Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
Question
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Chapter 5, Problem 5.14.2P
To determine

(a)

The design of bearing plate by LRFD.

To determine

(b)

The design of bearing plate by ASD.

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For the design of a shallow foundation, given the following: Soil: ' = 20° c' = 52 kN/m² Unit weight, y = 15 kN/m³ Modulus of elasticity, E, = 1400 kN/m² Poisson's ratio, μs = 0.35 Foundation: L=2m B=1m Df = 1 m Calculate the ultimate bearing capacity. Use the equation: 1 - qu = c' NcFcs Fcd Fcc +qNqFqsFqdFqc + ½√BN√Fãs F√dƑxc 2 For '=20°, Nc = 14.83, N₁ = 6.4, and N₁ = 5.39. (Enter your answer to three significant figures.) qu = kN/m²
A 2.0 m wide strip foundation carries a wall load of 350 kN/m in a clayey soil where y = 15 kN/m³, c' = 5.0 kN/m² and ' = 23°. The foundation depth is 1.5 m. For ' = 23°: Nc = 18.05; N₁ = 8.66; Ny = = = 8.20. Determine the factor of safety using the equation below. qu= c' NcFcs FcdFci+qNqFqsFq 1 F + gd. 'qi 2 ·BN√· FF γί Ysyd F (Enter your answer to three significant figures.) FS =
2P -1.8 m- -1.8 m- -B Wo P -1.8 m- C

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Steel Design (Activate Learning with these NEW titles from Engineering!)

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