Solve Example 20.1, if the soil is clay having an unconfined compressive strength of 70 kN/m^2 and a unit weight of 17 kN/m^3. Determine the maximum bending moment.

Question

Solve Example 20.1, if the soil is clay having an unconfined compressive strength of 70 [latex]kN / m ^{2}[/latex] and a unit weight of 17 [latex]kN / m ^{3}[/latex]. Determine the maximum bending moment.

Accepted Answer

The pressure distribution is assumed as shown in Fig. Ex. 20.3.

For [latex]\phi_{u}=0, \bar{p}_{a}=\gamma H-q_{u}=17 imes 6-70=32 kN / m ^{2}[/latex]

[latex]\begin{aligned}&z_{0}=\frac{q_{u}}{\gamma}=\frac{70}{17} m =4.12 m \&P_{a}=\frac{1}{2} \bar{p}_{a}\left(H-z_{0} ight)=\frac{1}{2} imes 32 imes(6.0-4.12)=30 kN / m ext { of wall } \&\bar{p}=2 q_{u}-\gamma H=2 imes 70-17 imes 6=38 kN / m ^{2} \&\bar{p}^{\prime}=2 q_{u}+\gamma H=2 imes 70+17 imes 6=242 kN / m ^{2}\end{aligned}[/latex]

[latex]\bar{y}=\frac{1}{3}\left(H-z_{0} ight)=\frac{1}{3}(6-4.12)=0.63 m[/latex]

For the determination of h, equate the summation of all horizontal forces to zero, thus

[latex]P_{a}-\bar{p} imes D+\frac{1}{2}\left(\bar{p}+\bar{p}^{\prime} ight) h=0[/latex]

or [latex]30-38 imes D+\frac{1}{2}(38+242) h=0[/latex]

Therefore [latex]h=\frac{3.8 D-3}{14}[/latex]

For the determination of D, taking moments of all the forces about the base of the wall, we have

[latex]P_{a} imes(D+\bar{y})-\bar{p} imes \frac{D^{2}}{2}+\left(\bar{p}+\bar{p}^{\prime} ight) imes \frac{h}{2} imes \frac{h}{3}=0[/latex]

or [latex]30(D+0.63)-38 imes \frac{D^{2}}{2}+(38+242) imes \frac{h^{2}}{6}=0[/latex]

Substituting for h we have,

[latex]3 D+1.89-1.9 D^{2}+4.7 \frac{3.8 D-3}{14}^{2}=0[/latex]

Simplifying, we have

[latex]D^{2}-1.57 D+1.35=0[/latex]

Solving D = 2.2 m; Increasing D by 40%, we have D = 1.4(2.2) = 3.1 m.

Maximum bending moment

From Eq. (20.20)

[latex]\bar{y}_{o}=\frac{P_{a}}{\bar{p}}[/latex] (20.20a)

[latex]M_{\max }=P_{a}\left(\bar{y}_{o}+\bar{y} ight)-\frac{\overline{p y_{o}^{2}}}{2}[/latex] (20.20b)

[latex]\begin{aligned}&M_{\max }=P_{a}\left(\bar{y}_{0}+\bar{y} ight)-\frac{\bar{p} \bar{y}_{0}^{2}}{2} \&\begin{aligned}&\bar{y}_{0} =\frac{P_{a}}{\bar{p}}=\frac{30}{38}=0.79 m \&\overline{ y }= 0.63 m \M_{\max } &=30(0.79+0.63)-\frac{38 imes(0.79)^{2}}{2} \&=42.6-11.9=30.7 kN - m / m ext { of wall }\end{aligned}\end{aligned}[/latex]

Question 20.3: Solve Example 20.1, if the soil is clay having an unconfined...

Related Answered Questions