NAME : ………………………………………………………………….. CLASS : ………………………………………………………………… KOLEJ MATRIKULASI KEJURUTERAAN JOHOR BASIC ENGINEERING BASIC ENGINEERING (EB025) SEMESTER II SESSION 2022/2023 MOCK PSPM II 1 hour 30 minutes Instructions: 1. This paper has FIVE (5) questions, please answer all the questions. 2. The use of electronic calculator is permitted. 1. a) A block is being pulled by two forces F1 and F2 with magnitude of 100kN and 120kN, respectively in 3D direction as shown in FIGURE 1.1.
SULIT MOCK PSPM - EB025 2 SULIT Based on FIGURE 1.1: (i) Resolve the magnitude of forces F1 and F2 along its component. [6 marks] (ii) Express F1 and F2 in Cartesian form. [2 marks] (iii) Find the magnitude and direction of the resultant force of F1 and F2. [8 marks] b) FIGURE 1.2 shows a beam supported by a pin at A and a cable at B is being pulled by a 50kg cart at C. FIGURE 1.1 z y x F1=100kN F2=120kN 30° 30° 45° 30° 75°
SULIT MOCK PSPM - EB025 3 SULIT FIGURE 1.2 Assuming the beam is in equilibrium and its weight is neglected: (i) Draw the Free Body Diagram of the beam. [5 marks] (ii) Calculate the support reaction forces of the pin and cable. [12 marks] 2. Answer the following questions: a) Explain the Kirchhoff’s Voltage Law and Kirchhoff’s Current Law. [5 marks] A 50kg 8m 2m B C
SULIT MOCK PSPM - EB025 4 SULIT FIGURE 2.1 b) By using Kirchhoff’s Law, Determine the value of i1, i2, and i3 of the circuit in FIGURE 2.1. [12 marks] 3. A series AC circuit in FIGURE 3.1 consists of the resistor, inductor and capacitor. a) With the given value, calculate : (i) Impedance of the circuit, Z. [4 marks] (ii) Current flow through the circuit, I. [2 marks] (iii)Voltage drop across R, L and C. [6 marks] b) Sketch phasor diagram to show the current I, voltage supply VS and the angle of the circuit.
SULIT MOCK PSPM - EB025 5 SULIT [4 marks] FIGURE 3.1 4. a) Civil engineering profession involves multiple sub-disciplines and deals with the design, construction and maintenance of the built environment involving nature and public safety. Explain TWO (2) problems and challenges faced by civil engineers other than environmental and safety. [4 marks] b) FIGURE 4 shows the deformation of a simply supported beam subjected to loads
SULIT MOCK PSPM - EB025 6 SULIT FIGURE 4 (i) Compare the properties of concrete and steel with respect to strength in tension and compression [4 marks] (ii) Briefly explain the consequence of strain in the upper and lower parts of the beam, and the function of reinforcing steel. [8 marks] 5. FIGURE 5 shows a simply supported beam loaded with point loads and uniformly distributed loads (UDL). From the figure above; a) Calculate the support reactions at point A and B. [6 marks] FIGURE 5 A B 4m 2m 1m 3m
SULIT MOCK PSPM - EB025 7 SULIT b) Sketch the shear force diagram of the beam. [4 marks] c) Sketch the bending moment diagram of the beam. [4 marks] d) What is maximum shear force and bending moment of the beam? [4 marks] END OF QUESTION PAPER KERTAS SOALAN TAMAT
SULIT MOCK PSPM - EB025 8 SULIT SKEMA MOCK PSPM II EB025 22/23
SULIT MOCK PSPM - EB025 9 SULIT = No Marking Scheme Marks 1 (a) a) i) Resolve the magnitude of forces F1 and F2 along its component F1 = 100kN F1x = 100{cos 75} i = 25.88i F1y = 100{cos 30} j = 86.6j F1z = 100{cos 45}k = 70.71k F2 = 120kN F2x = 120{- cos 30 cos 30} i = -90i F2y = 120{- cos 30 sin 30} j = -51.96j F2z = 120{sin 30}k = 60k Total = 6 marks 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark ii) F1 = {25.88i + 86.6j + 70.71k} kN F2 = {-90i - 51.96j + 60k} kN Total = 2 marks 1 mark 1 mark iii) FR= F1+ F2= FRX+ FRY+ FRZ FR= {25.88i + 86.6j + 70.71k} + {-90i - 51.96j + 60k} FR = {-64.12i +34.64j + 130.71k } kN |FR| == 2 + 2 + 2 |FR| = 149.6564. 12kN2 + 34. 642 + 130. 712 α = − 14964 . . 6512 = 115.37° β = 14936.64.65 = 75.83° γ = 130149..7165 29.14° Total = 8 marks ½ mark ½ mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark
SULIT MOCK PSPM - EB025 10 SULIT 1(b) Each force = ½ mark x 4 = 2 marks Dimensio n = 1 mark Angle = 1 mark Axis = 1 mark ii) Support reaction forces Total = 12 marks 1 mark 2 marks 1 mark 1 mark 1 mark 2 marks 1 mark 1 mark 1 mark 1 mark
SULIT MOCK PSPM - EB025 11 SULIT 2 a) Kirchhoff’s Current Law (KCL) Total current entering a node equal to Total current exiting the node. ∑IIN = ∑IOUT Kirchhoff’s Voltage Law (KVL) In a closed circuit, total voltage is equal to zero ∑V = 0 2 marks 1 mark 1 mark 1 mark b) ∑IIN = ∑IOUT i1 = i2 + i3 ……………………..(1) KVL Loop 1: 21V = 4i1 + 3i2 ………(2) 2 marks 1 mark
SULIT MOCK PSPM - EB025 12 SULIT Loop 2: 9V = 2i3 – 3i2 ………...(3) (1) into (2) 21 = 4(i2 + i3) + 3i2 21 = 7i2 + 4i3…………………(4) (4) – (3)x2 21 = 7i2 + 4i3 - [9 = 2i3 – 3i2 ] x2 3 = 13i2 I2 = 0.23A I2 into (3) 9 = 2i3 – 3(0.23) i3 = 4.845A i2 & i3 into (1) i1 = 0.23 + 4.845 i1 = 5.075A 1 mark 1 mark 1 mark 2 marks 2 marks 2 marks
SULIT MOCK PSPM - EB025 13 SULIT = = (0. 631 − 65. 36 )(318. 31 − 90) 3 a) (i)Given R = 220 Ω, L = 2 H, C = 10 μF = 2π1== 2π2π(50(501)(10)(µ2) )==318628..3132 = 2π Impedance, (ii ) = === 220220380+ . ++14 ( 54 (310628− . 64. 01. 32 ) − 318. 31) = 240.14− 12054 .64 = 380 (iii =)0. 631 − 65. 36 = ==(1380. 631. 82 −− 6565..3636 ) (2200) = ==(0396. 631. 524 −. 6465 .36 )(628. 3290) Total = 12 marks 0.5 m 0.5m 1m 1m 1m 1m 1m 1m 1m 1m 1m 1m 1m
SULIT MOCK PSPM - EB025 14 SULIT = 200. 85 − 155. 36 b) Total = 4 marks Correct Axis : 1m Current I (Arrow and label): 1m Voltage VS (Arrow and label): 1m Angle θ: 1m 4 a)
SULIT MOCK PSPM - EB025 15 SULIT b)
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SULIT MOCK PSPM - EB025 17 SULIT 5