where δ1 and δ2 are the radial displacements at the inner and outer radii of the strain gauge, r1 and r2 respectively. Equation 12.3 applies only to the center strand of a strain gauge and in general the equation must be augmented by similar quantities representing all the parallel strands [14]. This has been done for the results shown in Fig. 12.2. However, when making size comparisons, a realistic indication of the strain gauge response can be obtained simply from Eq. 12.3. This approach is taken here to allow conceptual and computational simplicity. Figure 12.3 shows the dimensionless calibration constants a and b for hole-drilling into finite thickness specimens. These results are computed using finite element calculations [13] using axi-symmetric elements for the a results and axi-harmonic elements with n ¼2 for the b results. For conceptual simplicity, Eq. 12.3 has been used to represent the strains that could be measured if strain gauges were used. The chosen radii are r1 ¼2a and r2 ¼4a. For an ASTM E837-13 standard strain gauge rosette, this corresponds to a hole radius 0.36 times the mean rosette radius. To generalize the graphical presentation, the hole depth h and specimen thickness W are both normalized by the hole radius a. In Fig. 12.3, the responses of a very thick material are represented by the line W/a ¼1. Both lines start from zero and reach a plateau at a hole depth of approximately one diameter (¼2a). Both graphs show that the response generally increases with reduction in material thickness, with increasingly rapid response for very thin specimens. This behavior is to be expected because reduction in specimen thickness reduces the amount of material immediately below the hole and therefore reduces the constraint on the surrounding material. This constraint reduction in turn allows the deformation response to increase more rapidly. All response curves reach the same terminal value, corresponding to the through-hole case when the drilled hole reaches the lower A prominent feature of the a curves in Fig. 12.3a is the peak that appears for finite thickness specimens. This occurs because, as shown in Fig. 12.4, a blind hole in a finite thickness specimen creates an unsymmetrical in-plane load on the specimen that causes local bending of the specimen and hence enlarged surface deformations. As the hole reaches the lower surface the load becomes symmetrical, the bending effect disappears and the deformation response reduces from its peak. For the axisymmetric case corresponding to a, the cross-section shape shown in Fig. 12.4 corresponds to a circular hill with the hole at the top. For the shear stress case corresponding to b, the surface is saddle shaped, with one diameter folded down and the perpendicular diameter folded up. The reversed curvatures of the two diameters tend to oppose each other, thereby reducing local bending effects and inhibiting the growth of bending peaks in the b curves in Fig. 12.3b. -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.0 0.5 1.0 1.5 2.0 Hole Depth, h/a b_bar = (u(4a)-u(2a)) / 2a W/a = 0.5 1.0 1.5 2.0 W/a = ∞ b -0.25 -0.20 -0.15 -0.10 -0.05 0.00 0.0 0.5 1.0 1.5 2.0 Hole Depth, h/a a_bar = (u(4a)-u(2a)) / 2a W/a = W/a = 0.5 1.0 1.5 2.0 3.0 a ∞ Fig. 12.3 Dimensionless calibration constants a and b for hole-drilling into finite thickness specimens Fig. 12.4 Deformation of a finite thickness specimen due to residual stress loads on the hole boundary, calculated using finite element analysis. The deformations are greatly exaggerated for pictorial clarity 92 G.S. Schajer and C. Abraham
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