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red/orange is corrosion.

we had taper roller bearing used in entry guide roll failed as shown in photograph, entry guide roll too has broken from one of it's collar . do suggest possible reasons. http://files.engineering.com/getfile.aspx?folder=f5637878-24df-4f0d-8b6a-66fd0ba08d3e&file=taper_roll.pptx

YES it's a corrected gear , with addendum modification coeff as = 0.33

3DDAVE PCD OF TWO PINS - 450 MM . SO RADIUS = 450/2 MM .

THANKS

PCD OF TWO PINS - 450 MM . SO RADIUS = 450/2 MM .

HOW TO FIND BASE TANGENT LENGTH OF DOUBLE HELICAL GEAR WITH 199 TEETH .NORMAl module - 8 , pressure angle - 20 degrees , formula and calculation thanks

HOW TO FIND BASE TANGENT LENGTH OF DOUBLE HELICAL GEAR WITH 199 TEETH .NORMAl module - 8 , pressure angle - 20 degrees , formula and calculation thanks

OK

SOME PHOTOGRAHS ARE SENT http://files.engineering.com/getfile.aspx?folder=7a65a1e5-62a5-49a6-8596-c7dd254a4cfa&file=SHEAR_PIN__BREAKAGE.pptx

oh,yes i was thinking about safety factor only ,i needed clarification, for the word effective influence . thanks and regards

rolmec what does effective influence of 1.5 means , which you have taken in your calculation .

Since the pin was very tight inside t he bush , with great difficulty it came out ,so freeting ? if their is looseness between the pin and bore than freeting will come into play , if i am not wrong .

Given nominal torque & geometrics etc.: 9550*400*2/(306*0,395*452,4) = 140 N/mm² What does 452.4 inidcates i nthe above equation ? do suggest and what does 140 N/sqmm indicates ? Given fy with 235 N/mm² & γF to 1,5 & shear: 235/(1,5*sqrt(3)) approx. 90 N/mm² give the specific formula ...

if we take endurance strength of material for ST -37 material ,it is = 222 N/ sqmm . Than breaking load on pin = stress x notch area = 222 x 3.14 /4 x 24 x 24 since notch dia is 24 mm = 81388.8 N =81.3888 KN...

Motor torque = 9550x KW X SF /RPM = 9550 X 400 X 1.5 /306 = 18725 Nm Since Max Torque = 110 % of mean torque ( as set from elect side ) SO MAX torque - 1.1 x 18725 - 20597.5 Nm Shear load on pin = T X 2/D = 20597.5...

actually pin is not taper throughout on one side pin dia is 32 mm and at the other side it is 35 mm . regarding marks on left side of pin ,it happened while taking out the pin .since pin was very tight inside bush .

I will send you the photographs of inside face of pin as well as that of coupling ,can't get the torque value as the drive system shows only motor load ,which was nominal at that time . as you have written,pin is bent , from where ? i am also interested in the mathematical part also ...

what extra information is needed do suggest http://files.engineering.com/getfile.aspx?folder=b1f114f0-ab29-4845-aa0e-2c686cb25638&file=DSC01117.JPG

PCD of gear coupling is 395 mm , it's a taper shear pin on one side dia is 35 mm and at the other side it is 32 mm . pin was tight inside t he bush, so no question of radial play inside the bush ,but still it failed .