cb68
Mechanical
- Jul 5, 2003
- 24
I originally posted this on 'Mechanical engineering other topics'. I have been asked to now post on this forum instead. Apologies if anyone has already read.
I have a slender square aluminium (6082 T6) column that is 44mm square and 160mm long. A load of 2250N is taken through the centre of the column in the 160mm direction. I have clearance holes for M5 A2 Class 70 stainless steel studding in the 4 corners of the column drilled the full length of the 160mm. The centre of each hole is mounted 5.4mm in from each respective edge.
The column is mounted on a flat stainless steel 316L plate with four tapped holes of 12mm full thread depth into which the studding is screwed and held with loctite in position. The other end is stainless steel flat washer and A2 nut.
I have assumed:
1. I am torquing the nuts to 3Nm.
2. Each stud carries 1/4 of the load ie 562.5N
3. That the stiffness of the studding is greater than the member stiffness. In fact I have said that the stud carries the full load.
So
Using T=0.2Pd where T is tightening torque, P is tension induced by torque and d is nominal diameter I end up with a P of 3000N. Add the external load of 562.5N to 3000N = 3562.5N. The tensile stress area for an M5 is 14.2mm2. Therefore stress is 250MPa. Now A2 stainless has a Proof Stress of 450MPa giving me 56% of proof stress.
Now I have heard rumours that for a joint that is dissassembled you can typically run up to 75% of proof. Can anybody shed some light on this in particular with relation to A2 stainless?
Next I thought would the studding strip from the 316L plate. Now Juvinall - Fundamentals of Machine Component Design presents a formula F=pi.d(0.75t)(0.58Sy) which is the tensile force to strip a nut of thickness t & nominal diameter d. With an Sy of 240MPa for 316L I end up with an F of 19,700N which gives a big safety factor of 5.5.
As I am not an experienced stress engineer do these assumptions and calculations seem respectable? Is there any other failure mode I have missed (eg should I be checking the nut)?
Thanks
I have a slender square aluminium (6082 T6) column that is 44mm square and 160mm long. A load of 2250N is taken through the centre of the column in the 160mm direction. I have clearance holes for M5 A2 Class 70 stainless steel studding in the 4 corners of the column drilled the full length of the 160mm. The centre of each hole is mounted 5.4mm in from each respective edge.
The column is mounted on a flat stainless steel 316L plate with four tapped holes of 12mm full thread depth into which the studding is screwed and held with loctite in position. The other end is stainless steel flat washer and A2 nut.
I have assumed:
1. I am torquing the nuts to 3Nm.
2. Each stud carries 1/4 of the load ie 562.5N
3. That the stiffness of the studding is greater than the member stiffness. In fact I have said that the stud carries the full load.
So
Using T=0.2Pd where T is tightening torque, P is tension induced by torque and d is nominal diameter I end up with a P of 3000N. Add the external load of 562.5N to 3000N = 3562.5N. The tensile stress area for an M5 is 14.2mm2. Therefore stress is 250MPa. Now A2 stainless has a Proof Stress of 450MPa giving me 56% of proof stress.
Now I have heard rumours that for a joint that is dissassembled you can typically run up to 75% of proof. Can anybody shed some light on this in particular with relation to A2 stainless?
Next I thought would the studding strip from the 316L plate. Now Juvinall - Fundamentals of Machine Component Design presents a formula F=pi.d(0.75t)(0.58Sy) which is the tensile force to strip a nut of thickness t & nominal diameter d. With an Sy of 240MPa for 316L I end up with an F of 19,700N which gives a big safety factor of 5.5.
As I am not an experienced stress engineer do these assumptions and calculations seem respectable? Is there any other failure mode I have missed (eg should I be checking the nut)?
Thanks