Design of Corbel for Precast Girder, G1 ENGR. THOMAS JOHN D. AGRABIO

Design Specifications (fc’ = 28 MPa, fy = 276 MPa)

Design a bracket that projects from a 500 mm square column to support dead and live loads of

60.08 kN and 33.75 KN, respectively. Assuming a horizontal force of 0.3V u due to creep, shrinkage and

temperature effects. Note: φ shall be taken equal to 0.75 in all design calculations according to NSCP

411.10.3.1

1. Factored Loads

Factored shear, V u= 1.2(60.08) + 1.6(33.75) = 126.10 kN

Horizontal force, Nuc=0.3(126.10) = 37.83 kN > (0.2Vu = 23.79 kN) (NSCP 411.10.3.4)

2. Preliminary bracket size. The shear span a is dependent on the bearing length required to support the

reaction on the concrete.

Bearing strength, V u=φ (0 .85 f c' ) A plate=φ (0. 85 f c

' )bw L

Length of bearing plate, L=

V u

φ 0 . 85 f c' bw

=1261000 .65(0 .85)(28 )(500)

=16 . 30mm

Use 75 mm plate length as the practical minimum (RCD, Wang & Salmon)

Shear span, av=25+ 2

3(75 )=75

mm

25 mm maximum clearance at the beam end. Beam reaction is assumed at third point of bearing

plate to simulate rotation of supported girder and triangular distribution of stress under bearing plate

(Simplified RCD, Gillesania)

3. Depth of bracket for shear

V n=V u

φ=126 . 10

0 .75=168 . 13

kN

For normal weight concrete, V n shall not exceed the smallest of: (NSCP 411.10.3.1)

V n=0 . 2 fc ' bw d=0 . 2(28)bw d=5 .6 bw dV n=(3 . 3+0 .08 fc ' )bw d=[3 .3+0. 08(28 )] bw d=5 .54 bw dV n=11bw d Governs!

V n=5 .54 bw d 168130=5 .54(500 )d

d= 60.7 mm minimum d based on shear requirements

4. Depth of bracket for flexure

M u=V u av+Nuc (h−d )=126 . 10(0 .075)+37 .83(0 .050)=11.35 kN-m (NSCP Section 411.10.3)

Estimating (h−d )=50 mm

Using Minimum reinforcement ratio, (NSCP Section 411.10.5)

min ρ=0 .04( f c'

f y)=0 . 04 (28

276)=0 .00406

Strength Ratio, (RCD, Wang & Salmon, 3.8.4a)

m= fy0 . 85 fc '

=2760 . 85(28)

=11. 60

Coefficient of resistance, (RCD, Wang & Salmon, 3.8.4b)

Ru=ρ fy (1− ρm2 )=0 . 00406(276 )(1−0 . 00406(11.60)

2 )=1. 094 MPa

Required effective depth, (RCD, Wang & Salmon, 3.8.4)

dreqd=√ M u

φRu b=√11. 35×106

0 . 75(1. 094 )(500)=166 .33

mm

5. Select bracket depth. Since the provisions of NSCP for brackets and corbel design apply only when a /d

does not exceed 1.0,

(a /d=0 . 45 )<1 .0 OK (NSCP Section 411.10.1)

min d=a=75 mm

Try h=350 mm , d≈350−25−8=317 mm

6. Determine shear-friction reinforcement Avf

Avf=V u

φ fy μ=126100

0. 75(276 )(1. 4 )=435 .13

mm2 (NSCP 411.8.4.1)

Whereμ=1.4 for monolithic concrete μ=1. 4 λ=1 . 4 (1 )=1 . 4

7. Determine the flexure reinforcement A f (ACI-11.9.3.3)

Coefficient of resistance, (RCD, Wang & Salmon, 3.8.4)

Ru=M u

φ bd2 =11.35 x 106

0 .75(500 )(317)2 =0 .301 MPa

Steel ratio,

ρ= 1m [1−√1−

2Ru mfy ]= 1

11.60 [1−√1−2(0.301)(11.60 )276 ]=0 .001098<( ρmin=0 .00406 )

43

ρ= 43

(0 .001098 )=0 . 001464

A f=ρ bd=0 .001464(500 )(317)=232.04 mm2

8. Determine additional reinforcement for axial tension An

An=Nuc

φ fy=37830

0. 75(276 )=182. 75

mm2 (NSCP 411.8.4.1)

9. Total main tension reinforcement A sc (NSCP 411.10.3)

Area of primary reinforcement tension reinforcement A sc shall not be less than the

larger of:

Asc=A f +An=232 .04+182.7=414 .74 mm2

A sc=23

Avf + An=23( 435. 13)+182 .75=472. 84

mm2 Governs!

Using 16 mm bars: Ab=

π4

(16 )2=201

n=472 . 84201

=2. 35, say 3-16 mm bars

10. Determine closed stirrup requirements (NSCP 411.10.4)

min Ah=0 .5( A sc−An )=0 .5(472 .84−182 .75 )=145 . 05

av =75mm

200

250

d=317

10 mm stirrups @ 100 mm o.c.

3-16 mm 75mm

h=350 mm

16 mm crossbar welded

25 mm plate

126.10 kN

37.83 kN

16 mm bars welded to underside of Steel plate

Welded

Using 10 mm ties: Ab=

π4

(10 )2(2 )=157 mm2

n=145. 05157

=0 .92, say 2-10 mm closed hoops

Spacing: s=2

3(317 )/2=106

mm, use 100 mm spacing

11. Overall bracket dimension. Assuming that a 25 mm thick bearing plate is to be welded to the main tension

reinforcement, the overall depth is

h = bearing plate thickness + bar radius + effective depth, d = 25 + 8 + 317 = 350 mm

Length of bracket projection = 50mm + 12 bearing plate + shear span,av = 162.5 mm, say 175 mm

= 50 + ½(75) +75 = 162.5 mm, say 200 mm

Minimum Depth of outer face of bracket =

12

h=12(350)=175

mm

Elastomeric bearing pads with minimum thickness 10 mm plate is used in the design (PCI 6.5.8.1).

Bearing pads are used to distribute concentrated loads and reactions over the bearing area

and to allow limited horizontal and rotational movements to provide stress relief.

12. Details