Recent content by Meson

For either a bore or a shaft alone, there is only maximum tolerance calculated with the formula given in the previous post. For a bore-shaft system, for example 60H7/r6, then there are maximum and minimum tolerance. In this case, the maximum tolerance is the maximum tolerance of the bore...

The maximum tolerance is ES - EI = IT that is the base tolerance 19 micrometer. It is the same as taking maximum diameter - minimum diameter = IT = 19 micrometer in this case.

yes, and to do so the maximum and minimum deviations have to be calculated as in ES and EI.

Normally, we have three tables: i) Base Tolerances Table ii) Base Deviations of Shafts Table iii) Base Deviations of Holes Table all are in micrometer and are of DIN-ISO 286 T1. Steps to determine the max and min tolerance: From 60P6, we know that it refers to a hole with inner...

This is because "Inner diameter of the socket (tube) = 145 unit = outer diameter of the shaft." Hence, there is no point repeating 145 twice. [SIZE="1"]By the way, why are you posting machine design problem here in Advanced Physics forum?:smile:

I don't have the formal definition with me, but for quick reference here we go: Capital letter refers to sockets or holes, while small letter refers to shafts. Remember this form: Socket/Shaft

They can be identify quite easily as the following: If there is a very long section after yielding point, which we call the plastic deformation section, existing in the curve then it is ductile. If the curve is basically just a short section of straigth line obeying Hooke's law followed...

Bingo. How did you obtain the W? The correct axial section modulus W is W=\frac{\pi*d^3}{32}, assuming that the cross section of the axle is not hollow. Setting \sigma_{all}=\frac{\sigma_{lim}}{n_{s}}=\frac{M_{b}}{W_{by}} and solve for the diameter d as you have found M_{b} in the first...

Your answer isn't correct. Try writing down the moment at C. Then we discuss the rest of the question.

Well, heaviside functions aren't needed here. Step 1: Determine the value of F_{1} using equilibrium condition along the vertical (y-axis). Step 2: By considering half part of the rotating machine (cut machine into two at the centre), write down the bending moment equation of either side...

You have to show your homework before anyone here discuss the solution with you. [SIZE="1"](You have been told to do so MANY times :smile:)

What do you think? Try to form the equation on your own and we will inspect it.

Be careful, recommended to draw the cross-section when evaluating the inner diameter for a clearer view. d_{i}=d_{a}-2(\frac{d_{a}}{12})=\frac{5}{6}d_{a}

Also worth mentioned that the bending stress distribution is best to be written in the form \sigma=\frac{y}{C}\sigma_{max} where the ratio \frac{y}{C} is emphasized. Thus y can takes the following range of values: -C < y < C with + indicating tensile, - indicating compression stress...

Good question. The limit stress, by definition is the stress a particular material can withstand before it a) exceeds the yield strength (yield point) OR b) exceeds the tensile strength (also known as breaking point, ultimate tensile strength: UTS) However, for most of the machines, we do...