CEE 377 HW6 SOLUTION

pdf

School

University of Washington *

*We aren’t endorsed by this school

Course

377

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by Alexkarlbrown

Homework 6 CEE 377 - Introduction to Structural Design Problem 1 We have been asked to redesign the roof and add an additional level to a portion of our Seattle hotel. A restaurant has been added to the top level around the elevator shaft, as shown below. The restaurant portion has a gable-shaped roof with a slope of 4-on-12 that is supported by four roof trusses along the same grid lines as the girders. The gable roof is constructed from 1/8-inch plywood and asphalt shingles. The lower tier of the roof outside of the restaurant will be converted into an outdoor patio area enclosed by a 3-foot-high parapet wall. The patio area and the restaurant will include a 1-inch cement finish floor, and the roof slab thickness has been increased to 6 inches. Assume that all rainwater on the restaurant roof is collected in gutters and deposited away from the building. For the patio, primary drainage is provided by a single 6-inch drain pipe. Secondary drainage occurs through a 6-inch circular scupper located 2 inches above the primary drainage system. The design 15-minute/100-year rainfall intensity for Seattle is i = 2.38 in./hr, and the ground snow load is p g = 25 psf. Given the updated layout, determine the rain load, R , on the lower portion of the roof. all relevant snow loads, S , on both the lower and upper tier of the roof, including any snow drift. the new live load for the restaurant and the new live roof load for the outdoor patio. the updated demand on girder B and beam A supporting the roof level. Use the tributary area approach. the demand on an internal roof truss (e.g., the truss along the line adjacent to girder B). 1

20 ′ 5 ′ 10 ′ 5 ′ 20 ′ 3 @ 9 ′ = 27 ′ (Typical) Opening Girders (Typ.) Beams (Typ.) Beam A Girder B Restaurant Patio Secondary Drainage Primary Drainage 3 @ 15 ′ = 45 ′ 15 ′ 3 @ 27 ′ = 81 ′ Ground Floor First Story Second Story Restaurant Patio Rain Load We assume that only the rain from the patio will accumulate and flow out through the drainage system. The drainage area is A = (54 ft)(60 ft) = 3240 ft 2 , and the design rainfall intensity is i = 2 . 38 in./hr, resulting in a flow rate of Q = 0 . 0104 Ai = 0 . 0104(3240)(2 . 38) = 80.2 gal/min. Using this data, for a 6-inch circular scupper, Table C8.3-5 allows us to interpolate between 3 inches (55 gal/min) and 4 inches (90 gal/min). For Q = 80.2 gal/min, this gives us d h = 3 . 72 inches. Since 2

the depth to the secondary drainage is d s = 2 inches, the rain load is R = 5 . 2 ( 2 + 3 . 72 ) = 29 . 74 psf Snow Loads Here we will need to consider both the balanced and unbalanced loads on the restaurant roof as well as the balanced and drift loads on the patio area. In both cases, we will start with a flat roof snow load. For a ground snow load of p g = 25 psf, assuming C e = 1 . 0 (Surface Roughness B, Partial Exposure), C t = 1 . 0 (No thermal modifications necessary), and I s = 1 . 0 (Risk Category II), we have p f = 0 . 7(1)(1)(1)(25) = 17 . 5 psf. – Restaurant Roof Balanced Load: p f = 17 . 5 psf Unbalanced Load: For p g = 25 psf and W = 13 . 5 ft, we have to use the special note in Figure 7.6-1 for low W . In this case, W is the minimum of 20 ft, so we have h d = 1 . 45 in. Per the note, h d cannot exceed p I s p g W/ 4 γ . The specific weight of the snow is γ = 0 . 13 p g + 14 = 17 . 25 pcf, thus h d = 1 . 45 in < q I s p g W/ 4 γ = p 1 . 0(25)(13 . 5) / (4 ∗ 17 . 25) = 2 . 21 in ⇐ ✓ We can use h d = 1 . 45 in. For a roof slope of 4 on 12, we have S = 3 . 0, thus the intensity of the drift surcharge is h d γ/ √ S = 14 . 4 psf and the horizontal extent is 8 √ Sh d / 3 = 6 . 7 ft. 17 . 5 psf 13 . 5 ft 13 . 5 ft Balanced Loading Condition 4 12 5 . 25 psf 31 . 9 psf 14 . 4 psf 13.5’ 6.7’ 6.8’ Unbalanced Loading Condition 4 12 – Patio Roof Balanced Load: We can use the same balanced snow load on the patio that we used on the restaurant roof, as none of our factors need to be modified. Thus, p f = 17 . 5 psf. Drift Load: For a specific weight of 17.25 pcf, we have h b = p f /γ = 1 . 01 ft, h c = 15 − 1 . 01 = 13 . 99 ft, h c /h b = 13 . 85 > 0 . 2, and drift loads must be considered. The drift height is then the maximum of the leeward drift ( l u = 27 ft , h d = 1 . 75 ft) and the windward drift ( l u = 54 ft , h d = 0 . 75(2 . 4) = 1 . 8 ft). Windward drift controls, and we have h d = 1 . 8 ft. Because h d < h c , the extent of the drift is w = 4 h d = 7 . 2 ft, and we have p d = 1 . 8(17 . 25) = 31 . 1 psf. ASCE 7-16 Section 7.8 reminds us to calculate drift surcharges next to the 3-foot parapet walls using the windward drift approach. For winds blowing right to left, we have l u = 54 ft and h d = 0 . 75(2 . 4) = 1 . 8 ft. For winds blow top to bottom or bottom to top, we have l u = 60 ft, and given the nature of Figure 7.6-1 this will be close enough that we can also use h d = 1 . 8 ft, corresponding to p d = 31 . 1 psf. 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

7 . 2 ′ 39 . 6 ′ 7 . 2 ′ 3 ′ 48 . 6 psf 17 . 5 psf 48 . 6 psf Live and Live Roof Loads The live and live roof loads are found in ASCE 7-16 Table 4.3-1 – Restaurant: Dining rooms and restaurants have a live load of L = 100 psf , which is nonre- duceable. There is a roof live load associated with the restaurant, which has an ordinary pitched roof live load of 20 psf. This can be reduced as well, with a tributary area of (27’)(20’) = 540 sq ft and roof slope of F = 4 inches of rise per foot of run. The corresponding live roof load is L r = (1 . 2 − 0 . 001(540))(1 . 0)(20 psf) = 13 . 2 psf. – Patio: The patio area of the roof requires a bit of thought, since there is not an item specifically calling out a patio. Looking at the roof loading, the most sensible option is to use ”Roof areas used for occupants”, since the idea would be that the patio would host restaurant guests and other visitors. In this case, we would again use L r = 100 psf . ASCE 7-16 Section 4.8.3 tells us that occupiable roofs are allowed to use reduced live loads in the same way that we would do for an interior floor load, but for values of 100 psf, we cannot reduce. Dead Loads While not explicitly stated, to determine the updated demands on girder B and beam A, as well as the roof truss, we will need to calculated dead loads. – Restaurant Roof: 1/8-inch plywood = 0.4 psf, asphalt shingles = 2 psf – Patio: 1-inch cement floor finish = 32 psf, concrete roof slab = (138 pcf)(6”)(1/12) = 69 psf For determining loads on beam A and girder B, we will have, in both cases, a simply supported beam subjected to distributed loads. Both members are on the patio area, where the dead load is 32 + 69 = 101 psf, the live roof load is 100 psf, and the rain load is 29.7 psf. Beam A has a tributary width of 9 feet, and girder B has a tributary width of 20 feet, and the corresponding uniformly distributed loads are shown below. Load Beam A Girder B Dead 0.909 k/ft 2.02 k/ft Live Roof 0.900 k/ft 2.00 k/ft Rain 0.267 k/ft 0.59 k/ft It is evident that the live roof load will control over the rain load, but we have not considered the snow load. That is because both members are situated within the drift surcharge, so we have to consider how we will calculate and distribute the additional load. Let’s start with Beam A. As shown below, a portion of the drift extends into the tributary width for Beam A. The easiest way to handle that is to smooth the triangular portion across the entire tributary width. w S = 17 . 5 psf(9 ft) + 0 . 5(11 . 7 psf)(2 . 7 ft) = 0 . 173 k / ft Beam A Trib Width 9 ft 2.7 ft 17.5 psf 29.2 psf 4

So we have four load types (dead, live roof, rain, snow), which requires us to consider two load combina- tions 1 . 4 w D = 1 . 4(0 . 909 k / ft) = 1 . 27 k / ft 1 . 2 w D + 1 . 6( w L r or w R or w S ) = 1 . 2 w D + 1 . 6 w L r = 1 . 2(0 . 909 k / ft) + 1 . 6(0 . 900 k / ft) = 2 . 53 k / ft ⇐ CONTROLS 20 ft 2 . 60 k / ft This results in shear and moment demands of V max = V support = wL 2 = 25 . 3 kip M max = M midspan = wL 2 8 = 127 k · ft We can take a similar approach for Girder B, but we have to consider the nonuniformity of the snow load. Our snow loading condition is distributed as shown below, including the 20-foot tributary width associated with girder B. 7 . 2 ft 19 . 8 ft 0 . 972 k / ft 0 . 350 k / ft We know that the maximum shear will still occur at a support, but the maximum moment no longer occurs at midspan. In fact, a static analysis of this beam will find that V max = 6 . 77 kip at the left support and M max = 34 . 6 k · ft approximately 12.9 feet from the left support. In this case, we compare the snow load demands with those of the live roof load (which we know will control over rain), and we find that the live roof load of 100 psf gives V max = wL 2 = 100 psf(20 ft)(27 ft) 2 = 27 kip M max = wL 2 8 = 100 psf(20 ft)(27 ft) 2 8 = 182 . 25 k · ft This is much higher than the maximum snow load, and occurs in the same location as the dead load demands, where, for the 2.02 k/ft dead load calculated above, we have V max = 27 . 3 kip and M max = 184 k · ft. Applying the appropriate load combination, we find: V max = 1 . 2 V D + 1 . 6 V L r = 1 . 2(27 . 3 kip) + 1 . 6(27 kip) = 76 . 0 kip M max = 1 . 2 M D + 1 . 6 M L r = 1 . 2(184 k · ft) + 1 . 6(182 k · ft) = 512 k · ft Lastly, we look to determine the demands on an interior roof truss. In this case, we consider the interior truss with a tributary width of 20 feet, similar to the girders. We can use a tributary area type approach to carry the dead, live roof, and snow loads to the 5 joints along the two diagonals. The dead load, live load, and balanced snow load are all distributed evenly uniformly across the roof. For a distributed load of 1 psf, each interior node gets (20 ft)(6.75 ft)(1 psf) = 0.135 kip, and each exterior node sees (20 ft)(3.375 ft)(1 psf) = 0.0675 kip. Loads at the exterior nodes go directly into the supports, so we only need the interior nodal values which can be scaled from the 1 psf load. The corresponding internal nodal values are P D = 0 . 125(2 . 4) = 0 . 30 kip, P L r = 0 . 125(13 . 2) = 1 . 65 kip, and P S = 2 . 19 kip. The corresponding truss loads can then be found: 5

Dead Loads, Live Roof, and Balanced Snow Loads P P P 4 . 5 P 4 . 5 P 4 . 5 P 4 . 5 P − 1 . 5 √ 10 P 0 − 0 . 5 √ 10 P − √ 10 P P − 1 . 5 √ 10 P 0 − 0 . 5 √ 10 P − √ 10 P 1 . 5 P 1 . 5 P For the unbalanced snow load, we will do a similar tributary area approach. In this case, we split the tributary areas as shown below and sum the total load within each 6.25-foot tributary width. Note that the unbalanced snow load can be unbalanced in either direction and because the structure is symmetric, we can simply flip the results about the center of the structure. 0 . 105 k / ft 0 . 638 k / ft 0 . 288 k / ft 13.5’ 6.7’ 6.8’ 6.75’ 6.75’ 6.75’ Unbalanced Snow Loads 0 . 71 kip 3 . 11 kip 2 . 51 kip 7 . 69 kip 7 . 69 kip 11 . 3 kip 11 . 3 kip − 8 . 11 kip 0 − 1 . 12 kip − 6 . 99 kip 1 . 91 kip − 11 . 9 kip 0 − 4 . 91 kip − 6 . 99 kip 2 . 57 kip 3 . 77 kip A B C D E F G H Let’s tabulate the results. The values in the table below represents axial forces in the truss members in kips, and the values in bold represent the maximum of the live roof and snow loads to use in the load combinations. Because the dead load is so small, we will neglect the 1.4D load combination. 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Member Dead Live Roof Snow, Balanced Snow, Unbalanced (max) 1.2 D + 1.6( L r or S ) AB 1.35 7.425 9.855 11.3 19.7 BC 1.35 7.425 9.855 11.3 19.7 CD 1.35 7.425 9.855 11.3 19.7 DE 1.35 7.425 9.855 11.3 19.7 AF -1.42 -7.83 -10.4 -11.9 -20.7 BF 0 0 0 0 0 CF -0.474 -2.61 -3.46 -4.91 -8.43 CG 0.3 1.65 2.19 1.91 3.86 CH -0.474 -2.61 -3.46 -4.91 -8.43 DH 0 0 0 0 0 EH -1.42 -7.83 -10.4 -11.9 -20.7 FG -0.949 -5.22 -6.93 -6.99 -12.3 GH -0.949 -5.22 -6.93 -6.99 -12.3 7

Problem 2 Using a derivation similar to what we did in lab and class, show that the effective length factor for a fixed- pinned column is K = 0 . 7. Note: The math on this problem is a bit tedious, and you will end up with a transcendental equation that must be solved graphically. Use your plotter of choice (Matlab, Mathematica, etc.) to help find your answer As always, we can start with the displacement equation and associated derivatives below. v ( x ) = A sin kx + B cos kx + Cx + D v ′ ( x ) = kA cos kx − kB sin kx + C v ′′ ( x ) = − k 2 A sin kx − k 2 B cos kx Let’s assume that the fixed end is located at x = 0. The boundary conditions for this problem are v (0) = v ′ (0) = 0 and v ( L ) = EIv ′′ ( L ) = 0. Because we know EI is not zero, the moment boundary condition can also be considered to be zero curvature, or v ′′ ( L ) = 0. We can apply each to solve our system. v ′ (0) = 0 ⇒ kA + C = 0 → C = − kA v ′′ ( L ) = 0 ⇒ − k 2 A sin kL − k 2 B cos kL = 0 → B = − A tan kL v (0) = 0 ⇒ B + D = 0 → D = − B = A tan kL v ( L ) = 0 ⇒ A sin kL + B cos kL + CL + D → A sin kL − A tan kL cos kL − AkL + A tan kL = A (tan kL − kL ) = 0 → tan kL = kL This is a transcendental equation that should be solved graphically. One method would be to plot kL vs f ( kL ) for both sides of the function and find where they overlap, and another would be to plot tan( kL ) − kL and look find the points where the function crosses zero. 2 4 6 8 10 2 4 6 8 10 We see that the two equation are equal at kL = 4 . 49 and kL = 7 . 73. We are interested in the lowest value, so we set kL = r P EI L = 4 . 49 ⇒ P = 20 . 16 EI L 2 = π 2 EI 0 . 4896 L 2 = π 2 EI (0 . 700 L ) 2 Here we see that K = 0 . 7, as expected. 8

Problem 3 The small building below supports a dead load of 100 psf and a live load of 50 psf (NOTE: treat the live load as a floor live load not a roof live load). It is supported by 6 columns, all having the cross-sectional properties shown, a yield stress of 50 ksi and modulus of elasticity of 29000 ksi. The braces can resist both tension and compression. Neglect other loads. Determine if the edge and corner columns are safe. Let’s start with the capacities. In the east-west direction, the corner column is pinned at each end while the edge column is pinned at both ends and the midpoint. In the north-south direction, both columns are part of a moment frame with pinned supports, thus we can idealize the east-west direction as a pin-slider connection. K = 1 . 0 K = 1 . 0 K = 1 . 0 K = 2 . 0 Corner, E-W Corner, E-W Both, N-S This lets us calculate the following critical loads: 9

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Corner Column, E-W (Weak Axis) P cr = π 2 EI y ( KL ) 2 = π 2 (29000)(36 . 6) (1 . 0(24)(12)) 2 = 126 . 3 kip ← CONTROLS Edge Column, E-W (Weak Axis) P cr = π 2 EI y ( KL ) 2 = π 2 (29000)(36 . 6) (1 . 0(12)(12)) 2 = 505 . 2 kip Both Columns, N-S (Strong Axis) P cr = π 2 EI x ( KL ) 2 = π 2 (29000)(171) (2 . 0(24)(12)) 2 = 147 . 5 kip Then we need to determine the demands on the columns. We can tabulate our answers below. Remember that, for columns, ASCE 7-16 Table 4.7-1 tells us K LL = 4. Member A T A I D L P D P L P u = 1 . 2 P D + 1 . 6 P L ϕP cr ft 2 ft 2 psf psf kip kip kip kip Corner Column 360 1440 100 32.3 36 11.6 61.8 113.7 Edge Column 720 2880 100 26.5 72 19.1 116.9 132.8 Here we see that both members have demands less than capacities and are thus safe. 10