EQUILIBRIUM STRESS SYSTEMS

Transcription

1 EQUILIBRIUM STRESS SYSTEMS Definition of stress The general definition of stress is: Stress = Force Area where the area is the cross-sectional area on which the force is acting. Consider the rectangular element of σ zz material shown, which is in equilibrium and acted on b forces at its si faces. The σ forces will in general act at oblique angles to the faces. We can resolve the forces to σ σ find the components normal σ to the faces; these give rise to normal stresses, obtained z numericall b dividing the component of force b the area of the face on which it σ zz acts. The normal stresses are denoted b σ, with subscripts to identif the component of stress relative to an ais sstem. The first component refers to the direction of the normal to the plane on which the force acts, and the second subscript to the direction of the force itself. With normal stresses, the two subscripts are alwas the same. We can also resolve the oblique force to give its components in the plane on which it acts. These components will be z z z shear forces, which we have alread met in beams. We z obtain the shear stress b dividing the shear force b the area of the face in which it acts. The smbol is usuall used for z shear stress, subscripted using the same rule as for normal stress; now, the two subscripts must alwas be different from each other.

2 The following diagram shows all the stress components acting. There are a total of nine stress components, though because of relationships between the shear stresses there are onl si independent components. σ zz z z z z σ σ z Sign conventions Normal stress With normal stresses, the sign of the stress is not related to the direction of arrows like the ones in the figure above. Rather, the convention is that, if a stress is tensile (i.e. such as to cause the material to stretch) the sign is positive. Otherwise, the stress is compressive and the sign us negative. Shear stress Consider a two-dimensional state of shear stress, as illustrated. According to the inde convention, the stress component acting along the top face must be, as shown. The element of material must be in equilibrium. Therefore, we can use equilibrium of horizontal forces to deduce that the stress component along the bottom horizontal edge is given b =. Also, there must be equilibrium of moments. Note that the two horizontal stress components would tend to cause a clockwise moment. To preserve equilibrium of moments, the two vertical components are necessar to suppl anticlockwise moment. The inde convention and the use of vertical equilibrium 2

3 show that these components are both of magnitude. The accompaning figure shows this. Note that, once we decided the direction of the initial horizontal arrow, the directions of the three other arrows were fied as a result of equilibrium. The state of shear stress corresponding to that shown here is b convention positive. B choosing the top horizontal arrow pointing left instead of right, the directions of all the other three arrows would be reversed. This alternative state of shear stress is b convention negative. Relationships between shear components We can use equilibrium of moments in the diagram above to stud the relationship between and. Assuming a unit thickness, the horizontal and vertical faces are of area d and d respectivel. Now take moments about the centre: 2 d d/2-2 d d/2 = 0 i.e. =. We can look at other planes the -z plane and the z- plane - and do similar eercises to show that z = z. and z = z. The eistence of these relations is the reason for there being onl si independent stress components. Presence of shear stresses The values of the components of stress depends on the ais set which is used. We can alwas choose an ais set in which the shear stress is zero. In nearl all stress sstems, we can choose an ais set in which shear stresses act. For instance, a tensile bar has shear stresses acting which become apparent when we choose an oblique ais set. Resolving forces shows that there is a shear force, and therefore a shear stress, in the plane. This is illustrated below. 3

4 The thin sphere. Here we naturall choose directions in which no shear stresses act. Stress in the shell of the sphere acts along a tangent. Assume this tangential stress is constant throughout and equal to σ. Now consider a thin sphere pressurised b an internal pressure p. We create a free bod b slicing the sphere in half. The resulting hemisphere shown is in equilibrium. The force in the shell is given b stress area. This must balance the force eerted b the pressure. Since the sphere is thin, the area of surface of the shell hemisphere is 2πrt, where t is the shell thickness and r the sphere radius. Equating the two forces gives σ2πrt = pπr 2. From this we get the epression for the stress pr σ =. 2t This completes the problem. The thin clinder. This time there are two sorts of stress tangential stress as with the sphere and also an aial stress, acting along the clinder ais. We look at two kinds of free bod to find the two stresses. For the aial stress σ the free bod is the clinder shown in (b) above. Using eactl the same argument as for the sphere, we obtain the same result: 4

5 σ = pr 2t where r is the clinder radius and t the shell thickness. For the tangential stress σ, we use the free bod (c). Assuming an aial length L, the total cross section of shell is 2tL, and the internal cross section is 2rL. Equilibrium then gives σ 2tL = p2rl to give σ = p r completing the problem. Eample A vertical ais clindrical storage tank contains water to a depth of 15m. Given that its radius is 4m, and the maimum allowable shell stress is 240 MNm -2, what is the required shell thickness? The specific weight of water is 9810 Nm -3. Shear eamples Eample The test below is designed to determine the transverse shear strength of animal bone. If the bone cross sectional area is 150 mm2, and a transverse force of 600N is required for fracture, what is the shear strength? 5

6 Eample Eample A pulle is keed (to prevent relative motion) to a 60 mm diameter shaft. The unequal belt pulls T1 and T2 on the two sides of the pulle give rise to a net turning moment of 120 Nm. The ke is 10 mm b 15 mm cross section and 75 mm long, as shown in Fig Dtermine the average shearing stress acting on a horizontal plane through the ke. Ans Mpa. 6