A simple light-framed wood building is subjected to the load conditions as specified. Using FBDs and equations of equilibrium, trace the loads through the building for the following elements: 1. Determine the equivalent (horizontally projected) load on the rafters. 2. Determine the load per foot on

Question

A simple light-framed wood building is subjected to the load conditions as specified. Using [latex]FBDs[/latex] and equations of equilibrium, trace the loads through the building for the following elements:
1. Determine the equivalent (horizontally projected) load on the rafters.
2. Determine the load per foot on the bearing wall.
3. Determine the load on the ridge beam.
4. Determine the column loads.
5. Determine the minimum width of the continuous
foundation.
6. Determine the size of the interior footings.

Load Conditions:

Soil Bearing Pressure = 2,000 psf

Flooring = 2 psf

Subfloor = 5 psf

Joists = 4 psf

LL (occupancy) = 40 psf

SL = 25 psf

Walls = 7 psf

Along Rafter Length:

Roofing = 5 psf

Sheathing = 3 psf

Rafters = 2 psf

Ceiling = 2 psf

Ridge beam spans 16 feet from column support to column support.

Accepted Answer

1. Rafters:

[latex]SL[/latex] = 25 psf: rafters spaced 16[latex]^{\prime \prime}[/latex] on center.

[latex] \omega_{SL} =(25 \# /ft.^{2})\left(\frac{16}{12} ight)^\prime =33.3 \# /ft.[/latex]

(Horizontal projection)

Roof [latex]DL[/latex] = 12 psf

[latex] \omega_{DL} =(12 \# /ft.^{2})\left(\frac{16}{12} ight)^\prime =16 \# /ft.[/latex]

(Along rafter length)

[latex] \omega^\prime_{DL} =\left(\frac{13}{12} ight) (16 \# /ft.)=17.3\# /ft.[/latex]

(Equivalent load horizontally projected)

[latex]\omega_{total}=\omega_{SL}+\omega^\prime_{DL}[/latex]

[latex]\omega_{total}=33.3 \# /ft.+17.3\# /ft.=50.6\# /ft.[/latex]

The reaction at each rafter support may be determined using equilibrium equations. When uniform loads on a simply supported member are present, a simple formula may be used where

[latex] R=\frac{ \omega L}{2} = \frac{(50.6\# /ft.)(14^{ \prime} ) }{2 }=354\#[/latex]

2. Bearing wall:

The reaction of the rafter onto the bearing wall is 354# every 16 inches. A conversion should be done to express the load at the top of the wall in pounds per lineal foot.

[latex]\omega =(354 \# / 16^{\prime \prime} )\left(\frac{12}{16} ight)^\prime =266 \# /ft.[/latex]

A strip of wall 1 foot wide and seven feet, six inches tall weighs

[latex]\omega_{wall} =7 \# /ft.^{2} (7.5^\prime) =52.5 \# /ft.[/latex]

[latex]\omega=266 \# /ft.+52.5\# /ft.=318.5\# /ft.[/latex]

3. Ridge beam:

Rafter reactions are equal to 354# per 16 inches, or 266 #/ft. Because the ridge beam is required to support rafters from both sides,

[latex]\omega=2 (266 \# /ft.)=532\# /ft.[/latex]

Note: The ridge beams are treated as two simple span beams, each 16 feet in length.

Exterior columns supporting the ridge beam carry

[latex]P_{ext} =(532\# /ft.)(8^\prime) =4,256\# [/latex]

Note: The eight feet represents the tributary beam length that is supported by the exterior columns.

Interior columns support a tributary beam length of 16 feet; therefore,

[latex]P_{int} =(532\# /ft.)(16^\prime) =8,512\# [/latex]

4. Floor joists:

Joists are spaced 16 inches on center, which also represents the tributary width of load assigned to each joist.

Loads: [latex]DL + LL = 11 \space psf + 40 \space psf = 51 \space psf[/latex]

(Floor Joist Loads and Reactions)

[latex]\omega_{D+L} =51 \#/ft.^{2}\left(\frac{16}{12} ight)^\prime =68 \# /ft.[/latex]

Foundation reactions may be obtained by

Foundation =[latex](68\# /ft.)(7^\prime [/latex] {tributary length})

[latex]=476\# /16^{\prime \prime } [/latex]

The central floor beam supports a floor joist reaction equal to

Beam =[latex](68\# /ft.)(14^\prime [/latex] {tributary length})

[latex]=952\# /16^{\prime \prime } [/latex]

Conversion of the floor joist reactions into load per foot results in

Foundation:[latex]\omega =(476\#/16 \space in.)\left(\frac{12}{16} ight)^\prime =357 \# [/latex]

Beam:[latex]\omega =(952\#/16 \space in.)\left(\frac{12}{16} ight)^\prime =714 \# [/latex]

5. Continuous foundation:

The stem wall measures eight inches thick and two feet tall. The footing base is eight inches thick and x wide.

Loads from the roof, wall, and floor are combined as a total load on top of the foundation stem:

[latex]\omega_{total}=318.5 \# /ft.+357\# /ft.=675.5 \# /ft.[/latex]

The foundation wall stem adds additional load on the footing equal to

Stem weight [latex] =\left(\frac{8}{12} ight)^\prime (2^\prime)(150 \#/ft.^{3})=200 \# /ft.[/latex]

Because the footing width x is unknown, the weight of the footing base must be computed in terms of pounds per square foot.

Footing weight [latex] =\left(\frac{8}{12} ight)^\prime (150 \#/ft.^{3})=100 \# /ft.^{2}[/latex]

In determining the footing width, examine a unit length [latex](1^\prime)[/latex] of foundation as a representation of the entire length.

[latex]q[/latex] = allowable soil bearing pressure = [latex]2,000 \space psf[/latex]

[latex]q_{net}=q-[/latex]footing weight= [latex]2,000 \space psf-100 \space psf[/latex]

[latex]=1,900 \space psf [/latex]

This value of [latex]q_{net}[/latex]represents the resistance of the soil available to safely support the loads of roof, walls, floor, and foundation stem.

[latex]\omega_{total}=675.5 \# /ft.+200\# /ft.=875.5\# /ft.[/latex]

The minimum required resistance area of the footing per unit length is

[latex]A=(1^\prime)(x)[/latex]

[latex]q_{net}= \frac{\omega}{x}[/latex] ; [latex]x= \frac{\omega}{q_{net}}[/latex]

[latex]x=\frac{875.5\# /ft. }{1,900 \# /ft.^{2}}=0.46^\prime \approx 6^{\prime \prime}[/latex]

Note that a footing base of six inches would be less than the thickness of the stem wall. A minimum footing width for a one-story, light-framed assembly should be 12 inches. If a 12 inch base width is provided, the actual pressure on the soil will be

Actual pressure=[latex]=\frac{\omega }{x=1^{ \prime}} =\frac{875.5 \# /ft.}{1^{ \prime}} [/latex]

[latex]=875.5 \# /ft.^{2} \lt q_{net}[/latex]

[latex] herefore OK.[/latex]

6. Interior spread footings:

A determination of individual post loads is necessary before footing sizes can be computed.

Assume the spread footings have a thickness of eight inches and [latex]q=2,000 \space psf [/latex]:

[latex]\omega_{footing} =\left(\frac{8}{12} ight)^\prime (150\# /ft.^{3})=100 \# /ft.^{2}[/latex]

[latex]q_{net}=q- [/latex]footing wt.=[latex]2,000 \space psf - 100 \space psf[/latex]

[latex]= 1,900 \space psf[/latex]

Critical center column:

[latex] A=x^{2} =\frac{P}{ q_{net}} =\frac{14,224 \# }{1,900\# /ft.^{2}}=7.5 ft.^{2}[/latex]

[latex] herefore x=2.74^{ \prime} \approx 2^{ \prime} 9^{ \prime \prime } [/latex] square

Other footing:

[latex] x^{2} = \frac{5,712 \# }{1,900\# /ft.^{2}}=3.00 ft.^{2}[/latex]

[latex] herefore x=1.73^{ \prime} \approx 1^{ \prime} 9^{ \prime \prime } [/latex] square

Question 4.6: A simple light-framed wood building is subjected to the load...

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