Jonathan Ochshorn - ARCH 264/564 Structural Elements Homework Assignment

1. Wood column analysis: A first-floor 6x8 Spruce-Pine-Fir South No.1 interior column supports a tributary area of 100 sq.ft. per floor in a residential structure with loads as follows: L = 40 psf (floors only); D = 15 psf (all floors and roof); S = 40 psf (roof only); and L_R = 30 psf (roof only). The column has an effective length of 8 feet. Ignore live load reduction for this problem. How many stories tall can this building be, with such a column at the first floor?

F_c = 625 psi (posts and timbers), C_F = 1.0 and C_D is unknown (assume 1.15 for know assuming that snow load governs; check later). The analysis of the column capacity proceeds as follows:

F*_c = 625(1.15) = 718.75 psi (using all adjustments except for C_P).

F_cE = 0.3(1,200,000) / [(8x12) / (5.5)]² = 360,000 / 304.66 = 1181.64

A = [1 + 1181.64 / 718.75] / (2 x 0.8) = 1.65; and B = (1181.64 / 718.75) / 0.8 = 2.055

C_P = 1.65 - sq.rt.(1.65² - 2.055) = 0.83

F'_c = F*_cC_P = 718.75 x 0.83 = 598.7 psi

Capacity, P = adjusted allowable stress x area = 598.7 x 41.25 = 24,697#
plan and section of building
Since each story provides a load of (40 + 15)(100) = 5,500# to the column, the capacity of 24,697 seems to allow for a 4-story building. But this needs to be checked based on a calculation of governing loads. The two candidates are:

[D + L] / 1.0 = [(15 x 100 x 4) + (40 x 100 x 3)] / 1.0 = [12,000 + 6,000] / 1.0 = 18,000#
[D + 0.75(L + S)] / 1.15 = [12,000 + 0.75(6,000 + 40x100)] / 1.15 = 15,662#

D + L governs, with C_D = 1.0. We need to check the capacity with C_D = 1.0 since we had assumed a different value initially.

F*_c = 625(1.0) = 625 psi; F_cE remains the same as before.

A = [1 + 1181.64 / 625] / (2 x 0.8) = 1.81; and B = (1181.64 / 625) / 0.8 = 2.36

C_P = 1.81 - sq.rt.(1.81² - 2.36) = 0.853

F'_c = F*_cC_P = 625 x 0.853 = 533 psi

Capacity, P = adjusted allowable stress x area = 533 x 41.25 = 21,988#

Since the capacity is greater than the governing load = 18,000#, a 4-story building will work. It is clear that an additional story will not work.

2. Wood column design: A triangular (gable) truss is being designed for a single concentrated load of 8,300#. This load consists of snow and dead load, so the duration of load factor, C_D = 1.15. The angled top chords consist of double 2x members, braced at the quarter points, so that their effective length can be taken as 1/4 of the distance from point "A" to point "C." Using Douglas Fir South No.2 lumber, and assuming an initial stability adjustment factor, C_P = 0.5, design the double 2x members. That is, find the required area based on the assumed adjustment factors; then check (analyze) the capacities of all likely candidates using the actual values of all adjustment factors (C_D, C_F, C_P) to see which one actually works best. Do not try the 2x3.

First, find the force in the compression member, as shown in the two steps below (first find reactions; then use the force or section method to find the internal force in the compression member). The vertical component of the unknown compressive force must be 4150# to equilibrate the reaction. Then, based on the geometry of the force triangle, with equal legs and a 45-degree angle, the force in the angled member, F = 4150 x sq.rt.(2) = 5869#.
truss analysis
Next, find the required area of the cross-section, based on an assumed value for C_P of 0.5:

F_c = 1300 psi (dimension lumber), C_F = 1.0 (assumed; check later), C_D = 1.15 (snow load governs), and C_P = 0.5 (assumed). The analysis of the column capacity proceeds as follows:

F'_c = 1300(1.0)(1.15)(0.5) = 747.5 psi.

required area = load/stress = 5869 / 747.5 = 7.85 in². Since there is a double 2x member here, divide the force in half to find the load on a single 2x member: 7.85 / 2 = 3.93 in².

The likely 2x candidates (see Table A-4.1) are:

2x4 with A = 5.25 in²; and if that doesn't work with the actual value of C_P,
2x6 with A = 8.25 in².

Check the 2x4 first:

F*_c = F_cC_FC_D = 1300(1.15)(1.15) = 1719.25 psi

To find F_cE, we need to determine the unbraced length, kL, for the member. The total length is the hypotenuse of a triangle with equal legs of 10 feet, or 10 x sq.rt.(2) = 14.14 feet. Since the member is braced at the quarter points, we can use an effective length of 14.14 / 4 = 3.54 feet.
F_cE = 0.3(1,200,000) / [(3.54x12) / (1.5)]² = 360,000 / 802 = 449

A = [1 + 449 / 1719.25] / (2 x 0.8) = 0.79; and B = (449 / 1719.25) / 0.8 = 0.33

C_P = 0.79 - sq.rt.(0.79² - 0.33) = 0.25

F'_c = F*_cC_P = 1719.25 x 0.25 = 426 psi

Capacity, P = adjusted allowable stress x area = 426 x (1.5 x 3.5) = 2237# for a single 2x4 and 2 x 2237 = 4473# for the double 2x4.

Since the capacity of 4473# is less than the actual compressive force of 5869#, the 2x4 doesn't work. It is necessary to check the 2x6.

Next check the 2x6:

F*_c = F_cC_FC_D = 1300(1.10)(1.15) = 1644.5 psi

F_cE = same as before = 449

A = [1 + 449 / 1644.5] / (2 x 0.8) = 0.796; and B = (449 / 1644.5) / 0.8 = 0.341

C_P = 0.796 - sq.rt.(0.796² - 0.341) = 0.255

F'_c = F*_cC_P = 1644.5 x 0.255 = 419.4 psi

Capacity, P = adjusted allowable stress x area = 419.4 x (1.5 x 5.5) = 3460.4# for a single 2x6 and 2 x 3460.4 = 6921# for the double 2x6.

Since the capacity of 6921# is greater than the actual compressive force of 5869#, the 2x6 works.

3. Steel column analysis:
a) Using the steel column analysis tables (A-7.3 or A-7.4) find the capacity of an "extra strong" steel pipe column with a 6" nominal diameter. See Table A-4.8 for section properties (note that the radius of gyration is given directly) and Table A-3.11 for guidance on yield stress (pay attention to note 2). The column is 20 feet high, and is modeled as pin-ended (hinged at both ends).

b) As a check, compute the radius of gyration for the section based on the equation: r = sq.rt.(I / A).

d) Will a W12x50 be weaker or stronger than the 6"-diamater pipe, given the same effective length and the same yield stress? Base your answer only on a comparison of radius of gyration for the two sections.

4. Steel column design: Design the steel column hidden inside the stone-clad supports for the superstructure of Uris Hall (see various photos below). We'll make some very approximate assumptions* about loads and dimensions: Assume that the load on the tributary area, shown in green on the plan below, is distributed equally to all 8 columns supporting the "Vierendeel" truss on the perimeter of the building (which we assume to be symmetrical); use a live load, L = 50 psf (for 3 floors) and consider live load reduction, a dead load, D = 47 psf (for 3 floors + the roof), and a snow load, S = 35 psf (see Tables A-2.1 - A-2.3 to confirm these loads) to find the governing load combination; assume that the effective length (height) of the column is 10 feet; and assume that the column supports three stories of live and dead load, plus the roof (dead load, and snow or construction live load). The building was constructed in the 1970s, but we will assume A-992 steel. Use design Table A-7.2. Architect: Skidmore, Owings & Merrill LLP.

* Plan dimensions and column height are based on blind guesses only; I didn't actually go into the building and measure anything.

In order to design the column, we must find the governing loads on the column. First, the tributary area for all 8 columns (the green shaded area on the plan above) is 6,600 sq.ft. per floor, so the tributary area on each column is 6,600 / 8 = 825 sq.ft. per floor. There are 3 floors of live load, so the reduction coefficient is:

0.25 + 15 / sq.rt.(4 x 825 x 3) = 0.4, which is also the lower limit for multi-story loading. The reduced live load is therefore: L = 0.4(50) = 20 psf.

We can now check the governing load combinations:

D + L = (47 x 825 x 4) + (20 x 825 x 3) = 155,100 + 49,500 = 204,600#
D + 0.75(L + S) = 155,100 + 0.75(49,500 + 35x825) = 213,881#

D + 0.75(L + S) governs, so the governing load = 213,881# = 214 k. The effective length, in feet, is 10 feet.

From Table A-7.2, we find the best (lightest) candidates:

W8x35
W10x33
W12x40

The W10x33 is the lightest column section that can support the governing load. Note that although the architectural expression (massive stone base) implies that these 8 columns provide stable points of support for the trusses above, a look inside the stone cladding shows that the actual lateral-force-resisting system is probably within the interior core of the building.