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Yes the engineer is making call that deeper testing is not required (so no revisit required) Does the investigation need to be in same spot as the intended piling location?

Hi, thanks for reply.. There was a failure with rig, looking at results it got to clay before failing at 1.5m (I mean the rig broke down!) Believe the pile will be double that.

Hi Electrical Engineer here, I have question regarding borehole sampling. What depth should borehole be tested to for piling? The expected depth of the pile?

I meant for a clevis type joint on the end

What is the recommended end connection for a tube that is taking a compressive load? It is better to thread the inside for a taper thread or use shear bolts? Is there any guidance on this? thanks John

no x is static, it can be moved for setting up, but after that it remains 'as is'. I know that angle thetaA is probably 5 to 10 degrees at the most so FAy is relatively small, similarly with thetaB.

ok rb1957, I have effectively done what you have said - I have summed the moments about B to arrive at a value for FC Removing FAx is to help me investigate the suitability of the crank pin. Again, I agree, you can't have an off axis load through a pin - I do learn rb1957! Have faith ;-) I do...

zekeman - The mechanism spends majority of its time in static equilibrium. Your last comment is bang on ist Euler buckling due to single compression, probably very worn cranks and friction - I just wanted to understand the magnitude forces in the rod. I understand Euler formula is for ideal...

rb1957 - not sure what you mean by 'datums of convenience', which datum should I use? Do you mean instead of referring to say FCy I should in fact state 'Fc Sin thetaB'? I have swapped the labels around for FCy and FCx - schoolboy error! Okay Terio and rb1957 I'll take it as read that FCy and...

hi MintJulep, I am lookihg for say a higher Young's modulus than steel as that will reduce the CSA, weight not so important. The materials being listed are probably too exotic for our use. Which is the best steel which is in general use?

hehe no, it is for work,

Terio - would I include FCy since it is not acting on B directly (B is a pivot), but is a moment around B?

Dear All Would you have any suggestions for a material with high Youngs modulus but is relatively light, mild steel has 210GPa but is roughly 7850kg/m3 - I suppose I am into Carbon Fibre territory? I need to keep the CSA fairly small. regards John

rb1957, is this okay for the FBD? http://files.engineering.com/getfile.aspx?folder=e04d6f73-a87f-4850-81da-5a9e9f275582&file=img003.jpg http://files.engineering.com/getfile.aspx?folder=e04d6f73-a87f-4850-81da-5a9e9f275582&file=img003.jpg

hi rb1957 - x is variable to suit the installation (a sleeve on a crank arm) I have the reactions for FA as FA Vertical and FA Horizontal similarly for FC (the rod) FCV and FCH - all relative to the crank. The moments (FCH and FCV) don't act on the centre pin. I have drawn the pin reaction...

Just a few obvious ones James Fails to maintain required pressure, cavity pressure >0.5 bar Fails to maintain required pressure, cavity pressure <0.5 bar If its off the shelf do you really want to go to component level of the regulator - if so why not ask the manufacturer? regards John

just to be a pedant, I need to point out in the summation of moments bit: one should be negative so that it becomes positive value when moves to the other side of the equation.

hi desertfox - yes thankyou I have studied the example and the sketch very carefully and tried to use the same principles to produce the free body diagram for the pivot of the first crank - which gives me the force in the connecting rod. hi rb1957 The Euler equation bears what we are seeing. I...

Yes it is bending and its due to Euler type compression forces, I have based calculations for sizing the new rod on the output force so say it delivers a maximum of 6kN to the back then that is what I have based my calcs on with a Factor of Safety of 4. I just need to develop the free body diagram.

thanks rb1957, you make a good point that the input and output forces are also not perpendicular to the crank arm, again it will be a case of resolving horizontal and vertical components relative to the crank arm, with the horizontal force being reacted back (lost) at the crank stud. I am with...