5a-ex. Torsion Member Examples • Ex.. 5a.1 • Ex. 5a.2 • Ex. 5a.3 Back | Index | Next

 Example 5a.1 Given: A stepped-shaft (polar moment of inertias 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: Determine the total angle of twist of the bar, qtotal. Sol'n: Step 1. Equilibrium. Solve for the reaction at Pt. A: R = 2T (clockwise about +y-axis, a negative torque). Step 2. Twist/Torque of each Segment. Break up the bar into segments over which all the values (torque, cross-section and modulus) are constant. The top segment, AB, has length, a, Polar Moment of Inertia, J1, modulus, G1, and supports torque 2T (a negative torque on the positive face of AB). With Cross-section A fixed, the rotation of Section B with respect to fixed Section A is: Taking each of the other segments with constant torque, section and modulus:           Step 3. Compatibility. The total angle of twist: Free Body Diagrams of shaft segments. Angular deflection of stepped-shaft.

 Example 5a.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: Determine the angle of twist of the imbedded length. Measure x from where in internal torque is zero, T=0, i.e., from the tip of the bit.Sol'n: Step 1. Equilibrium. Over dx, the torque, area and modulus are 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. Compatibility. Integrate 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)= –(Tx/L) 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 total rotation is found by integrating: The negative sign means that, as viewed along the positive x-axis, the left end rotates clockwise w.r.t. the right end. 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).

 Example 5a.3 Given: A drill requires 1/2 hp to turn a 3/8" 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 occurs at the end closest the drill and is given by: T = 39.4 lb-in. Step 2. Shear stress. The maximum shear stress for a solid shaft is: tmax = 0.535 ksi

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Updated: 05/24/09 DJD