How is the distance of the resultant from the toe found from the given data?

x = (SumM_R - SumM_O) / SumV = 21.667 m

What is the eccentricity and what does it show?

e = B/2 - x = 5.833 m < B/6 = 9.167 m, so there is no tension

e = x - B/2 = -5.83 m, so the dam overturns

e = B/2 - x = 5.833 m > B/6, so tension exists

Which expression gives the maximum vertical stress and where does it occur?

(SumV / B)(1 + 6e/B), at the toe = 1,785 kN/m^2

(SumV / B)(1 - 6e/B), at the toe

(SumV / B)(1 + 6e/B), at the heel

With no tail water and tan(phi_d) = 0.8, how is the principal stress at the toe obtained?

sigma_1 = sigma_max x (1 + 0.8^2) = 2,928 kN/m^2

sigma_1 = sigma_max x 0.8 = 1,428 kN/m^2

sigma_1 = sigma_max x (1 - 0.8^2)

2076 Chaitra · 2+2+2 marks

Foundation stresses from moments and vertical force

Short answer

From given moments and vertical force, resultant eccentricity is 5.83 m (< B/6). Maximum vertical stress at toe is 1785 kN/m², minimum at heel is 397 kN/m². Maximum principal stress at the toe is 2928 kN/m² (no tail water).

Determine the maximum and minimum vertical stresses to which the foundation of the dam will be subjected from the following data: total overturning moment about toe ( $\Sigma M_o$ ) = $1.2 \times 10^6$ kN-m; total resisting moment about toe ( $\Sigma M_R$ ) = $2.5 \times 10^6$ kN-m; total vertical force above the base ( $\Sigma V$ ) = $6 \times 10^4$ kN; base width of dam = 55 m; slope of d/s face = 0.8 : 1. Also calculate the maximum principal stress at the toe. Neglect tail water depth.

Position of the resultant

Net moment about the toe divided by the vertical load gives the distance of the resultant from the toe:

\bar x=\frac{\Sigma M_R-\Sigma M_O}{\Sigma V}=\frac{2.5\times10^6-1.2\times10^6}{6\times10^4}=21.667\ \text{m}

Eccentricity from the centre of the base:

e=\frac{B}{2}-\bar x=27.5-21.667=5.833\ \text{m}\quad (<B/6=9.167\ \text{m},\ \text{so no tension})

Maximum and minimum vertical stresses

\sigma_{max}=\frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right)=\frac{6\times10^4}{55}\left(1+\frac{6\times5.833}{55}\right)=1090.91\times1.6364=1,785.1\ \text{kN/m}^2\ (\text{at the toe})

\sigma_{min}=1090.91\times\left(1-0.6364\right)= 396.7\ \text{kN/m}^2\ (\text{at the heel})

Maximum principal stress at the toe

With no tail water, $p'=0$ and $\tan\phi_d=0.8$ :

\sigma_1=\sigma_{max}\sec^2\phi_d=\sigma_{max}(1+\tan^2\phi_d)=1,785.1\times(1+0.8^2)=2,927.6\ \text{kN/m}^2

Answer: σ_max = 1,785 kN/m² (toe), σ_min = 397 kN/m² (heel), principal stress at the toe σ₁ = 2,928 kN/m².

✓ Answer checked by Claude Opus 5.5 (high effort)

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