Thus, the given Tis a linear transformation. Find the matrix of T. Find the matrix of R. The coordinates are right-handed. Let n be a unit normal vector to the plane and let r be the position vector for a point in space.
Let the origin of the coordinates be on the axis of rotation and r be the position vector for a typical point in the body. Then, m, , p q forms an orthonormal set of vectors which rotates an angle of about the unit vector m. Find the matrix of the tensor S which first transforms every vector into its mirror image and then transforms them by a o 45 right- hand rotation about the 1 e -axis.
Let 1 2 and T T correspond to the reflection and the rotation respectively. Consider two successive small rotations 1 R and 2 R , show that the final result does not depend on the order of rotations. Thus, T T is a linear transformation, i. Show that the dot product is a scalar invariant with respect to orthogonal transformations of coordinates.
Prove that ijk R are the components of a third-order tensor. Prove that the decomposition is unique. Hint, assume that it is not true and show contradiction.
Thus , the decomposition is unique. Explain the direction of the eigenvectors corresponding to them for a proper orthogonal rotation tensor and for an improper orthogonal reflection tensor. But, since Tis antisymmetric, i. That is, for any given rotation tensor R, its axis of rotation and the angle of rotation can be obtained from the dual vector of A R and the first scalar invariant of R.
Find the axis of rotation and the angle of rotation. Use the result of the previous problem. Thus, we can obtain the direction of axis of rotation and the angle of rotation by obtaining the dual vector of A R.
On the other hand, the first scalar invariant of R is 0. That is, all vectors perpendicular to the axis of rotation are eigenvectors. Or, clearly one of the eigenvalue for [ ] S is 1 , which is not an eigenvalue for [ ] T , therefore the answer is NO. At 1,0,1? Find the unit vector normal to the surface at a given point , , x y z.
By writing the subscripted components form, verify the following identities. Let r v be a vector field. Thus, T T div div tr. Find the material derivative of , using both descriptions of the temperature. For mat. Similarly derivation gives kt y Ye. Obtain the acceleration field.
From D Dt t. Every particle moves only up and down in the 2 x direction. Thus, the given velocity field is that of a two dimensional source flow from the origin, the flow is purely radial with radial velocity inversely proportional to the radial distance from the origin.
Isotherms are circles. Unit elong. Decrease in angle between them is 4 12 2 10 E k radian. What can you say about the results? We see from this example that having the same principal scalar invariants is a necessary but not sufficient condition for the two tensors to be the same. The maximum elongation is 4k. Give reason s. The strain tensor Eis a real symmetric tensor, for which there always exists three principal directions, with respect to which, the matrix of Eis diagonal. That is, the non-diagonal elements, which give one-half of the change of angle between the elements which were along the principal directions, are zero.
Explain your answer. Therefore, there does not exist any element at the point which has a negative unit elongation i. If these unit elongation are designated by , , a b c respectively, what are the strain components 11 22 12 , and E E E?
The principal directions corresponding to the other two eigenvalues lie on the plane of 1 2 and e e. Nothing else changes. Use the formulas obtained in the previous problem.
Using the formulas drived in the previous problem, we have, [ ] 6 6 22 1 1 2 2 2 1 2 1. That is, the principal axes of D rotates with an angular velocity given by the dual vector of the spin tensor.
For an incompressible material, from the conservation of mass principle, obtain the most general form of the function , f r. The equation of continuity for an incompressible material is [see Eq. From the consideration of the principle of conservation of mass, find v. The equation of continuity for an incompressible fluid is [see Eq.
Thus, cos sin. Since the fluid is incompressible, therefore, 1 2 1 2 2 1 2 0 0 0 0 0 0. From the consideration of the principle of conservation of mass, a Find the density if it depends only on timet , i.
From the consideration of the principle of conservation of mass, determine how the fluid density varies with time, if in a spatial description, it is a function of time only. We note that the given ij E are linear in 1 2 3 , and X X X and the terms in the compatibility conditions all involve second derivatives with respect to i X , therefore these conditions are obviously satisfied by the given strain components.
The given strain components are not compatible. The displacement field obviously exists. In fact, the displacement field is given.
There is no need to check the compatibility conditions. Whenever a displacement field is given, there is never any problem of compatibility of strain components. This compatibility condition is not satisfied. Clearly, there is no way this equation can be satisfied, because the right side can not have terms of the form of 1 2 X X.
Show that for the strains to be compatible, 2 3 , f X X must be linear in 2 3 and X X. Thus, if 2 3 , f X X is a linear function of 2 3 and X X , then all compatibility equations are satisfied. By demanding that the net rate of inflow of mass must be equal to the rate of increase of mass inside the differential volume, obtain the equation of conservation of mass in cylindrical coordinates.
Check your answer with Eq. Thus, the total influx of mass through these three pairs of faces is: 1 r r z v v v v dr rd dz r r r z. Therefore, the conservation of mass principle gives, 1 r r z v v v v dr rd dz rd drdz r r r z t. This is the same as Eq. The only positive definite root. Obtain a the deformation gradient and F the right Cauchy-Green tensor C, b The eigenvalues and eigenvector of C, c the matrix of the stretch tensor 1 and.
U U with respect to the i e -basis and d the rotation tensor Rwith respect to the i e -basis. There is no change in angle. Let c ij A denote the cofactor of ij A , i.
From Eq. From 1 d d. Lai et al, Introduction to Continuum Mechanics Copyright , Elsevier Inc o 1 o 1 o 1 o o o o o o r z o 1 o 1 o 1 o o o r z Similarly, , ,. Now, from Eqs.
From Eqs. Obtain the components of the right Cauchy-Green Tensor Cwith respect to the basis at the reference configuration. Using Eqs. On each of the coordinate planes with normal in 1 2 3 , , e e e directions , a what is the normal stress and b what is the total shearing stress Ans.
MPa , on the 2 e plane is 4. MPa , and on the 3 e plane is 0. On the plane, there is a constant shearing stress in the 2 e direction and a linear distribution of normal stress 2 x , see figure. On the plane, there is a parabolic distribution of normal stress 2 2 x , see figure.
On the plane, there is a constant normal stress of and a linear distribution of shearing stress 3 2 x e , see figure. For an arbitrary state of stress defined at the pointP , show that the component of the stress vector m t in the n -direction is equal to the component of the stress vector n t in the mdirection. Show that the stress vector on any plane that contains the stress traction m t , lies in theM plane. Referring to the figure below, where mis perpendicular to the planeM , and m t is the stress vector for the plane.
LetN be any plane which contains the vector m t and let n be the unit vector perpendicular to the planeN. We wish to show that n t is perpendicular to m. Show that if k is a unit vector that determines a plane that contains m t and n t , then k t is perpendicular tomand n.
Since k is a unit vector that determines a plane that contains m t and n t , therefore,. Maybe true in some special cases. Therefore, t is normal to the plane, so that there is no shearing stress on the plane. That is, it is a principal plane. The first scalar invariant for the first matrix is MPa The first scalar invariant for the second matrix is MPa They are not the same, therefore, they can not represent the same stress tensor.
The maximum normal stress is MPa and the minimum normal stress is MPa. For 2 Thus, the maximum shearing stress is max Thus, max min 5. Thus, max 5 3 4. Find the maximum shearing stress and the plane on which it acts. The equations of equilibrium are 0 i j i j T B x. Yes, the equations of equilibrium are all satisfied. Thus, 1 2 and B B must be independent of 3 x. Thus, the equations of equilibrium are satisfied for any function 1 2 , x x which is continuous up to the third derivatives.
This is Eq. Again, 2 3 1 1 cos If the Cauchy stress tensor is given by: 11 That is, its undeformed plane is also 1 e plane. Give reason. The given equations are not acceptable as a physically acceptable deformation because it gives a negative ratio of deformed volume to the undeformed volume. What is the deformed area of the 1 e face of the cube?
Also, calculate the pseudo differential force for the same plane. These results are quite obvious from the geometry of the deformation. Thus, 1 2 Lai et al, Introduction to Continuum Mechanics Copyright , Elsevier Inc c The second PK stress tensor is: [ ] 1 o 1 0 0 0 0 1 0 0 0 0 0. Thus, 1 2 2. Hint: Assume that a is an isotropic vector, and use a simple change of basis to equate the primed and unprimed components. For an isotropic a , by definition, [ ] [ ] i i. Method I. In other words, the only isotropic vector is the null vector.
Method II. That is, a is an eigenvector for T Q for any orthogonal tensorQ. Open Preview See a Problem? Thomas Mase. Details if other :. Thanks for telling us about the problem. Return to Book Page. Get A Copy. Paperback , 50 pages. More Details Original Title. Friend Reviews. To see what your friends thought of this book, please sign up. Fck your stupid website? This question contains spoilers Lists with This Book. This book is not yet featured on Listopia.
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Sep 08, Mourya Tej marked it as to-read. View 1 comment. Feb 12, Jose added it. Nov 30, Pawan added it. Nov 03, Suprit S is currently reading it. Thus, T33 is either 5 or 1. Find the maximum shearing stress and the plane on which it acts. Yes, the equations of equilibrium are all satisfied. This is Eq. Give reason. What is the deformed area of the e1 face of the cube?
Also, calculate the pseudo differential force for the same plane. These results are quite obvious from the geometry of the deformation. The Cauchy stress vector has a smaller magnitude because the area is four times larger. Hint: Assume that a is an isotropic vector, and use a simple change of basis to equate the primed and unprimed components. In other words, the only isotropic vector is the null vector. Method II. That is, a is an eigenvector for QT for any orthogonal T tensor Q.
But clearly, there is no non-zero vector which is an eigenvector for all orthogonal tensors. From Table 5. Now, from Eq. It becomes determinate when the boundary condition s is are taken into account.
Let ei be the principal basis for E , then [ E ]e is diagonal. Thus the matrix of T is not diagonal with respect to the principal basis of E. Also find the bulk modulus.
Thus, all equations of equilibrium are satisfied. The given state of stress is not a possible equilibrium stress field. Obtain the differential equations which ui x2 , t must satisfy in order to be a possible motion in the absence of body forces. Under what condition s will the above motion satisfy this boundary condition for all time. What further conditions will be imposed on the above motion to satisfy this boundary condition for all time.
From the Navier equations of motion, Eq. That is, there is no reflected longitudinal wave. There is only a reflected transverse wave of the same amplitude which completely cancels out the incident transverse wave. What is the relation of the amplitudes, wavelengths, and direction of propagation of the incident and reflected wave?
Reflection of Plane Elastic Waves, Figure 5. As in Section 5. However, the relations between the amplitudes are different. We note that unlike the problem in Sect. If Young's moduli of the two portions are EY 1 and EY 2 , find how the applied force is distributed between the two halves. The allowable tensile stress is 0. If the allowable shearing stress is 0. Find the twisting moments transmitted to the circular shafts on either side of the disc. Figure 5P. Let M1 and M 2 be the twisting moments transmitted to the left and the right shaft respectively.
The hollow circular bar differs from the solid circular bar in that there is an inner lateral surface which is also traction free. However, the normal to the inner lateral surface differs from that to the outer surface only by a sign so that the zero surface traction in the inner surface is also satisfied since that for the outer surface is satisfied.
Both shafts are of the same material. Also compare the unit twist i. Assume that the maximum twisting moment which can be transmitted is controlled by the maximum shearing stress.
With C so obtained, verify that the other two lateral surfaces are also traction free. We know that when a rectangular membrane, fixed on its side, is subjected to a uniform pressure on one side of the membrane, the deformed surface has a maximum slope at the mid point of the longer side.
Thus, based on the membrane analogy discussed in Example 5. We have, [see Eq. Refer to Section 5. We have [Eq. The roots of this equation consists of two sets of double roots.
How would T11 be affected by the removal. From Section 5. Two independent solutions for f are given by C and r 2. The four independent solutions are: C , r 2 ,ln r and r 2 ln r. Using the results in Example 5. From the results of Example 5. Using Eq. That is, verify Eq. Thus the material derivative of an objective tensor T is not objective. Show 1 1 1 1 1 1 Dt that A 2 is objective. Reference source not found. From Prob. Show that the Jaumann derivative of T is objective.
In Prob. In Example 5. The gate is hinged at the upper edge A. Neglect the weight of the gate, find the reactional force at B. Take the gate AB as a free body. Neglect the weight of the gate, compute the water level h for which the gate will start to fall. Consider the gate plus the triangular region of water above the gate as the free body diagram. The liquids are immiscible. The area under WR is twice that under WL.
Figure P Ans. It is therefore equal to the weight of the water displaced by this left half. The line of action of Fy the buoyancy force passes through the centroid of the semi-circular area, i. Find the pressure at a point whose depth from the surface of the water is h.
Take the atmospheric pressure to be pa. Thus, the free surface is a plane. Determine the relation between a, l and h. Let z be pointing vertically upward with the origin at the lowest point of the free surface.
Find the pressure and density distribution for this case. Find the distribution of pressure and density in a polytropic atmosphere. At the point 1,2,1 m and on the plane whose normal is in the direction of e1 , a find the excess of the total normal compressive stress over the pressure p and b find the magnitude of the shearing stress.
For a steady flow, the velocity at every point on a streamline does not change with time. Therefore, any particle, which is at a point P on the streamline at a given time t, will move along the streamline at all time. That is, its pathline coincides with the streamline containing the point P.
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