Assignment #1
contact | homepage | spring 2008 index for ARCH 264/564 | current index for ARCH 264/564 | homework index | textIssued Jan. 31, 2008.
Due: Feb. 7, 2008
Analyze a 4-story 50-foot high rectangular building in Portland, ME for various loads. Assume 55' x 120' overall plan dimensions, with a 15-foot wide lateral-force-resisting truss on each end wall. Story heights above grade are as follows: 1st floor at grade; 2nd floor +15.0 feet; 3rd floor +30 feet; 4th floor +40 feet; and roof at +50 feet.
Assume residential occupancy with L = 40 psf; LR = 30 psf; S = 50 psf; and a dead load, D, consisting of floor slabs only (6"-thick reinforced concrete @ 150 pcf).
Find the governing design load on a typical interior 3rd-floor column, assuming a 30' x 30' column grid (I realize that this is not consistent with the sketch: even so, just assume 30 x 30 = 900 sq.ft. of tributary area per floor for the column), and:
a) allowable stress design, steel columns;
b) allowable stress design, wood columns (include duration of load calculations);
c) strength design (LRFD), reinforced concrete columns;
The tributary area for an interior column is 30 x 30 = 900 sq.ft. per floor. Testing for live load reduction, the coefficient = 0.25 + 15 / (sq.rt. 900 x 1 floor x 4) = 0.5; so the reduced live load, L = 0.5(40) = 20 psf. The dead load, D = 150 (6 / 12) = 75 psf.
a) Steel (ASD): From Table A-5.1, the two relevant load combinations to try are:
D + L = 75(900 x 2) + 20(900 x 1) = 135,000 + 18,000 = 153,000# or 153 k.
D + 0.75L + 0.75(LR or S) = 135,000 + 0.75(18,000) + 0.75(50 x 900) = 135,000 + 13,500 + 33,750 = 182,250# or 182 k.
The governing load is 182 k.
a) Wood (ASD): From Table A-5.1, the two relevant load combinations to try are the same as above, except that we divide each load combination by the appropriate duration of load factor for wood:
[D + L] / CD = [75(900 x 2) + 20(900 x 1)] / 1.0 = [135,000 + 18,000] / 1.0 = 153,000 / 1.0 = 153,000# or 153 k.
[D + 0.75L + 0.75(LR or S)] / CD = [135,000 + 0.75(18,000) + 0.75(50 x 900)] / 1.15 = [135,000 + 13,500 + 33,750] / 1.15 = 182,250 / 1.15 = 158,478# or 158 k.
We have used S instead of LR because, in this case, it is the larger value. Since the duration of load factor for LR is bigger than that for S, we need not try the alternate case using LR. However, if LR was larger than S, it would be prudent to try both cases. The governing load is not 158 k, but is 182,250# or 182 k (shown in red above). The duration of load factor of 1.15 will be used in the actual design or analysis process. Note that the value of 158 k was used only to determine the governing load condition, but is not itself the governing load.
c) Reinforced concrete (strength design or LRFD): From Table A-5.1, the three relevant "factored" or "design" load combinations to try are:
1.4D = 1.4(135,000) = 189,000# = 189 k.
1.2D + 1.6L + 0.5(LR or S) = 1.2(135,000) + 1.6(18,000) + 0.5(50 x 900) = 162,000 + 28,800 + 22,500 = 213,300# or 213 k.
1.2D + 1.6(LR or S) + (0.5L or 0.8W) = 1.2(135,000) + 1.6(50 x 900) + 0.5(18,000) = 162,000 + 72,000 + 9,000 = 243,000# or 243 k.
Note that we are permitted to use 0.5L in the second equation since the live load does not exceed 100 psf and is not a garage or place of public assembly. The wind load, W, is not a factor here, so it can be taken as zero.
The governing design load, Pu = 243 k.
Find and sketch the distribution of wind loads (external design pressures) on the windward and leeward surfaces of the building, assuming a wind direction as shown in the sketch above. Assume exposure "B" with Kt = I = Kd = 1.0.
From Table A-2.3, the basic wind speed (velocity) in Portland, ME is 100 mph. The building height (mean roof height for a flat roof) is 50 feet. The external design wind pressure, pe = 0.00256(K)(Kd)(Kt)(G)(Cp)(I)(V2). Values for K and Cp are found in Table A-2.4; the other values are given in the problem statement: Kt = Kd = I = 1.0
Windward wall:
| height | coef. | K | Kd | Kt | G | Cp | I | V2 | pe (psf) |
|---|---|---|---|---|---|---|---|---|---|
| 0-15 | 0.00256 | 0.57 | 1.0 | 1.0 | 0.85 | 0.8 | 1.0 | 10,000 | 9.92 |
| 20 | 0.00256 | 0.62 | 1.0 | 1.0 | 0.85 | 0.8 | 1.0 | 10,000 | 10.79 |
| 30 | 0.00256 | 0.70 | 1.0 | 1.0 | 0.85 | 0.8 | 1.0 | 10,000 | 12.19 |
| 50 | 0.00256 | 0.81* | 1.0 | 1.0 | 0.85 | 0.8 | 1.0 | 10,000 | 14.10 |
*Interpolate, if necessary to find value of K at mean roof height (not required in this case).
Leeward wall:
We use the same equation for pe, but only one value is needed based on K at the mean roof height and Cp = -0.5 (Table A-2.4) for the plan dimensions shown:
pe = 0.00256(K)(Kd)(Kt)(G)(Cp)(I)(V2) = 0.00256(0.81)(1.0)(1.0)(0.85)(-0.5)(1.0)(1002) = -8.81 psf.
Note that the "sign" of the external pressure, pe, is defined as follows: positive means the wind is "pushing" on the surface; negative means the wind is "pulling" (suction) on the surface. For that reason, the positive windward pressure and the negative leeward pressure are both shown acting in the same direction.
Distribution of wind pressure (psf) on windward and leeward surfaces
Find and sketch the distribution of earthquake forces on the building, assuming ground acceleration in the direction shown in the sketch above. Assume that the lateral force resisting system consists of a "special steel concentrically braced frame." As before, I = 1.0. Assume that the building site is characterized by "soft soil" conditions (site class "E").
Follow the methodology in text example 2.5 on page 37 (or class notes): Assume 50'-high building, on "soft soil" (Site class "E"). Plan dimensions are 55' x 120' with effective seismic weight of 75 psf (assumed for all floors and roof — see dead load calculation above). Floors are at heights of 15', 30', and 40' with roof at 50' above grade. Assume "special steel concentrically braced frame," so R = 6 (this is found in Table A-2.5). Also from Table A-2.5, Ct = 0.02, and x = 0.75.
The "solution overview" is as follows: Find the effective seismic building weight, W; compute the base shear, V; compute the story forces, Fx.
The specific problem solution is as follows:
1. Find W: The effective seismic weight for each story (floor or roof) is 75(55 x 120) = 495,000# = 495 k. Therefore, the effective seismic weight for the building, W = 495(4) = 1980 k. In this calculation we include floors 2, 3, 4, and the roof. The first floor weight (at grade) is not included since it has no impact on the seismic behavior of the structure.
2. Find V:
a) From Table A-2.3, find Ss and S1 for Portland, ME: Ss = 0.37; S1 = 0.10.
b) From Table A-2.5, find site coefficients Fa and Fv: For Site Class "E" [soft soil] and the values of Ss and S1 found above, Fa = 2.12; and Fv = 3.5. The first value is found by interpolating: (Fa - 2.5) / (1.7 - 2.5) = (0.37 - 0.25) / (0.5 - 0.25), and so on.
c) From Table A-2.5, find the design elastic response accelerations SDS and SD1: SDS = 2/3 (FaSs) = 0.523; SD1 = 2/3 (FvS1) = 0.233.
d) From Table A-2.5 find Cs: First, we need to find R = 6 (see above); I = 1.00 (for normal occupancies like office buildings); and the fundamental period T (for which we need to know Ct = 0.02, x = 0.75, and the building height, hn = 50 feet — these values can be found above). Since T = Cthnx, we get T = 0.02(50)0.75 = 0.38, i.e., a fundamental period of vibration just over a third of a second.
Then, Cs = SDS / (R / I) = 0.523 / (6 / 1.0) = 0.087, at least provisionally: we still need to make sure this value falls between an upper and lower bound.
The upper bound for Cs is SD1 / (TR / I) = 0.233 / (0.38 x 6 / 1.0) = 0.102.
The lower bound for Cs is 0.044SDSI = 0.044(0.523 x 1.0) = 0.023.
Only for buildings with SD1 > 0.75 is there another lower bound criteria, not applicable in this case.
Since Cs falls within the upper and lower bounds, we use Cs = 0.087.
3. From equation 2.13 (page 36 in the text), we can find the seismic "story" forces assumed to act at each floor (and roof) level:
Fx = (V)wxhxk / (Σwihik). Since k = 1.0 for buildings with periods less than or equal to 0.5 seconds (see p.36 in text for explanation), we get the simpler equation for this problem:
Fx = (V)wxhx / (Σwihi).
See text example 2.5 for calculation when the exponent, k, is not equal to 1.0. For this problem however, we can write the equation as:
Fx = 172(495)hx / [495(15 + 30 + 40 + 50)] = (1.274)hx. Notice that for equal weights on each floor, the per-floor weight drops out of the equation. The distribution of story forces is best calculated in tabular form:
| story level | hx (feet) | Fx = (1.274)hx (kips) |
|---|---|---|
| roof | 50 | 63.70 |
| 4 | 40 | 50.96 |
| 3 | 30 | 38.22 |
| 2 | 15 | 19.11 |
| total | 172.00 | |
Distribution of seismic "story forces" (kips) on building
Notice that the sum of the story forces equals the base shear, V, or 172 k. Also notice that the distribution of forces, in this case, is linear because the exponent in the force equation is 1.0.
Statics review: For the determinate lateral-force-resisting trusses shown in the sketch above, find the maximum tensile force based on the distribution of seismic forces found above. Only include the seismic loads in the calculations, and assume allowable stress design (i.e., do not use any load factors). Note that the seismic story forces should be divided in two, with each half assigned to one of the trusses. Also note that the truss should be assumed to be supported at grade by a hinge and a roller. The seismic loads are transferred through the floor beams to the truss nodes at each floor and roof level.
See Structural Elements text for statics review (p.22 for truss analysis method) or, for more detail, see: free truss chapter [PDF] from McGraw-Hill.
First, model the truss with hinge and roller, and find the reactions:
Note that the earthquake loads on the truss are found by taking half of the forces computed above in problem #3. To find reactions, take moments about one of the constraints, for example, the roller labeled "B." Assuming a counter-clockwise positive sign convention, we get:
ΣMB = -VL(15) - 9.56(15) - 19.11(30) - 25.48(40) - 31.85(50) = 0; from which: VL = -221.9 k.
The negative sign for VL indicates that the assumed direction of the force is wrong; it actually points down, indicating that the joint at "A" is subjected to tension. Next, we take a free-body diagram (FBD) cut through the vertical bar at "A" (which we determine to be the bar with the greatest tensile force, by analogy to a cantilever beam, whose greatest moment is at the support — see sketch at right, below).
By taking moments about point "C," we can find the tensile bar force, labeled "T" in the diagram:
ΣMC = -T(15) +221.9(15) - 86(15) = 0; from which: T = 135.9 k. This is the so-called "method of sections."
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Rivited double-angle truss, Teagle Hall Fitness Center, Cornell University, photo by J. Ochshorn
© 2008 Jonathan Ochshorn. First posted 21 January 2008. Last updated: 11 February 2008