Question

450 300 300 270 300 The motion of backhoe bucket is controlled by hydraulic cylinders CB, EC and GF. A 12 kN force W is exerted on the bucket vertically, as shown. To determine: 300 450 150 d) Draw FBD of all parts. e) Hydraulic force CB f) Hydraulic force EC g) Hydraulic force GF imensions in mm 630 B@ Hints: Two force member (2 decimals number for calculation)

Answer 1

Solution: (d) F.BD of all separated parts will be: 300 270 ht -FX A For 300 fuso + 150 200 *Fx 300 ,F goo mor. E FEC FEE sinus 450 FEC Y 45 fig(a) FEC Cos45° 450 300 2- 630 mm DE=480²turo? DE =45012 Fig (6) A FCB W=124N Ау fig (c) B A Ax 1270 300

In fin (c) Taking Moment about Aj considering 2 (Feb * 270) -( 12x300) =0 Fcb = 12x300 270 13.33 un Fag {fy=0; -FEB-12 +Ay=0 Ay = 13.33 112 Ay= y = 25.33 UN (t in fiale) it in fig (6) Ax=0 no hori Zontal force is acting this part. If x=0; since on (8) In fig (6) Talling Moment about fi considering ? - Ayx ( 300 1270) - (Fec Signs x 300) + (Fee Cos45x900) + FB x 300 =0 Fee (goo cosus -sin 45 300) = (25.33x570) FEL - (13:23*300) (2633x570) - (300x1333) (900-300) 1 / FEL = 24.61 KN

{fx=0; - Fx + FEccosus o Ex = 24.61 cosuso (& in fig (6) so in fing (a)) Ify=0, Fy + FB - FEC sinus - Ay=o Fy= 24.61 sinusot 25:33 -13.33 Fy = (24.61 sinus + 12 ) ( in fig (b) & infig(a)) 8) In fig (a) Taking moment about Dj considering to (FECT DE) x750) - (Fy x750) - (Fxx750) (24.61 sinus + 12) x750 +(24.61605 45x750) 24.61x45012 + (For + FLE 750 FGF = 25.92 UN (Answer)