Example 5.1.1

Given: A stepped-shaft (Polar Moment of Inertia J₁, J₂) made of two materials (shear moduli G₁, G₂). The shaft is fixed (no rotation) at the top. Torques are applied about the shaft's axis: 3T (counterclockwise about the +y-axis at Pt. B) and T (clockwise about the +y-axis at Pt. E).

Req'd: Total rotation of the bar, q_total.

Sol'n:

Step 1. Equilibrium:
Solve for the reaction at Pt. A: R = 2T (clockwise about +y-axis, "negative" torque).

Step 2. Twist/Torque of Elements:
Break up the bar into lengths over which all the values (torque, cross-section and modulus) are constant.

The top length of the shaft, AB, has length, a, Polar Moment of Inertia, J₁, modulus, G₁, and carries torque 2T (negative torque on the positive face of AB).

With Cross-section (Section) A fixed, the rotation of Section B with respect to Section A is:

Taking each of the other lengths with constant torque, section and modulus:

Step 3. Compatability:
The total rotation is then:

Free Body Diagrams of Shaft.

Angular Deflection of Stepped Shaft.

Example 5.1.2

Given: A drill bit is imbedded a length, L, into a piece of wood. The drill applies a torque of T. The surrounding wood applies a uniform shear force per unit length of the bit acting on the surface of the bit. Assume that the drill bit is a simple cylinder of radius, R and length, L.

Req'd: The rotation of the imbedded length. Again, measure x from where T=0.

Sol'n:

Step 1. Equilibrium:
Over dx, the torque, area and modulus are essentially constant. The torque at cross-section x is T(x) = -T(x/L), where x is measured from the tip of the drill bit.

Step 2. Twist/Torque of Elements:
The rotation, dq, of each dx is:

Step 3. Compatability
Integrate the dq to get the total rotation.

• J(x) = pR⁴/2 and G(x)=G are constant over L.
• Torque T(x) varies from 0 to T, linearly. T(x)= -T(x/L)

The total rotation is found by integrating:

Drilling into a piece of wood.

 Element dx long extracted from shaft.

Because a negative torque acts on the positive x-face:

T(x) = -T (x/L)

The negative sign means that as viewed along the positive x-axis, the left end rotates clockwise w.r.t. the right end. In other words, viewed in the negative x-direction, the rotation of the right end w.r.t. the left is counterclockwise (positive rotation on a negative face).

The shear force is keeping the business-end of the bit from rotating with the drill end... the right end lags the left end.

Example 5.1.3

Given: A certain drill motor requires 1/2 hp to turn a 3/8" in drill bit at 800 rpm into a block of wood.

Req'd: The maximum shear stress in the drill bit. Model the bit as a solid shaft.

Sol'n:

Step 1. Torque:
The maximum torque in the bit will occur at the end closest the collet and is given by:

T =

33,000 P[hp] 2pN[rpm]

=

33,000 (0.5 hp) 2p(800 rpm)

= 3.28 ft-lb = 39.4 in-lb

Step 2. Shear stress:
The maximum shear stress for a solid shaft is given by:

tmax =

16 T pd3

=

16(39.4 in-lb) p(.375 in)3

= 535 psi = 0.54 ksi