Determination of Maximum Pressure Assignment Help

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Determination of Maximum Pressure

It is essential to establish the points of action of the resultant normal forces F1, F2 and F3 on the corresponding faces for calculating the maximum pressure. The distance among the point of action of normal force F1 on the flat slideway II and the center of the carriage is denoted by xA . This is shown in diagram. The distance among the force F2 acting upon the vertical face of flat slideway II and the center of the carriage is indicated by xB , and the distance between force F3 acting upon horizontal face of flat slideway II and the center of the carriage is denoted by xc . For ascertaining xA , xB and xC , we have two equilibrium conditions :

Equation

∑ My  = 0

Fx h + Fz xp  - F1 xA  - F3 xc  + RzQ  = 0

Equation

∑ MZ  = 0

Fx yp + Fy xp  - RyQ  - F2 xB  +  μF2 ((l + u)/2) + μF3l = 0

An additional eq. may be written by assuming that the moment of reactive forces that are F1 and F3 about the Y-axis is proportional to the width of the slideway face, that is

Equation

F1 xA / F2 xc  = v/u

On solving above three Equations we get, the values of xA , xB and xc.

The ratio of  xA /L, xB /L and xc /L computes the shape of the pressure distribution diagram and the maximum pressure on a specific face of the slideway. The process for determining the maximum pressure on flat slideway I is being subjected to the normal force F1 that is described below. The most common case of pressure distribution along the length of contact L corresponds to a trapezoid as displayed in diagram.

1011_Determination of Maximum Pressure 1.png

Figure: Trapezoidal Pressure Distribution along Slideway Length

Force F1 acts at the center of gravity of trapezoid. The distance yc at the center of gravity from the larger arm of the trapezoid can be ascertained as:

Equation

1267_Determination of Maximum Pressure 2.png

∴ yc = (L/3) ((pmax + 2 pmin) / (pmax + pmin))

As a result,

Equation

xA = L/2 - yc

∴ XA = L/6 ((pmax + pmin) / (pmax + pmin))

Now xA / L < 1/6

In this case the pressure distribution diagram presents a trapezoid. This is described with the help of following illustration.

Let (xA / L) < (1/10)

From equation we attain,

Pmax = pmin = 6/10 (pmax + pmin)

Pmin = 0.4 / 1.6 pmax

From above equation, it is clear that pmin is a positive, non-zero value. Hence pressure distribution diagram must be a trapezoid.

Equation

∴ pmax   + pmin  = 2 pav

From Eq. (16), we get,

Equation 23

Pmax = Pmin = (12xA / L) Pav

On solving above two Equation

Pmax = pav [1+ (6xA / L)]

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