HPGE April 2024 PDF Past Paper

## Civil Engineering Licensure Examinations ### H-169 **1. CE Board April 2024** **CE Board April 2023** **CE Board Nov. 2022** **CE Board May 2022** **CE Board May 2019** **CE Board May 2018** **CE Board Nov. 2017** A layer of clay 7.6 m thick is overlain by a deposit of sand 15.2 m thick to ground surface. When the water table is at ground surface level, the saturated unit weight of the sand was determined to be 21.2 kN/m³ for which its dry unit weight of 18.2 kN/m³. The saturated unit weight of the clay is 18.8 kN/m³. When the ground water table lowered by 10 m, the degree of saturation of the sand above the water table was lowered to 20%. At the mid-height of the clay layer, **Solution:** a) Effective pressure at the mid height of clay before lowering the water table * Sand * $Y_{sat}$ = 21.2 kN/m³ * $Y_{drj}$ = 18.2 kN/m³ * Height = 15.2 m * Clay * $Y_{sat}$ = 18.8 kN/m³ * Height = 7.6 m $σ_A = (21.2 – 9.81)(15.2) + (18.8 – 9.81)(3.8)$ $σ_A = 207.29 kPa$ **b) Effective soil pressure at mid-height of clay after lowering the water table by 10 m** * Sand * $S$ = 20% * Height = 5.2 m * Clay * $Y_{sat}$ = 18.8 kN/m³ * Height = 3.8 m $Y_{sat}=\frac{(Gs+e)Yw}{1+e}$ $Y_{sat}=\frac{(Gs+e)Yw}{1+e}+\frac{eYw}{1+e} $ $Y_{sat}=Y_{dry}+\frac{e(9.81)}{1+e}$ $21.2 = 18.2 + \frac{9.81e}{1+e}$ $3 = \frac{9.81e}{1+e}$ $3+3e = 9.81e$ $e= 0.44$ $Y_{dry}=\frac{GsYw}{1+e}$ $3 = \frac{Gs(9.81)}{1+0.44}$ $Gs = 2.67$ $Y_{wet}$ of sand with degree of saturation of 20% $Y_{we}=\frac{(Gs+Se)Yw}{1+e}$ $Y_{wet}=\frac{[2.67 +0.2(0.44)]9.81}{1+0.44}$ $Y_{wet} = 18.79 kN/m^3$ $σ_A = 18.79(10) + (21.2 – 9.81)(5.2) + (18.8-9.81)(3.8)$ $σ_A = 281.29 kPa$ **c) Effective pressure when there is no water in the sand layer but remain in moist condition at the same degree of saturation** * Sand * $Y_{wet}$ = 18.79 kN/m³ * Height = 15.2 m * Clay * $Y_{sat}$ = 18.8 kN/m³ * Height = 3.8 m $σ_A = 18.79(15.2) + (18.8 – 9.81)(3.8)$ $σ_A = 319.77 kPa$ ### H-170 **2. CE Board April 2024** The total weight of soil when fully saturated is 1600 g and the weight of the soil after drying is 1000 g. If the specific gravity of the soil grains is 2.67. **Solution:** a) Evaluate the moisture content in percent $ω = \frac{ γ - γ_{dry}}{γ_{dry}} * 100$ $1000 = \frac{1600}{1+ω}$ $ω =0.6 or 60%$ **b) Evaluate the void ratio** $S = \frac{Gςω}{e}$ $100=\frac{2.67(60)}{e}$ $e=1.60$ **c) Evaluate the porosity** $n = \frac{e}{1+e}$ $n = \frac{1.6}{1+1.6}$ $n = 0.615$ $n = 61.5%$ ### H-171 **3. CE Board April 2024** The results of a consolidated drained tri-axial soil test conducted on a consolidated clay are as follows: Chamber confining pressure = 240 kPa Deviator stress at failure = 450 kPa **Solution:** a) Angle of friction of the soil sample * $σ_n = 240 kPa$ * $σ_f =450 kPa$ * $τ = 450 - 240 = 210 kPa$ $sin φ = \frac{τ}{σ_n} = \frac{225}{465}$ $φ = 28.9°$ **b) Shear stress on the failure plane** $τ = σ_n cos φ $ $τ =225 cos 28.9°$ $τ = 197 kPa$ **c) Normal stress on the plane of max. shear** $σ _n= 465 kPa $ ### H-172 **4. CE Board April 2024** A circular plate of radius r = 300 mm is vertically submerged in water with its top edge 2 m below the liquid surface. **Solution:** a) Total force acting on one side of the gate $h = 2 + (0.3 - 2.3) m$ $h = 2300 mm$ $P = yhA$ $P = 9.81(2.3)(π)(0.3)2$ $P = 6.38 kN$ **b) Location of force from the center of gravity of the plate** $e = \frac{I_g}{Ay}$ $A = π(0.3)2 = 0.283 m2$ $y = 2.3$ $I_g = \frac{π(0.6)4}{64}$ = 0.00636 $e = \frac{0.00636}{0.283(2.3)}$ $e = 0.00978 m$ $e = 9.78 mm$ **c) Location of force from the liquid surface** $D =h+e$ $D = 2.3+0.00978$ $D =2.30978$ $D =2309.78 mm$ $D=2310 mm$ ### H-173 **5. CE Board April 2024** **CE Board Nov. 2016** According to the Boussinesq's theory, the vertical stress at a point below the surface of a semi-infinite, homogeneous, isotropic soil mass due to a point load Q applied at the ground surface is given by the equation: $p = \frac{0.477QN}{z^2}$ $N = \frac{1}{[1 + (\frac{r}{z})^2]^{2.5}}$ Evaluate the vertical stress in kPa at a point below the ground for Q = 2500 kN. Solution: a) if the point is 3 m directly below the point of application of the load. $N = \frac{1}{[ 1 + (\frac{0}{3})^2 ]^{2.5}}$ $N = 1$ $p=\frac{0.477QN}{z^2}$ $p = \frac{0.477(2500)(1)}{(3)^2}$ $p = 132.5 kPa$ b) if the point is 6 m directly below the point of application of the load. $p = \frac{0.477QN}{z^2}$ $p = \frac{0.477(2500)((1)}{(6)^2}$ $p = 33.1 kPa$ c) if the point is 6 m below the load but 3 m horizontally from its line of application $N= \frac{1}{[1+(\frac{3}{6})^2]^{2.5}}$ $N = \frac{1}{[1+ (\frac{1}{2})^2]^{2.5}}$ $N = 0.572$ $p = \frac{0.477QN}{z^2}$ $p = \frac{0.477(2500)(0.572)}{(6)^2}$ $p = 19 kPa$ ### H-174 **6. CE Board April 2024** **CE Board Nov. 2023** According to the Westergaard theory, the vertical stress at a point below the surface of a semi-infinite, homogeneous, isotropic soil mass due to a point load Q applied at the ground surface is given by the equation: $p = \frac{0.318QN}{z^2}$ $where N = \frac{1}{[1 + (\frac{r}{z})^2]^{1.5}}$ Calculate the vertical stress in kPa at a point below the ground for Q = 2500 kN and the point is 6 m below the load but 3 m horizontally from the line of the load application. **Solution:** $N= \frac{1}{[1 + (\frac{3}{6})^2]^{1.5}}$ $N = \frac{1}{[1+ (\frac{1}{2})^2]^{1.5}} $ $N = 0.544$ $p = \frac{0.318QN}{z^2}$ $p = \frac{0.318(2500)(0.544)}{(6)^2}$ $p = 12 kPa$ **7. CE Board April 2024** **CE Board Nov. 2017** An open cylindrical vessel 1.3 m in diameter and 2.1 m high is 2/3 full of water. If rotated about the vertical axis at a constant angular speed of 87 rpm, determine the amount of water in liters that will be spilled out. **Solution:** * ω = 87 rpm * ω = 87(2π)/60 = 9.11 rad/sec. $y = \frac{ω²r²}{2g}$ $y =\frac{(9.11)²(0.65)²}{2(9.81)}$ $y = 1.79 m$ Amount of water spilled out : $V = \frac{π(0.65)²( 1.79) - π(0.65)²(1.4) }{2}$ $V = 0.259 cu.m.$ $V = 259 liters$ ### H-175 **8. CE Board April 2024** **CE Board Nov. 2023** A closed cylindrical tank placed in a horizontal position on a truck 1.5 m. in diameter and 5 m long. It is completely filled with water. If the truck travels horizontally forward at an acceleration of 4 m/s², determine the total force acting on the rear end of the container in kN. **Solution:** * $tan θ = y/a$ $tan θ = \frac{5}{4}$ * $a = \frac{y}{tan θ} =\frac{5}{ \frac{5}{4}}$ $a = 4 m/s2$ * $y = \frac{a}{g} 5 = \frac{4}{9.81} 5$ $y = 204 m$ * $h = 2.04 + 0.75 = 2.79 m$ $F = yh A$ $F = (9.81)(2.79)(π)( 1.5)2$ $F = 48.37 kN$ **9. CE Board April 2024** The water content of soil could a) be less than 0% b) never be greater than 100% c) be greater than 100% d) have values only from 0% to 100% **10. CE Board April 2024** A granular soil ’deposit is considered loose if the blow count of a standard penetration test is between: **Solution:** Relative Density of Sands According to Results of Standard Penetration Test SPT N-value | Relative Density | | --------------- | | 0-4 | Very loose | | 4-10 | Loose | | 10-30 | Medium | | 30-50 | Dense | | Over 50 | Very dense | The blow count of a standard penetration test for loose soil is between 4 to 10. **11. CE Board April 2024** A sample of saturated soil has 30% water content and the specific gravity of the soil grains is 2.6. Determine the dry density of the soil mass in g/cm³. **Solution:** $S = \frac{ωG_s}{e}$ $100 = \frac{30(2.6)}{e}$ $e = 0.78$ $Y_{dry}=\frac{GsYw}{1+e}$ $Y_{dry} = \frac {2.6(1)}{1+0.78}$ $Y_{dry} = 1.46 gr/cm^3$ **12. CE Board April 2024** The admixture of coarser particles like sand and silt to clay causes: a) Decrease in both liquid limit and plasticity index. b) Decrease in liquid limit and increase in plasticity index. c) Increase in both liquid limit and plasticity index. d) Decrease in liquid limit and no change in plasticity index. **13. CE Board April 2024** The shear strength of a cohesion soil is: a) Inversely proportional to the angle of shearing resistance. b) None of the given choices. c) Proportional to the angle of shearing stress. d) Proportional to the tangent of the angle of shearing resistance. **14. CE Board April 2024** Water flows at 20 liters/sec. through 500 m of 150 mm diameter pipeline having a coefficient of friction of 0.02. Calculate the head loss for that length of pipeline. **Solution:** $h_f = \frac{0.0826 fLQ^2}{D^5}$ $h_f = \frac{0.0826(0.02 )(500)(0.02)^2}{(0.15)^5}$ $h_f = 4.35 m$ ### H-176 **15. CE Board April 2024** A rectangular plate 1.2 m wide by 2 m long is vertically submerged half in oil having sp.gr. of 0.84 and half fresh water such that its shorter edge is flushed with the top surface of the upper liquid and parallel to it. Obtain the total force acting on the lower half of the plate in kN. **Solution:** $P= yhA$ * $h = 0.84 + 0.50$ * $h = 1.34$ $P = 9.81(1.34)(1.2)(1)$ $P = 15.8 kN$ **16. CE Board April 2024** **CE Board Nov. 2022** **CE Board Nov. 2019** **CE Board May 2015** A 450 kN load is transmitted by a column footing on the surface through a rectangular footing 1.5 m x 2.5 m. Assuming that the pressure on the underlying soil formation spreads on 2 vertical to 1 horizontal. Evaluate the pressure exerted by the footing on a soil 2.7 m below the footing. **Solution:** $P = \frac{450}{5.2(4.2)}$ $P = 20.6 kPa$ **17. CE Board April 2024** **CE Board May 2017** **CE Board May 2016** An irrigation canal with trapezoidal cross-section has the following dimensions: Bottom width = 2.50 m, depth of water = 0.90 m, side slope = 1.5 horizontal to 1 vertical, slope of the canal bed = 0.001, coefficient of roughness = 0.025. The canal will serve clay-loam Riceland for which the duty of water per hectare is 3.0 liters/sec. Using Manning's Formula, estimate the number of hectares served by the irrigation canal. ### H-177 **18. CE Board April 2024** **CE Board May 2017** A ship, with vertical sides near the waterline, weighs 40 MN including its cargo and has a draft of 6.7 meters in seawater (s.g. = 1.026). Unloading 2 MN of its cargo, the draft decreases to 6.4 m. With its cargo reduced, the ship enters a harbor of fresh water. Evaluate the draft of the ship in fresh water, in meters. **Solution:** * $40000 + W = BF_1$ * $40000 + W = 9.81(6.7)(A) (1.026)$ * $40000+ W 67.44A$ * $W=67.44A-40000$ * $38000 + W = BF_2$ * $38000 + W = 9.81(6.4)(A)(1.026)$ * $38000+67.44A – 40000 = 64.42A$ * $3.02A = 2000$ * $A = 662.25 m²$ * $W = 67.44(662.25) - 40000$ * $W = 4662.14$ * $38000 + W = BF_3$ * $38000 + 4662.14 = 9.81(d)(622.25)$ * $d = 6.56 m$ ### H-178 **19. CE Board April 2024** **CE Board Nov. 2019** **CE Board Nov. 2018** **CE Board May 2018** **CE Board May 2015** An improvised barometer uses a liquid which was observed to weigh 0.80 times of mercury. At the base of the mountain the barometer reads 950 mm. At the same instant, another barometer of the same land at the top of the mountain reads 600 mm. Determine the height of the mountain in kilometers. **Solution:** * $P_1 = P_2 + yh$ * $P_1 - P_2 = yh$ * $y = 12 N/m³ for air$ $P_1-P_2 = 12h$ $9810(13.6)(0.8)(0.95) – 9810(13.6)(0.8)(0.6) = 12h$ $h = 3113 m$ $h = 3.11 km$ **20. CE Board April 2024** An engineer wanted to determine the weight of a piece of wood having a rectangular cross section and certain length with only a carpenter's tape available and a curing tank containing water 1.8 m deep. He recorded the cross section of the wood to be 50 mm x 100 mm and the length is 1.5 m. Placing the wood vertically into the curing tank, he noticed that 0.50 m projects out of the water as it floats. Solve for the weight of the wood in Newtons. **Solution:** * $BF = W$ * $BF = 9810(0.05)(0.10)(1)$ * $BF = 49.05 N$ * $W = 49.05 N$ ### H-179 **21. CE Board April 2024** A vertical rectangular gate 1.5 m wide and 3 m high is hinged at the bottom with the shorter edge horizontal. On one side of the gate, water stands level with its top of the gate, on the other side, water stands 2 m. below the top. Evaluate the force applied horizontally at the top of the gate to open it. **Solution:** * $P_1 = γ_w h A_1$ * $P_1 = 9.81(1.5)(3)(1.5)$ * $P_1 = 66.22 kN$ * $P_2 = γ_w h A_2$ * $P_2 = 9.81(0.5)(1)(1.5)$ * $P_2 = 7.36 kN$ $∑M_A = 0$ $T(3) + P_33) = P_1(1)$ $3T +7.36(3) = 66.22(1)$ $T = 21.26 kN$ **22. CE Board April 2024** A rectangular irrigation canal is 5.4 m wide and 1.2 m deep has a hydraulic slope of 0.001 and a roughness coefficient of 0.013. Evaluate the appropriate depth of the canal in meters using the more appropriated dimensions but adhering to the same discharge and slope. **Solution:** * $A = 1.2(5.4) = 6.48$ * $P = 1.2+5.4+ 1.2$ * $P = 7.8$ $R = \frac{A}{P} $ $R =\frac{6.48}{7.8}$ $R = 0.83$ $V=R^{2/3}S^{1/2}/n$ $V=\frac{1}{0.013}(0.83)^{2/3}(0.001)^{1/2}$ $V= 2.148 m/s$ $Q = AV$ $Q = 6.48( 2.148)$ $Q = 13.92 m^3/sec.$ * $b = 2d$ * $A = bd$ * $A = 2dd = 2d^2$ * $P = b + 2d$ * $P = 2d + 2d$ * $P = 4d$ $13.93 = 2d^2( 2.43 )( \frac{2}{3})$ $2.86 = \frac{d^2}{2d^2} (\frac{2}{3})^{2/3}$ $4.54 = d^{8/3}$ $d = 1.76 m$ ### H-180 **23. CE Board April 2024** The sector gate shown consists of a cylindrical surface of radius R, of which AB is the base. Evaluate the total vertical force on the gate in kN if $R = 8 m$ and the length of the plate perpendicular to the paper is 10 m. **Solution:** * $h = 8 sin 60°$ * $h = 6.93$ $F_x = yhA$ $F_x = 9.81 (6.93)(6.93)(10)$ $F_x = 2355.6 kN$ **24. CE Board April 2024** The sector gate shown consists of a cylindrical surface of radius R, of which AB is the base. Evaluate the total vertical force on the gate in kN if R = 8 m and the length of the plate perpendicular to the paper is 10 m. **Solution:** * $h = 8 sin 60°$ * $h = 6.93$ Area ABD = Area of trapezoid ADBOA - Area of sector ABO * $OC = 8 cos 60°$ * $OC = 4 m$ * $AC=8-44 m$ Area of trapezoid ADBOA $A₁ = (4+8)(6.93)$ $A₁ = 41.58 m²$ Area of sector ABO: $A_2 = \frac{π(8)²(60)}{360}$ $A_2 =33.51 m²$ Area ABD = $A_1 - A_2$ Area ABD = 41.58-33.51 Area ABD = 8.07 m² $F_y = 9.81(8.07)(10)$ $F_y = 791.7$ ### H-181 **25. CE Board April 2024** A water tank has a sloping inclined at 45° with the horizontal. The total depth of water in the tank is 8 m. A water jet issues from an orifice located on the inclined side of the tank under a hydrostatic head of 5 m. or that orifice is located 3 m. vertically above the bottom of the tank. Coefficient of velocity is 1.0. Neglecting air resistance on the jet. Determine the horizontal distance on the ground traveled by the jet from the center of the orifice in meters. **Solution:** Vertical height in meters that the jet rises above the level of the orifice: * $V = C \sqrt{2gH}$ * $V = 1.0 \sqrt{2(9.81)(5)}$ * $V = 9.9045 m/s$ * $V_1 = 9.9045 Sin45°= 7$ * $V_2^2 = V_1^2 -2gh $ * $0 = (7)^2 -2(9.81)h $ * $h = 2.5 m$ Horizontal distance on the ground traveled by the jet from the center of the orifice in meters: $x=Vt$ $x = 9.9045 Cos 45°(1.88)$ $x = 13.2 m$ **26. CE Board April 2024** A column is to be supported by a square footing, 2.0 m on a side on a founding depth of 1.0 m into a cohesionless soil deposit. The unit weight of the soil is 16 kN/m³ and the angle of friction is 25 degrees. Using Terzaghi's formula for general shear failure, evaluate the concentric load in kN, that the footing can safely support using a factor of safety of 3.0 against bearing capacity failure. **Solution:** $q_{ult} = 1.3 c N_c + q N_a + 0.40 γ B N_y$ * $N_c = 25.1$ * $N_q = 12.7$ * $N_y = 8.34$ $q_{ult} = 1.3(0)(25.1) + 1(12.7) + 0.40(16)(2)(8.34)$ $q_{ult} = 119.452$ $q_{all} = \frac{119.452}{3} = 39.82$ $P = q_{all} x Area$ $P = 39.82(2)(2)$ $P = 159.27 kN$ ### H-182 **27. CE Board April 2024** A three layer soil has the following properties from top to bottom. Estimate the rate of flow in the vertical direction per square meter of layer in liters/hr. if the hydraulic gradient is 0.50. **Solution:** | Layer | k (cm/sec) | Thickness (m) | | ----- | ----------- | -------------- | | 1 | 1 x 104| 3 | | 2 | 3.2 x 10 -2 | 4 | | 3 | 4.1 x 10-5| 6 | Equivalent vertical coeff. of permeability: $K_{eq} = \frac{H}{H_1 + \frac{K_1K_2}{K_3} + \frac{H_3}{K_3}}$ $K_{eq} = \frac{13} {3 + \frac {1*10^{-4} * 3.2 * 10^{-2}}{4.1 * 10^{-5}} + \frac{6}{4.1 * 10^{-5}}}$ $K_{eq} = 7.37 * 10^{-5} cm/sec$ $K_{eq} = \frac {7.37 * 10^{-5}(3600)}{100}$ $K_{eq} = 2.65 * 10^{-3} m/hour$ Rate of flow: $Q=K_{eq}iA$ $Q = 2.65 * 10^{-3} (0.5)(1)$ $Q =1.325 *10^{-3} m^3/hr$ $Q = 1.325 liters/hr.$ **28. CE Board April 2024** **CE Board Nov. 2023** **CE Board Nov. 2018** **CE Board May 2016** A layer of soil having an initial void ratio of 1.0 is 8 m thick. Under a compressive load applied above it, the void ratio decreased by one-half. Evaluate the reduction in the thickness of layer. **Solution:** * $ΔH = \frac{H(e_1-e_2)}{1+e}$ * $e_1 = 1.0$ * $e_2 = \frac{1}{2}(1.0)$ * $e_2 = 0.5$ $ΔH = \frac{H(e_1-e_2)}{1+e}$ $ΔH = \frac{8(1.0 - 0.5)}{1+1}$ $ΔH = 2 m$ **29. CE Board April 2024** The quantity of seepage of water through soils is proportional to: a) Coefficient of permeability of soil. b) Neither coefficient of permeability and total head loss through the soil. c) Both coefficient of permeability and total head loss through the soil. d) Total head loss through the soil. **30. CE Board April 2024** **CE Board May 2015** The unit weight of a liquid is variable and is given by the relation $y = 12 + 0.5h$, where $y$ in kN/m³ and h is the depth of the liquid from the surface in meters. Determine the gage pressure in kPa at a depth of 3 m. **Solution:** * $y = 12 + 0.5h$ * $dP = y dh$ * $dP = (12 +0.5h) dh$ $P = 12h+ \frac{0.5h^2}{2}$ $P = 12(3) + \frac{0.5(3)^2}{2}$ $P = 38.25 kPa$ **31. CE Board April 2024** A prism of dimensions 1.5 m x 1.0 m x 2.0 m weighs 29.4 kN in water, determine its specific gravity. **Solution:** $BF = (1.5)(1)(2)(sp.gr.)(9.81)$ $29.4 = 1.5(2)(9.81)sp.gr.$ $sp.gr. = 1.0$ **32. CE Board April 2024** A soil has a bulk density of 2.3 g/cm³ and water content 15%, determine the dry density of the sample. **Solution:** $Y_{dry} = \frac{bulk density}{1+ ω}$ $Y_{dry} = \frac{2.3}{1+0.15}$ $Y_{dry} = 2 gr/cm^3$ **33. CE Board April 2024** A layer of soft clay having an initial void ratio of 1.0 is 9 m thick. Under a compressive load applied above it, the void ratio decreases by the fourth. Evaluate the reduction in the thickness of the clay layer in meters. **Solution:** * $ΔH = \frac{H(e_1-e_2)}{1+e}$ * $e_2 = \frac{1}{4} e_1$ $ΔH = \frac{9(1- \frac{1}{4} e_1)}{1+ e_1}$ $ΔH

HPGE April 2024 PDF Past Paper

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