LESSON 8: COLUMN BUCKLING I

Transcription

1 "lways, there is football." Cristal Choi, former exchange student who lived in Prof. Kurtz s house, explaining all merican holidays to a new exchange student. LESSON 8: COLUMN BUCING I Wednesday, September 12, 2018 LESSON OBJECTIVES 1. Determine the controlling slenderness /r for a pinned-braced column. 2. Calculate the Euler Elastic Buckling Stress F e for a column, considering both axes of buckling. 3. Describe the conditions for which the Euler expression accurately predicts column buckling. 4. Calculate the nominal buckling load P cr for a column per ISC equations, considering the controlling axis of buckling. TODY S SCOPE THE SIMPLE BRCED FRME Today s scope is a column in a simple-braced frame. simple braced frame uses pinned connections and is stabilized by the truss-action of diagonal bracing. Bracing prevents the relative translation of the top and the bottom of the column. In contrast, unbraced frames are characterized by the relative translation of the top and the bottom of the column and will be covered in a subsequent lesson. SOME NOMENCLTURE (for more, see the INDEX tab in the steel manual) In structural steel design, the letter F indicates a stress, P indicates a compressive force, T indicates a tensile force. F cr critical buckling stress (ksi). When columns are very long (long enough to remain elastic when buckling occurs), then F cr is based on F e, the Euler buckling stress F e Euler buckling stress (ksi). The Euler equation predicts the buckling stress, assuming that the column is elastic (generally true for very long columns). F y yield strength (ksi). This is the maximum elastic stress that steel can sustain, in tension. I x moment of inertia about the member s x-axis (in 4 ) I y moment of inertia about the member s y-axis (in 4 ) x effective column length (inches) for buckling about the x-axis this is the distance over which 0 to of a sine curve is completed by the column s buckled shape. y effective column length (inches) for buckling about the y-axis this is the distance over which 0 to of a sine curve is completed by the column s buckled shape. r x radius of gyration about the member s x-axis (inches) r y radius of gyration about the member s y-axis (inches) P cr critical buckling load (kips). This equals the Euler buckling load if the column is very slender and remains elastic during buckling. 2 2 P e Euler buckling load, EI E Pe 2 g 2 r P n nominal column capacity (kips). This is the same as P cr because all columns in compression fail by buckling. REDING The online version of this handout contains 8 pages of reading Textbook (Segui), pp. 109 to 121. REFERENCE ISC Steel Manual Chapter E in the specifications ( to ) ISC Steel Manual Part One (Table 1-1) ISC Table 4-22

2

3 HOMEWORK (Due Friday. Standard assignment. Presentation counts) Terms: P n is the nominal buckling strength (units = kips). This is the actual predicted buckling (failure) load of the column. P n / is the (SD) llowable load, where is the safety factor ( =1.67); i.e., an engineer would not want the applied (actual) load to exceed the allowable load, as this would be unsafe. P n is the (LRFD) Design Strength (it is not an allowable load!!! If your column were to actually have this load on it, then you should get the Hell out of the building fast), whereis the strength reduction factor (=0.9 for column buckling); i.e., the engineer would not want the factored load to exceed the design strength, as this would be unsafe. 0. (Do this but do not hand in). Go to the steel manual, Part 2 and flip through the pages; this is where you will find the most general information about steel design (LRFD, SD, fabrication tolerances, materials, etc.). Find page Go to ISC Tables 2-4 and 2-5 to answer the following: a. What is the preferred STM material specification for W shapes? What is the minimum yield strength for this? What is the minimum tensile strength for this? b. What is the preferred STM material specification for L shapes? What is the minimum yield strength for this? What is the minimum tensile strength for this? Note: from now on, unless stated otherwise, you are to assume the preferred STM material specification for all shapes. 0. (Do this but do not hand in). For a 24-ft long W12x96 column, two possible bracing ideas are shown, and B. Circle the bracing idea that is smart. 12 ft. 12 ft. 24 ft. Brace 24 ft. Brace 12 ft. 12 ft. 0. (Do this but do not hand in). Prove to yourself that Braces are blind to gravity loads ; i.e., the brace does not have a force in it, due to the vertical load. a. Do equilibrium on Point. From this, determine the force in B (remember zero-force members?) b. Knowing the force in B, do equilibrium on Joint B. From this determine the force in the brace. 100 k 100 k B B 0. (Do this but do not hand in). Let s suppose that we know that the actual applied load on a column is 100 kips. Let s also suppose that we know that the P n of the column is 105 kips. Is the column safe? 0. (Do this but do not hand in). Let s suppose that we know that the factored load on a column is 100 kips. Let s also suppose that we know that the P n of the column is 105 kips. Is the column safe?

4 1. Compute the allowable buckling stress F cr / for a column that has a slenderness of 50 if F y =50ksi. Then, go to Table 4-14 in the ISC manual and find the allowable buckling stress F cr / for a column that has a slenderness of 50 (note: from now on, Table 4-14 is a good table to use, to save yourself time. Put a tab there. void hand-calculating F cr unless F y is an odd specification, not found in the table.) 2. The 6 diameter Extra-Strong Pipe section (Pipe 6 x-strong) is subjected to axial forces P Dead = 65 kips and P Live = 55 kips. Determine if the section is adequate per SD. Then determine if the section is adequate per LRFD. Use the preferred material specification (if you don t know what this means, then you should go back to problem 0). Note x-strong pipe properties are found in Part 1 of the steel manual (flip through part 1 until you find Pipes ). 3. Determine if an HSS4x3x3/8 column is adequate per SD. Then determine if the section is adequate per LRFD. The single-story 1-bay building frame has a Roof Dead Load pressure of 25psf. This includes the proper allowance for all roof weights: fill beams, girders, the roof deck, the roofing material, etc.(it s all included in the 25 psf number) Snow Load pressure: 50psf (note: that s a lot. This building must be in Canada) The story height is 22 but pay close attention to the bracing and note the way it braces the column s weak-axis. The bay lengths are 30, each. Note: the nominal dimensions of the HSS ( hollow structural shape ) are 4x3x3/8, but use the real dimensions 1 and properties given in the ISC Part One Tables. s described in the previous problems (refer to problem 0, above), use the preferred material specification for rectangular HSS. 22 Braces HSS4x3x3/8 Column 1 lways use the decimal dimensions of cross-sections, for calculations. The fractional dimensions are for architects, only. Example: a W16x26 has a depth of 15.7 or 15 ¾. Engineers use the rchitects use the less exact 15 ¾.

5 SUPPLEMENTRY REDING: Objective: Calculate the Euler Elastic Buckling Load P e for a column and State the limitations of the Euler Expression. If an increasing axial force is applied to a column that is very short and stocky, then it will eventually fail because of material failure such as concrete crushing (concrete columns) or steel yielding (steel columns). However, if the column is somewhat slender, then increasing axial force will result in failure due to buckling at the buckling stress F cr, rather than by material failure. The buckling strength of a column depends upon how slender the column is. Columns that are very slender buckle at stresses that are lower than columns that are shorter and stockier. Column slenderness is quantified by the slenderness ratio, where is the effective length of the column and r is the radius of gyration of its cross section. Pn Pn Buckled Shape Buckled Shape Very Slender Column Long Length, Low Cross Section I Low P n Less Slender Column Shorter Length, Higher Cross Section I Higher P n THREE RNGES OF SLENDERNESS, For most engineering materials, three general ranges of column slenderness can be identified: Short Columns. Columns that are very short and stocky do not buckle. The maximum stress that a short column can reach in axial compression is the stress corresponding to the material s ultimate strength F u. Slender or Elastic Columns. Columns that are very slender will buckle in an elastic manner when subjected to their maximum load; that is, the buckling stress is sufficiently low so that the material does not exhibit any yielding or damage. Upon unloading, these columns will elastically rebound to their unloaded position, showing no damage. Elastic column buckling stress is reasonably predicted by the Euler expression: Where: Figure 1 E = Elastic Modulus (psi or ksi) = Effective column length (inches or feet) r = Radius of Gyration (inches or feet) = Slenderness Ratio (dimensionless) Note that the strength of the material (e.g., material yield stress F y or material ultimate strength F u ) is not present in the Euler expression. For elastic buckling, the strength of the material is irrelevant because buckling occurs at a stress that is well below the material strength. Intermediate or Inelastic Columns. These columns have slenderness that is in between short columns and elastic columns. For these columns, buckling is influenced by the strength of the material and this buckling occurs as part of the column cross section undergoes yielding or damage in some way.

6 Typical Column Buckling Strength Curve Buckling Stress F cr (ksi) Short Columns Intermediate or Inelastic Columns Slender or Elastic Columns Column Slenderness (/r) Figure 2 The effective length will be covered in greater depth with subsequent objectives and you will see that the effective length depends on the support conditions of the column (pinned, fixed, free, etc.). However, columns that are pinned on both ends are the simple case that is handled with this objective because the effective length is simply the physical length of the column L. Pn Pin Buckled Shape = The radius of gyration where I is the moment of inertia of the cross section and is the area of the cross section. Note that: Pin Pinned-Pinned Condition K = 1 = L Radius of gyration r is NOT the same as the radius r (be careful with this distinction, when analyzing circular cross sections such as pipes).

7 Because I x is not generally equal to I y, r x is not generally the same as r y. To: Calculate the Euler Elastic Buckling Load P e for a column and State the limitations of the Euler Expression Determine the effective length(s), (with this objective s examples, the effective lengths x and y are equal, but with subsequent objectives, you will see that x is not generally equal to y ) Determine the moment(s) of inertia I of the cross section (with this objective s examples, I x is equal to I y, but with subsequent objectives, you will see that I x is not generally equal to I y ). Based on the I and of the cross section, compute the radius of gyration using Compute the Euler stress, using consistent units. xial force = (axial stress) x (area). Therefore, the Euler buckling load P e = F e. Know that the Euler stress is: o good predictor of buckling stress when the column is very slender so that the buckling stress is substantially less than the ultimate or yield stress of the material. rule of thumb is that, if F e is less than one third of the material s ultimate or yield stress, than the Euler stress is an accurate predictor of buckling. o poor predictor of buckling stress when the column is very stocky. Note the fact that the F e goes to infinity when /r 0. Clearly, the Euler stress is a very poor predictor when /r is small. Example 1 Problem: Determine the Euler buckling load F e for the timber column and state whether or not the Euler buckling load is a reasonable and accurate predictor of the actual buckling load. Given: Timber column cross section has nominal dimensions of 6 x6, with actual dimensions of 5¼ x 5¼. The modulus of the elasticity E = 1500 ksi, while the ultimate strength of the timber is 6000 psi. The column is 8 ft long, pin supported on each end. Solution: Being careful with consistent units! = 8 = 96 Pe =? ¼ " 1.516" ft 5 ¼ Cross-Section View - 5 ¼ x 5 ¼ Timber Soln. Elevation View (Side View) Pin-Ended 8-ft-long Column However, the Euler buckling stress F e =3.692ksi is more than 60% of the material s ultimate stress of 6 ksi. Therefore, the Euler buckling stress and load are not accurate/reasonable predictors of the actual buckling load.

8 Example 2 Problem: Determine the Euler buckling load F e for the round steel HSS 6 x0.5 ( hollow structural shape ) column and state whether or not the Euler buckling load is a reasonable and accurate predictor of the actual buckling load. Given: The HSS 6 x0.5 column has a 6 outer diameter and a nominal wall thickness of 0.5, but an actual wall thickness of (refer to ISC Table 1 13 for other properties). The modulus of the elasticity of steel is ksi. The yield strength of the steel is 42ksi. The column is 25 ft long, pin supported on each end. Solution: Being careful with consistent units! " For a circle: Pe =? ft Cross-Section View - 6x½ HSS 300" 1.964" Soln. Elevation View (Side View) Pin-Ended 25-ft-long Column The Euler buckling stress F e =12.26ksi is about 30% of the material s yield stress of 42 ksi. Therefore, the Euler buckling stress and load are probably accurate/reasonable predictors of the actual buckling stress and load. Objective: Determine the controlling slenderness /r for a column. Ordinary column buckling is also known as flexural buckling due to the fact that buckling is resisted by the flexural stiffness of the column, which will be greater if the EI of the column is greater. For a fixed column area and column material, the EI of the section is maximized if the cross sectional area is spread out as much as possible, as is the case when a hollow pipe section is used as a column. Whereas the pipe section shown in Figure 3a below has the same area as the solid circular section shown in Figure 3b, the hollow pipe section is a vastly superior column because its moment of inertia is so much larger. The result is a column with far greater axial load carrying capability. y y () x (B) x Hollow Pipe Section Efficient Column Choice Illustration: Outside Diameter = 5 Wall Thickness = rea, = 3.14 in 2 Moment of Inertia, I = 9.02 in 4 Radius of Gyration, r = 1.69 in. Solid Circular Section Inefficient Column Choice Illustration: Diameter = 2 rea, = 3.14 in 2 Moment of Inertia, I = 1.57 in 4 Radius of Gyration, r = in. Both sections have the same area. They use the same amount of material. But the hollow section has an I that is 5.7 times larger and a radius of gyration that is 2.4 times larger. Figure 3

9 XES WEK XIS ND STRONG XIS The previous figure illustrates that a column shape in which material is spread out as much as possible will be the most efficient column shape, because it maximizes the flexural stiffness, while using the same amount of material. In the previous figure, the moments of inertia for the x and y axes were equal: I x = I y. In general, this is not the case for all columns. For the majority of columns, one axis of bending has a higher moment of inertia than the other. The I shaped section below illustrates this. The I shape has I x that is 10 times larger I y. Though the column could buckle such that it causes bending about either axis, the weak axis will control. L L 1 View 1 View 1 Buckling causes bending about the strong axis View 2 Buckling causes bending about the weak axis Two Views of the Same Column 2 View 2 Cross-Section View - Wide-Flange Section Conclusion: The column is weaker with respect to the y-y axis, so this axis will control Figure 4 If the shape previously shown were to be subjected to an increasing axial force until buckling occurs, the shape would fail with bending about the y y axis because its bending resistance is much less about this axis. For this reason, the y y axis is commonly called the weak axis and the x x axis is called the strong axis. One would also say that the weak axis controls. PINNED SUPPORTS = BRCES is the effective length of a column. For a pin ended column, the effective length IS the length of the column L. Though we may represent these as lines with pin symbols on both ends, P What does this symbol really represent? The symbol of a pin ended column really represents a column that is braced by pin ended trussaction, which prevents the translation of the column at the braced point. Because the truss action is pin ended, however, the bracing does not prevent the rotation of the column at the braced point. In order for the rotation of the column to also be prevented, the columns would have to be rigidly connected (as opposed to pin connected) to large, stiff elements such as a deep girder. When a frame is braced and consists exclusively of pin connected beams, columns, and braces, it is known as a simple braced frame. s shown in Figure 4, all of the columns in a simple braced frame have effective length equal to the length of the column between the braced points. SYMBOL FOR PIN- ENDED COLUMN

10 D E L M T C Brace F Column Beam K N S B G J O R H I P Q Truss action in this bay Simple-Braced Frame Prevents horizontal movement of all joints ll joints are braced. No joints can move horizontally Figure 4 XES X ND Y The previous figure showed the bracing for one planar view of a simple braced frame and illustrated that the effective length is the distance between the braces. However, the bracing is not necessarily the same for other planes in the structure. For example, Figure 6 illustrates that the designer may elect to brace the structure so that for one axis of buckling is different from of the other axis of buckling. This is usually an intentional strategy employed by the designer, as the use of additional bracing is used to strengthen the weak axis of the column. Two views of the same structure. Note that, for weak-axis bending the braces are closer 6 ft. 12 Figure 6 Column Weak xes y=6 ft CONTROLLING SLENDERNESS -. You have seen that x is not generally equal to y and that r x is not generally equal to r y. You also know that the buckling strength of a column depends on the slenderness ratio. While a column can buckle about either its x axis or its y axis, it will be controlled by whichever axis has the greater slenderness ratio. To Determine the controlling slenderness /r for a column: 1. Determine x and y for the column. 2. Determine r x and r y for the column cross section. 3. Compute the slendernesses. The greater, either, will control buckling. Example: If , then the column will fail by buckling about the x axis and the buckling stress will be determined based on 100. Column Strong xes =12 ft

11

12 Example Determine the controlling axis of buckling for one of the base columns. Two planar views are shown for the same structure. Given: The base column is a W12x72. Refer to ISC Table 1 1 for required properties. Solution: x = 14 ft = 168 y = 7 ft = 84 From ISC Table 1 1 r x =5.31 inches r y = 3.04 inches Compute slendernesses: Two views of the same structure. Note that, for weak-axis bending the braces are closer The controlling slenderness is Buckling will occur about the x axis. The buckling load can be computed using ft. ISC BUCING EQUTIONS s illustrated in Figure 2, the Euler Expression is valid for long slender columns that buckle elastically; i.e., no yielding of the steel takes place so that the column would elastically rebound once the load is removed. For somewhat shorter, stockier columns, buckling is not accurately predicted by the Euler Expression. Instead, the column behavior is influenced by the yielding of the material. Hence, less slender columns buckle such that they exhibit partial yielding and permanent deformation. The ISC column buckling expressions are shown below. Note that they incorporate the Euler Expression, F e. If 2.25 ELSTIC BUCING Base Col. Column Weak xes y=6 ft 14 Base Col. Column Strong xes x=12 ft If 2.25 INELSTIC BUCING Based on these, the nominal column strength P n is simply F cr, where is obviously the cross-sectional area of the column.