A 395 mm thick slab supports a 363 mm thick wall carrying uniform service dead loadof 211.5 kN/m and service live load of 157.4 kN/m. The base of the wall footing slabis 1.4 m from the ground surface. Design parameters are as follows: YsoilkN/m, Ycond = 24 kN/m³, q, = 207.92 kPa, f. 27 MPa and f, = 414 MPa.16.5Calculate the required center to center spacing of 16 mm bars for flexure.Note: Present the width (B) in multiples of 100 mm and present the exact requiredspacing in 2 decimal places (do not round off).

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A 395 mm thick slab supports a 363 mm thick wall carrying uniform service dead load of 211.5 kN/m and service live load of 157.4 kN/m. The base of the wall footing slab is 1.4 m from the ground surface. Design parameters are as follows: Ysoil = 16.5 kN/m, Yconcl = 24 kN/m°, qa = 207.92 kPa, f'.- 27 MPa and f, = 414 MPa. Calculate the required center to center spacing of 16 mm bars for flexure. Note: Present the width (B) in multiples of 100 mm and present the exact required spacing in 2 decimal places (do not round off).

A 395 mm thick slab supports a 363 mm thick wall carrying uniform service dead load of 211.5 kN/m and service live load of 157.4 kN/m. The base of the wall footing slab is 1.4 m from the ground surface. Design parameters are as follows: Ysoil = 16.5 kN/m, Yconcl = 24 kN/m°, qa = 207.92 kPa, f'.- 27 MPa and f, = 414 MPa. Calculate the required center to center spacing of 16 mm bars for flexure. Note: Present the width (B) in multiples of 100 mm and present the exact required spacing in 2 decimal places (do not round off).

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter17: Retaining Walls

Section: Chapter Questions

Problem 17.2P

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