Question

1400 4 25 The hoist is comprised of three separate parts: a continuous material from A to Bacon material from B to Cand a hydraulic cylinder that connects these two members at points and E. You can treat point B as a pin joint and point as a fired joint (it can't move The hydraulic cylinder has an outer diameter of 15 mm and a thickness of 3 mm (Figure 1). An average car engine weighs 150 kg Joe requires the following information for an average car engine. Joe's livelihood depends on your calculations and has asked for you to show your full working so he can verify your calculations with another engineer. 1. Draw the FBDs of ADB and BEC (simplify any two force members) (2 marks) 2. Calculate the reactions at pin B (horizontal and vertical) and at the fixed point C. (1.5 marks) 3. What is the stress in the hydraulic cylinder DE? (0.5 marks) 4. The hydraulic cylinder is made from AL 5052 which has a cy=176 MPa, what is the safety factor? (0.5 marks) More questions or the page Joe is servicing the car of a very special client. This client drives a Bugatti Veyron which sa W16 engine. This engine weighs 436 kg Joe wants to be sure that his engine host is safe couch to hold this engine that is worth over $20,000 5. Given this added weight, would you recommend that Joe uses the host to lift the engine? (1 mark) Justify your answer by calculating the stress in the linder and the safety factor. Use who yowahusay factors wer e recommendation 6. Comp l etion List the ways that you could increase the safety factor of a member in your frame machine without decreasing the force it experiences. (5 marks) o ur

Answer 1

Page 1 Saathi Date 0. 1 of BD of ADB 8 K 1400mm — Sk 1925- A DoT 1985 B RBX 425 mm i D po 75017 B 105-0/15 (mxg) = 150x9.81 = 1471,5 N Broso 710mm 425 using sine Rule, ADEB, BE = DB I 710 = 425 Sino sin (180-75-0) sino sin (105-0) SinC 105-0)-710 → Sin 105 C050 – COS105 Sin = 710 Sino 425 ? sino sin (105)ceto – cos (105) = 210 => [coto= 1,462 0 = 34.380 do FBD of BEC 425 So 1 FFD No // 710mmy 759 1019 mm E 345mm 333.2 89.29 183.76 mm mm Reaction at B & c. from FBD of ADB ] { fx=0; Efy=0; {Mam 70 point Page No.

Date Page ① Gaathi Eme=0 Sum of moments atpoint B=0 (Roxsino.x425) - (FAX 1825) 20 - @ R Since) Rsid(34.38°) 425] - 1491.5x 1825] =0. ( Roc 1471.5 X1825_ 11190.06N 425xsin (34.38 M S & RD=11190.06 N {Fx=0; – Rp cos(34.38) + Rgx=0 : Rex = 11190,06 COS(34.88) Page = 9235.31 Efy zo ; -1471.5+ Rosin (34.38) - Roy=0 Roy = -14771.5 +611190,06 376 34.38) Roy = 4847.39 Resultant & Reaction at point B = √ Retro y = 9235,3²+ 4847.8² Re10430,10 e from FBD OF BEC A s fo=fo=11190.06 ry EM@ =0; -fesino )*(83.76)+ Fyx(273.05) +R (1019)=0 Fry = 11190.06X183.76x Sin34.38) * - Rox(1019) 22.273.05 209 L es 273.05 I fcy=4252.48 – 3.73 Rcx - ③ 21:20 Efx=0; -Foxt Ecoso - Rea 20 - Fox + (11190,06 x COS(34.38)) - 9235.3=0 Fex=0 M substitute in ean . Fey a 4252.48M a Reaction at Point Rc =4252.48N). Page No.

Date __ I_ I_ Page 6 Gaathi New Rose Byt - 176 -0.6119 287.59 New fost=0.6119 ie 6 x6 which is Tor Not Theofore Bugati Veyron Engine - sate Design should not be lifted by hast as the man stress induced in Cylinder DE is greater them but (Allowable strength , so it will fail the Component immegladly. Recommendation - 0 Increase the cylinter Area Use High strength material 5 Minimum Ros should be @ » As we in increase Acea, stress ces'll chosate because 6 = force 62 acca As we in increase strength of makcial, we can increase dos a because, a induced balconhte.