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1. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ =k2=0.67. Determine the factor of safety against Sliding and overturning. 15m-D T -18 kN/m dj-M7 5=0 Section Area Weight/unit length Moment arm H,-0.458 m from point C no. (m²) (kN/m) (m) Moment (kN-m/m) 1 6 x 0.5-3 70.74 1.15 81.35 2 (0.2)60.6 14.15 0.833 11.79 3 4 x 0.7 = 2.8 66.02 2.0 132.04 4 6 x 2.6 = 15.6 280.80 2.7 758.16 H.-6m 5 (2.6)(0.458) = 0.595 10.71 3.13 33.52 P 10 P₁ = 28.0 ΣΕ = 470.42 4.0 112.0 1128.86 = EMR H,=0.7 m C 07f0.72.6m 12-19 kN/m³ -20° -40 kN/m² 2. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ = k2 = 0.67, e = 0.406 m. Find q heel 0.5m 1.5 m = D 0.7m H,-0.458 m Section no." Area (m²) Weight/unit length (kN/m) Moment arm from point C (m) Moment (kN-m/m) 7-18 kN/m 4;=30 6 x 0.5=3 70.74 1.15 81.35 -0 2 (0.2)6 = 0.6 14.15 0.833 11.79 H-6m 3 4 x 0.7 = 2.8 66.02 2.0 132.04 P 4 10 6 x 2.6 = 15.6 280.80 2.7 758.16 5 (2.6)(0.458) -0.595 10.71 3.13 33.52 P=28.0 4.0 112.0 ΣΤ = 470.42 1128.86 = ΣΜ. 0.7m+0.7 2.6 m H-6.7 m Y:- 19 AN/m 4-20 c4 kN/m

1. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ =k2=0.67. Determine the factor of safety against Sliding and overturning. 15m-D T -18 kN/m dj-M7 5=0 Section Area Weight/unit length Moment arm H,-0.458 m from point C no. (m²) (kN/m) (m) Moment (kN-m/m) 1 6 x 0.5-3 70.74 1.15 81.35 2 (0.2)60.6 14.15 0.833 11.79 3 4 x 0.7 = 2.8 66.02 2.0 132.04 4 6 x 2.6 = 15.6 280.80 2.7 758.16 H.-6m 5 (2.6)(0.458) = 0.595 10.71 3.13 33.52 P 10 P₁ = 28.0 ΣΕ = 470.42 4.0 112.0 1128.86 = EMR H,=0.7 m C 07f0.72.6m 12-19 kN/m³ -20° -40 kN/m² 2. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ = k2 = 0.67, e = 0.406 m. Find q heel 0.5m 1.5 m = D 0.7m H,-0.458 m Section no." Area (m²) Weight/unit length (kN/m) Moment arm from point C (m) Moment (kN-m/m) 7-18 kN/m 4;=30 6 x 0.5=3 70.74 1.15 81.35 -0 2 (0.2)6 = 0.6 14.15 0.833 11.79 H-6m 3 4 x 0.7 = 2.8 66.02 2.0 132.04 P 4 10 6 x 2.6 = 15.6 280.80 2.7 758.16 5 (2.6)(0.458) -0.595 10.71 3.13 33.52 P=28.0 4.0 112.0 ΣΤ = 470.42 1128.86 = ΣΜ. 0.7m+0.7 2.6 m H-6.7 m Y:- 19 AN/m 4-20 c4 kN/m

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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1. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ =k2=0.67. Determine the factor of safety against Sliding and overturning. 15m-D T -18 kN/m dj-M7 5=0 Section Area Weight/unit length Moment arm H,-0.458 m from point C no. (m²) (kN/m) (m) Moment (kN-m/m) 1 6 x 0.5-3 70.74 1.15 81.35 2 (0.2)60.6 14.15 0.833 11.79 3 4 x 0.7 = 2.8 66.02 2.0 132.04 4 6 x 2.6 = 15.6 280.80 2.7 758.16 H.-6m 5 (2.6)(0.458) = 0.595 10.71 3.13 33.52 P 10 P₁ = 28.0 ΣΕ = 470.42 4.0 112.0 1128.86 = EMR H,=0.7 m C 07f0.72.6m 12-19 kN/m³ -20° -40 kN/m² 2. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ = k2 = 0.67, e = 0.406 m. Find q heel 0.5m 1.5 m = D 0.7m H,-0.458 m Section no." Area (m²) Weight/unit length (kN/m) Moment arm from point C (m) Moment (kN-m/m) 7-18 kN/m 4;=30 6 x 0.5=3 70.74 1.15 81.35 -0 2 (0.2)6 = 0.6 14.15 0.833 11.79 H-6m 3 4 x 0.7 = 2.8 66.02 2.0 132.04 P 4 10 6 x 2.6 = 15.6 280.80 2.7 758.16 5 (2.6)(0.458) -0.595 10.71 3.13 33.52 P=28.0 4.0 112.0 ΣΤ = 470.42 1128.86 = ΣΜ. 0.7m+0.7 2.6 m H-6.7 m Y:- 19 AN/m 4-20 c4 kN/m
Transcribed Image Text:1. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ =k2=0.67. Determine the factor of safety against Sliding and overturning. 15m-D T -18 kN/m dj-M7 5=0 Section Area Weight/unit length Moment arm H,-0.458 m from point C no. (m²) (kN/m) (m) Moment (kN-m/m) 1 6 x 0.5-3 70.74 1.15 81.35 2 (0.2)60.6 14.15 0.833 11.79 3 4 x 0.7 = 2.8 66.02 2.0 132.04 4 6 x 2.6 = 15.6 280.80 2.7 758.16 H.-6m 5 (2.6)(0.458) = 0.595 10.71 3.13 33.52 P 10 P₁ = 28.0 ΣΕ = 470.42 4.0 112.0 1128.86 = EMR H,=0.7 m C 07f0.72.6m 12-19 kN/m³ -20° -40 kN/m² 2. Considering the figure and table below, if you know that H' = 7.158 m, Ph = 158.75 kN/m, Pp = 215 kN/m, k₁ = k2 = 0.67, e = 0.406 m. Find q heel 0.5m 1.5 m = D 0.7m H,-0.458 m Section no." Area (m²) Weight/unit length (kN/m) Moment arm from point C (m) Moment (kN-m/m) 7-18 kN/m 4;=30 6 x 0.5=3 70.74 1.15 81.35 -0 2 (0.2)6 = 0.6 14.15 0.833 11.79 H-6m 3 4 x 0.7 = 2.8 66.02 2.0 132.04 P 4 10 6 x 2.6 = 15.6 280.80 2.7 758.16 5 (2.6)(0.458) -0.595 10.71 3.13 33.52 P=28.0 4.0 112.0 ΣΤ = 470.42 1128.86 = ΣΜ. 0.7m+0.7 2.6 m H-6.7 m Y:- 19 AN/m 4-20 c4 kN/m
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