9 Kg
14.3 A . The extension of a 100 mm gauge length of 14.33 mm diameter bar was found to be 0.15 mm when a tensile load of 50 kN was applied. A torsion specimen of the same specification was made with a 19 mm diameter and a 200 mm gauge length. On test it twisted 0.502 degree under the action of a torque of 45 N m. Calculate , G, K and v. 206.7, 80.9, 155 GN m2 0.278. 14.4 A . A rectangular steel bar of 25 mm x 12 mm cross-section deflects 6 mm when simply supported on its 25 mm face over a span...
3 Wl
If the spring is constructed from strips and placed one on top of the other as shown in Fig. 12.9, uniform stress conditions are retained, since if the strips are cut along XX and replaced side by side, the equivalent leaf spring is obtained as shown. Fig. 12.9. Semi-elliptic carriage spring showing initial pre-forming. Fig. 12.9. Semi-elliptic carriage spring showing initial pre-forming. Such a spring is then termed a carriage spring with n strips of width b, i.e. B nb. Therefore the bending...
1 Fem
max. strain at notch nominal strain at notch Stress concentration factor Kp in presence of plastic flow is related to Ke by Neuber's rule The design of components subjected to contact, i.e. local compressive stress, is extremely important in such engineering applications as bearings, gears, railway wheels and rails, cams, pin-jointed.links, etc. Whilst in most other types of stress calculation it is usual to neglect local deflection at the loading point when deriving equations for stress...
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...
1415 Analytical determination of principal strains from rosette readings
The values of the principal strains associated with the three strain readings taken from a strain gauge rosette may be found by calculation using eqn. 14.14 , i.e. ee i e, e, i ex - e, cos 28 yx, sin 26 This equation can be applied three times for the three values of 8 of the rosette gauges. Thus with three known values of eg for three known values of 6, three simultaneous equations will give the unknown strains ex, ey and yxy. The principal strains can then be determined from eqn. 14.16 . or...
1 Ril
5.8 B C . Show that for the symmetrical section shown in Fig. 5.13 there is no stress in the central web. Show also that the shear stress in the remainder of the section has a value of T 4tb2. 5.9 C . A washing machine agitator of the cross-section shown in Fig. 5.14 acts as a torsional member subjected to a torque T. The central tube is 100 mm internal diameter and 12 mm thick the rectangular bars are 50 mm x 18 mm section. Assuming that the total torque carried by the member is given by...
21
This equation, therefore, gives the direction of the principal axes. To determine the second moments of area about these axes, J v2 dA J y cos 9 x sin 9 2 dA cos2 9 J y2dA sm29 J x2 dA - 2 cos 8 sin 9 J xydA I xx cos2 9 Iyy sin2 8 Ixy sin 29 Substituting for Ixy from eqn. 1.4 , Iu U cos 29 IXX 1 1 - COS 29 1 cos 28 1 xx j l - cos 28 1 yy - 5 1 cos28 1 xx 5 1 - COS28 1 yy - 5 sec28 1 yy - Ixx cos28 1 vv - Ixx Iyy Ixx - Iyy COS 29 - v gt , - lxx sec 26 w - Ixx cos 28 lu lxx cos2 0 Iyy sin2 9 Ixy...
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...
208
a A steel sleeve of 150 mm outside diameter is to be shrunk on to a solid steel shaft of 100 mm diameter. If the shrinkage pressure set up is 15 MN m2, find the initial difference between the inside diameter of the sleeve and the outside diameter of the shaft. b What percentage error would be involved if the shaft were assumed to be incompressible For steel, E 208 GN m2 v 0.3. a Treating the sleeve as a thick cylinder with internal pressure 15 MN m2, at r 0.05, ar - 15 MN m2 and at r 0.075, lt...
05 03
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...
2 I 2
When loading has been continued until the stress distribution is as in Fig. 3.3 c assumed , the beam with collapse. The moment required to produce this fully plastic state can be obtained from eqn. 3.2 , since d is then zero, i.e. fully plastic moment, MFP x 3D This is the moment therefore which produces a plastic hinge in a rectangular-section beam. 32. Shape factor - symmetrical sections The shape factor is defined as the ratio of the moments required to produce fully plastic and maximum...
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...
2
The B.M. at any point C, distance x from A, is given by These relationships are the basis of the rules stated in the summary, the proofs of which are as follows a The maximum or minimum B.M. occurs where dM dx - 0 Thus where S.F. is zero B.M. is a maximum or minimum. b The slope of the B.M. diagram dM dx Q. Thus where Q 0 the slope of the B.M. diagram is zero, and the B.M. is therefore constant. c Also, since Q represents the slope of the B.M. diagram, it follows that where the S.F. is positive...
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...
Ei Bnt
-- x 0.4 -- --3.72 x 1 2.952 - 3.33 0.661 - 0.032 - 3.72 2.952 The beam therefore is deflected downwards at the given position. Calculate the slope and deflection of the beam loaded as shown in Fig. 5.40 at a point 1.6 m from the left-hand end. El 1.4 MNm2. Since, by symmetry, the point of zero slope can be located at C a solution can be obtained conveniently using Mohr's method. This is best applied by drawing the B.M. diagrams for the separate effects of a the 30 kN loads, and b the 20 kN...
T Oww
Fig. 3.15. 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...
E 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...
P7
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...
d0
Fig. 8.19. a Element showing stresses which contribute to equilibrium in the radial and circumferential directions, b Radial components of hoop stresses.
74
i.e. maximum bending stress r 12.3 Fig. 12.2. Close-coiled helical spring subjected to axial torque T. Under the action of an axial torque the deflection of the spring becomes the wind-up angle of the spring, i.e. the angle through which one end turns relative to the other. This will be equal to the total change of slope along the wire, which, according to Mohr's area-moment theorem see 5.7 , is the area of the M EI diagram between the ends.
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...
2 Atb
3 v - R - RXR2 - R,R2 The complete radial and hoop stress distributions are indicated in Fig. 4.4. Fig. 4.4. Hoop and radial stress distribution in a rotating hollow disc. Fig. 4.4. Hoop and radial stress distribution in a rotating hollow disc. 4.4. Rotating thick cylinders or solid shafts In the case of rotating thick cylinders the longitudinal stress aL must be taken into account and the longitudinal strain is assumed to be constant. Thus, writing the equations for the strain in three...
30
It may be observed at this stage that the S.F. diagram can be obtained very quickly when working from the left-hand side, since after plotting the S.F. value at the support all subsequent steps are in the direction of and equal in magnitude to the applied loads, e.g. 10 kN up at A, down 10 kN at B, up 20 kN at C, etc., with horizontal lines joining the steps to show that the S.F. remains constant between points of application of concentrated loads. The S.F. and B.M. values at the left-hand...
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...
5w
This is sometimes written in the form where m the bending moment resulting from a unit load only in the place of W. This method of solution is then termed the unit load method. Castigliano's theorem also applies to angular movements If the total strain energy expressed in terms of the external moments be partially differentiated with respect to one of the moments, the result is the angular deflection in radians of the point of application of that moment and in its direction where M, is the...
Ixx
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...
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...
V 1
c Stress system at common surface c Stress system at common surface Fig. 10.17. Uniform heating of compound cylinders constructed from tubes of different materials-in this case, steel and brass. p, is the radial pressure introduced at the common interface by virtue of the differential thermal expansions. Therefore, as for the compound bar treatment of 2.3 compression of steel compression of brass difference in free lengths where d is the initial nominal diameter of the mating surfaces, ag and...
A Hng
11.26 B . A steel ring, of 250 mm diameter, has a width of 50 mm and a radial thickness of 5 mm. It is split to leave a narrow gap 5 mm wide normal to the plane of the ring. Assuming the radial thickness to be small compared with the radius of ring curvature, find the tangential force that must be applied to the edges of the gap to just close it. What will be the maximum stress in the ring under the action of this force E 200 GN m2. 11.27 B . Determine, for the cranked member shown in Fig....
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...
2 2
Here, a H and ar are the hoop and radial stresses at the inside radius and ay is the allowable yield stress of the material taking into account any safety factors which may be introduced by the company concerned. For brittle materials such as cast iron the Rankine maximum principal stress theory is used. In this case failure is deemed to occur when 10.17. Plastic yielding - auto-frettage It has been shown that the most highly stressed part of a thick cylinder is at the inside radius. It...
Stress Concentration Crack Tip
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 1
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...
Pl Ei
B.M. at D M - yE-2yD yc 3 L WL EI B.M. at C Mc P.-T- iyA-2yc yD At point A it is necessary to introduce the mirror image of the beam giving point C' to the left of A with a deflection c yc in order to produce the fourth equation.