W L

Consider now the uniformly loaded strut of Fig. 2.13 with the origin again selected at the centre but y measured from the maximum deflected position. y 5 A cos nx + B sin nx --x--- When x LI2, y S A - sec Thus the maximum deflection S, when y 0 and x 0, is given by In the case of a member carrying a tensile load (i.e. a tie) together with a uniformly distributed load, the above procedure applies with the sign for P reversed. The relevant differential expression then becomes i.e. (D2 n2)y in...

3

Therefore substituting in (2) and (1), the stress conditions at the inside surface are oH 420a> 2 +412 co2- 4.43 w2 827w2 with Or 0 and at the outside ah 420a + 11.42m2 - 159co2 272& > 2 The most severe stress conditions therefore occur at the inside radius where the maximum shear stress is greatest

V

Therefore, neglecting second-order small quantities, Assuming now that plane sections remain plane, i.e. the longitudinal strain is constant across the wall of the cylinder, It is also assumed that the longitudinal stress aL is constant across the cylinder walls at points remote from the ends. Multiplying through by r and rearranging, Therefore, integrating, arr2 - Ar2 constant B (say) The above equations yield the radial and hoop stresses at any radius r in terms of constants A and B. For any...

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 9 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...

1416 Alternative representations of strain distributions at a point

Alternative forms of representation for the distribution of stress at a point were presented in 13.7 the directly equivalent representations for strain are given below. The values of the direct strain ee and shear strain ye for any inclined plane 0 are given by equations (14.14) and (14.15) as Ee 2 (ex + ey) + 2 ( * - y) cos 20 + yxy sin 20 2 le - i (e* - y) sin 20 - bxy COS 20 Plotting these values for the uniaxial stress state on Cartesian axes yields the curves of Fig. 14.23 which can then...

2

Thus the maximum hoop stress is 33.5 MN m2 Example 10.9 (C) A thick cylinder of internal and external radii 300 mm and 500 mm respectively is subjected to a gradually increasing internal pressure P. Determine the value of P when (a) the material of the cylinder first commences to yield (b) yielding has progressed to mid-depth of the cylinder wall (c) the cylinder material suffers complete collapse. Solution See Chapter 3 from Mechanics of Materials From eqn. (3.35) the initial yield pressure...

12

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...

151 Maximum principal stress theory

This theory assumes that when the maximum principal stress in the complex stress system reaches the elastic limit stress in simple tension, failure occurs. The criterion of failure is thus It should be noted, however, that failure could also occur in compression if the least principal stress a3 were compressive and its value reached the value of the yield stress in compression for the material concerned before the value of ayi was reached in tension. An additional criterion is therefore

0949

Considerable care is necessary in the design of bearings when selecting appropriate ball and bearing race radii. If the radii are too similar the area of contact is large and excessive wear and thermal stress (from frictional heating) results. If the radii are too dissimilar then the contact area is very small, local compressive stresses become very high and the load capacity of the bearing is reduced. As a compromise between these extremes the radius of the race is normally taken to be between...

0333 0333

Timoshenko, Strength of Materials, Part I, Elementary Theory and Problems, Van Nostrand. New York. ' S. Timoshenko, Strength of Materials, Part I, Elementary Theory and Problems, Van Nostrand. New York. The essential difference between the shear stress distributions in circular and rectangular members is illustrated in Fig. 5.1, where the shear stress distribution along the major and minor axes of a rectangular section together with that along a radial line to the corner of the section are...

Info

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...

M

First moment of area of diagram about one end 0 EI first moment of area of B.M. diagram about one end 0 (6.2) To make use of these equations it is convenient to break down the B.M. diagram for the built-in beam into two parts (a) that resulting from the loading, assuming simply supported ends, and known as the free-moment diagram (b) that resulting from the end moments or fixing moments which must be applied at the ends to keep the slopes zero and termed the fixing-moment diagram. 6.1. Built-in...

Contents

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...

3 Ei

Therefore component of deflection perpendicular to the V axis _ WVL3 _ 10 000 cos 47 5T x 23 3EIV 3 x 200 x 109 x 2.27 x 106 and component of deflection perpendicular to the U axis _ WUL _ 10000sin47 51' x 23 3EIU 3 x 200 x 109 x 50.43 x 10 6 1.96 x 103 1.96 mm The total deflection is then given by yj 82u + S2V) 103a (39.42 + 1.962) 39.45 x 103 39.45 mm Alternatively, since bending actually occurs about the N.A., the deflection can be found from its direction being normal to the N.A. From...

56

Nominal stress at fracture - z 280 MN m2 Actual stress at fracture -- 856.9 MN m2 1.1 (A). A 25 mm square-cross-section bar of length 300 mm carries an axial compressive load of 50 kN. Determine the stress set up in the bar and its change of length when the load is applied. For the bar material 200GN m2. 80 MN m2 0.12mm. 1.2 (A). A steel tube, 25 mm outside diameter and 12 mm inside diameter, carries an axial tensile load of 40 kN. What will be the stress in the bar What further increase in...

425

For complete accuracy one or two intermediate values should be obtained along each u.d.l. portion of the beam, e.g. B.M. midway between A and B (42.5 x 1) (10 x 1 x 42.5-5 37.5 kNm Similarly, B.M. midway between C and D 45 kN m B.M. midway between D and E - 39 kN m The B.M. and S.F. diagrams are then as shown in Fig. 3.9. A point of contraflexure is a point where the curvature of the beam changes sign. It is sometimes referred to as a point of inflexion and will be shown later to occur at the...

6

This is a linear expression in h and k producing the line SS in Fig. 4.13. If the load is now applied in each of the other three quadrants the total limiting area within which P must be applied to produce zero tension in the section is obtained. This is the diamond area shown shaded in Fig. 4.14 with diagonals of b 3 and d 3 and hence termed the middle third. For circular sections of diameter d, whatever the position of application of P, an axis of symmetry will pass through this position so...

N Je

- 2(6 2 - 9X ) + (sin 26 2 - sin 20, ) 2 n The stress function concept described above was developed over 100 years ago. Despite this, however, the ideas contained are still of relevance today in providing a series of classical solutions to otherwise intractable problems, particularly in the study of plates and shells. At a point in a material subjected to a three-dimensional stress system the cartesian stress coordinates are oxx 100 MN m2 Gyy, 80 MN m2 azz 150 MN m2 oxy 40 MN m2 < jyZ -30...

1517

Co 1033 rad s 9860 rev min Example 42 A steel rotor disc which is part of a turbine assembly has a uniform thickness of 40 mm. The disc has an outer diameter of 600 mm and a central hole of 100 mm diameter. If there are 200 blades each of mass 0.153 kg pitched evenly around the periphery of the disc at an effective radius of 320 mm, determine the rotational speed at which yielding of the disc first occurs according to the maximum shear stress criterion of elastic failure. For steel, E 200 GN...

850

It is unlikely that the Tresca theory will give such a high value in practice since the axial working stress (ignored in this calculation) may well become the major principal stress o in the working condition and increase the magnitude of the denominator to reduce the resulting value of n. 3.1 (A B). Determine the shape factors for the beam cross-sections shown in Fig. 3.45, in the case of section (c) treating the section both with and without the dotted area. 1.23, 1.81, 1.92, 1.82.

1

A compound tube is made by shrinking one tube of 100 mm internal diameter and 25 mm wall thickness on to another tube of 100 mm external diameter and 25 mm wall thickness. The shrinkage allowance, based on radius, is 0.01 mm. If both tubes are of steel (with E 208 GN m2), calculate the radial pressure set up at the junction owing to shrinkage. Let p be the required shrinkage pressure, then for the inner tube At r 0.025, ar 0 and at r 0.05, ar - p

4g

Such a shaft is 0.06 m in diameter and has a flywheel of mass 30 kg and radius of gyration 0.25 m situated at a distance of 1.2 m from a bearing. The flywheel is rotating at 200 rev min when the bearing suddenly seizes. Calculate the maximum shear stress produced in the shaft material and the instantaneous angle of twist under these conditions. Neglect the shaft inertia. For the shaft material G 80 GN m2. B.P. 196.8 MN m2, 5.64 . 11.7 (A B). A solid shaft carrying a flywheel of mass 100 kg and...

100

Ws WE and the maximum stress is given by A horizontal steel beam of I-section rests on a rigid support at one end, the other end being supported by a vertical steel rod of 20 mm diameter whose upper end is rigidly held in a support 2.3 m above the end of the beam (Fig. 11.17). The beam is a 200 x 100 mm B.S.B. for which the relevant -value is 23 x 10 6 m4 and the distance between its two points of support is 3 m. A load of 2.25 kN falls on the beam at mid-span from a height of 20 mm above the...

Pr1

If the initial tensile stress in the wire is T the final tensile hoop stress in the winding at any radius r is less than T by an amount equal to the compressive hoop stress set up by the effective external pressure caused by the winding, Using the same analysis outlined in 10.2, r_T ff > 2+* ) r V+R2_r2+R2 Multiplying through by -j T and rearranging, (r Ki) (r2-R2)dr '(r2-R2)2r (r2-RJ) d T r2 1_ Tr drl -R2)*' (r2 R ) Therefore substituting in eqn. (10.20), 02 R ) r 2 e (Rl Rj)

4g K

It should be noted that in the four types of loading case considered above the strain energy expressions are all identical in form, 2 x product of two related constants the constants being related to the type of loading considered. In bending, for example, the relevant constants which appear in the bending theory are E and , whilst for torsion G and J are more applicable. Thus the above standard equations for strain energy should easily be remembered. 11.5. Strain energy of a three-dimensional...

A A

Where A is the cross-sectional area of the ring. Now with unit length assumed, m A is the mass of the ring material per unit volume, i.e. the density p. 9.3. Thin spherical shell under internal pressure Because of the symmetry of the sphere the stresses set up owing to internal pressure will be two mutually perpendicular hoop or circumferential stresses of equal value and a radial stress. As with thin cylinders having thickness to diameter ratios less than 1 20, the radial stress is assumed...

Introduction

This text is a revised and extended third edition of the highly successful text initially published in 1977 intended to cover the material normally contained in degree and honours degree courses in mechanics of materials and in courses leading to exemption from the academic requirements of the Engineering Council. It should also serve as a valuable reference medium for industry and for post-graduate courses. Published in two volumes, the text should also prove valuable for students studying...

1 TlL f 2 WG J t

The angle of twist of the tube is therefore given by For tubes of constant thickness this reduces to where s is the perimeter of the median line. The above equations must be used with care and do not apply to cases where there are abrupt changes in thickness or re-entrant corners. For closed sections which have constant thickness over specified lengths but varying from one part of the perimeter to another Si S2 S3 + + + etc. t h 3 5.7. Use of equivalent J for torsion of non-circular sections...

036

No need to complete these columns for this example With reference to 9.7, the element stresses are obtained from With reference to 9.7, the element stresses are obtained from o(e) w i(e) where, for element a, the displacement column matrix is s(a) (m, V 9 2 v2 2 in which 112 0 in this example. Substituting for element a and letting superscript i denote extreme inner fibres and superscript o denote extreme outer fibres, gives

3 T

There are many cases, particularly in civil engineering applications, where rolled steel or extruded alloy sections are used where some element of torsion is involved. In most cases the sections consist of a combination of rectangles, and the relationships given in eqns. (5.1) and (5.2) can be adapted with reasonable accuracy provided that (a) the sections are open, i.e. angles, channels. T-sections, etc., as shown in Fig. 5.2 (b) the sections are thin compared with the other dimensions. Fig....

De de

3 la 3 (< 7 aM) Similarly, in the e direction, the relevant equilibrium equation reduces to 3 1 o< M) 3 2 re n These are, then, the stress equations of equilibrium in cylindrical coordinates and in their most general form. Clearly these are difficult to memorise and, fortunately, very few problems arise in which the equations in this form are required. In many cases axial symmetry exists and circular sections remain concentric and circular throughout loading, i.e. Ore 0.

4

XxXx MjLx M2Li E , _ A2x2 M3L2 M2L2 E( 2 -) - - - --(- --1---- This is the full three-moment equation it can be greatly simplified if the beam is uniform, i.e. 11 I2 I, as follows If the supports are on the same level, i.e. S2 0, - Mi L, - 2M2 Lt + L2 - M3 L2 6 j + J This is the form in which Clapeyron's three-moment equation is normally used. The following standard results for j are very useful Wab a 2a Wab b I b -x- x--1--x- a + -

J V

Which, for the case of a prismatic element with constant stress and strain over the volume, becomes The formulation of a rod element will be considered using two approaches, namely the use of fundamental equations, based on equilibrium, compatibility and constitutive (i.e. stress strain law), arguments and use of the principle of virtual work equation. 9.7.1. Formulation of a rod element using fundamental equations Consider the structure shown in Fig. 9.18, for which the deformations (derived...

P3r

2 (Ri-R2p) i.e. the radial pressure at the elastic interface is Thus from eqns. (3.29) and (3.30) the stresses in the plastic zone are given by The pressure required for complete plastic collapse of the cylinder is given by eqn. (3.29) when r Ri and Rp R2 with P3 P2 0 (at the outside edge). With a knowledge of this collapse pressure the design pressure can be determined by dividing it by a suitable load factor as described in 3.8. The pressure at initial yield is found from eqn. (3.31) when Rp...

Ei

And is thus directly related to the value of the quantity EI. Since the radius of curvature is a direct indication of the degree of flexibility of the beam (the larger the value of R, the smaller the deflection and the greater the rigidity) the quantity EI is often termed the flexural rigidity or flexural stiffness of the beam. The relative stiffnesses of beam sections can then easily be compared by their EI values. It should be observed here that the above proof has involved the assumption of...

0026

A thick cylinder of 100 mm external diameter and 50 mm internal diameter is wound with steel wire of 1 mm diameter, initially stressed to 20 MN m2 until the outside diameter is 120 mm. Determine the maximum hoop stress set up in the cylinder if an internal pressure of 30 MN m2 is now applied. To find the stresses resulting from internal pressure only the cylinder and wire may be treated as a single thick cylinder of 50 mm internal diameter and 120 mm external diameter. Now < rr - 30 MN m2 at...

20

A 50 mm x 20 mm rectangular-section beam is used simply supported over a span of 2 m. If the beam is used with its long edges vertical, determine the value of the central concentrated load which must be applied to produce initial yielding of the beam material. If this load is then increased by 10 determine the depth to which yielding will take place at the centre of the beam span. Over what length of beam has yielding taken place What are the maximum deflections for each load case Take...

Introduction To The Finite Element Method

So far in this text we have studied the means by which components can be analysed using so-called Mechanics of Materials approaches whereby, subject to making simplifying assumptions, solutions can be obtained by hand calculation. In the analysis of complex situations such an approach may not yield appropriate or adequate results and calls for other methods. In addition to experimental methods, numerical techniques using digital computers now provide a powerful alternative. Numerical techniques...

E E E E

< r1 v< r2 vff3 ay (15.2) This theory is contradicted by the results obtained from tests on flat plates subjected to two mutually perpendicular tensions. The Poisson's ratio effect of each tension reduces the strain in the perpendicular direction so that according to this theory failure should occur at a higher load. This is not always the case. The theory holds reasonably well for cast iron but is not generally used in design procedures these days. 15.4. Maximum total strain energy per...

16

Thus the safe moment which the beam can carry within both limiting stress values is 69 kN m. (b) For this part of the question the dimensions of the new beam are required and it is necessary to assume a critical or economic section. The position of the N.A. is then determined from eqn. (4.19) by consideration of the proportions of the stress distribution (i.e. assuming that the maximum stresses in the streel and concrete occur together). Thus from eqn. (4.19) Substituting for 0.46 and solving...

R22R2l

Now, substituting for R2 R1+t and D 2Rlt Thus for various D t ratios the stress values from the two theories may be plotted and compared this is shown in Fig. 10.7. Also indicated in Fig. 10.7 is the percentage error involved in using the thin cylinder theory. It will be seen that the error will be held within 5 if D t ratios in excess of 15 are used. i i i i i i i i i i 2 4 6 8 10 12 14 re 18 20 Fig. 10.7. Comparison of thin and thick cylinder theories for various diameter thickness ratios. i...

Abcd Beam Of 10 M Bc Is Supported Beam 4 M Apart

A beam of length 10 m is symmetrically placed on two supports 7 m apart. The loading is 15kN m between the supports and 20 kN at each end. What is the central deflection of the beam 210 GN m2 200x 106m4. 6.8mm. 5.2 (A B). Derive the expression for the maximum deflection of a simply supported beam of negligible weight carrying a point load at its mid-span position. The distance between the supports is L, the second moment of area of the cross-section is I and the modulus of elasticity...

A 300 Mm By 125 Mm I-beam Point Load Factor

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...

Horizontal Shaft 75mm Diameter Projects

R2 lL67 58-3Nm and Ti 621.7 Nm T1R1 621.7 x 25 xlO-3 T( , tmax - 25.33 x 106 N m2 Angle of twist for each portion - x 504 x 1012 x 80 x 109 32 8.1 (A). A solid steel shaft A of 50 mm diameter rotates at 250rev min. Find the greatest power that can be transmitted for a limiting shearing stress of 60 MN m2 in the steel. It is proposed to replace A by a hollow shaft B, of the same external diameter but with a limiting shearing stress of 75 MN m2. Determine the internal diameter of B to transmit...

If Sf Between Two Sections Is 140kn Then Bm Is

Rc x 6 10 x 3 x 1.5 80 5 x 6 x 48 x x 6 6Kc 45 80 288 Rc 27.2 kN Rc Rb 10 x 3 i x 6 x 48 30 144 174 R 146.8 kN At any distance x from C between C and B the shear force is given by w 8x kN 'm S.F.XX - Rc-jx 8xxx -Rc 4x2 -27.2 4x2 The S.F. diagram is then as shown in Fig. 3.18. 47.3 kNm B.M. at A and C 0 B.M. immediately to left of B - 10 x 3 x 1.5 -45kNm At the point of application of the applied moment there will be a sudden change in B.M. of 80 kN m. There will be no such discontinuity in the...

In A Certain Material Under Load A Plane Ab Carries A Tensile Direct

Summary of principal planes and maximum shear planes. The stress system is as shown in Fig. 13.32. The centre of the Mohr's circle is positioned midway between the two direct stresses given, and the radius is such that ay 100 MN m2. By measurement dt 75 30' to BC, the plane on which the 50 MN m2 stress acts In a certain material under load a plane AB carries a tensile direct stress of 30 MN m2 and a shear stress of 20 MN m2, while another plane BC carries a tensile direct stress of...

Experimental Stress Analysis

Brittle Lacquer Figure Strain Gauge

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...

Complex Stresses

The normal stress a and shear stress t on oblique planes resulting from direct loading are a ay sin2 8 and t joy sin 28 The stresses on oblique planes owing to a complex stress system are normal stress lt jx ay j ax - ay cos 20 zxy sin 26 shear stress j ax ay sin 26 zxy cos 26 The principal stresses i.e. the maximum and minimum direct stresses are then oi iK lt ry iy l ax - ay 2 4T2, i ax ay - - ay 2 4t2, and these occur on planes at an angle 6 to the plane on which ax acts, given by either...

T 1v

With the normally accepted value of Poisson's ratio for general steel work of 0.3, the thickness ratio becomes i.e. the thickness of the cylinder walls must be approximately 2.4 times that of the hemispherical ends for no distortion of the junction to occur. In these circumstances, because of the reduced wall thickness of the ends, the maximum stress will occur in the ends. For equal maximum stresses in the two portions the thickness of the cylinder walls must be twice that in the ends but some...

05

600 x log, 1.67 600 x 0.513 308 MN m2 10.1 B . A thick cylinder of 150 mm inside diameter and 200 mm outside diameter is subjected to an internal pressure of 15 MN m2. Determine the value of the maximum hoop stress set up in the cylinder walls. 10.2 B . A cylinder of 100 mm internal radius and 125 mm external radius is subjected to an external pressure of 14bar 1.4 MN m2 . What will be the maximum stress set up in the cylinder 7.8 MN m2. 10.3 B . The cylinder of Problem 10.2 is now subjected...

818

Stress Concentration

I.e. for equal lengths as is normally the case for parallel shafts Thus two equations are obtained in terms of the torques in each part of the composite shaft and these torques can therefore be determined. The maximum stresses in each part can then be found from It nil be shown in 13.2 that a state of pure shear as produced by the torsion of shafts is equivalent to a system of biaxial direct stresses, one stress tensile, one compressive, of equal value and at 45 to the shaft axis as shown in...

Triangular Udl Shear Diagram

5R 5x 1 7x4 2x6 4x5 x 2.5 and since RA RB 5 7 2 4 x 5 34 The S.F. diagram may now be constructed as described in 3.4 and is shown in Fig. 3.17. B.M. at D 2 x 2 19 x 1 4 x 1 x 13kNm B.M. at 15x l - 4x 1 x 13kNm The maximum B.M. will be given by the point or points at which dM dx i.e. the shear force is zero. By inspection of the S.F. diagram this occurs midway between D and E, i.e. at 1.5 m from E. B.M. at this point 2.5 x 15 - 5 x 1.5 -1 4 x 2.5 x 17.5 kNm There will also be local maxima at the...

T

S.F., B.M. and thrust diagrams for system of inclined loads. Fig. 3.15. S.F., B.M. and thrust diagrams for system of inclined loads. yield the values of the vertical reactions at the supports and hence the S.F. and B.M. diagrams are obtained as described in the preceding sections. In addition, however, there must be a horizontal constraint applied to the beam at one or both reactions to bring the horizontal components of the applied loads into equilibrium. Thus there will be a...

Osr

In the fully plastic condition, therefore, when the stress is equal throughout the section, the above equation reduces to y areas above N.A. areas below N.A. and in the special case shown in Fig. 3.5 the N.A. will have moved to a position coincident with the lower edge of the flange. Whilst this position is peculiar to the particular geometry chosen for this section it is true to say that for all unsymmetrical sections the N.A. will move from its normal position when the section is completely...

Cylinder Stress Element

Circumferential Riveted Joint

Change of internal volume of cylinder under pressure 4v V change of volume of contained liquid under pressure where K is the bulk modulus of the liquid. For thin rotating cylinders of mean radius R the tensile hoop stress set up when rotating at co rad s is given by aH pa gt 2R2. circumferential or hoop stress aH change of volume under pressure 1 v V Effects of end plates and joints-add joint efficiency factor ri to denominator of stress equations above. 9.1. Thin cylinders under internal...

Octahedral Plane

Octahedral Plane

A three-dimensional complex stress system has principal stress values of 280 MN m2, 50 MN m2 and - 120 MN m2. Determine a analytically and b graphically i the limiting value of the maximum shear stress ii the values of the octahedral normal and shear stresses. i The limiting value of the maximum shear stress is the greatest value obtained in any plane of the three-dimensional system. In terms of the principal stresses this is given by ii The octahedral normal stress is given by iii The...

Stress Distribution In Thick Cylinders

Thick Cylinder Stress

For compound tubes the resultant hoop stress is the algebraic sum of the hoop stresses resulting from shrinkage and the hoop stresses resulting from internal and external pressures. For force and shrink fits of cylinders made of different materials, the total interference or shrinkage allowance on radius is where eH and eH are the hoop strains existing in the outer and inner cylinders respectively at the common radius r. For cylinders of the same material this equation reduces to For a hub or...

Shear Stress For Cross Section

Taking moments about the top edge Fig. 7.12a , 100 x 12 x 6 10 9 10 x 12 x 17 10 9 100 x 12 10 x 12 h x 109 where h is the centroid of the shaded T-section, The distribution of shear stress due to bending is then shown in Fig. 7.12b, giving a maximum shear stress of tm,x 66MN m2. Now the mean shear stress across the section is _ shear force _ 100 x 103 area 3.912 x 10 3 At a certain section a beam has the cross-section shown in Fig. 7.13. The beam is simply supported at its ends and carries a...

Octahedral Shear Stress

Thus satisfying the identity l2p m2p n2p 1. Substitution of any principal stress value, again say cr,, 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 02 03. 8.19. Octahedral planes and stresses Any complex three-dimensional stress system produces three mutually...

158 Graphical solution of twodimensional theory of failure problems

Theory Failure With Diagram

The graphical representations of the failure theories, or yield loci, may be combined onto a single set of a, and lt r2 coordinate axes as shown in Fig. 15.12. Inside any particular locus or failure envelope elastic conditions prevail whilst points outside the loci suggest that yielding or fracture will occur. It will be noted that in most cases the maximum shear stress criterion is the most conservative of the theories. The combined diagram is particularly useful since it allows experimental...

Stress Concentration Crack

Stress Concentration Crack

Shear stress t in double shear 1.8 Fig. 1.14. a Single shear, b Double shear. 1.16. Allowable working stress-factor of safety The most suitable strength or stiffness criterion for any structural element or component is normally some maximum stress or deformation which must not be exceeded. In the case of stresses the value is generally known as the maximum allowable working stress. Because of uncertainties of loading conditions, design procedures, production methods, etc., designers generally...

300

Determine for each case the resultant stress at P on a plane through P whose normal is coincident with the X axis. 8.10 C . At a point in a material the stresses are 37.2 MN m2 ayy 78.4 MN m2 azz 149 MN m2 axy 68.0 MN m2 ayz -18.1 MN m2 azx 32 MN m2 Calculate the shear stress on a plane whose normal makes an angle of 48 with the X axis and 71 with the Y axis. 8.11 C . At a point in a stressed material the cartesian stress components are lt Jxx -40 MN m2 lt jyy 80 MN m2 aZ7 120 MN m2 axy 72 MN...