די L. די Learning Goal: To calculate the maximum axial loads for a column based on yielding and buckling about both axes, and allowing for factors of safety. A column has a 5.5 cm by 9.3 cm rectangular cross section and a height 4 m. The column is fixed at both ends and has a lateral support at mid-height for the weak axis. The material has B = 200 GPa and y = 250 MPa. The critical load P for buckling of a column depends on the vertical distance between supports x²EI and how the column is connected to those supports. The length in the formula P₁ = I² the distance between points of zero moment. For a pin-supported column, this is the full height of the column. For a column that is not pinned at both ends, the points of zero moment are not at the supports. The equation for the critical load can be modified to use an effective length L = KL. where K depends on the support conditions. For a column fixed at both ends, K=0.5. For a column fixed at one end and pinned at the other, K = 0.7 (Figure 1). The equation for the critical load then becomes Part A - Yielding The factor of safety for the column yielding is FS=1.5. What is the maximum load that can be applied based on the column yielding due to the normal stress? Assume the load is applied at the centroid of the section. Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) for Pelit A for Party do for Part.Credo for Part A reefor Part A keyboard shortcuts for Part A help for Part A Pa= ΠΕΙ (KL)² Pmax = Value Units A column can also have lateral supports along its height. A lateral support prevents lateral movement in one direction, which reduces the effective length, Le. Many column sections are not circular. Since the critical load depends on the moment of inertia, it also depends on the axis about which the bending occurs. The axis with the smallest moment of inertia is called the weak axis, and the axis with the largest moment of inertia is called the strong axis. A factor of safety is used to account for uncertainties in material properties and dimensions. The factor of safety is the ratio of the failure load to the allowable load, or the ratio of the maximum design load to the maximum expected load. Submit Request Answer Part B - Strong axis buckling The design is to have a factor of safety for buckling FS=2.5. What is the maximum load based on buckling in the strong axis of the column? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Pstrong for Rpirt B for Partido for Part redo for Part B reseror Part B keyboard shortcuts for Part B help for Part B Value Units Figure 1 of 1 Submit Part C - Weak axis buckling The design is to have a factor of safety for buckling FS=2.5. What is the maximum load based on buckling in the weak axis of the column? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Pweak for Rart for Part do for Part & redo for eart C resor Part C keyboard shortcuts for Part C help for Part C Value Units Submit

די L. די Learning Goal: To calculate the maximum axial loads for a column based on yielding and buckling about both axes, and allowing for factors of safety. A column has a 5.5 cm by 9.3 cm rectangular cross section and a height 4 m. The column is fixed at both ends and has a lateral support at mid-height for the weak axis. The material has B = 200 GPa and y = 250 MPa. The critical load P for buckling of a column depends on the vertical distance between supports x²EI and how the column is connected to those supports. The length in the formula P₁ = I² the distance between points of zero moment. For a pin-supported column, this is the full height of the column. For a column that is not pinned at both ends, the points of zero moment are not at the supports. The equation for the critical load can be modified to use an effective length L = KL. where K depends on the support conditions. For a column fixed at both ends, K=0.5. For a column fixed at one end and pinned at the other, K = 0.7 (Figure 1). The equation for the critical load then becomes Part A - Yielding The factor of safety for the column yielding is FS=1.5. What is the maximum load that can be applied based on the column yielding due to the normal stress? Assume the load is applied at the centroid of the section. Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) for Pelit A for Party do for Part.Credo for Part A reefor Part A keyboard shortcuts for Part A help for Part A Pa= ΠΕΙ (KL)² Pmax = Value Units A column can also have lateral supports along its height. A lateral support prevents lateral movement in one direction, which reduces the effective length, Le. Many column sections are not circular. Since the critical load depends on the moment of inertia, it also depends on the axis about which the bending occurs. The axis with the smallest moment of inertia is called the weak axis, and the axis with the largest moment of inertia is called the strong axis. A factor of safety is used to account for uncertainties in material properties and dimensions. The factor of safety is the ratio of the failure load to the allowable load, or the ratio of the maximum design load to the maximum expected load. Submit Request Answer Part B - Strong axis buckling The design is to have a factor of safety for buckling FS=2.5. What is the maximum load based on buckling in the strong axis of the column? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Pstrong for Rpirt B for Partido for Part redo for Part B reseror Part B keyboard shortcuts for Part B help for Part B Value Units Figure 1 of 1 Submit Part C - Weak axis buckling The design is to have a factor of safety for buckling FS=2.5. What is the maximum load based on buckling in the weak axis of the column? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Pweak for Rart for Part do for Part & redo for eart C resor Part C keyboard shortcuts for Part C help for Part C Value Units Submit