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Consider a stress field whose matrix components in the vector basis {e} are: ĂŹĆ’11 = 63 MPa ĂŹĆ’22 = 44 MPa ĂŹĆ’33 = -2 MPa Where the Cartesian coordinate variables x, y, and z are in meters (m). Assuming body forces are negligible, does this stress field satisfy the equilibrium equations? Determine the traction vector acting at a point X = 2e1 + 3e2 - ze3 on the plane 2x1 + x2 - x3 = 5 m. Note that the unit normal to the plane defined by a = b is ne = [4e1, 0, 0]. Determine the normal and projected shear tractions acting at this point on this plane. Determine the principal stresses and principal directions of stress at this point.
Sri K.
(a) Let us indicate the forces acting on the sphere and their points of application. Choose positive direction of $x$ and $\varphi$ (rotation angle) along the incline in downward direction and in the sense of $\vec{\omega}$ (for undirectional rotation) respectively. Now from equations of dynamics of rigid body i.e. $F_{x}=m w_{c x}$ and $N_{c z}=I_{c} \beta_{z}$ we get : $m g \sin \alpha-f_{r}=m w$ and But $f r \leq k m g \cos \alpha$ In addition, the absence of slipping provides the kinematical realtionship between the accelerations : $w=\beta R$ The simultaneous solution of all the four equations yields : $k \cos \alpha \geq \frac{2}{7} \sin \alpha$, or $k \geq \frac{2}{7} \tan \alpha$ (b) Solving Eqs. (1) and (2) [of part (a)], we get :
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Choosing the positive direction for $x$ and $\varphi$ as shown in Fig, let us we write the equation of motion for the sphere $F_{x}=m w_{c x}$ and $N_{c z}=I_{c} \beta_{z}$ $f r=m_{2} w_{2} ; f r r=\frac{2}{5} m_{2} r^{2} \beta$ $\left(w_{2}\right.$ is the acceleration of the C.M. of sphere.) For the plank from the Eq. $F_{x}=m w_{x}$ $F-f_{r}=m_{1} w_{1}$ In addition, the condition for the absence of slipping of the sphere yields the kinematical relation between the accelerations : $w_{1}=w_{2}+\beta r$ Simultaneous solution of the four equations yields : and $w_{2}=\frac{2}{7} w_{1}$
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