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. Thehaving 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