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Unformatted text preview: Midterm Exam. ME572 (Prof Pan) Name IDA/GE Pﬂfk . 3:40PM — 5:00 PM, October 24, 2007
A formula sheet is allowed. Box your answers and include units. 1. For a given Cartesian coordinate system, consider the stress tensor 11.1
6=1111VJPa
111 (a) Determine the principal stresses and the corresponding principal directions.
Comment on the stress state and the principal directions. (b) Determine the maximum shear stress for the stress tensor.‘ Make a sketch 'of a
material element with respect to the principal directions and indicate the plane
where the maximum shear stress is acting on. " 2. Consider a hexagonal closed packed (hcp) single crystal material element with the
/ \' coordinate axes as shown. The ratio of the length c to the length a is about 1.57 to
‘ 1.89. We consider the slips on the basal plane as shown. We use the usual vector
' notation to denote the slip directions (we do not use the MillerBravais indices here
I for simplicity). The material element is subject to the stress tensor with respect to the
1 . coordinate system as shown. The stress tensor is 100 100 100
0' = 100 100 100 MP3
100 100 100 The slip plane has the normal in the direction: of [0 0 1]. The slip direction sA is in the
direction of [2 1 0], the slip direction .53 is in the direction of [0 1 0], and the slip
direction sC is in the direction of [2 I 0]. (a) Determine the traction on the slip plane with the normal in the direction of [0 0 1].
(b) Determine the magnitude of the normal stress on the slip plane. (0) Determinethe shear stress on the slip plane as a vector. (d) Determine the ma nitude of stress ' e sA direction on the slip
plane.
' (e) Determine the magnitude of the resolvedshear stress in the sB direction on the
slip plane. .
\T (f) Determine the magnitude of the resolved shear stress in the sC direction on the
slip plane.
(g) Which direction of the slip will be ﬁrst activated if the stresses increase
proportionally? 
x3
\_ 3. A thinwalled rubber tube has the inner radius of Ri of 25 mm and the outer radius R, of 30 mm. The length of the rubber tube is long and denoted as L. The rubber tube is used to shrink (or expansion) ﬁt to a steel bar as shown. "The length of the
rubber does not change after the fit. The steel bar has the radius of R, of 26 mm. The rubber tube has Young’s modulus E of 3.5 MPa and Poisson’s ratio v of 0.5.
The steel has a much larger Young’s modulus than the rubber. We can treat the steel bar as rigid for this ﬁt. Based on the cylindric ordinate iysttanmasshownym (a) Determine the'outer radius r0 of the rubber tube after the ﬁt. (b) Determine the strain tensor of the rubber material on the outside surface of the
tube. Write the strain tensor with 6 components in the matrix form. (c) Determine the stress tensor of the rubber material on the outside surface of the
tube. Write the stress tensor with 6 components in the matrix form. ((1) Determine the maximum shear stress. Make a sketch of a material element and
indicate the planes of the maximum shear stress.  L4) Prfmcffaa Wges Z: alwectran.
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Iu= 0E1: = 3 MP“ _
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6L Per?”  W ml! I32 ml or 441 M —'=9 Cf/l’m/rl'mj State 071) stress,
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