CEE 377 HW3 SOLUTION

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Dec 6, 2023

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Homework 3 Solution Problem 1 A new downtown Seattle restaurant is framed as shown 25 ft 8 ft 12 ft 12 ft 12 ft A A B B 8 × 8 Timber Support Column (Typ.) Indoor Restaurant Area Outdoor Patio Plan View Elevation View Simple Connection (Typ.) Supporting 2 × 12 Supporting 2 × 12 Full Height Glass Window Full Height Glass Window 2 × 10 Wood Beam 2 × 10 Wood Beam The beam along line A that supports the second story may be analyzed as pin-supported at the left end and propped on a roller at the locaton of the 2 × 12 supporting beam. The dead and live loads are calculated as shown below. Using the LRFD load combinations from ASCE 7, determine the maximum positive and negative moments for use in design of the beam (assume both loads at in full) D = 1 . 5 kip / ft 25 ft 8 ft L = 0 . 8 kip / ft 25 ft 8 ft Since both beams have the same layout, we’ll start this problem by solving the moment diagram for an arbitrary uniform distributed load of w .

w 25 ft 8 ft Solving for reactions (and acknowledging that there are no forces in the x -direction), we get X M pin = 0 ⟲ + R roller (25 ft) − ( w )(33 ft)(33 ft / 2) = 0 → R roller = 21 . 78 w X F y = 0 ↑ + : R pin + R roller − w (33 ft) = 0 → R pin = 11 . 22 w Note that the units above are in feet, but once we add the dimensional load it will convert to a force. Now we can get the moment diagram. I’m going to work through the shear and moment diagram by observation from the free body diagram below. For the shear, we know (1) the reactions represent a discontinuity in the shear diagram, and (2) the distributed load represents to slope of the shear diagram. For the moment, we have zero moment at the pin support by definition, then we can use the magnitudes from the shear diagram to give us the slopes of the moment diagram. To determine the magnitudes of the moment diagram at a point x , we can take the area of the shear diagram from x = 0 to x = x . FBD 11 . 22 w 21 . 78 w 25 ft 8 ft w V ( x ) (kip) 11 . 22 w − 13 . 78 w 8 . 0 w 11 . 22 ft M ( x ) (k-ft) 62 . 9 w − 32 w So we see here that the maximum positive moment occurs 11.22 feet from the pin support and is going to be equal to 62.9 w , while the maximum negative moment occurs at the roller support with a magnitude of − 32 w . Specifically, we find that D = 1 . 5 kip / ft → M + D,max = 94 . 4 kip ft , M − D,max = − 48 kip ft D = 0 . 8 kip / ft → M + L,max = 50 . 4 kip ft , M − L,max = − 25 . 6 kip ft Finally we consider the load combinations. We only have a live and a dead load, so wwe will only need to consider two load combinations: 1 . 4 D and 1 . 2 D + 1 . 6 L . M + max : max [1 . 4 D = 1 . 4(94 . 4) = 132 . 2 kip ft , 1 . 2 D + 1 . 6 L = 1 . 2(94 . 4) + 1 . 6(50 . 4) = 193 . 9 kip ft] M − max : max [1 . 4 D = − 67 . 2 kip ft , 1 . 2 D + 1 . 6 L = − 98 . 6 kip ft] Thus, the maximum factored moments are M + max = 193 . 9 kip ft M − max = − 98 . 6 kip ft

If the live load may instead be placed either on the 25 ft or the 8 ft spans (cases 1 and 2 below), determine the maximum factored positive and negative moments. L = 0 . 8 kip / ft 25 ft 8 ft Live Load Case 1 L = 0 . 8 kip / ft 25 ft 8 ft Live Load Case 2 We are going to approach the problem the same way as we did previously by solving the moment diagram. The dead load does not change, so we need to consider the modified live load. – Live Load Case 1 X M pin = 0 ⟲ + R roller (25 ft) − (0 . 8 kip /ft )(25 ft)(25 ft / 2) = 0 → R roller = 10 kip X F y = 0 ↑ + : R pin + R roller − (0 . 8 kip / ft)(25 ft) = 0 → R pin = 10 kip FBD 10 kip 10 kip 25 ft 8 ft 0 . 8 kip / ft V ( x ) (kip) 10 − 10 12 . 5 ft M ( x ) (k-ft) 62 . 5 kip ft – Live Load Case 2 X M pin = 0 ⟲ + R roller (25 ft) − (0 . 8 kip /ft )(8 ft)(29 ft) = 0 → R roller = 7 . 42 kip X F y = 0 ↑ + : R pin + R roller − (0 . 8 kip / ft)(8 ft) = 0 → R pin = − 1 . 02 kip FBD 1 . 02 kip 7 . 42 kip 25 ft 8 ft 0 . 8 kip / ft V ( x ) (kip) − 1 . 02 6 . 4 M ( x ) (k-ft) − 25 . 6 kip ft

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Now we have the maximum positive and negative live loads from the modified scenarios – Live Load Case 1: M + L,max = 62 . 5 kip ft, M − L,max = 0 – Live Load Case 2: M + L,max = 0, M − L,max = − 25 . 6 kip ft It is obvious that Live Load Case 1 may modify the maximum positive moment while Live Load Case 2 may modify the maximum negative moment. The maximum negative moment is straightforward because the maximum negative moment still occurs at the roller support and has the same magntude as before, so nothing will change and M − max = − 98 . 6 kip ft. The maximum positive moment will certainly change, and the location will change as well since the two systems no longer provide the same location for the maximum moment. The most accurate way to do this is to combine the two systems and find the actual maximum moment. Below we see the free body diagram with the factored distributed loads. We can accomplish this by simply factoring the free body diagrams above with the appropriate load combinations. R pin = 1 . 2(11 . 22(1 . 5)) + 1 . 6(10) = 36 . 2 kip R roller = 1 . 2(21 . 78(1 . 5)) + 1 . 6(10) = 55 . 2 kip w D = 1 . 2(1 . 5) = 1 . 8 kip / ft w L = 1 . 6(0 . 8) = 1 . 28 kip ft FBD 36 . 2 kip 55 . 2 kip 25 ft 8 ft 3 . 08 kip / ft 1 . 8 kip / ft V ( x ) (kip) 36 . 2 − 40 . 8 14 . 4 11 . 75 ft M ( x ) (k-ft) 212 . 7 − 57 . 6 Looking at these results, we see that the maximum factored positive moment is thus M + max = 212 . 7 kip ft. If you were designing this wood beam that has the same resistance for positive and negative moment, what value of moment would you design it for? Clearly here we would be looking for the maximum magnitude of the moment if the beam provides equal tensile and compressive resistance. In this case, we would choose M max = 212 . 7 kip ft. It is interesting to point out that, as engineers, we often simplify our problem in ways that ensure a conservative solution (one that will make our design more safe). The answer above is th exact solution, but a more conservative approach would be to simply take the two maximum values and use them as they are while acknowledging that (1) you the engineer are aware that these values do not occur at the same location on the beam, and (2) given the distribution of the moment, the maximum values occur within a reasonable proximity to one another. In this case, we would have estimated that the maximum moment is simply 1 . 2(94 . 4)+1 . 6(62 . 5) = 213 . 3 kip ft. Given the exact solution, this results in an error of 0.28%. From an engineering perspective, these are the same.

Problem 2 The lateral force resisting system for the restaurant in Problem 1 is located along line B-B in the plan view shown above. An elevation view for that lateral force resisting system, which consists of tension-only steel strap bracing, is shown below. The beams at each story in this end-wall frame are assumed to be subject to the same dead and live load as the beam in Problem 1a. Wind loading is calculated and found to be 20 kips at the top story and 10 kips at the bottom story, and is distributed as shown. Assume that these are the only loads acting on this end-wall frame (you do not need to consider the load from the 2 × 12 supporting beam that runs perpendicular to the end-wall frame). This structure is statically determinate and can be solved by hand. Remember a couple of things. First, the tension-only straps do not resist compression, so only one strap will be loaded per floor at a time. Second, because the beams are simply supported and braces are treated like truss elements, the entire frame can be analyzed in a manner similar to a truss. 25 ft 12 ft 12 ft B B Simple Connection (Typ.) Simple Column Support (Typ.) Tension-Only Steel Strap Bracing (Typ.) 10 kip 5 kip 10 kip 5 kip Use the LRFD load combinations from ASCE 7-16 to determine: a. The maximum factored compression in the first and second story columns. b. The maximum factored uplift that the connection of frame to foundation must be designed to resist. c. The maximum factored tension in the steel strap bracing. Assume that the steel strap bracing does not carry any gravity loads. The combination of steel strap bracing, which behaves like a truss member, and the simple connections for the beams means that this frame will behave very much like a truss in that it will not transfer moment from the beams/braces to the columns. In fact, because we assume that all of the lateral load moves into the braces, no members are subject to any transverse loading, so we can model this as a truss. For a truss problem, it is often easier to solve the entire truss for each loading type and factor any required values at the end, so that’s what we will do here. Note: because the strap braces are tension only, we remove the braces in compression, as shown below in the dashed lines . Dead Load 1 . 5 kip / ft (typ.) Live Load 0 . 8 kip / ft (typ.) Wind Load 5 kip 10 kip 5 kip 10 kip We can solve each of these systems in a rather straightforward manner. For the dead and live loads, because they are gravity loads, they do not move into the braces and only travel through the beams and into the columns. As a result, the column compressive forces are equal to the reactions from the beams, and because of symmetry, each of the pin reactions resists half of the total force.

37 . 5 kip 37 . 5 kip 37.5 (C) 18.8 (C) 37.5 (C) 18.8 (C) 0 0 0 0 Dead Load 20 kip 20 kip 20 (C) 10 (C) 20 (C) 10 (C) 0 0 0 0 Live Load 24 kip 24 kip 9.6 (T) 0 24 (C) 9.6 (C) 25 (C) 10 (C) 33.3 (T) 22.2 (T) Wind Load Now we can answer our questions directly. We have three load types, so we will need load combinations for live, dead, and wind loads. The table below shows the factored load in the individual members, in kips. The controlling values are shown in bold. LC 1 LC 2 LC 3 LC 4 LC 5 1.4D 1.2D+1.6L 1.2D+(L or 0.5W) 1.2D+1.0W+L 0.9D+1.0W Column AC 52.5 77 65 55.4 24.2 Column BD 52.5 77 65 89 57.8 Column CE 26.3 38.5 32.5 32.5 16.9 Column DF 26.3 38.5 32.5 42.2 26.5 Uplift Force No Uplift No Upllift No Uplift No Uplift No Uplift ( R y,left ) ( R y > 0) ( R y > 0) ( R y > 0) ( R y > 0) ( R y > 0) 1st Story Brace 0 0 33.3 33.3 33.3 2nd Story Brace 0 0 22.2 22.2 22.2

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Problem 3 The beam below is the same beam that is statically indeterminate with a roller support at midspan. If the beam is a 40 foot W18x50 steel wide flange beam, determine the maximum compressive stress you would have to design for in the cross section. Assume that the roller support and the point load are both located at the center of the span. D = 1 . 15 kip / ft L = 1 . 85 kip / ft D = 8 . 0 kip L r = 4 . 8 kip S = 10 . 0 kip W = 15 . 0 kip E = 25 . 0 kip 20 ft 20 ft First, let’s consider what we are looking for. This problem asks us for the maximum compressive stress. The equation for the compressive stress for a beam under both transverse and axial loads is σ = My I + P A We can look up the geometric values that we need for a W18 × 50. Specifically, we have A = 14 . 7 in 2 , I = 800 in 4 and y = 9 . 0 in. In the case of a steel W section, y top = y bottom = y . We only need to use the loads and load combinations to find the maximum moment and axial force. Note that the point loads at midspan do not contribute at all to either value, since a point load directly above a support will move directly into the support reaction but does not cause any deformation in the beam. We can rigorously solve this problem with a lot of math, but there are simpler ways to do this. The axial force is constant through the entire structure. Beam tables tell us that a two span beam with span length L , subjected to a uniformly distributed load, w , has a maximum positive moment of 9 wL 2 / 128 and a maximum negative moment − wL 2 / 8. Because we have a symmetric section, the tensile and compressive stresses are equal within the section, thus we only neeed the higher magnitude of M max = wL 2 / 8. We can also consider load combinations 1 . 4 D 1 . 2 D + 1 . 6 L + 0 . 5( L r or S or R ) 1 . 2 D + 1 . 6( L r or S or R ) + ( L or 0 . 5 W ) 1 . 2 D + 1 . 0 W + L + 0 . 5( L r or S or R ) 0 . 9 D + 1 . 0 W 1 . 2 D + E v + E h + L + 0 . 2 S 0 . 9 D − E v + E h Combination w ( kip / ft) P ( kip) M ( kip ft) My/I ( ksi) P/A ( ksi) σ comp ( ksi) LC 1: 1.4D 1.61 0 80.5 10.9 0 10.9 LC 2: 1.2D + 1.6L 4.34 0 217 29.3 0 29.3 LC 3a: 1.2D + L 3.23 0 161.5 21.8 0 21.8 LC 3b: 1.2D + 0.5W 1.38 7.5 69 9.32 0.51 9.83 LC 4: 1.2D + 1.0W + L 3.23 7.5 161.5 21.8 0.51 22.3 LC 5: 0.9D + 1.0W 1.04 7.5 51.8 6.99 0.51 7.50 LC 6: 1.2D + E + L 3.23 12.5 161.5 21.8 0.85 22.7 LC 7: 0.9 + E 1.04 12.5 51.8 6.99 0.85 7.84 As we see, load combination 2 provides the maximum compressive stress of 29.3 ksi.