- #1
Terra_Bitz
- 3
- 0
- Homework Statement
- Hi,
I'm stuck on answering the attached question, I know how to find the axial within the column however I do know what the question means by justifying a chosen material, help would be much appreciated.
Thanks in advance!
Alfie
- Relevant Equations
- The way how I found out calculate axial load is here: https://www.hunker.com/13400857/how-to-calculate-the-axial-load
To find axial load , 5m / 6m = 0.83
Arc tangent , tan-1(0.83) = 39.69
Cosine of the force , Cos(39.69) = 0.77
Sine of the force , Sin(39.69) = 0.64
Axial load in the vertical direction , (200kN/m X 6 = 1200kN / 2 = 600kN + self weight of 6kN = 606kN) So 606kN X 0.77 = 466.62kN
Axial load in the horizontal direction , (200kN/m X 6 = 1200kN / 2 = 600kN + self weight of 6kN = 606kN) So 606kN X 0.64 = 387.84kN
Arc tangent , tan-1(0.83) = 39.69
Cosine of the force , Cos(39.69) = 0.77
Sine of the force , Sin(39.69) = 0.64
Axial load in the vertical direction , (200kN/m X 6 = 1200kN / 2 = 600kN + self weight of 6kN = 606kN) So 606kN X 0.77 = 466.62kN
Axial load in the horizontal direction , (200kN/m X 6 = 1200kN / 2 = 600kN + self weight of 6kN = 606kN) So 606kN X 0.64 = 387.84kN
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