Design of Curved Beam (Plan View Curve) Example.pdf

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