Title HPGE 17 Solutions.pdf ✅

17 Hydraulics: Dams Solutions SITUATION 1. A masonry dam of trapezoidal cross-section, with one end face vertical, has a thickness of 0.6 m at the top and 3 m at the bottom. It is 6.75 m high and the vertical face is subjected to water pressure, the water standing 4.6 m above the base at the upstream side. Assume that the hydrostatic uplift varies uniformly from 50% at the heel to zero at the toe. Use sg = 2.4 for concrete. [PRE-SOLUTION] Shown is the figure ▣ 1. Find the vertical component of the foundation reaction. [SOLUTION] Draw the FBD:

∑Fy = 0 Fu + Ry − W1 − W2 = 0 33.8455 kN + Ry − 95.3532 kN − 190.7064 kN = 0 Ry = 252.2151 kN ▣ 2. Where will the vertical reaction intersect the base measured from the toe? [SOLUTION] Summation of moment along the toe: ∑MToe = 0 −Fw ( 4.6 m 3 ) + W1 (3 m − 0.6 m 2 ) + W2 [ 2 3 (3 m − 0.6 m)] − Fu (2 m) − Ryd = 0 d = 1.33 m ▣ 3. What is the factor of safety against overturning? [SOLUTION] RM = W1 (3 m − 0.6 m 2 ) + W2 [ 2 3 (3 m − 0.6 m)] RM = 562.58 kNm OM = Fw ( 4.6 m 3 ) + Fu (2 m) OM = 226.83 kNm FS = RM OM FS = 2.48 ▣ 4. What is the factor of safety against sliding assuming μ = 0.6? [SOLUTION] Rx = μRy Rx = 151.329 kN FS = Ry Fw FS = 1.458 ▣ 5. What is the maximum pressure along the base due to the forces acting on the dam? [SOLUTION] Since d is closer to the toe, then qmaximum occurs at the toe: qmax = Ry B (1 + 6e B ) Solve for e
e = | B 2 − d| e = | 3 m 2 − 1.33 m| e = 0.169 m qmax = 252.2151 kN 3 m (1 + 6 × 0.169 m 3 m ) qmax = 112.45 kPa ▣ 6. What is the minimum pressure along the base due to the forces acting on the dam? [SOLUTION] qmin = Ry B (1 − 6e B ) qmin = 252.2151 kN 3 m (1 − 6 × 0.169 m 3 m ) qmin = 55.69 kPa