Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis I©The McGraw-Hil Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 77 I Figure 3-9 Equation (3-10)defines two particular values for the angle 2p,one of which defines the maximum normal stress o and the other,the minimum normal stress o2.These two stresses are called the principal stresses,and their corresponding directions,the princi- pal directions.The angle between the principal directions is 90.It is important to note that Eq.(3-10)can be written in the form -sin2中p-xycos20p=0 2 (a) Comparing this with Eq.(3-9),we see that r=0,meaning that the surfaces contain- ing principal stresses have zero shear stresses. In a similar manner,we differentiate Eg.(3-9),set the result equal to zero,and obtain tan 20 =0x -dy (3-11) 2txy Equation (3-11)defines the two values of 20,at which the shear stress t reaches an extreme value.The angle between the surfaces containing the maximum shear stresses is 90.Equation (3-11)can also be written as ax -ay cos 20p+ty sin 20p=0 2 (6) Substituting this into Eq.(3-8)yields a=Ox+ay (3-12) 2 Equation(3-12)tells us that the two surfaces containing the maximum shear stresses also contain equal normal stresses of (ox +oy)/2. Comparing Eqs.(3-10)and (3-11),we see that tan 20,is the negative reciprocal of tan 2p.This means that 2,and 2p are angles 90 apart,and thus the angles between the surfaces containing the maximum shear stresses and the surfaces contain- ing the principal stresses are +45. Formulas for the two principal stresses can be obtained by substituting the angle 2p from Eq.(3-10)in Eq.(3-8).The result is 01,02= 0x+ (3-13) 2
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 82 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 77 Equation (3–10) defines two particular values for the angle 2φp, one of which defines the maximum normal stress σ1 and the other, the minimum normal stress σ2. These two stresses are called the principal stresses, and their corresponding directions, the principal directions. The angle between the principal directions is 90°. It is important to note that Eq. (3–10) can be written in the form σx − σy 2 sin 2φp − τxy cos 2φp = 0 (a) Comparing this with Eq. (3–9), we see that τ = 0, meaning that the surfaces containing principal stresses have zero shear stresses. In a similar manner, we differentiate Eq. (3–9), set the result equal to zero, and obtain tan 2φs = −σx − σy 2τxy (3–11) Equation (3–11) defines the two values of 2φs at which the shear stress τ reaches an extreme value. The angle between the surfaces containing the maximum shear stresses is 90°. Equation (3–11) can also be written as σx − σy 2 cos 2φp + τxy sin 2φp = 0 (b) Substituting this into Eq. (3–8) yields σ = σx + σy 2 (3–12) Equation (3–12) tells us that the two surfaces containing the maximum shear stresses also contain equal normal stresses of (σx + σy )/2. Comparing Eqs. (3–10) and (3–11), we see that tan 2φs is the negative reciprocal of tan 2φp. This means that 2φs and 2φp are angles 90° apart, and thus the angles between the surfaces containing the maximum shear stresses and the surfaces containing the principal stresses are ±45◦. Formulas for the two principal stresses can be obtained by substituting the angle 2φp from Eq. (3–10) in Eq. (3–8). The result is σ1, σ2 = σx + σy 2 ± σx − σy 2 2 + τ 2 xy (3–13) x n y x xy dx ds dy y xy dx ds dy Figure 3–9
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill 83 Mechanical Engineering Companies,2008 Design,Eighth Edition 78 I Mechanical Engineering Desigr In a similar manner the two extreme-value shear stresses are found to be -Ov T,T2= 2 十品 (3-14) Your particular attention is called to the fact that an extreme value of the shear stress may not be the same as the actual maximum value.See Sec.3-7. It is important to note that the equations given to this point are quite sufficient for performing any plane stress transformation.However,extreme care must be exercised when applying them.For example,say you are attempting to determine the principal state of stress for a problem where ox=14 MPa,oy=-10 MPa,and txy =-16 MPa. Equation (3-10)yieldsp=-26.57 and 63.43 to locate the principal stress surfaces, whereas,Eq.(3-13)gives o1 =22 MPa and o2 =-18 MPa for the principal stresses. If all we wanted was the principal stresses,we would be finished.However,what if we wanted to draw the element containing the principal stresses properly oriented rel- ative to the x,y axes?Well,we have two values ofp and two values for the princi- pal stresses.How do we know which value of op corresponds to which value of the principal stress?To clear this up we would need to substitute one of the values ofp into Eq.(3-8)to determine the normal stress corresponding to that angle. A graphical method for expressing the relations developed in this section,called Mohr's circle diagram,is a very effective means of visualizing the stress state at a point and keeping track of the directions of the various components associated with plane stress.Equations (3-8)and(3-9)can be shown to be a set of parametric equations for o and t,where the parameter is 20.The relationship between o and t is that of a cir- cle plotted in the o,plane,where the center of the circle is located at C=(o.r)= [(ox +o)/2.0]and has a radius of R =/[(ox-o)/212+.A problem arises in the sign of the shear stress.The transformation equations are based on a positive o being counterclockwise,as shown in Fig.3-9.If a positive t were plotted above the o axis,points would rotate clockwise on the circle 2 in the opposite direction of rotation on the element.It would be convenient if the rotations were in the same direction.One could solve the problem easily by plotting positive t below the axis. However,the classical approach to Mohr's circle uses a different convention for the shear stress. Mohr's Circle Shear Convention This convention is followed in drawing Mohr's circle: Shear stresses tending to rotate the element clockwise (cw)are plotted above the o axis. Shear stresses tending to rotate the element counterclockwise(ccw)are plotted below the o axis. For example,consider the right face of the element in Fig.3-8b.By Mohr's circle con- vention the shear stress shown is plotted below the o axis because it tends to rotate the element counterclockwise.The shear stress on the top face of the element is plotted above the o axis because it tends to rotate the element clockwise. In Fig.3-10 we create a coordinate system with normal stresses plotted along the abscissa and shear stresses plotted as the ordinates.On the abscissa,tensile (positive) normal stresses are plotted to the right of the origin O and compressive(negative)nor- mal stresses to the left.On the ordinate,clockwise (cw)shear stresses are plotted up; counterclockwise (ccw)shear stresses are plotted down
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 83 Companies, 2008 78 Mechanical Engineering Design In a similar manner the two extreme-value shear stresses are found to be τ1, τ2 = ±σx − σy 2 2 + τ 2 xy (3–14) Your particular attention is called to the fact that an extreme value of the shear stress may not be the same as the actual maximum value. See Sec. 3–7. It is important to note that the equations given to this point are quite sufficient for performing any plane stress transformation. However, extreme care must be exercised when applying them. For example, say you are attempting to determine the principal state of stress for a problem where σx = 14 MPa, σy = −10 MPa, and τxy = −16 MPa. Equation (3–10) yields φp = −26.57◦ and 63.43° to locate the principal stress surfaces, whereas, Eq. (3–13) gives σ1 = 22 MPa and σ2 = −18 MPa for the principal stresses. If all we wanted was the principal stresses, we would be finished. However, what if we wanted to draw the element containing the principal stresses properly oriented relative to the x, y axes? Well, we have two values of φp and two values for the principal stresses. How do we know which value of φp corresponds to which value of the principal stress? To clear this up we would need to substitute one of the values of φp into Eq. (3–8) to determine the normal stress corresponding to that angle. A graphical method for expressing the relations developed in this section, called Mohr’s circle diagram, is a very effective means of visualizing the stress state at a point and keeping track of the directions of the various components associated with plane stress. Equations (3–8) and (3–9) can be shown to be a set of parametric equations for σ and τ , where the parameter is 2φ. The relationship between σ and τ is that of a circle plotted in the σ, τ plane, where the center of the circle is located at C = (σ, τ ) = [(σx + σy )/2, 0] and has a radius of R = [(σx − σy )/2]2 + τ 2 xy . A problem arises in the sign of the shear stress. The transformation equations are based on a positive φ being counterclockwise, as shown in Fig. 3–9. If a positive τ were plotted above the σ axis, points would rotate clockwise on the circle 2φ in the opposite direction of rotation on the element. It would be convenient if the rotations were in the same direction. One could solve the problem easily by plotting positive τ below the axis. However, the classical approach to Mohr’s circle uses a different convention for the shear stress. Mohr’s Circle Shear Convention This convention is followed in drawing Mohr’s circle: • Shear stresses tending to rotate the element clockwise (cw) are plotted above the σ axis. • Shear stresses tending to rotate the element counterclockwise (ccw) are plotted below the σ axis. For example, consider the right face of the element in Fig. 3–8b. By Mohr’s circle convention the shear stress shown is plotted below the σ axis because it tends to rotate the element counterclockwise. The shear stress on the top face of the element is plotted above the σ axis because it tends to rotate the element clockwise. In Fig. 3–10 we create a coordinate system with normal stresses plotted along the abscissa and shear stresses plotted as the ordinates. On the abscissa, tensile (positive) normal stresses are plotted to the right of the origin O and compressive (negative) normal stresses to the left. On the ordinate, clockwise (cw) shear stresses are plotted up; counterclockwise (ccw) shear stresses are plotted down
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 79 Figure 3-10 Mohr's circle diagram. (g-0) 5-" 2 y H g E 2中 D 0 (g,TW x+dy Using the stress state of Fig.3-8b,we plot Mohr's circle,Fig.3-10,by first look- ing at the right surface of the element containing o to establish the sign of or and the cw or ccw direction of the shear stress.The right face is called the x face where =0.If ox is positive and the shear stress txy is ccw as shown in Fig.3-8b,we can establish point A with coordinates (ox.ew)in Fig.3-10.Next,we look at the top y face,where =90,which contains oy,and repeat the process to obtain point B with coordinates (oy.)as shown in Fig.3-10.The two states of stress for the element are△p=90°from each other on the element so they will be2△中=l80°from each other on Mohr's circle.Points A and B are the same vertical distance from the o axis. Thus,AB must be on the diameter of the circle,and the center of the circle C is where AB intersects the o axis.With points A and B on the circle,and center C,the complete circle can then be drawn.Note that the extended ends of line AB are labeled x and y as references to the normals to the surfaces for which points A and B represent the stresses. The entire Mohr's circle represents the state of stress at a single point in a struc- ture.Each point on the circle represents the stress state for a specific surface intersect- ing the point in the structure.Each pair of points on the circle 180 apart represent the state of stress on an element whose surfaces are 90 apart.Once the circle is drawn,the states of stress can be visualized for various surfaces intersecting the point being ana- lyzed.For example,the principal stresses o and o2 are points D and E,respectively. and their values obviously agree with Eq.(3-13).We also see that the shear stresses are zero on the surfaces containing o and o2.The two extreme-value shear stresses,one clockwise and one counterclockwise,occur at F and G with magnitudes equal to the radius of the circle.The surfaces at F and G each also contain normal stresses of (ox+o)/2 as noted earlier in Eq.(3-12).Finally,the state of stress on an arbitrary surface located at an angle o counterclockwise from the x face is point H
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 84 © The McGraw−Hill Companies, 2008 Using the stress state of Fig. 3–8b, we plot Mohr’s circle, Fig. 3–10, by first looking at the right surface of the element containing σx to establish the sign of σx and the cw or ccw direction of the shear stress. The right face is called the x face where φ = 0◦. If σx is positive and the shear stress τxy is ccw as shown in Fig. 3–8b, we can establish point A with coordinates (σx , τ ccw xy ) in Fig. 3–10. Next, we look at the top y face, where φ = 90◦, which contains σy , and repeat the process to obtain point B with coordinates (σy , τ cw xy ) as shown in Fig. 3–10. The two states of stress for the element are φ = 90◦ from each other on the element so they will be 2φ = 180◦ from each other on Mohr’s circle. Points A and B are the same vertical distance from the σ axis. Thus, AB must be on the diameter of the circle, and the center of the circle C is where AB intersects the σ axis. With points A and B on the circle, and center C, the complete circle can then be drawn. Note that the extended ends of line AB are labeled x and y as references to the normals to the surfaces for which points A and B represent the stresses. The entire Mohr’s circle represents the state of stress at a single point in a structure. Each point on the circle represents the stress state for a specific surface intersecting the point in the structure. Each pair of points on the circle 180° apart represent the state of stress on an element whose surfaces are 90° apart. Once the circle is drawn, the states of stress can be visualized for various surfaces intersecting the point being analyzed. For example, the principal stresses σ1 and σ2 are points D and E, respectively, and their values obviously agree with Eq. (3–13). We also see that the shear stresses are zero on the surfaces containing σ1 and σ2. The two extreme-value shear stresses, one clockwise and one counterclockwise, occur at F and G with magnitudes equal to the radius of the circle. The surfaces at F and G each also contain normal stresses of (σx + σy )/2 as noted earlier in Eq. (3–12). Finally, the state of stress on an arbitrary surface located at an angle φ counterclockwise from the x face is point H. Load and Stress Analysis 79 x y (x – y) O x + y 2 x – y 2 F (y , xy cw) (x , ccw) y B C G D H E xy 2 y x 1 2 A 2p xy x cw ccw x – y 2 + 2 xy 2 xy Figure 3–10 Mohr’s circle diagram.
Budynas-Nisbett:Shigley's I.Basics 3.Load and Stress Analysis T©The McGraw-Hill Mechanical Engineering Companies,2008 Design,Eighth Edition 80 Mechanical Engineering Design At one time,Mohr's circle was used graphically where it was drawn to scale very accurately and values were measured by using a scale and protractor.Here,we are strictly using Mohr's circle as a visualization aid and will use a semigraphical approach,calculat- ing values from the properties of the circle.This is illustrated by the following example. EXAMPLE 3-4 A stress element has ox=80 MPa and txy=50 MPa cw,as shown in Fig.3-11a. (a)Using Mohr's circle,find the principal stresses and directions,and show these on a stress element correctly aligned with respect to the xy coordinates.Draw another stress element to show t and t2,find the corresponding normal stresses,and label the drawing completely. (b)Repeat part a using the transformation equations only. Solution (a)In the semigraphical approach used here,we first make an approximate freehand sketch of Mohr's circle and then use the geometry of the figure to obtain the desired information. Draw the o and t axes first(Fig.3-11b)and from the x face locate ox=80 MPa along the o axis.On the x face of the element,we see that the shear stress is 50 MPa in the cw direction.Thus.for the x face.this establishes point A (80,50)MPa. Corresponding to the y face,the stress is o=0 and t=50 MPa in the ccw direction. This locates point B(0,50)MPa.The line AB forms the diameter of the required cir- cle,which can now be drawn.The intersection of the circle with the o axis defines o and o2 as shown.Now,noting the triangle ACD,indicate on the sketch the length of the legs AD and CD as 50 and 40 MPa,respectively.The length of the hypotenuse AC is Answer t1=√(50)2+(40)2=64.0MPa and this should be labeled on the sketch too.Since intersection C is 40 MPa from the origin,the principal stresses are now found to be Answer o1=40+64=104 MPa and o2=40-64=-24MPa The angle 26 from the x axis cw to o is Answer 24p=tan-1器=51.3° To draw the principal stress element(Fig.3-11c),sketch the x and y axes parallel to the original axes.The angle p on the stress element must be measured in the same direction as is the angle 2p on the Mohr circle.Thus,fromx measure 25.7(half of 51.3)clockwise to locate the o axis.The o2 axis is 90 from the o axis and the stress element can now be completed and labeled as shown.Note that there are no shear stresses on this element. The two maximum shear stresses occur at points E and F in Fig.3-11b.The two normal stresses corresponding to these shear stresses are each 40 MPa,as indicated. Point E is 38.7 ccw from point A on Mohr's circle.Therefore,in Fig.3-11d,draw a stress element oriented 19.3(half of 38.7)ccw from x.The element should then be labeled with magnitudes and directions as shown. In constructing these stress elements it is important to indicate thex and y direc- tions of the original reference system.This completes the link between the original machine element and the orientation of its principal stresses
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGraw−Hill 85 Companies, 2008 80 Mechanical Engineering Design At one time, Mohr’s circle was used graphically where it was drawn to scale very accurately and values were measured by using a scale and protractor. Here, we are strictly using Mohr’s circle as a visualization aid and will use a semigraphical approach, calculating values from the properties of the circle. This is illustrated by the following example. EXAMPLE 3–4 A stress element has σx = 80 MPa and τxy = 50 MPa cw, as shown in Fig. 3–11a. (a) Using Mohr’s circle, find the principal stresses and directions, and show these on a stress element correctly aligned with respect to the xy coordinates. Draw another stress element to show τ1 and τ2, find the corresponding normal stresses, and label the drawing completely. (b) Repeat part a using the transformation equations only. Solution (a) In the semigraphical approach used here, we first make an approximate freehand sketch of Mohr’s circle and then use the geometry of the figure to obtain the desired information. Draw the σ and τ axes first (Fig. 3–11b) and from the x face locate σx = 80 MPa along the σ axis. On the x face of the element, we see that the shear stress is 50 MPa in the cw direction. Thus, for the x face, this establishes point A (80, 50cw) MPa. Corresponding to the y face, the stress is σ = 0 and τ = 50 MPa in the ccw direction. This locates point B (0, 50ccw) MPa. The line AB forms the diameter of the required circle, which can now be drawn. The intersection of the circle with the σ axis defines σ1 and σ2 as shown. Now, noting the triangle AC D, indicate on the sketch the length of the legs AD and C D as 50 and 40 MPa, respectively. The length of the hypotenuse AC is Answer τ1 = (50)2 + (40)2 = 64.0 MPa and this should be labeled on the sketch too. Since intersection C is 40 MPa from the origin, the principal stresses are now found to be Answer σ1 = 40 + 64 = 104 MPa and σ2 = 40 − 64 = −24 MPa The angle 2φ from the x axis cw to σ1 is Answer 2φp = tan−1 50 40 = 51.3◦ To draw the principal stress element (Fig. 3–11c), sketch the x and y axes parallel to the original axes. The angle φp on the stress element must be measured in the same direction as is the angle 2φp on the Mohr circle. Thus, from x measure 25.7° (half of 51.3°) clockwise to locate the σ1 axis. The σ2 axis is 90° from the σ1 axis and the stress element can now be completed and labeled as shown. Note that there are no shear stresses on this element. The two maximum shear stresses occur at points E and F in Fig. 3–11b. The two normal stresses corresponding to these shear stresses are each 40 MPa, as indicated. Point E is 38.7° ccw from point A on Mohr’s circle. Therefore, in Fig. 3–11d, draw a stress element oriented 19.3° (half of 38.7°) ccw from x. The element should then be labeled with magnitudes and directions as shown. In constructing these stress elements it is important to indicate the x and y directions of the original reference system. This completes the link between the original machine element and the orientation of its principal stresses
Budynas-Nisbett:Shigley's1.Basics 3.Load and Stress Analysis T©The McGraw--il Mechanical Engineering Companies,2008 Design,Eighth Edition Load and Stress Analysis 81 Figure 3-11 All stresses in MPa. (80.505 A 38.7° 150 0 9=0 /51.30 0 (a) 花 40 =80 (0.50r F 四 -24 Answer 1=64 r,=10 (c) (b)The transformation equations are programmable.From Eq.(3-10). =tan-1 2）=m(20） =-25.7°，64.3° Ox-v 80 From Eq.(3-8),for the first anglep=-25.7°， 80+080-0 0= 2 2 cos[2(-25.7)]+(-50)sin[2(-25.7)]=104.03MPa The shear on this surface is obtained from Eq.(3-9)as 2sim2(-25.71+(-50)cos2(-25.71=0MPa 80-0 T=- which confirms that 104.03 MPa is a principal stress.From Eq.(3-8),forp=64.3, 80+0,80-0 0= L 2 2 -cos2(64.3)1+(-50)sin2(64.3)1=-24.03MPa
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis 86 © The McGraw−Hill Companies, 2008 Load and Stress Analysis 81 (b) The transformation equations are programmable. From Eq. (3–10), φp = 1 2 tan−1 2τxy σx − σy = 1 2 tan−1 2(−50) 80 = −25.7◦ , 64.3◦ From Eq. (3–8), for the first angle φp = −25.7◦, σ = 80 + 0 2 + 80 − 0 2 cos[2(−25.7)] + (−50)sin[2(−25.7)] = 104.03 MPa The shear on this surface is obtained from Eq. (3–9) as τ = −80 − 0 2 sin[2(−25.7)] + (−50) cos[2(−25.7)] = 0 MPa which confirms that 104.03 MPa is a principal stress. From Eq. (3–8), for φp = 64.3◦, σ = 80 + 0 2 + 80 − 0 2 cos[2(64.3)] + (−50)sin[2(64.3)] = −24.03 MPa Figure 3–11 All stresses in MPa. cw ccw 1 x A C D E 38.7° 64 50 51.3° 40 40 2p 1 x = 80 y = 0 2 B F 2 y (b) (80, 50cw) (0, 50ccw) (c) y 2 2 = –24 1 = 104 25.7° x 1 (d) y = 40 2 = 64 1 = 64 19.3° = 40 x E F (a) y x 50 50 80 Answer