Strain Transformation 10 CHAPTER OBJECTIVES The transformation of strain at a point is similar to the transformation of stress, and as a result the methods of Chapter 9 will be applied in this chapter. Here we will also discuss various ways for measuring strain and develop some important material-property relationships, including a generalized form of Hooke’s law. At the end of the chapter, a few of the theories used to predict the failure of a material will be discussed. 10.1 Plane Strain As outlined in Sec. 2.2, the general state of strain at a point in a body is represented by a combination of three components of normal strain, Px , Py , Pz , and three components of shear strain gxy , gxz , gyz . These six components tend to deform each face of an element of the material, and like stress, the normal and shear strain components at the point will vary according to the orientation of the element. The strains at a point are often determined by using strain gauges, which measure normal strain in specified directions. For both analysis and design, however, engineers must sometimes transform this data in order to obtain the strain in other directions. 485 486 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N To understand how this is done, we will first confine our attention to a study of plane strain. Specifically, we will not consider the effects of the components Pz , gxz , and gyz . In general, then, a plane-strained element is subjected to two components of normal strain, Px , Py , and one component of shear strain, gxy . Although plane strain and plane stress each have three components lying in the same plane, realize that plane stress does not necessarily cause plane strain or vice versa. The reason for this has to do with the Poisson effect discussed in Sec. 3.6. For example, if the element in Fig. 10–1 is subjected to plane stress sx and sy , not only are normal strains Px and Py produced, but there is also an associated normal strain, Pz . This is obviously not a case of plane strain. In general, then, unless n = 0, the Poisson effect will prevent the simultaneous occurrence of plane strain and plane stress. z Py dy Px dx Pz dz sx x sy y Plane stress, sx , sy, does not cause plane strain in the x–y plane since Pz ≠ 0. 10.2 General Equations of Plane-Strain Fig. 10–1 Transformation y Py dy A dy It is important in plane-strain analysis to establish transformation equations that can be used to determine the x¿, y¿ components of normal and shear strain at a point, provided the x, y components of strain are known. Essentially this problem is one of geometry and requires relating the deformations and rotations of line segments, which represent the sides of differential elements that are parallel to each set of axes. gxy 2 gxy 2 B x O dx y¿ (a) Px dx y x¿ u x 10 (b) Positive sign convention Fig. 10–2 Sign Convention. Before the strain-transformation equations can be developed, we must first establish a sign convention for the strains. With reference to the differential element shown in Fig. 10–2a, normal strains Px and Py are positive if they cause elongation along the x and y axes, respectively, and the shear strain gxy is positive if the interior angle AOB becomes smaller than 90°. This sign convention also follows the corresponding one used for plane stress, Fig. 9–5a, that is, positive sx , sy , txy will cause the element to deform in the positive Px , Py , gxy directions, respectively. The problem here will be to determine at a point the normal and shear strains Px¿ , Py¿ , gx¿y¿ , measured relative to the x¿, y¿ axes, if we know Px , Py , gxy , measured relative to the x, y axes. If the angle between the x and x¿ axes is u, then, like the case of plane stress, u will be positive provided it follows the curl of the right-hand fingers, i.e., counterclockwise, as shown in Fig. 10–2b. 10.2 487 GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION Normal and Shear Strains. In order to develop the strain- y y¿ transformation equation for Px¿ , we must determine the elongation of a line segment dx¿ that lies along the x¿ axis and is subjected to strain components Px , Py , gxy . As shown in Fig. 10–3a, the components of the line dx¿ along the x and y axes are dx = dx¿ cos u dy = dx¿ sin u x¿ dy u dx ¿ x dx (10–1) Before deformation (a) When the positive normal strain Px occurs, the line dx is elongated Px dx, Fig. 10–3b, which causes line dx¿ to elongate Px dx cos u. Likewise, when Py occurs, line dy elongates Py dy, Fig. 10–3c, which causes line dx¿ to elongate Py dy sin u. Finally, assuming that dx remains fixed in position, the shear strain gxy , which is the change in angle between dx and dy, causes the top of line dy to be displaced gxy dy to the right, as shown in Fig. 10–3d. This causes dx¿ to elongate gxy dy cos u. If all three of these elongations are added together, the resultant elongation of dx¿ is then y y¿ x¿ Px dx cos u dx¿ dx Normal strain Px u x Px dx Px dx sinu (b) y dx¿ = Px dx cos u + Py dy sin u + gxy dy cos u From Eq. 2–2, the normal strain along the line dx¿ is Px¿ = dx¿>dx¿. Using Eq. 10–1, we therefore have Px¿ = Px cos2 u + Py sin2 u + gxy sin u cos u y¿ u Py dy (10–2) Py dy cos u x¿ u Py dy sin u dy dx¿ x Normal strain Py (c) y y¿ gxy dy sinu g dy xy x¿ gxy dy cos u u dy¿ dy 10 gxy dx ¿ dx The rubber specimen is constrained between the two fixed supports, and so it will undergo plane strain when loads are applied to it in the horizontal plane. Shear strain gxy (d) Fig. 10–3 x 488 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N y y¿ y gxy dy sinu g dy xy x¿ gxy dy cos u y¿ u dy¿ dy gxy dy ¿ dx ¿ dx b x dy ¿ dx ¿ dy ¿ x¿ a dx ¿ u Shear strain gxy x (e) (d) Fig. 10–3 (cont.) The strain-transformation equation for gx¿y¿ can be developed by considering the amount of rotation each of the line segments dx¿ and dy¿ undergo when subjected to the strain components Px , Py , gxy . First we will consider the rotation of dx¿, which is defined by the counterclockwise angle a shown in Fig. 10–3e. It can be determined by the displacement caused by dy¿ using a = dy¿>dx¿. To obtain dy¿, consider the following three displacement components acting in the y¿ direction: one from Px , giving -Px dx sin u, Fig. 10–3b; another from Py , giving Py dy cos u, Fig. 10–3c; and the last from gxy , giving - gxy dy sin u, Fig. 10–3d. Thus, dy¿, as caused by all three strain components, is dy¿ = -Px dx sin u + Py dy cos u - gxy dy sin u Dividing each term by dx¿ and using Eq. 10–1, with a = dy¿>dx¿, we have a = 1-Px + Py2 sin u cos u - gxy sin2 u (10–3) As shown in Fig. 10–3e, the line dy¿ rotates by an amount b. We can determine this angle by a similar analysis, or by simply substituting u + 90° for u into Eq. 10–3. Using the identities sin1u + 90°2 = cos u, cos1u + 90°2 = - sin u, we have b = 1 -Px + Py2 sin1u + 90°2 cos1u + 90°2 - gxy sin21u + 90°2 10 = - 1-Px + Py2 cos u sin u - gxy cos2 u Since a and b represent the rotation of the sides dx¿ and dy¿ of a differential element whose sides were originally oriented along the x¿ and y¿ axes, Fig. 10–3e, the element is then subjected to a shear strain of gx¿y¿ = a - b = - 21Px - Py2 sin u cos u + gxy1cos2 u - sin2 u2 (10–4) 10.2 GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION 489 y y y¿ y¿ x¿ dy ¿ x¿ u u dy¿ dx¿ dx ¿ x Positive normal strain, Px ¿ Positive shear strain, gx ¿y¿ (a) (b) x Fig. 10–4 Using the trigonometric identities sin 2u = 2 sin u cos u, cos2 u = (1 + cos 2u)>2, and sin2 u + cos2 u = 1, we can rewrite Eqs. 10–2 and 10–4 in the final form Px¿ = Px + Py 2 gx¿y¿ 2 = -¢ + Px - Py 2 Px - Py 2 cos 2u + ≤ sin 2u + gxy 2 gxy 2 sin 2u cos 2u (10–5) (10–6) These strain-transformation equations give the normal strain Px¿ in the x¿ direction and the shear strain gx¿y¿ of an element oriented at an angle u, as shown in Fig. 10–4. According to the established sign convention, if Px¿ is positive, the element elongates in the positive x¿ direction, Fig. 10–4a, and if gx¿y¿ is positive, the element deforms as shown in Fig. 10–4b. If the normal strain in the y¿ direction is required, it can be obtained from Eq. 10–5 by simply substituting 1u + 90°2 for u. The result is 10 Py¿ = Px + Py 2 - Px - Py 2 cos 2u - gxy 2 sin 2u (10–7) The similarity between the above three equations and those for planestress transformation, Eqs. 9–1, 9–2, and 9–3, should be noted. By comparison, sx , sy , sx¿ , sy¿ correspond to Px , Py , Px¿ , Py¿ ; and txy , tx¿y¿ correspond to gxy>2, gx¿y¿>2. 490 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N Principal Strains. Like stress, an element can be oriented at a point so that the element’s deformation is caused only by normal strains, with no shear strain. When this occurs the normal strains are referred to as principal strains, and if the material is isotropic, the axes along which these strains occur will coincide with the axes that define the planes of principal stress. From Eqs. 9–4 and 9–5, and the correspondence between stress and strain mentioned above, the direction of the x¿ axis and the two values of the principal strains P1 and P2 are determined from gxy tan 2up = P1,2 = Complex stresses are often developed at the joints where the cylindrical and hemispherical vessels are joined together. The stresses are determined by making measurements of strain. Px + Py 2 ; C¢ (10–8) Px - Py Px - Py 2 2 ≤ + ¢ gxy 2 2 ≤ (10–9) Maximum In-Plane Shear Strain. Using Eqs. 9–6, 9–7, and 9–8, the direction of the x¿ axis, and the maximum in-plane shear strain and associated average normal strain are determined from the following equations: tan 2us = - ¢ max g in-plane 2 = B a Px - Py gxy Px - Py Pavg = 2 2 ≤ b + a Px + Py 2 (10–10) gxy 2 b 2 (10–11) (10–12) Important Points • In the case of plane stress, plane-strain analysis may be used within the plane of the stresses to analyze the 10 data from strain gauges. Remember, though, there will be a normal strain that is perpendicular to the gauges due to the Poisson effect. • When the state of strain is represented by the principal strains, no shear strain will act on the element. • The state of strain at a point can also be represented in terms of the maximum in-plane shear strain. In this case an average normal strain will also act on the element. • The element representing the maximum in-plane shear strain and its associated average normal strains is 45° from the orientation of an element representing the principal strains. 10.2 491 GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION EXAMPLE 10.1 A differential element of material at a point is subjected to a state of plane strain Px = 500110-62, Py = - 300110-62, gxy = 200110-62, which tends to distort the element as shown in Fig. 10–5a. Determine the equivalent strains acting on an element of the material oriented at the point, clockwise 30° from the original position. SOLUTION The strain-transformation Eqs. 10–5 and 10–6 will be used to solve the problem. Since u is positive counterclockwise, then for this problem u = - 30°. Thus, Px¿ = Px + Py 2 = c 2 2 500 + 1-3002 + B gx¿y¿ Px - Py + 2 200110 2 -6 2 = -¢ = -c cos 2u + d110-62 + c gxy 2 y gxy 2 Py dy dy gxy 2 dx sin 2u 500 - 1-3002 2 x Px dx (a) y y¿ d110-62 cos121-30°22 R sin121-30°22 Px¿ = 213110-62 Px - Py 2 ≤ sin 2u + 500 - 1 -3002 2 gxy 2 Ans. u 60 cos 2u x d110 2 sin121-30°22 + -6 gx¿y¿ = 793110-62 200110-62 2 u 30 cos121 -30°22 (b) Ans. x¿ y¿ The strain in the y¿ direction can be obtained from Eq. 10–7 with u = - 30°. However, we can also obtain Py¿ using Eq. 10–5 with u = 60°1u = - 30° + 90°2, Fig. 10–5b. We have with Py¿ replacing Px¿ , Py¿ = Px + Py 2 = c + + Px - Py 2 500 + 1-3002 2 200110-62 2 cos 2u + d110-62 + c gxy 2 sin 2u 500 - 1- 3002 2 gx ¿y¿ 2 dy¿ Py¿dy¿ d110-62 cos12160°22 10 dx ¿ sin12160°22 Py¿ = - 13.4110-62 These results tend to distort the element as shown in Fig. 10–5c. Px¿dx ¿ Ans. (c) Fig. 10–5 gx ¿y¿ 2 x¿ 492 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N EXAMPLE 10.2 A differential element of material at a point is subjected to a state of plane strain defined by Px = - 350110-62, Py = 200110-62, gxy = 80110-62, which tends to distort the element as shown in Fig. 10–6a. Determine the principal strains at the point and the associated orientation of the element. y gxy 2 SOLUTION Orientation of the Element. From Eq. 10–8 we have Py dy tan 2up = dy Px - Py 80110-62 gxy 2 x Px dx dx = 1-350 - 2002110-62 Thus, 2up = - 8.28° and -8.28° + 180° = 171.72°, so that up = - 4.14° and 85.9° (a) y gxy Ans. Each of these angles is measured positive counterclockwise, from the x axis to the outward normals on each face of the element, Fig. 10–6b. y¿ Principal Strains. The principal strains are determined from Eq. 10–9. We have P1,2 = P1dy¿ 2 ; 85.9 4.14 (b) Fig. 10–6 Px - Py 2 1 -350 + 2002110 2 x 2 2 b + a ; B gxy 2 b 2 -350 - 200 2 80 2 b + a b R 110-62 B 2 2 = - 75.0110-62 ; 277.9110-62 P1 = 203110-62 x¿ P2dx ¿ B a -6 = 10 Px + Py a P2 = - 353110-62 Ans. We can determine which of these two strains deforms the element in the x¿ direction by applying Eq. 10–5 with u = - 4.14°. Thus, Px¿ = Px + Py 2 = a + + Px - Py 2 cos 2u + gxy 2 sin 2u -350 + 200 -350 - 200 b110-62 + a b110-62 cos 21 -4.14°2 2 2 80110-62 2 sin 21 - 4.14°2 Px¿ = - 353110-62 Hence Px¿ = P2 . When subjected to the principal strains, the element is distorted as shown in Fig. 10–6b. 10.2 493 GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION EXAMPLE 10.3 A differential element of material at a point is subjected to a state of plane strain defined by Px = - 350110-62, Py = 200110-62, gxy = 80110-62, which tends to distort the element as shown in Fig. 10–7a. Determine the maximum in-plane shear strain at the point and the associated orientation of the element. SOLUTION Orientation of the Element. From Eq. 10–10 we have tan 2us = - ¢ Px - Py gxy ≤ = - 1-350 - 2002110-62 y Py dy 80110 2 -6 gxy 2g A xy dy 2 Thus, 2us = 81.72° and 81.72° + 180° = 261.72°, so that B us = 40.9° and 131° x O dx Note that this orientation is 45° from that shown in Fig. 10–6b in Example 10.2 as expected. Px dx (a) y Maximum In-Plane Shear Strain. Applying Eq. 10–11 gives max g in-plane 2 = B = B a Px - Py 2 2 b + a gxy 2 b 2 y¿ -350 - 200 2 80 2 b + a b R 110-62 B 2 2 (gxy)max 2 (gxy)max 2 a max g in-plane = 556110-62 Pavgdy¿ Ans. dy ¿ dx ¿ Due to the square root, the proper sign of g max can be obtained by in-plane applying Eq. 10–6 with us = 40.9°. We have gx¿y¿ 2 = - Px - Py = -a 2 x¿ sin 2u + gxy 2 cos 2u Pavgdx ¿ 40.9 x (b) Fig. 10–7 80110-62 -350 - 200 b110-62 sin 2140.9°2 + cos 2140.9°2 2 2 gx¿y¿ = 556110-62 This result is positive and so g max tends to distort the element so in-plane that the right angle between dx¿ and dy¿ is decreased (positive sign convention), Fig. 10–7b. Also, there are associated average normal strains imposed on the element that are determined from Eq. 10–12: Px + Py - 350 + 200 Pavg = = 110-62 = - 75110-62 2 2 These strains tend to cause the element to contract, Fig. 10–7b. 10 494 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N *10.3 Mohr’s Circle—Plane Strain Since the equations of plane-strain transformation are mathematically similar to the equations of plane-stress transformation, we can also solve problems involving the transformation of strain using Mohr’s circle. Like the case for stress, the parameter u in Eqs. 10–5 and 10–6 can be eliminated and the result rewritten in the form 1Px¿ - Pavg22 + ¢ gx¿y¿ 2 2 ≤ = R2 (10–13) where Pavg = R = Px + Py 2 B a Px - Py 2 2 b + a gxy 2 b 2 Equation 10–13 represents the equation of Mohr’s circle for strain. It has a center on the P axis at point C1Pavg , 02 and a radius R. Procedure for Analysis The procedure for drawing Mohr’s circle for strain follows the same one established for stress. Construction of the Circle. Px Py 2 C gxy Pavg 10 R • Establish a coordinate system such that the abscissa represents 2 Px Py 2 Px g 2 P Px Py 2 2 Fig. 10–8 A u 0 gxy 2 2 • the normal strain P, with positive to the right, and the ordinate represents half the value of the shear strain, g>2, with positive downward, Fig. 10–8. Using the positive sign convention for Px , Py , gxy , as shown in Fig. 10–2, determine the center of the circle C, which is located on the P axis at a distance Pavg = 1Px + Py2>2 from the origin, Fig. 10–8. • Plot the reference point A having coordinates A1Px , gxy>22. This • point represents the case for which the x¿ axis coincides with the x axis. Hence u = 0°, Fig. 10–8. Connect point A with the center C of the circle and from the shaded triangle determine the radius R of the circle, Fig. 10–8. • Once R has been determined, sketch the circle. 10.3 P1 Principal Strains. • • When P1 and P2 are indicated as being positive as in Fig. 10–9a, F P2 • The principal strains P1 and P2 are determined from the circle as the coordinates of points B and D, that is where g>2 = 0, Fig. 10–9a. The orientation of the plane on which P1 acts can be determined from the circle by calculating 2up1 using trigonometry. Here this angle happens to be counterclockwise from the radial reference line CA to line CB, Fig. 10–9a. Remember that the rotation of up1 must be in this same direction, from the element’s reference axis x to the x¿ axis, Fig. 10–9b.* 495 MOHR’S CIRCLE—PLANE STRAIN Q D 2 up1 C 2 us1 A g 2 P gxy 2u P E Pavg B 2 u 0 (a) y¿ the element in Fig. 10–9b will elongate in the x¿ and y¿ directions as shown by the dashed outline. y Maximum In-Plane Shear Strain. • The average normal strain and half the maximum in-plane shear strain are determined from the circle as the coordinates of point E or F, Fig. 10–9a. (1 P2)dy ¿ x¿ up1 • The orientation of the plane on which g max and Pavg act can be in-plane determined from the circle by calculating 2us1 using trigonometry. Here this angle happens to be clockwise from the radial reference line CA to line CE, Fig. 10–9a. Remember that the rotation of us1 must be in this same direction, from the element’s reference axis x to the x¿ axis, Fig. 10–9c.* x (1 P1)dx ¿ (b) y y¿ Pavg dy ¿ Strains on Arbitrary Plane. • The normal and shear strain components Px¿ and gx¿y¿ for a plane oriented at an angle u, Fig. 10–9d, can be obtained from the circle using trigonometry to determine the coordinates of point P, Fig. 10–9a. x¿ • To locate P, the known angle u of the x¿ axis is measured on the circle as 2u. This measurement is made from the radial reference line CA to the radial line CP. Remember that measurements for 2u on the circle must be in the same direction as u for the x¿ axis.* x us1 Pavg dx ¿ (c) Py¿dy¿ y¿ y • If the value of Py¿ is required, it can be determined by 10 calculating the P coordinate of point Q in Fig. 10–9a. The line CQ lies 180° away from CP and thus represents a 90° rotation of the x¿ axis. x¿ u Px ¿dx ¿ *If the g>2 axis were constructed positive upwards, then the angle 2u on the circle would be measured in the opposite direction to the orientation u of the plane. (d) Fig. 10–9 x 496 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N EXAMPLE 10.4 The state of plane strain at a point is represented by the components Px = 250110-62, Py = - 150110-62, and gxy = 120110-62. Determine the principal strains and the orientation of the element. SOLUTION Construction of the Circle. The P and g>2 axes are established in Fig. 10–10a. Remember that the positive g>2 axis must be directed downward so that counterclockwise rotations of the element correspond to counterclockwise rotation around the circle, and vice versa. The center of the circle C is located on the P axis at D(P2, 0) B(P1, 0) 2up R 1 20 8.8 C 50 250 g (10 6 ) 2 60 A P (10 6 ) Pavg = 250 + 1-1502 2 110-62 = 50110-62 Since gxy>2 = 60110-62, the reference point A 1u = 0°2 has coordinates A1250110-62, 60110-622. From the shaded triangle in Fig. 10–10a, the radius of the circle is CA; that is, R = C 21250 - 5022 + 16022 D 110-62 = 208.8110-62 (a) Principal Strains. The P coordinates of points B and D represent the principal strains. They are y¿ P1 = 150 + 208.82110-62 = 259110-62 y P2 = 150 - 208.82110-62 = - 159110-62 Ans. The direction of the positive principal strain P1 is defined by the counterclockwise angle 2up1 , measured from the radial reference line CA 1u = 0°2 to the line CB. We have P2dy¿ 10 Ans. dy ¿ x¿ up1 8.35 x dx¿ (b) Fig. 10–10 P1dx¿ tan 2up1 = 60 1250 - 502 up1 = 8.35° Ans. Hence, the side dx¿ of the element is oriented counterclockwise 8.35° as shown in Fig. 10–10b. This also defines the direction of P1 . The deformation of the element is also shown in the figure. 10.3 497 MOHR’S CIRCLE—PLANE STRAIN EXAMPLE 10.5 The state of plane strain at a point is represented by the components Px = 250110-62, Py = - 150110-62, and gxy = 120110-62. Determine the maximum in-plane shear strains and the orientation of an element. SOLUTION The circle has been established in the previous example and is shown in Fig. 10–11a. Maximum In-Plane Shear Strain. Half the maximum in-plane shear strain and average normal strain are represented by the coordinates of point E or F on the circle. From the coordinates of point E, max 1gx¿y¿2in-plane 2 = 208.8110-62 F R C 50 max = 418110 2 1gx¿y¿2in-plane -6 2 us1 Ans. A 60 u 0 Pavg = 50110-62 E Pavg, To orient the element, we can determine the clockwise angle 2us1 measured from CA 1u = 0°2 to CE. 2us1 = 90° - 218.35°2 us1 = 36.7° P (10 6 ) 208 .8 Ans. gmax in–plane 2 250 g (10 6 ) 2 (a) Fig. 10–11 This angle is shown in Fig. 10–11b. Since the shear strain defined from point E on the circle has a positive value and the average normal strain is also positive, these strains deform the element into the dashed shape shown in the figure. y y¿ 10 x us1 36.7 (b) x¿ 498 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N EXAMPLE 10.6 The state of plane strain at a point is represented on an element having components Px = - 300110-62, Py = - 100110-62, and gxy = 100110-62. Determine the state of strain on an element oriented 20° clockwise from this reported position. SOLUTION Construction of the Circle. The P and g>2 axes are established in Fig. 10–12a. The center of the circle is on the P axis at Px ¿ gx¿y¿ 2 P 50 c 13.43 C 13.43 40f 8 11. Q gx¿y¿ 1 R A 2 Py ¿ 200 300 P (10 6 ) -300 - 100 b110-62 = - 200110-62 2 The reference point A has coordinates A1- 300110-62, 50110-622. The radius CA determined from the shaded triangle is therefore R = C 21300 - 20022 + 15022 D 110-62 = 111.8110-62 g (10 6 ) 2 (a) Pavg = a Strains on Inclined Element. Since the element is to be oriented 20° clockwise, we must establish a radial line CP, 2120°2 = 40° clockwise, measured from CA 1u = 0°2, Fig. 10–12a. The coordinates of point P 1Px¿ , gx¿y¿>22 are obtained from the geometry of the circle. Note that f = tan-1 a 50 b = 26.57°, 1300 - 2002 c = 40° - 26.57° = 13.43° Thus, y Px¿ = - 1200 + 111.8 cos 13.43°2110-62 y¿ = - 309110-62 gx¿y¿ 2 = - 1111.8 sin 13.43°2110-62 gx¿y¿ = - 52.0110-62 x 10 20 x¿ (b) Fig. 10–12 Ans. Ans. The normal strain Py¿ can be determined from the P coordinate of point Q on the circle, Fig. 10–12a. Why? Py¿ = - 1200 - 111.8 cos 13.43°2110-62 = - 91.3110-62 Ans. As a result of these strains, the element deforms relative to the x¿, y¿ axes as shown in Fig. 10–12b. 10.3 MOHR’S CIRCLE—PLANE STRAIN 499 PROBLEMS 10–1. Prove that the sum of the normal strains in perpendicular directions is constant. 10–2. The state of strain at the point has components of Px = 200 110-62, Py = - 300 110-62, and gxy = 400(10-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. 10–5. The state of strain at the point on the arm has components Px = 250110-62, Py = -450110-62, gxy = - 825110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. y y x x Prob. 10–2 10–3. A strain gauge is mounted on the 1-in.-diameter A-36 steel shaft in the manner shown. When the shaft is rotating with an angular velocity of v = 1760 rev>min, the reading on the strain gauge is P = 800110-62. Determine the power output of the motor. Assume the shaft is only subjected to a torque. 60 Prob. 10–5 10–6. The state of strain at the point has components of Px = - 100110-62, Py = 400110-62, and gxy = - 300110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 60° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. 10–7. The state of strain at the point has components of Px = 100110-62, Py = 300110-62, and gxy = - 150110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented u = 30° clockwise. Sketch the deformed element due to these strains within the x–y plane. 10 Prob. 10–3 *10–4. The state of strain at a point on a wrench has components Px = 120110-62, Py = - 180110-62, gxy = 150110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. y x Probs. 10–6/7 500 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N *10–8. The state of strain at the point on the bracket has components Px = - 200110-62, Py = - 650110-62, gxy - 175110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 20° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. 10–10. The state of strain at the point on the bracket has components Px = 400110-62, Py = - 250110-62, gxy 310110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 30° clockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. y y x x Prob. 10–8 Prob. 10–10 10–9. The state of strain at the point has components of Px = 180110-62, Py = - 120110-62, and gxy = - 100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. 10–11. The state of strain at the point has components of Px = - 100110-62, Py = - 200110-62, and gxy = 100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. y x y 10 x Prob. 10–9 Prob. 10–11 10.3 *10–12. The state of plane strain on an element is given by Px = 500110-62, Py = 300110-62, and gxy = - 200110-62. Determine the equivalent state of strain on an element at the same point oriented 45° clockwise with respect to the original element. MOHR’S CIRCLE—PLANE STRAIN 501 10–14. The state of strain at the point on a boom of an hydraulic engine crane has components of Px = 250110-62, Py = 300110-62, and gxy = - 180110-62. Use the straintransformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case, specify the orientation of the element and show how the strains deform the element within the x–y plane. y y x Pydy dy gxy 2 gxy 2 dx x Pxdx Prob. 10–12 10–13. The state of plane strain on an element is Px = - 300110-62, Py = 0, and gxy = 150110-62. Determine the equivalent state of strain which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element. Prob. 10–14 ■10–15. Consider the general case of plane strain where Px , Py , and gxy are known. Write a computer program that can be used to determine the normal and shear strain, Px¿ and gx¿y¿ , on the plane of an element oriented u from the horizontal. Also, include the principal strains and the element’s orientation, and the maximum in-plane shear strain, the average normal strain, and the element’s orientation. *10–16. The state of strain at a point on a support has components of Px = 350110-62, Py = 400110-62, gxy = - 675110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. 10 y gxy dy 2 •10–17. x gxy 2 dx Prob. 10–13 Pxdx Solve part (a) of Prob. 10–4 using Mohr’s circle. 10–18. Solve part (b) of Prob. 10–4 using Mohr’s circle. 10–19. Solve Prob. 10–8 using Mohr’s circle. *10–20. Solve Prob. 10–10 using Mohr’s circle. •10–21. Solve Prob. 10–14 using Mohr’s circle. 502 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N z x *10.4 Absolute Maximum Shear Strain y (1 P1)dx (1 P2)dy In Sec. 9.5 it was pointed out that in the case of plane stress, the absolute maximum shear stress in an element of material will occur out of the plane when the principal stresses have the same sign, i.e., both are tensile or both are compressive. A similar result occurs for plane strain. For example, if the principal in-plane strains cause elongations, Fig. 10–13a, then the three Mohr’s circles describing the normal and shear strain components for elements oriented about the x¿, y¿, and z¿ axes are shown in Fig. 10–13b. By inspection, the largest circle has a radius R = 1gx¿z¿2max>2. Hence, xy plane strain (a) g abs = 1gx¿z¿2max = P1 max P2 P P1 (gy z )max 2 P1 and P2 have the same sign This value gives the absolute maximum shear strain for the material. Note that it is larger than the maximum in-plane shear strain, which is 1gx¿y¿2max = P1 - P2 . Now consider the case where one of the in-plane principal strains is of opposite sign to the other in-plane principal strain, so that P1 causes elongation and P2 causes contraction, Fig. 10–14a. Mohr’s circles, which describe the strains on each element’s orientation about the x¿, y¿, z¿ axes, are shown in Fig. 10–14b. Here (gx y)max 2 (gx z )max 2 g 2 (b) Fig. 10–13 z = 1gx¿y¿2in-plane = P1 - P2 g abs max max P1 and P2 have opposite signs x (10–14) (1 P1)dx (1 – P2)dy xy plane strain y (10–15) We may therefore summarize the above two cases as follows. If the in-plane principal strains both have the same sign, the absolute maximum shear strain will occur out of plane and has a value of g abs = Pmax . max However, if the in-plane principal strains are of opposite signs, then the absolute maximum shear strain equals the maximum in-plane shear strain. (a) Important Points 10 P2 P1 (gyz)max 2 (gxz)max 2 (gxy)max g 2 2 (b) Fig. 10–14 P • The absolute maximum shear strain will be larger than the maximum in-plane shear strain whenever the in-plane principal strains have the same sign. When this occurs the absolute maximum shear strain will act out of the plane. • If the in-plane principal strains are of opposite signs, then the absolute maximum shear strain will equal the maximum in-plane shear strain. 10.4 ABSOLUTE MAXIMUM SHEAR STRAIN 503 EXAMPLE 10.7 The state of plane strain at a point is represented by the strain components Px = - 400110-62, Py = 200110-62, gxy = 150110-62. Determine the maximum in-plane shear strain and the absolute maximum shear strain. 75 P2 A P1 9 30 R P(106) 100 (gx¿y¿)max in–plane 2 400 g (106) 2 Fig. 10–15 SOLUTION Maximum In-Plane Shear Strain. We will solve this problem using Mohr’s circle. From the strain components, the center of the circle is on the P axis at -400 + 200 110-62 = - 100110-62 Pavg = 2 Since gxy>2 = 75110-62, the reference point A has coordinates 1 -400110-62, 75110-622. As shown in Fig. 10–15, the radius of the circle is therefore R = C 21400 - 10022 + 17522 D 110-62 = 309110-62 Calculating the in-plane principal strains from the circle, we have P1 = 1-100 + 3092110-62 = 209110-62 P2 = 1-100 - 3092110-62 = - 409110-62 Also, the maximum in-plane shear strain is g max = P1 - P2 = [209 - 1 -4092]110-62 = 618110-62 Ans. in-plane Absolute Maximum Shear Strain. From the above results, we have P1 = 209110-62, P2 = - 409110-62. The three Mohr’s circles, plotted for element orientations about each of the x, y, z axes, are also shown in Fig. 10–15. It is seen that since the principal in-plane strains have opposite signs, the maximum in-plane shear strain is also the absolute maximum shear strain; i.e., g abs = 618110-62 max Ans. 10 504 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N 10.5 Strain Rosettes b When performing a tension test on a specimen as discussed in Sec. 3.1, the normal strain in the material is measured using an electricalresistance strain gauge, which consists of a wire grid or piece of metal foil bonded to the specimen. For a general loading on a body, however, the strains at a point on its free surface are determined using a cluster of three electrical-resistance strain gauges, arranged in a specified pattern. This pattern is referred to as a strain rosette, and once the normal strains on the three gauges are measured, the data can then be transformed to specify the state of strain at the point. Since these strains are measured only in the plane of the gauges, and since the body is stress-free on its surface, the gauges may be subjected to plane stress but not plane strain. Although the strain normal to the surface is not measured, realize that the out-of-plane displacement caused by this strain will not affect the in-plane measurements of the gauges. In the general case, the axes of the three gauges are arranged at the angles ua, ub, uc shown in Fig. 10–16a. If the readings Pa, Pb, Pc are taken, we can determine the strain components Px, Py, gxy at the point by applying the strain-transformation equation, Eq. 10–2, for each gauge. We have a ub ua uc x c (a) c b 45 45 x a 45 strain rosette (b) Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub c (10–16) Pc = Px cos uc + Py sin uc + gxy sin uc cos uc 2 b 60 60 x a 60 strain rosette (c) Fig. 10–16 2 The values of Px, Py, gxy are determined by solving these three equations simultaneously. Strain rosettes are often arranged in 45° or 60° patterns. In the case of the 45° or “rectangular” strain rosette shown in Fig. 10–16b, ua = 0°, ub = 45°, uc = 90°, so that Eq. 10–16 gives Px = Pa Py = Pc gxy = 2Pb - 1Pa + Pc2 And for the 60° strain rosette in Fig. 10–16c, ua = 0°, ub = 60°, uc = 120°. Here Eq. 10–16 gives Px = Pa 10 1 12Pb + 2Pc - Pa2 3 2 = 1Pb - Pc2 13 Py = gxy Typical electrical resistance 45° strain rosette. (10–17) Once Px, Py, gxy are determined, the transformation equations of Sec. 10.2 or Mohr’s circle can then be used to determine the principal in-plane strains and the maximum in-plane shear strain at the point. 10.5 505 STRAIN ROSETTES EXAMPLE 10.8 The state of strain at point A on the bracket in Fig. 10–17a is measured using the strain rosette shown in Fig. 10–17b. Due to the loadings, the readings from the gauges give Pa = 60110-62, Pb = 135110-62, and Pc = 264110-62. Determine the in-plane principal strains at the point and the directions in which they act. c A SOLUTION We will use Eqs. 10–16 for the solution. Establishing an x axis as shown in Fig. 10–17b and measuring the angles counterclockwise from the +x axis to the centerlines of each gauge, we have ua = 0°, ub = 60°, and uc = 120°. Substituting these results, along with the problem data, into the equations gives 60110 2 = Px cos 0° + Py sin 0° + gxy sin 0° cos 0° = Px 135110-62 = Px cos2 60° + Py sin2 60° + gxy sin 60° cos 60° -6 2 (a) c 2 = 0.25Px + 0.75Py + 0.433gxy b (1) 120 60 (2) x 264110 2 = Px cos 120° + Py sin 120° + gxy sin 120° cos 120° (3) = 0.25Px + 0.75Py - 0.433gxy -6 b a 2 a 2 (b) Using Eq. 1 and solving Eqs. 2 and 3 simultaneously, we get Px = 60110-62 Py = 246110-62 gxy = - 149110-62 60 These same results can also be obtained in a more direct manner from Eq. 10–17. The in-plane principal strains can be determined using Mohr’s circle. The reference point on the circle is at A [60110-62, -74.5110-62] and the center of the circle, C, is on the P axis at Pavg = 153110-62, Fig. 10–17c. From the shaded triangle, the radius is R = C 21153 - 602 + 174.52 D 110 2 = 119.1110 2 2 2 -6 -6 A R P2 11 9 2up2 .2 74.5 C P1 153 P(106) g (106) 2 (c) The in-plane principal strains are thus P1 = 153110-62 + 119.1110-62 = 272110-62 P2 = 153110-62 - 119.1110-62 = 33.9110-62 74.5 2up2 = tan-1 = 38.7° 1153 - 602 up2 = 19.3° Ans. y¿ Ans. x¿ Ans. NOTE: The deformed element is shown in the dashed position in Fig. 10–17d. Realize that, due to the Poisson effect, the element is also subjected to an out-of-plane strain, i.e., in the z direction, although this value will not influence the calculated results. up2 19.3 x (d) Fig. 10–17 10 506 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N PROBLEMS 10–22. The strain at point A on the bracket has components Px = 300110-62, Py = 550110-62, gxy = -650110-62. Determine (a) the principal strains at A in the x –y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. *10–24. The strain at point A on the pressure-vessel wall has components Px = 480110-62, Py = 720110-62, gxy = 650110-62. Determine (a) the principal strains at A, in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. y A x y A x Prob. 10–22 Prob. 10–24 •10–25. 10–23. The strain at point A on the leg of the angle has components Px = - 140110-62, Py = 180110-62, gxy = -125110-62. Determine (a) the principal strains at A in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. The 60° strain rosette is mounted on the bracket. The following readings are obtained for each gauge: Pa = - 100110-62, Pb = 250110-62, and Pc = 150110-62. Determine (a) the principal strains and (b) the maximum in-plane shear strain and associated average normal strain. In each case show the deformed element due to these strains. b 10 c A 60 60 Prob. 10–23 a Prob. 10–25 10.5 10–26. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. 507 STRAIN ROSETTES *10–28. The 45° strain rosette is mounted on the link of the backhoe. The following readings are obtained from each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and associated average normal strain. a 45 45 b b c a 30 30 c 60 Prob. 10–28 Prob. 10–26 10–27. The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each gauge: Pa = 300110-62, Pb = - 250110-62, and Pc = - 450110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. 10–29. Consider the general orientation of three strain gauges at a point as shown. Write a computer program that can be used to determine the principal in-plane strains and the maximum in-plane shear strain at the point. Show an application of the program using the values ua = 40°, Pa = 160110-62, ub = 125°, Pb = 100110-62, uc = 220°, Pc = 80110-62. b ub uc a ua x b c 45 45 a c Prob. 10–27 Prob. 10–29 10 508 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N 10.6 Material-Property Relationships In this section we will present some important relationships involving a material’s properties that are used when the material is subjected to multiaxial stress and strain. To do so we will assume that the material is homogeneous and isotropic and behaves in a linear-elastic manner. Generalized Hooke’s Law. If the material at a point is subjected to a state of triaxial stress, sx, sy, sz, Fig. 10–18a, associated normal strains Px, Py, Pz will be developed in the material. The stresses can be related to these strains by using the principle of superposition, Poisson’s ratio, Plat = - nPlong, and Hooke’s law, as it applies in the uniaxial direction, P = s>E. For example, consider the normal strain of the element in the x direction, caused by separate application of each normal stress. When sx is applied, Fig. 10–18b, the element elongates in the x direction and the strain Pxœ is Pxœ = sx E Application of sy causes the element to contract with a strain Pxfl, Fig. 10–18c. Here Pxfl = - n sy E Likewise, application of sz, Fig. 10–18d, causes a contraction such that PxÔ = - n sz E sz sz 10 ⴝ ⴙ ⴙ sy sy sx sx (a) (b) (c) Fig. 10–18 (d) 10.6 MATERIAL-PROPERTY RELATIONSHIPS 509 When these three normal strains are superimposed, the normal strain Px is determined for the state of stress in Fig. 10–18a. Similar equations can be developed for the normal strains in the y and z directions. The final results can be written as Px = 1 [s - n1sy + sz2] E x Py = 1 [s - n1sx + sz2] E y Pz = 1 [s - n1sx + sy2] E z (10–18) These three equations express Hooke’s law in a general form for a triaxial state of stress. For application tensile stresses are considered positive quantities, and compressive stresses are negative. If a resulting normal strain is positive, it indicates that the material elongates, whereas a negative normal strain indicates the material contracts. If we now apply a shear stress txy to the element, Fig. 10–19a, experimental observations indicate that the material will deform only due to a shear strain gxy; that is, txy will not cause other strains in the material. Likewise, tyz and txz will only cause shear strains gyz and gxz, Figs. 10–19b and 10–19c, and so Hooke’s law for shear stress and shear strain can be written as gxy = 1 t G xy gyz = 1 t G yz gxz = 1 t G xz (10–19) 10 tyz txy tzx (a) (b) Fig. 10–19 (c) 510 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N Relationship Involving E, N, and G. In Sec. 3.7 it was stated that the modulus of elasticity E is related to the shear modulus G by Eq. 3–11, namely, y txy G = x (a) x¿ y smin txy smax txy up1 45 x (b) Fig. 10–20 dx sz txy (10–21) 11 + n2 E This strain, which deforms the element along the x¿ axis, can also be related to the shear strain gxy. To do this, first note that since sx = sy = sz = 0, then from the first and second Eqs. 10–18, Px = Py = 0. Substituting these results into the strain transformation Eq. 10–9, we get gxy P1 = Pmax = 2 Pmax = dV = 11 + Px211 + Py211 + Pz2 dx dy dz - dx dy dz (1 Pz)dz 10 sy (1 Px)dx (1 Py)dy One way to derive this relationship is to consider an element of the material to be subjected to pure shear 1sx = sy = sz = 02, Fig. 10–20a. Applying Eq. 9–5 to obtain the principal stresses yields smax = txy and smin = - txy. This element must be oriented up1 = 45° counterclockwise from the x axis as shown in Fig. 10–20b. If the three principal stresses smax = txy, sint = 0, and smin = - txy are substituted into the first of Eqs. 10–18, the principal strain Pmax can be related to the shear stress txy. The result is Dilatation and Bulk Modulus. When an elastic material is subjected to normal stress, its volume will change. For example, consider a volume element which is subjected to the principal stresses sx, sy, sz. If the sides of the element are originally dx, dy, dz, Fig. 10–21a, then after application of the stress they become 11 + Px2 dx, 11 + Py2 dy, 11 + Pz2 dz, Fig. 10–21b. The change in volume of the element is therefore (a) sx (10–20) By Hooke’s law, gxy = txy>G, so that Pmax = txy>2G. Substituting into Eq. 10–21 and rearranging terms gives the final result, namely, Eq. 10–20. dz dy E 211 + n2 (b) Fig. 10–21 Neglecting the products of the strains since the strains are very small, we have dV = 1Px + Py + Pz2 dx dy dz The change in volume per unit volume is called the “volumetric strain” or the dilatation e. It can be written as e = dV = Px + Py + Pz dV (10–22) By comparison, the shear strains will not change the volume of the element, rather they will only change its rectangular shape. 10.6 MATERIAL-PROPERTY RELATIONSHIPS Also, if we use Hooke’s law, as defined by Eq. 10–18, we can write the dilatation in terms of the applied stress. We have 1 - 2n 1sx + sy + sz2 e = (10–23) E When a volume element of material is subjected to the uniform pressure p of a liquid, the pressure on the body is the same in all directions and is always normal to any surface on which it acts. Shear stresses are not present, since the shear resistance of a liquid is zero. This state of “hydrostatic” loading requires the normal stresses to be equal in any and all directions, and therefore an element of the body is subjected to principal stresses sx = sy = sz = - p, Fig. 10–22. Substituting into Eq. 10–23 and rearranging terms yields p E = (10–24) e 311 - 2n2 511 sz p sy p sx p Hydrostatic stress Fig. 10–22 Since this ratio is similar to the ratio of linear elastic stress to strain, which defines E, i.e., s>P = E, the term on the right is called the volume modulus of elasticity or the bulk modulus. It has the same units as stress and will be symbolized by the letter k; that is, k = E 311 - 2n2 (10–25) Note that for most metals n L 13 so k L E. If a material existed that did not change its volume then dV = e = 0, and k would have to be infinite. From Eq. 10–25 the theoretical maximum value for Poisson’s ratio is therefore n = 0.5. During yielding, no actual volume change of the material is observed, and so n = 0.5 is used when plastic yielding occurs. Important Points • When a homogeneous isotropic material is subjected to a state of triaxial stress, the strain in each direction is influenced by the strains produced by all the stresses.This is the result of the Poisson effect, and results in the form of a generalized Hooke’s law. • Unlike normal stress, a shear stress applied to homogeneous isotropic material will only produce shear strain in the same plane. • The material constants E, G, and n are related mathematically. • Dilatation, or volumetric strain, is caused only by normal strain, • not shear strain. The bulk modulus is a measure of the stiffness of a volume of material. This material property provides an upper limit to Poisson’s ratio of n = 0.5, which remains at this value while plastic yielding occurs. 10 512 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N EXAMPLE 10.9 The bracket in Example 10–8, Fig. 10–23a, is made of steel for which Est = 200 GPa, nst = 0.3. Determine the principal stresses at point A. c A b a (a) Fig. 10–23 SOLUTION I From Example 10.8 the principal strains have been determined as P1 = 272110-62 P2 = 33.9110-62 Since point A is on the surface of the bracket for which there is no loading, the stress on the surface is zero, and so point A is subjected to plane stress. Applying Hooke’s law with s3 = 0, we have P1 = s1 n - s2 ; E E 272110-62 = s1 200110 2 9 - 0.3 s2 20011092 54.411062 = s1 - 0.3s2 P2 = 10 (1) s2 s2 n 0.3 - s1 ; 33.9110-62 = s1 E E 20011092 20011092 6.7811062 = s2 - 0.3s1 (2) Solving Eqs. 1 and 2 simultaneously yields s1 = 62.0 MPa Ans. s2 = 25.4 MPa Ans. 10.6 MATERIAL-PROPERTY RELATIONSHIPS 513 29.4 A R 11.46 18 s2 .3 C s1 s (MPa) 43.7 t (MPa) Fig. 10–23 (cont.) (b) SOLUTION II It is also possible to solve the problem using the given state of strain, Px = 60110-62 Py = 246110-62 gxy = - 149110-62 as specified in Example 10.8. Applying Hooke’s law in the x–y plane, we have 0.3sy sx sx n Px = - sy ; 60110-62 = 9 E E 200110 2 Pa 20011092 Pa sy sy 0.3sx n Py = - sx ; 246110-62 = 9 E E 200110 2 Pa 20011092 Pa sx = 29.4 MPa sy = 58.0 MPa The shear stress is determined using Hooke’s law for shear. First, however, we must calculate G. G = Thus, txy = Ggxy ; E 200 GPa = = 76.9 GPa 211 + n2 211 + 0.32 txy = 76.911092[-149110-62] = - 11.46 MPa The Mohr’s circle for this state of plane stress has a reference point A129.4 MPa, -11.46 MPa2 and center at savg = 43.7 MPa, Fig. 10–23b. The radius is determined from the shaded triangle. R = 2143.7 - 29.422 + 111.4622 = 18.3 MPa Therefore, 10 s1 = 43.7 MPa + 18.3 MPa = 62.0 MPa Ans. s2 = 43.7 MPa - 18.3 MPa = 25.4 MPa Ans. NOTE: Each of these solutions is valid provided the material is both linear elastic and isotropic, since then the principal planes of stress and strain coincide. 514 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N EXAMPLE 10.10 The copper bar in Fig. 10–24 is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, width b = 50 mm, and thickness t = 20 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take Ecu = 120 GPa, ncu = 0.34. t 500 MPa 800 MPa a b 800 MPa 500 MPa Fig. 10–24 SOLUTION By inspection, the bar is subjected to a state of plane stress. From the loading we have sx = 800 MPa sy = - 500 MPa txy = 0 sz = 0 The associated normal strains are determined from the generalized Hooke’s law, Eq. 10–18; that is, Px = sx n - 1sy + sz2 E E 800 MPa 0.34 1-500 MPa + 02 = 0.00808 12011032 MPa 12011032 MPa sy n - 1sx + sz2 Py = E E 0.34 - 500 MPa = 1800 MPa + 02 = - 0.00643 3 120110 2 MPa 12011032 MPa sz n - 1sx + sy2 Pz = E E 0.34 = 0 1800 MPa - 500 MPa2 = - 0.000850 12011032 MPa The new bar length, width, and thickness are therefore = 10 a¿ = 300 mm + 0.008081300 mm2 = 302.4 mm b¿ = 50 mm + 1- 0.006432150 mm2 = 49.68 mm t¿ = 20 mm + 1-0.0008502120 mm2 = 19.98 mm Ans. Ans. Ans. 10.6 MATERIAL-PROPERTY RELATIONSHIPS 515 EXAMPLE 10.11 If the rectangular block shown in Fig. 10–25 is subjected to a uniform pressure of p = 20 psi, determine the dilatation and the change in length of each side. Take E = 600 psi, n = 0.45. c 3 in. a 4 in. b 2 in. Fig. 10–25 SOLUTION Dilatation. The dilatation can be determined using Eq. 10–23 with sx = sy = sz = - 20 psi. We have e = = 1 - 2n 1sx + sy + sz2 E 1 - 210.452 600 psi [31 -20 psi2] = - 0.01 in3>in3 Ans. Change in Length. The normal strain on each side can be determined from Hooke’s law, Eq. 10–18; that is, P = = 1 [s - n1sy + sz2] E x 1 [-20 psi - 10.4521 -20 psi - 20 psi2] = - 0.00333 in.>in. 600 psi Thus, the change in length of each side is da = - 0.0033314 in.2 = - 0.0133 in. Ans. db = - 0.0033312 in.2 = - 0.00667 in. Ans. dc = - 0.0033313 in.2 = - 0.0100 in. Ans. The negative signs indicate that each dimension is decreased. 10 516 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N PROBLEMS 10–30. For the case of plane stress, show that Hooke’s law can be written as sx = E E 1Px + nPy2, sy = 1Py + nPx2 2 11 - n 2 11 - n22 10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the stress-transformation equations, Eqs. 9–1 and 9–2. 10–37. Determine the bulk modulus for each of the following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and (b) glass, Eg = 811032 ksi, ng = 0.24. 10–38. The principal stresses at a point are shown in the figure. If the material is A-36 steel, determine the principal strains. 12 ksi *10–32. A bar of copper alloy is loaded in a tension machine and it is determined that Px = 940110-62 and sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of elasticity, Ecu, and the dilatation, ecu, of the copper. ncu = 0.35. •10–33. The principal strains at a point on the aluminum fuselage of a jet aircraft are P1 = 780110-62 and P2 = 400110-62. Determine the associated principal stresses at the point in the same plane. Eal = 1011032 ksi, nal = 0.33. Hint: See Prob. 10–30. 10–34. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the absolute maximum shear strain in the rod at a point on its surface. 10–35. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the principal strains at a point on the surface of the rod. 700 N 700 N 20 ksi 8 ksi Prob. 10–38 10–39. The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est = 200 GPa and nst = 0.3. Probs. 10–34/35 20 mm *10–36. The steel shaft has a radius of 15 mm. Determine the torque T in the shaft if the two strain gauges, attached to the surface of the shaft, report strains of Px¿ = - 80110-62 and Py¿ = 80110-62. Also, compute the strains acting in the x and y directions. Est = 200 GPa, nst = 0.3. 10 y T y¿ x¿ 45 x T Prob. 10–36 Prob. 10–39 10.6 *10–40. The strain in the x direction at point A on the steel beam is measured and found to be Px = - 100110-62. Determine the applied load P. What is the shear strain gxy at point A? Est = 2911032 ksi, nst = 0.3. 3 in. 0.5 in. A P y 0.5 in. 8 in. 0.5 in. *10–44. A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = - 200(10-6). Determine the principal strains in the pipe at point A. The pipe has an outer diameter of 2 in. and an inner diameter of 1 in. and is made of A-36 steel. 6 in. x 3 ft 4 ft 517 10–43. A single strain gauge, placed on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = - 200(10-6). Determine the horizontal force P if the pipe has an outer diameter of 2 in. and an inner diameter of 1 in. The pipe is made of A-36 steel. 3 in. A MATERIAL-PROPERTY RELATIONSHIPS 7 ft 1.5 ft Prob. 10–40 •10–41. The cross section of the rectangular beam is subjected to the bending moment M. Determine an expression for the increase in length of lines AB and CD. The material has a modulus of elasticity E and Poisson’s ratio is n. P 2.5 ft 30 C A D B h Probs. 10–43/44 A 10–45. The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the bending stress that would occur if flat ends were used. The bending stresses at the seam where the caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder, respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder. Show that this ratio is tc>th = 12 - n2>11 - n2. Assume that the vessel is made of the same material and both the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of 0.5 in., what is the required thickness of the hemispheres? Take n = 0.3. M Prob. 10–41 b 10–42. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains. 26 ksi tc 10 th r 15 ksi 10 ksi Prob. 10–42 Prob. 10–45 518 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N 10–46. The principal strains in a plane, measured experimentally at a point on the aluminum fuselage of a jet aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is a case of plane stress, determine the associated principal stresses at the point in the same plane. Eal = 10(103) ksi and nal = 0.33. 10–47. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains. •10–49. Initially, gaps between the A-36 steel plate and the rigid constraint are as shown. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 100°F . To solve, add the thermal strain a¢T to the equations for Hooke’s Law. y 0.0015 in. 3 ksi 6 in. 0.0025 in. 8 in. x Prob. 10–49 8 ksi 4 ksi 10–50. Two strain gauges a and b are attached to a plate made from a material having a modulus of elasticity of E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the intensities of the uniform distributed load wx and wy acting on the plate. The thickness of the plate is 25 mm. Prob. 10–47 *10–48. The 6061-T6 aluminum alloy plate fits snugly into the rigid constraint. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 50°C . To solve, add the thermal strain a¢T to the equations for Hooke’s Law. 10–51. Two strain gauges a and b are attached to the surface of the plate which is subjected to the uniform distributed load wx = 700 kN>m and wy = - 175 kN>m. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the modulus of elasticity E, shear modulus G, and Poisson’s ratio n for the material. wy y 400 mm 10 b 45 y 300 mm a x Prob. 10–48 z x Probs. 10–50/51 wx 10.6 *10–52. The block is fitted between the fixed supports. If the glued joint can resist a maximum shear stress of tallow = 2 ksi, determine the temperature rise that will cause the joint to fail. Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with an additional strain term of a¢T (Eq. 4–4). 40 MATERIAL-PROPERTY RELATIONSHIPS 519 *10–56. A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it is subjected to an internal pressure p, show that the increase in its inner radius is dr = rP1 = pr211 - 12 n2>Et and the increase in its length is ¢L = pLr112 - n2>Et. Using these results, show that the change in internal volume becomes dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are small quantities, show further that the change in volume per unit volume, called volumetric strain, can be written as dV>V = pr12.5 - 2n2>Et. 10–57. The rubber block is confined in the U-shape smooth rigid block. If the rubber has a modulus of elasticity E and Poisson’s ratio n, determine the effective modulus of elasticity of the rubber under the confined condition. P Prob. 10–52 •10–53. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is covered and the temperature is increased by 200°F, determine the stress components sx , sy , and sz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4). 10–54. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is not covered and the temperature is increased by 200°F, determine the strain components Px , Py , and Pz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4). z Prob. 10–57 10–58. A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that Px = 0 and Py = 0, determine the factor by which the modulus of elasticity will be increased when a load is applied if n = 0.3 for the material. 0.012 in. z 4 in. 4 in. 6 in. P y 10 y x Probs. 10–53/54 x 10–55. A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p. Show that the increase in the volume within the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain analysis. Prob. 10–58 520 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N *10.7 Theories of Failure When an engineer is faced with the problem of design using a specific material, it becomes important to place an upper limit on the state of stress that defines the material’s failure. If the material is ductile, failure is usually specified by the initiation of yielding, whereas if the material is brittle, it is specified by fracture. These modes of failure are readily defined if the member is subjected to a uniaxial state of stress, as in the case of simple tension; however, if the member is subjected to biaxial or triaxial stress, the criterion for failure becomes more difficult to establish. In this section we will discuss four theories that are often used in engineering practice to predict the failure of a material subjected to a multiaxial state of stress. No single theory of failure, however, can be applied to a specific material at all times, because a material may behave in either a ductile or brittle manner depending on the temperature, rate of loading, chemical environment, or the way the material is shaped or formed. When using a particular theory of failure, it is first necessary to calculate the normal and shear stress at points where they are the largest in the member. Once this state of stress is established, the principal stresses at these critical points are then determined, since each of the following theories is based on knowing the principal stress. Ductile Materials 45 Lüder’s lines on mild steel strip 10 Fig. 10–26 Maximum-Shear-Stress Theory. The most common type of yielding of a ductile material such as steel is caused by slipping, which occurs along the contact planes of randomly ordered crystals that make up the material. If we make a specimen into a highly polished thin strip and subject it to a simple tension test, we can actually see how this slipping causes the material to yield, Fig. 10–26. The edges of the planes of slipping as they appear on the surface of the strip are referred to as Lüder’s lines. These lines clearly indicate the slip planes in the strip, which occur at approximately 45° with the axis of the strip. The slipping that occurs is caused by shear stress. To show this, consider an element of the material taken from a tension specimen, when it is subjected to the yield stress sY, Fig. 10–27a. The maximum shear stress can be determined by drawing Mohr’s circle for the element, Fig. 10–27b. The results indicate that tmax = sY 2 (10–26) 10.7 Furthermore, this shear stress acts on planes that are 45° from the planes of principal stress, Fig. 10–27c, and these planes coincide with the direction of the Lüder lines shown on the specimen, indicating that indeed failure occurs by shear. Using this idea, that ductile materials fail by shear, in 1868 Henri Tresca proposed the maximum-shear-stress theory or Tresca yield criterion. This theory can be used to predict the failure stress of a ductile material subjected to any type of loading. The theory states that yielding of the material begins when the absolute maximum shear stress in the material reaches the shear stress that causes the same material to yield when it is subjected only to axial tension. Therefore, to avoid failure, it is required that tabs in the material must be less than or equal to sY> 2, max where sY is determined from a simple tension test. For application we will express the absolute maximum shear stress in terms of the principal stresses. The procedure for doing this was discussed in Sec. 9.5 with reference to a condition of plane stress, that is, where the out-of-plane principal stress is zero. If the two in-plane principal stresses have the same sign, i.e., they are both tensile or both compressive, then failure will occur out of the plane, and from Eq. 9–13, T sY Axial tension (a) T s1 sY s2 0 s1 2 90 savg y¿ If instead the in-plane principal stresses are of opposite signs, then failure occurs in the plane, and from Eq. 9–14, sY 2 tmax sY 2 (b) x¿ sY 2 savg 45 (c) Using these equations and Eq. 10–26, the maximum-shear-stress theory for plane stress can be expressed for any two in-plane principal stresses s1 and s2 by the following criteria: s1 , s2 have same signs sY 2 x s1 - s2 tabs = max 2 ƒ s1 ƒ = sY r ƒ s2 ƒ = sY s A(0, 0) max = tabs max 521 THEORIES OF FAILURE Fig. 10–27 s2 sY (10–27) ƒ s1 - s2 ƒ = sY6 s1 , s2 have opposite signs A graph of these equations is given in Fig. 10–28. Clearly, if any point of the material is subjected to plane stress, and its in-plane principal stresses are represented by a coordinate (s1, s2) plotted on the boundary or outside the shaded hexagonal area shown in this figure, the material will yield at the point and failure is said to occur. sY sY sY Maximum-shear-stress theory Fig. 10–28 s1 10 522 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N Maximum-Distortion-Energy Theory. It was stated in Sec. 3.5 that s3 an external loading will deform a material, causing it to store energy internally throughout its volume. The energy per unit volume of material is called the strain-energy density, and if the material is subjected to a uniaxial stress the strain-energy density, defined by Eq. 3–6, becomes s1 u = s2 1 sP 2 (10–28) If the material is subjected to triaxial stress, Fig. 10–29a, then each principal stress contributes a portion of the total strain-energy density, so that (a) ⴝ u = 1 1 1 s1P1 + s2P2 + s3P3 2 2 2 Furthermore, if the material behaves in a linear-elastic manner, then Hooke’s law applies. Therefore, substituting Eq. 10–18 into the above equation and simplifying, we get savg u = savg savg (b) ⴙ (s3 savg ) 10 (s1 savg ) (s2 savg ) 1 C s 2 + s22 + s32 - 2n1s1s2 + s1s3 + s3s22 D 2E 1 This strain-energy density can be considered as the sum of two parts, one part representing the energy needed to cause a volume change of the element with no change in shape, and the other part representing the energy needed to distort the element. Specifically, the energy stored in the element as a result of its volume being changed is caused by application of the average principal stress, savg = 1s1 + s2 + s32>3, since this stress causes equal principal strains in the material, Fig. 10–29b. The remaining portion of the stress, 1s1 - savg2, 1s2 - savg2, 1s3 - savg2, causes the energy of distortion, Fig. 10–29c. Experimental evidence has shown that materials do not yield when subjected to a uniform (hydrostatic) stress, such as savg discussed above. As a result, in 1904, M. Huber proposed that yielding in a ductile material occurs when the distortion energy per unit volume of the material equals or exceeds the distortion energy per unit volume of the same material when it is subjected to yielding in a simple tension test. This theory is called the maximum-distortion-energy theory, and since it was later redefined independently by R. von Mises and H. Hencky, it sometimes also bears their names. To obtain the distortion energy per unit volume, we will substitute the stresses 1s1 - savg2, 1s2 - savg2, and 1s3 - savg2 for s1 , s2 , and s3 , respectively, into Eq. 10–29, realizing that savg = 1s1 + s2 + s32>3. Expanding and simplifying, we obtain (c) Fig. 10–29 (10–29) ud = 1 + n C 1s1 - s222 + 1s2 - s322 + 1s3 - s122 D 6E 10.7 In the case of plane stress, s3 = 0, and this equation reduces to ud = 523 THEORIES OF FAILURE s2 1 + n A s12 - s1s2 + s22 B 3E sY For a uniaxial tension test, s1 = sY , s2 = s3 = 0, and so sY 1 + n 2 1ud2Y = s 3E Y Since the maximum-distortion-energy theory requires ud = 1ud2Y , then for the case of plane or biaxial stress, we have s12 - s1s2 + s22 = sY2 (10–30) This is the equation of an ellipse, Fig. 10–30. Thus, if a point in the material is stressed such that (s1, s2) is plotted on the boundary or outside the shaded area, the material is said to fail. A comparison of the above two failure criteria is shown in Fig. 10–31.Note that both theories give the same results when the principal stresses are equal, i.e., s1 = s2 = sY, or when one of the principal stresses is zero and the other has a magnitude of sY. If the material is subjected to pure shear, t, then the theories have the largest discrepancy in predicting failure. The stress coordinates of these points on the curves can be determined by considering the element shown in Fig. 10–32a. From the associated Mohr’s circle for this state of stress, Fig. 10–32b, we obtain principal stresses s1 = t and s2 = - t. Thus, with s1 = - s2, then from Eq. 10–27, the maximumshear-stress theory gives 1sY >2, -sY >22, and from Eq. 10–30, the maximum-distortion-energy theory gives 1sY > 23, -sY > 232, Fig.10–31. Actual torsion tests, used to develop a condition of pure shear in a ductile specimen, have shown that the maximum-distortion-energy theory gives more accurate results for pure-shear failure than the maximum-shear-stress theory. In fact, since 1sY> 132>1sY>22 = 1.15, the shear stress for yielding of the material, as given by the maximumdistortion-energy theory, is 15% more accurate than that given by the maximum-shear-stress theory. t s2 t s1 t 90 A (t, 0) t (b) (a) Fig. 10–32 s sY s1 sY Maximum-distortion-energy theory Fig. 10–30 s2 Pure shear sY (sY, sY) sY (sY,sY) sY sY s1 sY , sY 3 3 sY , sY 2 2 Fig. 10–31 10 524 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N Brittle Materials Maximum-Normal-Stress Theory. It was previously stated that brittle materials, such as gray cast iron, tend to fail suddenly by fracture with no apparent yielding. In a tension test, the fracture occurs when the normal stress reaches the ultimate stress sult , Fig. 10–33a. Also, brittle fracture occurs in a torsion test due to tension since the plane of fracture for an element is at 45° to the shear direction, Fig. 10–33b. The fracture surface is therefore helical as shown.* Experiments have further shown that during torsion the material’s strength is somewhat unaffected by the presence of the associated principal compressive stress being at right angles to the principal tensile stress. Consequently, the tensile stress needed to fracture a specimen during a torsion test is approximately the same as that needed to fracture a specimen in simple tension. Because of this, the maximum-normal-stress theory states that a brittle material will fail when the maximum tensile stress, s1, in the material reaches a value that is equal to the ultimate normal stress the material can sustain when it is subjected to simple tension. If the material is subjected to plane stress, we require that Failure of a brittle material in tension (a) ƒ s1 ƒ = sult 45 ƒ s2 ƒ = sult These equations are shown graphically in Fig. 10–34. Therefore, if the stress coordinates 1s1 , s22 at a point in the material fall on the boundary or outside the shaded area, the material is said to fracture. This theory is generally credited to W. Rankine, who proposed it in the mid-1800s. Experimentally it has been found to be in close agreement with the behavior of brittle materials that have stress–strain diagrams that are similar in both tension and compression. 45 Failure of a brittle material in torsion (b) Mohr’s Failure Criterion. In some brittle materials tension and Fig. 10–33 s2 sult 10 sult sult (10–31) s1 compression properties are different. When this occurs a criterion based on the use of Mohr’s circle may be used to predict failure. This method was developed by Otto Mohr and is sometimes referred to as Mohr’s failure criterion. To apply it, one first performs three tests on the material. A uniaxial tensile test and uniaxial compressive test are used to determine the ultimate tensile and compressive stresses 1sult2t and 1sult2c , respectively. Also a torsion test is performed to determine the material’s ultimate shear stress tult. Mohr’s circle for each of these stress sult Maximum-normal-stress theory Fig. 10–34 *A stick of blackboard chalk fails in this way when its ends are twisted with the fingers. 10.7 conditions is then plotted as shown in Fig. 10–35. These three circles are contained in a “failure envelope” indicated by the extrapolated colored curve that is drawn tangent to all three circles. If a plane-stress condition at a point is represented by a circle that has a point of tangency with the envelope, or if it extends beyond the envelope’s boundary, then failure is said to occur. We may also represent this criterion on a graph of principal stresses s1 and s2. This is shown in Fig. 10–36. Here failure occurs when the absolute value of either one of the principal stresses reaches a value equal to or greater than 1sult2t or 1sult2c or in general, if the state of stress at a point defined by the stress coordinates 1s1 , s22 is plotted on the boundary or outside the shaded area. Either the maximum-normal-stress theory or Mohr’s failure criterion can be used in practice to predict the failure of a brittle material. However, it should be realized that their usefulness is quite limited. A tensile fracture occurs very suddenly, and its initiation generally depends on stress concentrations developed at microscopic imperfections of the material such as inclusions or voids, surface indentations, and small cracks. Since each of these irregularities varies from specimen to specimen, it becomes difficult to specify fracture on the basis of a single test. 525 THEORIES OF FAILURE Failure envelope s (sult)c (sult)t tult t Fig. 10–35 s2 (sult)t (sult)c (sult)t s1 (sult)c Mohr’s failure criterion Fig. 10–36 Important Points • If a material is ductile, failure is specified by the initiation of yielding, whereas if it is brittle, it is specified by fracture. • Ductile failure can be defined when slipping occurs between the crystals that compose the material. This slipping is due to shear stress and the maximum-shear-stress theory is based on this idea. • Strain energy is stored in a material when it is subjected to normal stress. The maximum-distortion-energy theory depends on the strain energy that distorts the material, and not the part that increases its volume. • The fracture of a brittle material is caused only by the maximum tensile stress in the material, and not the compressive stress. This is the basis of the maximum-normal-stress theory, and it is applicable if the stress–strain diagram is similar in tension and compression. • If a brittle material has a stress–strain diagram that is different in tension and compression, then Mohr’s • failure criterion may be used to predict failure. Due to material imperfections, tensile fracture of a brittle material is difficult to predict, and so theories of failure for brittle materials should be used with caution. 10 526 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N EXAMPLE 10.12 The solid cast-iron shaft shown in Fig. 10–37a is subjected to a torque of T = 400 lb # ft. Determine its smallest radius so that it does not fail according to the maximum-normal-stress theory. A specimen of cast iron, tested in tension, has an ultimate stress of 1sult2t = 20 ksi. tmax s2 T 400 lbft s1 s T 400 lbft tmax r t (a) (b) Fig. 10–37 SOLUTION The maximum or critical stress occurs at a point located on the surface of the shaft. Assuming the shaft to have a radius r, the shear stress is tmax = 1400 lb # ft2112 in.>ft2r Tc 3055.8 lb # in. = = 4 J 1p>22r r3 Mohr’s circle for this state of stress (pure shear) is shown in Fig. 10–37b. Since R = tmax , then s1 = - s2 = tmax = 3055.8 lb # in. r3 The maximum-normal-stress theory, Eq. 10–31, requires ƒ s1 ƒ … sult 10 3055.8 lb # in. … 20 000 lb>in2 r3 Thus, the smallest radius of the shaft is determined from 3055.8 lb # in. = 20 000 lb>in2 r3 r = 0.535 in. Ans. 10.7 527 THEORIES OF FAILURE EXAMPLE 10.13 The solid shaft shown in Fig. 10–38a has a radius of 0.5 in. and is made of steel having a yield stress of sY = 36 ksi. Determine if the loadings cause the shaft to fail according to the maximum-shear-stress theory and the maximum-distortion-energy theory. SOLUTION The state of stress in the shaft is caused by both the axial force and the torque. Since maximum shear stress caused by the torque occurs in the material at the outer surface, we have -15 kip P = = - 19.10 ksi sx = 15 kip A p10.5 in.22 txy 3.25 kip # in. 10.5 in.2 Tc = = = 16.55 ksi p 4 J 2 10.5 in.2 0.5 in. A 3.25 kipin. (a) The stress components are shown acting on an element of material at point A in Fig. 10–38b. Rather than using Mohr’s circle, the principal stresses can also be obtained using the stress-transformation Eq. 9–5. s1,2 = = sx + sy 2 ; B a sx - sy 2 16.55 ksi 19.10 ksi 2 b + txy2 - 19.10 - 0 2 -19.10 + 0 ; a b + 116.5522 2 B 2 = - 9.55 ; 19.11 (b) Fig. 10–38 s1 = 9.56 ksi s2 = - 28.66 ksi Maximum-Shear-Stress Theory. Since the principal stresses have opposite signs, then from Sec. 9.5, the absolute maximum shear stress will occur in the plane, and therefore, applying the second of Eqs. 10–27, we have ƒ s1 - s2 ƒ … sY ƒ 9.56 - 1 -28.662 ƒ … 36 ? 38.2 7 36 Thus, shear failure of the material will occur according to this theory. Maximum-Distortion-Energy Theory. Applying Eq. 10–30, we have A s12 - s1s2 + s22 B … sY2 C 19.5622 - 19.5621 -28.662 + 1-28.6622 D … 13622 ? 1187 … 1296 Using this theory, failure will not occur. 10 528 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N PROBLEMS 10–59. A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of sx , sy , and txy . *10–60. A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms of sx , sy , and txy . Assume that the principal stresses are of different algebraic signs. *10–68. The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N # m and an axial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory. The ultimate stress of the concrete is sult = 28 MPa. 2 kN 500 Nm •10–61. An aluminum alloy 6061-T6 is to be used for a solid drive shaft such that it transmits 40 hp at 2400 rev>min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-shear-stress theory. 500 Nm 10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory. 2 kN Prob. 10–68 10–63. An aluminum alloy is to be used for a drive shaft such that it transmits 25 hp at 1500 rev>min. Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-distortion-energy theory. sY = 3.5 ksi. *10–64. A bar with a square cross-sectional area is made of a material having a yield stress of sY = 120 ksi. If the bar is subjected to a bending moment of 75 kip # in., determine the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with respect to yielding. •10–69. Cast iron when tested in tension and compression has an ultimate strength of 1sult2t = 280 MPa and 1sult2c = 420 MPa, respectively. Also, when subjected to pure torsion it can sustain an ultimate shear stress of tult = 168 MPa. Plot the Mohr’s circles for each case and establish the failure envelope. If a part made of this material is subjected to the state of plane stress shown, determine if it fails according to Mohr’s failure criterion. 120 MPa 100 MPa •10–65. Solve Prob. 10–64 using the maximum-shearstress theory. 10–66. Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular cross 10 section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. 10–67. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. 220 MPa Prob. 10–69 10–70. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same maximum shear stress as the combination of an applied moment M and torque T. Assume that the principal stresses are of opposite algebraic signs. 10.7 529 THEORIES OF FAILURE 10–71. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-shearstress theory. 10–75. If the A-36 steel pipe has outer and inner diameters of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-shear-stress theory. *10–72. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. *10–76. If the A-36 steel pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-distortion-energy theory. 900 N 60 MPa 200 mm 150 mm A 40 MPa 100 mm 200 mm 70 MPa Probs. 10–71/72 900 N •10–73. If the 2-in. diameter shaft is made from brittle material having an ultimate strength of sult = 50 ksi for both tension and compression, determine if the shaft fails according to the maximum-normal-stress theory. Use a factor of safety of 1.5 against rupture. 10–74. If the 2-in. diameter shaft is made from cast iron having tensile and compressive ultimate strengths of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively, determine if the shaft fails in accordance with Mohr’s failure criterion. Probs. 10–75/76 •10–77. The element is subjected to the stresses shown. If sY = 36 ksi, determine the factor of safety for the loading based on the maximum-shear-stress theory. 10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory. 12 ksi 4 ksi 8 ksi Probs. 10–77/78 30 kip 4 kip · ft Probs. 10–73/74 10–79. The yield stress for heat-treated beryllium copper is sY = 130 ksi. If this material is subjected to plane stress and elastic failure occurs when one principal stress is 145 ksi, what is the smallest magnitude of the other principal stress? Use the maximum-distortion-energy theory. 10 530 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N *10–80. The plate is made of hard copper, which yields at sY = 105 ksi. Using the maximum-shear-stress theory, determine the tensile stress sx that can be applied to the plate if a tensile stress sy = 0.5sx is also applied. •10–85. The state of stress acting at a critical point on a machine element is shown in the figure. Determine the smallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stress theory. •10–81. Solve Prob. 10–80 using the maximum-distortionenergy theory. sy 0.5sx sx 10 ksi 4 ksi 8 ksi Probs. 10–80/81 10–82. The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based on the maximumshear-stress theory. Prob. 10–85 10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory. 10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t, and s3 = 0. If the yield stress is sY, determine the maximum value of p based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory. 25 ksi 80 ksi 10–87. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is tallow = 116>pd322M2 + T2. Assume the principal stresses to be of opposite algebraic signs. *10–88. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is sallow = 116>pd321M + 2M2 + T22. 10 Probs. 10–82/83 *10–84. A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of sY = 150 ksi. If the bar is subjected to a torque of 30 kip # in. and a bending moment of 56 kip # in., determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding. T T M M Probs. 10–87/88 10.7 •10–89. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-shear-stress theory. Use a factor of safety of 1.5 against yielding. THEORIES OF FAILURE 531 *10–92. The gas tank has an inner diameter of 1.50 m and a wall thickness of 25 mm. If it is made from A-36 steel and the tank is pressured to 5 MPa, determine the factor of safety against yielding using (a) the maximum-shear-stress theory, and (b) the maximum-distortion-energy theory. 10–90. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-distortion-energy theory. Use a factor of safety of 1.5 against yielding. A 80 mm B T C 80 mm 100 mm Prob. 10–92 T •10–93. Probs. 10–89/90 10–91. The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb # ft, a bending moment of 1500 lb # ft, and an axial thrust of 2500 lb. If the yield points for tension and shear are sY = 100 ksi and tY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shearstress theory. The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding. 10 2300 lbft 1500 lbft 2500 lb Prob. 10–91 Prob. 10–93 532 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N CHAPTER REVIEW When an element of material is subjected to deformations that only occur in a single plane, then it undergoes plane strain. If the strain components Px , Py , and gxy are known for a specified orientation of the element, then the strains acting for some other orientation of the element can be determined using the plane-strain transformation equations. Likewise, the principal normal strains and maximum in-plane shear strain can be determined using transformation equations. Px + Py Px¿ = Px + Py Py¿ = gx¿y¿ 2 P1,2 = gmax in-plane 2 Pavg = 2 2 2 B Px - Py Px - Py Px + Py = 2 - 2 = -¢ Px - Py + 2 a ; cos 2u gxy ≤ sin 2u + B Px - Py 2 cos 2u + a 2 Px - Py 2 2 b + a gxy 2 gxy 2 2 b sin 2u cos 2u b + a gxy 2 sin 2u gxy 2 b 2 2 Px + Py 2 Strain transformation problems can also be solved in a semi-graphical manner using Mohr’s circle. To draw the circle, the P and g>2 axes are established and the center of the circle C [1Px + Py2>2, 0] and the “reference point” A 1Px , gxy>22 are plotted. The radius of the circle extends between these two points and is determined from trigonometry. Px P y 2 C gxy Pavg g 2 R If P1 and P2 have the same sign then the absolute maximum shear strain will be out of plane. 10 In the case of plane strain, the absolute maximum shear strain will be equal to the maximum in-plane shear strain provided the principal strains P1 and P2 have opposite signs. gabs = P1 max in-plane = P gmax 1 - P2 gabs = P1 - P2 max 2 Px Py 2 Px Px Py 2 2 A u 0 gxy 2 2 P CHAPTER REVIEW If the material is subjected to triaxial stress, then the strain in each direction is influenced by the strain produced by all three stresses. Hooke’s law then involves the material properties E and n. If E and n are known, then G can be determined. Px = 1 [s - n1sy + sz2] E x Py = 1 [s - n1sx + sz2] E y Pz = 1 [s - n1sx + sy2] E z G = E 211 + n2 The dilatation is a measure of volumetric strain. e = 1 - 2n 1sx + sy + sz2 E The bulk modulus is used to measure the stiffness of a volume of material. k = E 311 - 2n2 533 If the principal stresses at a critical point in the material are known, then a theory of failure can be used as a basis for design. Ductile materials fail in shear, and here the maximum-shear-stress theory or the maximum-distortion-energy theory can be used to predict failure. Both of these theories make comparison to the yield stress of a specimen subjected to a uniaxial tensile stress. Brittle materials fail in tension, and so the maximum-normal-stress theory or Mohr’s failure criterion can be used to predict failure. Here comparisons are made with the ultimate tensile stress developed in a specimen. 10 534 C H A P T E R 10 S T R A I N T R A N S F O R M AT I O N REVIEW PROBLEMS 10–94. A thin-walled spherical pressure vessel has an inner radius r, thickness t, and is subjected to an internal pressure p. If the material constants are E and n, determine the strain in the circumferential direction in terms of the stated parameters. 10–95. The strain at point A on the shell has components Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. •10–97. The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is sY = 650 MPa. 340 MPa 65 MPa 55 MPa y x A Prob. 10–95 *10–96. The principal plane stresses acting at a point are shown in the figure. If the material is machine steel having a yield stress of sY = 500 MPa, determine the factor of safety with respect to yielding if the maximum-shear-stress theory is considered. Prob. 10–97 10–98. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 600110-62, Pb = - 700110-62, and Pc = 350110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. a 60 60 100 MPa b 60 10 150 MPa c Prob. 10–96 Prob. 10–98 535 REVIEW PROBLEMS 10–99. A strain gauge forms an angle of 45° with the axis of the 50-mm diameter shaft. If it gives a reading of P = - 200110-62 when the torque T is applied to the shaft, determine the magnitude of T. The shaft is made from A-36 steel. 10–102. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = - 300110-62. Determine the equivalent state of strain on an element at the same point oriented 30° clockwise with respect to the original element. Sketch the results on the element. y T 45 Pydy dy gxy 2 T gxy 2 dx Prob. 10–99 *10–100. The A-36 steel post is subjected to the forces shown. If the strain gauges a and b at point A give readings of Pa = 300110-62 and Pb = 175110-62, determine the magnitudes of P1 and P2. P1 P2 a 2 in. 2 ft A 1 in. b 45 A A c 4 in. c x Pxdx Prob. 10–102 10–103. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding element at the point with respect to the original element. Sketch the results on the element. Section c– c 1 in. y Pydy Prob. 10–100 dy 10–101. A differential element is subjected to plane strain that has the following components: Px = 950110-62, Py = 420110-62, gxy = - 325110-62. Use the strain-transformation equations and determine (a) the principal strains and (b) the maximum in-plane shear strain and the associated average strain. In each case specify the orientation of the element and show how the strains deform the element. 10 gxy 2 gxy 2 dx Prob. 10–103 x Pxdx Beams are important structural members that are used to support roof and floor loadings. 10–1. Prove that the sum of the normal strains in perpendicular directions is constant. ex¿ = ey¿ = ex + ey 2 ex + ey 2 + - ex - ey 2 ex - ey 2 cos 2u + cos 2u - gxy 2 gxy 2 sin 2u (1) sin 2u (2) Adding Eq. (1) and Eq. (2) yields: ex¿ + ey¿ = ex + ey = constant QED 738 10–2. The state of strain at the point has components of Px = 200 110-62, Py = -300 110-62, and gxy = 400(10-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. y x In accordance to the established sign convention, ex = 200(10 - 6), ex¿ = ex + ey 2 = c + ey = -300(10 - 6) ex - ey 2 cos 2u + gxy 2 gxy = 400(10 - 6) u = 30° sin 2u 200 + (-300) 200 - (-300) 400 + cos 60° + sin 60° d (10 - 6) 2 2 2 = 248 (10 - 6) gx¿y¿ 2 = -a Ans. ex - ey 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = e - C 200 - ( -300) D sin 60° + 400 cos 60° f(10 - 6) = -233(10 - 6) ey¿ = ex + ey = c 2 - Ans. ex - ey 2 cos 2u - gxy 2 sin 2u 200 - (-300) 200 + ( -300) 400 cos 60° sin 60° d(10 - 6) 2 2 2 = -348(10 - 6) Ans. The deformed element of this equivalent state of strain is shown in Fig. a 739 10–3. A strain gauge is mounted on the 1-in.-diameter A-36 steel shaft in the manner shown. When the shaft is rotating with an angular velocity of v = 1760 rev>min, the reading on the strain gauge is P = 800110-62. Determine the power output of the motor. Assume the shaft is only subjected to a torque. v = (1760 rev>min)a 60⬚ 2p rad 1 min ba b = 184.307 rad>s 60 sec 1 rev ex = ey = 0 ex¿ = ex + ey 2 + ex - ey 2 800(10 - 6) = 0 + 0 + cos 2u + gxy 2 gxy 2 sin 2u sin 120° gxy = 1.848(10 - 3) rad t = G gxy = 11(103)(1.848)(10 - 3) = 20.323 ksi t = Tc ; J 20.323 = T(0.5) p 2 (0.5)4 ; T = 3.99 kip # in = 332.5 lb # ft P = Tv = 0.332.5 (184.307) = 61.3 kips # ft>s = 111 hp Ans. 740 *10–4. The state of strain at a point on a wrench has components Px = 120110-62, Py = - 180110-62, gxy = 150110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. ex = 120(10 - 6) e1, 2 = a) ey = -180(10 - 6) gxy = 150(10 - 6) Ex - Ey 2 gxy 2 ex + ey ; a b + a b 2 A 2 2 120 + (-180) 120 - ( -180) 2 150 2 -6 ; a b + a b d 10 2 A 2 2 e1 = 138(10 - 6); e2 = -198(10 - 6) = c Ans. Orientation of e1 and e2 gxy 150 = tan 2up = = 0.5 ex - ey [120 - (-180)] up = 13.28° and -76.72° Use Eq. 10.5 to determine the direction of e1 and e2 ex¿ = ex + ey + 2 ex - ey 2 cos 2u + gxy 2 sin 2u u = up = 13.28° ex¿ = c 120 + (-180) 120 - ( -180) 150 + cos (26.56°) + sin 26.56° d 10 - 6 2 2 2 = 138 (10 - 6) = e1 Therefore up1 = 13.3° ; g b) max in-plane = 2 in-plane ex + ey 2 A ex - ey b + a 2 gxy b Ans. 2 2 2 150 2 120 - ( -180) 2 -6 -6 = 2c a b + a b d10 = 335 (10 ) A 2 2 gmax eavg = a up2 = -76.7° = c 120 + (-180) d 10 - 6 = -30.0(10 - 6) 2 Ans. Ans. Orientation of gmax tan 2us = -(ex - ey) gxy = -[120 - ( -180)] = - 2.0 150 us = -31.7° and 58.3° Ans. gmax Use Eq. 10–6 to determine the sign of in-plane gx¿y¿ ex - ey gxy = sin 2u + cos 2u 2 2 2 u = us = -31.7° gx¿y¿ = 2c - 120 - (-180) 150 sin (-63.4°) + cos (-63.4°) d10 - 6 = 335(10 - 6) 2 2 741 10–5. The state of strain at the point on the arm has components Px = 250110-62, Py = -450110-62, gxy = -825110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. ex = 250(10 - 6) ey = -450(10 - 6) y gxy = -825(10 - 6) x a) ex + ey e1, 2 = ; 2 = c A ex - ey a 2 2 b + a gxy 2 b 2 250 - 450 250 - ( -450) 2 -825 2 -6 ; a b + a b d(10 ) 2 A 2 2 e1 = 441(10 - 6) Ans. e2 = -641(10 - 6) Ans. Orientation of e1 and e2 : gxy tan 2up = ex - ey up = -24.84° -825 250 - (-450) = up = 65.16° and Use Eq. 10–5 to determine the direction of e1 and e2: ex¿ = ex + ey + 2 ex - ey 2 cos 2u + gxy 2 sin 2u u = up = -24.84° ex¿ = c 250 - (-450) 250 - 450 -825 + cos ( -49.69°) + sin (-49.69°) d(10 - 6) = 441(10 - 6) 2 2 2 Therefore, up1 = -24.8° Ans. up2 = 65.2° Ans. b) g max in-plane 2 g max in-plane eavg = = A = 2c a ex - ey A 2 a ex + ey 2 b + a 2 gxy 2 b 2 250 - (-450) 2 -825 2 -6 -3 b + a b d(10 ) = 1.08(10 ) 2 2 = a 250 - 450 b (10 - 6) = -100(10 - 6) 2 Ans. Ans. 742 10–6. The state of strain at the point has components of Px = -100110-62, Py = 400110-62, and gxy = -300110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 60° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. y x In accordance to the established sign convention, ex = -100(10 - 6) ex¿ = ex + ey 2 = c + ey = 400(10 - 6) ex - ey 2 gxy cos 2u + 2 gxy = -300(10 - 6) u = 60° sin 2u -100 - 400 -300 -100 + 400 + cos 120° + sin 120° d(10 - 6) 2 2 2 = 145(10 - 6) gx¿y¿ 2 = -a Ans. ex - ey 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = c -(-100 - 400) sin 120° + (-300) cos 120° d (10 - 6) = 583(10 - 6) ey¿ = ex + ey = c 2 - Ans. ex - ey 2 cos 2u - gxy 2 sin 2u -100 + 400 -100 - 400 -300 cos 120° sin 120° d(10 - 6) 2 2 2 = 155 (10 - 6) Ans. The deformed element of this equivalent state of strain is shown in Fig. a 743 10–7. The state of strain at the point has components of Px = 100110-62, Py = 300110-62, and gxy = -150110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented u = 30° clockwise. Sketch the deformed element due to these strains within the x–y plane. y x In accordance to the established sign convention, ex = 100(10 - 6) ex¿ = ex + ey 2 = c + ey = 300(10 - 6) ex - ey 2 cos 2u + gxy = -150(10 - 6) gxy 2 u = - 30° sin 2u 100 - 300 -150 100 + 300 + cos (-60°) + sin ( -60°) d (10 - 6) 2 2 2 = 215(10 - 6) gx¿y¿ 2 = -a Ans. ex - ey 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = c -(100 - 300) sin ( -60°) + ( -150) cos ( -60°) d(10 - 6) = -248 (10 - 6) ey¿ = ex + ey = c 2 - Ans. ex - ey 2 cos 2u - gxy 2 sin 2u 100 - 300 -150 100 + 300 cos ( -60°) sin (-60°) d (10 - 6) 2 2 2 = 185(10 - 6) Ans. The deformed element of this equivalent state of strain is shown in Fig. a 744 *10–8. The state of strain at the point on the bracket has components Px = -200110-62, Py = -650110-62, gxy ⫽ -175110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 20° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. ex = -200(10 - 6) ex¿ = ex + ey 2 = c + ey = -650(10 - 6) ex - ey 2 cos 2u + gxy 2 y x gxy = -175(10 - 6) u = 20° sin 2u (-200) - (-650) (-175) -200 + (-650) + cos (40°) + sin (40°) d (10 - 6) 2 2 2 = -309(10 - 6) ey¿ = ex + ey 2 = c - Ans. ex - ey 2 cos 2u - gxy 2 sin 2u -200 - ( -650) ( -175) -200 + (-650) cos (40°) sin (40°) d(10 - 6) 2 2 2 = -541(10 - 6) gx¿y¿ 2 = - ex - ey 2 Ans. sin 2u + gxy 2 cos 2u gx¿y¿ = [-(-200 - (-650)) sin (40°) + (-175) cos (40°)](10 - 6) = -423(10 - 6) Ans. 745 10–9. The state of strain at the point has components of Px = 180110-62, Py = -120110-62, and gxy = -100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. y x a) In accordance to the established sign convention, ex = 180(10 - 6), ey = -120(10 - 6) and gxy = -100(10 - 6). ex + ey e1, 2 = ; 2 = b a A ex - ey 2 2 b + a gxy 2 b 2 180 + (-120) 180 - ( -120) 2 -100 2 -6 ; c d + a b r (10 ) 2 A 2 2 = A 30 ; 158.11 B (10 - 6) e1 = 188(10 - 6) tan 2uP = e2 = -128(10 - 6) gxy ex - ey Ans. -100(10 - 6) = C 180 - (-120) D (10 - 6) uP = -9.217° and = -0.3333 80.78° Substitute u = -9.217°, ex + ey ex¿ = 2 = c + ex - ey 2 cos 2u + gxy 2 sin 2u 180 + ( -120) 180 - ( -120) -100 + cos (-18.43°) + sin (-18.43) d(10 - 6) 2 2 2 = 188(10 - 6) = e1 Thus, (uP)1 = -9.22° (uP)2 = 80.8° Ans. The deformed element is shown in Fig (a). gmax ex - ey 2 gxy 2 in-plane = b) a b + a b 2 A 2 2 gmax in-plane tan 2us = - a = b2 A ex - ey gxy c 180 - ( -120) 2 -100 2 -6 -6 d + a b r (10 ) = 316 A 10 B 2 2 b = -c C 180 - (-120) D (10 - 6) us = 35.78° = 35.8° and -100(10 - 6) Ans. s = 3 -54.22° = -54.2° Ans. 746 10–9. Continued gmax The algebraic sign for in-plane when u = 35.78°. ex - ey gxy gx¿y¿ = -a b sin 2u + cos 2u 2 2 2 gx¿y¿ = e - C 180 - ( -120) D sin 71.56° + ( -100) cos 71.56° f(10 - 6) eavg = -316(10 - 6) ex + ey 180 + (-120) = = c d(10 - 6) = 30(10 - 6) 2 2 Ans. The deformed element for the state of maximum In-plane shear strain is shown is shown in Fig. b 747 10–10. The state of strain at the point on the bracket has components Px = 400110-62, Py = -250110-62, gxy ⫽ 310110-62. Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 30° clockwise from the original position. Sketch the deformed element due to these strains within the x–y plane. ex = 400(10 - 6) ex¿ = ex + ey 2 = c + ey = -250(10 - 6) ex - ey 2 cos 2u + gxy 2 gxy = 310(10 - 6) y x u = - 30° sin 2u 400 - ( -250) 400 + (-250) 310 + cos (-60°) + a b sin ( -60°) d(10 - 6) 2 2 2 = 103(10 - 6) ey¿ = ex + ey 2 = c - Ans. ex - ey 2 cos 2u - gxy 2 sin 2u 400 - (-250) 400 + (-250) 310 cos (60°) sin (-60°) d (10 - 6) 2 2 2 = 46.7(10 - 6) gx¿y¿ 2 = - ex - ey 2 Ans. sin 2u + gxy 2 cos 2u gx¿y¿ = [-(400 - (-250)) sin (-60°) + 310 cos ( -60°)](10 - 6) = 718(10 - 6) 748 Ans. 10–11. The state of strain at the point has components of Px = -100110-62, Py = -200110-62, and gxy = 100110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. In accordance to the established ey = -200(10 - 6) and gxy = 100(10 - 6). ex + ey e1, 2 = ; 2 = b A a ex - ey 2 b + a 2 gxy 2 b sign y x convention, ex = -100(10 - 6), 2 -100 + (-200) 100 2 -100 - (-200) 2 -6 ; c d + a b r (10 ) 2 A 2 2 = A -150 ; 70.71 B (10 - 6) e1 = -79.3(10 - 6) tan 2uP = gxy e2 = - 221(10 - 6) 100(10 - 6) = ex - ey C -100 - ( -200) D (10 - 6) uP = 22.5° Ans. = 1 -67.5° and Substitute u = 22.5, ex + ey ex¿ = + ex - ey cos 2u + gxy sin 2u 2 2 2 -100 + (-200) -100 - (-200) 100 = c + cos 45° + sin 45° d(10 - 6) 2 2 2 = -79.3(10 - 6) = e1 Thus, (uP)1 = 22.5° (uP)2 = -67.5° Ans. The deformed element of the state of principal strain is shown in Fig. a gmax ex - ey 2 gxy 2 in-plane = a b + a b 2 A 2 2 gmax in-plane = b2 tan 2us = - a -100 - ( -200) 2 100 2 -6 -6 d + a b r (10 ) = 141(10 ) A 2 2 c ex - ey gxy b = -c us = -22.5° The algebraic sign for gx¿y¿ 2 = -a ex - ey 2 C -100 - ( -200) D (10 - 6) 100(10 - 6) and gmax in-plane b sin 2u + Ans. s = -1 Ans. 67.5° when u = -22.5°. gxy 2 cos 2u gx¿y¿ = - C -100 - (-200) D sin ( -45°) + 100 cos (-45°) eavg = 141(10 - 6) ex + ey -100 + ( -200) = = c d(10 - 6) = -150(10 - 6) 2 2 Ans. The deformed element for the state of maximum In-plane shear strain is shown in Fig. b. 749 10–11. Continued *10–12. The state of plane strain on an element is given by Px = 500110-62, Py = 300110-62, and gxy = - 200110-62. Determine the equivalent state of strain on an element at the same point oriented 45° clockwise with respect to the original element. y Pydy dy gxy 2 Strain Transformation Equations: ex = 500 A 10 - 6 B ey = 300 A 10 - 6 B gxy = -200 A 10 - 6 B u = -45° We obtain ex¿ = ex + ey 2 = c + ex - ey 2 cos 2u + gxy 2 sin 2u 500 + 300 500 - 300 -200 + cos (-90°) + a b sin ( -90°) d A 10 - 6 B 2 2 2 = 500 A 10 - 6 B gx¿y¿ 2 = -a Ans. ex - ey 2 b sin 2u + gxy 2 cos 2u gx¿y¿ = [-(500 - 300) sin ( -90°) + (-200) cos ( -90°)] A 10 - 6 B = 200 A 10 - 6 B ey¿ = ex + ey = c 2 - Ans. ex - ey 2 cos 2u - gxy 2 sin 2u 500 + 300 500 - 300 -200 cos ( -90°) - a b sin (-90°) d A 10 - 6 B 2 2 2 = 300 A 10 - 6 B Ans. The deformed element for this state of strain is shown in Fig. a. 750 gxy 2 dx x Pxdx 10–13. The state of plane strain on an element is Px = -300110-62, Py = 0, and gxy = 150110-62. Determine the equivalent state of strain which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element. y gxy dy 2 x In-Plane Principal Strains: ex = -300 A 10 - 6 B , ey = 0, and gxy = 150 A 10 - 6 B . We obtain ex + ey e1, 2 = 2 = C ; C ¢ ex - ey 2 2 ≤ + ¢ gxy 2 ≤ 2 -300 - 0 2 150 2 -300 + 0 ; ¢ ≤ + ¢ ≤ S A 10 - 6 B 2 C 2 2 = (-150 ; 167.71) A 10 - 6 B e1 = 17.7 A 10 - 6 B e2 = - 318 A 10 - 6 B Ans. Orientation of Principal Strain: tan 2up = gxy ex - ey = 150 A 10 - 6 B (-300 - 0) A 10 - 6 B = -0.5 uP = -13.28° and 76.72° Substituting u = -13.28° into Eq. 9-1, ex¿ = ex + ey = c + 2 ex - ey 2 cos 2u + gxy 2 sin 2u -300 + 0 -300 - 0 150 + cos (-26.57°) + sin (-26.57°) d A 10 - 6 B 2 2 2 = -318 A 10 - 6 B = e2 Thus, A uP B 1 = 76.7° and A uP B 2 = - 13.3° Ans. The deformed element of this state of strain is shown in Fig. a. Maximum In-Plane Shear Strain: gmax ex - ey 2 gxy 2 in-plane = ¢ ≤ + ¢ ≤ 2 C 2 2 gmax in-plane -300 - 0 2 150 2 -6 -6 b + a b R A 10 B = 335 A 10 B A 2 2 = B2 a Ans. Orientation of the Maximum In-Plane Shear Strain: tan 2us = - ¢ ex - ey gxy ≤ = -C (-300 - 0) A 10 - 6 B 150 A 10 - 6 B S = 2 us = 31.7° and 122° Ans. 751 gxy 2 dx Pxdx 10–13. Continued The algebraic sign for gx¿y¿ 2 = -¢ ex - ey 2 gmax in-plane ≤ sin 2u + when u = us = 31.7° can be obtained using gxy 2 cos 2u gx¿y¿ = [-(-300 - 0) sin 63.43° + 150 cos 63.43°] A 10 - 6 B = 335 A 10 - 6 B Average Normal Strain: eavg = ex + ey 2 = a -300 + 0 b A 10 - 6 B = -150 A 10 - 6 B 2 Ans. The deformed element for this state of strain is shown in Fig. b. 752 10–14. The state of strain at the point on a boom of an hydraulic engine crane has components of Px = 250110-62, Py = 300110-62, and gxy = - 180110-62. Use the straintransformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case, specify the orientation of the element and show how the strains deform the element within the x–y plane. y a) In-Plane Principal Strain: Applying Eq. 10–9, ex + ey e1, 2 = ; 2 = B a A ex - ey 2 b + a 2 gxy 2 b 2 250 + 300 -180 2 250 - 300 2 -6 ; a b + a b R A 10 B 2 A 2 2 = 275 ; 93.41 e1 = 368 A 10 - 6 B e2 = 182 A 10 - 6 B Ans. Orientation of Principal Strain: Applying Eq. 10–8, gxy tan 2uP = ex - ey = -180(10 - 6) (250 - 300)(10 - 6) uP = 37.24° and = 3.600 -52.76° Use Eq. 10–5 to determine which principal strain deforms the element in the x¿ direction with u = 37.24°. ex¿ = ex + ey = c 2 + ex - ey 2 cos 2u + gxy 2 sin 2u 250 + 300 250 - 300 -180 + cos 74.48° + sin 74.48° d A 10 - 6 B 2 2 2 = 182 A 10 - 6 B = e2 Hence, uP1 = -52.8° and uP2 = 37.2° Ans. b) Maximum In-Plane Shear Strain: Applying Eq. 10–11, g max ex - ey 2 gxy 2 in-plane = a b + a b 2 A 2 2 g max in-plane = 2B 250 - 300 2 -180 2 -6 b + a b R A 10 B A 2 2 a = 187 A 10 - 6 B Ans. 753 x 10–14. Continued Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10, tan 2us = - ex - ey gxy us = -7.76° and The proper sign of gx¿y¿ 2 = - = - ex - ey 2 g max in-plane 250 - 300 = -0.2778 -180 Ans. 82.2° can be determined by substituting u = -7.76° into Eq. 10–6. sin 2u + gxy 2 cos 2u gx¿y¿ = {-[250 - 300] sin (-15.52°) + ( -180) cos (-15.52°)} A 10 - 6 B = -187 A 10 - 6 B Normal Strain and Shear strain: In accordance with the sign convention, ex = 250 A 10 - 6 B ey = 300 A 10 - 6 B gxy = -180 A 10 - 6 B Average Normal Strain: Applying Eq. 10–12, eavg = ex + ey 2 = c 250 + 300 d A 10 - 6 B = 275 A 10 - 6 B 2 Ans. 754 *10–16. The state of strain at a point on a support has components of Px = 350110-62, Py = 400110-62, gxy = -675110-62. Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane. a) e1, 2 = = ex + ey ; 2 B a ex -ey 2 b + a 2 gxy 2 b 2 350 - 400 2 -675 2 350 + 400 ; a b + a b 2 A 2 2 e1 = 713(10 - 6) Ans. e2 = 36.6(10 - 6) Ans. tan 2uP = gxy ex - ey = -675 (350 - 400) uP = 42.9° Ans. b) (gx¿y¿)max 2 (gx¿y¿)max 2 = = A A a ex - ey a 350 - 400 2 -675 2 b + a b 2 2 2 b + a 2 gxy 2 b 2 (gx¿y¿)max = 677(10 - 6) eavg = ex + ey tan 2us = 2 = 350 + 400 = 375(10 - 6) 2 -(ex - ey) gxy Ans. = Ans. 350 - 400 675 us = -2.12° Ans. 755 •10–17. Solve part (a) of Prob. 10–4 using Mohr’s circle. ex = 120(10 - 6) ey = -180(10 - 6) gxy = 150(10 - 6) A (120, 75)(10 - 6) C (-30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) e1 = (-30 + 167.71)(10 - 6) = 138(10 - 6) Ans. e2 = (-30 - 167.71)(10 - 6) = - 198(10 - 6) Ans. tan 2uP = a Ans. 75 b, 30 + 120 uP = 13.3° 756 10–18. Solve part (b) of Prob. 10–4 using Mohr’s circle. ex = 120(10 - 6) ey = -180(10 - 6) gxy = 150(10 - 6) A (120, 75)(10 - 6) C ( -30, 0)(10 - 6) R = C 2[120 - (-30)]2 + (75)2 D (10 - 6) = 167.71 (10 - 6) gxy max 2 in-plane gxy = R = 167.7(10 - 6) max in-plane = 335(10 - 6) Ans. eavg = -30 (10 - 6) tan 2us = 120 + 30 75 Ans. us = -31.7° Ans. 757 10–19. Solve Prob. 10–8 using Mohr’s circle. ex = -200(10 - 6) ey = -650(10 - 6) gxy = -175(10 - 6) gxy 2 = -87.5(10 - 6) u = 20°, 2u = 40° A(-200, -87.5)(10 - 6) C(-425, 0)(10 - 6) R = [2(-200 - (-425))2 + 87.52 ](10 - 6) = 241.41(10 - 6) tan a = 87.5 ; -200 - (-425) a = 21.25° f = 40 + 21.25 = 61.25° ex¿ = (-425 + 241.41 cos 61.25°)(10 - 6) = -309(10 - 6) Ans. ey¿ = ( -425 - 241.41 cos 61.25°)(10 - 6) = -541(10 - 6) Ans. -gx¿y¿ 2 = 241.41(10 - 6) sin 61.25° gx¿y¿ = -423(10 - 6) Ans. 758 *10–20. Solve Prob. 10–10 using Mohr’s circle. ex = 400(10 - 6) A(400, 155)(10 - 6) ey = -250(10 - 6) gxy = 310(10 - 6) gxy 2 = 155(10 - 6) C(75, 0)(10 - 6) R = [2(400 - 75)2 + 1552 ](10 - 6) = 360.1(10 - 6) tan a = 155 ; 400 - 75 a = 25.50° f = 60 + 25.50 = 85.5° ex¿ = (75 + 360.1 cos 85.5°)(10 - 6) = 103(10 - 6) Ans. ey¿ = (75 - 360.1 cos 85.5°)(10 - 6) = 46.7(10 - 6) Ans. gx¿y¿ 2 = (360.1 sin 85.5°)(10 - 6) gx¿y¿ = 718(10 - 6) Ans. 759 u = 30° •10–21. Solve Prob. 10–14 using Mohr’s circle. Construction of the Circle: In accordance with the sign convention, ex = 250 A 10 - 6 B , gxy ey = 300 A 10 - 6 B , and = -90 A 10 - 6 B . Hence, 2 eavg = ex + ey 2 = a 250 + 300 b A 10 - 6 B = 275 A 10 - 6 B 2 Ans. The coordinates for reference points A and C are A(250, -90) A 10 - 6 B C(275, 0) A 10 - 6 B The radius of the circle is R = a 2(275 - 250)2 + 902 b A 10 - 6 B = 93.408 In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (275 + 93.408) A 10 - 6 B = 368 A 10 - 6 B Ans. e2 = (275 - 93.408) A 10 - 6 B = 182 A 10 - 6 B Ans. Orientation of Principal Strain: From the circle, tan 2uP2 = 90 = 3.600 275 - 250 2uP2 = 74.48° 2uP1 = 180° - 2uP2 uP1 = 180° - 74.78° = 52.8° (Clockwise) 2 Ans. Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane 2 g = -R = -93.408 A 10 - 6 B max in-plane = -187 A 10 - 6 B Ans. Orientation of the Maximum In-Plane Shear Strain: From the circle, tan 2us = 275 - 250 = 0.2778 90 us = 7.76° (Clockwise) Ans. 760 10–22. The strain at point A on the bracket has components Px = 300110-62, Py = 550110-62, gxy = -650110-62. Determine (a) the principal strains at A in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 300(10 - 6) ey = 550(10 - 6) A(300, -325)10 - 6 gxy = -650(10 - 6) y gxy 2 = -325(10 - 6) A C(425, 0)10 - 6 R = C 2(425 - 300)2 + (-325)2 D 10 - 6 = 348.2(10 - 6) a) e1 = (425 + 348.2)(10 - 6) = 773(10 - 6) Ans. e2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6) Ans. b) g max in-plane = 2R = 2(348.2)(10 - 6) = 696(10 - 6) Ans. 773(10 - 6) ; 2 Ans. c) gabs max = 2 gabs max = 773(10 - 6) 10–23. The strain at point A on the leg of the angle has components Px = - 140110-62, Py = 180110-62, gxy = -125110-62. Determine (a) the principal strains at A in the x– y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = -140(10 - 6) A( -140, -62.5)10 - 6 ey = 180(10 - 6) gxy = -125(10 - 6) A gxy 2 = -62.5(10 - 6) C(20, 0)10 - 6 R = A 2(20 - (-140))2 + (-62.5)2 B 10 - 6 = 171.77(10 - 6) a) e1 = (20 + 171.77)(10 - 6) = 192(10 - 6) Ans. e2 = (20 - 171.77)(10 - 6) = - 152(10 - 6) Ans. (b, c) gabs max = g max in-plane = 2R = 2(171.77)(10 - 6) = 344(10 - 6) Ans. 761 x *10–24. The strain at point A on the pressure-vessel wall has components Px = 480110-62, Py = 720110-62, gxy = 650110-62. Determine (a) the principal strains at A, in the x–y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 480(10 - 6) ey = 720(10 - 6) A(480, 325)10 - 6 C(600, 0)10 - 6 gxy = 650(10 - 6) y A gxy 2 = 325(10 - 6) R = (2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6) a) e1 = (600 + 346.44)10 - 6 = 946(10 - 6) Ans. e2 = (600 - 346.44)10 - 6 = 254(10 - 6) Ans. b) g max in-plane = 2R = 2(346.44)10 - 6 = 693(10 - 6) Ans. 946(10 - 6) ; 2 Ans. c) gabs max 2 = gabs max = 946(10 - 6) 762 x •10–25. The 60° strain rosette is mounted on the bracket. The following readings are obtained for each gauge: Pa = -100110-62, Pb = 250110-62, and Pc = 150110-62. Determine (a) the principal strains and (b) the maximum inplane shear strain and associated average normal strain. In each case show the deformed element due to these strains. b c 60⬚ 60⬚ This is a 60° strain rosette Thus, ex = ea = -100(10 - 6) 1 A 2eb + 2ec - ea B 3 ey = 1 C 2(250) + 2(150) - (-100) D (10 - 6) 3 = = 300(10 - 6) gxy = 2 23 (eb - ec) = 2 23 (250 - 150)(10 - 6) = 115.47(10 - 6) In accordance to the established sign convention, ex = -100(10 - 6), ey = 300(10 - 6) gxy and = 57.74(10 - 6). 2 Thus, eavg = ex + ey 2 = a -100 + 300 b(10 - 6) = 100(10 - 6) 2 Ans. Then, the coordinates of reference point A and Center C of the circle are A( -100, 57.74)(10 - 6) C(100, 0)(10 - 6) Thus, the radius of the circle is R = CA = a 2(-100 - 100)2 + 208.16 b(10 - 6) = 208.17(10 - 6) Using these result, the circle is shown in Fig. a. The coordinates of points B and D represent e1 and e2 respectively. e1 = (100 + 208.17)(10 - 6) = 308(10 - 6) Ans. e2 = (100 - 208.17)(10 - 6) = -108(10 - 6) Ans. Referring to the geometry of the circle, tan 2(uP)2 = 57.74(10 - 6) (100 + 100)(10 - 6) = 0.2887 A uP B 2 = 8.05° (Clockwise) Ans. The deformed element for the state of principal strain is shown in Fig. b. 763 a 10–25. Continued gmax The coordinates for point E represent eavg and gmax in-plane 2 in-plane 2 . Thus, = R = 208.17(10 - 6) gmax in-plane = 416(10 - 6) Ans. Referring to the geometry of the circle, tan 2us = 100 + 100 57.74 us = 36.9° (Counter Clockwise) Ans. The deformed element for the state of maximum In-plane shear strain is shown in Fig. c. 764 10–26. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 200110-62, Pb = - 450110-62, and Pc = 250110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. b a 30⬚ 30⬚ c With ua = 60°, ub = 120° and uc = 180°, ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 200(10 - 6) = ex cos2 60° + ey sin2 60° + gxy sin 60° cos 60° gxy = 200(10 - 6) 0.25ex + 0.75ey + 0.4330 (1) eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub -450(10 - 6) = ex cos2 120° + ey sin2 120° + gxy sin 120° cos 120° gxy = -450(10 - 6) 0.25ex + 0.75ey - 0.4330 (2) ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc 250(10 - 6) = ex cos2 180° + ey sin2 180° + gxy sin 180° cos 180° ex = 250(10 - 6) Substitute this result into Eqs. (1) and (2) and solve them, ey = -250 (10 - 6) gxy = 750.56 (10 - 6) In accordance to the established sign convention, ex = 250(10 - 6), ey = -250(10 - 6), gxy and = 375.28(10 - 6), Thus, 2 eavg = ex + ey 2 = c 250 + ( -250) d(10 - 6) = 0 2 Ans. Then, the coordinates of the reference point A and center C of the circle are A(250, 375.28)(10 - 6) C(0, 0) Thus, the radius of the circle is R = CA = A 2(250 - 0)2 + 375.282 B (10 - 6) = 450.92(10 - 6) Using these results, the circle is shown in Fig. a. The coordinates for points B and D represent e1 and e2, respectively. Thus, e1 = 451(10 - 6) e2 = -451(10 - 6) Ans. Referring to the geometry of the circle, tan 2(uP)1 = 375.28 = 1.5011 250 (uP)1 = 28.2° (Counter Clockwise) Ans. The deformed element for the state of principal strains is shown in Fig. b. 765 60⬚ 10–26. Continued gmax in-plane The coordinates of point E represent eavg and . Thus, 2 gmax in-plane gmax = R = 450.92(10 - 6) = 902(10 - 6) in-plane 2 Ans. Referring to the geometry of the circle, tan 2us = 250 = 0.6662 375.28 us = 16.8° (Clockwise) Ans. 766 10–27. The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each gauge: Pa = 300110-62, Pb = -250110-62, and Pc = -450110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. b c 45⬚ With ua = 45°, ub = 90° and uc = 135°, ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 300(10 - 6) = ex cos2 45° + ey sin2 45° + gxy sin 45° cos 45° ex + ey + gxy = 600(10 - 6) (1) eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub -250(10 - 6) = ex cos2 90° + ey sin2 90° + gxy sin 90° cos 90° ey = -250(10 - 6) ec = ex cos2 uc + ey sin2 uc + gxy sin uc cos uc -450(10 - 6) = ex cos2 135° + ey sin2 135° + gxy sin 135° cos 135° ex + ey - gxy = -900(10 - 6) (2) Substitute the result of ey into Eq. (1) and (2) and solve them ex = 100(10 - 6) gxy = 750(10 - 6) In accordance to the established sign convention, ex = 100(10 - 6), ey = -250(10 - 6) gxy and = 375(10 - 6). Thus, 2 eavg = ex + ey 2 = c 100 + (-250) d(10 - 6) = -75(10 - 6) 2 Ans. Then, the coordinates of the reference point A and the center C of the circle are A(100, 375)(10 - 6) C(-75, 0)(10 - 6) Thus, the radius of the circle is R = CA = a 2 C 100 - ( -75) D 2 + 3752 b (10 - 6) = 413.82(10 - 6) Using these results, the circle is shown in Fig. a. The Coordinates of points B and D represent e1 and e2, respectively. Thus, e1 = A -75 + 413.82 B (10 - 6) = 339(10 - 6) Ans. e2 = A -75 - 413.82 B (10 - 6) = - 489(10 - 6) Ans. Referring to the geometry of the circle tan 2(uP)1 = 375 = 2.1429 100 + 75 (uP)1 = 32.5° (Counter Clockwise) Ans. 767 45⬚ a 10–27. Continued The deformed element for the state of principal strains is shown in Fig. b. gmax The coordinates of point E represent eavg and gmax in-plane 2 = R = 413.82(106) in-plane gmax 2 in-plane . Thus = 828(10 - 6) Ans. Referring to the geometry of the circle tan 2us = -100 + 75 = 0.4667 375 us = 12.5° (Clockwise) Ans. 768 *10–28. The 45° strain rosette is mounted on the link of the backhoe. The following readings are obtained from each gauge: Pa = 650110-62, Pb = -300110-62, Pc = 480110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and associated average normal strain. a 45⬚ b ea = 650(10 - 6); ua = 180°; eb = -300(10 - 6); ub = 225° c uc = 270° Applying Eq. 10–16, e = ex cos2 u + ey sin2 u + gxy sin u cos u 650(10 - 6) = ex cos2 (180°) + ey sin2 (180°) + gxy sin (180°) cos (180°) ex = 650 (10 - 6) 480 (10 - 6) = ex cos2 (270°) + ey sin2 (270°) + gxy sin (270°) cos (270°) ey = 480 (10 - 6) -300 (10 - 6) = 650 (10 - 6) cos2 (225°) + 480 (10 - 6) sin2 (225°) + gxy sin (225°) cos (225°) gxy = -1730 (10 - 6) Therefore, ex = 650 (10 - 6) gxy 2 ey = 480 (10 - 6) gxy = -1730 (10 - 6) = -865 (10 - 6) Mohr’s circle: A(650, -865) 10 - 6 C(565, 0) 10 - 6 R = CA = C 2(650 - 565)2 + 8652 D 10 - 6 = 869.17 (10 - 6) (a) (b) e1 = [565 + 869.17]10 - 6 = 1434 (10 - 6) Ans. e2 = [565 - 869.17]10 - 6 = -304 (10 - 6) Ans. g max in-plane 45⬚ ec = 480(10 - 6) = 2 R = 2(869.17) (10 - 6) = 1738 (10 - 6) Ans. eavg = 565(10 - 6) Ans. 769 10–30. For the case of plane stress, show that Hooke’s law can be written as sx = E E 1Px + nPy2, sy = 1Py + nPx2 11 - n22 11 - n22 Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18, ex = 1 A sx - v sy B E vEex = A sx - v sy B v vEex = v sx - v2 sy ey = [1] 1 (sy - v sx) E E ey = -v sx + sy [2] Adding Eq [1] and Eq.[2] yields. vE ex - E ey = sy - v2 sy sy = E A vex + ey B 1 - v2 (Q.E.D.) Substituting sy into Eq. [2] E ey = -vsx + sx = = = E A v ex + ey B 1 - v2 E A v ex + ey B v (1 - v ) 2 - Eey v E v ex + E ey - E ey + Eey v2 v(1 - v2) E (ex + v ey) 1 - v2 (Q.E.D.) 770 10–31. Use Hooke’s law, Eq. 10–18, to develop the straintransformation equations, Eqs. 10–5 and 10–6, from the stress-transformation equations, Eqs. 9–1 and 9–2. Stress transformation equations: sx + sy sx¿ = 2 tx¿y¿ = sy¿ = + sx - sy 2 2 cos 2u + txy sin 2u (1) sin 2u + txy cos 2u 2 sx + sy sx - sy - sx - sy 2 (2) cos 2u - txy sin 2u (3) Hooke’s Law: ex = v sy sx E E (4) ey = sy -v sx + E E (5) txy = G gxy G = (6) E 2 (1 + v) (7) From Eqs. (4) and (5) ex + ey = ex - ey = (1 - v)(sx + sy) (8) E (1 + v)(sx - sy) (9) E From Eqs. (6) and (7) txy = E g 2 (1 + v) xy (10) From Eq. (4) ex¿ = v sy¿ sx¿ E E (11) Substitute Eqs. (1) and (3) into Eq. (11) ex¿ = (1 - v)(sx - sy) 2E + (1 + v)(sx - sy) 2E cos 2u + (1 + v)txy sin 2u E (12) By using Eqs. (8), (9) and (10) and substitute into Eq. (12), ex¿ = ex + ey 2 + ex - ey 2 cos 2u + gxy 2 QED sin 2u 771 10–31. Continued From Eq. (6). gx¿y¿ = G gx¿y¿ = E gx¿y¿ 2 (1 + v) (13) Substitute Eqs. (13), (6) and (9) into Eq. (2), E (ex - ey) E E = g sin 2u + g cos 2u 2 (1 + v) x¿y¿ 2 (1 + v) 2 (1 + v) xy gx¿y¿ 2 = - (ex - ey) 2 sin 2u + gxy 2 QED cos 2u *10–32. A bar of copper alloy is loaded in a tension machine and it is determined that Px = 940110-62 and sx = 14 ksi, sy = 0, sz = 0. Determine the modulus of elasticity, Ecu, and the dilatation, ecu, of the copper. ncu = 0.35. ex = 1 [sx - v(sy + sz)] E 940(10 - 6) = ecu = 1 [14(103) - 0.35(0 + 0)] Ecu Ecu = 14.9(103) ksi Ans. 1 - 2(0.35) 1 - 2v (14 + 0 + 0) = 0.282(10 - 3) (sx + sy + sz) = E 14.9(103) Ans. •10–33. The principal strains at a point on the aluminum fuselage of a jet aircraft are P1 = 780110-62 and P2 = 400110-62. Determine the associated principal stresses at the point in the same plane. Eal = 1011032 ksi, nal = 0.33. Hint: See Prob. 10–30. Plane stress, s3 = 0 See Prob 10-30, s1 = = s2 = = E (e1 + ve2) 1 - v2 10(103) 1 - 0.332 (780(10 - 6) + 0.33(400)(10 - 6)) = 10.2 ksi Ans. E (e2 + ve1) 1 - v2 10(103) 1 - 0.332 (400(10 - 6) + 0.33(780)(10 - 6)) = 7.38 ksi Ans. 772 10–34. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the absolute maximum shear strain in the rod at a point on its surface. 700 N Normal Stress: For uniaxial loading, sy = sz = 0. sx = P = A p 4 700 = 2.228 MPa (0.022) Normal Strain: Applying the generalized Hooke’s Law. ex = = 1 C sx - v A sy + sz B D E 1 C 2.228 A 106 B - 0 D 73.1(109) = 30.48 A 10 - 6 B ey = = 1 C s - v(sx + sz) D E y 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109) = -10.67 A 10 - 6 B ez = = 1 C sz - v A sx + sy B D E 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109) = -10.67 A 10 - 6 B Therefore. emax = 30.48 A 10 - 6 B emin = -10.67 A 10 - 6 B Absolute Maximum Shear Strain: gabs max = emax - emin = [30.48 - (-10.67)] A 10 - 6 B = 41.1 A 10 - 6 B Ans. 773 700 N 10–35. The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter of 20 mm, determine the principal strains at a point on the surface of the rod. 700 N Normal Stress: For uniaxial loading, sy = sz = 0. sx = P = A p 4 700 = 2.228 MPa (0.022) Normal Strains: Applying the generalized Hooke’s Law. ex = = 1 C sx - v A sy + sz B D E 1 C 2.228 A 106 B - 0 D 73.1(109) = 30.48 A 10 - 6 B ey = = 1 C sy - v(sx + sz) D E 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109) = -10.67 A 10 - 6 B ez = = 1 C sz - v A sx + sy B D E 1 C 0 - 0.35 A 2.228 A 106 B + 0 B D 73.1(109) = -10.67 A 10 - 6 B Principal Strains: From the results obtained above, emax = 30.5 A 10 - 6 B eint = emin = -10.7 A 10 - 6 B Ans. 774 700 N *10–36. The steel shaft has a radius of 15 mm. Determine the torque T in the shaft if the two strain gauges, attached to the surface of the shaft, report strains of Px¿ = -80110-62 and Py¿ = 80110-62. Also, compute the strains acting in the x and y directions. Est = 200 GPa, nst = 0.3. ex¿ = -80(10 - 6) y T ex = ey = 0 Ans. ex¿ = ex cos2 u + ey sin2 u + gxy sin u cos u u = 45° -80(10 - 6) = 0 + 0 + gxy sin 45° cos 45° gxy = -160(10 - 6) Ans. Also, u = 135° 80(10 - 6) = 0 + 0 + g sin 135° cos 135° gxy = -160(10 - 6) 200(109) E = = 76.923(109) 2(1 + V) 2(1 + 0.3) t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa p 12.308(106) A B (0.015)4 2 = 65.2 N # m 0.015 Ans. 10–37. Determine the bulk modulus for each of the following materials: (a) rubber, Er = 0.4 ksi, nr = 0.48, and (b) glass, Eg = 811032 ksi, ng = 0.24. a) For rubber: Kr = Er 0.4 = = 3.33 ksi 3 (1 - 2 vr) 3[1 - 2(0.48)] Ans. b) For glass: Kg = Eg 3 (1 - 2 vg) = x T Pure shear tJ T = = c x¿ 45⬚ ey¿ = 80(10 - 6) G = y¿ 8(103) = 5.13 (103) ksi 3[1 - 2(0.24)] Ans. 775 10–38. The principal stresses at a point are shown in the figure. If the material is A-36 steel, determine the principal strains. 12 ksi e1 = 1 1 e 12 - 0.32 C 8 + (-20) D f = 546 (10-6) C s1 - v(s2 + s3) D = E 29.0(103) e2 = 1 1 e 8 - 0.32 C 12 + (-20) D f = 364 (10-6) C s2 - v(s1 + s3) D = E 29.0(103) e3 = 20 ksi 8 ksi 1 1 C s3 - v(s1 + s2) D = C -20 - 0.32(12 + 8) D = -910 (10-6) E 29.0(103) emax = 546 (10 - 6) eint = 346 (10 - 6) emin = -910 (10 - 6) Ans. 10–39. The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est = 200 GPa and nst = 0.3. 20 mm r 1000 = = 100 7 10, the thin wall analysis is valid to t 10 determine the normal stress in the wall of the spherical vessel. This is a plane stress Normal Stresses: Since problem where smin = 0 since there is no load acting on the outer surface of the wall. smax = slat = pr p(1000) = = 50.0p 2t 2(10) [1] Normal Strains: Applying the generalized Hooke’s Law with emax = elat = 0.012 = 0.600 A 10 - 3 B mm>mm 20 emax = 1 C smax - V (slat + smin) D E 0.600 A 10 - 3 B = 1 [50.0p - 0.3 (50.0p + 0)] 200(104) p = 3.4286 MPa = 3.43 MPa Ans. From Eq.[1] smax = slat = 50.0(3.4286) = 171.43 MPa Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the result, the state of stress is the same consisting of two normal stresses with zero shear stress regardless of the orientation of the element. t = 0 Ans. smax - smin 171.43 - 0 = = 85.7MPa 2 2 Ans. max in-plane Absolute Maximum Shear Stress: tabs max = 776 *10–40. The strain in the x direction at point A on the steel beam is measured and found to be Px = -100110-62. Determine the applied load P. What is the shear strain gxy at point A? Est = 2911032 ksi, nst = 0.3. P y 3 in. A 3 ft 1 1 (5.5)(83) = 129.833 in4 (6)(9)3 12 12 Ix = QA = (4.25)(0.5)(6) + (2.75)(0.5)(2.5) = 16.1875 in3 s = Eex = 29(103)(100)(10 - 6) = 2.90 ksi My , I s = 2.90 = 1.5P(12)(1.5) 129.833 P = 13.945 = 13.9 kip tA = VQ 0.5(13.945)(16.1875) = = 1.739 ksi It 129.833(0.5) G = 29(103) E = = 11.154(103) ksi 2(1 + v) 2(1 + 0.3) gxy = txy G = Ans. 1.739 = 0.156(10 - 3) rad 11.154(103) Ans. 777 x 4 ft 7 ft •10–41. The cross section of the rectangular beam is subjected to the bending moment M. Determine an expression for the increase in length of lines AB and CD. The material has a modulus of elasticity E and Poisson’s ratio is n. C D B h For line AB, sz = ey = ¢LAB A M 12My My My = = 1 3 I b h3 12 b h v sz = E b 12 v My E b h3 h 2 h 2 12 v M = ey dy = y dy E b h3 L0 L0 = 3vM 2Ebh Ans. For line CD, sz = ex = - M h2 Mc 6M = - 2 = - 1 3 I bh b h 12 v sz E = 6vM E b h2 ¢LCD = ex LCD = = 6vM (b) E b h2 6vM E h2 Ans. 10–42. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains. 26 ksi ex = 1 1 (sx - v(sy + sz)) = (10 - 0.33(-15 - 26)) = 2.35(10 - 3) E 10(103) ey = 1 1 (sy - v(sx + sz)) = (-15 - 0.33)(10 - 26)) = -0.972(10 - 3)Ans. E 10(103) ez = Ans. 1 1 (sz - v(sx + sy)) = (-26 - 0.33(10 - 15)) = -2.44(10 - 3) Ans. E 10(103) 778 15 ksi 10 ksi 10–43. A single strain gauge, placed on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = -200(10-6). Determine the horizontal force P if the pipe has an outer diameter of 2 in. and an inner diameter of 1 in. The pipe is made of A-36 steel. 1.5 ft Using the method of section and consider the equilibrium of the FBD of the pipe’s upper segment, Fig. a, ©Fz = 0; Vz - p = 0 Vz = p ©Mx = 0; Tx - p(1.5) = 0 Tx = 1.5p ©My = 0; My - p(2.5) = 0 My = 2.5p 30⬚ A The normal strees is due to bending only. For point A, z = 0. Thus sx = My z Iy = 0 The shear stress is the combination of torsional shear stress and transverse shear stress. Here, J = p2 (14 - 0.54) = 0.46875 p in4. Thus, for point A tt = 1.5p(12)(1) 38.4 p Txc = = p J 0.46875p Referring to Fig. b, (QA)z = y1œ A1œ - y2œ A2œ = 4 (1) p 2 4(0.5) p c (1 ) d c (0.52) d 3p 2 3p 2 = 0.5833 in3 Iy = p 4 (14 - 0.54) = 0.234375 p in4 Combine these two shear stress components, t = tt + tv = P 2.5 ft 38.4P 2.4889P 40.8889P + = p p p Since no normal stress acting on point A, it is subjected to pure shear which can be represented by the element shown in Fig. c. For pure shear, ex = ez = 0, ea = ex cos3 ua + ez sin2 ua + gxz sin ua cos ua -200(10 - 6) = 0 + 0 + gxz sin 150° cos 150° gxz = 461.88(10 - 6) Applying the Hooke’s Law for shear, txz = G gxz 40.8889P = 11.0(103) C 461.88(10 - 6) D p P = 0.3904 kip = 390 lb Ans. 779 *10–44. A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 30° to the axis of the pipe, gives a reading at point A of Pa = -200(10-6). Determine the principal strains in the pipe at point A. The pipe has an outer diameter of 2 in. and an inner diameter of 1 in. and is made of A-36 steel. 1.5 ft P 2.5 ft Using the method of sections and consider the equilibrium of the FBD of the pipe’s upper segment, Fig. a, ©Fz = 0; Vz - P = 0 Vz = P ©Mx = 0; Tx - P(1.5) = 0 Tx = 1.5P ©My = 0; My - P(2.5) = 0 My = 2.5P By observation, no normal stress acting on point A. Thus, this is a case of pure shear. For the case of pure shear, ex = ez = ey = 0 ea = ex cos2 ua + ez sin2 ua + gxz sin ua cos ua -200(10 - 6) = 0 + 0 + gxz sin 150° cos 150° gxz = 461.88(10 - 6) e1, 2 = ex + ez = B 2 + A a ex - ez 2 b + a 2 gxz 2 b 2 0 + 0 461.88 2 0 - 0 2 -6 ; a b + a b R (10 ) 2 A 2 2 e1 = 231(10 - 6) e2 = - 231(10 - 6) Ans. 780 30⬚ A 10–45. The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the bending stress that would occur if flat ends were used. The bending stresses at the seam where the caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder, respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder. Show that this ratio is tc>th = 12 - n2>11 - n2.Assume that the vessel is made of the same material and both the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of 0.5 in., what is the required thickness of the hemispheres? Take n = 0.3. tc th r For cylindrical vessel: s1 = pr ; tc e1 = 1 [s1 - v (s2 + s3)] E = s2 = pr 2 tc s3 = 0 vpr pr 1 pr 1 b = a1 - v b a E tc 2 tc E tc 2 d r = e1 r = p r2 1 a1 - v b E tc 2 (1) For hemispherical end caps: s1 = s2 = e1 = = pr 2 th 1 [s1 - v (s2 + s3)] ; E s3 = 0 vpr pr 1 pr b = (1 - v) a E 2 th 2 th 2 E th d r = e1 r = p r2 (1 - v) 2 E th (2) Equate Eqs. (1) and (2): p r2 p r2 1 a1 - vb = (1 - v) E tc 2 2 E th 2 (1 - 12 v) tc 2 - v = = th 1 - v 1 - v th = QED (1 - v) tc (1 - 0.3) (0.5) = = 0.206 in. 2 - v 2 - 0.3 Ans. 781 10–46. The principal strains in a plane, measured experimentally at a point on the aluminum fuselage of a jet aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is a case of plane stress, determine the associated principal stresses at the point in the same plane. Eal = 10(103) ksi and nal = 0.33. Normal Stresses: For plane stress, s3 = 0. Normal Strains: Applying the generalized Hooke’s Law. e1 = 1 C s1 - v (s2 + s3) D E 630 A 10 - 6 B = 1 [s1 - 0.33(s2 + 0)] 10(103) 6.30 = s1 - 0.33s2 e2 = [1] 1 C s2 - v (s1 + s3) D E 350 A 10 - 6 B = 1 C s2 - 0.33(s1 + 0) D 10(103) 3.50 = s2 - 0.33s1 [2] Solving Eqs.[1] and [2] yields: s1 = 8.37 ksi s2 = 6.26 ksi Ans. 10–47. The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal = 1011032 ksi and nal = 0.33, determine the principal strains. 3 ksi e1 = 1 1 e 8 - 0.33 C 3 + (-4) D f = 833 (10 - 6) C s1 - v(s2 + s3) D = E 10(103) e2 = 1 1 e 3 - 0.33 C 8 + (-4) D f = 168 (10 - 6) C s2 - v(s1 + s3) D = E 10(103) e3 = 1 1 C s3 - v(s1 + s2) D = C -4 - 0.33(8 + 3) D = -763 (10 - 6) E 10(103) Using these results, e1 = 833(10 - 6) e2 = 168(10 - 6) e3 = -763(10 - 6) 782 8 ksi 4 ksi *10–48. The 6061-T6 aluminum alloy plate fits snugly into the rigid constraint. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 50°C. To solve, add the thermal strain a¢T to the equations for Hooke’s Law. y 400 mm 300 mm x Generalized Hooke’s Law: Since the sides of the aluminum plate are confined in the rigid constraint along the x and y directions, ex = ey = 0. However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex = 0 = 1 c sx - v A sy + sz B d + a¢T E 1 68.9 A 109 B csx - 0.35 A sy + 0 B d + 24a10 - 6 b(50) sx - 0.35sy = -82.68 A 106 B ey = 0 = (1) 1 C sy - v A sx + sz B D + a¢T E 1 68.9a 109 b C sy - 0.35(sx + 0) D + 24 A 10 - 6 B (50) sy - 0.35sx = -82.68 A 106 B (2) Solving Eqs. (1) and (2), sx = sy = -127.2 MPa = 127.2 MPa (C) Ans. Since sx = sy and sy 6 sY, the above results are valid. 783 •10–49. Initially, gaps between the A-36 steel plate and the rigid constraint are as shown. Determine the normal stresses sx and sy developed in the plate if the temperature is increased by ¢T = 100°F. To solve, add the thermal strain a¢T to the equations for Hooke’s Law. y 0.0015 in. 6 in. 8 in. 0.0025 in. x Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid constraint, the plate is allowed to expand before it comes in contact with the constraint. dy dx 0.0025 0.0015 ey = = = = 0.3125 A 10 - 3 B and = 0.25 A 10 - 3 B . Thus, ex = Lx 8 Ly 6 However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex = 1 csx - v A sy + sz B d + a¢T E 0.3125a 10 - 3 b = 1 29.0 a10 b C sx - 0.32 A sy + 0 B D + 6.60 A 10 - 6 B (100) 3 sx - 0.32sy = -10.0775 ey = (1) 1 C s - v A sx + sz B D + a¢T E y 0.25 A 10 - 3 B = 1 29.0 A 103 B C sy - 0.32(sx + 0) D + 6.60 A 10 - 6 B (100) sy - 0.32sx = -11.89 (2) Solving Eqs. (1) and (2), sx = -15.5 ksi = 15.5 ksi (C) Ans. sy = -16.8 ksi = 16.8 ksi (C) Ans. Since sx 6 sY and sy 6 sY, the above results are valid. 784 10–50. Two strain gauges a and b are attached to a plate made from a material having a modulus of elasticity of E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the intensities of the uniform distributed load wx and wy acting on the plate. The thickness of the plate is 25 mm. wy b 45⬚ y a Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0 and ub = 45°, we have ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0°+0 ex = 450 A 10 - 6 B eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub 100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0 ey = -250 A 10 - 6 B Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0. ex = 1 C sx - v A sy + sz B D E 450 A 10 - 6 B = 1 70 A 109 B C sy - 0.35 A sy + 0 B D sx - 0.35sy = 31.5 A 106 B ey = (1) 1 C s - v A sx + sz B D E y -250 A 10-6 B = 1 70 A 109 B C sy - 0.35 A sy + 0 B D sy - 0.35sx = -17.5 A 106 B (2) Solving Eqs. (1) and (2), sy = -7.379 A 106 B N>m2 sx = 28.917 A 106 B N>m2 Then, wy = syt = -7.379 A 106 B (0.025) = -184 N>m Ans. wx = sxt = 28.917 A 106 B (0.025) = 723 N>m Ans. 785 z x wx 10–51. Two strain gauges a and b are attached to the surface of the plate which is subjected to the uniform distributed load wx = 700 kN>m and wy = -175 kN>m. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the modulus of elasticity E, shear modulus G, and Poisson’s ratio n for the material. wy b 45⬚ y Normal Stress and Strain: The normal stresses along the x, y, and z axes are sx = 700 A 103 B 0.025 sy = - a = 28 A 106 B N>m2 175 A 103 B 0.025 = -7 A 106 B N>m2 z sz = 0 (plane stress) Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0° and ub = 45°, we have ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0 ex = 450 A 10 - 6 B eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub 100 A 10 - 6 B = 450 A 10 - 6 B cos2 45°+ ey sin2 45° + 0 ey = -250 A 10 - 6 B Generalized Hooke’s Law: ex = 1 C sx - v A sy + sz B D E 450 A 10 - 6 B = 1 B 28 A 106 B - v C -7 A 106 B + 0 D R E 450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B ey = (1) 1 [s - v(sx + sz)] E y -250 A 10 - 6 B = 1 b -7 A 106 B - v C 28 A 106 B + 0 D r E 250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B (2) Solving Eqs. (1) and (2), E = 67.74 A 109 B N>m2 = 67.7 GPa Ans. v = 0.3548 = 0.355 Ans. Using the above results, G = 67.74 A 109 B E = 2(1 + v) 2(1 + 0.3548) = 25.0 A 109 B N>m2 = 25.0 GPa Ans. 786 x wx *10–52. The block is fitted between the fixed supports. If the glued joint can resist a maximum shear stress of tallow = 2 ksi, determine the temperature rise that will cause the joint to fail. Take E = 10 (103) ksi, n = 0.2, and Hint: Use Eq. 10–18 with an additional strain term of a¢T (Eq. 4–4). 40⬚ Normal Strain: Since the aluminum is confined along the y direction by the rigid frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ey = 0 = 1 C sy - v(sx + sz) D + a¢T E 1 C sy - 0.2(0 + 0) D + 6.0 A 10 - 6 B (¢T) 10.0(103) sy = -0.06¢T Construction of the Circle: In accordance with the sign convention. sx = 0, sy = -0.06¢T and txy = 0. Hence. savg = sx + sy 2 = 0 + ( -0.06¢T) = -0.03¢T 2 The coordinates for reference points A and C are A (0, 0) and C( -0.03¢T, 0). The radius of the circle is R = 2(0 - 0.03¢T)2 + 0 = 0.03¢T Stress on The inclined plane: The shear stress components tx¿y¿, are represented by the coordinates of point P on the circle. tx¿y¿ = 0.03¢T sin 80° = 0.02954¢T Allowable Shear Stress: tallow = tx¿y¿ 2 = 0.02954¢T ¢T = 67.7 °F Ans. 787 z •10–53. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is covered and the temperature is increased by 200°F, determine the stress components sx , sy , and sz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4). 0.012 in. 4 in. 4 in. 6 in. y Normal Strains: Since the aluminum is confined at its sides by a rigid container and 0.012 allowed to expand in the z direction, ex = ey = 0; whereas ez = = 0.002. 6 Applying the generalized Hooke’s Law with the additional thermal strain, ex = 0 = 1 C sx - v(sy + sz) D + a¢T E 1 C sx - 0.35 A sy + sz B D + 13.1 A 10 - 6 B (200) 10.0(103) 0 = sx - 0.35sy - 0.35sz + 26.2 ey = 0 = [1] 1 C sy - v(sx + sz) + a¢T E 1 C sy - 0.35(sx + sz) D + 13.1 A 10 - 6 B (200) 10.0(103) 0 = sy - 0.35sx - 0.35sz + 26.2 ez = 0.002 = [2] 1 C sz - v A sx + sy B D + a¢T E 1 C sz - 0.35 A sx + sy B D + 13.1 A 10 - 6 B (200) 10.0(103) 0 = sz - 0.35sx - 0.35sy + 6.20 [3] Solving Eqs.[1], [2] and [3] yields: sx = sy = -70.0 ksi sz = -55.2 ksi Ans. 788 x z 10–54. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.012 in. from the top of the cavity. If the top of the cavity is not covered and the temperature is increased by 200°F, determine the strain components Px , Py , and Pz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4). 0.012 in. 4 in. 4 in. 6 in. y Normal Strains: Since the aluminum is confined at its sides by a rigid container, then ex = ey = 0 Ans. and since it is not restrained in z direction, sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ex = 0 = 1 C sx - v A sy + sz B D + a¢T E 1 C sx - 0.35 A sy + 0 B D + 13.1 A 10 - 6 B (200) 10.0(103) 0 = sx - 0.35sy + 26.2 ey = 0 = [1] 1 C s - v(sx + sz) D + a¢T E y 1 C sy - 0.35(sx + 0) D + 13.1 A 10 - 6 B (200) 10.0(103) 0 = sy - 0.35sx + 26.2 [2] Solving Eqs. [1] and [2] yields: sx = sy = -40.31 ksi ez = = 1 C sz - v A sx + sy B D + a¢T E 1 {0 - 0.35[-40.31 + (-40.31)]} + 13.1 A 10 - 6 B (200) 10.0(103) = 5.44 A 10 - 3 B Ans. 789 x 10–55. A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p. Show that the increase in the volume within the vessel is ¢V = 12ppr4>Et211 - n2. Use a small-strain analysis. pr 2t s1 = s2 = s3 = 0 e1 = e2 = 1 (s1 - vs2) E e1 = e2 = pr (1 - v) 2t E e3 = 1 (-v(s1 + s2)) E e3 = V = v pr tE 4pr3 3 V + ¢V = 4p 4pr3 ¢r 3 (r + ¢r)3 = (1 + ) r 3 3 where ¢V V V, ¢r V r V + ¢V eVol = ¢r 4p r3 b a1 + 3 r 3 ¢V ¢r = 3a b V r Since e1 = e2 = eVol = 3e1 = 2p(r + ¢r) - 2p r ¢r = r 2p r 3pr (1 - v) 2t E ¢V = VeVol = 2pp r4 (1 - v) Et QED 790 *10–56. A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it is subjected to an internal pressure p, show that the increase in its inner radius is dr = rP1 = pr211 - 12 n2>Et and the increase in its length is ¢L = pLr112 - n2>Et. Using these results, show that the change in internal volume becomes dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are small quantities, show further that the change in volume per unit volume, called volumetric strain, can be written as dV>V = pr12.5 - 2n2>Et. Normal stress: pr ; t s1 = s2 = pr 2t Normal strain: Applying Hooke’s law e1 = = 1 [s1 - v (s2 + s3)], E vpr pr 1 1 pr a b = a1 - vb E t 2t Et 2 d r = et r = e2 = = s3 = 0 p r2 1 a1 - v b Et 2 1 [s2 - v (s1 + s3)], E QED s3 = 0 vpr pr 1 1 pr a b = a - vb E 2t t Et 2 ¢L = e2 L = pLr 1 a - vb Et 2 V¿ = p(r + e1 r)2 (L + e2L) ; QED V = p r2 L dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L QED (1 + e1)2 = 1 + 2 e1 neglect e21 term (1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term dV = 1 + e2 + 2 e1 - 1 = e2 + 2 e1 V = pr 1 2pr 1 a - vb + a1 - v b Et 2 Et 2 = pr (2.5 - 2 v) Et QED 791 10–57. The rubber block is confined in the U-shape smooth rigid block. If the rubber has a modulus of elasticity E and Poisson’s ratio n, determine the effective modulus of elasticity of the rubber under the confined condition. P Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We have ex = 0 = 1 C sx - v A sy + sz B D E 1 (sx - vsz) E sx = vsz (1) ez = 1 C s - v A sx + sy B D E z ez = 1 [s - v(sx + 0)] E z ez = 1 (sz - vsx) E (2) Substituting Eq. (1) into Eq. (2), ez = sz E A 1 - v2 B The effective modulus of elasticity of the rubber block under the confined condition can be determined by considering the rubber block as unconfined but rather undergoing the same normal strain of ez when it is subjected to the same normal stress sz, Thus, sz = Eeff ez Eeff = sz ez = sz sz E A 1 - v2 B = E 1 - v2 Ans. 792 z 10–58. A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that Px = 0 and Py = 0, determine the factor by which the modulus of elasticity will be increased when a load is applied if n = 0.3 for the material. P x Normal Strain: Since the material is confined in a rigid cylinder. ex = ey = 0. Applying the generalized Hooke’s Law, ex = 1 C sz - v(sy + sx) D E 0 = sx - v(sy + sz) ey = [1] 1 C sy - v(sx + sz) D E 0 = sy - v(sx + sz) [2] Solving Eqs.[1] and [2] yields: sx = sy = v sz 1 - v Thus, ez = = = 1 C sz - v(sx + sy) D E v v 1 csz - va sz + sz b d E 1 - v 1 - v sz E c1 - 2v2 d 1 - v = sz 1 - v - 2v2 c d E 1 - v = sz (1 + v)(1 - 2v c d E 1 - v Thus, when the material is not being confined and undergoes the same normal strain of ez, then the requtred modulus of elasticity is E¿ = sz ez = The increased factor is k = 1 - v E (1 - 2v)(1 + v) E¿ 1 - v = E (1 - 2v)(1 + v) = 1 - 0.3 [1 - 2(0.3)](1 + 0.3) = 1.35 Ans. 793 y 10–59. A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of sx , sy , and txy . Maximum distortion energy theory: (s21 - s1 s2 + s22) = s2Y s1,2 = sx + sy 2 Let a = ; sx + sy 2 s1 = a + b; A a (1) sx - sy 2 and b = A a 2 2 b + txy sx - sy 2 2 2 b + txy s2 = a - b s21 = a2 + b2 + 2 a b; s22 = a2 + b2 - 2 a b s1 s2 = a2 - b2 From Eq. (1) (a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = s2y (a2 + 3 b2) = s2Y (sx + sy)2 4 + 3 (sx - sy)2 4 + 3 t2xy = s2Y s2x + s2y - sxsy + 3 t2xy = s2Y Ans. *10–60. A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms of sx , sy , and txy . Assume that the principal stresses are of different algebraic signs. Maximum shear stress theory: |s1 - s2| = sY s1,2 = sx + sy 2 ` s1 - s2 ` = 2 (1) ; A a A a sx - sy 2 sx - sy 2 2 2 b + txy 2 b + txy 2 From Eq. (1) 4 ca sx - sy 2 2 b + t2xy d = s2Y 2 (sx - sy) + 4 t2xy = s2Y Ans. 794 •10–61. An aluminum alloy 6061-T6 is to be used for a solid drive shaft such that it transmits 40 hp at 2400 rev>min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-shear-stress theory. v = a2400 T = rev 2p rad 1 min ba ba b = 80 p rad>s min rev 60s 40 (550) (12) P 3300 # lb in. = = p v 80 p Tc J Applying t = t = A 3300 p B c p 2 = c4 6600 p3 c3 The principal stresses: s1 = t = 6600 ; p2 c3 s2 = -t = 6600 p2 c3 Maximum shear stress theory: Both principal stresses have opposite sign, hence, ` s1 - s2 ` = 2a sY ; F.S. 37 (103) 6600 b = ` ` 2 p2c3 c = 0.4166 in. d = 0.833 in. Ans. 10–62. Solve Prob. 10–61 using the maximum-distortionenergy theory. v = a2400 T = 2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s 40 (550) (12) P 3300 = = lb.in. p v 80 p Applying t = t = A 3300 p B c p 2 4 c = Tc J 6600 p2 c3 The principal stresses: s1 = t = 6600 ; p2 c3 s2 = - t = - 6600 p2 c3 The maximum distortion-energy theory: s21 - s1 s2 + s22 = a 3B sY 2 b F.S. 37(103) 2 6600 2 = a b R 2 p2 c3 c = 0.3971 in. d = 0.794 in. Ans. 795 10–63. An aluminum alloy is to be used for a drive shaft such that it transmits 25 hp at 1500 rev>min. Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-distortion-energy theory. sY = 3.5 ksi. 1500(2p) = 50p 60 T = P v T = 25(550)(12) 3300 = p 50p t = Tc , J t = 3300 p c p 4 2c s1 = v = J = = p 4 c 2 6600 p2c3 6600 p2c3 s2 = s21 - s1 s2 + s22 = a 3a -6600 p2c3 sY 2 b F.S. 3.5(103) 2 6600 2 b = a b 2.5 p2c3 c = 0.9388 in. d = 1.88 in. Ans. *10–64. A bar with a square cross-sectional area is made of a material having a yield stress of sY = 120 ksi. If the bar is subjected to a bending moment of 75 kip # in., determine the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with respect to yielding. Normal and Shear Stress: Applying the flexure formula, s = 75 A a2 B 450 Mc = 1 4 = 3 I a a 12 In-Plane Principal Stress: Since no shear stress acts on the element s1 = sx = 450 a3 s2 = sy = 0 Maximum Distortion Energy Theory: s21 - s1 s2 + s22 = s2allow a 120 2 450 2 b 0 + 0 = a b 1.5 a3 a = 1.78 in. Ans. 796 •10–65. Solve Prob. 10–64 using the maximum-shearstress theory. Normal and Shear Stress: Applying the flexure formula, s = 75 A a2 B Mc 450 = 1 4 = 3 I a a 12 In-Plane Principal Stress: Since no shear stress acts on the element. s1 = sx = 450 a3 s2 = sx = 0 Maximum Shear Stress Theory: |s2| = 0 6 sallow = 120 = 80.0 ksi 1.5 (O.K!) |s1| = sallow 120 450 = 1.5 a3 a = 1.78 in. Ans. 10–66. Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. t = Te c J Principal stress: s1 = tx ¿ ud = s2 = -t 1 + v 2 (s1 - s1 s2 + s22) 3E (ud)1 = 1 + v 1 + v 3 T2x c2 ( 3 t2) = a b 3E 3E J2 Bending moment and torsion: s = Mc ; I t = Tc J Principal stress: s1, 2 = s1 = s - 0 2 s + 0 2 ; a b + t 2 A 2 s s2 + + t2 ; 2 A4 s2 = s s2 + t2 2 A4 797 10–66. Continued Let a = s 2 b = s2 + t2 A4 s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s21 - s1 s2 + s22 = 3 b2 + a2 1 + v 2 (s1 - s1 s2 + s22) 3E ud = (ud)2 = = 1 + v 1 + v 3 s2 s2 (3 b2 + a2) = a + 3t2 + b 3E 3E 4 4 c2(1 + v) M2 1 + v 2 3 T2 b (s + 3 t2) = a 2 + 3E 3E I J2 (ud)1 = (ud)2 c3(1 + v) 3 Tx 2 c2(1 + v) M2 3 T2 = b a 2 + 2 3E 3E J I J2 For circular shaft J = I p 3 p 4 c4 c4 =2 Te = J2 M2 + T2 A I2 3 Te = 4 2 M + T2 A3 Ans. 10–67. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T. Principal stresses: s1 = Me c ; I ud = 1 + v 2 (s1 - s1 s2 + s22) 3E (ud)1 = s2 = 0 1 + v M2e c2 a 2 b 3E I (1) 798 10–67. Continued Principal stress: s + 0 s - 0 2 3 ; a b + t 2 A 2 s1, 2 = s s2 + + t2; 2 A4 s1 = s2 = s s2 + t2 2 A4 Distortion Energy: s s2 ,b = + t2 A4 2 Let a = s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s22 - s1 s2 + s22 = 3 b2 + a2 Apply s = Mc ; I t = Tc J 1 + v 1 + v s2 3s2 (3 b2 + a2) = a + + 3 t2 b 3E 3E 4 4 (ud)2 = 1 + v M2 c2 3 T2 c2 1 + v 2 b (s + 3 t2) = a 2 + 3E 3E I J2 = (2) Equating Eq. (1) and (2) yields: (1 + v) Me c2 1 + v M2 c2 3T2 c2 b a 2 b = a 2 + 3E 3E I I J2 M2e 2 I = M1 3 T2 + 2 I J2 M2e = M1 + 3 T2 a I 2 b J For circular shaft I = J p 4 p 2 c4 4 c = 1 2 1 2 Hence, M2e = M2 + 3 T2 a b 2 Me = A M2 + 3 2 T 4 Ans. 799 *10–68. The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N # m and an axial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory. The ultimate stress of the concrete is sult = 28 MPa. A = p (0.05)2 = 1.9635(10 - 3) m2 4 J = p (0.025)4 = 0.61359(10 - 4) m4 2 2 kN 500 N⭈m 500 N⭈m 2 kN 3 s = 2(10 ) P = 1.019 MPa = A 1.9635(10 - 3) t = 500(0.025) Tc = = 20.372 MPa J 0.61359(10 - 6) sx = 0 sy = -1.019 MPa sx + sy s1, 2 = s1,2 = 2 ; A a sx - sy 2 txy = 20.372 MPa 2 b + txy 2 0 - 1.018 0 - (-1.019) 2 2 ; a b + 20.372 2 A 2 s1 = 19.87 MPa s2 = -20.89 MPa Failure criteria: |s1| 6 salt = 28 MPa OK |s2| 6 salt = 28 MPa OK No. Ans. •10–69. Cast iron when tested in tension and compression has an ultimate strength of 1sult2t = 280 MPa and 1sult2c = 420 MPa, respectively. Also, when subjected to pure torsion it can sustain an ultimate shear stress of tult = 168 MPa. Plot the Mohr’s circles for each case and establish the failure envelope. If a part made of this material is subjected to the state of plane stress shown, determine if it fails according to Mohr’s failure criterion. 120 MPa 100 MPa 220 MPa s1 = 50 + 197.23 = 247 MPa s2 = 50 - 197.23 = -147 MPa The principal stress coordinate is located at point A which is outside the shaded region. Therefore the material fails according to Mohr’s failure criterion. Yes. Ans. 800 10–69. Continued 10–70. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same maximum shear stress as the combination of an applied moment M and torque T. Assume that the principal stresses are of opposite algebraic signs. Bending and Torsion: Mc 4M Mc ; = p 4 = I p c3 4 c s = t = Tc 2T Tc = p 4 = J p c3 2c The principal stresses: s1, 2 = = tabs max sx + sy 2 ; A a sx - sy 2 2 b + t2xy = 4M pc3 + 0 2 ; Q ¢ 4M p c3 - 0 2 2 ≤ + a b 3 pc 2T 2 2M 2 ; 2M2 + T2 p c3 p c3 = s1 - s2 = 2 c 2 2M2 + T2 d p c3 (1) Pure bending: s1 = tabs max 4 Me Me c Mc ; = p 4 = I c p c3 4 = s1 - s2 = s2 = 0 4 Me (2) p c3 Equating Eq. (1) and (2) yields: 4 Me 4 2M2 + T2 = 3 pc p c3 Me = 2M2 + T2 Ans. 801 10–71. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-shearstress theory. 60 MPa 40 MPa In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and txy = 40 MPa. s1, 2 = = sx + sy 2 ; A a sx - sy 2 70 MPa 2 b + txy 2 70 + (-60) 70 - ( -60) 2 2 ; c d + 40 2 A 2 = 5 ; 25825 s1 = 81.32 MPa s2 = -71.32 MPa In this case, s1 and s2 have opposite sign. Thus, |s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sy = 250 MPa Based on this result, the steel shell does not yield according to the maximum shear stress theory. *10–72. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. 60 MPa 40 MPa In accordance to the established sign convention, sx = 70 MPa, sy = -60 MPa and txy = 40 MPa. s1, 2 = = sx + sy 2 ; A a sx - sy 2 2 b + txy 2 70 + ( -60) 70 - ( -60) 2 2 ; c d + 40 2 A 2 = 5 ; 25825 s1 = 81.32 MPa s2 = -71.32 MPa s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sy 2 = 62500 Based on this result, the steel shell does not yield according to the maximum distortion energy theory. 802 70 MPa •10–73. If the 2-in. diameter shaft is made from brittle material having an ultimate strength of sult = 50 ksi for both tension and compression, determine if the shaft fails according to the maximum-normal-stress theory. Use a factor of safety of 1.5 against rupture. 30 kip 4 kip · ft Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = pin2 J = p 4 p A 1 B = in4 2 2 The normal stress is caused by axial stress. s = 30 N = -9.549 ksi = p A The shear stress is contributed by torsional shear stress. t = 4(12)(1) Tc = 30.56 ksi = p J 2 The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = -9.549 ksi, sy = 0 and txy = -30.56 ksi. We have s1, 2 = = sx + sy 2 ; A a sx - sy 2 2 b + txy 2 -9.549 - 0 2 -9.549 + 0 2 ; a b + (-30.56) 2 A 2 = (-4.775 ; 30.929) ksi s1 = 26.15 ksi s2 = -35.70 ksi Maximum Normal-Stress Theory. sallow = sult 50 = = 33.33 ksi F.S. 1.5 |s1| = 26.15 ksi 6 sallow = 33.33 ksi (O.K.) |s2| = 35.70 ksi 7 sallow = 33.33 ksi (N.G.) Based on these results, the material fails according to the maximum normal-stress theory. 803 10–74. If the 2-in. diameter shaft is made from cast iron having tensile and compressive ultimate strengths of 1sult2t = 50 ksi and 1sult2c = 75 ksi, respectively, determine if the shaft fails in accordance with Mohr’s failure criterion. Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are A = p A 12 B = p in2 J = p 4 p A 1 B = in4 2 2 The normal stress is contributed by axial stress. s = N 30 = - 9.549 ksi = p A The shear stress is contributed by torsional shear stress. t = 4(12)(1) Tc = = 30.56 ksi p J 2 The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. sx = -9.549 ksi, sy = 0, and txy = -30.56 ksi. We have s1, 2 = = sx + sy 2 ; A a sx - sy 2 2 b + txy 2 -9.549 - 0 2 -9.549 + 0 2 ; a b + ( -30.56) 2 A 2 = ( -4.775 ; 30.929) ksi s1 = 26.15 ksi s2 = -35.70 ksi Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which represent the principal stresses, are located inside the shaded region. Therefore, the material does not fail according to Mohr’s failure criteria. 804 30 kip 4 kip · ft 10–75. If the A-36 steel pipe has outer and inner diameters of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-shear-stress theory. 900 N 150 mm A 100 mm 200 mm Internal Loadings. Considering the equilibrium of the free - body diagram of the post’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0 Vy = 0 T = - 360 N # m ©Mx = 0; T + 900(0.4) = 0 ©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz = p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4 J = p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2 Normal Stress and Shear Stress. The normal stress is contributed by bending stress. Thus, sY = - MyA 90(0.015) = = -42.31MPa Iz 10.15625p A 10 - 9 B The shear stress is contributed by torsional shear stress. t = 360(0.015) Tc = 84.62 MPa = J 20.3125p A 10 - 9 B The state of stress at point A is represented by the two - dimensional element shown in Fig. b. In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 = = sx + sz 2 ; A a sx - sz 2 2 b + txz 2 -42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 2 A 2 = (-21.16 ; 87.23) MPa s1 = 66.07 MPa s2 = -108.38 MPa 805 200 mm 900 N 10–75. Continued Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires |s1 - s2| = sallow 66.07 - (-108.38) = sallow sallow = 174.45 MPa The factor of safety is F.S. = sY 250 = = 1.43 sallow 174.45 Ans. 806 *10–76. If the A-36 steel pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-distortion-energy theory. 900 N 150 mm A 100 mm 200 mm Internal Loadings: Considering the equilibrium of the free - body diagram of the pipe’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0 Vy = 0 T = - 360 N # m ©Mx = 0; T + 900(0.4) = 0 ©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz = p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4 J = p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2 Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus, sY = - MyA 90(0.015) = = -42.31MPa Iz 10.15625p A 10 - 9 B The shear stress is caused by torsional stress. t = 360(0.015) Tc = 84.62 MPa = J 20.3125p A 10 - 9 B The state of stress at point A is represented by the two -dimensional element shown in Fig. b. In - Plane Principal Stress. sx = -42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 = = sx + sz 2 ; A a sx - sz 2 2 b + txz 2 -42.31 - 0 2 -42.31 + 0 2 ; a b + 84.62 2 A 2 = (-21.16 ; 87.23) MPa s1 = 66.07 MPa s2 = -108.38 MPa 807 200 mm 900 N 10–76. Continued Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 66.072 - 66.07(-108.38) + (-108.38)2 = sallow 2 sallow = 152.55 MPa Thus, the factor of safety is F.S. = sY 250 = = 1.64 sallow 152.55 Ans. 808 •10–77. The element is subjected to the stresses shown. If sY = 36 ksi, determine the factor of safety for the loading based on the maximum-shear-stress theory. sx = 4 ksi s1, 2 = = sy = - 12 ksi sx + sy 2 ; A a sx - sy 2 txy = -8 ksi 4 ksi 2 b + txy 2 8 ksi 4 - (-12) 2 4 - 12 2 ; a b + (-8) 2 A 2 s1 = 7.314 ksi s2 = -15.314 ksi tabsmax = 7.314 - (-15.314) s1 - s2 = = 11.314 ksi 2 2 tallow = sY 36 = = 18 ksi 2 2 F.S. = 12 ksi tallow 18 = = 1.59 abs tmax 11.314 Ans. 10–78. Solve Prob. 10–77 using the maximum-distortionenergy theory. sx = 4 ksi s1, 2 = = sy = -12 ksi sx + sy 2 ; A a sx - sy txy = -8 ksi 4 ksi 2 b + txy 2 8 ksi 4 - (-12) 2 4 - 12 2 ; a b + (-8) 2 A 2 s1 = 7.314 ksi s2 = -15.314 ksi s1 2 - s1 s2 + s2 2 = a F.S. = 2 12 ksi sY 2 b F.S. 362 = 1.80 A (7.314) - (7.314)(-15.314) + (-15.314)2 Ans. 2 809 10–79. The yield stress for heat-treated beryllium copper is sY = 130 ksi. If this material is subjected to plane stress and elastic failure occurs when one principal stress is 145 ksi, what is the smallest magnitude of the other principal stress? Use the maximum-distortion-energy theory. Maximum Distortion Energy Theory : With s1 = 145 ksi, s21 - s1 s2 + s22 = s2Y 1452 - 145s2 + s22 = 1302 s22 - 145s2 + 4125 = 0 s2 = -(-145) ; 2(-145)2 - 4(1)(4125) 2(1) = 72.5 ; 33.634 Choose the smaller root, s2 = 38.9 ksi Ans. sy ⫽ 0.5sx *10–80. The plate is made of hard copper, which yields at sY = 105 ksi. Using the maximum-shear-stress theory, determine the tensile stress sx that can be applied to the plate if a tensile stress sy = 0.5sx is also applied. s1 = sx sx 1 s2 = sx 2 |s1| = sY sx = 105 ksi Ans. sy ⫽ 0.5sx •10–81. Solve Prob. 10–80 using the maximum-distortionenergy theory. s1 = sx s2 = sx 2 sx s21 - s1 s2 + s22 = s2Y s2x - s2x s2x + = (105)2 2 4 sx = 121 ksi Ans. 810 10–82. The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based on the maximumshear-stress theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi sy = 0 25 ksi txy = 25 ksi 80 ksi In - Plane Principal Stress: Applying Eq. 9-5. s1,2 = = sx + sy 2 ; A a s x - sy 2 2 b + txy 2 80 - 0 2 80 + 0 2 ; a b + 25 2 A 2 = 40 ; 47.170 s1 = 87.170 ksi s2 = -7.170 ksi Maximum Shear Stress Theory: s1 and s2 have opposite signs so |s1 - s2| = sY |87.170 - (-7.170)| = sY sY = 94.3 ksi Ans. 10–83. Solve Prob. 10–82 using the maximum-distortionenergy theory. Normal and Shear Stress: In accordance with the sign convention. sx = 80 ksi sy = 0 txy = 25 ksi In - Plane Principal Stress: Applying Eq. 9-5. s1,2 = = sx + sy 2 ; A a 25 ksi sx - s 2 b + txy 2 2 80 ksi 80 - 0 2 80 + 0 2 ; a b + 25 2 A 2 = 40 ; 47.170 s1 = 87.170 ksi s2 = -7.170 ksi Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2Y 87.1702 - 87.170(-7.170) + (-7.170)2 = s2Y sY = 91.0 ksi Ans. 811 *10–84. A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of sY = 150 ksi. If the bar is subjected to a torque of 30 kip # in. and a bending moment of 56 kip # in., determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding. Normal and Shear Stresses: Applying the flexure and torsion formulas. s = t = 56 A d2 B Mc 1792 = = p d 4 I pd3 4 A2B 30 A d2 B Tc 480 = = 3 p d 4 J pd 2 A2B The critical state of stress is shown in Fig. (a) or (b), where sx = 1792 pd3 sy = 0 txy = 480 pd3 In - Plane Principal Stresses : Applying Eq. 9-5, s1,2 = = = s1 = sx + sy 2 1792 3 pd + 0 2 ; ; A D a ¢ sx - sy 2 1792 3 pd - 0 2 2 b + txy 2 2 ≤ + a 480 2 b pd3 896 1016.47 ; pd3 pd3 1912.47 pd3 s2 = - 120.47 pd3 Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2allow a 1912.47 2 1912.47 120.47 120.47 2 150 2 b - a bab + ab = a b 3 3 3 3 2 pd pd pd pd d = 2.30 in. Ans. 812 •10–85. The state of stress acting at a critical point on a machine element is shown in the figure. Determine the smallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stress theory. 10 ksi The Principal stresses: s1,2 = = sx + sy 2 ; A 4 ksi a sx - sy 2 2 b + txy 2 8 ksi 8 - (-10) 2 8 - 10 2 ; a b + 4 2 A 2 s1 = 8.8489 ksi s2 = -10.8489 ksi Maximum shear stress theory: Both principal stresses have opposite sign. hence, |s1 - s2| = sY 8.8489 - (-10.8489) = sY sY = 19.7 ksi Ans. 10–86. The principal stresses acting at a point on a thinwalled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t, and s3 = 0. If the yield stress is sY, determine the maximum value of p based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory. a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then |s2| = sg 2 pr 2 = sg 2t p = 2t s r g |s1| = sg 2 pr 2 = sg t p = t s (Controls!) r g Ans. b) Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2g a pr 2 pr pr pr 2 b - a b a b + a b = s2g t t 2t 2t p = 2t 23r sg Ans. 813 10–87. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximumshear-stress theory the maximum allowable shear stress is tallow = 116>pd322M2 + T2. Assume the principal stresses to be of opposite algebraic signs. T T M M Section properties : I = p d 4 pd4 a b = ; 4 2 64 J = p d 4 pd4 a b = 2 2 32 Thus, s = M(d2 ) Mc 32 M = = p d4 I pd3 64 t = Tc = J T (d2 ) p d4 32 = 16 T pd3 The principal stresses : s1,2 = = sx + sy 2 ; A a sx - sy 2 2 b + txy 2 16 M 2 16 M 16 T 2 16 M 16 ; + a ; 2M2 + T2 a b = 3 3 b A pd pd p d3 pd3 p d3 Assume s1 and s2 have opposite sign, hence, tallow = 2 C 163 2M2 + T2 D s1 - s2 16 pd 2M2 + T2 = = 2 2 pd3 QED *10–88. If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the maximum-normalstress theory the maximum allowable principal stress is sallow = 116>pd321M + 2M2 + T22. T M M Section properties : I = p d4 ; 64 p d4 32 J = Stress components : s = M (d2 ) Mc 32 M ; = p 4 = I p d3 64 d t = T(d2 ) Tc 16 T = p 4 = J p d3 32 d The principal stresses : s1,2 = = sx + sy 2 ; A a sx - sy 2 b + 2 t2xy = 32 M 3 pd + 0 2 32 M ; 3 D ¢ pd - 0 2 2 ≤ + a 16 T 2 b p d3 16 M 16 ; 2M2 + T2 p d3 p d3 Maximum normal stress theory. Assume s1 7 s2 sallow = s1 = = 16 M 16 + 2M2 + T2 p d3 p d3 16 [M + 2M2 + T2] p d3 QED 814 T •10–89. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-shear-stress theory. Use a factor of safety of 1.5 against yielding. A B T C Shear Stress: This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h = T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B For the solid segment, Js = (tmax)s = p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2 T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 = = sx + sy 2 ; C ¢ sx - sy 2 2 ≤ + t2xy 0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2 s1 = 9947.18T s2 = -9947.18T Maximum Shear Stress Theory. sallow = 80 mm sY 250 = = 166.67 MPa F.S. 1.5 Since s1 and s2 have opposite sings, |s1 - s2| = sallow 9947.18T - ( -9947.18T) = 166.67 A 106 B T = 8377.58 N # m = 8.38 kN # m Ans. 815 80 mm 100 mm T 10–90. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum-distortion-energy theory. Use a factor of safety of 1.5 against yielding. A B T C Shear Stress. This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = A 0.054 - 0.044 B = 1.845p A 10 - 6 B m4. Thus, 2 (tmax)h = T(0.05) Tch = = 8626.28T Jh 1.845p A 10 - 6 B For the solid segment, Js = (tmax)s = p A 0.044 B = 1.28p A 10 - 6 B m4. Thus, 2 T(0.04) Tcs = = 9947.18T Js 1.28p A 10 - 6 B By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 = = sx + sy 2 ; C ¢ sx - sy 2 2 ≤ + t2xy 0 - 0 2 0 + 0 ; ¢ ≤ + (9947.18T)2 2 C 2 s1 = 9947.18T s2 = -9947.18T Maximum Distortion Energy Theory. sallow = 80 mm sY 250 = = 166.67 MPa F.S. 1.5 Then, s1 2 - s1s2 + s2 2 = sallow 2 (9947.18T)2 - (9947.18T)(-9947.18T) + (-9947.18T)2 = C 166.67 A 106 B D 2 T = 9673.60 N # m = 9.67 kN # m Ans. 816 80 mm 100 mm T 10–91. The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb # ft, a bending moment of 1500 lb # ft, and an axial thrust of 2500 lb. If the yield points for tension and shear are sY = 100 ksi and tY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shearstress theory. 2300 lb⭈ft 2500 lb p I = c4 4 A = pc 2 sA = p J = c4 2 1500(12)(c) P 2500 72 000 Mc 2500 + b = -a + b + = -a 4 2 2 pc A I pc pc p c3 4 tA 2300(12)(c) Tc 55 200 = = = 4 p c J p c3 2 s1,2 = sx + sy = -a 2 ; A a sx - sy 2 2 b + txy 2 2500c + 72 000 2 2500 c + 72 000 55200 2 b ; a b + a b 3 3 A 2p c 2p c p c3 (1) Assume s1 and s2 have opposite signs: |s1 - s2| = sg 2500c + 72 000 2 55 200 2 3 b + a b = 100(10 ) 3 A 2p c p c3 2 a (2500c + 72000)2 + 1104002 = 10 000(106)p2 c6 6.25c2 + 360c + 17372.16 - 10 000p2 c6 = 0 By trial and error: c = 0.750 57 in. Substitute c into Eq. (1): s1 = 22 193 psi s2 = -77 807 psi s1 and s2 are of opposite signs OK Therefore, d = 1.50 in. Ans. 817 1500 lb⭈ft *10–92. The gas tank has an inner diameter of 1.50 m and a wall thickness of 25 mm. If it is made from A-36 steel and the tank is pressured to 5 MPa, determine the factor of safety against yielding using (a) the maximum-shear-stress theory, and (b) the maximum-distortion-energy theory. (a) Normal Stress. Since 0.75 r = = 30 7 10, thin - wall analysis can be used.We have t 0.025 s1 = sh = pr 5(0.75) = = 150 MPa t 0.025 s2 = slong = pr 5(0.75) = = 75 MPa 2t 2(0.025) Maximum Shear Stress Theory. s1 and s2 have the sign. Thus, |s1| = sallow sallow = 150 MPa The factor of safety is F.S. = sY 250 = = 1.67 sallow 150 Ans. (b) Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 1502 - 150(75) + 752 = sallow 2 sallow = 129.90 MPa The factor of safety is F.S. = sY 250 = = 1.92 sallow 129.90 Ans. 818 •10–93. The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum-shear-stress theory, and (b) maximumdistortion-energy theory. Apply a factor of safety of 1.5 against yielding. (a) Normal Stress. Assuming that thin - wall analysis is valid, we have s1 = sh = 5 A 106 B (0.75) 3.75 A 106 B pr = = t t t s2 = slong = 5 A 106 B (0.75) 1.875 A 106 B pr = = 2t 2t t Maximum Shear Stress Theory. sallow = 250 A 106 B sY = = 166.67 A 106 B Pa FS. 1.5 s1 and s2 have the same sign. Thus, |s1| = sallow 3.75 A 106 B = 166.67 A 106 B t t = 0.0225 m = 22.5 mm Since Ans. r 0.75 = = 33.3 7 10, thin - wall analysis is valid. t 0.0225 (b) Maximum Distortion Energy Theory. sallow = 250 A 106 B sY = = 166.67 A 106 B Pa F.S. 1.5 Thus, s1 2 - s1s2 + s2 2 = sallow 2 C 3.75 A 106 B t 3.2476 A 106 B t 2 S - C 3.75 A 106 B t SC 1.875 A 106 B t S + C 1.875 A 106 B t 2 S = c166.67 A 106 B d = 166.67 A 106 B t = 0.01949 m = 19.5 mm Since 2 Ans. r 0.75 = = 38.5 7 10, thin - wall analysis is valid. t 0.01949 819 10–94. A thin-walled spherical pressure vessel has an inner radius r, thickness t, and is subjected to an internal pressure p. If the material constants are E and n, determine the strain in the circumferential direction in terms of the stated parameters. s1 = s2 = pr 2t e1 = e2 = e = e = 1 (s - vs) E pr 1 - v 1 - v pr s = a b = (1 - v) E E 2t 2Et Ans. 10–95. The strain at point A on the shell has components Px = 250(10 - 6), Py = 400(10 - 6), gxy = 275(10 - 6), Pz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain. ex = 250(10 - 6) A(250, 137.5)10 - 6 ey = 400(10 - 6) gxy = 275(10 - 6) y A gxy 2 = 137.5(10 - 6) C(325, 0)10 - 6 R = a 2(325 - 250)2 + (137.5)2 b10 - 6 = 156.62(10 - 6) a) e1 = (325 + 156.62)10 - 6 = 482(10 - 6) Ans. e2 = (325 - 156.62)10 - 6 = 168(10 - 6) Ans. b) g max in-plane = 2R = 2(156.62)(10 - 6) = 313(10 - 6) Ans. c) gabs max 2 gabs max = 482(10 - 6) 2 = 482(10 - 6) Ans. 820 x *10–96. The principal plane stresses acting at a point are shown in the figure. If the material is machine steel having a yield stress of sY = 500 MPa, determine the factor of safety with respect to yielding if the maximum-shear-stress theory is considered. 100 MPa 150 MPa Have, the in plane principal stresses are s1 = sy = 100 MPa s2 = sx = -150 MPa Since s1 and s2 have same sign, F.S = sy = |s1 - s2| 500 = 2 |100 - (-150)| Ans. •10–97. The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory. The yield stress for the steel is sY = 650 MPa. 340 MPa 65 MPa 55 MPa sx = -55 MPa s1, 2 = = sx + sy 2 sy = 340 MPa ; A a sx - sy 2 txy = 65 MPa 2 b + txy 2 -55 - 340 2 -55 + 340 2 ; a b + 65 2 A 2 s1 = 350.42 MPa s2 = -65.42 MPa (s1 2 - s1s2 + s2 ) = [350.422 - 350.42(-65.42) + ( -65.42)2] = 150 000 6 s2Y = 422 500 OK No. Ans. 821 10–98. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: Pa = 600110-62, Pb = -700110-62, and Pc = 350110-62. Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains. a 60⬚ 60⬚ b Strain Rosettes (60º): Applying Eq. 10-15 with ex = 600 A 10 - 6 B , c eb = -700 A 10 - 6 B , ec = 350 A 10 - 6 B , ua = 150°, ub = -150° and uc = -90°, 350 A 10 - 6 B = ex cos2 (-90°) + ey sin2( -90°) + gxy sin (-90°) cos ( -90°) ey = 350 A 10 - 6 B 600 A 10 - 6 B = ex cos2 150° + 350 A 10 - 6 B sin2 150° + gxy sin 150° cos 150° 512.5 A 10 - 6 B = 0.75 ex - 0.4330 gxy [1] -700 A 10 - 6 B = ex cos2 ( -150°) + 350 A 10 - 6 B sin2(-150°) + gxy sin (-150°) cos (-150°) -787.5 A 10 - 6 B = 0.75ex + 0.4330 gxy [2] Solving Eq. [1] and [2] yields ex = -183.33 A 10 - 6 B gxy = -1501.11 A 10 - 6 B Construction of she Circle: With ex = -183.33 A 10 - 6 B , ey = 350 A 10 - 6 B , and gxy = -750.56 A 10 - 6 B . 2 eavg = ex + ey 2 = a -183.33 + 350 b A 10 - 6 B = 83.3 A 10 - 6 B 2 Ans. The coordinates for reference points A and C are A(-183.33, -750.56) A 10 - 6 B C(83.33, 0) A 10 - 6 B The radius of the circle is R = a 2(183.33 + 83.33)2 + 750.562 b A 10 - 6 B = 796.52 A 10 - 6 B a) In-plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (83.33 + 796.52) A 10 - 6 B = 880 A 10 - 6 B Ans. e2 = (83.33 - 796.52) A 10 - 6 B = -713 A 10 - 6 B Ans. Orientation of Principal Strain: From the circle, tan 2uP1 = 750.56 = 2.8145 183.33 + 83.33 2uP2 = 70.44° 2uP1 = 180° - 2uP2 uP = 180° - 70.44° = 54.8° (Clockwise) 2 Ans. 822 60⬚ 10–98. Continued b) Maximum In - Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in-plane 2 g max in-plane = -R = -796.52 A 10 - 6 B = -1593 A 10 - 6 B Ans. Orientation of Maximum In-Plane Shear Strain: From the circle. tan 2uP = 183.33 + 83.33 = 0.3553 750.56 uP = 9.78° (Clockwise) Ans. 10–99. A strain gauge forms an angle of 45° with the axis of the 50-mm diameter shaft. If it gives a reading of P = -200110-62 when the torque T is applied to the shaft, determine the magnitude of T. The shaft is made from A-36 steel. T 45⬚ Shear Stress. This is a case of pure shear, and the shear stress developed is p contributed by torsional shear stress. Here, J = A 0.0254 B = 0.1953125p A 10 - 6 B m4. 2 Then 0.128 A 106 B T T(0.025) Tc = = t = p J 0.1953125p A 10 - 6 B T The state of stress at points on the surface of the shaft can be represented by the element shown in Fig. a. Shear Strain: For pure shear ex = ey = 0. We obtain, ea = ex cos2ua + ey sin2ua + gxysin ua cos ua -200 A 10 - 6 B = 0 + 0 + gxy sin 45° cos 45° gxy = -400 A 10 - 6 B Shear Stress and Strain Relation: Applying Hooke’s Law for shear, txy = Ggxy - 0.128 A 106 B T p = 75 A 109 B C -400 A 10 - 6 B D T = 736 N # m Ans. 823 *10–100. The A-36 steel post is subjected to the forces shown. If the strain gauges a and b at point A give readings of Pa = 300110-62 and Pb = 175110-62, determine the magnitudes of P1 and P2. P1 P2 a A Internal Loadings: Considering the equilibrium of the free - body diagram of the 1 in. post’s segment, Fig. a, P2 - V = 0 V = P2 + c ©Fy = 0; N - P1 = 0 N = P1 a + ©MO = 0; M + P2(2) = 0 M = 2P2 Section Properties: The cross - sectional area and the moment of inertia about the bending axis of the post’s cross - section are A = 4(2) = 8 in2 I = 1 (2) A 43 B = 10.667 in4 12 Referring to Fig. b, A Qy B A = x¿A¿ = 1.5(1)(2) = 3 in3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. sA = 2P2(12)(1) MxA P1 N + = + = 2.25P2 - 0.125P1 A I 8 10.667 The shear stress is caused by transverse shear stress. tA = VQA P2(3) = = 0.140625P2 It 10.667(2) Thus, the state of stress at point A is represented on the element shown in Fig. c. Normal and Shear Strain: With ua = 90° and ub = 45°, we have ea = ex cos2ua + ey sin2ua + gxysin ua cos ua 300 A 10 - 6 B = ex cos2 90° + ey sin2 90° + gxysin 90° cos 90° ey = 300 A 10 - 6 B eb = ex cos2ub + ey sin2 ub + gxysin ub cos ub 175 A 10 - 6 B = ex cos2 45° + 300 A 10 - 6 B sin2 45° + gxy sin 45°cos 45° ex + gxy = 50 A 10 - 6 B (1) 824 A 1 in. b 45⬚ c + ©F = 0; : x 2 in. 2 ft A 4 in. c Section c– c 10–100. Continued Since sy = sz = 0, ex = -vey = - 0.32(300) A 10 - 6 B = -96 A 10 - 6 B Then Eq. (1) gives gxy = 146 A 10 - 6 B Stress and Strain Relation: Hooke’s Law for shear gives tx = Ggxy 0.140625P2 = 11.0 A 103 B C 146 A 10 - 6 B D P2 = 11.42 kip = 11.4 kip Ans. Since sy = sz = 0, Hooke’s Law gives sy = Eey 2.25(11.42) - 0.125P1 = 29.0 A 103 B C 300 A 10 - 6 B D P1 = 136 kip Ans. 825 10–101. A differential element is subjected to plane strain that has the following components: Px = 950110-62, Py = 420110-62, gxy = -325110-62. Use the strain-transformation equations and determine (a) the principal strains and (b) the maximum in-plane shear strain and the associated average strain. In each case specify the orientation of the element and show how the strains deform the element. e1, 2 = ex + ey ; 2 = c A a ex - ey 2 2 b + gxy 2 950 + 420 -325 2 950 - 420 2 -6 ; a b + a b d (10 ) 2 A 2 2 e1 = 996(10 - 6) Ans. e2 = 374(10 - 6) Ans. Orientation of e1 and e2 : gxy tan 2uP = ex - ey -325 950 - 420 = uP = -15.76°, 74.24° Use Eq. 10.5 to determine the direction of e1 and e2. ex¿ = ex + ey + 2 ex - ey 2 cos 2u + gxy 2 sin 2u u = uP = -15.76° ex¿ = b (-325) 950 + 420 950 - 420 + cos (-31.52°) + sin (-31.52°) r (10 - 6) = 996(10 - 6) 2 2 2 uP1 = -15.8° Ans. uP2 = 74.2° Ans. b) gmax in-plane 2 gmax in-plane eavg = = A = 2c a ex - ey 2 2 gxy 2 b 2 950 - 420 2 -325 2 -6 -6 b + a b d(10 ) = 622(10 ) A 2 2 a ex + ey 2 b + a = a Ans. 950 + 420 b (10 - 6) = 685(10 - 6) 2 Ans. 826 10–101. Continued Orientation of gmax : -(ex - ey) tan 2uP = gxy -(950 - 420) -325 = uP = 29.2° and uP = 119° Ans. Use Eq. 10.6 to determine the sign of gx¿y¿ 2 = - ex - ey sin 2u + 2 gxy 2 g max in-plane : cos 2u u = uP = 29.2° gx¿y¿ = 2c -(950 - 420) -325 sin (58.4°) + cos (58.4°) d(10 - 6) 2 2 gxy = -622(10 - 6) 10–102. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain on an element at the same point oriented 30° clockwise with respect to the original element. Sketch the results on the element. y Pydy dy Stress Transformation Equations: ex = 400 A 10 - 6 B ey = 200 A 10 - 6 B gxy = -300 A 10 - 6 B u = - 30° gxy 2 dx We obtain, ex¿ = ex + ey 2 = B + ex - ey 2 cos 2u + gxy 2 sin 2u 400 + 200 400 - 200 -300 + cos (-60°) + a b sin (-60°) R A 10 - 6 B 2 2 2 = 480 A 10 - 6 B gx¿y¿ 2 = -¢ Ans. ex - ey 2 ≤ sin 2u + gxy 2 cos 2u gx¿y¿ = [-(400 - 200) sin ( -60°) + (-300) cos ( -60°)] A 10 - 6 B = 23.2 A 10 - 6 B ey¿ = ex + ey = B 2 - Ans. ex - ey 2 gxy 2 cos 2u - gxy 2 sin 2u 400 + 200 400 - 200 -300 cos ( -60°) - a b sin (-60°) R A 10 - 6 B 2 2 2 = 120 A 10 - 6 B Ans. 827 x Pxdx 10–103. The state of plane strain on an element is Px = 400110-62, Py = 200110-62, and gxy = -300110-62. Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding element at the point with respect to the original element. Sketch the results on the element. Construction of the Circle: ex = 400 A 10 -6 y Pydy dy B , ey = 200 A 10 B , and -6 gxy 2 = -150 A 10 -6 B. Thus, eavg = ex + ey 2 = a 400 + 200 b A 10 - 6 B = 300 A 10 - 6 B 2 Ans. The coordinates for reference points A and the center C of the circle are A(400, -150) A 10 - 6 B C(300, 0) A 10 - 6 B The radius of the circle is R = CA = 2(400 - 300)2 + (-150)2 = 180.28 A 10 - 6 B Using these results, the circle is shown in Fig. a. In - Plane Principal Stresses: The coordinates of points B and D represent e1 and e2, respectively. Thus, e1 = (300 + 180.28) A 10 - 6 B = 480 A 10 - 6 B Ans. e2 = (300 - 180.28) A 10 - 6 B = 120 A 10 - 6 B Ans. Orientation of Principal Plane: Referring to the geometry of the circle, tan 2 A up B 1 = 150 = 1.5 400 - 300 A up B 1 = 28.2° (clockwise) Ans. The deformed element for the state of principal strains is shown in Fig. b. Maximum In - Plane Shear Stress: The coordinates of point E represent eavg and gmax . Thus in-plane gmax = -R = -180.28 A 10 - 6 B in-plane 2 gmax in-plane = - 361 A 10 - 6 B Ans. Orientation of the Plane of Maximum In - Plane Shear Strain: Referring to the geometry of the circle, tan 2us = 400 - 300 = 0.6667 150 uS = 16.8° (counterclockwise) Ans. The deformed element for the state of maximum in - plane shear strain is shown in Fig. c. 828 gxy 2 gxy 2 dx x Pxdx 10–103. Continued 829