Introduction To Advanced Elasticity Theory
Any element of material may be subjected to three independent types of stress. Two of these have been considered in detai previously, namely direct stresses and shear stresses, and need not be considered further here. The third type, however, has not been specifically mentioned previously although it has in fact been present in some of the loading cases considered in earlier chapters these are the so-called body-force stresses. These body forces arise by virtue of the bulk of the material,...
T Ots
The Bredt-Batho theory developed in the previous section may be applied to the solution of problems involving cellular sections of the type shown in Fig. 5.7. Fig. 5.7. Thin-walled cellular section. Fig. 5.7. Thin-walled cellular section. Assume the length RSMN is of constant thickness 11 and subjected therefore to a constant shear stress t . Similarly, NOPR is of thickness t2 and stress t2 with NR of thickness i3 and stress Z3. Considering the equilibrium of complementary shear stresses on a...
U 1
Evaluation of all the integrals of eqn. 9.84 leads to the beam element flexural stiffness matrix 12 L2 6 L -12 L2 6 L 6 L 4 -6 L 2 -12 L2 -6 L 12 L2 -6 L 6 L 2 -6 L 4 J which is identical to the stiffness matrix of eqn. 9.68 derived using fundamental equations. The same arguments made in Section 9.8.1 apply with regard to including axial terms to give the force displacement relation, eqn. 9.69 , and corresponding element stiffness matrix, eqn. 9.70 . Element stress matrix in local coordinates...
Info Koy
Eliminating rvv between eqns 4 and 1 we obtain 1 d2 jxx v 32ctvv 1 d2Oyy v d2oxx 1 v d2ox A similar development for cylindrical coordinates yields the stress equation of compability which in the case of axial symmetry where stresses are independent of 9 reduces to From the earlier work of this chapter it should now be evident that in elastic stress analysis there are generally fifteen unknown quantities to be determined six stresses, six strains and three displacements. These are functions of...
Info Fgu
Fig. 6.17. Schematic arrangement of a typical carrier frequency system. Merrow. Fig. 6.17. Schematic arrangement of a typical carrier frequency system. Merrow. The previous discussion has related entirely to the electrical resistance type of strain gauge and, indeed, this is by far the most extensively used type of gauge in industry today. It should be noted, however, that many other forms of strain gauge are available. They include a mechanical gauges or extensometers using optical or...
T Cme
3.6 B . A cantilever is to be constructed from a 40 mm x 60 mm T-section beam with a uniform thickness of 5 mm. The cantilever is to carry a u.d.l. over its complete length of 1 m. Determine the maximum u.d.l. that the cantilever can carry if yielding is permitted over the lower part of the web to a depth of 10 mm. ay 225 MN m. 3.7 B . A 305 mm x 127 mm symmetrical I-section has flanges 13 mm thick and a web 5.4 mm thick. Treating the web and flanges as rectangles, calculate the bending moment...
V Dyv
Therefore the criterion of failure becomes Oy 1 en - ai 1 a2 - lt r3 2 - or, 2 1'2 i.e. 2a2 lt n - a2 2 lt t2 - lt r3 2 lt r3 - atf 8.59 This is clearly the same criterion as that referred to earlier as the Maxwell von Mises distortion or shear strain energy theory. It is sometimes convenient to consider stresses with reference to some false zero, i.e. to measure their values above or below some selected datum stress value, and not their absolute values. This is particularly useful in advanced...
Info Qmp
A circular disc 150 mm diameter and 12 mm thickness is clamped around the periphery and built into a piston of diameter 60 mm at the centre. Assuming that the piston remains rigid, determine the maximum deflection of the disc when the piston carries a load of 5 kN. For the material of the disc E 208 GN m2 and v 0.3. From eqn. 7.29 the deflection of the disc is given by y j -Voger- -j- C2 oger C3 1 slope 0 2 log, r 1 H I---- 2 Now slope 0 at r 0.03 m. Therefore from eqn. 2 0 --- 2 log, 0.03- 1...
Info Ntv
where B is a second convenient constant of integration, aH A - 1 3v r 8 For a solid disc the stress at the centre is given when r 0. With r equal to zero the above equations will yield infinite stresses whatever the speed of rotation unless B is also zero, i.e. B 0 and hence B r2 0 gives the only finite solution. Now at the outside radius R the radial stress must be zero since there are no external forces to provide the necessary balance of equilibrium if o gt were not zero. Therefore from eqn....
Info Mct
Substituting for the element loads column matrix using a relation of the form of eqn. 9.13 gives a h k'M s' e H'M s' e which is the same form as eqn. 9.16 and H e is the stress matrix with respect to local coordinates. Evaluating h e k' lt e gives the stress matrix as r-1 6 t L 41 1 6 t L It L -2b Transformation of displacements and loads Relations of similar form to those of eqns. 9.18 - 9.23 but with additional rotational dof. terms, previously not included in the rod element transformation,...
Info Hmj
thus satisfying the identity l2p m2p n2p 1. Substitution of any principal stress value, again say 01, into the above equations together with the given cartesian stress components allows solution of the determinants and yields values for a 1, b and c , hence k and hence l , m and n , the desired eigen vectors. The process can then be repeated for the other principal stress values o2 03. 8.19. Octahedral planes and stresses Any complex three-dimensional stress system produces three mutually...
Experimental Stress Analysis
We live today in a complex world of manmade structures and machines. We work in buildings which may be many storeys high and travel in cars and ships, trains and planes we build huge bridges and concrete dams and send mammoth rockets into space. Such is our confidence in the modern engineer that we take these manmade structures for granted. We assume that the bridge will not collapse under the weight of the car and that the wings will not fall away from the aircraft. We are confident that the...
Info Tuo 1
The distributions of these stresses are shown in Fig. 8.34. They are similar to that for the pure moment application. The simple bending a My I result is also shown. As in the previous case it is noted that the simple approach underestimates the stresses on the inner fibre. 8.27.8. Case 5-The asymmetric cases n 2-stress concentration at a circular hole in a tension field The example chosen to illustrate this category concerns the derivation of the stress concentration due to the presence of a...
Membrane analogy
It has been stated earlier that the mathematical solution for the torsion of certain solid and thin-walled sections is complex and beyond the scope of this text. In such cases it is extremely fortunate that an analogy exists known as the membrane analogy, which provides a very convenient mental picture of the way in which stresses build up in such components and allows experimental determination of their values. It can be shown that the mathematical solution for elastic torsion problems...
W Mvf
Determine the value of the constants a, b, c and e and hence show that the stresses are a P 2d 3My 2d' - 3Wxy 2d lt rvv 0 8.50 C . A cantilever of unit width length L and depth 2a is loaded by a linearly distributed load as shown in Fig. 8.51, such that the load at distance x is qx per unit length. Proceeding from the sixth order polynomial derive the 25 constants using the boundary conditions, overall equilibrium and the biharmonic equation. Show that the stresses are Examine the state of...
Info Xty
i.e. 1 cm on the stress diagram represents 60 MN m2. The two principal stresses in the plane of the surface are then o, 5.25 cm 315 MN m2 ct2 2.0 cm 120 MN m2 The third principal stress, normal to the free unloaded surface, is zero, i.e. lt 73 0 The directions of the principal stresses are also obtained from the stress circle. With reference to the 0 gauge direction, o lies at 9 15 clockwise cr2 lies at 15 90 105 clockwise with 03 normal to the surface and hence to the plane of o and cr2. N.B....
Info Dny
85.3, 19.8, -25.1 MN m2,0.9592,0.2206,0.1771. 8.43 C . A hollow steel shaft is subjected to combined torque and internal pressure of unknown magnitudes. In order to assess the strength of the shaft under service conditions a rectangular strain gauge rosette is mounted on the outside surface of the shaft, the centre gauge being aligned with the shaft axis. The strain gauge readings recorded from this gauge are shown in Fig. 8.47. If E for the steel 207 GN m2 and v 0.3, determine a the principal...
Info Mug
In this case, therefore, the approximate method yields the same answer for maximum B.M. as the full solution. The maximum stress will then also be equal to that obtained above, i.e. 75.6 MN m2. A hollow circular steel strut with its ends fixed in position has a length of 2 m, an outside diameter of 100 mm and an inside diameter of 80 mm. Assuming that, before loading, there is an initial sinusoidal curvature of the strut with a maximum deflection of 5 mm, determine the maximum stress set up due...
Circular Plates And Diaphragms
The slope and deflection of circular plates under various loading and support conditions are given by the fundamental deflection equation where y is the deflection at radius r dy dr is the slope 0 at radius r Q is the applied load or shear force per unit length, usually given as a function of r D is a constant termed the flexural stiffness or flexural rigidity Ep 12 1 v2 and t is the plate thickness. For applied uniformly distributed load i.e. pressure q the equation becomes Q - and the...
Info Jyt
This equation, therefore, gives the direction of the principal axes. To determine the second moments of area about these axes, cos2 9 J y2dA sm29 J x2 dA - 2 cos 9 sin 9 J xydA Ixx cos2 9 Iyy sin2 9 Ixy sin 29 Substituting for Ixy from eqn. 1.4 , 1 cos29 IXX 1 - cos 29 1 1 cos 29 1 xx j l - cos 29 1 yy - 5 1 cos26 1 xx 5 1 - COS26 1 yy - 5 sec26 1 yy - 4- A cos26 1 yy - Ixx 5 I Iyy gt d - Iyy cos 20 vy sec 20 I w - Ixx cos 26 Iu Jxx cos2 0 Iyy sin2 9 Ixy sin 29 1 cos 29 1 XX 1 - cos 29 1 yy -...
C Hdl
Fig. 2.10. Strut with eccentric load pinned ends Applying a similar procedure to that used previously This is a second-order differential equation, the solution of which is as follows
Unsymmetrical Bending
Similarly, repeating the process with OQ perpendicular to OQ gives the result sin 6 Iv sin 6 cos 9 - I - sin 26 Ixy Thus the construction shown in Fig. 1.8 can be used to determine the second moments of area and the product second moment of area about any set of perpendicular axis at a known orientation to the principal axes. Consider again the general plane surface of Fig. 1.7 having radii of gyration ku and kv about the U and V axes respectively. An ellipse can be constructed on the principal...
Disc With Hole Rotating Radial Displacement
1.1 Product second moment of area 3 1.2 Principal second moments of area 4 1.3 Mohr's circle of second moments of area 6 1.4 Land's circle of second moments of area 7 1.5 Rotation of axes determination of moments of area in terms of the principal values 8 1.6 The ellipse of second moments of area 9 1.8 Stress determination 11 1.9 Alternative procedure for stress determination 11 1.10 Alternative procedure using the momental ellipse 13 1.11 Deflections 15 Examples 16 Problems 24 2.2 Equivalent...
W
The properties of the section are as follows J 19 mm, v 45 mm,IXX 4 x 106 m4. vv 1.1 x 106 m4, .,.,, 1.2 x 106 m4. a the magnitude of the principal second moments of area together with the inclination of their axes relative to XX b the position of the neutral plane N-N and the magnitude of Inn', c the end deflection of the centroid G in magnitude, direction and sense. 444 x 108 m4, 66 x 108 m4, -19 51' to XX, 47 42' to XX, 121 x 10 8 m4, 8.85 mm at -42 18' to XX. 1.7 B . An extruded aluminium...
K 1
Alternatively, since bending always occurs about the N.A., the deflection equation can be written in the form where n.a. is the second moment of area about the N.A. and W' is the component of the load perpendicular to the N.A. The value of n.a. may be found either graphically using Mohr's circle or the momental ellipse, or by calculation using n.a. Wu , du - , cos2 1.28 where au is the angle between the N.A. and the principal U axis. E.J. Hearn, Mechanics of Materials , Butterworth-Heinemann,...
Euler And Rankine-gordon
Having derived the result for the buckling load of a strut with pinned ends the Euler loads for other end conditions may all be written in the same form, where I is the equivalent length of the strut and can be related to the actual length of the strut depending on the end conditions. The equivalent length is found to be the length of a simple bow half sine-wave in each of the strut deflection curves shown in Fig. 2.6. The buckling load for each end condition shown is then readily obtained. The...
Unsymmetrical Bending
The second moments of area of a section are given by The product second moment of area of a section is defined as which reduces to Ixy Ahk for a rectangle of area A and centroid distance h and k from the X and Y axes. The principal second moments of area are the maximum and minimum values for a section and they occur about the principal axes. Product second moments of area about principal axes are zero. With a knowledge of Iyy and Ixy for a given section, the principal values may be determined...