A catalog record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data. A catalog record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data.
PRODUCTION PROCESSES
Alternative upper and lower bound analyzes of the forming loads for metal forming are given in Chapter 7. A graphical analysis of the plasticity induced by longitudinal impact of bars is given in Chapter 13.
LIST OF SYMBOLS
P Hfj, C m
- Shear Stress
- Direction Cosines
- Principal Stress Directions
- Principal Stresses as Co-ordinates
The abbreviated expressions in parentheses show that the equivalent equations appear in the respective concepts of a direct tensor (ie the dot product), exponential tensor components, and a matrix. Example 1 3 Given the principal stresses at = 7.5, at = 3.1 and c% = 1.4 (MPa) find (i) the maximum shear stresses and their directions and (ii) the magnitude and direction of normal- and the shear stresses for the plane octahedra.
Now, from eq(1.44a), the tensor S', with components S¥\ is written in terms of the Cauchy stress tensor as. Find the principal stresses, the largest shear stress, and the state of stress in the octahedral plane.
Introduction
- Distortion and Rotation
Further strain-displacement relations, similar in form to Eq.(2.1a,b,c), will apply to normal and shear deformations in the x-z and y-z planes. STRAIN ANALYSIS 35 Consequently, the full deformation of a ¥olume element & x < fy x dfe can be expressed as the sum of corresponding stretches and rotations in the matrix form.
2 I &? + 3y
It is possible to determine the correspondence between r i n voltage eq( 1.12c) and ffl in eq(2.8b). Equation (2.8b) will then give the shear strain for the original perpendicular directions: one parallel to the normal hex a and the other aligned with the direction of the resultant shear stress r in Figure ii. The technical shear factor f i is determined from eq( 2.8b).
Finite shear stress refers to an angular deformation similar to the infinitesimal shear stress expression (2,1c). Finite shear deformation is identified with the cosine of the angle $ between P'Q' and P'R' in the deformed configuration (see Figure 2.4b).
Polar Decomposition
This means that the elongation ratios for the principal directions are the principal values of U'. The principal values of B* (i.e. eigenvalues) are the three real and positive roots Pi (i' = 1,2,3) of the characteristic equation for B. For infinitesimal deformations, the differential products in equations (2.27b) and (2.28b) are can ignore. The Eulerian and Langrangian finite strain tensors E and L will then reduce to equation (2.3), which means that the displacement between the material and spatial coordinates is negligible.
Determine the magnitude and direction of principal stresses, maximum shear and octahedral stresses. Determine the components of the displacement vector u = x - X (see Fig. 2.6) in both material and space coordinates. Find R, U and V by polar decomposition and show that the principal values of U and V are equal.
YIELD CRITERIA
- Introduction
- Yielding of Ductile Isotropic Materials
- Influence of the Third Invariant (a) Symmetrical Functions
- Superimposed Hydrostatic Stress
- Distorted Yield Loci
- Introduction
- Classical Theories of Plasticity
- Prandtl-Reuss
- Combined Tension-Torsion with Constant Shear Strain
- Combined Tension-Torsion with Constant Axial Strain
Subtracting Eq.(3.5) from Eq.(3.4) leads to the shear stress energy associated with the deviatoric stress, in Figs. The tensile yield stress O] is used within the denominator of the second term in Eq.(3.16) to ensure homogeneity in stress. Also, with the exception of equation (3.16), they show yield stresses of equal magnitude regardless of the sense of tension.
That is , / i n eq(3.22) is associated with the von Mises criterion of yield/= WzOfj/o-/, where the stress deviator o^1 provides plastic incompressibility. The integration assumes that the components of the voltage deviation tensor a$ in equation (4.3d) increase: (i) proportionally, or (ii) stepwise. Note that, contrary to Prandtl-Reuss flow theory, the Hencky eq(4.2a) has no specific identity with the elastic and plastic components of the total stress.
Neutral Loading Under Internal Pressure Combined with Torsion
The theoretical analyzes each use the von Mises yield criterion (4.8), (a) Hencky. The total stress components are re-expressed from Eq. C4.9a,b) by assuming that any radial and hoop stresses arising from P and C are negligibly small. Eliminating ^ between equations (4.8) and (4.9), the normalized axial stress 5 = rfF, for the plastic region, is found from the solution for . The voltages corresponding to each pv value can be found from Eq.(4.43a,b), given F and E for the material.
Hencky solutions are shown within the enlarged scales of Figs. 4.8c and d compared. This applies to an elastic-plastic state {pv = 0.4) and to a fully plastic state (p^ — 0). The comparison shows.. negligible error in the voltage magnitudes for v=¥i. Within the plastic region (p9 s p s pa) the total axial and shear strain increments, eq.(4.11 a,b), are written as Since the tube is fully plastic, it follows from the corresponding forms of eq. (4.41a,b) and (4.42a,b) that.
Hencky versus Prandtl-Reuss
The shear stress is then held constant as the cylinder is subjected to increasing stress. Derive expressions for the subsequent variation in axial strain with the axial and shear stresses (a, r) according to Hencky and Prandtl-Reuss theories. The axial tension is then held constant as the cylinder is subjected to increasing torque.
Derive expressions for the subsequent change in shear stress with axial and shear stresses (a, r) according to the Hencky and Prandtl-Reuss theories. Show, with progressively increasing circumferential strain, that the Prandtl-Reuss prediction for the subsequent circumferential strain is given by. Compare the predictions with deformation derived from Henky's incompressible theory (v = Vi) and the Prandtl-Reuss theory with v=.
ELASTIC-PERFECT PLASTICITY
Introduction
Elastic-Plastic Bending of Beams
Under an intermediate elastic-plastic moment Mv in Fig. 5.1c, entered the plastic zone by the amount h, from the upper and lower surfaces. The cross section's moment of resistance is composed of elastic and plastic components, such that: .. where s = d - h is the distance between the net force F = Ybh, in each plastic region. Id, the plastic zone has fully penetrated both the tensile and compressive sides of the NA.
Residual Bending Stresses
When an elastic-plastic beam is completely unloaded from the moment M^, a state of residual stress 0R will remain in the cross-section after recovery of bending elastic stresses erE. It is assumed that purely elastic discharge takes place from the elastic-plastic stress state ff which exists below M . Bending theory Within the elastic zone y lies in the inner regions Q <,y &± (df2 - ft).
ZTJ53
Influence of Hardening
It is possible to take into account the effect of linear stiffening on the moment capacity of the profile. If the yield stress is constant at F = 278 MPa, find the applied moment, the stress at the top of the flange, and the position of the neutral axis. Elastic-plastic buckling of a circular shaft is similar to elastic-plastic bending of beams in that three conditions apply to the shaft: fully elastic, elastic-plastic, and fully plastic.
Assuming that the cross-section is elastic, the torsion theory will describe torsion, shear stress and strain. The angle of twist defl is found by applying elastic torsion theory to the core. When the elastic-plastic rod is fully unloaded, elastic stresses are restored to leave a residual shear stress distribution tR.
C^ICyXCyL r o )
A residual angular rotation exists after reading from a torque exceeding Cr. The ratio 0Jdr increases to infinity below Cuk. Cy, elastic torsion theory describes the torque versus twist line for the unit slope, since C fCr = Ql 0r = (Cro)/(Jk). Again, an elastic-perfectly plastic material is assumed, where the plastic zone widens below a constant yield stress value k.
The combination of equations (5.39b) and (5.40) gives the dimensionless elastic-plastic torque Cv for boundary radii in the range rs< r^ s r0;.
- Thick-Walled, Pressurised Cylinder with Closed-Ends
Eq(5.45a) can be normalized by twisting at the yield strength for the outer radius, 0t = MIGrB. To show this, compare the expression for the volume V of a cone of sand that would rest on the end of an upright circular rod (see Table 5.1) with eq(5.30a). Thick-walled cylinders can be safely used even if the internal pressure causes a partially plastic state in their wall.
As with the elastic-perfectly plastic beam and torsion bar, the non-hardening cylinder is also a statically determined problem. These are sufficient to ensure that a strain compatibility condition exists at the elastic-plastic interface. Here we want to determine the internal pressure which penetrates a plastic zone to a radial depth rm, where rti r^s, rB.
5.53) Substituting eq(5.53) into eqs(5.52a,b,e) provides the triaxial stresses in the elastic zone;
- Fully Plastic Disc
- Tresca Versus von Mises
The condition that ar is common to both zones at r = rv is used to redefine the constant A in equation (5.68a) for an elastic-plastic disk. In particular, the solution 0 = 0t for r = rs, when substituted into equation (5.68c), gives the internal pressure pt. When the elastic-plastic pressure pt in equation (5.68c) is released, the elastic stress fffi is assumed to recover for all radii according to the Lame equations (5.47a,b).
If eq(5.73a) predicts am > Y, the implication is that reverse flow of the bore has occurred, invalidating our assumption of an elastic recovery after pressure release. The equilibrium eq(5.55) for an elastic-plastic disc is modified for the centrifugal force within the inner plastic core O s r s rr This gives. The elastic zone stresses are again given by eq(5.79a,b) for the same boundary condition: ar = 0 for r = r0.
SLIP LINE FIELDS
- Introduction
The image point P2 reveals that a translation from P, to P2 can be obtained by rolling the original circle without sliding along fee ordinate r = + k. The geometric interpretation of eqn{6.1 la) is that the pole will trace a cycloid P, P2. The stress level construction in Fig. However, 6.7b shows that a and ft are not uniquely defined. The line connecting vA to vB in Fig, 6.1 lb is the orthogonal image of the slip line between A and B in the physical plane (Fig. 6.1 la).
Where the a-slip line is curved between A and B (see Figure 6.12a), it can be inferred that the curve connecting vA and vB is in the figure. lines (see Fig. 6.19b), the required horizontal stress component is aH. The slip line connecting points 0 and 5 on the physical plane (see Figure 6.23a) is assumed to be straight.
6.19) where Y=2k is the constant, plane stain yield stress, assumed within a deforming zone under
Extrusion Jbr 2<R<3
However, it can be seen that the 5' end point for the cycloid placing an orthogonal line » -, cannot be fixed to the container wall without a field extension. The requirement is that another line a, originating at point 6 and extending to the string line from 5 to 9, must both meet the wall at 45°. The cycloid 2'6' is found by rotating a circle with pole point, 2' and radius k, which rotates along r = - k.
