Pedestals supporting bridge bearings are subjected to large concentrated loads. When the projection of the pedestal is relatively small compared to its effective depth, the pedestal is designed as a reinforced concrete corbel rather than a cantilever beam.

This article presents a complete numerical example for the design of an RCC pedestal as a corbel.

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Problem Statement

Design an RCC pedestal supporting a bridge bearing using the following data.

Pedestal Dimensions

Parameter	Value
Length (L)	1000 mm
Width (W)	1000 mm
Height	470 mm

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Design Data

Geometry

Depth of corbel,

h = 1.00 m

Cover,

Cover = 50 mm

Effective depth,$$d=h-cover$$$$=1000-50$$$$=950\;mm$$

Therefore,

d = 950 mm

Width,

b =1000;mm

Shear span,

a =420;mm

Cantilever length,

Lc =470;mm

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Material Properties

Concrete$$f’_c=400\;kg/cm^2$$

Steel$$f_y=5000\;kg/cm^2$$

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Design Loads

Ultimate shear force,$$V_u=55\;ton$$

Horizontal force,$$N_u=11\;ton$$

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Step 1 – Check Whether Member is a Corbel

Code requirement$$a<0.95d$$

Calculation$$0.95d = 0.95\times0.95 = 0.9025\;m$$

Since$$a=0.42m$$

Therefore,$$0.42<0.9025$$

Hence,

The pedestal qualifies to be designed as a corbel.

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Step 2 – Horizontal Load Requirement

According to corbel design provisions,$$N_u\ge0.2V_u$$

Calculation$$0.2\times55 = 11\;ton$$

Given,$$N_u=11\;ton$$

Hence,$$11=11$$

Requirement is satisfied.

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Step 3 – Effective Ledge Section

Effective area$$A_c = b\times d$$$$=100\times95$$$$=9500cm^2$$

or$$A_c=0.95m^2$$

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Nominal Shear Capacity

From the design sheet,$$V_n = 535\;ton$$

Design shear strength$$\phi V_n = 0.75\times535 = 401\;ton$$

Applied shear$$V_u = 55\;ton$$

Check$$401>55$$

Hence,

Safe in shear transfer.

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Step 4 – Shear Friction Reinforcement

For monolithic concrete,

Coefficient of friction,$$\mu=1.4$$

Concrete factor,$$\lambda=1.0$$

Required reinforcement,$$A_{vf} = \frac{V_u}{\mu f_y}$$

From calculation,$$A_{vf} = 10.5cm^2$$

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Step 5 – Direct Tension Reinforcement

Direct tension steel,$$A_n = \frac{N_u}{\phi f_y}$$

Substituting,$$A_n = 2.9cm^2$$

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Step 6 – Flexural Reinforcement

Ultimate bending moment,$$M_u = V_ua+N_u(h-d)$$

Substituting,$$= 55(0.42)+11(1.0-0.95)$$$$= 23.1+0.55$$$$= 23.65 \approx24\;ton\cdot m$$

Required flexural steel,$$A_f = 5.9cm^2$$

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Step 7 – Primary Tension Reinforcement

Option 1

$$A_f+A_n = 5.9+2.9 = 8.8cm^2$$

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Option 2

$$\frac{2A_{vf}}3+A_n$$$$= \frac{2\times10.5}{3}+2.9$$$$= 7.0+2.9 = 9.9cm^2$$

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Minimum Reinforcement

Minimum steel,$$A_{s(min)} = 0.04 \left( \frac{f’_c}{f_y} \right) bd$$

From calculation,$$A_{s(min)} = 30.4cm^2$$

Therefore,$$A_s = \max(9.9,\;30.4) = 30.4cm^2$$

Hence,

Required main steel$$A_s=30.4cm^2$$

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Step 8 – Shear Reinforcement

Required shear reinforcement,$$A_h = 0.5(A_s-A_n)$$$$= 0.5(30.4-2.9)$$$$= 13.75 \approx13.7cm^2$$

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Step 9 – Main Reinforcement Provided

Provide

8 Nos. 25 mm diameter bars

Area of one 25 mm bar,$$A = \frac{\pi}{4}(25)^2 = 490.9mm^2 = 4.91cm^2$$

Total steel,$$8\times4.91 = 39.28cm^2$$

Provided$$39.25cm^2$$

Required$$30.4cm^2$$

Hence,$$39.25>30.4$$

Safe

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Step 10 – Shear Reinforcement Provided

Provide

4-legged 12 mm stirrups

Area of one 12 mm bar,$$= 1.13cm^2$$

Total provided according to spacing,$$18.09cm^2$$

Required$$13.7cm^2$$

Therefore,$$18.09>13.7$$

Safe.

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Final Reinforcement

Item	Provided	Required	Status
Main Reinforcement	8–25 mm bars	30.4 cm²	✔ OK
Steel Area	39.25 cm²	30.4 cm²	✔ OK
Shear Reinforcement	4-leg 12 mm stirrups	13.7 cm²	✔ OK
Shear Steel Area	18.09 cm²	13.7 cm²	✔ OK

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Final Conclusion

The pedestal satisfies all the design requirements for a reinforced concrete corbel:

• Corbel geometry check: ✔ $a < 0.95d$a<0.95d
• Horizontal force check: ✔ $N_u \ge 0.2V_u$Nu​≥0.2Vu​
• Shear transfer capacity: ✔ Safe
• Flexural reinforcement: ✔ Adequate
• Primary tension reinforcement: ✔ Adequate
• Main reinforcement provided: ✔ 39.25 cm² > 30.4 cm²
• Shear reinforcement provided: ✔ 18.09 cm² > 13.7 cm²

Hence, the RCC pedestal is safe and satisfactory for the applied loading conditions.