Title Apr 2025 PB1 - PSAD Kippap Solutions.pdf ✅

The minimum velocity so that no water can be spilled occurs when the centrifugal force alone can fully counteract the weight. From the figure, +↑ ∑Fv = 0: CF − W = 0 mv 2 r − mg = 0 v = √gr v = √9.81(1.1) = 3.285 m/s For the angular velocity, v = ωr 3.285 = ω(1.1) ω = 2.986 rad s or 28. 517 rpm Problem 3. RCD – One-way slabs What should be the thickness of a one-way slab with a simple span of 3 m? Use fc ′ = 24.5 MPa, fy = 245 MPa, steel cover = 40 mm. Refer to PSAD Code 1. a. 100 mm c. 130 mm b. 120 mm d. 150 mm Solution: h = 3000 20 (0.4 + 245 700) = 112.5 mm, say 120 mm Problem 4. Composite Members A 250 mm x 400 mm wooden beam is reinforced by two 250 x 20 mm steel plates. If the allowable stress of wood is 3 MPa and the allowable stress of steel is 50 MPa, determine the allowable moment. Use n = 15. a. 119.3 kNm c. 131.9 kNm b. 120.5 kNm d. 105.2 kNm Solution:
Computing the moment of inertia of the transformed section, using MODE-STAT-SD with frequency on, X Freq 400 2 + 20 2 = 210 mm nAs = 15(250)(20) = 75000 0 Aw = 250(400) = 100000 − 400 2 − 20 2 = −210 mm nAs = 15(250)(20) = 75000 The moment of inertia will be I̅ x = 250(4003 ) 12 + 2(15)(250)(203 ) 12 + n(xσn) 2 = 7953.333 × 109mm4 From the allowable stress of steel, fs n = Mcs I 50 15 = M(220) 7953.333 × 109 M = 120.505 × 106 Nmm or 120.505 kNm From the allowable stress of wood, fw = Mcs I 3 = M(200) 7953.333 × 109 M = 119.300 × 106 Nmm or 119. 300 kNm Therefore, the allowable moment is 119.3 kNm. Problem 5. Helical Springs A spring is made by coiling a 3 mm wire with 20 turns into a mean radius of 20 mm. How much stress is developed when it is compressed by 10 mm? Use G = 70 GPa. Hint: δ = 64PR 3n Gd 4 τ = 16PR πd 3 (1 + d 4R ) a. 21.7 MPa c. 24.3 MPa b. 27.1 MPa d. 23.4 MPa Solution: For the developed load of the spring, δ = 64PR 3n Gd 4 10 = 64P(203 )(20) 70000(3 4) P = 5.537 N For the stress in the spring, τ = 16PR πd 3 (1 + d 4R ) τ = 16(5.537)(20) π(3) 3 [1 + 3 4(20) ] τ = 21. 672 MPa
Problem 6. Statically Indeterminate Members A rigid bar is held by two rods at A and B as shown. Find the displacement of point C when P = 100 kN. a. 3.5 mm c. 1.8 mm b. 4.4 mm d. 2.6 mm Solution: Using summation of moments about the pin support, ↷ +∑Mpin = 0: TA (2) + TB (4) − 5(100) = 0 TA + 2TB = 250 eqn. 1 Using ratio and proportion on the deformed shape, δA 2 = δB 4 = δC 5 1 2 [ TA (2000) 500(200) ] = 1 4 [ TB (2500) 600(96) ] TA − 1.085TB = 0 Solving the equations simultaneously, TA = 87.925 kN, TB = 81.037 kN. From the equations, δB 4 = δC 5 1 4 [ (81.037)(2500) 600(96) ] = δC 5 δC = 4. 397 mm