CEE 377 HW5 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 5 CEE 377 - Introduction to Structural Design Problem 1 This problem involves some additional calculations involving PACCAR Hall. Consider again the W24 × 55 beam examined in Lab 3 (i.e., beam B.8/11-B.8/13). a. In lab we determined that the equivalent uniform D + L load corresponding to the 38 k reactions for this beam would be 1.73 k/ft. Use the dead and live floor load values from the floor load map to calculate the dead load portion of this 1.73 k/ft loading (i.e., D/ ( D + L )). Don’t forget to subtract the member’s self weight first, then add it back in later once you’ve done a proportional split on the dead and live floor loads. The weight of the beam is 55 lb/ft, so we can first remove that from the total load, leaving w = 1.675 kip/ft. In the lab exercises, we determined that the total dead load consisted of 58.1 psf from the 4.5 inch thick slab and 30 psf from the combined ceiling (15 psf) and floor finish (15 psf) associated with dead load label ”4”, and that the live load is 100 psf associated with lobby/assembly/stairs, labeled ”F”. If the ratio of dead to total load is D/(D+L) = 88/188 = 0.468, we can assume that the dead load contributes 0.468(1.675 kip/ft) = 0.784 kip/ft. Adding back the self-weight of the beam, we arrive at a distributed dead load of w D = 0 . 839 kip/ft . b. Use this uniform dead load, w D , to determine the maximum displacement at the center of this beam due to dead load only, again assuming the connections act like simple supports. To determine the displacement, we need the stiffness terms E = 29,000 ksi and I = 1350 in 4 . For a simply supported beam under uniform loading, the maximum deflection occurs at midspan and is of magnitude v max = 5wL 4 384EI = ( 0 . 839 k / ft × 1 ft / 12 in)( 44 ft × 12 in / 1 ft) 4 384 ( 29000 ksi)( 1350 in 4 ) = 1 . 81 in c. In a simple sense, camber can be used to accommodate dead loads so that floors are close to flat prior to live loading. Compare your dead load displacement to the specified camber of this beam. It won’t match particularly well in this case, so see if you can identify some reasons why this might be so. The camber for the beam of interest, as shown in the plans, is 1 inch, much lower that what we see here. This is most likely due to overestimation of our loads and the conser- vatism associated with the tributary area method we assumed. It could also be a function of connection type and the system stiffness not matching a perfectly pinned-pinned as- sumption. d. The 38 kip end reactions shown for the W24 × 55 girder in the plans (shown again in the figure below). What end reactions would you get using each of the three methods presented in class (tributary area method and alternative methods 1 and 2)? Recall that for this floor area, using ASD load combinations, D + L = 188 psf.

Because we have a simply supported beam, we can quickly determine the end reactions by summing the total area for which the load will travel through the girders. The three options are shown below. Tributary Area Method 1 (Two-Way Slab) Method 2 (One-Way Slab) Now we can calculate the total area and subsequent total load that will move into the girder by multiplying the associated areas by 188 psf: Method A blue , ft 2 D + L , kip Reaction, kip Tributary Area 484 91.0 45.5 Method 1 393.25 73.9 37.0 Method 2 363 68.2 34.1

e. Find the end reactions using the area shown in the figure below as the tributary area. Note that the floor surface on the right is about 6 inches narrower due to the wall thickness. Does this get you closer to the 38 kip value shown in the plans? This part of the problem is essentially the same as we solved for Method 1 above. In this case, we got 37.0 kips, which nearly matches the 38 kip reactions provided. If we add the self weight back, we end up with an additional reaction of (0.055 kip/ft)(44 ft)/2 = 1.2 kips, and that basically gets us to the 38 kip reactions. The 6 inch wall thickness would make the spacing in the right most span equal to 10.5, which would reduce the total area by [( 5 . 5 ft)( 11 ft) + ( 0 . 5 )( 5 . 5 ft)( 5 . 5 ft)] − [( 5 . 25 ft)( 11 ft) + 0 . 5 ( 5 . 25 ft)( 5 . 25 ft)] = 4 . 1 ft 2 This results in a reduction of approximately 0.4 kips in the reactions. All of this gets us to 37.8 kips, which is a pretty close result.

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

f. Instead of ASD, assume that you are using LRFD to check the design of the W24 × 55 girder. Using the tributary area provided in part [e.] and the dead and live loads you found in part [a.], determine i. The maximum shear in the beam ii. The maximum moment in the beam iii. The maximum bending stress in the beam iv. The maximum deflection in the beam Here we find that the distributed area load 1.2D + 1.6L = 1.2(88 psf) + 1.6(100 psf) = 266 psf. Applying the area as a distributed load, we get 1.46 klf 2.93 klf Analyzing this beam, we can find that the end reactions are R left = R right = 1 2 [ 0 . 5 ( 1 . 46 k / ft)( 11 ft) + ( 2 . 93 k / ft)( 33 ft)] = 52 . 4 kip We know that, because this is a simply supported member, the maximum shear occurs at the pinned supports, thus we have V max = 52 . 4 kip . We also know that, for a simply supported beam with symmetric loading, the maximum moment will occur at midspan. We can calculate the maximum moment as M max = ( 52 . 4 kip)( 22 ft) − ( 0 . 5 )( 1 . 46 k / ft)( 5 . 5 ft)( 18 . 333 ft) − ( 2 . 93 k / ft)( 16 . 5 ft)( 8 . 25 ft) = 680 k · ft The maximum bending stress in the beam can then be calculated by finding the seciton modulus for the W24 × 55, which is S = 114 in 3 , thus the maximum bending stress is σ max = M max / S = 71 . 6 ksi . This is very high, way over the typical yield stress for a beam of 50 ksi. There are a few reasons for this, the primary one being that this is composite construction. In composite construction, the moment capacity can grow quite a bit as the steel beam and the concrete slab work together. If you notice, there are 24 shear studs on the W24 × 55, which justifies our assumption. The actual analysis here is quite complex and will be left for future study. Finally, for the deflection, this is a complicated beam to solve, and I am happy if you used some judgment. The exact displacment calculation, based on dead and live service loads, is 4.235 inches. A more reasonable approach would be to use the conservative tributary approach, which gives us an upper bound of 5wL 4 / 384EI = 4 . 46 inches, which represents a total displacement of L / 118, a value that is going to be too high for any reasonable limit state. Again, however, the composite nature of the beam will help with this tremendously, and that’s how the beam can be properly designed.

Problem 2 For the frame below, assume that a unit load can move from point A to point E along the horizontal member. Draw the influence lines for the vertical reactions at the supports D and F and for the shear and bending moment at point C . A B C D E F 10 ft 15 ft 15 ft 10 ft 20 ft Here we first need to determine the influence lines for the reactions. The reaction at F will behave the same as if it were at point B because this is a statically determinate frame and the force in the direction is zero. We can use the same concept where the influence lines for the reactions equal one directly above the joint in question and zero at other joints, with a straight line connecting all points. Once we solve for the reactions, we consider the scenarios when the point load is to the left and to the right of point C: Unit load to the right of point C (cutting and looking to the left) V C = − R D M C = 15 R D Unit load to the left of point C (cutting and looking to the right) V C = R F M C = 15 R F This gives us the following influence lines: 1.333 -0.333 ( IL ) R D -0.333 1.333 ( IL ) R F 0.333 -0.333 -0.5 0.5 ( IL ) V C -5 -5 7.5 ( IL ) M C

Problem 3 Using the influence lines you developed in Problem 2, determine the maximum shear and moment at point C assuming a uniformly distributed dead load of 1.25 kips/ft from points A to E and both a 0.640 kips/ft uniformly distributed live load and a 40 kip concentrated live load placed to maximize shear, positive moment, and negative moment at point C . We can use the influence lines directly to solve this problem. For maximum shear, the dead load provides zero shear at point C due to symmetry so we can maximize either positive or negative shear. To maximize positive shear, we place the point load at midspan and the distributed loads on the left cantilever and to the right of point C within the simple span. For maximizing positive moment, the point load goes at midspan and the distributed load is between the supports. For maximizing negative moment, the point load goes at either free end and the distributed load spans both cantilever sections. We can check the values for both load combinations to see which controls. We can use the following values to help us use the influence lines to determine the shear and moment: A total A positive A negative ( IL ) + max ( IL ) − max V C , kips 0 5.417 -5.417 0.5 -0.5 M C , kip · ft 62.5 112.5 -50 7.5 -5 LC1: 1.4D – w D = 1 . 4(1 . 25 k / ft) = 1 . 75 k / ft – w L = P L = 0 – V C = w D A total + w L A positive + P L ( IL max ) = (1 . 75)(0) + (0)(5 . 417) + (0)(0 . 5) = 0 – M + C = w D A total + w L A positive + P L ( IL + max ) = (1 . 75)(62 . 5) + (0)(112 . 5) + (0)(7 . 5) = 109 k · ft – M − C = 0 because the dead load only results in positive moment at C LC2: 1.2D + 1.6L – w D = 1 . 2(1 . 25 k / ft) = 1 . 50 k / ft – w L = 1 . 6(0 . 64 k / ft) = 1 . 024 k / ft – P L = 1 . 6(40 kip) = 64 kip – V C = w D A total + w L A positive + P L ( IL max ) = (1 . 50)(0) + (1 . 024)(5 . 417) + (64)(0 . 5) = 37 . 5 kip – M + C = w D A total + w L A positive + P L ( IL + max ) = (1 . 50)(62 . 5)+(1 . 024)(112 . 5)+(64)(7 . 5) = 689 k · ft – M − C = w D A total + w L A negative + P L ( IL − max ) = (1 . 50)(62 . 5)+(1 . 024)( − 50)+(64)( − 5) = − 277 k · ft And the 1.2D + 1.6L combination controls for all three values.

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

Problem 4 Consider the hoist crane shown. Stringers transfer the moving crane load to panel points on the truss. Determine the influence line diagrams for Member AC and CD . If the max concentrated live load is 25 k, and the normal stress is not to exceed 36 ksi, determine the truss diameter assuming each member is circular rod with the same area. Hint: this problem can be simplified by using the method of sections to determine the force in CD as a function of the reaction at the right pin, then using the method of joints at joint C to calculate the force in AC as a function of the same reaction. C D A B 4 @ 20 ft = 80 ft 15 ft 25 kip The influence lines for the reactions will be the same as we have had before since the truss is simply supported. The influence lines and corresponding equations are shown below. 1 ( IL ) R left = 1 − x 80 1 ( IL ) R right = x 80 Now we can solve the two member forces as functions of the reactions. Using the method of sections, we can cut the truss as shown below, then C D A B 1 R left R right When the unit load is to the left of member CD , we can look to the right, and we get X M B = 40 R right − 15 F CD = 0 ⇒ F CD = 8 3 R right = x 30 X F y = R right − 3 5 F BC = 0 ⇒ F BC = 5 3 R right = x 48 When the unit load is to the right of member CD , we can look to the left, and we get X F y = R left + 3 5 F BC = 0 ⇒ F BC = − 5 3 R left = − 5 3 1 − x 80 = x − 80 48 X M A = − 20 R left + 15 F CD + 15 4 5 F BC = 0 ⇒ F CD = 1 15 (20 + 20) R left = 8 3 R left = 8 3 1 − x 80 = 80 − x 30 Then we separately consider member AC by solving the method of joints at joint C . In this case, we will consider the unit load to the left or to the right of point C . When the unit load is to the left of joint C , we can look to the right, and we get X F y = F AC + 3 5 F BC = 0 ⇒ F AC = − 3 5 F BC = − R right = − x 80 When the unit load is to the right of joint C , we can look to the left, and we get X F y = F AC + 3 5 F BC = 0 ⇒ F AC = − 3 5 F BC = R left = 1 − x 80 So this gets us here:

1 ( IL ) F AC 1 ( IL ) F CD We have two things left to check. First, we can connect the dots for the force in member CD . We need to actually think about what is happening in the missing section. For a truss, the load can only appear at a joint, which has the implication that there is no discontinuity as shown in the result for member CD . For the influence line for member AC , we have an inconsistency because this member is in the same direction as the loads. The values shown in the figure above are just to the right or just to the left of point C , but in fact we cannot be just to the left or right, only directly at point C . If we place the load at point C , then our assumption that the ”load is to the left of member CD so we look to the right” neglects to actually include the unit load at point C , so if we add it back, we get X F y = F AC + 3 5 F BC − 1 = 0 ⇒ F AC = − 3 5 F BC = 1 − R right = 1 − x 80 = 0 . 75 This confirms that the correct selection for the influence line is 0.75 at that point. Thus, our final influence lines are shown below. 0.75 ( IL ) F AC 4/3 ( IL ) F CD

Problem 5 Determine the moment envelope for the AASHTO design tandem (two 25 kip loads spaced 4 ft apart) traveling across a 30 foot simply supported beam. We did a problem very similar to this in the examples. A B 30 ft 25 k 25 k 4 ′ 1 . 0 ( I.L. ) R A 1 . 0 ( I.L. ) R B We will construct an influence line for the moment at an arbitrary point C , located a distance a from the left support. Taking cuts at point C and looking to either side as the load moves from one side to the other, we again find that M C = bR B when the unit load is to the left of point C and M C = aR A when the unit load is to the right. , L = 30 ft a b A B C bR B aR A ( I.L. ) M C ab/L C It doesn’t matter which load we choose to place at point C , so I will place the left load there. Again, we will take advantage of symmetry and it does not matter which direction we assume for our analysis here. The magnitude of the moment influence line needs to be found at each load. We can use the right side of the influence line, where we find: ( I.L. ) M C = a ( I.L. ) R A = a 1 − x L @ x = a : ( I.L. ) M C = a 1 − a L = a ( L − a ) L = ab L @ x = a + 4 : ( I.L. ) M C = a 1 − a + 4 L = a ( L − a − 4) L = a ( b − 4) L bR B aR A ( I.L. ) M C ab/L a ( b − 4) /L C 25 k 25 k a 4 ′ b − 4 ft This allows us to calculate the moment at point C directly as a function of a M C = ab L (25 kip) + a ( b − 4) L (25 kip) = 25 a ( L − a ) L + 25 a ( L − a − 4) L Now we can plug in L = 30 feet and simplify, and we get M C = 5 a 6 (2(30) − 2 a − 4) = 5 a 3 (28 − a ) = − 5 a 2 3 + 140 a 3 This represents the equation for the maximum moment at each location a for all potential load locations. We can find the location of the absolute maximum moment by taking the derivative of M C with respect to a and finding the local maximum dM C dx = 140 − 10 a 3 = 0 ⇒ a = 14 ft

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

This tells us that the maximum moment occurs 14 feet from the left support, or just slightly to the left of midspan. The corresponding maximum value of the moment at C for a = 14 ft is M C = 327 k · ft. Because of symmetry, we can also invert this equation and find that a second location provides the same maximum moment 14 feet from the right support. The entire moment envelope is shown below. Note that the dashed lines inside of the envelope represent the moment envelope for the truck driving in a direction that does not control at that location, and the two curves represent the two directions of travel. M max = 327 k · ft As a note, in the case of the simply supported beam, when one load is larger than all others in a load train, the absolute maximum moment occurs when the load train is placed so the midspan of the beam is halfway between the largest load and the centroid. For the problem above, we calculate the centroid of the load train as directly in the middle, or ¯ x = 2 ft. Placing point C so that it is halfway between the centroid and the load train and the maximum load, we would place the load where an axle is one foot from point C .