cm 2φ f ys A 0 2 ( 0.85 ) 4200 ( 0.85 ⋅1909.44 ) S 3.5 ( 40 ) 3.5bw AT = = 0.0167 cm 2 / cm = 2 ( 4200 ) S min 2f ys AV AT 0 + 0.086 = 0.086cm 2 / cm = + S S A 1.13 Use φ12mm @12.5cm (closed stirrups) with= = 0.0904cm 2 / cm S 12.5 AT A l = = (184.8) 15.98cm 2 p h 0.086= S 1.3 f c' Acp
A l= ,min
fy
A − T S
1.3 350 ( 40 ⋅ 70 ) − 15.98 = 0.234 cm 2 ph = 4200
15.98 = 5.33cm 2 3 Use 2φ 20 top and 4φ14skin reinforcement
= Al
(A (A
) )
s , +ve total s , −ve total
= 5.33 + 8.44 = 13.77cm 2 use 6φ18mm = 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
