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The critical load of the column is determined. In the case considered, the column is divided into two equal parts, joined by a torsional spring.You Are asked to analyze a similar system. This time, the upper section is one quarter of the column length, and the lower section is three quarters of the column length.a.What is the critical load?

The critical load of the column is determined. In the case considered, the column is divided into two equal parts, joined by a torsional spring.You Are asked to analyze a similar system. This time, the upper section is one quarter of the column length, and the lower section is three quarters of the column length.a.What is the critical load?

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Page 692 to 694.In these pages,the problem of Fig.10.3 is analyzed.The critical load of the column is determined. In the case considered, the column is divided into two equal parts, joined by a torsional spring.You Are asked to analyze a similar system. This time, the upper section is one quarter of the column length, and the lower section is three quarters of the column length.a.What is the critical load?b.Briefly give your thoughts regarding comparison of the answer in the book and your answer



If the two rods and forces P and P' are perfectly aligned, the system will remain in the position of equilibrium shown in Fig.10.4a as long as it is not disturbed. But suppose we move C slightly to the right so that each rod forms a small angle A0 with the vertical (Fig. 10.4b). Will the system return to its original equilibrium position, or will it move further away? In the first case, the system is stable; in the second, it is unstable. To determine whether the two-rod system is stable or unstable, con- sider the forces acting on rod AC (Fig. 10.5). These forces consist of the couple formed by P and P' of moment P(L/2) sin A0, which tends to move the rod away from the vertical, and the couple M exerted by the spring, which tends to bring the rod back into its original vertical position. Since the angle of deflection of the spring is 2 A0, the moment of couple M is M = K(2 A0). If the moment of the second couple is larger than the moment of the first couple, the system tends to return to its original equi- librium position; the system is stable. If the moment of the first couple is larger than the moment of the second couple, the system tends to move 694 Columns Be (a) (b) Fig. 10.6 (0) Model column in buckled position, (b) free-body diagram of rod AC. B Fig. 10.4 Free-body diagram of model column (a) perfectly aligned, (b) point C moved slightly out of alignment. A or A0 M PL 0 4K sin 8 away from its original equilibrium position; the system is unstable. The load when the two couples balance each other is called the critical load, Pa, which is given as 1/2 Fig. 10.5 Free-body diagram of rod AC in unaligned position. Par (L/2)sin A0 = K(2 A0) (10.1) or since sin A0 A0, when the displacement of C is very small (at the immediate onset of buckling), P₁ = 4K/L Clearly, the system is stable for P < P and unstable for P> Par Assume that a load P > Per has been applied to the two rods of Fig. 10.3 and the system has been disturbed. Since P> Per, the system will move further away from the vertical and, after some oscillations, will settle into a new equilibrium position (Fig. 10.6a). Considering the equilibrium of the free body AC (Fig. 10.6b), an equation similar to Eq. (10.1) but involving the finite angle 8, is P(L/2) sin = K(20) (10.2) (10.3) The value of corresponding to the equilibrium position in Fig. 10.6 is obtained by solving Eq. (10.3) by trial and error. But for any positive value of 0, sin 0 <0. Thus, Eq. (10.3) yields a value of 0 different from zero only when the left-hand member of the equation is larger than one. Recalling Eq. (10.2), this is true only if P> P. But, if P < Pa, the second equilibrium position shown in Fig. 10.6 would not exist, and the only possible equilibrium Fig. 10.1 (repeated). Fig. 10.2 (repeated). position would be the one corresponding to 0 = 0. Thus, for P < Pa, the position where 0 = 0 must be stable. This observation applies to structures and mechanical systems in general and is used in the next section for the stability of elastic columns.
Transcribed Image Text:If the two rods and forces P and P' are perfectly aligned, the system will remain in the position of equilibrium shown in Fig.10.4a as long as it is not disturbed. But suppose we move C slightly to the right so that each rod forms a small angle A0 with the vertical (Fig. 10.4b). Will the system return to its original equilibrium position, or will it move further away? In the first case, the system is stable; in the second, it is unstable. To determine whether the two-rod system is stable or unstable, con- sider the forces acting on rod AC (Fig. 10.5). These forces consist of the couple formed by P and P' of moment P(L/2) sin A0, which tends to move the rod away from the vertical, and the couple M exerted by the spring, which tends to bring the rod back into its original vertical position. Since the angle of deflection of the spring is 2 A0, the moment of couple M is M = K(2 A0). If the moment of the second couple is larger than the moment of the first couple, the system tends to return to its original equi- librium position; the system is stable. If the moment of the first couple is larger than the moment of the second couple, the system tends to move 694 Columns Be (a) (b) Fig. 10.6 (0) Model column in buckled position, (b) free-body diagram of rod AC. B Fig. 10.4 Free-body diagram of model column (a) perfectly aligned, (b) point C moved slightly out of alignment. A or A0 M PL 0 4K sin 8 away from its original equilibrium position; the system is unstable. The load when the two couples balance each other is called the critical load, Pa, which is given as 1/2 Fig. 10.5 Free-body diagram of rod AC in unaligned position. Par (L/2)sin A0 = K(2 A0) (10.1) or since sin A0 A0, when the displacement of C is very small (at the immediate onset of buckling), P₁ = 4K/L Clearly, the system is stable for P < P and unstable for P> Par Assume that a load P > Per has been applied to the two rods of Fig. 10.3 and the system has been disturbed. Since P> Per, the system will move further away from the vertical and, after some oscillations, will settle into a new equilibrium position (Fig. 10.6a). Considering the equilibrium of the free body AC (Fig. 10.6b), an equation similar to Eq. (10.1) but involving the finite angle 8, is P(L/2) sin = K(20) (10.2) (10.3) The value of corresponding to the equilibrium position in Fig. 10.6 is obtained by solving Eq. (10.3) by trial and error. But for any positive value of 0, sin 0 <0. Thus, Eq. (10.3) yields a value of 0 different from zero only when the left-hand member of the equation is larger than one. Recalling Eq. (10.2), this is true only if P> P. But, if P < Pa, the second equilibrium position shown in Fig. 10.6 would not exist, and the only possible equilibrium Fig. 10.1 (repeated). Fig. 10.2 (repeated). position would be the one corresponding to 0 = 0. Thus, for P < Pa, the position where 0 = 0 must be stable. This observation applies to structures and mechanical systems in general and is used in the next section for the stability of elastic columns.
B Fig. 10.1 Pin-ended Fig. 10.2 Buckled axially loaded column. pin-ended column. -> umns, the Load and Resistance Factor Design method. The design of columns under an eccentric axial load is covered in Sec. 10.4. 10.1 STABILITY OF STRUCTURES Consider the design of a column AB of length L to support a given load P (Fig. 10.1). The column is pin-connected at both ends, and P is a centric axial load. If the cross-sectional area A is selected so that the value 6 = P/A of the stress on a transverse section is less than the allowable stress all for the mate- rial used and the deformation & PL/AE falls within the given specifications, we might conclude that the column has been properly designed. However, it may happen that as the load is applied, the column buckles (Fig. 10.2). Instead of remaining straight, it suddenly becomes sharply curved such as shown in Photo 10.1 Laboratory test showing a buckled column. Courtesy of Fritz Engineering Laboratory, Lehigh University Photo 10.1. Clearly, a column that buckles under the load it is to support is not properly designed. We can gain insight into the stability of elastic columns by consider- L/2 L/2 AG Be 10.1 Stability of Structures Fig. 10.3 Model column made of two rigid rods joined by a torsional spring at C. P - constant K ΔΘ AB B 240 693 с
Transcribed Image Text:B Fig. 10.1 Pin-ended Fig. 10.2 Buckled axially loaded column. pin-ended column. -> umns, the Load and Resistance Factor Design method. The design of columns under an eccentric axial load is covered in Sec. 10.4. 10.1 STABILITY OF STRUCTURES Consider the design of a column AB of length L to support a given load P (Fig. 10.1). The column is pin-connected at both ends, and P is a centric axial load. If the cross-sectional area A is selected so that the value 6 = P/A of the stress on a transverse section is less than the allowable stress all for the mate- rial used and the deformation & PL/AE falls within the given specifications, we might conclude that the column has been properly designed. However, it may happen that as the load is applied, the column buckles (Fig. 10.2). Instead of remaining straight, it suddenly becomes sharply curved such as shown in Photo 10.1 Laboratory test showing a buckled column. Courtesy of Fritz Engineering Laboratory, Lehigh University Photo 10.1. Clearly, a column that buckles under the load it is to support is not properly designed. We can gain insight into the stability of elastic columns by consider- L/2 L/2 AG Be 10.1 Stability of Structures Fig. 10.3 Model column made of two rigid rods joined by a torsional spring at C. P - constant K ΔΘ AB B 240 693 с
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