Structural Steel Design (6th Edition)
Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 2, Problem 2.11PFS
To determine

Theuniform load per foot using ASD expressions supported by each beam.

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A gravity retaining wall is shown in the figure below. Calculate the factor of safety with respect to overturning and sliding, given the following data: Wall dimensions: H = 6 m, x₁ = 0.6 m, x2 = 2 m, x3 = 2m, x4 0.5 m, x5 = 0.75 m, x6 = 0.8 m, D= 1.5 m Soil properties: 71 = 14 kN/m³, ₁ = 32°, 72 = 18 kN/m³, 2=22°, c₂ = 40 kN/m² Y₁ c₁ = 0 H Φί x5 x6 Use the Rankine active earth pressure in your calculation. Use Yconcrete = 23.08 kN/m³. Also, use k₁ = k₂ = 2/3 and Pp = 0 in the equation FS (sliding) (ΣV) tan(k102) + Bk2c2 + Pp Pa cos a (Enter your answers to three significant figures.) FS (overturning) FS (sliding) =
For the cantilever retaining wall shown in the figure below, let the following data be given: Wall dimensions: H = 8 m, x1 = 0.4 m, x2 = 0.6 m, x3 = 1.5 m, x4 3.5 m, x5 = 0.96 m, D= 1.75 m, a = 10° Soil properties: 71 = 14.8 kN/m³, ₁ = 32°, Y₂ = 1 2 = 28°, c = 30 kN/m² 17.6 kN/m³, The value of Ka is 0.3210. For 2 = 28°: N = 25.80; N₁ = 14.72; N₁ = 16.72. c=0 H Χς Calculate the factor of safety with respect to overturning, sliding, and bearing capacity. Use Yconcrete = 21.58 kN/m³. Also, use k₁ = k₂ = 2/3 and P = 0 in the equation FS (sliding) (ΣV) tan(k₁₂) + Bk2C + Pp Pa cosa (Enter your answers to three significant figures.) FS (overturning) FS (sliding) FS (bearing) =
Question 2 The following strains are obtained by a 0-60-120 strain rosette: ε0 = 300 x 10-6, 60 = 200 x 10-6 and 120= 150 x 10-6. i. Determine strains Ex, Ey and Yxy ii. Determine the strains for 0 = 40° iii. Calculate principal strains, maximum shear strain and the orientation of principal strains iv. Determine normal stresses (σx, σy) and shear stress (Txy), if E = 200kPa and v = 0.25. (Hint: You may use stress-strain relationship for plane strain, summarised in matric format as follows: E σχ бу 1-v v 0 Ex = v 1-v 0 Ey txy. (1+v)(1 − 2v) 0 0 0.5 v
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