A simple carport is framed using knee-braced frames spaced at five feet on center. Assuming wind pressure at 20 psf, analyze a typical interior frame.

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Each interior knee-braced frame is required to resist a load applied to a tributary wall area of 20 square feet at 20 psf.

[latex]F=p imes A=20 \# /ft.^{2} imes 20 ft.^{2}=400\# [/latex]

Notice that in the [latex]FBD[/latex] of the frame, a total of four support reactions develop at [latex]A[/latex] and [latex]B[/latex]. Because only three equations of equilibrium are permitted, all support reactions cannot be solved unless an assumption(s) is made about the frame or its load distribution characteristics. In this case, only one assumption is necessary, because the external support condition is indeterminate to the first degree.

Assume

[latex]A_{x}=B_{x}[/latex]

Then,

[latex] \left[\sum{F_{x}=0 } ight] A_{x}+B_{x}=400\# [/latex]

[latex] herefore A_{x}=B_{x}=200\# [/latex]

Writing the other equations of equilibrium, we get

[latex]\left[\sum{M_{A}=0 } ight] -400 \# (8^\prime )+B_{y}(10^\prime )=0 [/latex]

[latex]B_{y}=320\# [/latex]

[latex] \left[\sum{F_{y}=0 } ight] -A_{y}+B_{y}=0[/latex]

[latex]A_{y}=320\# [/latex]

Once the support reactions have been determined, pass a vertical section a-a through one knee-brace. Isolate the left column, and draw an [latex]FBD[/latex].

[latex]\left[\sum{M_{D}=0 } ight] +CE_{x}(2^\prime ) -200 \# (8^\prime )=0 [/latex]

[latex] herefore CE_{x}=800\# [/latex]

[latex] CE=800\sqrt{2} \# [/latex]

[latex]CE_{y}=800\# [/latex]

[latex] \left[\sum{F_{x}=0 } ight]+400\# -A_{x}+CE_{x}-DE_{x} =0[/latex]

Substituting for [latex]A_{x}[/latex] and [latex]CE_{x}[/latex] ,

[latex] DE_{x}=400\# -200\#+800\#[/latex]

[latex] DE_{x} =+1,000\#[/latex]

[latex] \left[\sum{F_{y}=0 } ight]- DE_{y}+ CE_{y}-A_{y} =0[/latex]

[latex] DE_{y}=800\# -320 \# [/latex]

[latex]DE_{y}=+480\# [/latex]

In a similar manner, isolate the right column using section cut b-b.

[latex]\left[\sum{M_{G}=0 } ight] -200 \# (8^\prime ) +FH_{x}(2^\prime ) =0[/latex]

[latex]FH_{x}=+800\# [/latex]

[latex]FH=800\sqrt{2} \# [/latex]

[latex]FH_{y}=800\# [/latex]

[latex] \left[\sum{F_{y}=0 } ight]+ FG_{y}- FH_{y}+B_{y} =0[/latex]

[latex] FG_{y}=800\# -320\# =480\# [/latex]

[latex]\left[\sum{F_{x}=0 } ight]-FG_{x}+ FH_{x}-B_{x} =0 [/latex]

[latex] FG_{x}=800\# -200\# =600\# [/latex]

Question 4.8: A simple carport is framed using knee-braced frames spaced a...

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