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An aluminum column with a length of L and a rectangular cross section has a fixed end B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry of the column but allow it to move in the other plane. (a) Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling. (b) Design the most efficient cross section for the column, knowing that L = 20 in, E = 10.1 x 105 psi, P = 5 kips, and a factor of safety of 2.5 is required.

An aluminum column with a length of L and a rectangular cross section has a fixed end B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry of the column but allow it to move in the other plane. (a) Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling. (b) Design the most efficient cross section for the column, knowing that L = 20 in, E = 10.1 x 105 psi, P = 5 kips, and a factor of safety of 2.5 is required.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Page701.ConsiderSampleProblem10.1.Solve the same problem, but this time replace thepinned boundary condition on top with a fixed boundary condition

Sample Problem 10.1 An aluminum column with a length of L and a rectangular cross section has a fixed end B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry of the column but allow it to move in the other plane. (a) Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling. (b) Design the most efficient cross section for the column, knowing that L = 20 in., E = 10.1 x 106 psi, P = 5 kips, and a factor of safety of 2.5 is required. STRATEGY: The most efficient design is that for which the critical stresses corresponding to the two possible buckling modes are equal. This occurs if the two critical stresses obtained from Eq. (10.13b) are the same. Thus for this problem, the two effective slenderness ratios in this equation must be equal to solve part a. Use Fig. 10.18 to determine the effective lengths. The design data can then be used with Eq. (10.13b) to size the cross section for part b. MODELING: Buckling in xy Plane. Referring to Fig. 10.18c, the effective length of the column with respect to buckling in this plane is L, = 0.7L. The radius of gyration of the cross section is obtained by and since I = Ar?, 1₁=ba A = ab ba A ab 10.1 Stability of Structures 701 L₂ 0.7L = T₂a/√12 The effective slenderness ratio of the column with respect to buckling in the xy plane is r₂ = a/√12 42L = r, b/√12 (1) Buckling in xz Plane. Referring to Fig. 10.18a, the effective length of the column with respect to buckling in this plane is L, = 2L, and the corre- sponding radius of gyration is r, = b/√12. Thus, (2) (continued)
Transcribed Image Text:Sample Problem 10.1 An aluminum column with a length of L and a rectangular cross section has a fixed end B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry of the column but allow it to move in the other plane. (a) Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling. (b) Design the most efficient cross section for the column, knowing that L = 20 in., E = 10.1 x 106 psi, P = 5 kips, and a factor of safety of 2.5 is required. STRATEGY: The most efficient design is that for which the critical stresses corresponding to the two possible buckling modes are equal. This occurs if the two critical stresses obtained from Eq. (10.13b) are the same. Thus for this problem, the two effective slenderness ratios in this equation must be equal to solve part a. Use Fig. 10.18 to determine the effective lengths. The design data can then be used with Eq. (10.13b) to size the cross section for part b. MODELING: Buckling in xy Plane. Referring to Fig. 10.18c, the effective length of the column with respect to buckling in this plane is L, = 0.7L. The radius of gyration of the cross section is obtained by and since I = Ar?, 1₁=ba A = ab ba A ab 10.1 Stability of Structures 701 L₂ 0.7L = T₂a/√12 The effective slenderness ratio of the column with respect to buckling in the xy plane is r₂ = a/√12 42L = r, b/√12 (1) Buckling in xz Plane. Referring to Fig. 10.18a, the effective length of the column with respect to buckling in this plane is L, = 2L, and the corre- sponding radius of gyration is r, = b/√12. Thus, (2) (continued)
ANALYSIS: a. Most Efficient Design. The most efficient design is when the criti- cal stresses corresponding to the two possible modes of buckling are equal. Referring to Eq. (10.13b), this is the case if the two values obtained above for the effective slenderness ratio are equal. and solving for the ratio a/b, 0.7L a/√12 Using a = 0.35b, A = ab 0.35b² and % = ²E (L./r)² b. Design for Given Data. Since F.S. = 2.5 is required, 2L b/√12 P ===== A a P = (F.S.)P = (2.5) (5 kips) = 12.5 kips 12,500 lb 0.35b² 12,500 lb 0.35b² 0.7 2 Making L = 20 in. in Eq. (2), L./r, = 138.6/b. Substituting for E, L/r, and into Eq. (10.13b) gives ローb (10.1 x 10º psi) (138.6/b)² b = 1.620 in. = 0.35 a = 0.35b = 0.567 in. REFLECT and THINK: The calculated critical Euler buckling stress can never be taken to exceed the yield strength of the material. In this problem, you can readily determine that the critical stress = 13.6 ksi; though the specific alloy was not given, this stress is less than the tensile yield strength by values for all aluminum alloys listed in Appendix D.
Transcribed Image Text:ANALYSIS: a. Most Efficient Design. The most efficient design is when the criti- cal stresses corresponding to the two possible modes of buckling are equal. Referring to Eq. (10.13b), this is the case if the two values obtained above for the effective slenderness ratio are equal. and solving for the ratio a/b, 0.7L a/√12 Using a = 0.35b, A = ab 0.35b² and % = ²E (L./r)² b. Design for Given Data. Since F.S. = 2.5 is required, 2L b/√12 P ===== A a P = (F.S.)P = (2.5) (5 kips) = 12.5 kips 12,500 lb 0.35b² 12,500 lb 0.35b² 0.7 2 Making L = 20 in. in Eq. (2), L./r, = 138.6/b. Substituting for E, L/r, and into Eq. (10.13b) gives ローb (10.1 x 10º psi) (138.6/b)² b = 1.620 in. = 0.35 a = 0.35b = 0.567 in. REFLECT and THINK: The calculated critical Euler buckling stress can never be taken to exceed the yield strength of the material. In this problem, you can readily determine that the critical stress = 13.6 ksi; though the specific alloy was not given, this stress is less than the tensile yield strength by values for all aluminum alloys listed in Appendix D.
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