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Unformatted text preview: mall strains. The net consequence of these approximations is that the
shear strain along length AB1 is uniform, as can be seen by the angle between any vertical line and line AB1 at any point along the line.
2. The shear stress is assumed uniform across the cross section because of thin bars, but it is also uniform along the length because of the
approximations described in comment 1.
3. The shear stress acts on a surface with outward normal in the direction of the length of the bar, which is also the axis of the disc. The
shear force acts in the tangent direction to the circle of radius r. If we label the direction of the axis x, and the tangent direction θ, then
the shear stress is represented by τxθ, as in Section 1.2
4. The sum in Equation (E7) can be rewritten as 2 ∑i =1 r τ Δ Ai , where τ is the shear stress acting at the radius r, and Δ Ai is the cross-sec- tional area of the i th bar. If we had n bars attached to the disc at the same radius, then the total torque would be given by n ∑i =1 r τ Δ Ai . As we increase the number of bars n to infinity, the assembly approaches a continuos body. The cross-sectional area Δ Ai becomes the
infinitesimal area dA, and the summation is replaced by an integral. We will formalize the observations in Section 5.1.1.
5. In this example we visualized a circular shaft as an assembly of bars. The next two examples further develop this idea. EXAMPLE 5.2
A rigid disc of 20-mm diameter is attached to a circular shaft made of hard rubber, as shown in Figure 5.5. The left end of the shaft is
fixed into a rigid wall. The rigid disc was rotated counterclockwise by 3.25°. Determine the average shear strain at point A. 3.25
A Figure 5.5 Geometry in Example 5.2. 200 mm PLAN
We can visualize the shaft as made up of infinitesimally thick bars of the type shown in Example 5.1. We relate the shear strain in the bar
to the rotation of the disc, as we did in Example 5.1. SOLUTION
We consider one line on the bar, as shown in Figure 5.6. Point B moves to point B1. The right angle between AB and AC changes, and the change represents the shear strain γ. As in Example 5.1, we obtain the shear strain shown in Equation (E2):
Δφ = ---------------- = 0.05672 rad
180 ° BB 1 = r Δ φ = ( 10 mm ) Δ φ = 0.5672 mm (E1) BB 1 0.5672 mm
tan γ = γ = --------- = --------------------------- = 0.002836 rad
200 mm (E2)
ANS. Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm (a) (b)
A C r 10 mm
B1 Figure 5.6 Deformed shape: (a) 3-D; (b) End view. γ = 2836 μ rad 200 mm Δφ rΔφ O B1 COMMENTS
1. As in Example 5.1, we assumed that the line AB remains straight. If the assumption were not valid, then the shear strain would vary in
the axial direction.
2. The change of right angle that is being measured by the shear strain is the angle between a line in the axial direction and the tangent at
any point. If we designate the axial direction x and the tangent direction θ (i.e., use polar coordinates), then the shear strain with subscripts will be γx θ.
January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 208 3. The value of the shear strain does not depend on the angular position as the problem is axisymmetric.
4. If we start with a rectangular grid overlaid on the shaft, as shown in Figure 5.7a, then each rectangle will deform by the same amount,
as shown in Figure 5.7b. Based on the argument of axisymmetry, we will deduce this deformation for any circular shaft under torsion
in the next section. (a) (b) Figure 5.7 Deformation in torsion of (a) an un-deformed shaft. (b) a deformed shaft. EXAMPLE 5.3
Three cylindrical shafts made from hard rubber are securely fastened to rigid discs, as shown in Figure 5.8. The radii of the shaft sections
are rAB = 20 mm, rCD = 15 mm, and rEF = 10 mm. If the rigid discs are twisted by the angles shown, determine the average shear strain in
each section assuming the lines AB, CD, and EF remain straight.
2.5 1.5 1.5
3.25 A B Figure 5.8 Shaft geometry in Example 5.3. C 200 mm D 160 mm E F 120 mm METHOD 1: PLAN
Each section of the shaft will undergo the deformation pattern shown in Figure 5.6, but now we need to account for the rotation of the
disc at each end. We can analyze each section as we did in Example 5.2. In each section we can calculate the change of angle between the
tangent and a line drawn in the axial direction at the point where we want to know the shear strain. We can then determine the sign of the
shear strain using the definition of shear strain in Chapter 3. SOLUTION
Label the left most disc as disc 1 and the rightmost disc, disc 4. The rotation of each disc in radians is as follows:
° ° 2.5
φ 1 = ----------- ( 3.142 rad ) = 0.0436 rad
180 ° 1.5
φ 2 = ----------- ( 3.142 rad ) = 0.0262 rad
180 ° ° (E1) ° 1.5
φ 3 = ----------- ( 3.142 rad ) = 0.0262 rad
180 ° 3.25
φ 4 = ----------- ( 3.142 rad ) = 0.0567 rad
180 ° Figure 5.9 shows approximate deformed shapes of the three segments, (a) (b)
B AB A AB B1
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This note was uploaded on 04/08/2010 for the course ENGR 232 taught by Professor Smith during the Spring '10 term at Aarhus Universitet.
- Spring '10