Bending of a Drill Bit This example illustrates the full problem-solving process from Section 3.2, which incorporates the principles of dimensional analysis from Sections 3.3–3.5. A drill press holds sharpened bits in a rotating chuck and is used to boreholes in a workpiece. The steel drill bit has

Question

Bending of a Drill Bit

This example illustrates the full problem-solving process from Section 3.2, which incorporates the principles of dimensional analysis from Sections 3.3–3.5.

A drill press holds sharpened bits in a rotating chuck and is used to boreholes in a workpiece . The steel drill bit has a diameter d = 6 mm and length L = 65 mm. The bit is accidentally bent as the workpiece shifts during a drilling operation, and it is subjected to the side force of F = 50 N. As derived in mechanical engineering courses on stress analysis, the sideways deflection of the tip is calculated by using the equation

[latex]\Delta x =\frac{64FL^{3} }{3\pi Ed^{4} } [/latex]

where the terms have the following units:
Δx (length) the deflection of the tip
F (force) the magnitude of the force applied at the tip
L (length) the drill bit’s length
E (force/length2) a property of the drill bit’s material, called the elastic modulus
d (length) the drill bit’s diameter

By using the numerical value[latex] E = 200 × 10^{9}[/latex] Pa for steel, calculate the amount Δx that the tip deflects. (See Figure 3.6.)

Approach
We are tasked to solve for the deflection at the tip of the steel drill bit given the applied force. We first make a number of assumptions about the system:

• The curved flutes on the bit are small and can be neglected in the analysis
• The force is perpendicular to the primary bending axis of the bit
• The channels that spiral along the bit have minimal impact on the bending and can be ignored
We will first combine the units of each quantity in the given equation according to the rules of algebra and verify that the units appearing on each side of the equation are identical. Then we will insert the known quantities, including the bit length, diameter, elastic modulus, and applied force, to solve for the deflection.

Accepted Answer

The quantity 64/3π is a dimensionless scalar, and it therefore has no units to influence dimensional consistency. The units of each quantity in the given equation are cancelled:

[latex](length)=\frac{(force)(length)^{3} }{(force/(length)^{2})(length)^{4}} [/latex]

=[latex](length)[/latex]

The equation is indeed dimensionally consistent. The tip moves sideways by the amount

[latex]\Delta x =\frac{64(50N)(0.065m)^{3}}{3\pi (200 imes 10^{9} Pa)(6 imes 10^{-3}m)^4} \longleftarrow \left[\Delta x =\frac{64FL^3}{3\pi Ed^4} ight] [/latex]

Next, we combine the numerical values and dimensions

[latex]\Delta x =3.6 imes 10^{-4}\frac{N .m^{3} }{Pa.m^4} [/latex]

and then expand the derived unit pascal according to its definition in Table 3.2:

Table 3.2 Certain Derived Units in the SI

Quantity	SI Derived Unit	Abbreviation	Definition
Length	micrometer or micron	[latex]\mu m[/latex]	[latex]1\mu m = 10^{-6} m[/latex]
Volume	liter	L	[latex]1 L = 0.001 m^{3}[/latex]
Force	newton	N	[latex]1 N = 1 (kg . m)/s^{2}[/latex]
Torque, or moment of a force	newton-meter	N . m	—
Pressure or stress	pascal	Pa	[latex]1 Pa = 1 N/m^{2}[/latex]
Energy, work, or heat	joule	J	1 J = 1 N . m
Power	watt	W	1 W = 1 J/s
Temperature	degree Celsius	°C	°C = K – 273.15
Although a change in temperature of 1 Kelvin equals a change of 1 degree Celsius, numerical values are converted using the formula.

[latex]\Delta x =3.6 imes 10^{-4}\frac{N .m^{3} }{(N/m^2)(m^4)} [/latex]

Finally, we cancel units in the numerator and denominator to obtain

[latex]\Delta x =3.6 imes 10^{-4}m[/latex]

Discussion
First, we evaluate the order of magnitude of the solution. For a steel bit of this length, a large deflection is not expected. Therefore the solution’s order of magnitude is reasonable. Second, we revisit our assumptions to make sure they are reasonable. While the curved flutes and channels on the bit may slightly influence the bending mechanics, we must assume that their impact is negligible for this application. Also, the force may not remain perfectly perpendicular to the bit, but it is reasonable to assume this at the moment of deflection. Third, we draw conclusions from the solution and explain its physical meaning. Drill bits undergo many forces during operation, and it makes sense that most bits would be made of steel to minimize deflection. Because the numerical value has a large negative exponent, we convert it to standard form by using the SI prefix “milli” to represent a factor of [latex]10^{−3}[/latex]. The tip moves by the amount Δx = 0.36 mm, just over one-third of a millimeter.

Δx = 0.36 mm

Question 3.6: Bending of a Drill Bit This example illustrates the full pro...

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