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Timoshenko (26 31)

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Strength of Materials-2
22
ension, Compression, and Shear: iation of Stress with Aspect of Cross-Sectio In the case of axial tension of a prismatic bar, Fig. 2.1a, the stress on a normal cross- section mn is uniform and has the magnitude σ = P / A as discussed in Art. 1.2. Let us consider now the state of stress on an oblique cross- section pq cutting the bar at an angle with the normal cross-section mn. First, we isolate that portion of the bar to the left of the oblique
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  • Tension, Compression, and Shear: 2.1 Variation of Stress with Aspect of Cross-Section In the case of axial tension of a prismatic bar, Fig. 2.1a, the stress on a normal cross-section mn is uniform and has the magnitude = P / A as discussed in Art. 1.2. Let us consider now the state of stress on an oblique cross-section pq cutting the bar at an angle with the normal cross-section mn. First, we isolate that portion of the bar to the left of the oblique section pq as a free body and represent the action of the removed portion on this free body by the stress resultant S as shown in Fig. 2.1b. From the equilibrium condition, this internal force S must

  • be equal, opposite, and collinear with the external force P, as shown. Resolving the force S into components N and Q, normal and tangential, respectively, to the plane pq, we find N = P cos ; Q = P sin . Then since the area of the oblique section pq is A' = A/cos , the corresponding stresses are (a)(2.1)

  • These are called, respectively, the normal stress and the shear stress on the oblique section pq, the aspect of which is defined by . We see that when = 0 and the section pq coincides with the normal section mn, eqs (2.1) give (b)and = 0, as they should. However, as is increased, the normal stress n diminishes, until when = /2, n = 0. Thus there is seen to be no normal lateral stress between the longitudinal fibers of a prismatic bar in tension. On the other hand, with increase in the angle , the shear stress increases to a maximum value

  • when = /4 and then diminishes to = 0 when = /2. 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(b)

  • 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. 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.

  • 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 (2.1)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

  • (d)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. 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, themoments of which must balance each other. Thus T(dsdz) ds' = T'(ds'dz) ds,from which = ' where ' has already been represented as negative in Fig. 2.6.FIG 2.6

  • EXAMPLE 1. A short steel bar having a 1-in. 1-in. square cross-section is subjected to compressive forces P = 25,000 lb axially applied as shown in Fig. 2.7.

  • Compute the complete set of complementary stresses on the sides of the rectangular element A oriented as shown.SOLUTION. Taking P/A = -25,000 psi and = 30 in eqs. (2.1)n = -25,000 (0.866)2 = -18,750psi, = -12,500 0.866 = -10,820 psi. Similarly, from eqs. (2.1') 'n = -25,000 XX (0.50)2 = -6250 psi, = - (-12,500) 0.866 = + 10,820 psi.

  • PROBLEM SET 2.1 A mild steel tensile-test specimen having a circular cross-section of diameter d = 0.505 in. shows an elongation reading of 0.00200 in. over a gauge length of 2 in. Calculate the maximum shear stress in the material, assuming that E = 30(10)6 psi. Ans. max = 15,000 psi.A brass wire of diameter d = 1/l6 in. and length l = 24 in. is tightly stretched between fixed points A and B at its ends so that it is under a tension of 31 lb. If the temperature of the wire subsequently drops 50F, what is the maximum shear stress in the material? The coefficient of thermal expansion for brass is

  • b = 10.4(10)-6 in./in./oF and the modulus of elasticity is Eb = 14(10)6 psi. Ans. max = 8640 psi.A prismatic bar carrying an axial tensile stress x is cut by an oblique section pq as shown in Fig. A. If the normal and shear stresses, respectively, on this section are n = 12,000 psi and = 4000 psi, find the value of x and the angle defining the aspect of the section pq. Ans. x = 13,333 psi; = 18 26'. Referring to the case of axial tension of a prismatic bar as shown in Fig. 2.5, we are given the following numerical data: A = 0.785 in.2, P = 10,000 lb, = 20. Calculate the stresses n, 'n, ,', for sections pq and p'q'. Ans. n = 11,250 psi, 'n = 1490 psi, = - ' = 4095 psi.

  • FIG. AFIG. B

  • A steel rod of circular cross-section is to carry a tensile load P = 30 kips. The allowable working stress in shear for the rod is = 8000 psi. Find the required diameter d for the rod. Ans. d = 1.55 in. A prismatic bar is subjected to axial tension. Find the aspect angle which defines an oblique section on which the normal and shearing stresses are equal. Am. = 45. Referring to Fig. A, assume that the angle = 30 and that n = 10,000 psi. In such case, what is the shear stress ? Ans. = 5780 psi. The normal stresses on the edges of the element A in Fig. 2.7 are given as follows: n = -14,500 psi, 'n = -9500 psi. What is the

  • aspect angle defining the orientation of the element and what is the axial stress P/A to which the bar is subjected? Ans. = 39, P/A = -24,000 psi. A concrete test cylinder having length l = 12 in. and diameter d = 6 in. is subjected to axial compressive forces P in a testing machine. If the maximum shear stress in the concrete is not to exceed 2000 psi, what is the safe value for the axial load P? Ans. P = 113,200 lb.


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