Uniaxial Stress system (Stress due to axial loading)Prismatic member is subjected to an axial load P as shown in Fig. 1 Figure No. 1

The internal forces at any cross-section with in the length of he member is determined by passing a section perpendicular to the longitudinal axis of the member and set it to equilibrium as shown in Fig. 2. Figure No. 2

The internal force N at the cross-section of intrest is equal in magnitude to the applied load P. The stress also acts normal to the cut section as shown in Fig. 3 and is equal to Figure No. 3

Where = Normal stressP = Axial forceA = Area of cross-section perpendicular to longitudinal axisHowever there is no shear stress. Now if we consider a section which is inclined at an angle w.r.t. horizontal or longitudinal axis (shown in Fig. 4). Figure No. 4

Then the force P is decomposed into components N and V. (shown in Fig. 5) Figure No. 5 The force, N leads to system of normal stresses (n) and force V leads to shear stresses () as shown in Fig. 6

Figure No. 6Following expressions are obtained by applying the equation of equilibrium along n and t axis as shown in Fig. 7. PNPcosPsinn-axist-axist-axisvn-axisFigure No. 7

A'AFigure No. 8 So substituting value of A' in (2).

The above formulae may also be expressed in terms of double angles.

Graphical representation of these is shown in figure (9). Figure No. 9

045901351802252703153601010100000

It would prove helpful in visualizing more readily the variation in normal and shearing stress with respect to each other and with angle .The graph shows that normal stresses are maximum at 0o and 180o and minimum at 90o and 270o. It can be noted that where normal stresses are maximum or minimum the shear stress is always zero. These maximum or minimum normal stresses are called Principal stresses and the Plane on which they act (Plane of zero shear stress) are known as Principal Planes.Also it can be noted that maximum values of shear stress occurs when is 45o or 135o and also that

Therefore maximum stresses are given by the following expression. The maximum and minimum values of and can also be determined by using calculus i.e. by differentiating them with and equating it zero.

Importance of Tension Test These observations lead us to consider more carefully the question of the strength of a bar in simple tension. If the bar is made of a material that if much weaker in shear than it is in cohesion, it may happen that failure will take place due to relative slipping between two parts of the bar along a 45- plane where the shear stress is a maximum, rather than due to direct rupture across a normal section where the normal stress is a maximum. For example, a short wood post loaded in axial compression, as shown in Fig. 2.2a, may actually fail by shearing along a jagged plane inclined roughly by 45 to the axis of the post. In such case, we may still specify the value of P/A at which this failure occurs

as the ultimate strength of the wood in compression, even though the failure is not a true compression failure of the material. Similarly, during a tensile test of a flat bar of low-carbon steel with polished surfaces, it is possible to observe a very interesting phenomenon. At a certain value of the tensile stress = P/A I visible slip bands approximately inclined by 45 to the axis of the bar will appear on the flat sides of the specimen as shown in Fig. 2.2b. These lines, called Lueders' lines, indicate that the material is failing in shear, even though the bar is being loaded in simple tension. This relative sliding along 45-planes causes the specimen to elongate axially, and after unloading it will not return to its original length. Such apparent stretching of the bar due to this slip phenomenon is called plastic

yielding. Again, the axial tensile stress y.p. = P/A at which this occurs may be designated as the yield stress in tension, even though the failure is not a true tension failure of the material. These matters will be discussed further in the next article. Laboratory experiments indicate that both shearing and normal stress under axial loading are important since a brittle material loaded in tension will fail in tension on transverse plane whereas a ductile material loaded in tension will fail in shear on the 45o plane.Equations (2.18) and (2.19) illustrate that the normal and tangential components are each less than the stress of Fig. 2.12b. However, there are physical situations when one of the components acting on the inclined surface might produce a more critical stress

situation. The following example is conceptually elementary and yet illustrates a typical engineering analysis problem. A short wooden compression member is shown in Fig. 2.13. The photograph of the actual test specimen (Fig. 2.13e) illustrates the failure mechanism. The failure surface is inclined approximately 58 degrees from the vertical, and a shear-type failure is readily observed as the top section appears to slide relative to the lower section. The material is weaker in shear than in compression and the failure will be explained using the foregoing analysis. Shear stress and normal stress are computed using Eqs. (2.18) and (2.19). Substituting the actual values of , b, and h gives

The normal and shear stresses are shown idealized as they act on the cut section in Fig. 2.13d. The shear stress that is producing failure is shown in Figs. 2.13e and f. For = 90 degrees, a failure surface normal to the axis of the post, the stress would be

But T = 179.74 and N = 287.64 and the failure is a shear failure; hence, the failure stress is 45 percent of the axial normal stress. This result is due to actual material behavior that is an integral part of the study of mechanics of materials. The wood test specimen represents a material that is less resistant to failure in shear than in compression and the grain structure of wood lends itself to the failure mechanism of Fig. 2.13e. The calculation of stress is independent of material behavior. However, if stress is to be related to deformation, the mechanical properties of the material must be considered. Equations (2.18) and (2.19) illustrate the earlier statement that stress is not a vector quantity. The force vectors of Fig. 2.12d have components described by a single trigonometric function.

FIGURE 2.13Failure of an actual wood test specimen.

Sign convention Formulas (2.1), derived for the case of axial tension, can be used also for axial compression, simply by changing the sign of P/A. We then obtain negative values for both the normal stress n and the shear stress. The complete state of stress on a thin element between two parallel oblique sections for axial tension and axial compression are compared in Fig. 2.3. The directions of these stresses associated with axial tension (Fig. 2.3a) will be considered as positive; those associated with axial compression (Fig. 2.3b), as negative. Thus n is positive when it is a tensile stress and negative when it is a compressive stress. By reference to Fig. 2.3, the rule for

sign of shear stress will be as follows: The shear stress on any face of the element will be considered positive when it has a clockwise moment with respect to a center inside the element (Fig. 2.3a). If the moment is counterclockwise with respect to a center inside the element, the shear stress is negative. Stated in a different way, the shear stress on any surface of a body will be considered to be of positive sign if it points in a direction corresponding to clockwise rotation about a center inside the body, otherwise of negative sign. Several examples of both positive and negative shear stress are shown in Fig. 2.4. These sign conventions, while arbitrary, must nonetheless be carefully observed to avoid confusion.At this point, we shall adopt a sign convention for these

stresses: If the stress points in a direction that corresponds to a clockwise rotation, the stress is positive; a negative shear stress is associated with counterclockwise rotation. As previously mentioned, the normal stresses are considered positive if there is a tensile load acting on that face; it is negative for a compressive load. Referring to Figure 1-20, stress xy is, according to our convention, positive, yx is negative, 1 and 2 are positive.

Returning to the case of a bar in axial tension, let us consider now the stresses on an oblique section p' q' at right angles to the section pq, as shown in Fig. 2.5. To obtain the stresses 'n and ' on this section, we need

only to replace q, by 90 + in eqs. (2.1). Then remembering that sin (90 + ) = cos q" while cos(90 + ) = - sin this gives These stresses on the plane p' q' act as shown in Fig. 2.5b. The complete set of stresses given by eqs. (2.1) and (2.1') are called complementary stresses because they occur on mutually perpendicular planes. Comparing the two sets of formulas, we observe that

Thus the sum of normal stresses n and 'n on any two mutually perpendicular sections of a bar in axial tension is constant and equal to P / A, the normal stress on the normal section mn. Also, complementary shear stresses are always equal in magnitude but opposite in sign. One observes that when of Fig. 1-7 is greater than 90, the sign of the shearing stress in Eq. b changes. The magnitude of the shearing stress for any angle , however, is the same as that for 90 + . The sign change merely indicates that the shear force vector changes

sense, being directed toward the top of the element instead of toward the bottom as in Fig. 1-6. Normal and shearing stresses on planes having aspects 1 and 90 + 1 are shown in Fig. 1-8.

The equality of complementary shear stresses such as and ' on the faces of a rectangular element (Fig. 2.6) also can be established from the equilibrium conditions of the element itself, as follows: Let dz denote the thickness of the element normal to the plane of the paper and ds, ds', the lengths of its edges. Then the areas on which and ' act will be, respectively, dsdz and ds'dz. Multiplying the shear stresses by the areas on which they act, we obtain two counteracting couples, the moments of which must balance each other. Thus (dsdz) ds' = '(ds'dz) ds,Alternative

from which = ', where ' has already been represented as negative in Fig. 2.6. Fig. 2.6