TORQUE AND THRUST FORCE IN DRILLING

CIVIL_ENGINEERING

INTRODUCTION:

Drilling is probably the most important conventional mechanical process associated with chipboard processing. In the furniture industry, for instance, large quantities of holes have to be drilled due to the use of connections, handles and hinges. A considerable part of the current research effort in this field is still being devoted to major process-optimization issues such as the most appropriate cutting parameters or tool geometries. Chipboard drilling require different process  parameter optimization approaches: in the former process, the smoothness of the surface processed and tool wear are equally important; in chipboard drilling, the former parameter is prioritized over the latter given the difficulty to drill laminate without producing unacceptable cracks.

 

A suitable model would assist in a focused selection of the most appropriate feed rates, spindle speeds and geometrical cutting tool shapes. A detailed review of dynamic cutting models is provided in Ehmann et al. [1]. The study of drilling has often presented some difficulties which are linked to the complex geometry of the twist drill (Fig. ). In practice, generally  empirical equations are used to calculate thrust force and torque. These equations are very approximate, because they do not take all the cutting parameters into account. They often use only the feed speed and the diameter of the drill

View showing geometric data of a twist drill.

 

Few theoretical works have been undertaken on drilling. Bera and Bhattacharya described the first attempt to use a cutting model to determine torque and thrust in drilling.

 

They analyzed the whole drill and considered that the chisel edge acted as an indenting tool and the lip as a cutting tool. They assumed that the resultant force per unit length of the lip is constant.

 

They assumed that the resultant force per unit length of the lip is constant. Williams [3] recognised the significance of the feed on the resultant velocity and on altering the cutting geometry. In making predictions of torque and thrust, Williams argued that a portion of the drill acted as an orthogonal cutting edge because the cutting velocity is assumed to be perpendicular to the cutting edge.

 

In 1972 Armarego and Cheng [4] proposed an approach to predict thrust and torque during drilling for a conventional drill and a  modified drill in order to simplify the calculations. The method of calculation used the orthogonal cutting model and the oblique cutting model, and was also used in 1979 by Wiriyacosol and Armarego [5]. Basically, this method consists of dividing the cutting edges  into a limited number of cutting elements.

 

These elements were assumed to be oblique cutting edges on the cutting lip and orthogonal cutting edges on the chisel edge. The calculation used empirical equations established from orthogonal cutting tests. In most of the methods mentioned above, the major problem was to choose the number of cutting elements, and to  determine the empirical equations for some cutting parameters.

 

More recently Watson initially used practically the same method, with a different geometry. He developed a model for the chisel edge and the lip from the orthogonal cutting model and the oblique cutting model, respectively. The author initially used the same principle which consisted of dividing drill edges into a number of elementary cutting edges. Watson  recognised that the chips front the lips and the chisel edges are continuous across their width and that continuity imposes a restriction on the possible variation of the chip flow angle across those edges. Other works have been interested in particular drilling operations, such as deep hole drilling , using an experimental model,  and drilling with a three-cutting-edge drill the models for drilling presented above were based on experimental measurements.
 

MODEL OF DRILLING

CIVIL_ENGINEERING

where  Kr is the half point angle of the tool. ¥ _r ,  ω and  η are intermediate angles, and are calculated from the geometrical parameters of the drill by:

where w is half of the distance between the two lips, and δ is the helix angle at point M, given by:

and δo is the helix angle at the periphery of the tool. All the parameters quoted above depend only on radius r of point M from the drill axis and on the global drill geometry at this point. By using r and the geometry of the drill, an expression for the length of the lip can be obtained,

where d1 is the length of the chisel edge and τ is the angle between the chisel edge amid the lip d1 and τ are shown in Figure The differential element dl for the length of the cutting lip  is given by

View of forces on the lip

At each cutting point M, the cutting depth is obtained in terms of the feed speed f and the cutting geometry at this point. The following formula for t1 can be established:

Determination of the increments  dF_l and  dMpr at point M (Fig. ) is undertaken using the oblique cutting model established by Oxley. The basis of this model is to analyze the stresses along the shear plane and the tool/chip interface so that the resultant force transmitted by the shear plane and the interface are in equilibrium. The thrust force element dF_l and the torque dMpr are determined in terms of the differential force dF_C in the direction parallel to the cutting speed Vc, the differential force dF_T in the direction perpendicular to the cutting speed and to the cutting edge at point  M, and the differential force  dF_R in the direction perpendicular to dFC and dF_Tas shown by Fig.  The differential element of shear force dFs is given for an element of the lip dl by:

SYMMETRICAL TWO HINGED ARCH

CIVIL_ENGINEERING

Solving for H , yields

 

 

Using the above equation, the horizontal reaction  H for any two-hinged symmetrical arch may be calculated. The above equation is valid for any general type of loading. Usually the above equation is further simplified. The second term in the numerator is small compared with the first terms and is neglected in the analysis. Only in case of very accurate analysis second term s considered. Also for flat arched,  θ ≅1cos as θ is small. The equation is now written as,  

 

As axial rigidity is very high, the second term in the denominator may also be neglected. Finally the horizontal reaction is calculated by the equation 

 

 

For an arch with uniform cross section  EI is constant and hence,

 

 

 

 

In the above equation, is the bending moment at any cross section of the arch when one of the hinges is replaced by a roller support.  is the height of the arch
as shown in the figure. If the moment of inertia of the arch rib is not constant, then equation must be used to calculate the horizontal reaction M0 y ~ H .  

TEMPERATURE EFFECT

CIVIL_ENGINEERING

Introduction:

Consider an unloaded two-hinged arch of span L . When the arch undergoes a uniform temperature change of  T°C , then its span would increase by  α TL if it
were allowed to expand freely .  α is the co-efficient of thermal expansion of the arch material. Since the arch is restrained from the horizontal movement, a horizontal force is induced at the support as the temperature is increased.

 

 

 

Now applying the Castigliano’s first theorem,

 

 

Solving for H ,

 

 

The second term in the denominator may be neglected, as the axial rigidity is quite high. Neglecting the axial rigidity, the above equation can be written as

 

ANALYSIS OF TWO HINGED ARCH

CIVIL_ENGINEERING

Introduction:

A typical two-hinged arch is shown in Fig.. In the case of two-hinged arch, we have four unknown reactions, but  there are only three equations of equilibrium available. Hence, the degree of statical indeterminacy is one for twohinged arch.

The fourth equation is written considering deformation of the arch. The unknown redundant reaction  is calculated by noting that the horizontal displacement of
hinge Hb B is zero. In general the horizontal  reaction in the two hinged arch is evaluated by straightforward application of the theorem of least work , which states that the partial derivative of the strain energy of a statically indeterminate structure with respect to statically indeterminate action should vanish. Hence to obtain, horizontal reaction, one must develop an expression for strain energy. Typically, any section of the arch  is subjected to shear forceV , bending moment  M and the axial compression . The strain energy due to bending  is calculated from the following expression.

The above expression is similar to the one used in the case of straight beams. However, in this case, the integration needs to be evaluated along the curved
arch length. In the above equation, s is the length of the centerline of the arch,  I is the moment of inertia of the arch cross section,  E is the Young’s modulus of
the arch material. The strain energy due to shear is small as compared to the strain energy due to bending and is usually neglected in the analysis. In the case of flat arches, the strain energy due to axial compression can be appreciable and is given by,

The total strain energy of the arch is given by, 

Now, according to the principle of least work

where H is chosen as the redundant reaction.

Solving equation the horizontal reaction  H is evaluated.

INFLUENCE LINE DIAGRAM

CIVIL_ENGINEERING

Introduction:

In engineering, an influence line graphs the variation of a function (such as the shear felt in a structure member) at a specific point on a beam or truss caused by a unit load placed at any point along the structure.Some of the common functions studied with influence lines include reactions (the forces that the structure’s supports must apply in order for the structure to remain static), shear, moment, and deflection. Influence lines are important in the designing beams and trusses used in bridges, crane rails, conveyor belts, floor girders, and other structures where loads will move along their span.[The influence lines show where a load will create the maximum effect for any of the functions studied.

 

 

Influence lines are both scalar and additive.[5] This means that they can be used even when the load that will be applied is not a unit load or if there are multiple loads applied. To find the effect of any non-unit load on a structure, the ordinate results obtained by the influence line are multiplied by the magnitude of the actual load to be applied. The entire influence line can be scaled, or just the maximum and minimum effects experienced along the line. The scaled maximum and minimum are the critical magnitudes that must be designed for in the beam or truss.

 

 

In cases where multiple loads may be in effect, the influence lines for the individual loads may be added together in order to obtain the total effect felt by the structure at a given point. When adding the influence lines together, it is necessary to include the appropriate offsets due to the spacing of loads across the structure. For example, if it is known that load A will be three feet in front of load B, then the effect of A at x feet along the structure must be added to the effect of B at (x – 3) feet along the structure—not the effect of B at x feet along the structure Many loads are distributed rather than concentrated. Influence lines can be used with either concentrated or distributed loadings. For a concentrated (or point) load, a unit point load is moved along the structure.

 

For a distributed load of a given width, a unit-distributed load of the same width is moved along the structure, noting that as the load nears the ends and moves off the structure only part of the total load is carried by the structure. The effect of the distributed unit load can also be obtained by integrating the point load’s influence line over the corresponding length of the structures.

(a) This simple supported beam is shown with a unit load placed a distance x from the left end. Its influence lines for four different functions: (b) the reaction at the left support (denoted A), (c) the reaction at the right support (denoted C), (d) one for shear at a point B along the beam, and (e) one for moment also at point B.

MULLER – BRESLAU PRINCIPLE

CIVIL_ENGINEERING

The procedure for applying the Muller-Breslau principle is as follows:

 

 

• Remove the constraint at the point of interest for the function of interest. This means if the influence line for a reaction is asked for simply start by pretending the beam is no longer attached to the reaction in question and is free to rotate about the other support. If the influence line for a moment is desired, pretend the point in question is a hinge and the subsequent two sides can rotate about their supports. If the influence line for shear is desired, again pretend the point in question is a shear release, again where both sides can rotate about their supports.

 

• Consider the remaining portion of the beam to have infinite rigidity, so it is a straight line free to rotate about the support.

 

• Lastly rotate whatever is free to rotate in its positive direction, but only enough to creat a deflection of 1 unit total. This means if the moment IL is in question and an imaginary hinge is splitting the beam in two pieces, the two angles created between each rotated side and the original beam must add to equal 1. Similarly if the shear IL is in question the two sides will have opposite directions of rotation. So at the shear release the right side will typically be rotated upwards and the left side will be rotated downward, as this is the sign convention for shear. The total displacement between the two sides of the shear release must equal 1.

 

Part (a) of the figure to the right shows a simply supported beam with a unit load traveling across it. The structure is   statically determinate. Therefore, all influence lines will be straight lines.

 

Parts (b) and (c) of the figure shows the influence lines for the reactions in the y-direction. Releasing the vertical reaction for A allows the beam to rotate to Δ. Likewise for part (c). Δ is typically taken as positive upwards.

 

Part (d) of the figure shows the influence line for shear at point B. Using the  beam sign conventionand cutting the beam at B, we can deduce the figure shown.

 

Part (e) of the figure shows the influence line for the bending moment at point B. Again making a cut through the beam at point B and using the beam sign convention, we can deduce the figure shown.

INCORPORATION OF MOMENT DUE TO REACTION

CIVIL_ENGINEERING

Description:

As mentioned before, for continuous beams  prestressing generates reactions at the supports.  The reactions at the intermediate supports cause moment at a section of the continuous beam.  This moment is linear between the supports and is in addition to the moment due to the eccentricity of the prestressing force.  The concept is explained by a simple hypothetical two-span beam in the  following figure.  The beam is prestressed with a parabolic tendon in each span, with zero eccentricity of the CGS at the supports.   The moment diagram due to the eccentricity of the prestressing force and neglecting the intermediate support is denoted as the M1 diagram.  This diagram is obtained as M1 = Pe, where, P is the prestressing force (P0 at transfer and Pe at service) and  e is the eccentricity of the CGS with respect to CGC.  Neglecting the variation of P along the length due to frictional losses, the value of M1 is proportional to e.  Hence, the shape of the M1 diagram is similar to the cable profile.

Profile of the CGS

Free body diagram of concrete

Next, the moment diagram due to the prestressing force and including the effect of the intermediate support is denoted as the  M2 diagram.  This is obtained by structural analysis of the continuous beam subjected to the upward thrust.  Since the profile of the tendon is parabolic in each span, the upward thrust is uniform and is given as .  The downward thrust at the location of the central kink is not considered as it directly goes to the intermediate support.  The hold down force at the intermediate support neglecting the downward thrust is 10wupl/8 = 10Pe/l.  The downward forces at the ends are from the anchorages.  The moment diagram due to wup alone (without the support) is added to that due to the hold down force.  The resultant  M2 diagram is similar to the previous  M1 diagram, but shifted linearly from an end support to the intermediate support. For a general case, the resultant moment (M2) at a location due  to the prestressing force can be written as follows.

In the above equation,

M₁  = moment due to the eccentricity of the prestressing force neglecting the intermediate supports .

M1^/ = moment due to the reactions at intermediate supports.   
P_e  = effective prestress  

e   = eccentricity of CGS with respect to CGC.

M₁ is the primary moment and M1/is the secondary moment.

The moment due to the external loads (M0) that is obtained from the envelop moment diagrams is added to M2 to get the resultant moment (M3) at a location.

The variation of M3 along the length of the beam (M3 diagram) can be calculated as follows.  
1) The M₀ diagram is available from the envelop moment diagram.   
2) Plot M₁ diagram which is similar to the profile of the CGS.  The variation of Pe along the length due to friction may be neglected.  
3)  Plot the shear force (V) diagram corresponding to the  M1  diagram from the relationship V = dM1/dx.
4) Plot the equivalent load (weq) diagram corresponding to the V diagram from the relationship weq = dV/dx.  Note, over the supports weq can be downwards.   Also, a singular moment needs to be included at an end when the eccentricity of the CGS is not zero at the end.

STEPS INVOLVED IN MOMENT DISTRIBUTION METHOD

CIVIL_ENGINEERING

The steps involved in the moment distribution method are as follows:

(1) Calculate fixed end moments due to applied loads following the same sign convention and procedure, which was adopted in the slope-deflection method.

(2) Calculate relative stiffness.

(3) Determine the distribution factors for various members framing into a particular joint.

(4) Distribute the net fixed end moments at the joints to various members by multiplying the net moment by their respective distribution factors in the first cycle.

(5) In the second and subsequent cycles, carry-over moments from the far ends of the same member (carry-over moment will be half of the distributed moment).

(6) Consider this carry-over moment as a fixed end moment and determine the balancing moment. This procedure is repeated from second cycle onwards till convergence.

For the previous given loaded beam, we attempt the problem in a tabular form..

 

VECTOR OF KINEMATICAL CONDITION

CIVIL_ENGINEERING

Vector of kinematical conditions ur:

 

A vector of kinematical conditions  ur for a given statically indeterminate structure is known a priori. Determination of this vector requires only the knowledge of a choice of redundants, support conditions at locations and in  directions where support reactions are chosen as redundants, and continuity conditions at locations where the internal forces are chosen as redundants.

 

 If the support reaction Ri  is chosen as one of redundants, the corresponding kinematical condition uir  vanishes if there is no support settlement in the direction of Ri  and uir  vanishes  if the support settlement takes place in the direction of Ri  with an amount of α.

 

If the internal force Ri  is chosen as one of redundants, the  corresponding kinematical condition uir  vanishes if there is no discontinuity (e.g. relative rotation, gap, overlapping) taking place in the direction of Ri and uir = α if the discontinuity takes  place in the direction of Ri  with an amount of α . Note that such discontinuity occurs only at a point where the internal constraint cannot completely be developed, e.g. a flexible rotational point and an extensible point.

 

 

Vector of kinematical quantities u_o:
As evident from the definition described above, a vector uo contains the displacement components, rotations, or internal discontinuity (e.g. relative rotation, gap, overlapping) of the primary structure at the locations and in the directions of released redundants.

 

Since the primary structure is statically determinate, computation of the vector uo can readily be achieved by using various techniques such as the method of curvature area, the conjugate structure analogy, and the principle of complementary virtual work or the unit load method once the static analysis for the internal forces of such structure is completed.