HW2 Solutions.pdf - Homework 2 Solutions CEE 452 Problem 1...

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Unformatted text preview: Homework 2 Solutions CEE 452 Problem 1 The cantilever retaining wall shown below is in an active state of failure (Figure 1). Plane A-C is a slip plane. a) Sketch the vertical effective stress profile along plane AB, provide numerical values. Sketch the lateral earth pressure distribution along plane AB (provide numerical values). b) Compute the magnitude of the lateral earth pressure force on plane AB. c) Show the point of application and the orientation of the lateral earth pressure force on the plane AB. (1) What method of analysis did you use in computing lateral earth pressures? What are the assumptions made for this method? Discuss how well these assumptions match the problem solved, and if there are deviations, specify exactly what the deviations are and discuss the significance with regard to your lateral earth pressure computations. Clean Sand 25’ Yd I 1.10 pcf ¢'=4{}° Sand beneath the wall has. the same properties as the backfill. Grmmd Mite-r table is very Inn A\ b) d) Homework 2 Solutions CEE 452 A 25 ft Ha 8.33 ft B lateral earth pressure 1 — sin ¢' 1 — sin 40° Ka - = 0.217 = 1+sin¢' _ 1+sin40° Active Pressure = Pa = Kaydh = (0.217)(110 pcf)(25 ft) = 596. 75 psf 1 1 Active Force = Ha = —Pah = —(596.75 ps )(25) = 74-59. 38 lb/ t 2 2 point of application of the lateral earth pressure 1 1 H = 3h = g (25 ft) = 8. 33 ft from the bottom of the wall *see part a) The simplest approach to this problem is to find the active pressure and force using the Rankine Method. Rankine assumes that there is no adhesion or friction between the wall and the soil (the wall is assumed to be smooth). Wall failure is assumed to occur in the form of a sliding wedge along an assumed failure plane as a function of the soil's effective angle of internal friction. Because 45+(¢/2)=65°, plane AB is principal plane and Rankine’s assumptions apply. Homework 2 Solutions CEE 452 Problem 2 Text Book (Liu and Evett) Problem 11.8 [11*8) Eq. (11-14): It. . (1- sin 30")!“ + sin 30°) I 0.333 Eq. (11-10}: P. I [mHHSHZSJainBI-Ifl - 11.990 lb per foot of will Polntofapplicatlonforfia HIB-ZS/J-Sfifiahonbmofmll Ell. Eli-20): P' I {snouzsnuaasa} I um I: per foot 01' wall Pohtoflppllutlonfan’I l-lfl-ZSIZ- IZSOROMbIs-Oofmll {I} Total active ell'th pressure unamuno I 16,150lbperfootofmll (bl Latthe point ofapplhflonofdnetoulacfln earth pressure he 'h'ftabovc but ofwall. Take moments of forces at has: of wall. 16.15111 - (nastiness) + (4170}[1250] h I 9.41 ft Thus totalactweearth mnactatSAlftabowbaseofwalL Problem 3 Text Book (Liu and Evett) Problem 11.1 (next page) Homework 2 Solutions CEE 452 {11-1} Equl—E]: Ku=1~5in3flu:fl.5 in} hunt-1 um nun:- i I I . 4' {LEHHJIIHI} : 1:an mm? =- I J— “.El 1- {H.EI {15.5” -' 3.31111} .- 13.35 “In: [his lawn: nu: m J- .L {3.311121} - !.I1 kiln: {almntmlml I ...
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