Chapter 13 Shear Strength of Soil Solutions for Principles of Geotechnical Engineering

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Chapter 13 Shear Strength of Soil Solutions for Principles of Geotechnical Engineering c. Referring to the diagram in Part a, 1 1 147.8 kN/m = aP − .3( )(91 )82 = ( .1)(23 )09 2 2 1 d. 160.3 kN/m = aP = .3( )(91 )82 2 .3 91 Location of the resultant from the bottom of the wall: m 1.3= 3 ′ 28 φ ⎞ 13.21 2 tan 45 2 tan 45 .0 ;361 .6.0 ⎛ − ⎛ ⎞ = 2 2 =⎟ ⎠ =⎟ ⎠ = − K ⎜ ⎝ ⎜ ⎝ K Eq. (13.44): a a 1 ′ 2 2 c = 2 − ′ + 2 γ P HK HcK a a a 2 γ 1 2 )(2( )750 3502.18 lb/ft = .0( 2 − 361 )( 122 )( )33 )(6.0)(2( + 750 )( )33 = 2 122 13.22 Use Eqs. (13.53) and (13.54). α = 0; θ = 12º; φ′ = 34º; γ = 119 lb/ft 3 ; H = 32 ft Part δ′ Ka 2 1 HK P = γ a a 2 (deg) [Eq. (13.54)] [Eq. (13.53)] 1 14 0.3511 21,392 lb/ft 2 21 0.3509 21,381 lb/ft Pa is located at a vertical distance of 32/3 = 10.67 ft above the bottom of the wall and is inclined at an angle δ′ to the normal drawn to the back face of the wall. 13.23 a. φ′ = 38º; = −− = ° ′−− = 90 90 5 20 65 ψ θ δ 1 Weight of wedge ABC = ()1.7)(6.11( )128 5271 lb/ft 5.271 kip/ft = = 2 γ The above chart shows that for any backfill inclination α and any wall height H, the active force Pa per unit length of the wall decreases as the soil friction angle (or the compacted unit weight) increases. For a desired level of Pa (at a given α), a compaction unit weight could be estimated from the chart for field specification.

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Uploaded

20 February 2026
Updated

20 February 2026
Category

Engineering
Item Type

questions & answers
Tags

geotechnical engineering

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