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