| Introduction | ||
 Building Canopy with New Hoist | The manager at the loading dock thought it would be a good idea to install an electric hoist under the existing canopy. If installed on a rail system, then large loads can be easily lifted and moved to another truck. The existing canopy is constructed from 1.5" O.D. pipes that are connected with ball joints. There is one truss for every 5 feet of canopy. The canopy roof needs to carry at least 60 lb/ft2 for possible dead and live loads (wind, weight, snow, etc.). What is known:
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| Question | ||
| What is the maximum weight that the hoist can lift (ignore weight of the hoist itself) for the existing canopy with a factor of safety of 2.0? | ||
 3D Truss Example at the Milwaukee Airport | ||
| Approach | ||
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| Column Buckling | |||||
 Basic Pinned Column before and after Buckling | In addition to understanding stress and material failure due to tension and compression, buckling needs to be investigated. While buckling can occur in plates and shells, column buckling is most common and easiest understood. If an object buckles, it does not necessarily mean it has failed, but a buckled structure will experience significant lateral deflection. Furthermore, a buckled column structure will not sustain any significant additional load. Column buckling is not an exact science, and many issues can cause pre-mature buckling, such as misaligned loads, material imperfections, and varying structural dimensions. However, for even perfect systems, buckling will be significantly lower than the crushing or compression limit of the material. | ||||
 Load-Deflection Diagram for Column Buckling Click to view movie (61k) | When any long slender object is loaded in the axial direction, it will buckle when the load reaches a particular load. This load, called the critical load, Pcr, is when the object will deflect significantly in the lateral direction (perpendicular to the load). After buckling starts, the structure will not be able to sustain any additional loads. For an ideal column, buckling will not occur until Pcr has been reached. Thus the column will have no lateral deflection up to Pcr, and then all at one time, the member will deflect. This is plotted in the diagram at the left. The beam's internal resistance to bending keeps the beam from buckling. However, at some point, the potential moment generated by the axial load will be greater than the internal resisting moment, and the column will buckle. For columns, it is generally assumed that if buckling occurs, the structure has failed. Thus, a column structure should be designed not to buckle. | ||||
| Euler's Formula for Simply Supported Column (pinned-pinned) | |||||
 Simply Supported (pinned-pinned) Column in Buckled Mode  Free-Body Diagram of Lower Column Section | The theoretical buckling load, Pcr, for a basic column can be determined and the resulting equation(s) are called Euler's formula. The simplest column to develop the buckling equations is when both ends are simply supported by a pin joint (also called a pinned-pinned column). This means that it can not deflect at the joint, but it can rotate as shown in the diagram at the left. If the column is loaded and starts to deflect laterally, then the moment in the column must equal the moment cased by the load P at each end. This condition can be better understood by taking a section cut at any point along the column and constructing a free-body diagram. Summing the moments at any point along the lower column section gives, ΣM = 0 M + Pv = 0 The moment M, is drawn in the positive direction, even though it may be actually act in the opposite direction. The distance v is the lateral deflection at the cut location, x, from the bottom pin joint. | ||||
 Column Rotated - Similar to Beam | If the column is rotated 90o, it becomes a simply supported beam. As such, the beam bending equations are still valid and can be used. Recall from beam deflections, the moment is related to the deflection as Combining the two previous equations gives  This is a standard differential equation which has a general solution of  where C1 and C2 are unknown constants that depend on the boundary conditions of the differential equation. In this particular case, both ends of the beam or column are pinned so that v = 0 when x = 0 and x = L. The first condition, vx=0 = 0 gives | ||||
| | 0 = C1 (0) + C2 (1) => C2 = 0 The second condition, vx=L = 0 gives  Since C1 cannot be zero (would be trivial solution, v = 0), the sin function must be equal to zero which requires,  | ||||
 First Three Column Buckling Modes | The lowest load is when n equals 1, and is referred to as the critical load Pcr. Euler's formula for a pinned-pinned column is
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 Simply Supported Column Possible Buckling Directions | Buckling Direction | ||||
| Simply supported columns (pinned at each end) will buckle around the axis with the lowest moment of inertia. For example, for a rectangular cross sectional column, as shown at the left, the column will buckle around the z-axis since Iz will be less than Iy. If it is an unsymmetric cross section, then it may not be obvious and may even buckle at a particular angle between the y- and z-axis. In this eBook, only symmetric cross section members are considered. When calculating buckling, check both directions to find the lowest moment of inertia. Then use that for the critical buckling load calculations | |||||
 Basic Canopy Truss Joint Labels and Forces | The canopy currently exists and a new hoist needs to be installed under the canopy to move packages and equipment around on the dock. The logical location for the hoist is to attach it to the center truss node as shown in the diagram. The actual load W is not known and needs to be determined. The roof load is given, 60 lb/ft2, and the truss configuration is known. The question is what is the maximum load W before one of the members buckle? The canopy requires a factor of safety of 2 which needs to be included in the calculations. There is a canopy truss every 5 feet along the length of the dock, so each truss needs to support at least 5 feet of the roof load. The first step will be to solve for the loads in the truss members using both the joint method and the section method from statics. | |
| Tension and Compression Members | ||
 Tension (blue) and Compression (red) Members | Before calculating the force in every member, some members can be eliminated from consideration. Recall, the problem is only concerned about buckling so just the compression members are of interest. The top two members, AB and BC, will be in tension as the canopy is pushed downward by the roof and the hoist loads. Next, the full hoist load will be carried in tension by member BE since they are collinear at joint E. Likewise, from inspection members DB, DE, and EC will be in compression resisting the roof and hoist loads. If DB is in compression then AD will be in tension. This leaves only members DB, DE, and EC. | |
| Roof Load Distribution | ||
 Roof Load Distribution on Truss | Since the trusses are spaced every 5 feet along the length of the dock, each truss will need to carry the roof load over 5 feet. The total load will be Truss Total Load = (60 lb/ft2)(8 ft)(5 ft) = 2,400 lb The roof is connected to the truss through the joint points and so the roof load is distributed to each of the joints as concentrated forces. However, the distribution is not equal. The outside two joints (A and C) will only carry half of what the center joint will carry. FA = FC = (60 lb/ft2)(2 ft)(5 ft) = 600 lb FB = (60 lb/ft2)(4 ft)(5 ft) = 1,200 lb The angle θ can be determined from the geometry, tanθ = 3/8 => θ = 20.56o | |
| Member Loads in EC and DE | ||
 Joint E Free Body Diagram | Notice that all members at joint E are perpendicular. This requires the force in member DE equal the force in member EC. Thus, FDE = FEC There are many ways to find the load in truss members, but one useful method is the method of joints. This method requires one to analyze a single joint with only two unknown member forces. This can be done at joint C. | |
 Joint C Free Body Diagram | The equilibrium equations can be applied to joint C, giving ΣFy = 0 -600 + FBC sin20.56 = 0 FBC = 1,709 lb ΣFx = 0 -1,709 cos20.56 - FEC = 0 FEC = -1,600 lb The negative value means it is under compression, as originally assumed. Notice, that both members are not effected by the hoist load. In other words, none of the hoist load will be supported by members BC or EC. Therefore, members BC or EC are not critical when the hoist is used. | |
| Member Load in DB | ||
 Truss Section | The other compression member is DB. The load in DB can best be determined using the method of sections. Making a section cut through members DE, DB, and AB allows solving for all three members. Taking moments about joint C will give one equation and one unknown, ΣMC = 0 W (4) +1,200 (4) + FDBcosθ (1.5) + FDBsinθ (4) = 0 FDB = -1.424W - 1,709 lb The compression force in FDB is a function of the hoist weight W. | |
| Buckling load of Member DB | ||
The buckling load of member DB can be determined using Euler's Formula for simply supported (pinned-pinned ends) columns. The moment of inertia for the 1.5 inch pipe with a wall thickness of 1/16 in is  = 0.07304 in4 The length of member DB is L = (42 + 1.52)1/2 = 4.272 ft = 51.26 in The objective is to find the largest hoist load, W, that can be applied without buckling with a a factor of safety of 2.0. This gives,  W = 1,593 lb Thus, the hoist and its load can be a maximum of 1,593 lb. This should be sufficient to assist in moving most packages and equipment around on the dock. | ||
| Example | ||
 | A 20 foot high steel column is constructed from an I-beam W8×31. In order to strengthen the section, the engineer decided to cover each flange with a 10 in by 1 in steel plate as shown in the diagram. What is the increase in critical load capacity of the new column? Assume the steel modulus of elasticity, Esteel, is 29,000 ksi and the yield stress, σyld, is 36 ksi. | |
| Solution | ||
 Basic W8x31 I-Beam | A simply supported column (pinned at each end) will buckle around the axis with the lowest moment of inertia. The moment of inertia of a standard W8×31 can be found in the Structure Shapes appendix as, Ixx = 110 in4 Iyy = 37.1 in4 A = 9.13 in2 The lowest critical load for this I-beam is around the y axis, Pcr = π2 EIyy / L2 = π2 (29,000 ksi)(37.1 in4) / {(20 ft) (12)}2 = 184.4 kip The critical stress, σcr = Pcr / A = 184.4 / 9.13 = 20.20 ksi This is less than the yield stress of 36 ksi so application of Euler's equation is appropriate. | |
 W8x31 I-Beam with Cover Plates | To increase the critical load of the section, the I-Beam is covered with a plate 10 in × 1 in. For this new section, the moment of inertia and area are,  The improved section Iyy still has a lower value than Ixx. So, it will buckle around y axis. Critical load with the cover plates is Pcr = π2 EIyy / L2 = π2 (29,000 ksi)(203.8 in4) / {(20 ft)(12)}2 = 1,013 kip The critical stress, σcr = Pcr / A = 1013 / 29.13 = 34.78 ksi This stress is still below the yield stress so Euler's equation is still appropriate. The increase in critical load capacity, δPcr = (1013 - 184.4) kip = 828.6 kip or nearly 450% increase. | |
