## 3

-3Mb- 14MC+ (40 x 103) (66.67 + 48)103 - 3Mb- 14Mc 74.67 x 103 (2) x 16 3 - 16Mb - 74.67MC 398.24 x 103 (3) (1) -71.67MC 313.66 x 103 -16Mb - 3( - 4.37 x 103) 84.58 x 103 5 Ra ( 4.47 + 12.5) 103 Ra 1.61 kN Ra x 8 - (1 x 103 x 5 x 5.5) + (RB x 3) - (20 x 103 x 1) - 4.37 x 103 3Rb - 4.37 x 103 + 27.5 x 103 + 20 x 103 - 8 x 1.61 x 103 3Rb 30.3 x 103 (-10 x 103 x 5) + 4Rd - (3 x 103 x 4 x 2) - 4.37 x 103 4Rd ( 4.37 + 50 + 24)103 Rd 17.4 kN Ra + Rb + Rc + Rd 47 kN 1.61 + 10.1 +RC+ 17.4 47 This value...

## 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...

## 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...

## 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...

## 1

.82.8 2 x 1902 850 x 106 log,(r 82.8) - 0.405 and < 7ge 850 x 106 log,(r 82.8) + 0.595 .-. At the bore surface where r 62.5 mm the stresses due to autofrettage are Orr -580 MN m2 and om 266.7 MN m2. Residual stresses are then obtained by elastic unloading of the autofrettage pressure, i.e. by applying arr +580 MN m2 at the bore in eqns (10.5) and (10.6) i.e. by proportions arr 580 MN m2 and am -298.3 x - -721 MN m2.

## 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...

## 2

The state of stress on an inclined plane through any given point in a three-dimensional cartesian Fig. 8.3. The state of stress on an inclined plane through any given point in a three-dimensional cartesian ABC and YZ are given by the angle between the normals to both planes n and x, etc.) For convenience, let the plane ABC initially be some perpendicular distance h from Q so that the cartesian stress components actually acting at Q can be shown on the sides of the tetrahdedron element...

## 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...

## 6

(70 x 50 x 12 + 2 x 100 x 50 x 12)109 x 10-6 70 J70 + 200)_15*103 3.89 kN force in brass 15-3.89 11.11 kN load 11.11 xlO3 area 2x50xl2xKT6 9.26 MN m2 (compressive) area 50xl2xl0-6 6.5 MN m2 (compressive) These stresses represent the changes in the stresses owing to the applied load. The total or resultant stresses owing to combined applied loading plus temperature effects are, therefore, A 25 mm diameter steel rod passes concentrically through a bronze tube 400 mm long, 50 mm external diameter...

## 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

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

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

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

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

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

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

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...