Time left 0:36:40Question 5Answer savedMarked out of 7.00Flag questionQuestion textProblem 2:A steel pin-jointed truss is shown in Figure P2. A load of P kN is applied to joint C resulting in the following internal axial member forces;FAC = FCB = kN and FAB = kN where a negative value indicates a compression force.Assume the member lengths L = mmYou are required to carry out a strength limit state design of the three members. Note the diagonal members will be the same as they experience the same axial force. The diagonal members, BC and AC are to be designed as circular hollow sections of outer radius 50mm while the member AB is to be a solid circular section. It is known that load intensity may increase over time so you are to use a factor of safety of in your design.Assume values for elastic modulus E=200,000MPa and yield strength = 300MPa. When doing this question only consider in-plane failure.
Question
Time left 0:36:40Question 5Answer savedMarked out of 7.00Flag questionQuestion textProblem 2:A steel pin-jointed truss is shown in Figure P2. A load of P kN is applied to joint C resulting in the following internal axial member forces;FAC = FCB = kN and FAB = kN where a negative value indicates a compression force.Assume the member lengths L = mmYou are required to carry out a strength limit state design of the three members. Note the diagonal members will be the same as they experience the same axial force. The diagonal members, BC and AC are to be designed as circular hollow sections of outer radius 50mm while the member AB is to be a solid circular section. It is known that load intensity may increase over time so you are to use a factor of safety of in your design.Assume values for elastic modulus E=200,000MPa and yield strength = 300MPa. When doing this question only consider in-plane failure.
Solution
I'm sorry, but you didn't provide a specific question related to the problem. However, I can guide you on how to approach this problem.
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Identify the forces: The problem states that the forces in members AC and BC are equal, and there's a different force in member AB. The load P is applied at joint C.
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Determine the stress: Stress is force divided by area. For the hollow sections (AC and BC), the area is π(Outer radius^2 - Inner radius^2). For the solid section (AB), the area is π(radius^2).
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Apply the factor of safety: The problem asks for a strength limit state design, which means the design should be safe under the maximum expected load. This is where the factor of safety comes in. Multiply the stress by the factor of safety to get the design stress.
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Check against yield strength: The design stress should not exceed the yield strength of the material. If it does, the member will deform permanently.
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If the design stress is too high, you may need to choose a larger section or a stronger material.
Remember to keep units consistent throughout your calculations.
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