In this example, the load trace will involve the framing for a small deck addition to a residence. Once post reactions have been determined, a preliminary footing size will be designed assuming the soil capacity of 3,000 psf is known from a geotechnical investigation. Loads: Live Load = 60 #/ft.^2

Question

In this example, the load trace will involve the framing for a small deck addition to a residence. Once post reactions have been determined, a preliminary footing size will be designed assuming the soil capacity of 3,000 psf is known from a geotechnical investigation.

Loads:

Live Load =60 [latex]\# /ft.^{2} [/latex]

Dead Loads:

Decking = 5 [latex]\# /ft.^{2} [/latex]

Beams = 5 [latex]\# /ft. [/latex]

Girder = 10 [latex]\# /ft. [/latex]

[latex]\gamma_{concrete}[/latex] = 150[latex]\# /ft.^{3} [/latex] (density)

For this load-trace problem, we will investigate the following:

1. Draw an [latex]FBD[/latex] of the typical beam with its load condition shown.

2. Draw an [latex]FBD[/latex] of the girder with its load conditions shown.

3. Determine the load in each post.

4. Determine the size x of the critical pier footing
(account for the weight of the concrete).

Accepted Answer

1. Beam (typical interior):

Dead Load:

[latex]5 \space psf (4^{\prime})[/latex]=[latex]20\# /ft. [/latex]

Beam wt. =[latex]5\# /ft. [/latex]

[latex]\omega _{DL} [/latex] =[latex]25\# /ft. [/latex]

Live Load:

[latex]60 \space psf (4^{\prime})[/latex] =[latex]240\# /ft. [/latex]

[latex]\omega _{DL+ LL} [/latex] =[latex]265\# /ft. [/latex]

2. Beam (typical exterior):

Dead Load:

[latex]5 \space psf (2^{\prime})[/latex]=[latex]10\# /ft. [/latex]

Beam wt. =[latex]5\# /ft. [/latex]

[latex]\omega _{DL} [/latex] =[latex]15\# /ft. [/latex]

Live Load:

[latex]60 \space psf (2^{\prime})[/latex] =[latex]120\# /ft. [/latex]

[latex]\omega _{DL+ LL} [/latex] =[latex]135\# /ft. [/latex]

3/4. Girder and post:

[latex]\left[\sum{M_{B}=0 } ight] - 3,066 \# (4 ^{\prime} ) -3,066 \# (8^{\prime} )-1,562\# (12^{\prime} )-(10\# /ft.) (12^{\prime} ) (6^{\prime} )+A_{y}(8^{\prime} )=0[/latex]

[latex] herefore A_{y}=7,032\# [/latex]

[latex]\left[\sum{F_{y}=0 } ight]- \frac{3,066\# }{2}- 3,066\# -3,066\# -1,562\# - (10\# /ft.) (12^{\prime} ) +B_{y}(right \space side)+7,032\# =0[/latex]

[latex] herefore B_{yR}=2,315\#[/latex]

The total reaction at post B is the sum of the reactions:

[latex]B_{yR}+B_{yL}=4,630\# [/latex]

5. Critical footing:

The soil is capable of resisting a total bearing pressure of [latex]3,000\# /ft.^{2}[/latex]

Note: pressure =[latex]\frac{load}{area}; \space \space q=\frac{P}{A} [/latex]

By setting [latex]q=3,000\# /ft.^{2}[/latex]. (allowable capacity of the soil), we need to deduct the weight of the footing itself to determine the footing’s capacity to resist applied load from above. Therefore,

[latex]q_{net}=q-[/latex] footing weight (as pressure measured in psf)

Footing weight can be solved by converting the density of concrete ([latex]\gamma_{concrete} =150 \# /ft.^{3}[/latex]) into equivalent pounds per square foot units by multiplying:

footing wt. (psf)=([latex]\gamma_{concrete}[/latex]) (thickness of concrete in feet)

[latex] herefore [/latex] footing wt. = [latex](150 \# /ft.^{3})(10^{\prime \prime} /12 \space in. /ft. )[/latex]

[latex]=125\# /ft.^{2}[/latex]

The remaining soil capacity to resist point loads is expressed as

[latex] q_{net} =3,000\# /ft.^{2}-125 \# /ft.^{2} =2,875 \# /ft.^{2}[/latex]

Because pressure =[latex]\frac{load}{area}[/latex]

[latex]q_{net}=\frac{P}{A} =\frac{P}{x^{2}}[/latex]

[latex] herefore x^{2} =\frac{P}{q_{net}} =\frac{7,032\# }{2,875 \# /ft.^{2}} =2.45 ft.^{2}[/latex]

[latex] herefore x =1.57^{\prime } =1^{\prime }7^{\prime \prime }[/latex] square footing (theoretical size)

Practical size may be [latex] x =2^{\prime }0^{\prime \prime }[/latex]

Question 4.4: In this example, the load trace will involve the framing for...

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