5a-ex
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5a-ex. Torsion Member Examples• Ex.. 5a.1 • Ex. 5a.2 • Ex. 5a.3
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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.
Drilling into a piece of wood.
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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 isT(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)
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:
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