Explanation Given information: The height ( H ) of retaining wall is 6.0 m. The top width ( x 1 ) of the stem is 0.6 m. The bottom width ( x 2 ) of the section 3 is 0.2 m. The bottom width ( x 3 ) of the section 1 is 2 m. The width ( x 4 ) of the toe is 0.5 m. The width ( x 5 ) of the heel is 0.75 m. The thickness ( x 6 ) of base slab is 0.8 m. The depth ( D ), to the bottom of the base slab is 1.5 m. The backfill angle ( α ) is 0 . The cohesion ( c ′ 1 ) of backfill soil is 0. The unit weight ( γ 1 ) of the backfill soil is 16.5 kN/m 3 . The friction angle ( ϕ ′ 1 ) of the backfill soil is 32 ° . The unit weight of soil ( γ 2 ) in front of the heel and under the base is 18 kN/m 3 . The friction angle ( ϕ ′ 2 ) of the soil in front of the heel and under the base is 22 ° . The cohesion ( c ′ 2 ) of soil in front of the heel and under the base is 40 kN/m 2 . The angle ( δ ' ) between the normal to the retaining wall and the coulomb’s active force is δ ' = 2 3 ϕ ′ 1 . Calculation: Consider point C as the left end of the toe base. Sketch the retaining wall with dimension and section as shown in Figure 1. Refer Figure 1. Find the height H ′ . H ′ = H + x 5 Substitute 6.0 m for H and 0.80 m for x 5 . H ′ = 6.0 + 0.8 = 6.80 m Find the distance of point of pressure from base of the slab using the equation: H ¯ = H ′ 3 Substitute 6.80 m for H ′ . H ¯ = 6.80 3 = 2.27 m Find the angle ( δ ' ) between the normal to the retaining wall and the coulomb’s active force using the equation: δ ' = 2 3 ϕ ′ 1 Substitute 32 ° for ϕ ′ 1 . δ ' = 2 3 × 32 ° = 21.33 ° Consider the triangle 1 (section 1). Find the angle θ using Figure 1. tan θ = x 3 H Substitute 2 m for x 3 and 6 m for H . tan θ = 2 6 θ = tan − 1 ( 2 6 ) θ = 18.43 ° Find the distance d using trigonometry relation: tan ( 90 ° − θ ) = H ¯ − x 6 d d = H ¯ − x 6 tan ( 90 ° − θ ) Substitute 2.27 m for H ¯ , 0.8 m for x 6 , and 18.43 ° for θ . d = 2.27 − 0.8 tan ( 90 ° − 18.43 ° ) = 1.47 3 = 0.489 m Find the value of coefficient of Coulomb’s active earth pressure ( K a ) . Refer Table 14.4, “Values of K a ”. Take the value of K a with corresponding value of friction angle of backfill ϕ ′ 1 of 32 ° and the angle between the sloping backfill with the horizontal α of 0 ° is 0.45. Find the Coulomb’s active force per unit length ( P a ) using the equation: P a = 1 2 γ 1 ( H ' ) 2 K a Substitute 16.50 kN/m 3 for γ 1 , 6.80 m for H ' , and 0.45 for K a . P a = 1 2 × 16.50 × ( 6.80 ) 2 × 0.45 = 171.67 kN/m Find the horizontal component ( P h ) of Coulomb’s active force per unit length using the equation: P h = P a × cos ( δ ' + θ ) Substitute 171.67 kN/m for P a , 18.43 ° for θ , and 21.33 ° for δ ' . P h = 171.67 × cos ( 21.33 ° + 18.43 ° ) = 131.97 kN/m Find the vertical component ( P v ) of Coulomb’s active force per unit length using the equation: P v = P a × sin ( δ ' + θ ) Substitute 171.67 kN/m for P a , 18.43 ° for θ , and 21.33 ° for δ ' . P v = 171.67 × sin ( 21.33 ° + 18.43 ° ) = 109.80 kN/m Find the moment arm or lever arm [ ( M A ) P v ] of vertical component ( P v ) of Coulomb’s active force per unit length using the equation ( M A ) P v = x 4 + x 2 + x 1 + x 3 − d Substitute 0.5 m for x 4 , 0.2 m for x 2 , 0.6 m for x 1 , 2 m for x 3 , and 0.489 m for d . ( M A ) P v = 0.5 + 0.2 + 0.6 + 2 − 0.489 = 2.811 m Find the area of section 1 ( A 1 ) . A 1 = 1 2 x 3 × H Substitute 2 m for x 3 and 6.0 m for H . A 1 = 1 2 ( 2 ) ( 6 ) = 6 m 2 Consider unit weight of concrete ( γ c ) is 23.58 kN/m 3 . Find the weight of section 1 ( W 1 ) . W 1 = A 1 γ c Substitute 6 m 2 for A 1 and 23.58 kN/m 3 for γ c . W 1 = 6 × 23.58 = 141.48 kN/m Find the moment arm or lever arm [ ( M A ) 1 ] of section 1 from point C . ( M A ) 1 = x 4 + x 2 + x 1 + 1 3 ( x 3 ) Substitute 0.5 m for x 4 , 0.2 m for x 2 , 0.6 m for x 1 , and 2 m for x 3 . ( M A ) 1 = 0.5 + 0.2 + 0.6 + 1 3 ( 2 ) = 1.97 m Find the moment about point C ( M C , 1 ) due to the self-weight of section 1: M C , 1 = W 1 × ( M A ) 1 Substitute 141.48 kN/m for W 1 and 1.97 m for ( M A ) 1 . M C , 1 = 141.48 × 1.97 = 278.72 kN-m/m Find the area of section 2 ( A 2 ) . A 2 = x 1 × H Substitute 0.6 m for x 1 and 6 m for H . A 2 = 0.6 × 6 = 3.6 m 2 Find the weight of section 2 ( W 2 )