School of Civil, Environmental & Mining Engineering Practice test 1: Axial loading & Torsion (Chapters 1-3) CEME 2001: Strength of Materials Important note: This is a practice test to prepare for Test 1 on 9 April 2021. There should be no assumptions that the same types of questions will be asked in Test 1. INSTRUCTIONS TO CANDIDATES 1. This is an OPEN book test. 2. This is a 60-min test (including the reading time). 30 minutes are added to take into account the time (and any potential minor issues) for downloading questions, scanning and uploading solutions on MyUni. 3. All questions carry equal marks. 4. Clearly state your procedure and show all the calculations. A correct answer without any details will not result in any marks. 5. If you believe that additional information is required, clearly state your assumption and proceed with your solution. 6. Solutions should be uploaded as a single PDF file on MyUni. You can use CamScanner to convert photos taken using your smart phone (either Android or iPhone) to a single PDF file with good quality. The free version is sufficient for this purpose. Please install and try it a few times before the quiz. Each question will be marked following the below criteria: x Correct approach and answer: 100% mark x Correct approach and answer but wrong unit: 81% – 90% x Correct approach but wrong answer: 31% – 80% mark x Incorrect approach and answer but demonstrating a certain level of understanding: 1% – 30% mark x Incorrect approach and answer, and demonstrating no understanding: 0% mark Formulae you may find useful: Axial Loading: A P V , L GH , EA PL G Torsion: T J UW , GW J , L UIJ , JG TL I , For a solid circular cross-section with radius r: 2 4rJ S For a hollow circular cross section: 4 41 outside inside2J r rS Page 2 of 2 Question 1 (50 marks total): The rigid beam ACE is subjected to the load P and suspended at points A, C, and E by three suspender bars AB, CD, and EF (see Figure 1). These three suspender bars are made of the same steel and have the same circular cross section with the same diameter of d=30 mm. The ultimate stress of steel is Vu=450 MPa and Young modulus is E=200 GPa. (a) Using a factor of safety FS=1.5, determine the maximum allowable load P so that the stresses in all three suspender bars will not exceed the allowable stress Vall (Vall = Vu / FS) AND the maximum deflecti
on will not exceed 2 mm. (40 marks). (b) If bars AB, CD, and EF have different diameters, what should you do to determine the maximum allowable load P Explain without calculation. (10 marks). Question 2 (50 marks total): An engineer is in charge of the design of a cylinder (length L=150mm; Figure 2) under torsion. The cylinder is fully fixed at one end and subjected to a torque T = 5 kNm at its free end (Figure 2). The shear modulus of the material is G = 200 GPa and its ultimate shear stress is Wu=280 MPa. Preliminary design: At first a solid circular cylinder with radius of 25mm is used (Figure 2). Could you help the engineer determine: (a) the maximum shearing stress, and the maximum angle of twist in the cylinder. Determine the factor of safety (FS=Wu/Wmax). Sketch the variation of angle of twist, and shear strain along the length of the cylinder. (15 marks). (b) the shearing stress at point A (see Figure 2: point A lies on a 10-mm-radius circle drawn on the free end of the cylinder). (5 marks). (c) the percent of the torque carried by the portion of the cylinder within the 10mm radius. (10 marks). Optimisation: (d) The engineer realises that the factor of safety in question 2(a) is still relatively high and hence he (she) wants to reduce it to an acceptable value of FS=1.1 to save material, by using a hollow circular cylinder with the same outer radius of 25mm. Could you help the engineer determine the inner radius How much stress increase (in percentage) compared to the percentage of material saving What are your conclusions (15 marks). (e) Still not happy with the material saving, the engineer wants to use a different hollow circular cylinder with different outer radius, so that when subjected to the above torque of 5 kNm the maximum angle of twist and maximum shearing stress are still equal to those calculated in question 2(a). Is it possible Explain without calculation. (5 marks). 1 m 1 m 2 m P 0.5 m 0.5 m A B C D E F Figure 1 T = 5 kNm Point A Figure 2 AB-EF-ZCD-2EFgzty-nl-Fg-FEF.pe 。 DA-EE.cnEMA-P-5-IFDXltFEFXZ-t-DXSAB.SEF -矼 千 繓 2獸獸翺 丽 那 后 号 -_- FEF 邮 anmna 阳开么 Y 的Bfty 崝 __i P=363 kNAO.se znn 譽 t7FZ42KN.TMPa- ep.hr a) t =等 ,TmaxzTJ-fnc4.T-5XHN.mn _嗤 叫” ” 0 狒 加 90万 = 6- 6.4 5 i , =等 , 经 1mm C) Jitǚ t ‘是 , 丽 在11 2 景 T.nu哥 , Tzh.zhy.sc modswyi 些 xlin.zixlznuy.im/,TMz-Ta1l1td.-