NCEES problem #108

[chess5329]

Could somebody explain me how you get to this answer of AXIAL force resisted by the brace= 2896 ........why the Rx= 2,048 has to be multiplied by sq root of 2?

 

[Jacob_PE]

Could somebody explain me how you get to this answer of AXIAL force resisted by the brace= 2896 ........why the Rx= 2,048 has to be multiplied by sq root of 2?

When you consider the moments about A, the horizontal force Fx 10' above ground is what you are solving to obtain. The axial force, the force along the alignment of the brace (at 45 deg.) is the Resultant of it's x and y components. Remember that R=(x^2+y^2)^.5. Anyway, just need to translate the Fx of 2048 into R. If you draw a triangle, cos 45 = adj/hyp = Fx/R .... R=Fx/cos45. As for the sqr root of 2. A 45 deg slope is a 1 to 1 slope, x is 1, y is 1, so what's R? It's square root of 2. So if you have Fx and need to translate it to R, the axial force, Fx is x is 1 so translate it by multiplying it by square root of 2.

 

[chess5329]

Could somebody explain me how you get to this answer of AXIAL force resisted by the brace= 2896 ........why the Rx= 2,048 has to be multiplied by sq root of 2?

When you consider the moments about A, the horizontal force Fx 10' above ground is what you are solving to obtain. The axial force, the force along the alignment of the brace (at 45 deg.) is the Resultant of it's x and y components. Remember that R=(x^2+y^2)^.5. Anyway, just need to translate the Fx of 2048 into R. If you draw a triangle, cos 45 = adj/hyp = Fx/R .... R=Fx/cos45. As for the sqr root of 2. A 45 deg slope is a 1 to 1 slope, x is 1, y is 1, so what's R? It's square root of 2. So if you have Fx and need to translate it to R, the axial force, Fx is x is 1 so translate it by multiplying it by square root of 2.

Thanks Jacob, where I read "pay attention to details" arty-smiley-048: