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Engineering Civil Engineering Principles of Foundation Engineering (MindTap Course List)

The factor of safety with respect to overturning, sliding and bearing capacity. 3.41, 1.51 and 5.39 Given data: γ c o n c r e t e = 23.58 k N / m 3 k 1 = k 2 = 2 3 P p = 0 Wall dimensions: H = 8 m x 1 = 0.4 m x 2 = 0.6 m x 3 = 1.5 m x 4 = 3.5 m x 5 = 0.96 m D = 1.75 m α = 10 o Soil properties: γ 1 = 16.8 k N / m 3 ϕ 1 ' = 32 o γ 2 = 17.6 k N / m 3 ϕ 2 ' = 28 o c 2 ' = 30 k N / m 2 Explanation: The given figure is shown below: The cantilever-wall is shown below: The active earth pressurecoefficient is K a = cos α cos α − cos 2 α − cos 2 ϕ ' cos α + cos 2 α − cos 2 ϕ ' K a = cos 10 o cos 10 o − cos 2 10 o − cos 2 32 o cos 10 o + cos 2 10 o − cos 2 32 o K a = 0.3216 The additional height because of the sloping part is x ' = x 4 tan α w h e r e , x ' i s e x t r a h e i g h t x 4 i s p r o t r u s i o n o f t h e b a s e int o t h e a c t i v e s i d e Substituting the values x ' = x 4 tan α x ' = 3.4 × tan 10 o x ' = 0.617 m Consider the sec(1) The section area is A 1 = 0.4 × 8 A 1 = 3.2 m 2 The weight per unit span of the concrete section 1 stand-in vertically downside is W 1 = A 1 × γ c W 1 = 1.8 × 23.58 W 1 = 42.44 k N / m The moment arm of section (1) about toe point is Moment arm = x 3 + x 2 − x 1 2 Moment arm = 1.5 + 0.6 − 0.4 2 Moment arm = 1.9 m The moments and forces for stabilizing are presented below: Portion Area m 2 γ ( k N m 3 ) , V w = ( A × γ ) ( k N m ) X t o e ( m ) M t o e ( K N − m m ) 1 = 0.4 × 8 = 3.2 23.58 75.46 = 1.5 + 0.6 − 0.4 2 = 1.9 143.37 2 = 1 2 × ( 0.6 − 0.4 ) × 8 = 0.8 23.58 18.86 = 1.5 + [ 2 3 × ( 0.6 − 0.4 ) ] = 1.633 30.8 3 = ( 1.5 + 0.6 + 3.5 ) × 0.96 = 5.376 23.58 126.77 = 1.5 + 0.6 + 3.5 2 = 2.8 354.96 4 = ( 8 + 8.617 2 ) × 3.5 = 29.08 16.8 488.54 = 1.5 + 0.6 + 3.5 2 = 3.85 1880.88 Total ∑ V w = 709.63 ∑ M t o e = 2410 The active force in total behind the wall is P a = 1 2 γ ( H + x 5 ) 2 K a where, P a is total active force behind the wall H is height of stem x 5 is thickness of base slab K a is coefficient of active earth pressure Substituting the values P a = 1 2 γ ( H + x 5 ) 2 K a P a = 1 2 × 16.8 × ( 8 + 0.96 + 0.617 ) 2 × 0.3216 P a = 247.8 k N / m The destabilizing force’s horizontal constituent is P a h = P a × cos α P a h = 247.8 × cos 10 o P a h = 244 k N / m The destabilizing force’s vertical constituent is P a v = P a × sin α P a v = 247.8 × sin 10 o P a v = 43 k N / m The entire stabilizing moment about toe point is M R = ∑ M t o e + P a v × x a v M R = 2410 + 43 × 5.6 M R = 2650.8 k N m / m The overturning is M o = P a h × ( H + x 5 3 ) M o = 244 × ( 8 + 0.617 + 0.96 3 ) M o = 778.9 k N m / m The entireforce in the vertical direction on the cantilever wall is ∑ V = ∑ V w + P a v ∑ V = 709.63 + 43 ∑ V = 752.6 k N / m The FOS(factor of safety) with respect to overturning is F . S O = M R M O F . S O = 2650.8 778.9 F . S O = 3.41 The FOS(factor of safety) with respect to sliding is F . S S l i d i n g = ( ∑ V ) tan ( k 1 ϕ 2 ' ) + B k 2 c 2 ' + P p P a cos α where, F . S S l i d i n g is factor of safety with respect to sliding k 1 and k 2 are constants ∑ V is total vertical force ϕ 2 ' is friction angle of soil in front of toe and under base slab B is width of base slab of retaining wall P p is passive earth pressure c 2 ' is cohesion of soil in front of toe and under base slab Substituting the values F . S S l i d i n g = ( ∑ V ) tan ( k 1 ϕ 2 ' ) + B k 2 c 2 ' + P p P a cos α F . S S l i d i n g = 752.6 × tan ( 2 3 × 28 o ) + 5.6 × 2 3 × 30 + 0 247.8 × cos 10 o F . S S l i d i n g = 1.51 The eccentricity of the moment caused due to the resultant force is e = B 2 − x ¯ e = B 2 − ∑ M R − ∑ M O ∑ V e = 5.6 2 − 2650.8 − 778.9 752.6 e = 0.313 m The maximum pressure at the base of the foundation is q max = ∑ V B ( 1 + 6 e B ) q max = 752.6 5.6 × ( 1 + 6 × 0.313 5.6 ) q max = 179.4 k N / m 2 The bearing capacity factors are N q = tan 2 ( 45 o + ϕ 2 ' 2 ) e π tan ϕ 2 ' N q = tan 2 ( 45 o + 28 o 2 ) e π tan 28 o N q = 14.72 N c = ( N q − 1 ) cot ϕ 2 ' N c = ( 14.72 − 1 ) cot 28 o N c = 25.8 N γ = 2 ( N q + 1 ) × tan ϕ 2 ' N γ = 2 ( 14.72 + 1 ) × tan 28 o N γ = 16.72 The shape factors are F q d = 1 + [ 2 { tan ( ϕ 2 ' ) } { 1 − sin ( ϕ 2 ' ) } 2 D B ' ] F q d = 1 + [ 2 { tan ( ϕ 2 ' ) } { 1 − sin ( ϕ 2 ' ) } 2 D ( B − 2 e ) ] F q d = 1 + [ 2 × { tan ( 28 o ) } × { 1 − sin ( 28 o ) } 2 × 1.75 ( 5.6 − 2 × 0.313 ) ] F q d = 1.105 F c d = F q d − ( 1 − F q d N c tan ϕ 2 ' ) F c d = 1.105 − ( 1 − 1.105 25.8 × tan 28 o ) F c d = 1.113 F γ d = 1.0 F q i = F c i F q i = ( 1 − β 90 o ) 2 β = tan − 1 ( P a h ∑ V ) β = tan − 1 ( 244 752.6 ) β = 17.96 o F q i = ( 1 − β 90 o ) 2 F q i = ( 1 − 17.96 o 90 o ) 2 F q i = 0.64 F γ i = ( 1 − β ϕ 2 ' ) 2 F γ i = ( 1 − 17.96 o 28 o ) 2 F γ i = 0.13 The ultimate bearing capacity is q u = c 2 ' N c F c d F c i + q N q F q d F q i + 1 2 γ 2 B ' N γ F γ d F γ i q u = c 2 ' N c F c d F c i + ( γ 2 D ) N q F q d F q i + 1 2 γ 2 ( B − 2 e ) N γ F γ d F γ i q u = ( 30 × 25.8 × 1.113 × 0.64 ) + ( 17.6 × 1.75 ) × 14.72 × 1.105 × 0.64 + 1 2 × 4.974 × 17.6 × 16.72 × 1 × 0.13 q u = 967.1 k N / m 2 The factor of safety with respect to bearing capacity is F . S B C = q u q max F . S B C = 967.1 179.4 F . S B C = 5.39 Conclusion: The FOS(factor of safety) with respect to overturning, sliding, and bearing capacity is 3.41, 1.51, and 5.39 respectively.

The factor of safety with respect to overturning, sliding and bearing capacity. 3.41, 1.51 and 5.39 Given data: γ c o n c r e t e = 23.58 k N / m 3 k 1 = k 2 = 2 3 P p = 0 Wall dimensions: H = 8 m x 1 = 0.4 m x 2 = 0.6 m x 3 = 1.5 m x 4 = 3.5 m x 5 = 0.96 m D = 1.75 m α = 10 o Soil properties: γ 1 = 16.8 k N / m 3 ϕ 1 ' = 32 o γ 2 = 17.6 k N / m 3 ϕ 2 ' = 28 o c 2 ' = 30 k N / m 2 Explanation: The given figure is shown below: The cantilever-wall is shown below: The active earth pressurecoefficient is K a = cos α cos α − cos 2 α − cos 2 ϕ ' cos α + cos 2 α − cos 2 ϕ ' K a = cos 10 o cos 10 o − cos 2 10 o − cos 2 32 o cos 10 o + cos 2 10 o − cos 2 32 o K a = 0.3216 The additional height because of the sloping part is x ' = x 4 tan α w h e r e , x ' i s e x t r a h e i g h t x 4 i s p r o t r u s i o n o f t h e b a s e int o t h e a c t i v e s i d e Substituting the values x ' = x 4 tan α x ' = 3.4 × tan 10 o x ' = 0.617 m Consider the sec(1) The section area is A 1 = 0.4 × 8 A 1 = 3.2 m 2 The weight per unit span of the concrete section 1 stand-in vertically downside is W 1 = A 1 × γ c W 1 = 1.8 × 23.58 W 1 = 42.44 k N / m The moment arm of section (1) about toe point is Moment arm = x 3 + x 2 − x 1 2 Moment arm = 1.5 + 0.6 − 0.4 2 Moment arm = 1.9 m The moments and forces for stabilizing are presented below: Portion Area m 2 γ ( k N m 3 ) , V w = ( A × γ ) ( k N m ) X t o e ( m ) M t o e ( K N − m m ) 1 = 0.4 × 8 = 3.2 23.58 75.46 = 1.5 + 0.6 − 0.4 2 = 1.9 143.37 2 = 1 2 × ( 0.6 − 0.4 ) × 8 = 0.8 23.58 18.86 = 1.5 + [ 2 3 × ( 0.6 − 0.4 ) ] = 1.633 30.8 3 = ( 1.5 + 0.6 + 3.5 ) × 0.96 = 5.376 23.58 126.77 = 1.5 + 0.6 + 3.5 2 = 2.8 354.96 4 = ( 8 + 8.617 2 ) × 3.5 = 29.08 16.8 488.54 = 1.5 + 0.6 + 3.5 2 = 3.85 1880.88 Total ∑ V w = 709.63 ∑ M t o e = 2410 The active force in total behind the wall is P a = 1 2 γ ( H + x 5 ) 2 K a where, P a is total active force behind the wall H is height of stem x 5 is thickness of base slab K a is coefficient of active earth pressure Substituting the values P a = 1 2 γ ( H + x 5 ) 2 K a P a = 1 2 × 16.8 × ( 8 + 0.96 + 0.617 ) 2 × 0.3216 P a = 247.8 k N / m The destabilizing force’s horizontal constituent is P a h = P a × cos α P a h = 247.8 × cos 10 o P a h = 244 k N / m The destabilizing force’s vertical constituent is P a v = P a × sin α P a v = 247.8 × sin 10 o P a v = 43 k N / m The entire stabilizing moment about toe point is M R = ∑ M t o e + P a v × x a v M R = 2410 + 43 × 5.6 M R = 2650.8 k N m / m The overturning is M o = P a h × ( H + x 5 3 ) M o = 244 × ( 8 + 0.617 + 0.96 3 ) M o = 778.9 k N m / m The entireforce in the vertical direction on the cantilever wall is ∑ V = ∑ V w + P a v ∑ V = 709.63 + 43 ∑ V = 752.6 k N / m The FOS(factor of safety) with respect to overturning is F . S O = M R M O F . S O = 2650.8 778.9 F . S O = 3.41 The FOS(factor of safety) with respect to sliding is F . S S l i d i n g = ( ∑ V ) tan ( k 1 ϕ 2 ' ) + B k 2 c 2 ' + P p P a cos α where, F . S S l i d i n g is factor of safety with respect to sliding k 1 and k 2 are constants ∑ V is total vertical force ϕ 2 ' is friction angle of soil in front of toe and under base slab B is width of base slab of retaining wall P p is passive earth pressure c 2 ' is cohesion of soil in front of toe and under base slab Substituting the values F . S S l i d i n g = ( ∑ V ) tan ( k 1 ϕ 2 ' ) + B k 2 c 2 ' + P p P a cos α F . S S l i d i n g = 752.6 × tan ( 2 3 × 28 o ) + 5.6 × 2 3 × 30 + 0 247.8 × cos 10 o F . S S l i d i n g = 1.51 The eccentricity of the moment caused due to the resultant force is e = B 2 − x ¯ e = B 2 − ∑ M R − ∑ M O ∑ V e = 5.6 2 − 2650.8 − 778.9 752.6 e = 0.313 m The maximum pressure at the base of the foundation is q max = ∑ V B ( 1 + 6 e B ) q max = 752.6 5.6 × ( 1 + 6 × 0.313 5.6 ) q max = 179.4 k N / m 2 The bearing capacity factors are N q = tan 2 ( 45 o + ϕ 2 ' 2 ) e π tan ϕ 2 ' N q = tan 2 ( 45 o + 28 o 2 ) e π tan 28 o N q = 14.72 N c = ( N q − 1 ) cot ϕ 2 ' N c = ( 14.72 − 1 ) cot 28 o N c = 25.8 N γ = 2 ( N q + 1 ) × tan ϕ 2 ' N γ = 2 ( 14.72 + 1 ) × tan 28 o N γ = 16.72 The shape factors are F q d = 1 + [ 2 { tan ( ϕ 2 ' ) } { 1 − sin ( ϕ 2 ' ) } 2 D B ' ] F q d = 1 + [ 2 { tan ( ϕ 2 ' ) } { 1 − sin ( ϕ 2 ' ) } 2 D ( B − 2 e ) ] F q d = 1 + [ 2 × { tan ( 28 o ) } × { 1 − sin ( 28 o ) } 2 × 1.75 ( 5.6 − 2 × 0.313 ) ] F q d = 1.105 F c d = F q d − ( 1 − F q d N c tan ϕ 2 ' ) F c d = 1.105 − ( 1 − 1.105 25.8 × tan 28 o ) F c d = 1.113 F γ d = 1.0 F q i = F c i F q i = ( 1 − β 90 o ) 2 β = tan − 1 ( P a h ∑ V ) β = tan − 1 ( 244 752.6 ) β = 17.96 o F q i = ( 1 − β 90 o ) 2 F q i = ( 1 − 17.96 o 90 o ) 2 F q i = 0.64 F γ i = ( 1 − β ϕ 2 ' ) 2 F γ i = ( 1 − 17.96 o 28 o ) 2 F γ i = 0.13 The ultimate bearing capacity is q u = c 2 ' N c F c d F c i + q N q F q d F q i + 1 2 γ 2 B ' N γ F γ d F γ i q u = c 2 ' N c F c d F c i + ( γ 2 D ) N q F q d F q i + 1 2 γ 2 ( B − 2 e ) N γ F γ d F γ i q u = ( 30 × 25.8 × 1.113 × 0.64 ) + ( 17.6 × 1.75 ) × 14.72 × 1.105 × 0.64 + 1 2 × 4.974 × 17.6 × 16.72 × 1 × 0.13 q u = 967.1 k N / m 2 The factor of safety with respect to bearing capacity is F . S B C = q u q max F . S B C = 967.1 179.4 F . S B C = 5.39 Conclusion: The FOS(factor of safety) with respect to overturning, sliding, and bearing capacity is 3.41, 1.51, and 5.39 respectively.

Solution Summary: The author explains the factor of safety with respect to overturning, sliding and bearing capacity.

Principles of Foundation Engineering (MindTap Course List)

Principles of Foundation Engineering (MindTap Course List)

8th Edition

ISBN: 9781305081550

Author: Braja M. Das

Publisher: Cengage Learning

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Principles of Foundation Engineering (MindTap Course List)

Ch. 13 - Prob. 13.1P Ch. 13 - Prob. 13.2P Ch. 13 - Prob. 13.3P Ch. 13 - Prob. 13.7P Ch. 13 - Prob. 13.8P Ch. 13 - Prob. 13.9P Ch. 13 - Prob. 13.10P

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