MODULE F: SIMPLE CONNECTIONS

Transcription

1 MODULE F: SIMPLE CONNECTIONS This module of CIE 428 covers the following subjects Connector characterization Failure modes of bolted shear connections Detailing of bolted connections Bolts: common and high-strength Shear strength of high-strength bolts Slip-critical connections High-strength bolts in tension Combined shear and tension in fasteners Welded connections Fillet welds READING: Chapter 7 of Segui AISC LRFD Manual of Steel Construction, 3rd Ed. 8/21/ :12 PM 1

2 INTRODUCTION For the purpose of this class, connections are characterized as either Simple Eccentric The figure below from Segui illustrates different types of bolted and welded connections. Those connections in parts a and b of the figure are termed simple Line of action of resultant force passes through the center of gravity of the connection All parts of the connection share equally in resisting the load The other connections are termed eccentric and are discussed in the following module. Eccentric connections are shown in parts c and d of the figure below. Chapter J of the LRFD Specification addresses the design of connections. The following connectors are covered Bolts Rivets Welds Only bolted and welded connections are considered in CIE /21/ :12 PM 2

3 FAILURE MODES OF BOLTED SHEAR CONNECTIONS Two types of bolted connector failure are considered in this section Failure of the connector Failure of the connected parts 8/21/ :12 PM 3

4 Connector failure Consider first the figure shown below from Segui that presents joints in single (one shear plane) and double (two shear planes) shear. In part a of the figure (single shear connection), the load (shear) on the fastener is P and the connector failure is in the fastener. Loading is not concentric on the fastener, the eccentricity is small and it will be ignored. The load resisted by the fastener can be written as P = f A v where f v is the average shearing stress and A is the cross-sectional area. In part b of the figure, the connector is in double shear and the load is equal to P = 2 v f A 8/21/ :12 PM 4

5 Why? Failure of the connected parts The failure of the connected parts can be separated into two categories. Failure resulting from excessive tension, shear, or bending in the parts being connected For a tension member must consider tension on the net area, tension on the gross area, and block shear For beam-beam or beam-column connections, must consider block shear Gusset plates and framing angles must be checked for P, M, and V Failure of the connected part because of bearing exerted by the fastener If the hole is slightly larger than the fastener and the fastener is assumed to be placed loosely in the hole (rarely the case), contact between the fastener and the connected part will exist over approximately 50% of the circumference of the fastener. See the figure below for information. 8/21/ :12 PM 5

6 The average bearing stress is f p = P dt where terms are defined in the figure. If the maximum bearing pressure is known, the maximum load is easily calculated by manipulating the above equation The bearing problem is complicated a little by the edge distance and bolt spacing; subjects that are discussed below. See the figure below from Segui for an illustration of these complications. 8/21/ :12 PM 6

7 DETAILING BOLTED CONNECTIONS Bearing strength, bolt spacing and bolt edge distance are considered in this section. Bearing strength limits One possible failure mode resulting from excessive bearing is shown in the figure below from Segui. In this figure, the actual failure surface is replaced by a failure surface in part b, which simplifies the calculation. Noting the total failure load is equal to the sum of the failure loads on the two surfaces shown in part b of the figure: where R = 2(0.6 F ) L t = 1.2F L t n u c u c 8/21/ :12 PM 7

8 0.6F u = Shear fracture stress of the connected part L c = Distance from the edge of the hole to the edge of the connected part t = Thickness of the connected part The tear-out described above can also occur between bolt holes in the direction of the applied load as shown in the figure below, which provides an alternate definition for L c. To prevent excessive elongation of the hole, an upper limit is placed on the bearing load. This limit is where C = Constant d = Bolt diameter Rn = CFu ( dt ) t = Thickness of the connected part The values assigned to C are 2.4, 3.0, and 2.0, depending on the hole type and the acceptability of hole ovaling at service load. See Equations J3-2a, -2b, and 2c of the LRFD Specification for 8/21/ :12 PM 8

9 details. Considering only standard (not slotted) holes, if excessive deformation is a concern, C = 2.4. In this case, If L 2 d, R = 1.2L tf c n c u If L > 2 d, R = 2.4dtF c n u The design bearing strength is equal to φ Rn, where φ = Spacing and edge-distance requirements Section J3.3 of the LRFD Specification sets a minimum center-tocenter spacing (s) of bolts equal to 2.67d (and preferably not less than 3d), where d is the fastener diameter. Such distances are needed to maintain clearances between bolt nuts and maintain clearance for wrench sockets. See the figure from Segui below for details. Section J3.4 of the LRFD Specification gives minimum edge distances ( L e ) as a function of bolt-size and type of edge (sheared, rolled, or gas cut). The table referenced in that section (Table J3.4) is presented below. 8/21/ :12 PM 9

10 The example below from Segui illustrates the detailing requirements introduced above. 8/21/ :12 PM 10

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12 In the above example, the calculated bolt spacings and edge distances are the same in the gusset and tension member. Only the thickness will control in this case, and because the thickness of the tension member is less than that of the gusset, the gusset will control. BOLTS: COMMON AND HIGH STRENGTH Common bolts Common bolts differ from high-strength bolts by Material Clamping force not accounted for with common bolts Common bolts are designated as ASTM A307, with an ultimate shearing stress of 24 ksi (see Table J3.2 of the AISC LRFD Specification). The design shear strength of A307 bolts is φ Rn, where φ is 0.75, and the nominal shear strength is R = F A = 24A n v b b 8/21/ :12 PM 12

13 where F v = Ultimate shearing stress A b = Cross-sectional area of the unthreaded part of the bolt (nominal bolt area) t = Thickness of the connected part 8/21/ :12 PM 13

14 The example below from Segui illustrates the calculation of the design strength of a tension connection with A307 bolts. 8/21/ :12 PM 14

15 High-strength bolts High-strength bolts for structural joints come in two grades, namely ASTM A325 ASTM A490 Two conditions of bolt installation are used with high-strength bolts Snug-tight (producing a bearing connection) Few impacts of an impact wrench Full effort of a worker with an ordinary spud wrench 8/21/ :12 PM 15

16 Tensioned (producing a slip-critical connection) Turn-of-nut method: specified number of rotations of the nut from snug tight (nut rotations correlated to bolt elongation) Calibrated wrench tightening Alternate design bolts: specially design bolts whose tops twist off when the proper tension has been achieved Direct tension indicators: compress washer (under bolt head or nut) with protrusions to a gap that is correlated to bolt tension 8/21/ :12 PM 16

17 When high-strength bolts are to be tensioned, minimum limits are set on the bolt tension. See Table J3.1 in the LRFD Specification. Tension equal to 70% of the minimum tensile strength of the bolt Purpose of tensioning is to achieve the clamping force shown in the figure below from Segui 8/21/ :12 PM 17

18 In reference to the figure above, as the bolt is tensioned and elongates, the connected parts undergo compression. The total compressive force on the parts is equal to the bolt tension. If an external force P is applied, a friction force will develop between the connected parts, with a maximum value of F = µ N where µ is the coefficient of static friction between the connected parts (a function of surface condition of the steel parts) and N is the normal force shown in the figure above. Each bolt in the connection is then capable of resisting a force P = F before slippage. Prior to slippage, there is no bearing or shear. SHEAR STRENGTH OF HIGH-STRENGTH BOLTS The design shear strength of high strength bolts, like A307 bolts, is φ R n, where φ is Unlike A307 bolts, the shear strength of A325 and A490 bolts depends on whether the threads are in the shear plane (threads not excluded from the shear plane) or not (threads excluded from the shear plane). A325-N: threads included in the shear plane A325-X: threads excluded from the shear plane Rather than using a reduced cross-sectional area when the threads are not excluded from the shear plane, the ultimate shear stress is multiplied by a factor of 0.75 in the LRFD Specification. Results are presented in the last column of Table J3.2 of the Specification, which was reproduced above. In summary, 8/21/ :12 PM 18

19 Fastener Threads Rn = FvAb A325 Included in shear plane 48A b A325 Excluded from shear plane 60A b A490 Included in shear plane 60A b A490 Excluded from shear plane 75A b The calculation of the design strength of a connection is illustrated below. The bolts are 7/8 inch in diameter; threads are not in the shear plane, and the steel is Grade 50 steel. 8/21/ :12 PM 19

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21 SLIP-CRITICAL CONNECTIONS As noted earlier, a connection with high-strength bolts is classified as either a bearing or slip-critical connection. In bearing connections, the bolt is brought to a snug-tight condition so that the surfaces of the connected parts are in firm contact. Slippage is acceptable Shear and bearing on the connector In slip-critical connections, no slippage is permitted and the friction force described earlier must not be exceeded. Slippage is not acceptable Slip-critical strength based on factored loads herein. The LRFD Specification writes rules for basing the strength on service loads but such an approach is not discussed in CIE 428. Refer to the Specification for details. Proper installation and tensioning is key Connector not subjected to shear and bearing (in theory) 8/21/ :12 PM 21

22 Must have sufficient shear and bearing strength in the event of overload that causes slip. See Section J3.8 of the LRFD Specification for details. AISC writes that slip-critical bolted (or welded) connections be used in specific circumstances. Part of Section J1.11 of the LRFD Specification that lists these circumstances is reproduced below. As introduced earlier, the resistance to slip is a function of the product of the Normal force (equal to the bolt tension) Coefficient of static friction between the contact surfaces The LRFD Specification writes that the design slip resistance per bolt, φ rstr, shall equal or exceed the required force due to factored loads, where r = 1.13µ T N str b s 8/21/ :12 PM 22

23 where 1.13 is a factor that accounts for the expected 13% increase above the minimum specified preload provided by calibrated torque wrench tightening procedures, and µ = Mean slip coefficient of friction for Class A, B, or C surfaces T b = Minimum fastener tension (see Table J3.1 of the Specification) N s = Number of slip planes (described earlier) Guidance is provided for the mean slip coefficient (static coefficient of friction) associated with the three classes of surface noted above. For unpainted clean mill scale steel surfaces (Class A), µ = 0.33 Assumed surface for CIE 428 u.n.o. For unpainted blast-cleaned steel surfaces (Class B), µ = 0.50 For hot-dip galvanized and roughened steel surfaces (Class C), µ = 0.35 The resistance factor varies as a function of the type of hole and its orientation. The LRFD Specification writes that For standard holes, φ = 1.0 Assumed hole geometry for CIE 428 u.n.o. 8/21/ :12 PM 23

24 For oversized and short-slotted holes, φ = 0.85 For long-slotted holes transverse to the direction of load, φ = 0.70 For long-slotted holes parallel to the direction of load, φ = 0.60 The two examples from Segui below illustrate aspects of the checking or design of high-strength, slip-critical bolted connections. 8/21/ :12 PM 24

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26 The following example presents a design procedure for slip-critical connections. The C6x13 shown below is to resist a factored load of 108 kips. Bolts to be used are 7/8-inch diameter, A325; threads are in the shear plane and slip of the connection is permitted. Determine the number and layout of bolts so that h is a minimum. A36 steel is used. 8/21/ :12 PM 26

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30 Based on the above example, what are the steps in the design of a slip-critical connection? Below is an example from Segui of the design of a tension member and its connection. This example is an extension of the last example where the size of the tension member was given. In the example below, the tension member must also be sized, based in part on the connection details. Is this an appropriate consideration for design? 8/21/ :12 PM 30

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36 HIGH-STRENGTH BOLTS IN TENSION When a tensile force Q is applied to a non-tensioned (snug-tight) bolt, the resulting tensile force in the bolt is equal to the applied tensile force, Q. When a tensile force Q is applied to a tensioned bolt with pretension T 0, the resulting tensile force in the bolt is approximately equal to T 0, where the applied force relieves the compression (clamping forces) on the connected parts. An example of a high-strength bolt in tension is shown below (from Segui). Consider now the circled bolt in tension. 8/21/ :12 PM 36

37 Consider the bolt and the connected part shown in the figure below from Segui. Part a of the figure shows free-body diagrams of the parts after tensioning. Note that the bolt and the connected part are in equilibrium. Shown in part a of the figure is the bolt tension T 0 and the normal clamping force N 0 (equal in magnitude to T 0). 8/21/ :12 PM 37

38 When an external tensile force F is applied (to one bolt), the forces are those shown in part b of the figure above. Summing the forces in part c of the figure gives the total tensile force in the bolt as T = F + N The application of the force F will increase the bolt tension and cause it to elongate by an amount δ b. Compression in the flange of the structural tee (assumed connection detail) will be reduced, resulting in a distortion δ fl in the same sense as δ b. Assuming that the clamping force N is applied over a uniform area of flange A fl, and that the connected parts (the flanges) do not separate AE b bδ AflE b flδ fl A A b fl F = T N = = Eδb ( ) L L L L b fl b fl where L b is the length of the bolt (in tension) and L fl is the thickness of the flange. Note that these lengths are of the same order and that A fl is much greater than A b: N T and F N The ratio of T to N is approximately 0.05 to Most of the applied load relieves the compression on the connected parts. To estimate the magnitude of the applied load required to overcome the clamping effect (for the parts to separate), consider the figure below from Segui. 8/21/ :12 PM 38

39 For the connected parts to separate, the change in the compressive force must equal the initial pre-compression, namely, N = N0. When the parts have separated, F = T + T = T N = T + 0.1N = 1.1T That is, at the point of separation, the bolt tension is approximately 10% larger than the pretension. What happens once the parts have separated? Any increase in external load F will be resisted entirely by an increase in bolt tension. To avoid separation (and use the design strengths given in Table J3.2 of the LRFD Specification), high-strength bolts subject to direct tension must be pretensioned to the values given in Table J3.1, regardless of whether the connection is slip-critical or not. Calculate the tension force in the bolt assuming that there is no initial tension. Prying Action In connections in which the fasteners are subjected to tension forces, the flexibility of the connected parts can lead to deformations that increase the tensile force applied to the fastener. 8/21/ :12 PM 39

40 Force is known as prying force Shown in the figure below from Segui Before the external load is applied, the normal compressive force N is centered on the bolt. As the external load is applied and if 0 the flange is sufficiently flexible to deform as shown above, the compressive forces will migrate towards the edges of the flanges. This redistribution will change the relationship between all of the forces and the bolt tension will increase. If the flange is rigid, no redistribution will occur. How stiff is rigid in this case? The maximum prying force will be reached when only the corners of the flange remain in contact with the connected part. See the figure above for details. 8/21/ :12 PM 40

41 For connections of the type shown above, Section J3.6 of the LRFD Specification writes that the applied load shall be the sum of the factored loads and any tension resulting from prying action produced by deformation of the connected parts. A procedure for determining prying forces is given in Part 9 of the LRFD Specification. The presentation of Segui is similar and gives identical results. Consider the part of a structural tee shape shown below. The presentation is for one fastener only. The force T is the external factored tension force applied to one bolt, Q is the prying force applied to one bolt, and B is the total bolt force. The prying c force has shifted to the tip of the flange of the tee and is at its maximum value. 8/21/ :12 PM 41

42 From consideration of force and moment equilibrium, the following equations are derived: Tb M = Qa ; M = Qa ; B = T + Q a a b b c These equations can be combined into a single equation for the bolt force, including the effects of prying. First, a variable is defined as the ratio of the moment per unit length along the bolt line to the moment per unit length at the face of the stem M α = δ M b b a a where δ = Net area at bolt line divided by gross area at stem = 1 d / p d = Diameter of the bolt hole p = Length of flange tributary to one bolt M a a = Design strength at a-a = 2 pt F b f y φ ( /4) The total bolt force B can thus be calculated as c B c δα b = T[1 + ] (1 + δα ) a At this level of loading, the resultant of the tensile force in the bolt does not coincide with the axis of the bolt. For better agreement with test results, the force resultant is shifted toward the stem of the tee by d/2, where d is the bolt diameter, that is 8/21/ :12 PM 42

43 b = b 0.5 d; a = a+ 0.5d and B c δα b = T[1 + ] (1 + δα) a If the bolt force B c is set equal to the design tensile strength, denoted B, Two limit states are possible [( B/ T) 1]( a / b ) α = δ {1 [ B / T) 1]( a / b )} Tensile failure of the bolt Bending failure of the tee Plastic moments form at lines a-a and b-b producing a mechanism If the absolute value of α is less than 1.0, the moment at the bolt line per unit length is less than that at the face of the stem, indicating that the mechanism has not formed Tensile failure of the bolt controls, and B c = B If the absolute value of α is greater than or equal to 1.0, plastic hinges will have formed at both a-a and b-b Controlling limit state is bending failure of the tee 8/21/ :12 PM 43

44 Since the moments at a-a and b-b are limited to the plastic moment, α should be set equal to 1.0. The three equilibrium equations presented at the top of page 42 can be combined into a single equation to establish the required flange thickness, t, as follows: and for φ = 0.90, b f Tb M = M a a b b = δαm a a 2 f pt F y = δα( φ M ) = δα( φ ) b p b 4 t f = 4.444Tb pf (1 + δα) y The design of connections subjected to prying forces involves iteration. Always make an allowance for prying force in design Select a trial thickness and bolt geometry Calculate t and f B using the above equations c If actual t f is different from the required value, the actual values of α and B will be different from those c previously calculated 8/21/ :12 PM 44

45 If the actual bolt force, which includes the prying force, is desired, α must be recomputed. Using b instead of b, and setting M φ M =, then a a b p Tb α = ( 1) δ pt F The total bolt force can now be found using the boxed equation on page 43. What if the flange thickness is inadequate? 2 f Increase the flange thickness by using another tee shape Use more bolts to reduce T The example below from Segui illustrates most of the concepts introduced above. y 8/21/ :12 PM 45

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48 COMBINED SHEAR AND TENSION IN FASTENERS Most connections involving simultaneous shear and tension are eccentric connections, which will be discussed in the next module. One exception is the connection shown below, where, because the line of action passes through the center of gravity of the connection, each fastener can be assumed to resist an equal share of each component (V and T) of the load. An elliptical interaction curve is used for the case of combined loading in bearing-type connections, as shown in the figure below. The equation for this curve is given by the following equation where Tu 2 Vu 2 [ ] [ ] 1.0 ( φr ) + ( φr ) = n t n v 8/21/ :12 PM 48

49 T u = Factored tensile load on the bolt ( φ R n ) t = Design strength of the bolt in tension V u = Factored tensile load on the bolt ( φ R n ) v = Design strength of the bolt in shear For performance checking, the above equation is written as Tu 2 Vu 2 [ ] [ ] 1.0 ( φr ) + ( φr ) n t This equation can be re-packaged to permit calculation of the required bolt area in a connection as follows A b n v Tu V ( ) + ( ) φf φf t 2 u 2 The 3rd Ed of the LRFD presents the interaction equations in two forms. The interaction equations are listed in Table A-J3.1 of the Specification. The body of the specification uses a different format and parses the elliptical interaction curve into three straight-line segments as shown in the figure below from the Specification. v 8/21/ :12 PM 49

50 The Specification writes that in combined shear and tension in a bearing type connection, the design strength of the bolt is given by where φ FA t b F t = Nominal tension stress computed using the equations in Table J3.5 of the LRFD Specification (see below) as a function of the shear stress produced by the factored loads. Note that the design shear strength, φ Fv, must equal or exceed the shear stress f v. A b = Nominal unthreaded body area of the bolt φ = 0.75 For slip-critical connections in which the bolts are subjected to shear and tension, the effect of the applied tensile force is to reduce the clamping force, thereby reducing the allowable friction force. 8/21/ :12 PM 50

51 The LRFD Specification reduces the slip-critical shear strength in this case by multiplying the slip-critical shear strength ( φ r ) by where Tu [1 ] 1.13T N T u = Factored tensile load on the bolt T m = Prescribed initial bolt tension N b = Number of bolts in the connection The uses of some of the above design equations are illustrated below using an example from Segui. m b str 8/21/ :12 PM 51

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53 WELDED CONNECTIONS Structural welding is a process whereby the parts to be connected are heated and fused with a molten filler metal. The figure below from Segui illustrates two fillet-welded connections. 8/21/ :12 PM 53

54 Upon cooling, the structural steel (parent metal) and weld or filler metal will act as one continuous part. The filler metal is deposited from a special electrode. A number of welding processes are used, depending on the application Field welds Shop welds Shown below is a figure from Segui that illustrates shielded metal arc welding (SMAW): Current arcs across the gap between the electrode and the base metal Connected parts are heated and part of the filler metal is deposited into the molten base metal Coating on the electrode vaporizes and forms a protective gaseous shield, preventing the molten metal from oxidizing before it solidifies 8/21/ :12 PM 54

55 The electrode is moved across the joint and a weld bead is deposited Size of the weld bead depends on the rate of travel As the weld cools, impurities rise to the surface and form a coating called slag Slag must be removed before the next pass or the weld is painted Shielded metal arc welding is normally done manually and is widely used for field welding Self-shielded flux core Gas shielded flux core These and other processes are used for shop welding. Shop processes are automated or semi-automated in many cases. Other processes include Submerged arc welding: end of the electrode and the arc are submerged in a granular flux that melts and forms a gaseous shield. Electroslag As noted earlier, the two common types of welds are Fillet welds Welds placed in a corner formed by two parts in contact See figure on page 54 8/21/ :12 PM 55

56 Groove welds Welds deposited in a gap between two parts; see figure below Single bevel groove welds shown in the figure Backing bar One of both of the parts must be prepared Plug welds are shown in the figure below from Segui Circular or slotted hole that is filled with weld metal Used sometimes when more weld length is needed than is available 8/21/ :12 PM 56

57 Panel zone doubler plates of seismic moment frames Of the two key types of welds, fillet welds are the most common and are considered in the following section. Fillet welds are cheaper than groove welds. Why? The design of complete joint penetration (CJP) groove welds is generally straightforward. The filler metal is equally or overmatched to the parent metal (equal strength or greater) Connected parts considered continuous through the weld FILLET WELDS The design and analysis of fillet welds is based on the assumption that the geometry of the weld is a 45-degree right triangle as shown below in the figure from Segui. Shown in the above figure are the leg length (w) and the throat thickness (t). 8/21/ :12 PM 57

58 Standard weld sizes are expressed in sixteenths of an inch. Failure of fillet welds is assumed to occur in shear on the throat. That failure plane is also shown in the figure above. The strength of a fillet weld depends on the strength of the filler or electrode metal used. The strength of an electrode is given in terms of its tensile strength in ksi. Strengths of 60, 70, 80, 90, 100, 110, and 120 ksi are available. The standard notation for an electrode is E**XX where ** indicate the tensile strength in ksi and XX denotes the type of coating used. Usually XX is the focus of design E70XX is an electrode with a tensile strength of 70 ksi Electrodes should be chosen to match the base metal. Use E70XX electrodes for use with steels that have a yield stress less than 60 ksi Use E80XX electrodes that have a yield stress of 60 ksi or 65 ksi The critical shearing stress on a weld of length L is given by f v = P 0.707wL where all terms are defined above. If the ultimate shearing stress in the weld is termed F W, the nominal design strength of the weld can be written as 8/21/ :12 PM 58

59 φr = wl( φf ) = wl(0.75[0.6 F ]) = 0.32wLF n w EXX EXX For E70XX and E80XX electrodes, the design stresses are φ F, or W 31.5 ksi and 36 ksi, respectively. In addition, the factored load shear on the base metal shall not produce a stress in excess of φ F, where F BM BM is the nominal shear strength of the connected material. The factored load on the connection is thus subjected to the limit of φr = φf A = 0.90(0.6 F ) A = 0.54F A n BM g y g y g where A is the area subjected to shear. When the load is in the g same direction as the axis of the weld, such as that shown in the figure below from Segui, the base metal must also be examined using the above equation. Putting aside the eccentricity of the connection, the design strength of the welds (2 in total) is 2(0.707 wφ F ) and the shear capacity of W the bracket is tφ F. BM The following table from the LRFD Specification presents the design strength of welds. The example on the page following the table, from Segui, illustrates the design approaches presented above. 8/21/ :12 PM 59

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62 Practical detailing of welded connections requires a consideration of minimum weld size, maximum weld size, and length. The minimum size of a fillet weld is a function of the thickness of the thicker connected part. See Table J2.4 from the LRFD Specification, reproduced below, for details. The maximum size of a fillet weld is as follows Along the edge of a connected part less than ¼-inch thick, the maximum fillet weld size (w) equals the plate thickness For other values of plate thickness, t, the maximum weld size is t 1/16 in. The minimum permissible length of a fillet weld is 4 times its size. If only a shorter length is available, w= L/4. For the welds in the connection shown below, L W to address shear lag in such connections. 8/21/ :12 PM 62

63 When a weld extends to the corner of a member, it must be continued around the corner (an end return) as shown in the figure below from Segui. Prevent stress concentrations at the corner of the weld Minimum length of return is 2w Welds are shown on structural drawings by standard symbols. Some standard symbols are shown below in a figure from Segui. Symbols from the LRFD Specification are shown on the next page. Near side, far side, weld size, weld length 8/21/ :12 PM 63

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65 Below is an example from Segui that illustrates some of the concepts introduced above. 8/21/ :12 PM 65

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