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Investigation of strain on diamond crystal mounted on two parallel wires Brent Evans, University of Connecticut (presented by Richard Jones)

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Presentation on theme: "Investigation of strain on diamond crystal mounted on two parallel wires Brent Evans, University of Connecticut (presented by Richard Jones)"— Presentation transcript:

1 Investigation of strain on diamond crystal mounted on two parallel wires Brent Evans, University of Connecticut (presented by Richard Jones)

2 Brent Evans September 12, 2002 2 Bending and Shearing in the Diamond Radiator  Fluctuations in the position of the coherent bremsstrahlung peak were seen during g8 test runs with a thin radiator.  One proposed explanation: bending of the diamond radiator by its mount.  The 20  m-thick crystal is mounted on two parallel tungsten wires with 25  m  diameters. çIf the wires were twisted when the diamond was glued on, they would exert torques on the crystal.  Can twists of a few degrees on the wires between the mount and the crystal produce enough warping of the crystal to explain the observation?

3 Brent Evans September 12, 2002 3    5mm x 5mm x 20  m  diamond wafer 25  m dia. tungsten wire The Diamond Mount y x

4 Brent Evans September 12, 2002 4 Torques Produced by the Diamond Mount  There are two contributions to torque in the tungsten wires, that are distinguished by imagining the wire to be a bundle of fibers.  Shear: fibers shift parallel to each other but do not change length.  Stretch: fibers change length but do not shift parallel to each other. where  is the shear modulus, Y is Young’s modulus,  the angular twist, R the wire radius and L the wire length.  Since the radius is small, we ignore the stretch part, which is nonlinear.

5 Brent Evans September 12, 2002 5 Warping Effects in the Crystal  Next, we calculated the vertical displacement of the crystal at a position x as a function of the torques exerted by each wire.  As with the wires, there is shearing, where fibers stay parallel, and bending, where they do not.  Combining the two, the solution is where l is the spacing between the wires (x direction), w its width along the wires (y direction), h its thickness (z direction),  its shearing modulus, and Y its Young’s modulus.

6 Brent Evans September 12, 2002 6 Warping Effects in the Crystal  Consider the case where the torsions on the wire are equal and opposite in sign, so that the crystal deforms to an arc, not an S-shape. The maximum vertical displacement occurs at x = l /2, and is, as a function of wire twist,  z max is about 30  m for a 10-degree twist.

7 Brent Evans September 12, 2002 7 Warping Effects in the Crystal 1.Animation: 20mm crystal (local applet)20mm crystal local applet 2.Animation: 50mm crystal (local applet)50mm crystal local applet


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