Tiêu đề Erroneous Problems Post-test Solutions (1).pdf ✅

If the aquifer is unconfined, Q = πk(h1 2 − h2 2 ) ln R1 R2 69 ( 1 m3 1000 L ) ( 86400 s day ) = πk[(t −0.5) 2 − (t − 1.1) 2 ] ln 95 35 5961.6 = πk[(t − 0.5) 2 −(t − 1.1) 2 ] ln 95 35 Trying to solve for the thickness for each coefficient of permeability in the choices, a. If k = 61.72 m,t = 26.39 m b. If k = 60.27 m,t = 27.00 m c. If k = 70.26 m,t = 23.27 m d. If k = 72.60 m,t = 22.55 m Thus, the most possible thickness of the aquifer is 27 m from option B if the aquifer is unconfined. Problem 4. Find the direction of the vector represented by <-5, -5> a. D/4 c. 5D/4 b. 3D/4 d. 7D/4 Solution: For the direction of the vector, θ = Arctan −5 −5 = 5π 4 In the choices, option C has 5/4. Thus, π may have been misrepresented as “D”. SITUATION. An 8 m simply supported beam has the following dimensions: b = 250 mm, h = 600 mm, d = 500 mm. It is reinforced with 3- 26 mm bars for tension and 12 mm stirrups. It carries a uniform dead load of 12 kN/m and a live load of 15 kN/m. Use f’c = 28 MPa and fy = 276 MPa. Problem 5. Find the ultimate shearing force at critical section (kN). a. 134.4 c. 130.6 b. 153.6 d. 94.5 Solution: Finding the reaction, wu = 1.2(12) + 1.6(15) = 38.4 kN m R = 38.4(8) 2 = 153.60 kN At d-distance from the support, Vu = 153.60− 38.4(0.5) = 134. 4 kN Problem 6. Find the ultimate shearing strength by concrete (kN). a. 71.9 c. 66.7 b. 77.3 d. 70.2
Solution: For the ultimate strength by concrete, Vnc = 0.17λ√fc ′bwd Vnc = 0.17(1)√28(250)(500) Vnc = 112444.43 N or 112.444 kN Vuc = 0.75(112.444) = 84.333 kN However, this value is not in the choices. Trying to divide the choices by this value, a. 71.9 84.333 = 0.85 b. 77.3 84.333 = 0.92 c. 76.7 84.333 = 0.79 d. 70.2 84.333 = 0.83 The value in option A is the factor for sand-lightweight concrete. Thus, the detail that the concrete is sand-lightweight may have been the case in the problem. Problem 7. Find the ultimate shearing strength by steel (kN). a. 64.2 c. 57.7 b. 57.1 d. 62.5 Solution: Vus = 134.4 − 71.9 = 62. 5 kN SITUATION. A masonry wall weigh 9 kN/m is carried by a wall footing 0.6 m wide. The vertical stress induced by a line load can be estimated by P = 0.637q N where: q = line load in kN/m, N = z (1+ ( r z ) 2 ) 2 r = horizontal distance in m from the footing, z = depth in m below the footing Problem 8. Find the foundation pressure at the base of the footing. a. 12 kPa c. 14 kPa b. 13 kPa d. 15 kPa Solution: f = q B = 9 0.6 = 15 kPa Problem 9. Find the pressure increase at a point twice the depth of the footing. a. 7.69 kPa c. 9.67 kPa b. 7.96 kPa d. 9.76 kPa Solution: N = 1.2 [1+ ( 0 1.2 ) 2 ] 2 = 1.2