Toy models for the falling chimney Gabriele Varieschi* and Kaoru Kamiya Department of Physics, Loyola Marymount University One LMU Drive, Los Angeles, CA 90045 (Dated: May 22, 2006) Abstract In this paper we review the theory of the "falling chimney", which deals with the breaking in mid-air of tall structures, when they fall to the ground. We show that these ruptures can be caused by either shear forces, typically developing near the base, or by the bending of the structure, which is caused primarily by the internal bending moment In the latter case the breaking is more likely to occur between one third and one half of the height of the chimney. Small scale toy models are used to reproduce the dynamics of the falling chimney By examining photos taken during the fall of these models we test the adequacy of the outlined theory. This type of experiment, easy to perform and conceptually challenging, can become part of a rotational mechanics lab for undergraduate student
arXiv:physics/0210033 v2 17 Feb 2003 Toy models for the falling chimney Gabriele Varieschi ∗ and Kaoru Kamiya † Department of Physics, Loyola Marymount University, One LMU Drive, Los Angeles, CA 90045 (Dated: May 22, 2006) Abstract In this paper we review the theory of the “falling chimney”, which deals with the breaking in mid-air of tall structures, when they fall to the ground. We show that these ruptures can be caused by either shear forces, typically developing near the base, or by the bending of the structure, which is caused primarily by the internal bending moment. In the latter case the breaking is more likely to occur between one third and one half of the height of the chimney. Small scale toy models are used to reproduce the dynamics of the falling chimney. By examining photos taken during the fall of these models we test the adequacy of the outlined theory. This type of experiment, easy to perform and conceptually challenging, can become part of a rotational mechanics lab for undergraduate students. 1
I. INTRODUCTION One of the most interesting demonstrations for an introductory mechanics course is the Falling Chimney-Free Fall Paradox, as it was named by Sutton in his classical book Demonstration Experiments in Physics. In the original version of this demonstration a ball is placed at one end of a uniform stick, which is pivoted at the other end and makes initially an angle of about 30 with the horizontal. The elevated end of the stick is suddenly dropped together with the ball, thus showing a very counter-intuitive behavior. The falling end of the stick accelerates at a greater rate than the free-falling ball, proving that its acceleration is greater than g, the acceleration of gravity A simplified version of the experiment can be performed just with a meter stick and a coin. The stick is supported in the horizontal position by two fingers, placed near the two ends. a coin is set on the stick near one end, which is suddenly released. The effect is similar to the previous demonstration: the falling end of the rotating stick eventually acquires ar acceleration greater than that of the freely falling coin, which loses contact with the stick surface and lags behind the falling stick a description of the first version of the experiment can be found in almost every book of physics demonstrations, ,3,4 sometimes with different names("Falling Stick","Hinged Stick can be found on-line in several web-pages (see our web-page cn and Falling Ball", and others). Photographic descriptions or even video-clips of this demo or a collection of related links In addition, countless papers exist in the literature; we have traced several of these, from the 1930's to the present. Some of the earliest discussions can be found in Constantinides and Ludeke(as well in the book by Sutton), followed by many others. 8. 9, 10, 11,2These concentrate mostly on the simple explanation of the effect, which relies on the concept of "center of percussion"of the rotating stick(a simple introduction to this concept can be found in Bloomfield). This particular point of the stick (located at a distance from the hinged end equal to two thirds of the length, for a uniform stick) is moving with the same acceleration as a particle under gravity, constrained to move along the same circular path. Points on the stick beyond the center of percussion descend with accelerations greater than that of particles freely moving under gravity, on their respective circular paths. As a consequence of this, if the initial angle formed by the stick with the horizontal is less than about 35, the end point will possess at all times a vertical component of the acceleration greater than g, producing the effect described above Several variations of the basic demonstration also exist, 14, 15, 16, 17, 18, 19 the majority of which suggest attaching an additional mass to the rotating stick at different positions. The effect for the student or the viewer is even less intuitive than the original version: an additional mass placed near the end of the stick actually reduces the acceleration of the end point affecting substantially the outcome of the experiment. In general the addition of a mass at any point on the stick will increase both the total torque on the system(thus increasing th rotational acceleration) and the moment of inertia of the system around the axis of rotation (resulting in a decreased rotational acceleration). The center of percussion of the stick still plays a key role: if the additional mass is placed beyond it, the effect of the increased Momen of inertia dominates and the acceleration of the rotational motion will be reduced. If the placed before the center of percussion, the increase in the torque will dominate and the rotational motion will be enhanced. The effect is null if the mass is placed exactly at the center of percussion(a complete discussion of this effect can be found in Bartlettand
I. INTRODUCTION One of the most interesting demonstrations for an introductory mechanics course is the “Falling Chimney - Free Fall Paradox,” as it was named by Sutton in his classical book Demonstration Experiments in Physics.1 In the original version of this demonstration a ball is placed at one end of a uniform stick, which is pivoted at the other end and makes initially an angle of about 30◦ with the horizontal. The elevated end of the stick is suddenly dropped, together with the ball, thus showing a very counter-intuitive behavior. The falling end of the stick accelerates at a greater rate than the free-falling ball, proving that its acceleration is greater than g, the acceleration of gravity. A simplified version of the experiment can be performed just with a meter stick and a coin. The stick is supported in the horizontal position by two fingers, placed near the two ends. A coin is set on the stick near one end, which is suddenly released. The effect is similar to the previous demonstration: the falling end of the rotating stick eventually acquires an acceleration greater than that of the freely falling coin, which loses contact with the stick surface and lags behind the falling stick. A description of the first version of the experiment can be found in almost every book of physics demonstrations,2,3,4 sometimes with different names (“Falling Stick”, “Hinged Stick and Falling Ball”, and others). Photographic descriptions or even video-clips of this demo can be found on-line in several web-pages (see our web-page,5 for a collection of related links). In addition, countless papers exist in the literature; we have traced several of these, from the 1930’s to the present. Some of the earliest discussions can be found in Constantinides6 and Ludeke7 (as well in the book by Sutton1 ), followed by many others.8,9,10,11,12 These concentrate mostly on the simple explanation of the effect, which relies on the concept of “center of percussion” of the rotating stick (a simple introduction to this concept can be found in Bloomfield13). This particular point of the stick (located at a distance from the hinged end equal to two thirds of the length, for a uniform stick) is moving with the same acceleration as a particle under gravity, constrained to move along the same circular path. Points on the stick beyond the center of percussion descend with accelerations greater than that of particles freely moving under gravity, on their respective circular paths. As a consequence of this, if the initial angle formed by the stick with the horizontal is less than about 35◦ , the end point will possess at all times a vertical component of the acceleration greater than g, producing the effect described above. Several variations of the basic demonstration also exist,14,15,16,17,18,19 the majority of which suggest attaching an additional mass to the rotating stick at different positions. The effect for the student or the viewer is even less intuitive than the original version: an additional mass placed near the end of the stick actually reduces the acceleration of the end point, affecting substantially the outcome of the experiment. In general the addition of a mass at any point on the stick will increase both the total torque on the system (thus increasing the rotational acceleration) and the moment of inertia of the system around the axis of rotation (resulting in a decreased rotational acceleration). The center of percussion of the stick still plays a key role: if the additional mass is placed beyond it, the effect of the increased moment of inertia dominates and the acceleration of the rotational motion will be reduced. If the mass is placed before the center of percussion, the increase in the torque will dominate and the rotational motion will be enhanced. The effect is null if the mass is placed exactly at the center of percussion (a complete discussion of this effect can be found in Bartlett17 and 2
Haber-Schaim) The next logical step is to analyze the behavior of a real falling chimney. almost in- variably a tall chimney, falling to the ground like the stick in the previous discussion, will break in mid-air at some characteristic height. This is well documented in several photos reproduced in the literature, such as the one which appeared on the cover of the September 1976 issue of The Physics Teacher(other photos can be found in Bundy20 and Bartlett, 21 or also on our web-page ). The causes of such breaking, the height of the rupture point and the angle at which the breaking is most likely to occur, are the most natural questions which arise The first analysis- of this problem compared the fall of the real chimney to the fall of the hinged stick, but wrongly identified the center of percussion(at about two thirds of the height)as the probable point of rupture. Reynolds s first identified the possible causes of the breaking with the shear forces and the bending moment originating within the structure of the toppling chimney. More detailed analyses were given by Bundy20 and Madsen24 (the most complete papers we found on the subject)while simplified explanations are also reported 2125, 26 It even appears in graduate student study guides,27, 28 although the chimney is shown bending the wrong way in one of these books In this paper we review the theory of the real falling chimney, outlined by Madsen, 24 iming for a complete and clear explanation of this phenomenon in Sects. II and Ill. Then in Sect. IV, we propose simple ways of using small scale models (literally toy models- made with toy blocks and bricks) to test effectively the outlined theory. More information on these toy models can also be found on our web-site, together with photos and movie clips of the experiments we have performed ROTATIONAL MOTION OF THE FALLING CHIIMNEY The rotational motion of a falling chimney under gravity is equivalent to that of the falling hinged stick of Sect. I. We can describe it as in Fig. 1, where we use polar coordinates r and 6(with er and ee as unit vectors) for the position of an arbitrary point A on the longitudinal axis of the chimney, measuring the angle 0 from the vertical direction. We treat the chimney as a uniform rigid body of mass m and height H, under the action of its weight w=mg, applied to the center of gravity(basically the center of mass -CM-of the body ), and a force F exerted by the ground on the base of the chimney, assumed to act at a single point (we neglect air resistance, or any other applied force). In plane polar coordinates W=Wrer+weee=-mg cos Ber + mg sin eee F=Fer+ (1b) The moment of inertia of the chimney can be approximated with the one for a uniform thin rod, with rotation axis perpendicular to the length and passing through one end Applying the torque equation 10= T2, for a rotation around the origin, with an external torque given by T2 =mgH sin e, we find the angular acceleration T2 3 9 H
Haber-Schaim19). The next logical step is to analyze the behavior of a real falling chimney. Almost invariably a tall chimney, falling to the ground like the stick in the previous discussion, will break in mid-air at some characteristic height. This is well documented in several photos reproduced in the literature, such as the one which appeared on the cover of the September 1976 issue of The Physics Teacher (other photos can be found in Bundy20 and Bartlett,21 or also on our web-page5 ). The causes of such breaking, the height of the rupture point and the angle at which the breaking is most likely to occur, are the most natural questions which arise. The first analysis22 of this problem compared the fall of the real chimney to the fall of the hinged stick, but wrongly identified the center of percussion (at about two thirds of the height) as the probable point of rupture. Reynolds23 first identified the possible causes of the breaking with the shear forces and the bending moment originating within the structure of the toppling chimney. More detailed analyses were given by Bundy20 and Madsen24 (the most complete papers we found on the subject) while simplified explanations are also reported.21,25,26 It even appears in graduate student study guides,27,28 although the chimney is shown bending the wrong way in one of these books. In this paper we review the theory of the real falling chimney, outlined by Madsen,24 aiming for a complete and clear explanation of this phenomenon in Sects. II and III. Then, in Sect. IV, we propose simple ways of using small scale models (literally toy models - made with toy blocks and bricks) to test effectively the outlined theory. More information on these toy models can also be found on our web-site,5 together with photos and movie clips of the experiments we have performed. II. ROTATIONAL MOTION OF THE FALLING CHIMNEY The rotational motion of a falling chimney under gravity is equivalent to that of the falling hinged stick of Sect. I. We can describe it as in Fig. 1, where we use polar coordinates r and θ (with ebr and beθ as unit vectors) for the position of an arbitrary point A on the longitudinal axis of the chimney, measuring the angle θ from the vertical direction. We treat the chimney as a uniform rigid body of mass m and height H, under the action of its weight W = mg, applied to the center of gravity (basically the center of mass -CM- of the body), and a force F exerted by the ground on the base of the chimney, assumed to act at a single point (we neglect air resistance, or any other applied force). In plane polar coordinates: W = Wrebr + Wθbeθ = −mg cos θber + mg sin θbeθ (1a) F = Frebr + Fθbeθ. (1b) The moment of inertia of the chimney can be approximated with the one for a uniform thin rod, with rotation axis perpendicular to the length and passing through one end:35 I = 1 3 mH2 . (2) Applying the torque equation I .. θ = τz, for a rotation around the origin, with an external torque given by τz = mg H 2 sin θ, we find the angular acceleration .. θ = τz I = 3 2 g H sin θ. (3) 3
H FIG. 1: The falling chimney described as a rotating uniform stick. The external forces are the weight of the body applied to the center of gravity, and the constraint force at the base A simple integration, using 0= d= d0=2g sin 8, gives the angular velocity cOS suming that the chimney starts moving from rest and is initially in the vertical direction A further integration of Eg. 4, can lead to g(t) in terms of elliptic integrals We recall that the acceleration in polar coordinates can be written as a=r=(r-re er+ (r8+ 2re)ee, so that, for a point A at a fixed distance r from the origin, it becomes a=aer +agee=-ro er treee For a point at two thirds of the height, r=zH, combining Eqs. 3 and 5 we get aB(r 3H)=3H0= gsin 0, proving that this particular point is the center of percussion of the body, as already mentioned in Sect. I The torque equation allowed us to determine the angular acceleration of the motion in Eq. 3. We can use this result and Newton's second law for the motion of the center of mass (CM) of the whole chimney to determine the unknown force F at the base. The vectorial luation Is mraM=W+F
er e 0 W A H r X Y F FIG. 1: The falling chimney described as a rotating uniform stick. The external forces are the weight of the body applied to the center of gravity, and the constraint force at the base. A simple integration, using .. θ = d . θ dt = d . θ dθ . θ = 3 2 g H sin θ, gives the angular velocity . θ 2 = 3 g H (1 − cos θ), (4) assuming that the chimney starts moving from rest and is initially in the vertical direction. A further integration of Eq. 4, can lead to θ(t) in terms of elliptic integrals. We recall that the acceleration in polar coordinates can be written as a= .. r=(.. r −r . θ 2 )ber + (r .. θ + 2 . r . θ)beθ, so that, for a point A at a fixed distance r from the origin, it becomes a = arber + aθbeθ = −r . θ 2 ber + r .. θbeθ. (5) For a point at two thirds of the height, r = 2 3H, combining Eqs. 3 and 5 we get aθ(r = 2 3H) = 2 3H .. θ = g sin θ, proving that this particular point is the center of percussion of the body, as already mentioned in Sect. I. The torque equation allowed us to determine the angular acceleration of the motion in Eq. 3. We can use this result and Newton’s second law for the motion of the center of mass (CM) of the whole chimney to determine the unknown force F at the base. The vectorial equation is m .. rCM = W + F (6) 4
人N FIG. 2: The forces acting on the lower portion of the chimney, due to the upper part and the action of the constraint at the base, are shown here. The two insets explain the definition of the bending moment in terms of a couple of forces. The resulting deformation of the structure is also shown for the two possible cases which, for r=H, splits into radial and angular equations, H m Fr- mg cos 8 (7 mo-b= Fo+mg sin g Using egs. 3 and 4, the two components of the force F are easily determined 5 3 F Fe 'ng sin (8b) III INTERNAL FORCES AND BENDING MOMENT We now move to the analysis of the internal forces which develop inside the structure of the falling chimney. The resulting stresses and bending moment are the causes of the rupture of the toppling chimney. Consider, as in Fig. 2, an arbitrary lower portion of the himney of height r(as opposed to the total height H)and the forces acting on this part of the structure due to the upper portion and the base The weight of the lower portion is now
b a Nb -f f -f f Nb Nb X Y S 0 W(r) r F P FIG. 2: The forces acting on the lower portion of the chimney, due to the upper part and the action of the constraint at the base, are shown here. The two insets explain the definition of the bending moment in terms of a couple of forces. The resulting deformation of the structure is also shown for the two possible cases. which, for r = H 2 , splits into radial and angular equations, −m H 2 . θ 2 = Fr − mg cos θ (7a) m H 2 .. θ = Fθ + mg sin θ. (7b) Using Eqs. 3 and 4, the two components of the force F are easily determined: Fr = 5 2 mg cos θ − 3 5 (8a) Fθ = − 1 4 mg sin θ. (8b) III. INTERNAL FORCES AND BENDING MOMENT We now move to the analysis of the internal forces which develop inside the structure of the falling chimney. The resulting stresses and bending moment are the causes of the rupture of the toppling chimney. Consider, as in Fig. 2, an arbitrary lower portion of the chimney of height r (as opposed to the total height H) and the forces acting on this part of the structure due to the upper portion and the base. The weight of the lower portion is now 5
W(r)= mgf(assuming again a uniform structure, so that the weight is proportional to the height of the considered portion and it is applied to the center of gravity of this lower portion(at a distance 5 from the origin). In polar coordinates W(r)=Wr(r)er+ Wo(r)ee=-mgg cos er + mgw sin eee The force F, applied at the base is still the same as in eqs. &a, &b, but we have to add the action of the upper part on the lower portion. We follow here the general analysis of the internal forces and moments which can be found in every textbook on Statics(see for example, su) and which can be easily adapted to our case. The distribution of all the internal forces. at the cross section being considered. can be equivalently described by a resultant force and a resultant moment acting at a specific point of the cross section(typically the "centroid"of the sectioned area, in our case simply the central point of the section, on the longitudinal axis). In particular, the resultant force can be decomposed into a transverse shearing force S= Seee, and a longitudinal stress force (tension or compression)P= Prer, applied as in Fig. 2, at the cross section between the upper and lower portions, and assumed positive in the e, er direction respectively In addition we have to consider the resultant moment of the forces at the cross section which is usually called the "bending moment " Nb, because its effect will ultimately result in bending the structure. It is represented in the picture by the curved arrow. Since we treat this as a plane problem, the bending moment can only have a component perpendicular to the plane of the figure, i.e., in the z direction. No other components are considered here, in particular we assume that no torsional moment exist in the structure, which would tend to twist the chimney around its longitudinal axis. The bending moment Nb= Nbez can be thought as originating from a couple of forces f and -f, that can be regarded as applied to the leading and trailing edge of the structure at the cross section considered. This couple of forces is shown explicitly in the papers by Bundy 2 and Madsen, but we prefer to use directly Nb in our treatment, because the bending moment is the result of the whole distribution of forces at the cross section considered. The two small insets inside Fig. 2 explain the definition of the bending moment in terms of the couple of forces f and -f. We also show the resulting deformation of the structure due to a"(diagram a), or a"counter-clockwise"bending moment diagram b). The latter case will be the actual deformation of the falling chimney. Nb will be assumed to be positive if it acts as in the figure, i.e., a positive component of the torque in the z direction(we assume here the use of a right-handed system of coordinate axis). In the following we will refer to Nb as the bending moment, acting on the lower portion of the hiney Again, we will consider the torque equation and the second law for the motion of the center of mass (located at 5)for just the lower portion of the chimney(of mass fm). It is better to analyze the CM motion first. The vector equation w(r)+F+P+S, will split into the radial and angular directions (1-cos6) mg cos 0+5mg cos6-3 +P(11a) 2H 4mg sin e-=mgT sinAI 4mg sin 8+ Se, (11b)
W(r) = mg r H (assuming again a uniform structure, so that the weight is proportional to the height of the considered portion) and it is applied to the center of gravity of this lower portion (at a distance r 2 from the origin). In polar coordinates: W(r) = Wr(r)ber + Wθ(r)beθ = −mg r H cos θber + mg r H sin θbeθ. (9) The force F, applied at the base, is still the same as in Eqs. 8a, 8b, but we have to add the action of the upper part on the lower portion. We follow here the general analysis of the internal forces and moments which can be found in every textbook on Statics (see for example29,30) and which can be easily adapted to our case. The distribution of all the internal forces, at the cross section being considered, can be equivalently described by a resultant force and a resultant moment acting at a specific point of the cross section (typically the “centroid” of the sectioned area, in our case simply the central point of the section, on the longitudinal axis). In particular, the resultant force can be decomposed into a transverse shearing force S = Sθbeθ, and a longitudinal stress force (tension or compression) P = Prber, applied as in Fig. 2, at the cross section between the upper and lower portions, and assumed positive in the beθ, ber direction respectively. In addition, we have to consider the resultant moment of the forces at the cross section, which is usually called the “bending moment” Nb, because its effect will ultimately result in bending the structure. It is represented in the picture by the curved arrow. Since we treat this as a plane problem, the bending moment can only have a component perpendicular to the plane of the figure, i.e., in the z direction. No other components are considered here, in particular we assume that no torsional moment exist in the structure, which would tend to twist the chimney around its longitudinal axis. The bending moment Nb = Nbbez can be thought as originating from a couple of forces, f and −f, that can be regarded as applied to the leading and trailing edge of the structure, at the cross section considered. This couple of forces is shown explicitly in the papers by Bundy20 and Madsen,24 but we prefer to use directly Nb in our treatment, because the bending moment is the result of the whole distribution of forces at the cross section considered. The two small insets inside Fig. 2 explain the definition of the bending moment in terms of the couple of forces f and −f. We also show the resulting deformation of the structure due to a “clockwise” (diagram a), or a “counter-clockwise” bending moment (diagram b). The latter case will be the actual deformation of the falling chimney. Nb will be assumed to be positive if it acts as in the figure, i.e., a positive component of the torque in the z direction (we assume here the use of a right-handed system of coordinate axis). In the following we will refer to Nb as the bending moment, acting on the lower portion of the chimney.36 Again, we will consider the torque equation and the second law for the motion of the center of mass (located at r 2 ) for just the lower portion of the chimney (of mass r H m). It is better to analyze the CM motion first. The vector equation m r H .. rCM = W(r) + F + P + S, (10) will split into the radial and angular directions, − m r 2 2H . θ 2 = − 3 2 mg(1 − cos θ) r 2 H2 = −mg r H cos θ + 5 2 mg cos θ − 3 5 + Pr (11a) m r 2 2H .. θ = 3 4 mg sin θ r 2 H2 = mg r H sin θ − 1 4 mg sin θ + Sθ, (11b) 6
having used Eqs. 3, 4, 8a, 8b, and 9. We can solve for the longitudinal and transverse forces r 5+3 0-3(1+ (12a) H ng s 723H3 which depend on the fraction of height t, the angle of rotation 0, and also the total weight mg. Following the analysis by Bundy, 20 we plot these two forces in Figs. 3 and 4 respectively normalized to the total weight mg, as a function of the height fraction, for several angles From Fig. 3 we see that P is negative(a compression) for smaller angles, but eventually becomes positive(a tension) for angles greater than about 45. Pr also depends critically on h(for 6=00, Pr represents simply the compression due to the weight of the upper part acting on the lower part ). This longitudinal force will be combined later with the bending moment to determine the total stress at the leading and trailing edges, which is the most typical cause of the rupture In Fig. 4 we plot the(transverse) shear force Se, which can be the other leading cause of rupture. It is easily seen that, for any considered angle, the magnitude of the shear force Sel has an absolute maximum at f=0(and a positive value), meaning that large shear forces, in the ee direction, usually originate near the base. The shear force is always zero at one third of the height, and Sel also has a(relative) maximum at 2H (with a negative value, therefore Se is in the -ee direction), but this value is smaller than the one near the From this analysis, it is typically concluded that if the structure breaks from shear stress alone, this is usually more likely to happen near the base. This can be seen for example in the already mentioned cover photo of the September 1976 issue of The Physics Teacher showing the fall of a chimney in Detroit. The two ruptures at the bottom are likely due to shear forces, while a third rupture can be seen at about r=0.47H, and this is due to the combination of bending moment and longitudinal force Pr, as we will explain in the following. More photos and detailed pictorial descriptions of chimney ruptures can be found in the paper by bundy ,20 The "bending moment" Nb can be calculated from the torque equation I(r)8=T2, where now I(r)=imEr is the moment of inertia of just the lower part. 0 will come from Eq. 3 and the total external torque is now T= 5We(r)+rSe+Nb. Using Eqs. 9 and 12b, we can solve the torque equation for the bending moment, obtainin sine r or in a non-dimensional form =0(1-方) which is plotted in Fig. 5, as a function of the height fraction and for various angles Nb is always negative, showing that it is actually directed in the opposite way of Fig. 2 (or as in diagram b of Fig. 2). This particular direction of the bending moment will induce a tension in the leading edge of the chimney and a compression in the trailing edge. The
having used Eqs. 3, 4, 8a, 8b, and 9. We can solve for the longitudinal and transverse forces Pr = − 1 2 mg 1 − r H h5 + 3 r H cos θ − 3 1 + r H i (12a) Sθ = 3 4 mg sin θ r 2 H2 − 4 3 r H + 1 3 , (12b) which depend on the fraction of height r H , the angle of rotation θ, and also the total weight mg. Following the analysis by Bundy,20 we plot these two forces in Figs. 3 and 4 respectively, normalized to the total weight mg, as a function of the height fraction, for several angles. From Fig. 3 we see that Pr is negative (a compression) for smaller angles, but eventually becomes positive (a tension) for angles greater than about 45 ◦ . Pr also depends critically on r H (for θ = 0◦ , Pr represents simply the compression due to the weight of the upper part acting on the lower part). This longitudinal force will be combined later with the bending moment to determine the total stress at the leading and trailing edges, which is the most typical cause of the rupture. In Fig. 4 we plot the (transverse) shear force Sθ, which can be the other leading cause of rupture. It is easily seen that, for any considered angle, the magnitude of the shear force |Sθ| has an absolute maximum at r H = 0 (and a positive value), meaning that large shear forces, in the beθ direction, usually originate near the base. The shear force is always zero at one third of the height, and |Sθ| also has a (relative) maximum at 2 3H (with a negative value, therefore Sθ is in the −beθ direction), but this value is smaller than the one near the base. From this analysis, it is typically concluded that if the structure breaks from shear stress alone, this is usually more likely to happen near the base. This can be seen for example in the already mentioned cover photo of the September 1976 issue of The Physics Teacher,5,21 showing the fall of a chimney in Detroit. The two ruptures at the bottom are likely due to shear forces, while a third rupture can be seen at about r = 0.47H, and this is due to the combination of bending moment and longitudinal force Pr, as we will explain in the following. More photos and detailed pictorial descriptions of chimney ruptures can be found in the paper by Bundy.5,20 The “bending moment” Nb can be calculated from the torque equation I(r) .. θ = τz, where now I(r) = 1 3m r H r 2 is the moment of inertia of just the lower part. .. θ will come from Eq. 3, and the total external torque is now τz = r 2Wθ(r) + rSθ + Nb. Using Eqs. 9 and 12b, we can solve the torque equation for the bending moment, obtaining Nb = − 1 4 mg sin θ r 1 − r H 2 , (13) or, in a non-dimensional form Nb mgH = − 1 4 sin θ r H 1 − r H 2 , (14) which is plotted in Fig. 5, as a function of the height fraction and for various angles. Nb is always negative, showing that it is actually directed in the opposite way of Fig. 2 (or as in diagram b of Fig. 2). This particular direction of the bending moment will induce a tension in the leading edge of the chimney and a compression in the trailing edge. The 7
1.5 90 75 0.5 60 45 0.5 30 20 09 r=H/3 2H/3 Oo 02 0.4 0.6 0.8 r/h FIG. 3: The longitudinal stress force per unit weight of the chimney is shown as a function of the height fraction for several angles. Positive values indicate tensions, while negative values represent compressions
FIG. 3: The longitudinal stress force per unit weight of the chimney is shown as a function of the height fraction for several angles. Positive values indicate tensions, while negative values represent compressions. 8
0.3 r=H/3 Ir=2H/3 90° 0.2 75 60° 5 0.1 300 20 10 0 0.1 02 0.4 0.6 0.8 r/h FIG. 4: The transverse shear force per unit weight of the chimney is shown as a function of the height fraction for several angles. Positive values are for forces in the ee direction
FIG. 4: The transverse shear force per unit weight of the chimney is shown as a function of the height fraction for several angles. Positive values are for forces in the ebθ direction. 9
0 10 0.01 20 0.02 30 r=H/ i r=2H/3 45° 0.03 60 5 90 -0.04 02 0.4 0.6 0.8 r/h FIG. 5: The bending moment, divided by the weight and the height of the chimney, is shown as a function of the height fraction for several angle
FIG. 5: The bending moment, divided by the weight and the height of the chimney, is shown as a function of the height fraction for several angles. 10