Question

solve for the reactions in each journal bearing below

Solve for the reactions in each Journal Bearing below. AM 400 N 0.6 m CB 600ND 0.6 m 0.6 m 0.4 m

Answer 1

Given data in the form of free body diagram which means removing all supports & representing their respective reactions on the figure. ARBZ AMAZ A orsun TRAY ARAY 660 TO 0.6m RBX (6.6m XRcz 0.4m by RCY Rcx Initially consider all reactions in positive sence Here Ray , Roy and Roz are axial forces and these are neglected as it is assumed as rigid body. Now Applying equilaborum equations, All horizontal in direction of x - Ifx=0 L RAXT RBX TRcx=0 In direction of Y Ify = 0 Rcy +400 =0 Đ Rcy=-400NANE ② In direction of z Ef=0 RAZT RRZ +60o=0

considering moment equiliborum equatron Along x-directron, ie Mx=0 Moment acurs due perpendicular forces so consides the forces which are perpendiculat to x-axis and Ilene of action donot pass through the origin ] for this above equation forces applicable are RB2, GOON, Rey ZMx = 0 Cclockwise negative RBZ XO 6+600x 1.2 + Rey xo.420 Anticlockwise positive RBYX0-6 + 600 X 1-2 - 400x0.4 20 Roz=-933.33N Ans from ③ RAZt-933.33) +600 =0 RAZ = 333.33 N Ans Ft

Here RBX as Now along y direction Applicable forces are Goon, Rex Myco - 600 XO 6 -Rex X0-4 = 0 - Rcx = -900N ans perpendicular but there is no datance so no moment generated by -Yana you Now along z-direction Applicable forces are RBX , Rex , Rcy - RBX XO.6 - Rcx x 1.2 + RcYXO-6=0 - RBX X0-6-6-900)*12 + (-400) X0 62 à Roy = 1400 N ANS from Raxt RByt Rex=0 Por Rex

from Raxt Rext Rex = 0 Rox+ 1400 + (-900) = 0 - Rox=-500N ans Reaction are RAX=250 ON RB X= 140ON RREX = -900N RAZ= 333.33 N RBZ = -933.331. Rcy = -400 N RAY, RBY , RCZ are zero since is it is rigid.