The aircraft shown is being tested to determine how the forces due to lift would be distributed over the wing. This chapter deals with stresses and

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1 The aircraft shown is being tested to determine how the forces due to lift would be distributed over the wing. This chapter deals with stresses and strains in structures and machine components. 436

2 H P T E R Transformations of Stress and Strain 437

3 hapter 7 Transformations of Stress and Strain 7. Introduction 7. Transformation of Plane Stress 7.3 Principal Stresses; Maimum Shearing Stress 7.4 Mohr s ircle for Plane Stress 7.5 General State of Stress 7.6 pplication of Mohr s ircle to the Three-Dimensional nalsis of Stress *7.7 Yield riteria for Ductile Materials under Plane Stress *7.8 Fracture riteria for rittle Materials under Plane Stress 7.9 Stresses in Thin-Walled Pressure Vessels *7.0 Transformation of Plane Strain *7. Mohr s ircle for Plane Strain *7. Three-Dimensional nalsis of Strain *7.3 Measurements of Strain; Strain Rosette 7. INTRDUTIN We saw in Sec.. that the most general state of stress at a given point Q ma be represented b si components. Three of these components, s, s, and s, define the normal stresses eerted on the faces of a small cubic element centered at Q and of the same orientation as the coordinate aes (Fig. 7.a), and the other three, t, t, and t, the components of the shearing stresses on the same element. s we remarked at the time, the same state of stress will be represented b a different set of components if the coordinate aes are rotated (Fig. 7.b). We propose in the first part of this chapter to determine how the components of stress are transformed under a rotation of the coordinate aes. The second part of the chapter will be devoted to a similar analsis of the transformation of the components of strain. Q ' '' '' ' ' Q '' '' '' '' ' ' (a) Fig. 7. General state of stress at a point. ' (b) F F 3 Fig. 7. Plane stress. ur discussion of the transformation of stress will deal mainl with plane stress, i.e., with a situation in which two of the faces of the cubic element are free of an stress. If the ais is chosen perpendicular to these faces, we have s 5 t 5 t 5 0, and the onl remaining stress components are s, s, and t (Fig. 7.). Such a situation occurs in a thin plate subjected to forces acting in the midplane of the plate (Fig. 7.3). It also occurs on the free surface of a structural element or machine component, i.e., at an point of the surface of that element or component that is not subjected to an eternal force (Fig. 7.4). F F 4 F F 6 F F 5 Fig. 7.3 Eample of plane stress. Fig. 7.4 Eample of plane stress. 438 We recall that t 5 t, t 5 t, and t 5 t.

4 onsidering in Sec. 7. a state of plane stress at a given point Q characteried b the stress components s, s, and t associated with the element shown in Fig. 7.5a, ou will learn to determine the components s 9, s 9, and t 99 associated with that element after it has been rotated through an angle u about the ais (Fig. 7.5b). In Sec. 7.3, ou will determine the value u p of u for which the stresses s 9 and s 9 are, respectivel, maimum and minimum; these values of the normal stress are the principal stresses at point Q, and the faces of the corresponding element define the principal planes of stress at that point. You will also determine the value u s of the angle of rotation for which the shearing stress is maimum, as well as the value of that stress. 7. Introduction 439 ' ' '' ' Q Q ' ' (a) Fig. 7.5 Transformation of stress. (b) In Sec. 7.4, an alternative method for the solution of problems involving the transformation of plane stress, based on the use of Mohr s circle, will be presented. In Sec. 7.5, the three-dimensional state of stress at a given point will be considered and a formula for the determination of the normal stress on a plane of arbitrar orientation at that point will be developed. In Sec. 7.6, ou will consider the rotations of a cubic element about each of the principal aes of stress and note that the corresponding transformations of stress can be described b three different Mohr s circles. You will also observe that, in the case of a state of plane stress at a given point, the maimum value of the shearing stress obtained earlier b considering rotations in the plane of stress does not necessaril represent the maimum shearing stress at that point. This will bring ou to distinguish between in-plane and outof-plane maimum shearing stresses. Yield criteria for ductile materials under plane stress will be developed in Sec To predict whether a material will ield at some critical point under given loading conditions, ou will determine the principal stresses s a and s b at that point and check whether s a, s b, and the ield strength s Y of the material satisf some criterion. Two criteria in common use are: the maimum-shearing-strength criterion and the maimum-distortion-energ criterion. In Sec. 7.8, fracture criteria for brittle materials under plane stress will be developed in a similar fashion; the will involve the principal stresses s a and s b at some critical point and the ultimate strength s U of the

5 440 Transformations of Stress and Strain material. Two criteria will be discussed: the maimum-normal-stress criterion and Mohr s criterion. Thin-walled pressure vessels provide an important application of the analsis of plane stress. In Sec. 7.9, we will discuss stresses in both clindrical and spherical pressure vessels (Photos 7. and 7.). Photo 7. lindrical pressure vessel. Photo 7. Spherical pressure vessel. Sections 7.0 and 7. will be devoted to a discussion of the transformation of plane strain and to Mohr s circle for plane strain. In Sec. 7., we will consider the three-dimensional analsis of strain and see how Mohr s circles can be used to determine the maimum shearing strain at a given point. Two particular cases are of special interest and should not be confused: the case of plane strain and the case of plane stress. Finall, in Sec. 7.3, we discuss the use of strain gages to measure the normal strain on the surface of a structural element or machine component. You will see how the components P, P, and g characteriing the state of strain at a given point can be computed from the measurements made with three strain gages forming a strain rosette. 7. TRNSFRMTIN F PLNE STRESS Let us assume that a state of plane stress eists at point Q (with s 5 t 5 t 5 0), and that it is defined b the stress components s, s, and t associated with the element shown in Fig. 7.5a. We propose to determine the stress components s 9, s 9, and t 99 associated with the element after it has been rotated through an angle u about

6 ' 7. Transformation of Plane Stress 44 ' '' ' Q Q ' ' (a) Fig. 7.5 (repeated) (b) the ais (Fig. 7.5b), and to epress these components in terms of s, s, t, and u. In order to determine the normal stress s 9 and the shearing stress t 99 eerted on the face perpendicular to the 9 ais, we consider a prismatic element with faces respectivel perpendicular to the,, and 9 aes (Fig. 7.6a). We observe that, if the area of the ' ' '' cos ' ( cos ) ( cos ) ' ' sin ( sin ) (a) (b) ( sin ) Fig. 7.6 oblique face is denoted b D, the areas of the vertical and horiontal faces are respectivel equal to D cos u and D sin u. It follows that the forces eerted on the three faces are as shown in Fig. 7.6b. (No forces are eerted on the triangular faces of the element, since the corresponding normal and shearing stresses have all been assumed equal to ero.) Using components along the 9 and 9 aes, we write the following equilibrium equations: gf 5 0: s s cos u cos u t cos u sin u s sin u sin u t sin u cos u 5 0 gf 5 0: t s cos u sin u t cos u cos u s sin u cos u t sin u sin u 5 0

7 44 Transformations of Stress and Strain Solving the first equation for s 9 and the second for t 99, we have s 5 s cos u s sin u t sin u cos u (7.) t 5s s sin u cos u t cos u sin u (7.) Recalling the trigonometric relations sin u 5 sin u cos u cos u 5 cos u sin u (7.3) and cos u 5 cos u sin cos u u 5 (7.4) we write Eq. (7.) as follows: or s 5 s cos u s cos u t sin u s 5 s s s s cos u t sin u (7.5) Using the relations (7.3), we write Eq. (7.) as t 5 s s sin u t cos u (7.6) The epression for the normal stress s 9 is obtained b replacing u in Eq. (7.5) b the angle u 908 that the 9 ais forms with the ais. Since cos (u 808) 5 cos u and sin (u 808) 5 sin u, we have s 5 s s s s cos u t sin u (7.7) dding Eqs. (7.5) and (7.7) member to member, we obtain s s 5 s s (7.8) Since s 5 s 9 5 0, we thus verif in the case of plane stress that the sum of the normal stresses eerted on a cubic element of material is independent of the orientation of that element. f. first footnote on page 97.

8 7.3 PRINIPL STRESSES; MXIMUM SHERING STRESS The equations (7.5) and (7.6) obtained in the preceding section are the parametric equations of a circle. This means that, if we choose a set of rectangular aes and plot a point M of abscissa s 9 and ordinate t 99 for an given value of the parameter u, all the points thus obtained will lie on a circle. To establish this propert we eliminate u from Eqs. (7.5) and (7.6); this is done b first transposing (s s )/ in Eq. (7.5) and squaring both members of the equation, then squaring both members of Eq. (7.6), and finall adding member to member the two equations obtained in this fashion. We have Setting as s s b t 5 a s s b t (7.9) 7.3 Principal Stresses; Maimum Shearing Stress 443 s ave 5 s s and R 5 a s s b t (7.0) we write the identit (7.9) in the form s s ave t 5 R (7.) which is the equation of a circle of radius R centered at the point of abscissa s ave and ordinate 0 (Fig. 7.7). It can be observed that, due to the smmetr of the circle about the horiontal ais, the same result would have been obtained if, instead of plotting M, we had plotted a point N of abscissa s 9 and ordinate t 99 (Fig. 7.8). This propert will be used in Sec '' ' D '' min ave R M '' ' ave R N '' ' ma E ' Fig. 7.7 ircular relationship of transformed stresses. Fig. 7.8 Equivalent formation of stress transformation circle. The two points and where the circle of Fig. 7.7 intersects the horiontal ais are of special interest: Point corresponds to the maimum value of the normal stress s 9, while point corresponds

9 444 Transformations of Stress and Strain to its minimum value. esides, both points correspond to a ero value of the shearing stress t 99. Thus, the values u p of the parameter u which correspond to points and can be obtained b setting t in Eq. (7.6). We write tan u p 5 t s s (7.) This equation defines two values u p that are 808 apart, and thus two values u p that are 908 apart. Either of these values can be used to determine the orientation of the corresponding element (Fig. 7.9). ' min p ma ' ma Q p min Fig. 7.9 Principal stresses. The planes containing the faces of the element obtained in this wa are called the principal planes of stress at point Q, and the corresponding values s ma and s min of the normal stress eerted on these planes are called the principal stresses at Q. Since the two values u p defined b Eq. (7.) were obtained b setting t in Eq. (7.6), it is clear that no shearing stress is eerted on the principal planes. We observe from Fig. 7.7 that s ma 5 s ave R and s min 5 s ave R (7.3) Substituting for s ave and R from Eq. (7.0), we write s ma, min 5 s s 6 a s s b t (7.4) Unless it is possible to tell b inspection which of the two principal planes is subjected to s ma and which is subjected to s min, it is necessar to substitute one of the values u p into Eq. (7.5) in order to determine which of the two corresponds to the maimum value of the normal stress. Referring again to the circle of Fig. 7.7, we note that the points D and E located on the vertical diameter of the circle correspond to This relation can also be obtained b differentiating s 9 in Eq. (7.5) and setting the derivative equal to ero: ds 9 du 5 0.

10 the largest numerical value of the shearing stress t 99. Since the abscissa of points D and E is s ave 5 (s s ), the values u s of the parameter u corresponding to these points are obtained b setting s 9 5 (s s ) in Eq. (7.5). It follows that the sum of the last two terms in that equation must be ero. Thus, for u 5 u s, we write 7.3 Principal Stresses; Maimum Shearing Stress 445 or s s cos u s t sin u s 5 0 ' s s tan u s 5 (7.5) t This equation defines two values u s that are 808 apart, and thus two values u s that are 908 apart. Either of these values can be used to determine the orientation of the element corresponding to the maimum shearing stress (Fig. 7.0). bserving from Fig. 7.7 that the maimum value of the shearing stress is equal to the radius R of the circle, and recalling the second of Eqs. (7.0), we write ' ma ' Q ' s ma ' s Fig. 7.0 Maimum shearing stress. ' t ma 5 a s s b t (7.6) s observed earlier, the normal stress corresponding to the condition of maimum shearing stress is s 5 s ave 5 s s (7.7) omparing Eqs. (7.) and (7.5), we note that tan u s is the negative reciprocal of tan u p. This means that the angles u s and u p are 908 apart and, therefore, that the angles u s and u p are 458 apart. We thus conclude that the planes of maimum shearing stress are at 458 to the principal planes. This confirms the results obtained earlier in Sec.. in the case of a centric aial loading (Fig..38) and in Sec. 3.4 in the case of a torsional loading (Fig. 3.9.) We should be aware that our analsis of the transformation of plane stress has been limited to rotations in the plane of stress. If the cubic element of Fig. 7.5 is rotated about an ais other than the ais, its faces ma be subjected to shearing stresses larger than the stress defined b Eq. (7.6). s ou will see in Sec. 7.5, this occurs when the principal stresses defined b Eq. (7.4) have the same sign, i.e., when the are either both tensile or both compressive. In such cases, the value given b Eq. (7.6) is referred to as the maimum in-plane shearing stress. This relation ma also be obtained b differentiating t 99 in Eq. (7.6) and setting the derivative equal to ero: dt 99 du 5 0.

11 EXMPLE MPa 40 MPa 50 MPa Fig. 7. For the state of plane stress shown in Fig. 7., determine (a) the principal planes, (b) the principal stresses, (c) the maimum shearing stress and the corresponding normal stress. (a) Principal Planes. Following the usual sign convention, we write the stress components as s 550 MPa s 50 MPa t 540 MPa Substituting into Eq. (7.), we have tan u p 5 t 5 40 s s u p and u p and 6.6 (b) Principal Stresses. Formula (7.4) ields min 30 MPa ma 70 MPa s ma, min 5 s s 6 a s s b t p s ma MPa s min MPa Fig. 7. The principal planes and principal stresses are sketched in Fig. 7.. Making u in Eq. (7.5), we check that the normal stress eerted on face of the element is the maimum stress: min ma ' 45 Fig. 7.3 Fig. 7.4 ma p 6.6 s p ' 0 MPa ma 50 MPa ' 0 MPa p 8.4 s cos sin cos sin MPa 5 s ma (c) Maimum Shearing Stress. Formula (7.6) ields t ma 5 a s s b t MPa Since s ma and s min have opposite signs, the value obtained for t ma actuall represents the maimum value of the shearing stress at the point considered. The orientation of the planes of maimum shearing stress and the sense of the shearing stresses are best determined b passing a section along the diagonal plane of the element of Fig. 7.. Since the faces and of the element are contained in the principal planes, the diagonal plane must be one of the planes of maimum shearing stress (Fig. 7.3). Furthermore, the equilibrium conditions for the prismatic element require that the shearing stress eerted on be directed as shown. The cubic element corresponding to the maimum shearing stress is shown in Fig The normal stress on each of the four faces of the element is given b Eq. (7.7): s 5 s ave 5 s s MPa 446

12 8 in. 0 in.. in. 4 in. H P D SMPLE PRLEM 7. single horiontal force P of magnitude 50 lb is applied to end D of lever D. Knowing that portion of the lever has a diameter of. in., determine (a) the normal and shearing stresses on an element located at point H and having sides parallel to the and aes, (b) the principal planes and the principal stresses at point H. SLUTIN Force-ouple Sstem. We replace the force P b an equivalent forcecouple sstem at the center of the transverse section containing point H: T.7 kip in. P 50 lb P 5 50 lb T 5 50 lb8 in. 5.7 kip? in. M 5 50 lb0 in. 5.5 kip? in. H M.5 kip in. a. Stresses S, S, T at Point H. Using the sign convention shown in Fig. 7., we determine the sense and the sign of each stress component b carefull eamining the sketch of the force-couple sstem at point : s 5 0 s 5 Mc I t 5 Tc J.5 kip? in.0.6 in. 5 s 4p 0.6 in ksi b 5.7 kip? in.0.6 in. p 0.6 in. 4 t ksi b We note that the shearing force P does not cause an shearing stress at point H. b. Principal Planes and Principal Stresses. Substituting the values of the stress components into Eq. (7.), we determine the orientation of the principal planes: 8.84 ksi 7.96 ksi 0 tan u p 5 t s s u p 56.0 and u p and 59.5 b Substituting into Eq. (7.4), we determine the magnitudes of the principal stresses: s ma, min 5 s s 6 a s s b t H b ma 3.5 ksi a p 30.5 min 4.68 ksi a b s ma 53.5 ksi b s min ksi b onsidering face ab of the element shown, we make u p in Eq. (7.5) and find s ksi. We conclude that the principal stresses are as shown. 447

13 PRLEMS 7. through 7.4 For the given state of stress, determine the normal and shearing stresses eerted on the oblique face of the shaded triangular element shown. Use a method of analsis based on the equilibrium of that element, as was done in the derivations of Sec ksi 80 MPa 45 MPa 5 ksi MPa 60 6 ksi 75 7 MPa 5 ksi 8 MPa Fig. P7. Fig. P7. Fig. P7.3 Fig. P through 7.8 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. 7.9 through 7. For the given state of stress, determine (a) the orientation of the planes of maimum in-plane shearing stress, (b) the maimum in-plane shearing stress, (c) the corresponding normal stress. 40 MPa 50 MPa ksi ksi 35 MPa 5 ksi 60 MPa 0 MPa 4 ksi 8 ksi 5 MPa 5 ksi Fig. P7.5 and P7.9 Fig. P7.6 and P7.0 Fig. P7.7 and P7. Fig. P7.8 and P through 7.6 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 58 clockwise, (b) 08 counterclockwise. 8 ksi 90 MPa ksi 80 MPa 5 ksi 30 MPa 60 MPa 8 ksi 6 ksi 50 MPa Fig. P7.3 Fig. P7.4 Fig. P7.5 Fig. P

14 7.7 and 7.8 The grain of a wooden member forms an angle of 58 with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain. Problems MPa 400 psi 4 MPa P 5 5 T in. 4 Fig. P7.7 Fig. P steel pipe of -in. outer diameter is fabricated from 4-in.-thick plate b welding along a heli that forms an angle of.58 with a plane perpendicular to the ais of the pipe. Knowing that a 40-kip aial force P and an 80-kip? in. torque T, each directed as shown, are applied to the pipe, determine s and t in directions, respectivel, normal and tangential to the weld. 7.0 Two members of uniform cross section mm are glued together along plane a-a that forms an angle of 58 with the horiontal. Knowing that the allowable stresses for the glued joint are s kpa and t kpa, determine the largest centric load P that can be applied. Fig. P7.9 Weld.5 a a 5 50 mm P Fig. P Two steel plates of uniform cross section mm are welded together as shown. Knowing that centric 00-kN forces are applied to the welded plates and that b 5 58, determine (a) the in-plane shearing stress parallel to the weld, (b) the normal stress perpendicular to the weld. 7. Two steel plates of uniform cross section mm are welded together as shown. Knowing that centric 00-kN forces are applied to the welded plates and that the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) the angle b, (b) the corresponding normal stress perpendicular to the weld. 00 kn 80 mm 00 kn Fig. P7. and P7.

15 450 Transformations of Stress and Strain 6 in lb vertical force is applied at D to a gear attached to the solid -in. diameter shaft. Determine the principal stresses and the maimum shearing stress at point H located as shown on top of the shaft. in. D H 7.4 mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 4-lb force at, determine the principal stresses and the maimum shearing stress at point H located as shown on top of the 3 4-in. diameter shaft. Fig. P lb H E 6 in. 4 lb 0 in. Fig. P The steel pipe has a 0-mm outer diameter and a 6-mm wall thickness. Knowing that arm D is rigidl attached to the pipe, determine the principal stresses and the maimum shearing stress at point K. 00 mm 6 mm T D 5 mm 0 kn 0.5 m 0. m 3 kn H H K 50 mm Fig. P N m Fig. P7.6 3 kn 7.6 The ale of an automobile is acted upon b the forces and couple shown. Knowing that the diameter of the solid ale is 3 mm, determine (a) the principal planes and principal stresses at point H located on top of the ale, (b) the maimum shearing stress at the same point.

16 7.7 For the state of plane stress shown, determine (a) the largest value of t for which the maimum in-plane shearing stress is equal to or less than ksi, (b) the corresponding principal stresses. 8 ksi Problems 45 0 ksi Fig. P For the state of plane stress shown, determine the largest value of s for which the maimum in-plane shearing stress is equal to or less than 75 MPa. 0 MPa 60 MPa Fig. P Determine the range of values of s for which the maimum inplane shearing stress is equal to or less than 0 ksi. 5 ksi 8 ksi Fig. P For the state of plane stress shown, determine (a) the value of t for which the in-plane shearing stress parallel to the weld is ero, (b) the corresponding principal stresses. MPa 75 MPa Fig. P7.30

17 45 Transformations of Stress and Strain 7.4 MHR S IRLE FR PLNE STRESS ma ma min ma p (a) Y (, ) Fig. 7.5 Mohr s circle. b p ( ) (b) min min X (, ) a The circle used in the preceding section to derive some of the basic formulas relating to the transformation of plane stress was first introduced b the German engineer tto Mohr (835 98) and is known as Mohr s circle for plane stress. s ou will see presentl, this circle can be used to obtain an alternative method for the solution of the various problems considered in Secs. 7. and 7.3. This method is based on simple geometric considerations and does not require the use of specialied formulas. While originall designed for graphical solutions, it lends itself well to the use of a calculator. onsider a square element of a material subjected to plane stress (Fig. 7.5a), and let s, s, and t be the components of the stress eerted on the element. We plot a point X of coordinates s and t, and a point Y of coordinates s and t (Fig. 7.5b). If t is positive, as assumed in Fig. 7.5a, point X is located below the s ais and point Y above, as shown in Fig. 7.5b. If t is negative, X is located above the s ais and Y below. Joining X and Y b a straight line, we define the point of intersection of line XY with the s ais and draw the circle of center and diameter XY. Noting that the abscissa of and the radius of the circle are respectivel equal to the quantities s ave and R defined b Eqs. (7.0), we conclude that the circle obtained is Mohr s circle for plane stress. Thus the abscissas of points and where the circle intersects the s ais represent respectivel the principal stresses s ma and s min at the point considered. We also note that, since tan (X) 5 t (s s ), the angle X is equal in magnitude to one of the angles u p that satisf Eq. (7.). Thus, the angle u p that defines in Fig. 7.5a the orientation of the principal plane corresponding to point in Fig. 7.5b can be obtained b dividing in half the angle X measured on Mohr s circle. We further observe that if s. s and t. 0, as in the case considered here, the rotation that brings X into is counterclockwise. ut, in that case, the angle u p obtained from Eq. (7.) and defining the direction of the normal a to the principal plane is positive; thus, the rotation bringing into a is also counterclockwise. We conclude that the senses of rotation in both parts of Fig. 7.5 are the same; if a counterclockwise rotation through u p is required to bring X into on Mohr s circle, a counterclockwise rotation through u p will bring into a in Fig. 7.5a. Since Mohr s circle is uniquel defined, the same circle can be obtained b considering the stress components s 9, s 9, and t 99, corresponding to the 9 and 9 aes shown in Fig. 7.6a. The point X9 of coordinates s 9 and t 99, and the point Y9 of coordinates s 9 and t 99, are therefore located on Mohr s circle, and the angle X9 in Fig. 7.6b must be equal to twice the angle 9a in Fig. 7.6a. Since, as noted before, the angle X is twice the angle a, it follows that This is due to the fact that we are using the circle of Fig 7.8 rather than the circle of Fig. 7.7 as Mohr s circle.

18 the angle XX9 in Fig. 7.6b is twice the angle 9 in Fig. 7.6a. Thus the diameter X9Y9 defining the normal and shearing stresses s 9, s 9, and t 99 can be obtained b rotating the diameter XY through an angle equal to twice the angle u formed b the 9and aes in Fig. 7.6a. We note that the rotation that brings the diameter XY into the diameter X9Y9 in Fig. 7.6b has the same sense as the rotation that brings the aes into the 99 aes in Fig. 7.6a. 7.4 Mohr s ircle for Plane Stress 453 b min a Y' ( ', ' ' ) ma Y ' ' X '' X' ( ', ' ' ) ' (a) ' (b) Fig. 7.6 The propert we have just indicated can be used to verif the fact that the planes of maimum shearing stress are at 458 to the principal planes. Indeed, we recall that points D and E on Mohr s circle correspond to the planes of maimum shearing stress, while and correspond to the principal planes (Fig. 7.7b). Since the diameters and DE of Mohr s circle are at 908 to each other, it follows that the faces of the corresponding elements are at 458 to each other (Fig. 7.7a). e b ' ' d ma ' ave D min 45 a 90 ma ma E (a) (b) Fig. 7.7

19 454 Transformations of Stress and Strain The construction of Mohr s circle for plane stress is greatl simplified if we consider separatel each face of the element used to define the stress components. From Figs. 7.5 and 7.6 we observe that, when the shearing stress eerted on a given face tends to rotate the element clockwise, the point on Mohr s circle corresponding to that face is located above the s ais. When the shearing stress on a given face tends to rotate the element counterclockwise, the point corresponding to that face is located below the s ais (Fig. 7.8). s far as the normal stresses are concerned, the usual convention holds, i.e., a tensile stress is considered as positive and is plotted to the right, while a compressive stress is considered as negative and is plotted to the left. (a) lockwise bove (b) ounterclockwise elow Fig. 7.8 onvention for plotting shearing stress on Mohr s circle. EXMPLE 7.0 For the state of plane stress alread considered in Eample 7.0, (a) construct Mohr s circle, (b) determine the principal stresses, (c) determine the maimum shearing stress and the corresponding normal stress. (MPa) 0 Y 40 G 0 0 MPa (a) F (MPa) MPa 50 MPa 40 R X (a) onstruction of Mohr s ircle. We note from Fig. 7.9a that the normal stress eerted on the face oriented toward the ais is tensile (positive) and that the shearing stress eerted on that face tends to rotate the element counterclockwise. Point X of Mohr s circle, therefore, will be plotted to the right of the vertical ais and below the horiontal ais (Fig. 7.9b). similar inspection of the normal stress and shearing stress eerted on the upper face of the element shows that point Y should be plotted to the left of the vertical ais and above the horiontal ais. Drawing the line XY, we obtain the center of Mohr s circle; its abscissa is s ave 5 s s Since the sides of the shaded triangle are MPa F MPa and FX 5 40 MPa Fig. 7.9 (b) the radius of the circle is R 5 X MPa The following jingle is helpful in remembering this convention. In the kitchen, the clock is above, and the counter is below.

20 (b) Principal Planes and Principal Stresses. The principal stresses are s ma MPa s min MPa Recalling that the angle X represents u p (Fig. 7.9b), we write tan u p 5 FX F u p u p Since the rotation which brings X into in Fig. 7.0b is counterclockwise, the rotation that brings into the ais a corresponding to s ma in Fig. 7.0a is also counterclockwise. (c) Maimum Shearing Stress. Since a further rotation of 908 counterclockwise brings into D in Fig. 7.0b, a further rotation of 458 counterclockwise will bring the ais a into the ais d corresponding to the maimum shearing stress in Fig. 7.0a. We note from Fig. 7.0b that t ma 5 R 5 50 MPa and that the corresponding normal stress is s9 5 s ave 5 0 MPa. Since point D is located above the s ais in Fig. 7.0b, the shearing stresses eerted on the faces perpendicular to d in Fig. 7.0a must be directed so that the will tend to rotate the element clockwise. d ' 0 MPa e ' 0 MPa 50 MPa ma (MPa) Y ' ave 0 D 45 b a 70 MPa ma 90 ma 50 (MPa) p 53. p min 30 MPa min 30 X E R 50 ma 70 (a) (b) Fig

21 456 Transformations of Stress and Strain Mohr s circle provides a convenient wa of checking the results obtained earlier for stresses under a centric aial loading (Sec..) and under a torsional loading (Sec. 3.4). In the first case (Fig. 7.a), we have s 5 P, s 5 0, and t 5 0. The corresponding points X and Y define a circle of radius R 5 P that passes through the e d D P' P Y R X P' ' ma P E P/ (a) (b) Fig. 7. Mohr s circle for centric aial loading. (c) origin of coordinates (Fig. 7.b). Points D and E ield the orientation of the planes of maimum shearing stress (Fig. 7.c), as well as the values of t ma and of the corresponding normal stresses s9: t ma 5 s 5 R 5 P (7.8) In the case of torsion (Fig. 7.a), we have s 5 s 5 0 and t 5 t ma 5 TcJ. Points X and Y, therefore, are located on the t ais, T' ma T Y R ma Tc J T' b ma a T X min (a) (b) (c) Fig. 7. Mohr s circle for torsional loading. and Mohr s circle is a circle of radius R 5 TcJ centered at the origin (Fig. 7.b). Points and define the principal planes (Fig. 7.c) and the principal stresses: s ma, min 56R 56 Tc J (7.9)

22 60 MPa 00 MPa 48 MPa SMPLE PRLEM 7. For the state of plane stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components eerted on the element obtained b rotating the given element counterclockwise through 308. (MPa) min 8 MPa ave 80 MPa Y(60, 48) R F ma 3 MPa X(00, 48) p (MPa) m 5 MPa SLUTIN onstruction of Mohr s ircle. We note that on a face perpendicular to the ais, the normal stress is tensile and the shearing stress tends to rotate the element clockwise; thus, we plot X at a point 00 units to the right of the vertical ais and 48 units above the horiontal ais. In a similar fashion, we eamine the stress components on the upper face and plot point Y(60, 48). Joining points X and Y b a straight line, we define the center of Mohr s circle. The abscissa of, which represents s ave, and the radius R of the circle can be measured directl or calculated as follows: s ave 5 5 s s MPa R 5 F FX MPa (MPa) ' '' p 33.7 min 8 MPa ma 3 MPa X' X 60 p 67.4 (MPa) K L Y ' 30 Y' '.6 MPa a ' ' 48.4 MPa '' 4.3 MPa a. Principal Planes and Principal Stresses. We rotate the diameter XY clockwise through u p until it coincides with the diameter. We have tan u p 5 XF F u p i u p i b The principal stresses are represented b the abscissas of points and : s ma s min s ma 53 MPa b s min 5 8 MPa b Since the rotation that brings XY into is clockwise, the rotation that brings into the ais a corresponding to s ma is also clockwise; we obtain the orientation shown for the principal planes. b. Stress omponents on Element Rotated 308 l. Points X9 and Y9 on Mohr s circle that correspond to the stress components on the rotated element are obtained b rotating X Y counterclockwise through u We find f f b s 5 K 5 K cos 5.6 s MPa b s 5 L 5 L cos 5.6 s 5.6 MPa b t 5 K X 5 5 sin 5.6 t MPa b Since X9 is located above the horiontal ais, the shearing stress on the face perpendicular to 9 tends to rotate the element clockwise. 457

23 SMPLE PRLEM ksi state of plane stress consists of a tensile stress s ksi eerted on vertical surfaces and of unknown shearing stresses. Determine (a) the magnitude of the shearing stress t 0 for which the largest normal stress is 0 ksi, (b) the corresponding maimum shearing stress. 0 (ksi) SLUTIN min 8 ksi ksi ave 4 ksi 4 ksi 0 ma 0 ksi D X R s p F 0 ma (ksi) onstruction of Mohr s ircle. We assume that the shearing stresses act in the senses shown. Thus, the shearing stress t 0 on a face perpendicular to the ais tends to rotate the element clockwise and we plot the point X of coordinates 8 ksi and t 0 above the horiontal ais. onsidering a horiontal face of the element, we observe that s 5 0 and that t 0 tends to rotate the element counterclockwise; thus, we plot point Y at a distance t 0 below. We note that the abscissa of the center of Mohr s circle is s ave 5 s s ksi The radius R of the circle is determined b observing that the maimum normal stress, s ma 5 0 ksi, is represented b the abscissa of point and writing Y E s ma 5 s ave R 0 ksi 5 4 ksi R R 5 6 ksi 0 (a) 0 s 0.9 p 4. ave 4 ksi ma 6 ksi min ksi ma 0 ksi a d a. Shearing Stress t 0. onsidering the right triangle FX, we find cos u p 5 F X 5 F R 5 4 ksi 6 ksi t 0 5 FX 5 R sin u p 5 6 ksi sin 48. u p i u p 5 4. i t ksi b b. Maimum Shearing Stress. The coordinates of point D of Mohr s circle represent the maimum shearing stress and the corresponding normal stress. t ma 5 R 5 6 ksi t ma 5 6 ksi b u s 5 90 u p l u l min ksi ma 0 ksi The maimum shearing stress is eerted on an element that is oriented as shown in Fig. a. (The element upon which the principal stresses are eerted is also shown.) ma 6 ksi Note. If our original assumption regarding the sense of t 0 was reversed, we would obtain the same circle and the same answers, but the orientation of the elements would be as shown in Fig. b. (b) ave 4 ksi 458

24 PRLEMS 7.3 Solve Probs. 7.5 and 7.9, using Mohr s circle. 7.3 Solve Probs. 7.7 and 7., using Mohr s circle Solve Prob. 7.0, using Mohr s circle Solve Prob. 7., using Mohr s circle Solve Prob. 7.3, using Mohr s circle Solve Prob. 7.4, using Mohr s circle Solve Prob. 7.5, using Mohr s circle Solve Prob. 7.6, using Mohr s circle Solve Prob. 7.7, using Mohr s circle Solve Prob. 7.8, using Mohr s circle. 7.4 Solve Prob. 7.9, using Mohr s circle. 7.4 Solve Prob. 7.0, using Mohr s circle Solve Prob. 7., using Mohr s circle Solve Prob. 7., using Mohr s circle Solve Prob. 7.3, using Mohr s circle Solve Prob. 7.4, using Mohr s circle Solve Prob. 7.5, using Mohr s circle Solve Prob. 7.6, using Mohr s circle Solve Prob. 7.7, using Mohr s circle Solve Prob. 7.8, using Mohr s circle. 7.5 Solve Prob. 7.9, using Mohr s circle. 7.5 Solve Prob. 7.30, using Mohr s circle Solve Prob. 7.30, using Mohr s circle and assuming that the weld forms an angle of 608 with the horiontal. 459

25 460 Transformations of Stress and Strain 7.54 and 7.55 Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown. 7 ksi 45 4 ksi + 6 ksi 4 ksi Fig. P MPa 35 MPa + 40 MPa 30 Fig. P and 7.57 Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown Fig. P Fig. P7.57

26 7.58 For the state of stress shown, determine the range of values of u for which the magnitude of the shearing stress t 99 is equal to or less than 8 ksi. Problems 46 6 ksi 6 ksi ' '' ' Fig. P For the state of stress shown, determine the range of values of u for which the normal stress s 9 is equal to or less than 50 MPa. ' ' 90 MPa '' 60 MPa Fig. P7.59 and P For the state of stress shown, determine the range of values of u for which the normal stress s 9 is equal to or less than 00 MPa. 7.6 For the element shown, determine the range of values of t for which the maimum tensile stress is equal to or less than 60 MPa. 0 MPa 0 MPa Fig. P7.6 and P For the element shown, determine the range of values of t for which the maimum in-plane shearing stress is equal to or less than 50 MPa For the state of stress shown it is known that the normal and shearing stresses are directed as shown and that s 5 4 ksi, s 5 9 ksi, and s min 5 5 ksi. Determine (a) the orientation of the principal planes, (b) the principal stress s ma, (c) the maimum in-plane shearing stress. Fig. P7.63

27 46 Transformations of Stress and Strain 7.64 The Mohr s circle shown corresponds to the state of stress given in Fig. 7.5a and b. Noting that s 9 5 (X9) cos (u p u) and that t 99 5 (X9) sin (u p u), derive the epressions for s 9 and t 99 given in Eqs. (7.5) and (7.6), respectivel. [Hint: Use sin ( ) 5 sin cos cos sin and cos ( ) 5 cos cos sin sin.] ' Y Y' p X' '' X ' Fig. P (a) Prove that the epression s 9 s 9 t 99, where s 9, s 9, and t 99 are components of the stress along the rectangular aes 9 and 9, is independent of the orientation of these aes. lso, show that the given epression represents the square of the tangent drawn from the origin of the coordinates to Mohr s circle. (b) Using the invariance propert established in part a, epress the shearing stress t in terms of s, s, and the principal stresses s ma and s min. 7.5 GENERL STTE F STRESS Fig. 7.3 ( ) Q ( ) N ( ) In the preceding sections, we have assumed a state of plane stress with s 5 t 5 t 5 0, and have considered onl transformations of stress associated with a rotation about the ais. We will now consider the general state of stress represented in Fig. 7.a and the transformation of stress associated with the rotation of aes shown in Fig. 7.b. However, our analsis will be limited to the determination of the normal stress s n on a plane of arbitrar orientation. onsider the tetrahedron shown in Fig Three of its faces are parallel to the coordinate planes, while its fourth face,, is perpendicular to the line QN. Denoting b D the area of face, and b l, l, l the direction cosines of line QN, we find that the areas of the faces perpendicular to the,, and aes are, respectivel, (D)l, (D)l, and (D)l. If the state of stress at point Q is defined b the stress components s, s, s, t, t, and t, then the forces eerted on the faces parallel to the coordinate planes can be obtained b multipling the appropriate stress components b the area of each face (Fig. 7.4). n the other hand, the forces eerted on face consist of a normal force of magnitude s n D directed along QN, and of a shearing force of magnitude t D perpendicular to QN but of otherwise unknown direction. Note that, since Q,

28 Q, and Q, respectivel, face the negative,, and aes, the forces eerted on them must be shown with negative senses. We now epress that the sum of the components along QN of all the forces acting on the tetrahedron is ero. bserving that the component along QN of a force parallel to the ais is obtained b multipling the magnitude of that force b the direction cosine l, and that the components of forces parallel to the and aes are obtained in a similar wa, we write gf n 5 0: s n s l l t l l t l l t l l s l l t l l t l l t l l s l l 5 0 Dividing through b D and solving for s n, we have s n 5 s l s l s l t l l t l l t l l (7.0) We note that the epression obtained for the normal stress s n is a quadratic form in l, l, and l. It follows that we can select the coordinate aes in such a wa that the right-hand member of Eq. (7.0) reduces to the three terms containing the squares of the direction cosines. Denoting these aes b a, b, and c, the corresponding normal stresses b s a, s b, and s c, and the direction cosines of QN with respect to these aes b l a, l b, and l c, we write s n 5 s a l a s b l b s c l c (7.) The coordinate aes a, b, c are referred to as the principal aes of stress. Since their orientation depends upon the state of stress at Q, and thus upon the position of Q, the have been represented in Fig. 7.5 as attached to Q. The corresponding coordinate planes are known as the principal planes of stress, and the corresponding normal stresses s a, s b, and s c as the principal stresses at Q. Fig General State of Stress N n Q 463 b b Q c a a a c b c Fig. 7.5 Principal stresses. In Sec. 9.6 of F. P. eer and E. R. Johnston, Vector Mechanics for Engineers, 9th ed., McGraw-Hill ook ompan, 00, a similar quadratic form is found to represent the moment of inertia of a rigid bod with respect to an arbitrar ais. It is shown in Sec. 9.7 that this form is associated with a quadric surface, and that reducing the quadratic form to terms containing onl the squares of the direction cosines is equivalent to determining the principal aes of that surface. For a discussion of the determination of the principal planes of stress and of the principal stresses, see S. P. Timoshenko and J. N. Goodier, Theor of Elasticit, 3d ed., McGraw- Hill ook ompan, 970, Sec. 77.

29 464 Transformations of Stress and Strain 7.6 PPLITIN F MHR S IRLE T THE THREE- DIMENSINL NLYSIS F STRESS c b c Fig. 7.6 Q a If the element shown in Fig. 7.5 is rotated about one of the principal aes at Q, sa the c ais (Fig. 7.6), the corresponding transformation of stress can be analed b means of Mohr s circle as if it were a transformation of plane stress. Indeed, the shearing stresses eerted on the faces perpendicular to the c ais remain equal to ero, and the normal stress s c is perpendicular to the plane ab in which the transformation takes place and, thus, does not affect this transformation. We therefore use the circle of diameter to determine the normal and shearing stresses eerted on the faces of the element as it is rotated about the c ais (Fig. 7.7). Similarl, circles of diameter and can be used to determine the stresses on the element as it is rotated about the a and b aes, respectivel. While our analsis will be limited to rotations about the principal aes, it could be shown that an other transformation of aes would lead to stresses represented in Fig. 7.7 b a point located within the shaded area. Thus, the radius ma min ma Fig. 7.7 Mohr s circles for general state of stress. D of the largest of the three circles ields the maimum value of the shearing stress at point Q. Noting that the diameter of that circle is equal to the difference between s ma and s min, we write t ma 5 0s ma s min 0 (7.) min Fig. 7.8 Z E ma ma where s ma and s min represent the algebraic values of the maimum and minimum stresses at point Q. Let us now return to the particular case of plane stress, which was discussed in Secs. 7. through 7.4. We recall that, if the and aes are selected in the plane of stress, we have s 5 t 5 t 5 0. This means that the ais, i.e., the ais perpendicular to the plane of stress, is one of the three principal aes of stress. In a Mohr-circle diagram, this ais corresponds to the origin, where s 5 t 5 0. We also recall that the other two principal aes correspond to points and where Mohr s circle for the plane intersects the s ais. If and are located on opposite sides of the origin (Fig. 7.8),

30 the corresponding principal stresses represent the maimum and minimum normal stresses at point Q, and the maimum shearing stress is equal to the maimum in-plane shearing stress. s noted in Sec. 7.3, the planes of maimum shearing stress correspond to points D and E of Mohr s circle and are at 458 to the principal planes corresponding to points and. The are, therefore, the shaded diagonal planes shown in Figs. 7.9a and b. 7.6 pplication of Mohr s ircle to the Three- Dimensional nalsis of Stress 465 b b b b a a Q a Q a a a (a) b (b) b Fig. 7.9 If, on the other hand, and are on the same side of, that is, if s a and s b have the same sign, then the circle defining s ma, s min, and t ma is not the circle corresponding to a transformation of stress within the plane. If s a. s b. 0, as assumed in Fig. 7.30, we have s ma 5 s a, s min 5 0, and t ma is equal to the radius of the circle defined b points and, that is, t ma 5 s ma. We also note that the normals Qd9 and Qe9 to the planes of maimum shearing stress are obtained b rotating the ais Qa through 458 within the a plane. Thus, the planes of maimum shearing stress are the shaded diagonal planes shown in Figs. 7.3a and b. D' D Z E' min 0 ma a Fig ma a b b b 45 d' a b a a a Q a a Q 45 e' b b (a) (b) Fig. 7.3

31 EXMPLE ksi 3 ksi Q 6 ksi Fig ksi X 3 ksi F Y 3.5 ksi Fig For the state of plane stress shown in Fig. 7.3, determine (a) the three principal planes and principal stresses, (b) the maimum shearing stress. (a) Principal Planes and Principal Stresses. We construct Mohr s circle for the transformation of stress in the plane (Fig. 7.33). Point X is plotted 6 units to the right of the t ais and 3 units above the s ais (since the corresponding shearing stress tends to rotate the element clockwise). Point Y is plotted 3.5 units to the right of the t ais and 3 units below the s ais. Drawing the line XY, we obtain the center of Mohr s circle for the plane; its abscissa is s ave 5 s s ksi Since the sides of the right triangle FX are F ksi and FX 5 3 ksi, the radius of the circle is R 5 X ksi The principal stresses in the plane of stress are s a ksi s b ksi Since the faces of the element that are perpendicular to the ais are free of stress, these faces define one of the principal planes, and the corresponding principal stress is s 5 0. The other two principal planes are defined b points and on Mohr s circle. The angle u p through which the element should be rotated about the ais to bring its faces to coincide with these planes (Fig. 7.34) is half the angle X. We have tan u p 5 FX F u p i u p i b 8.00 ksi.50 ksi p.50 ksi 8.00 ksi a D' Fig ma (b) Maimum Shearing Stress. We now draw the circles of diameter and, which correspond respectivel to rotations of the element about the a and b aes (Fig. 7.35). We note that the maimum shearing stress is equal to the radius of the circle of diameter. We thus have t ma 5 s a ksi ksi Fig E' 8.00 ksi a Since points D9 and E9, which define the planes of maimum shearing stress, are located at the ends of the vertical diameter of the circle corresponding to a rotation about the b ais, the faces of the element of Fig can be brought to coincide with the planes of maimum shearing stress through a rotation of 458 about the b ais. 466

32 *7.7 YIELD RITERI FR DUTILE MTERILS UNDER PLNE STRESS Structural elements and machine components made of a ductile material are usuall designed so that the material will not ield under the epected loading conditions. When the element or component is under uniaial stress (Fig. 7.36), the value of the normal stress s that will cause the material to ield can be obtained readil from a tensile test conducted on a specimen of the same material, since the test specimen and the structural element or machine component are in the same state of stress. Thus, regardless of the actual mechanism that causes the material to ield, we can state that the element or component will be safe as long as s, s Y, where s Y is the ield strength of the test specimen. 7.7 Yield riteria for Ductile Materials under Plane Stress 467 P' P Fig Structural element under uniaial stress. n the other hand, when a structural element or machine component is in a state of plane stress (Fig. 7.37a), it is found convenient to use one of the methods developed earlier to determine the principal stresses s a and s b at an given point (Fig. 7.37b). The material can then be regarded as being in a state of biaial stress at that point. Since this state is different from the state of uniaial stress found in a specimen subjected to a tensile test, it is clearl not possible to predict directl from such a test whether or not the structural element or machine component under investigation will fail. Some criterion regarding the actual mechanism of failure of the material must first be established, which will make it possible to compare the effects of both states of stress on the material. The purpose of this section is to present the two ield criteria most frequentl used for ductile materials. Maimum-Shearing-Stress riterion. This criterion is based on the observation that ield in ductile materials is caused b slippage of the material along oblique surfaces and is due primaril to shearing stresses (cf. Sec..3). ccording to this criterion, a given structural component is safe as long as the maimum value t ma of the shearing stress in that component remains smaller than the corresponding value of the shearing stress in a tensile-test specimen of the same material as the specimen starts to ield. Recalling from Sec.. that the maimum value of the shearing stress under a centric aial load is equal to half the value of the corresponding normal, aial stress, we conclude that the maimum shearing stress in a tensile-test specimen is s Y as the specimen starts to ield. n the other hand, we saw in Sec. 7.6 that, for plane stress, the maimum value t ma of the shearing stress is equal to 0 s ma 0 if the principal stresses are either both positive or both negative, and to 0 s ma s min 0 if the maimum stress is positive and the b (a) (b) a Fig Structural element in state of plane stress. P P

33 468 Transformations of Stress and Strain minimum stress negative. Thus, if the principal stresses s a and s b have the same sign, the maimum-shearing-stress criterion gives b 0s a 0, s Y 0 s b 0, s Y (7.3) Y If the principal stresses s a and s b have opposite signs, the maimumshearing-stress criterion ields 0s a s b 0, s Y (7.4) Y Y Y Fig Tresca s heagon. a The relations obtained have been represented graphicall in Fig n given state of stress will be represented in that figure b a point of coordinates s a and s b, where s a and s b are the two principal stresses. If this point falls within the area shown in the figure, the structural component is safe. If it falls outside this area, the component will fail as a result of ield in the material. The heagon associated with the initiation of ield in the material is known as Tresca s heagon after the French engineer Henri Edouard Tresca (84 885). b Y Y Y D Y Fig Von Mises criterion. a Maimum-Distortion-Energ riterion. This criterion is based on the determination of the distortion energ in a given material, i.e., of the energ associated with changes in shape in that material (as opposed to the energ associated with changes in volume in the same material). ccording to this criterion, also known as the von Mises criterion, after the German-merican applied mathematician Richard von Mises ( ), a given structural component is safe as long as the maimum value of the distortion energ per unit volume in that material remains smaller than the distortion energ per unit volume required to cause ield in a tensile-test specimen of the same material. s ou will see in Sec..6, the distortion energ per unit volume in an isotropic material under plane stress is u d 5 6G s a s a s b s b (7.5) where s a and s b are the principal stresses and G the modulus of rigidit. In the particular case of a tensile-test specimen that is starting to ield, we have s a 5 s Y, s b 5 0, and u d Y 5 s Y 6G. Thus, the maimum-distortion-energ criterion indicates that the structural component is safe as long as u d, (u d ) Y, or s a s a s b s b, s Y (7.6) i.e., as long as the point of coordinates s a and s b falls within the area shown in Fig This area is bounded b the ellipse of equation s a s a s b s b 5 s Y (7.7) which intersects the coordinate aes at s a 56s Y and s b 56s Y. We can verif that the major ais of the ellipse bisects the first and third quadrants and etends from (s a 5 s b 5 s Y ) to (s a 5 s b 5 s Y ), while its minor ais etends from (s a 5 s b s Y ) to D (s a 5 s b s Y ). The maimum-shearing-stress criterion and the maimumdistortion-energ criterion are compared in Fig We note that the ellipse passes through the vertices of the heagon. Thus, for the states of stress represented b these si points, the two criteria give

34 the same results. For an other state of stress, the maimum-shearingstress criterion is more conservative than the maimum-distortionenerg criterion, since the heagon is located within the ellipse. state of stress of particular interest is that associated with ield in a torsion test. We recall from Fig. 7. of Sec. 7.4 that, for torsion, s min 5 s ma ; thus, the corresponding points in Fig are located on the bisector of the second and fourth quadrants. It follows that ield occurs in a torsion test when s a 5s b 560.5s Y according to the maimum-shearing-stress criterion, and when s a 5s b s Y according to the maimum-distortion-energ criterion. ut, recalling again Fig. 7., we note that s a and s b must be equal in magnitude to t ma, that is, to the value obtained from a torsion test for the ield strength t Y of the material. Since the values of the ield strength s Y in tension and of the ield strength t Y in shear are given for various ductile materials in ppendi, we can compute the ratio t Y s Y for these materials and verif that the values obtained range from 0.55 to Thus, the maimum-distortion-energ criterion appears somewhat more accurate than the maimum-shearing-stress criterion as far as predicting ield in torsion is concerned. *7.8 FRTURE RITERI FR RITTLE MTERILS UNDER PLNE STRESS s we saw in hap., brittle materials are characteried b the fact that, when subjected to a tensile test, the fail suddenl through rupture or fracture without an prior ielding. When a structural element or machine component made of a brittle material is under uniaial tensile stress, the value of the normal stress that causes it to fail is equal to the ultimate strength s U of the material as determined from a tensile test, since both the tensile-test specimen and the element or component under investigation are in the same state of stress. However, when a structural element or machine component is in a state of plane stress, it is found convenient to first determine the principal stresses s a and s b at an given point, and to use one of the criteria indicated in this section to predict whether or not the structural element or machine component will fail. Maimum-Normal-Stress riterion. ccording to this criterion, a given structural component fails when the maimum normal stress in that component reaches the ultimate strength s U obtained from the tensile test of a specimen of the same material. Thus, the structural component will be safe as long as the absolute values of the principal stresses s a and s b are both less than s U : 0s a 0, s U 0s b 0, s U (7.8) The maimum-normal-stress criterion can be epressed graphicall as shown in Fig If the point obtained b plotting the values s a and s b of the principal stresses falls within the square area shown in the figure, the structural component is safe. If it falls outside that area, the component will fail. The maimum-normal-stress criterion, also known as oulomb s criterion, after the French phsicist harles ugustin de oulomb Y Fig U 7.8 Fracture riteria for rittle Materials under Plane Stress Y b U U b 0.5 Y Y U Y a Fig. 7.4 oulomb s criterion Y Torsion a 469

35 470 Transformations of Stress and Strain ( ), suffers from an important shortcoming, since it is based on the assumption that the ultimate strength of the material is the same in tension and in compression. s we noted in Sec..3, this is seldom the case, because of the presence of flaws in the material, such as microscopic cracks or cavities, which tend to weaken the a a material in tension, while not appreciabl affecting its resistance to U compressive failure. esides, this criterion makes no allowance for b b UT effects other than those of the normal stresses on the failure mechanism of the material. (a) U (b) Fig UT b U UT a Mohr s riterion. This criterion, suggested b the German engineer tto Mohr, can be used to predict the effect of a given state of plane stress on a brittle material, when results of various tpes of tests are available for that material. Let us first assume that a tensile test and a compressive test have been conducted on a given material, and that the values s UT and s U of the ultimate strength in tension and in compression have been determined for that material. The state of stress corresponding to the rupture of the tensile-test specimen can be represented on a Mohrcircle diagram b the circle intersecting the horiontal ais at and s UT (Fig. 7.43a). Similarl, the state of stress corresponding to the failure of the compressive-test specimen can be represented b the circle intersecting the horiontal ais at and s U. learl, a state of stress represented b a circle entirel contained in either of these circles will be safe. Thus, if both principal stresses are positive, the state of stress is safe as long as s a, s UT and s b, s UT ; if both principal stresses are negative, the state of stress is safe as long as s a, s U and s b, s U. Plotting the point of coordinates s a and s b (Fig. 7.43b), we verif that the state of stress is safe as long as that point falls within one of the square areas shown in that figure. In order to anale the cases when s a and s b have opposite signs, we now assume that a torsion test has been conducted on the material and that its ultimate strength in shear, t U, has been determined. Drawing the circle centered at representing the state of stress corresponding to the failure of the torsion-test specimen (Fig. 7.44a), we observe that an state of stress represented b a circle entirel contained in that circle is also safe. Mohr s criterion is a logical etension of this observation: ccording to Mohr s criterion, a state of stress is safe if it is represented b a circle located entirel within the area bounded U U U b U U a nother failure criterion known as the maimum-normal-strain criterion, or Saint- Venant s criterion, was widel used during the nineteenth centur. ccording to this criterion, a given structural component is safe as long as the maimum value of the normal strain in that component remains smaller than the value P U of the strain at which a tensiletest specimen of the same material will fail. ut, as will be shown in Sec. 7., the strain is maimum along one of the principal aes of stress, if the deformation is elastic and the material homogeneous and isotropic. Thus, denoting b P a and P b the values of the normal strain along the principal aes in the plane of stress, we write 0P a 0, P U 0P b 0, P U (7.9) U Fig. 7.4 Saint-Venant s criterion. Making use of the generalied Hooke s law (Sec..), we could epress these relations in terms of the principal stresses s a and s b and the ultimate strength s U of the material. We would find that, according to the maimum-normal-strain criterion, the structural component is safe as long as the point obtained b plotting s a and s b falls within the area shown in Fig. 7.4 where n is Poisson s ratio for the given material.

36 b UT 7.8 Fracture riteria for rittle Materials under Plane Stress 47 U U UT a U UT U (a) Fig Mohr s criterion. (b) b the envelope of the circles corresponding to the available data. The remaining portions of the principal-stress diagram can now be obtained b drawing various circles tangent to this envelope, determining the corresponding values of s a and s b, and plotting the points of coordinates s a and s b (Fig. 7.44b). More accurate diagrams can be drawn when additional test results, corresponding to various states of stress, are available. If, on the other hand, the onl available data consists of the ultimate strengths s UT and s U, the envelope in Fig. 7.44a is replaced b the tangents and 99 to the circles corresponding respectivel to failure in tension and failure in compression (Fig. 7.45a). From the similar triangles drawn in that figure, we note that the abscissa of the center of a circle tangent to and 99 is linearl related to its radius R. Since s a 5 R and s b 5 R, it follows that s a and s b are also linearl related. Thus, the shaded area corresponding to this simplified Mohr s criterion is bounded b straight lines in the second and fourth quadrants (Fig. 7.45b). Note that in order to determine whether a structural component will be safe under a given loading, the state of stress should be calculated at all critical points of the component, i.e., at all points where stress concentrations are likel to occur. This can be done in a number of cases b using the stress-concentration factors given in Figs..60, 3.9, 4.7, and 4.8. There are man instances, however, when the theor of elasticit must be used to determine the state of stress at a critical point. Special care should be taken when macroscopic cracks have been detected in a structural component. While it can be assumed that the test specimen used to determine the ultimate tensile strength of the material contained the same tpe of flaws (i.e., microscopic cracks or cavities) as the structural component under investigation, the specimen was certainl free of an detectable macroscopic cracks. When a crack is detected in a structural component, it is necessar to determine whether that crack will tend to propagate under the epected loading condition and cause the component to fail, or whether it will remain stable. This requires an analsis involving the energ associated with the growth of the crack. Such an analsis is beond the scope of this tet and should be carried out b the methods of fracture mechanics. U b (a) U (b) R ' ' a UT b UT UT a U Fig Simplified Mohr s criterion.

37 SMPLE PRLEM MPa 80 MPa 5 MPa The state of plane stress shown occurs at a critical point of a steel machine component. s a result of several tensile tests, it has been found that the tensile ield strength is s Y 5 50 MPa for the grade of steel used. Determine the factor of safet with respect to ield, using (a) the maimum-shearingstress criterion, and (b) the maimum-distortion-energ criterion. SLUTIN 5 MPa Y 40 MPa b 80 MPa D R 0 MPa a F X m 5 MPa Mohr s ircle. We construct Mohr s circle for the given state of stress and find s ave 5 5 s s MPa t m 5 R 5 F FX MPa Principal Stresses s a MPa s b MPa a. Maimum-Shearing-Stress riterion. Since for the grade of steel used the tensile strength is s Y 5 50 MPa, the corresponding shearing stress at ield is For t m 5 65 MPa: t Y 5 s Y 5 50 MPa 5 5 MPa F.S. 5 t Y 5 MPa 5 t m 65 MPa F.S b Y 50 MPa 85 H T M Y 50 MPa a b. Maimum-Distortion-Energ riterion. Introducing a factor of safet into Eq. (7.6), we write s a s a s b s b 5 a s Y F.S. b For s a 5 85 MPa, s b 5 45 MPa, and s Y 5 50 MPa, we have a 50 F.S. b F.S. F.S. 5.9 omment. For a ductile material with s Y 5 50 MPa, we have drawn the heagon associated with the maimum-shearing-stress criterion and the ellipse associated with the maimum-distortion-energ criterion. The given state of plane stress is represented b point H of coordinates s a 5 85 MPa and s b 5 45 MPa. We note that the straight line drawn through points and H intersects the heagon at point T and the ellipse at point M. For each criterion, the value obtained for F.S. can be verified b measuring the line segments indicated and computing their ratios: a F.S. 5 T H 5.9 M b F.S. 5 H

38 PRLEMS 7.66 For the state of plane stress shown, determine the maimum shearing stress when (a) s 5 6 ksi and s 5 8 ksi, (b) s 5 4 ksi and s 5 ksi. (Hint: onsider both in-plane and out-of-plane shearing stresses.) σ 8 ksi σ Fig. P7.66 and P For the state of plane stress shown, determine the maimum shearing stress when (a) s 5 0 and s 5 ksi, (b) s 5 ksi and s 5 9 ksi. (Hint: onsider both in-plane and out-of-plane shearing stresses.) σ 80 MPa 7.68 For the state of stress shown, determine the maimum shearing stress when (a) s 5 40 MPa, (b) s 5 0 MPa. (Hint: onsider both in-plane and out-of-plane shearing stresses.) 7.69 For the state of stress shown, determine the maimum shearing stress when (a) s 5 0 MPa, (b) s 5 40 MPa. (Hint: onsider both in-plane and out-of-plane shearing stresses.) Fig. P7.68 and P MPa 7.70 and 7.7 For the state of stress shown, determine the maimum shearing stress when (a) s 5 4 ksi, (b) s 5 4 ksi, (c) s 5 0. ksi 0 ksi 6 ksi 6 ksi σ 7 ksi σ 5 ksi Fig. P7.70 Fig. P

39 474 Transformations of Stress and Strain 7.7 and 7.73 For the state of stress shown, determine the maimum shearing stress when (a) s 5 0, (b) s 5 45 MPa, (c) s 5 45 MPa. 0 MPa 70 MPa 75 MPa 75 MPa σ 00 MPa σ 50 MPa Fig. P7.7 Fig. P For the state of stress shown, determine two values of s for which the maimum shearing stress is 0 ksi. σ σ 8 ksi 48 MPa 4 ksi 50 MPa Fig. P7.74 Fig. P For the state of stress shown, determine two values of s for which the maimum shearing stress is 73 MPa For the state of stress shown, determine the value of t for which the maimum shearing stress is (a) 0 ksi, (b) 8.5 ksi. 3 ksi τ 5 ksi Fig. P7.76

40 7.77 For the state of stress shown, determine the value of t for which the maimum shearing stress is (a) 60 MPa, (b) 78 MPa. Problems MPa τ 00 MPa σ Fig. P For the state of stress shown, determine two values of s for which the maimum shearing stress is 80 MPa For the state of stress shown, determine the range of values of t for which the maimum shearing stress is equal to or less than 60 MPa. Fig. P MPa 60 MPa σ 00 MPa *7.80 For the state of stress of Prob. 7.69, determine (a) the value of s for which the maimum shearing stress is as small as possible, (b) the corresponding value of the shearing stress. 7.8 The state of plane stress shown occurs in a machine component made of a steel with s Y 5 35 MPa. Using the maimum-distortionenerg criterion, determine whether ield will occur when (a) s MPa, (b) s MPa, (c) s MPa. If ield does not occur, determine the corresponding factor of safet. 7.8 Solve Prob. 7.8, using the maimum-shearing-stress criterion. Fig. P7.79 σ 0 τ 00 MPa 60 MPa 7.83 The state of plane stress shown occurs in a machine component made of a steel with s Y 5 45 ksi. Using the maimum-distortionenerg criterion, determine whether ield will occur when (a) t 5 9 ksi, (b) t 5 8 ksi, (c) t 5 0 ksi. If ield does not occur, determine the corresponding factor of safet. Fig. P7.8 σ 0 ksi τ 36 ksi Fig. P Solve Prob. 7.83, using the maimum-shearing-stress criterion.

41 476 Transformations of Stress and Strain 7.85 The 36-mm-diameter shaft is made of a grade of steel with a 50-MPa tensile ield stress. Using the maimum-shearing-stress criterion, P determine the magnitude of the torque T for which ield occurs when P 5 00 kn. T 7.86 Solve Prob. 7.85, using the maimum-distortion-energ criterion. 36 mm 7.87 The.75-in.-diameter shaft is made of a grade of steel for which the ield strength is s Y 5 36 ksi. Using the maimum-shearingstress criterion, determine the magnitude of the force P for which ield occurs when T 5 5 kip? in. Fig. P in. T Fig. P Solve Prob. 7.87, using the maimum-distortion-energ criterion. P 00 MPa 60 MPa 0 MPa 7.89 and 7.90 The state of plane stress shown is epected to occur in an aluminum casting. Knowing that for the aluminum allo used s UT 5 80 MPa and s U 5 00 MPa and using Mohr s criterion, determine whether rupture of the casting will occur. 75 MPa 3 MPa Fig. P7.89 Fig. P and 7.9 The state of plane stress shown is epected to occur in an aluminum casting. Knowing that for the aluminum allo used s UT 5 0 ksi and s U 5 30 ksi and using Mohr s criterion, determine whether rupture of the casting will occur. 5 ksi 9 ksi 7 ksi ksi 8 ksi Fig. P7.9 Fig. P7.9

42 7.93 The state of plane stress shown will occur at a critical point in an aluminum casting that is made of an allo for which s UT 5 0 ksi and s U 5 5 ksi. Using Mohr s criterion, determine the shearing stress t 0 for which failure should be epected. Problems ksi 0 Fig. P The state of plane stress shown will occur at a critical point in a pipe made of an aluminum allo for which s UT 5 75 MPa and s U 5 50 MPa. Using Mohr s criterion, determine the shearing stress t 0 for which failure should be epected The cast-aluminum rod shown is made of an allo for which s UT 5 60 MPa and s U 5 0 MPa. Using Mohr s criterion, determine the magnitude of the torque T for which failure should be epected. Fig. P MPa 3 mm T Fig. P kn 0 T' 7.96 The cast-aluminum rod shown is made of an allo for which s UT 5 70 MPa and s U 5 75 MPa. Knowing that the magnitude T of the applied torques is slowl increased and using Mohr s criterion, determine the shearing stress t 0 that should be epected at rupture. T Fig. P machine component is made of a grade of cast iron for which s UT 5 8 ksi and s U 5 0 ksi. For each of the states of stress shown, and using Mohr s criterion, determine the normal stress s 0 at which rupture of the component should be epected Fig. P7.97 (a) (b) (c)

43 478 Transformations of Stress and Strain 7.9 STRESSES IN THIN-WLLED PRESSURE VESSELS Fig ssumed stress distribution in thin-walled pressure vessels. Thin-walled pressure vessels provide an important application of the analsis of plane stress. Since their walls offer little resistance to bending, it can be assumed that the internal forces eerted on a given portion of wall are tangent to the surface of the vessel (Fig. 7.46). The resulting stresses on an element of wall will thus be contained in a plane tangent to the surface of the vessel. ur analsis of stresses in thin-walled pressure vessels will be limited to the two tpes of vessels most frequentl encountered: clindrical pressure vessels and spherical pressure vessels (Photos 7.3 and 7.4). Photo 7.3 lindrical pressure vessels. Photo 7.4 Spherical pressure vessels. Fig Pressuried clindrical vessel. d p d d Fig Free bod to determine hoop stress. r t t r r t onsider a clindrical vessel of inner radius r and wall thickness t containing a fluid under pressure (Fig. 7.47). We propose to determine the stresses eerted on a small element of wall with sides respectivel parallel and perpendicular to the ais of the clinder. ecause of the aismmetr of the vessel and its contents, it is clear that no shearing stress is eerted on the element. The normal stresses s and s shown in Fig are therefore principal stresses. The stress s is known as the hoop stress, because it is the tpe of stress found in hoops used to hold together the various slats of a wooden barrel, and the stress s is called the longitudinal stress. In order to determine the hoop stress s, we detach a portion of the vessel and its contents bounded b the plane and b two planes parallel to the plane at a distance D from each other (Fig. 7.48). The forces parallel to the ais acting on the free bod defined in this fashion consist of the elementar internal forces s d on the wall sections, and of the elementar pressure forces p d eerted on the portion of fluid included in the free bod. Note that p denotes the gage pressure of the fluid, i.e., the ecess of the inside pressure over the outside atmospheric pressure. The resultant of the internal forces s d is equal to the product of s and of the cross-sectional area t D of the wall, while the resultant of the pressure forces p d is equal to the product of p and of the area r D. Writing the equilibrium equation of 5 0, we have

44 F 5 0: s t pr 5 0 and, solving for the hoop stress s, 7.9 Stresses in Thin-Walled Pressure Vessels 479 s 5 pr t (7.30) To determine the longitudinal stress s, we now pass a section perpendicular to the ais and consider the free bod consisting of the portion of the vessel and its contents located to the left of the section t d r p d Fig Free bod to determine longitudinal stress. (Fig. 7.49). The forces acting on this free bod are the elementar internal forces s d on the wall section and the elementar pressure forces p d eerted on the portion of fluid included in the free bod. Noting that the area of the fluid section is pr and that the area of the wall section can be obtained b multipling the circumference pr of the clinder b its wall thickness t, we write the equilibrium equation: of 5 0: s prt ppr 5 0 and, solving for the longitudinal stress s, s 5 pr t (7.3) We note from Eqs. (7.30) and (7.3) that the hoop stress s is twice as large as the longitudinal stress s : s 5 s (7.3) Using the mean radius of the wall section, r m 5 r t, in computing the resultant of the forces on that section, we would obtain a more accurate value of the longitudinal stress, namel, s 5 pr t t r (7.39) However, for a thin-walled pressure vessel, the term tr is sufficientl small to allow the use of Eq. (7.3) for engineering design and analsis. If a pressure vessel is not thin-walled (i.e., if tr is not small), the stresses s and s var across the wall and must be determined b the methods of the theor of elasticit.

45 480 Transformations of Stress and Strain Drawing Mohr s circle through the points and that correspond respectivel to the principal stresses s and s (Fig. 7.50), and D' recalling that the maimum in-plane shearing stress is equal to the radius of this circle, we have E' E ma Fig Mohr s circle for element of clindrical pressure vessel. D t ma in plane 5 s 5 pr (7.33) 4t This stress corresponds to points D and E and is eerted on an element obtained b rotating the original element of Fig through 458 within the plane tangent to the surface of the vessel. The maimum shearing stress in the wall of the vessel, however, is larger. It is equal to the radius of the circle of diameter and corresponds to a rotation of 458 about a longitudinal ais and out of the plane of stress. We have t ma 5 s 5 pr t (7.34) Fig. 7.5 Pressuried spherical vessel. t d r We now consider a spherical vessel of inner radius r and wall thickness t, containing a fluid under a gage pressure p. For reasons of smmetr, the stresses eerted on the four faces of a small element of wall must be equal (Fig. 7.5). We have s 5 s (7.35) To determine the value of the stress, we pass a section through the center of the vessel and consider the free bod consisting of the portion of the vessel and its contents located to the left of the section (Fig. 7.5). The equation of equilibrium for this free bod is the same as for the free bod of Fig We thus conclude that, for a spherical vessel, s 5 s 5 pr t (7.36) p d Fig. 7.5 Free bod to determine wall stress. D' ma Fig Mohr s circle for element of spherical pressure vessel. Since the principal stresses s and s are equal, Mohr s circle for transformations of stress within the plane tangent to the surface of the vessel reduces to a point (Fig. 7.53); we conclude that the in-plane normal stress is constant and that the in-plane maimum shearing stress is ero. The maimum shearing stress in the wall of the vessel, however, is not ero; it is equal to the radius of the circle of diameter and corresponds to a rotation of 458 out of the plane of stress. We have t ma 5 s 5 pr 4t (7.37) It should be observed that, while the third principal stress is ero on the outer surface of the vessel, it is equal to p on the inner surface, and is represented b a point (p, 0) on a Mohr-circle diagram. Thus, close to the inside surface of the vessel, the maimum shearing stress is equal to the radius of a circle of diameter, and we have t ma 5 s p 5 pr t a t r b For a thin-walled vessel, however, the term t/r is small, and we can neglect the variation of t ma across the wall section. This remark also applies to spherical pressure vessels.

46 8 ft SMPLE PRLEM in. compressed-air tank is supported b two cradles as shown; one of the cradles is designed so that it does not eert an longitudinal force on the tank. The clindrical bod of the tank has a 30-in. outer diameter and is fabricated from a 3 8-in. steel plate b butt welding along a heli that forms an angle of 58 with a transverse plane. The end caps are spherical and have a uniform wall thickness of 5 6 in. For an internal gage pressure of 80 psi, determine (a) the normal stress and the maimum shearing stress in the spherical caps. (b) the stresses in directions perpendicular and parallel to the helical weld. a SLUTIN a. Spherical ap. Using Eq. (7.36), we write b psi D' ma, p 5 80 psi, t in in., r in. s 5 s 5 pr 80 psi4.688 in. 5 s psi b t 0.35 in. We note that for stresses in a plane tangent to the cap, Mohr s circle reduces to a point (, ) on the horiontal ais and that all in-plane shearing stresses are ero. n the surface of the cap the third principal stress is ero and corresponds to point. n a Mohr s circle of diameter, point D9 represents the maimum shearing stress; it occurs on planes at 458 to the plane tangent to the cap. t ma psi t ma 5 5 psi b b. lindrical od of the Tank. We first determine the hoop stress s and the longitudinal stress s. Using Eqs. (7.30) and (7.3), we write a 700 psi p 5 80 psi, t in in., r in. s 5 pr 80 psi4.65 in psi s 5 t in. s psi s ave 5 s s psi R 5 s s psi 350 psi b Stresses at the Weld. Noting that both the hoop stress and the longitudinal stress are principal stresses, we draw Mohr s circle as shown. n element having a face parallel to the weld is obtained b rotating the face perpendicular to the ais b counterclockwise through 58. Therefore, on Mohr s circle we locate the point X9 corresponding to the stress components on the weld b rotating radius counterclockwise through u psi ave 565 psi 350 psi 50 w X' R R 755 psi w s w 5 s ave R cos cos 50 t w 5 R sin sin 50 s w 5440 psi b t w psi b Since X9 is below the horiontal ais, t w tends to rotate the element counterclockwise. ' w 440 psi w 344 psi Weld 48

47 PRLEMS 7.98 spherical gas container made of steel has a 5-m outer diameter and a wall thickness of 6 mm. Knowing that the internal pressure is 350 kpa, determine the maimum normal stress and the maimum shearing stress in the container The maimum gage pressure is known to be 8 MPa in a spherical steel pressure vessel having a 50-mm outer diameter and a 6-mm wall thickness. Knowing that the ultimate stress in the steel used is s U MPa, determine the factor of safet with respect to tensile failure basketball has a 9.5-in. outer diameter and a 0.5-in. wall thickness. Determine the normal stress in the wall when the basketball is inflated to a 9-psi gage pressure. 7.0 spherical pressure vessel of 900-mm outer diameter is to be fabricated from a steel having an ultimate stress s U MPa. Knowing that a factor of safet of 4.0 is desired and that the gage pressure can reach 3.5 MPa, determine the smallest wall thickness that should be used. 7.0 spherical pressure vessel has an outer diameter of 0 ft and a wall thickness of 0.5 in. Knowing that for the steel used s all 5 ksi, E psi, and n 5 0.9, determine (a) the allowable gage pressure, (b) the corresponding increase in the diameter of the vessel spherical gas container having an outer diameter of 5 m and a wall thickness of mm is made of steel for which E 5 00 GPa and n Knowing that the gage pressure in the container is increased from ero to.7 MPa, determine (a) the maimum normal stress in the container, (b) the corresponding increase in the diameter of the container. 300 m 750 mm Fig. P7.04 and P steel penstock has a 750-mm outer diameter, a -mm wall thickness, and connects a reservoir at with a generating station at. Knowing that the densit of water is 000 kg/m 3, determine the maimum normal stress and the maimum shearing stress in the penstock under static conditions steel penstock has a 750-mm outer diameter and connects a reservoir at with a generating station at. Knowing that the densit of water is 000 kg/m 3 and that the allowable normal stress in the steel is 85 MPa, determine the smallest thickness that can be used for the penstock The bulk storage tank shown in Photo 7.3 has an outer diameter of 3.3 m and a wall thickness of 8 mm. t a time when the internal pressure of the tank is.5 MPa, determine the maimum normal stress and the maimum shearing stress in the tank.

48 7.07 Determine the largest internal pressure that can be applied to a clindrical tank of 5.5-ft outer diameter and 5 8-in. wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safet of 5.0 is desired. 5 ft Problems clindrical storage tank contains liquefied propane under a pressure of.5 MPa at a temperature of 388. Knowing that the tank has an outer diameter of 30 mm and a wall thickness of 3 mm, determine the maimum normal stress and the maimum shearing stress in the tank. 48 ft h 7.09 The unpressuried clindrical storage tank shown has a 3 6-in. wall thickness and is made of steel having a 60-ksi ultimate strength in tension. Determine the maimum height h to which it can be filled with water if a factor of safet of 4.0 is desired. (Specific weight of water lb/ft 3.) Fig. P For the storage tank of Prob. 7.09, determine the maimum normal stress and the maimum shearing stress in the clindrical wall when the tank is filled to capacit (h 5 48 ft). 7. standard-weight steel pipe of -in. nominal diameter carries water under a pressure of 400 psi. (a) Knowing that the outside diameter is.75 in. and the wall thickness is in., determine the maimum tensile stress in the pipe. (b) Solve part a, assuming an etra-strong pipe is used, of.75-in. outside diameter and 0.5-in. wall thickness. 3 m.6 m 7. The pressure tank shown has a 8-mm wall thickness and butt-welded seams forming an angle b 5 08 with a transverse plane. For a gage pressure of 600 kpa, determine, (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld. Fig. P For the tank of Prob. 7., determine the largest allowable gage pressure, knowing that the allowable normal stress perpendicular to the weld is 0 MPa and the allowable shearing stress parallel to the weld is 80 MPa. 7.4 For the tank of Prob. 7., determine the range of values of b that can be used if the shearing stress parallel to the weld is not to eceed MPa when the gage pressure is 600 kpa. 7.5 The steel pressure tank shown has a 750-mm inner diameter and a 9-mm wall thickness. Knowing that the butt-welded seams form an angle b with the longitudinal ais of the tank and that the gage pressure in the tank is.5 MPa, determine, (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld. 7.6 The pressuried tank shown was fabricated b welding strips of plate along a heli forming an angle b with a transverse plane. Determine the largest value of b that can be used if the normal stress perpendicular to the weld is not to be larger than 85 percent of the maimum stress in the tank. Fig. P7.5 and P7.6

49 484 Transformations of Stress and Strain 7.7 The clindrical portion of the compressed-air tank shown is fabricated of 0.5-in.-thick plate welded along a heli forming an angle 0 in. b with the horiontal. Knowing that the allowable stress normal to the weld is 0.5 ksi, determine the largest gage pressure that can be used in the tank. 60 in. 7.8 For the compressed-air tank of Prob. 7.7, determine the gage pressure that will cause a shearing stress parallel to the weld of 4 ksi. 7.9 Square plates, each of 0.5-in. thickness, can be bent and welded together in either of the two was shown to form the clindrical portion of a compressed-air tank. Knowing that the allowable normal stress perpendicular to the weld is ksi, determine the largest allowable gage pressure in each case. Fig. P7.7 ft ft 0 ft 45 Fig. P7.9 (a) (b) 7.0 The compressed-air tank has an inner diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing that the gage pressure inside the tank is. MPa, determine the maimum normal stress and the maimum in-plane shearing stress at point a on the top of the tank. 750 mm 750 mm b a 500 mm D 5 kn Fig. P For the compressed-air tank and loading of Prob. 7.0, determine the maimum normal stress and the maimum in-plane shearing stress at point b on the top of the tank.

50 7. The compressed-air tank has a 50-mm outside diameter and an 8-mm wall thickness. It is fitted with a collar b which a 40-kN force P is applied at in the horiontal direction. Knowing that the gage pressure inside the tank is 5 MPa, determine the maimum normal stress and the maimum shearing stress at point K. 50 mm Problems In Prob. 7., determine the maimum normal stress and the maimum shearing stress at point L. 7.4 pressure vessel of 0-in. inner diameter and 0.5-in. wall thickness is fabricated from a 4-ft section of spirall-welded pipe and is equipped with two rigid end plates. The gage pressure inside the vessel is 300 psi and 0-kip centric aial forces P and P9 are applied to the end plates. Determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld. K L 600 mm P P' 4 ft 50 mm P Fig. P7. 35 Fig. P Solve Prob. 7.4, assuming that the magnitude P of the two forces is increased to 30 kips. 7.6 brass ring of 5-in. outer diameter and 0.5-in. thickness fits eactl inside a steel ring of 5-in. inner diameter and 0.5-in. thickness when the temperature of both rings is 508F. Knowing that the temperature of both rings is then raised to 58F, determine (a) the tensile stress in the steel ring, (b) the corresponding pressure eerted b the brass ring on the steel ring..5 in. 5 in. Fig. P7.6 STEEL t s 8 in. E s psi s /F RSS t b 4 in. E b psi s b /F 7.7 Solve Prob. 7.6, assuming that the brass ring is 0.5 in. thick and the steel ring is 0.5 in. thick.

51 486 Transformations of Stress and Strain *7.0 TRNSFRMTIN F PLNE STRIN Fied support Fied support Fig Plane strain eample: laterall restrained plate. Transformations of strain under a rotation of the coordinate aes will now be considered. ur analsis will first be limited to states of plane strain, i.e., to situations where the deformations of the material take place within parallel planes, and are the same in each of these planes. If the ais is chosen perpendicular to the planes in which the deformations take place, we have P 5 g 5 g 5 0, and the onl remaining strain components are P, P, and g. Such a situation occurs in a plate subjected along its edges to uniforml distributed loads and restrained from epanding or contracting laterall b smooth, rigid, and fied supports (Fig. 7.54). It would also be found in a bar of infinite length subjected on its sides to uniforml distributed loads since, b reason of smmetr, the elements located in a given transverse plane cannot move out of that plane. This idealied model shows that, in the actual case of a long bar subjected to uniforml distributed transverse loads (Fig. 7.55), a state of plane strain eists in an given transverse section that is not located too close to either end of the bar. s Q s s ( ) Q s ( ) Fig Plane strain element deformation. ' Q s s ' ' s ( ' ) '' '' Q s ( ' ) Fig Transformation of plane strain element. ' Fig Plane strain eample: bar of infinite length. Let us assume that a state of plane strain eists at point Q (with P 5 g 5 g 5 0), and that it is defined b the strain components P, P, and g associated with the and aes. s we know from Secs.. and.4, this means that a square element of center Q, with sides of length Ds respectivel parallel to the and aes, is deformed into a parallelogram with sides of length respectivel equal to Ds ( P ) and Ds ( P ), forming angles of p g and p g with each other (Fig. 7.56). We recall that, as a result of the deformations of the other elements located in the plane, the element considered ma also undergo a rigid-bod motion, but such a motion is irrelevant to the determination of the strains at point Q and will be ignored in this analsis. ur purpose is to determine in terms of P, P, g, and u the strain components P 9, P 9, and g 99 associated with the frame of reference 99 obtained b rotating the and aes through the angle u. s shown in Fig. 7.57, these new strain It should be observed that a state of plane strain and a state of plane stress (cf. Sec. 7.) do not occur simultaneousl, ecept for ideal materials with a Poisson ratio equal to ero. The constraints placed on the elements of the plate of Fig and of the bar of Fig result in a stress s different from ero. n the other hand, in the case of the plate of Fig. 7.3, the absence of an lateral restraint results in s 5 0 and P Z 0.

52 components define the parallelogram into which a square with sides respectivel parallel to the 9 and 9 aes is deformed. We first derive an epression for the normal strain P(u) along a line forming an arbitrar angle u with the ais. To do so, we consider the right triangle, which has for hpothenuse (Fig. 7.58a), and the oblique triangle 999 into which triangle is deformed (Fig. 7.58b). Denoting b Ds the length of, we epress the length of 99 as Ds [ P(u)]. Similarl, denoting b D and D the lengths of sides and, we epress the lengths of 99 and 99 as D ( P ) and D ( P ), respectivel. Recalling from Fig that the right angle at in Fig. 7.58a deforms into an angle equal to p g in Fig. 7.58b, and appling the law of cosines to triangle 999, we write 5 cos a p g b s 3 Pu4 5 P P P P cos a p g b (7.38) ut from Fig. 7.58a we have 5 s cos u 5 s sin u (7.39) and we note that, since g is ver small, cos a p g b 5sin g < g (7.40) Fig Transformation of Plane Strain ' s (a) ( ) s [ ( )] (b) ' ( ) ' 487 Substituting from Eqs. (7.39) and (7.40) into Eq. (7.38), recalling that cos u sin u 5, and neglecting second-order terms in P(u), P, P, and g, we write Pu 5 P cos u P sin u g sin u cos u (7.4) Equation (7.4) enables us to determine the normal strain P(u) in an direction in terms of the strain components P, P, g, and the angle u that forms with the ais. We check that, for u 5 0, Eq. (7.4) ields P(0) 5 P and that, for u 5 908, it ields P(908) 5 P. n the other hand, making u in Eq. (7.4), we obtain the normal strain in the direction of the bisector of the angle formed b the and aes (Fig. 7.59). Denoting this strain b P, we write Fig P 5 P45 5 P P g (7.4) Solving Eq. (7.4) for g, we have g 5 P P P (7.43) This relation makes it possible to epress the shearing strain associated with a given pair of rectangular aes in terms of the normal strains measured along these aes and their bisector. It will pla a fundamental role in our present derivation and will also be used in Sec. 7.3 in connection with the eperimental determination of shearing strains.

53 488 Transformations of Stress and Strain Recalling that the main purpose of this section is to epress the strain components associated with the frame of reference 99 of Fig in terms of the angle u and the strain components P, P, and g associated with the and aes, we note that the normal strain P 9 along the 9 ais is given b Eq. (7.4). Using the trigonometric relations (7.3) and (7.4), we write this equation in the alternative form P 5 P P P P cos u g sin u (7.44) Replacing u b u 908, we obtain the normal strain along the 9 ais. Since cos (u 808) 5 cos u and sin (u 808) 5 sin u, we have P 5 P P P P cos u g sin u (7.45) dding Eqs. (7.44) and (7.45) member to member, we obtain P P 5 P P (7.46) Since P 5 P 9 5 0, we thus verif in the case of plane strain that the sum of the normal strains associated with a cubic element of material is independent of the orientation of that element. Replacing now u b u 458 in Eq. (7.44), we obtain an epression for the normal strain along the bisector 9 of the angle formed b the 9 and 9 aes. Since cos (u 908) 5 sin u and sin (u 908) 5 cos u, we have P 5 P P P P sin u g cos u (7.47) Writing Eq. (7.43) with respect to the 9 and 9 aes, we epress the shearing strain g 99 in terms of the normal strains measured along the 9 and 9 aes and the bisector 9: g 5 P P P (7.48) Substituting from Eqs. (7.46) and (7.47) into (7.48), we obtain g 5P P sin u g cos u (7.49) Equations (7.44), (7.45), and (7.49) are the desired equations defining the transformation of plane strain under a rotation of aes in the plane of strain. Dividing all terms in Eq. (7.49) b, we write this equation in the alternative form g 5 P P sin u g cos u (7.499) and observe that Eqs. (7.44), (7.45), and (7.499) for the transformation of plane strain closel resemble the equations derived in Sec. 7. for the transformation of plane stress. While the former ma be obtained from the latter b replacing the normal stresses b the corresponding normal strains, it should be noted, however, that the shearing stresses t and t 99 should be replaced b half of the corresponding shearing strains, i.e., b g and g, respectivel. f. first footnote on page 97.

54 *7. MHR S IRLE FR PLNE STRIN Since the equations for the transformation of plane strain are of the same form as the equations for the transformation of plane stress, the use of Mohr s circle can be etended to the analsis of plane strain. Given the strain components P, P, and g defining the deformation represented in Fig. 7.56, we plot a point XP, g of abscissa equal to the normal strain P and of ordinate equal to minus half the shearing strain g, and a point YP, g (Fig. 7.60). Drawing the diameter XY, we define the center of Mohr s circle for plane strain. The abscissa of and the radius R of the circle are respectivel equal to P ave 5 P P and R 5 a P P b a g b (7.50) We note that if g is positive, as assumed in Fig. 7.56, points X and Y are plotted, respectivel, below and above the horiontal ais in Fig ut, in the absence of an overall rigid-bod rotation, the side of the element in Fig that is associated with P is observed to rotate counterclockwise, while the side associated with P is observed to rotate clockwise. Thus, if the shear deformation causes a given side to rotate clockwise, the corresponding point on Mohr s circle for plane strain is plotted above the horiontal ais, and if the deformation causes the side to rotate counterclockwise, the corresponding point is plotted below the horiontal ais. We note that this convention matches the convention used to draw Mohr s circle for plane stress. Points and where Mohr s circle intersects the horiontal ais correspond to the principal strains P ma and P min (Fig. 7.6a). We find P ma 5 P ave R and P min 5 P ave R (7.5) where P ave and R are defined b Eqs. (7.50). The corresponding value u p of the angle u is obtained b observing that the shearing strain is ero for and. Setting g in Eq. (7.49), we have g tan u p 5 P P (7.5) The corresponding aes a and b in Fig. 7.6b are the principal aes of strain. The angle u p, which defines the direction of the principal ais a in Fig. 7.6b corresponding to point in Fig. 7.6a, is equal to half of the angle X measured on Mohr s circle, and the rotation that brings into a has the same sense as the rotation that brings the diameter XY of Mohr s circle into the diameter. We recall from Sec..4 that, in the case of the elastic deformation of a homogeneous, isotropic material, Hooke s law for shearing stress and strain applies and ields t 5 Gg for an pair of rectangular and aes. Thus, g 5 0 when t 5 0, which indicates that the principal aes of strain coincide with the principal aes of stress. Y (, ) 7. Mohr s ircle for Plane Strain 489 X (, ) Fig Mohr s circle for plane strain. b min p Y ave ma s s ( ma ) p D E (a) (b) p X s ( min ) ma (in plane) Fig. 7.6 Principal strain determination. a

55 490 Transformations of Stress and Strain The maimum in-plane shearing strain is defined b points D and E in Fig. 7.6a. It is equal to the diameter of Mohr s circle. Recalling the second of Eqs. (7.50), we write g ma in plane 5 R 5 P P g (7.53) Finall, we note that the points X9 and Y9 that define the components of strain corresponding to a rotation of the coordinate aes through an angle u (Fig. 7.57) are obtained b rotating the diameter XY of Mohr s circle in the same sense through an angle u (Fig. 7.6). '' ' Q s s ' ' s ( ' ) '' Q s ( ' ) ' Y Y' X X' Fig (repeated) Fig. 7.6 EXMPLE mm 0 mm 0 mm 4m rad Fig () In a material in a state of plane strain, it is known that the horiontal side of a mm square elongates b 4 mm, while its vertical side remains unchanged, and that the angle at the lower left corner increases b rad (Fig. 7.63). Determine (a) the principal aes and principal strains, (b) the maimum shearing strain and the corresponding normal strain. (a) Principal es and Principal Strains. We first determine the coordinates of points X and Y on Mohr s circle for strain. We have P m m 5400 m P g 5 0 ` ` 5 00 m Since the side of the square associated with P rotates clockwise, point X of coordinates P and g is plotted above the horiontal ais. Since P 5 0 and the corresponding side rotates counterclockwise, point Y is plotted directl below the origin (Fig. 7.64). Drawing the diameter XY, we determine the center of Mohr s circle and its radius R. We have D X(400, 00) 5 P P 5 00 m Y 5 00 m R 5 Y 5 00 m 00 m 5 83 m Y(0, 00) p () The principal strains are defined b the abscissas of points and. We write P a 5 5 R 5 00 m 83 m m P b 5 5 R 5 00 m 83 m 583 m Fig E The principal aes a and b are shown in Fig Since 5 Y, the angle at in triangle Y is 458. Thus, the angle u p that brings XY into is 458i and the angle u p bringing into a is.58i.

56 (b) Maimum Shearing Strain. Points D and E define the maimum in-plane shearing strain which, since the principal strains have opposite signs, is also the actual maimum shearing strain (see Sec. 7.). We have g ma 5 R 5 83 m g ma m The corresponding normal strains are both equal to P m The aes of maimum shearing strain are shown in Fig Three-Dimensional nalsis of Strain 49 b e Fig p.5 a Fig d *7. THREE-DIMENSINL NLYSIS F STRIN b We saw in Sec. 7.5 that, in the most general case of stress, we can determine three coordinate aes a, b, and c, called the principal aes of stress. small cubic element with faces respectivel perpendicular to these aes is free of shearing stresses (Fig. 7.5); i.e., we have t ab 5 t bc 5 t ca 5 0. s recalled in the preceding section, Hooke s law for shearing stress and strain applies when the deformation is elastic and the material homogeneous and isotropic. It follows that, in such a case, g ab 5 g bc 5 g ca 5 0, i.e., the aes a, b, and c are also principal aes of strain. small cube of side equal to unit, centered at Q and with faces respectivel perpendicular to the principal aes, is deformed into a rectangular parallelepiped of sides P a, P b, and P c (Fig. 7.67). b Q a c b c Fig. 7.5 (repeated) c a a b b a Q c c a Fig Principal strains.

57 49 Transformations of Stress and Strain b b b a a Q Q c c c a c Fig (repeated) Fig If the element of Fig is rotated about one of the principal aes at Q, sa the c ais (Fig. 7.68), the method of analsis developed earlier for the transformation of plane strain can be used to determine the strain components P, P, and g associated with the faces perpendicular to the c ais, since the derivation of this method did not involve an of the other strain components. We can, therefore, draw Mohr s circle through the points and corresponding to the principal aes a and b (Fig. 7.69). Similarl, circles of diameters and can be used to anale the transformation of strain as the element is rotated about the a and b aes, respectivel. ma min ma Fig Mohr s circle for threedimensional analsis of strain. The three-dimensional analsis of strain b means of Mohr s circle is limited here to rotations about principal aes (as was the case for the analsis of stress) and is used to determine the maimum shearing strain g ma at point Q. Since g ma is equal to the diameter of the largest of the three circles shown in Fig. 7.69, we have g ma 5 0P ma P min 0 (7.54) where P ma and P min represent the algebraic values of the maimum and minimum strains at point Q. We note that the other four faces of the element remain rectangular and that the edges parallel to the c ais remain unchanged.

58 Returning to the particular case of plane strain, and selecting the and aes in the plane of strain, we have P 5 g 5 g 5 0. Thus, the ais is one of the three principal aes at Q, and the corresponding point in the Mohr-circle diagram is the origin, where P 5 g 5 0. If the points and that define the principal aes within the plane of strain fall on opposite sides of (Fig. 7.70a), the corresponding principal strains represent the maimum and minimum normal strains at point Q, and the maimum shearing strain is equal to the maimum in-plane shearing strain corresponding to points D and E. If, on the other hand, and are on the same side of (Fig. 7.70b), that is, if P a and P b have the same sign, then the maimum shearing strain is defined b points D9 and E9 on the circle of diameter, and we have g ma 5 P ma. We now consider the particular case of plane stress encountered in a thin plate or on the free surface of a structural element or machine component (Sec. 7.). Selecting the and aes in the plane of stress, we have s 5 t 5 t 5 0 and verif that the ais is a principal ais of stress. s we saw earlier, if the deformation is elastic and if the material is homogeneous and isotropic, it follows from Hooke s law that g 5 g 5 0; thus, the ais is also a principal ais of strain, and Mohr s circle can be used to anale the transformation of strain in the plane. However, as we shall see presentl, it does not follow from Hooke s law that P 5 0; indeed, a state of plane stress does not, in general, result in a state of plane strain. Denoting b a and b the principal aes within the plane of stress, and b c the principal ais perpendicular to that plane, we let s 5 s a, s 5 s b, and s 5 0 in Eqs. (.8) for the generalied Hooke s law (Sec..) and write Z 7. Three-Dimensional nalsis of Strain min min 0 D Z E ma D' E' ma E D a ma (a) ma (b) Fig Mohr s circle for plane strain. 493 P a 5 s a E ns b E P b 5 ns a E s b E (7.55) (7.56) P c 5 n E s a s b (7.57) dding Eqs. (7.55) and (7.56) member to member, we have P a P b 5 n E s a s b (7.58) Solving Eq. (7.58) for s a s b and substituting into Eq. (7.57), we write P c 5 n n P a P b (7.59) The relation obtained defines the third principal strain in terms of the in-plane principal strains. We note that, if is located between and on the Mohr-circle diagram (Fig. 7.7), the maimum shearing strain is equal to the diameter of the circle corresponding to a rotation about the b ais, out of the plane of stress. See footnote on page 486. D' E' D E ma Fig. 7.7 Mohr s circle strain analsis for plane stress.

59 EXMPLE 7.05 s a result of measurements made on the surface of a machine component with strain gages oriented in various was, it has been established that the principal strains on the free surface are P a in./in. and P b in./in. Knowing that Poisson s ratio for the given material is n , determine (a) the maimum in-plane shearing strain, (b) the true value of the maimum shearing strain near the surface of the component. (a) Maimum In-Plane Shearing Strain. We draw Mohr s circle through the points and corresponding to the given principal strains (Fig. 7.7). The maimum in-plane shearing strain is defined b points D and E and is equal to the diameter of Mohr s circle: g ma in plane rad (b) Maimum Shearing Strain. We first determine the third principal strain P c. Since we have a state of plane stress on the surface of the machine component, we use Eq. (7.59) and write P c 5 n n P a P b in./in. Drawing Mohr s circles through and and through and (Fig. 7.73), we find that the maimum shearing strain is equal to the diameter of the circle of diameter : g ma rad We note that, even though P a and P b have opposite signs, the maimum in-plane shearing strain does not represent the true maimum shearing strain. (0 6 rad) D (0 6 rad) D' 50 ma (in plane) (0 6 in./in.) ma 400 (0 6 in./in.) E 450 E' 550 Fig. 7.7 Fig *7.3 MESUREMENTS F STRIN; STRIN RSETTE The normal strain can be determined in an given direction on the surface of a structural element or machine component b scribing two gage marks and across a line drawn in the desired direction and measuring the length of the segment before and after the

60 load has been applied. If L is the undeformed length of and d its deformation, the normal strain along is P 5 dl. more convenient and more accurate method for the measurement of normal strains is provided b electrical strain gages. tpical electrical strain gage consists of a length of thin wire arranged as shown in Fig and cemented to two pieces of paper. In order to measure the strain P of a given material in the direction, the gage is cemented to the surface of the material, with the wire folds running parallel to. s the material elongates, the wire increases in length and decreases in diameter, causing the electrical resistance of the gage to increase. measuring the current passing through a properl calibrated gage, the strain P can be determined accuratel and continuousl as the load is increased. The strain components P and P can be determined at a given point of the free surface of a material b simpl measuring the normal strain along and aes drawn through that point. Recalling Eq. (7.43) of Sec. 7.0, we note that a third measurement of normal strain, made along the bisector of the angle formed b the and aes, enables us to determine the shearing strain g as well (Fig. 7.75): g 5 P P P (7.43) It should be noted that the strain components P, P, and g at a given point could be obtained from normal strain measurements made along an three lines drawn through that point (Fig. 7.76). Denoting respectivel b u, u, and u 3 the angle each of the three lines forms with the ais, b P, P, and P 3 the corresponding strain measurements, and substituting into Eq. (7.4), we write the three equations P 5 P cos u P sin u g sin u cos u P 5 P cos u P sin u g sin u cos u (7.60) P 3 5 P cos u 3 P sin u 3 g sin u 3 cos u 3 which can be solved simultaneousl for P, P, and g. The arrangement of strain gages used to measure the three normal strains P, P, and P 3 is known as a strain rosette. The rosette used to measure normal strains along the and aes and their bisector is referred to as a 458 rosette (Fig. 7.75). nother rosette frequentl used is the 608 rosette (see Sample Prob. 7.7). 7.3 Measurements of Strain; Strain Rosette Fig Electrical strain gage Fig L L 3 3 L 3 Fig Strain rosette. It should be noted that the free surface on which the strain measurements are made is in a state of plane stress, while Eqs. (7.4) and (7.43) were derived for a state of plane strain. However, as observed earlier, the normal to the free surface is a principal ais of strain and the derivations given in Sec. 7.0 remain valid.

61 SMPLE PRLEM in. clindrical storage tank used to transport gas under pressure has an inner diameter of 4 in. and a wall thickness of 3 4 in. Strain gages attached to the surface of the tank in transverse and longitudinal directions indicate strains of and in./in. respectivel. Knowing that a torsion test has shown that the modulus of rigidit of the material used in the tank is G psi, determine (a) the gage pressure inside the tank, (b) the principal stresses and the maimum shearing stress in the wall of the tank. SLUTIN a. Gage Pressure Inside Tank. We note that the given strains are the principal strains at the surface of the tank. Plotting the corresponding points and, we draw Mohr s circle for strain. The maimum in-plane shearing strain is equal to the diameter of the circle. (0 6 rad) 60 D E 55 ma (in plane) (0 6 in./in.) g ma in plane 5 P P rad From Hooke s law for shearing stress and strain, we have t ma in plane 5 Gg ma in plane psi rad 5 84 psi 5.84 ksi Substituting this value and the given data in Eq. (7.33), we write t ma in plane 5 pr 4t p in. 84 psi in. ma D' ma (in plane).84 ksi D E Solving for the gage pressure p, we have p psi b. Principal Stresses and Maimum Shearing Stress. Recalling that, for a thin-walled clindrical pressure vessel, s 5 s, we draw Mohr s circle for stress and obtain s 5 t ma in plane 5.84 ksi ksi s ksi s 5 s ksi s ksi The maimum shearing stress is equal to the radius of the circle of diameter and corresponds to a rotation of 458 about a longitudinal ais. t ma 5 s 5 s ksi t ma ksi 496

62 SMPLE PRLEM Q Using a 608 rosette, the following strains have been determined at point Q on the surface of a steel machine base: P 5 40 m P m P m Using the coordinate aes shown, determine at point Q, (a) the strain components P, P, and g, (b) the principal strains, (c) the maimum shearing strain. (Use n ) F b X 368 p R b a D' a 006 Y. a ma SLUTIN a. Strain omponents e, e, G. For the coordinate aes shown u 5 0 u 5 60 u Substituting these values into Eqs. (7.60), we have P 5 P P 0 g 0 P 5 P P g P 3 5 P P g Solving these equations for P, P, and g, we obtain P 5 P P 5 3 P P 3 P g 5 P P Substituting the given values for P, P, and P 3, we have P 5 40 m P P 5860 m g g m These strains are indicated on the element shown. b. Principal Strains. We note that the side of the element associated with P rotates counterclockwise; thus, we plot point X below the horiontal ais, i.e., X(40, 375). We then plot Y(860, 375) and draw Mohr s circle. P ave m 40 m m R m 40 m m tan u p m u p i u p 5. i 40 m Points and correspond to the principal strains. We have P a 5 P ave R m 556 m P a 506 m P b 5 P ave R m 556 m P b 5006 m Since s 5 0 on the surface, we use Eq. (7.59) to find the principal strain P c : P c 5 n n P a P b m 006 m P c 5368 m c. Maimum Shearing Strain. Plotting point and drawing Mohr s circle through points and, we obtain point D9 and write g ma m 368 m g ma m 497

63 PRLEMS ' Fig. P7.8 through P7.35 ' 7.8 through 7.3 For the given state of plane strain, use the method of Sec. 7.0 to determine the state of plane strain associated with aes 9 and 9 rotated through the given angle u. P P g u 7.8 and m 50m 0 58 l 7.9 and m 60m 50m 608 i 7.30 and m 450m 00m 58 i 7.3 and m 00m 308 l 7.3 through 7.35 For the given state of plane strain, use Mohr s circle to determine the state of plane strain associated with aes 9 and 9 rotated through the given angle u through 7.39 The following state of strain has been measured on the surface of a thin plate. Knowing that the surface of the plate is unstressed, determine (a) the direction and magnitude of the principal strains, (b) the maimum in-plane shearing strain, (c) the maimum shearing strain. (Use n 5 3) P P g m 60m 480m m 400m 350m m 480m 600m m 570m 70m 7.40 through 7.43 For the given state of plane strain, use Mohr s circle to determine(a) the orientation and magnitude of the principal strains, (b) the maimum in-plane strain, (c) the maimum shearing strain P P g m 40m 50m m 00m 375m m 60m 00m m 60m 35m Determine the strain P knowing that the following strains have been determined b use of the rosette shown: 498 Fig. P7.44 P 5480m P 50m P 3 580m

64 7.45 The strains determined b the use of the rosette shown during the test of a machine element are P 5600m P 5450m P 3 575m Problems 499 Determine (a) the in-plane principal strains, (b) the in-plane maimum shearing strain The rosette shown has been used to determine the following strains at a point on the surface of a crane hook: P in./in. P in./in. P in./in. Fig. P7.45 (a) What should be the reading of gage 3? (b) Determine the principal strains and the maimum in-plane shearing strain Fig. P The strains determined b the use of the rosette attached as shown during the test of a machine element are P in./in. P in./in. P in./in. Determine (a) the orientation and magnitude of the principal strains in the plane of the rosette, (b) the maimum in-plane shearing strain Fig. P Using a 458 rosette, the strains P, P, and P 3 have been determined at a given point. Using Mohr s circle, show that the principal strains are: P ma, min 5 P P 3 6 cp P P P 3 d / (Hint: The shaded triangles are congruent.) min ma Fig. P7.48

65 500 Transformations of Stress and Strain 7.49 Show that the sum of the three strain measurements made with a 608 rosette is independent of the orientation of the rosette and equal to P P P 3 5 3P avg where P avg is the abscissa of the center of the corresponding Mohr s circle Fig. P single strain gage is cemented to a solid 4-in.-diameter steel shaft at an angle b 5 58 with a line parallel to the ais of the shaft. Knowing that G psi, determine the torque T indicated b a gage reading of in./in. T' T in. Fig. P Solve Prob. 7.50, assuming that the gage forms an angle b with a line parallel to the ais of the shaft. 7.5 single strain gage forming an angle b 5 88 with a horiontal plane is used to determine the gage pressure in the clindrical steel tank shown. The clindrical wall of the tank is 6-mm thick, has a 600-mm inside diameter, and is made of a steel with E 5 00 GPa and n Determine the pressure in the tank indicated b a strain gage reading of 80m. Fig. P Solve Prob. 7.5, assuming that the gage forms an angle b with a horiontal plane.

66 7.54 The given state of plane stress is known to eist on the surface of a machine component. Knowing that E 5 00 GPa and G GPa, determine the direction and magnitude of the three principal strains (a) b determining the corresponding state of strain [use Eq. (.43) and Eq. (.38)] and then using Mohr s circle for strain, (b) b using Mohr s circle for stress to determine the principal planes and principal stresses and then determining the corresponding strains. 50 MPa Problems The following state of strain has been determined on the surface of a cast-iron machine part: 75 MPa P 570m P 5400m g 5660m Knowing that E 5 69 GPa and G 5 8 GPa, determine the principal planes and principal stresses (a) b determining the corresponding state of plane stress [use Eq. (.36), Eq. (.43), and the first two equations of Prob..7] and then using Mohr s circle for stress, (b) b using Mohr s circle for strain to determine the orientation and magnitude of the principal strains and then determine the corresponding stresses. Fig. P centric aial force P and a horiontal force Q are both applied at point of the rectangular bar shown. 458 strain rosette on the surface of the bar at point indicates the following strains: P in./in. P in./in. P in./in. Knowing that E psi and n , determine the magnitudes of P and Q. in. P Q in. 3 3 in in. Fig. P Solve Prob. 7.56, assuming that the rosette at point indicates the following strains: P in./in. P in./in. P in./in.

67 REVIEW ND SUMMRY The first part of this chapter was devoted to a stud of the transformation of stress under a rotation of aes and to its application to the solution of engineering problems, and the second part to a similar stud of the transformation of strain. ' ' '' ' Q Q ' ' Fig (a) (b) Transformation of plane stress ' min p ma ' ma Q p min Fig Principal planes. Principal stresses 50 onsidering first a state of plane stress at a given point Q [Sec. 7.] and denoting b s, s, and t the stress components associated with the element shown in Fig. 7.77a, we derived the following formulas defining the components s 9, s 9, and t 99 associated with that element after it had been rotated through an angle u about the ais (Fig. 7.77b): s 5 s s s 5 s s t 5 s s s s s s cos u t sin u (7.5) cos u t sin u (7.7) sin u t cos u (7.6) In Sec. 7.3, we determined the values u p of the angle of rotation which correspond to the maimum and minimum values of the normal stress at point Q. We wrote t tan u p 5 s s (7.) The two values obtained for u p are 908 apart (Fig. 7.78) and define the principal planes of stress at point Q. The corresponding values

68 of the normal stress are called the principal stresses at Q; we obtained s ma, min 5 s s 6 a s s b t (7.4) We also noted that the corresponding value of the shearing stress is ero. Net, we determined the values u s of the angle u for which the largest value of the shearing stress occurs. We wrote tan u s 5 s s (7.5) t The two values obtained for u s are 908 apart (Fig. 7.79). We also noted that the planes of maimum shearing stress are at 458 to the principal planes. The maimum value of the shearing stress for a rotation in the plane of stress is t ma 5 a s s b t (7.6) and the corresponding value of the normal stresses is s 5 s ave 5 s s (7.7) We saw in Sec. 7.4 that Mohr s circle provides an alternative method, based on simple geometric considerations, for the analsis of the Review and Summar ' ' ' s ma Q ma s ' ' ' Fig Maimum in-plane shearing stress Mohr s circle for stress b ma min ma ma p min a min Y (, ) p X (, ) Fig (a) (b) ( ) transformation of plane stress. Given the state of stress shown in black in Fig. 7.80a, we plot point X of coordinates s, t and point Y of coordinates s, t (Fig. 7.80b). Drawing the circle of diameter XY, we obtain Mohr s circle. The abscissas of the points of intersection and of the circle with the horiontal ais represent the principal stresses, and the angle of rotation bringing the diameter XY into is twice the angle u p defining the principal planes in Fig. 7.80a, with both angles having the same sense. We also noted that diameter DE defines the maimum shearing stress and the orientation of the corresponding plane (Fig. 7.8) [Eample 7.0, Sample Probs. 7. and 7.3]. ' ave D ma 90 E Fig. 7.8

69 504 Transformations of Stress and Strain onsidering a general state of stress characteried b si stress components [Sec. 7.5], we showed that the normal stress on a plane of General state of stress arbitrar orientation can be epressed as a quadratic form of the direction cosines of the normal to that plane. This proves the eistence of three principal aes of stress and three principal stresses at an given point. Rotating a small cubic element about each of the three principal aes [Sec. 7.6], we drew the corresponding Mohr s circles that ield the values of s ma, s min, and t ma (Fig. 7.8). In the particular case of ma plane stress, and if the and aes are selected in the plane of stress, point coincides with the origin. If and are located on opposite sides of, the maimum shearing stress is equal to the maimum in-plane shearing stress as determined in Secs. 7.3 or 7.4. If and are located on the same side of, this will not be the case. If s a. min s b. 0, for instance the maimum shearing stress is equal to s a and corresponds to a rotation out of the plane of stress (Fig. 7.83). Fig. 7.8 ma Z D' D ma a E' min 0 ma a Fig Yield criteria for ductile materials Yield criteria for ductile materials under plane stress were developed in Sec To predict whether a structural or machine component will fail at some critical point due to ield in the material, we first determine the principal stresses s a and s b at that point for the given loading condition. We then plot the point of coordinates s a and s b. If this point falls within a certain area, the component is safe; if it falls outside, the component will fail. The area used with the maimum-shearing-strength criterion is shown in Fig and the area used with the maimumdistortion-energ criterion in Fig We note that both areas depend upon the value of the ield strength s Y of the material. b b Y Y Y Y a Y Y a D Y Y Fig Fig. 7.85

70 Fracture criteria for brittle materials under plane stress were developed in Sec. 7.8 in a similar fashion. The most commonl used is Mohr s criterion, which utilies the results of various tpes of test available for a given material. The shaded area shown in Fig is used when the ultimate strengths s UT and s U have been determined, respectivel, from a tension and a compression test. gain, the principal stresses s a and s b are determined at a given point of the structural or machine component being investigated. If the corresponding point falls within the shaded area, the component is safe; if it falls outside, the component will rupture. Review and Summar 505 Fracture criteria for brittle materials b UT U UT a U Fig In Sec. 7.9, we discussed the stresses in thin-walled pressure vessels and derived formulas relating the stresses in the walls of the vessels and the gage pressure p in the fluid the contain. In the case of a clindrical vessel of inside radius r and thickness t (Fig. 7.87), we obtained the following epressions for the hoop stress s and the longitudinal stress s : s 5 pr t s 5 pr t (7.30, 7.3) We also found that the maimum shearing stress occurs out of the plane of stress and is t ma 5 s 5 pr t (7.34) lindrical pressure vessels Fig r t In the case of a spherical vessel of inside radius r and thickness t (Fig. 7.88), we found that the two principal stresses are equal: Spherical pressure vessels s 5 s 5 pr t (7.36) gain, the maimum shearing stress occurs out of the plane of stress; it is t ma 5 s 5 pr 4t (7.37) Fig. 7.88

71 506 b Transformations of Stress and Strain min p Fig Transformation of plane strain Y ave Mohr s circle for strain ma s s ( ma ) p D E (a) (b) p X s ( min ) ma (in plane) Strain gages. Strain rosette L a The last part of the chapter was devoted to the transformation of strain. In Secs. 7.0 and 7., we discussed the transformation of plane strain and introduced Mohr s circle for plane strain. The discussion was similar to the corresponding discussion of the transformation of stress, ecept that, where the shearing stress t was used, we now used g, that is, half the shearing strain. The formulas obtained for the transformation of strain under a rotation of aes through an angle u were P 5 P P P P cos u g sin u (7.44) P 5 P P P P cos u g sin u (7.45) g 5P P sin u g cos u (7.49) Using Mohr s circle for strain (Fig. 7.89), we also obtained the following relations defining the angle of rotation u p corresponding to the principal aes of strain and the values of the principal strains P ma and P min : where P ave 5 P P g tan u p 5 (7.5) P P P ma 5 P ave R and P min 5 P ave R (7.5) and R 5 a P P b a g b (7.50) The maimum shearing strain for a rotation in the plane of strain was found to be g ma in plane 5 R 5 P P g (7.53) Section 7. was devoted to the three-dimensional analsis of strain, with application to the determination of the maimum shearing strain in the particular cases of plane strain and plane stress. In the case of plane stress, we also found that the principal strain P c in a direction perpendicular to the plane of stress could be epressed as follows in terms of the in-plane principal strains P a and P b : P c 5 n n P a P b (7.59) Finall, we discussed in Sec. 7.3 the use of strain gages to measure the normal strain on the surface of a structural element or machine component. onsidering a strain rosette consisting of three gages aligned along lines forming respectivel, angles u, u, and u 3 with the ais (Fig. 7.90), we wrote the following relations among the measurements P, P, P 3 of the gages and the components P, P, g characteriing the state of strain at that point: L 3 3 Fig L P 5 P cos u P sin u g sin u cos u P 5 P cos u P sin u g sin u cos u (7.60) P 3 5 P cos u 3 P sin u 3 g sin u 3 cos u 3 These equations can be solved for P, P, and g, once P, P, and P 3 have been determined.

72 REVIEW PRLEMS 7.58 Two wooden members of mm uniform rectangular cross section are joined b the simple glued scarf splice shown. Knowing that b 5 8 and that the maimum allowable stresses in the joint are, respectivel, 400 kpa in tension (perpendicular to the splice) and 600 kpa in shear (parallel to the splice), determine the largest centric load P that can be applied. P' 0 mm 80 mm P Fig. P7.58 and P Two wooden members of mm uniform rectangular cross section are joined b the simple glued scarf splice shown. Knowing that b 5 58 and that centric loads of magnitude P 5 0 kn are applied to the members as shown, determine (a) the in-plane shearing stress parallel to the splice, (b) the normal stress perpendicular to the splice The centric force P is applied to a short post as shown. Knowing that the stresses on plane a-a are s 5 5 ksi and t 5 5 ksi, determine (a) the angle b that plane a-a forms with the horiontal, (b) the maimum compressive stress in the post. 7.6 Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown. P a Fig. P7.60 a Fig. P

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