Limit on Slenderness Ratio

Limit on Slenderness Ratio
According to AISC–D1, there is no maximum slenderness limit for design of members in tension. Even though stability is not a criterion in the design of tension members, it is still preferable to limit their length in order to prevent a member from becoming too flexible both during erection and final use of the structure. Two main factors controlling slenderness ratio in tensions members are: Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Steel Structures by Dr. Zahid Ahmad Siddiqi
Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Tension members that are too long may sag excessively due to their own weight. They may vibrate when subjected to wind forces as in an open truss or when supporting vibrating equipment such as fans or compressors. For members whose design is based on tensile force, the slenderness ratio L/r preferably should not exceed 300 where L is the actual and not the effective length. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

The above limitation does not apply to rods in tension where L/d may be kept up to 500 (corresponding to slenderness ratio less than or equal to 2000). However, this second value is not an AISC limit. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

The advisory upper limits on slenderness contained in AISC-D1 are based on professional judgment and practical considerations of economics, ease of handling and care required to minimized inadvertent damage during fabrication, transport and erection. Out-of-straightness within reasonable tolerances does not affect the strength of the tension member. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Design for Repeated Loading / Fatigue Strength
In the above mentioned reference, AISC-D1, AISC means that the Specification is given by American Institute of Steel Construction, D is the chapter number and 1 is the article number of that chapter. Repeated loading and unloading may result in failure at a stress level lesser than the yield stress. Design for Repeated Loading / Fatigue Strength Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

The term fatigue means reduction in material strength and hence failure under cyclic loading. The effect is more pronounced when repeating loads have tensile extreme value. The fatigue strength is mainly governed by three variables: The number of cycles of loading. The presence and initial size of any microscopic discontinuities/flaws within the metal structure. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

The range of variation of service load stress. This range is calculated by taking the difference between the maximum and minimum stress during the history of loading. If stress-reversal occurs, the range becomes equal to the sum of the maximum magnitude of tension and maximum magnitude of compression in a cycle and the condition becomes more critical. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

According to AISC, the stress range is defined as the magnitude of the change in stress due to application or removal of the service live load. The stress range will be equal to sum of magnitudes of two extreme stresses if these are acting in the opposite sense (like compressive and tensile). Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

In Appendix-3, AISC Specifications prescribe no fatigue effect for fewer than 20,000 cycles, which is approximately two applications a day for 25 years. Since, most loadings in buildings are in this category, fatigue is generally not considered. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Design of Tension Members
Following notations are used in the design: Tu = factored or ultimate tensile load Øt = resistance factor related with the tensile strength, 0.9 when failure occurs by yielding and 0.75 when failure occurs by fracture Tn = nominal strength of a tension member Øt Tn = expected strength to be used in design Design Equation Tu ≤ Øt Tn (LRFD) Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

The design strength Øt Tn according to AISC-D2 is the smaller of that based on: Yielding in the gross section (Yielding limit state) Tn=Fy Ag and Øt=0.90 (LRFD) For LRFD design, Tu = Øt Tn= Øt Fy Ag / 1000 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Fracture in the net section (Fracture Limit State) Tn=Fu Ae and Øt=0.75 (LRFD) For LRFD design, Tu = Øt Tn= Øt Fu U R Ag /1000 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Yielding in the net section is not a failure but yielding on the gross section is a failure. The reason is that the net section is limited in length and hence elongation due to yielding may not be excessive. However, gross area is present nearly all along the length and the elongation limit state may be exceeded. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Tearing Failure at Bolt Holes/Block Shear Failure Mode
In block shear failure, a part of the failure plane is transverse subjected to tension while the other part is longitudinal subjected to shear. In Figure 2.13 (a), ab part is subjected to shear and bc part is having tension. The failure plane abcd, as shown in Figure 2.13 (b) and (c), consists of a plane subjected to tension denoted by bc and two planes subjected to shear shown as ab and dc. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

The tearing out failure is either a fracture failure on both the tension resisting and shear resisting sections together or shear yielding combined with tension fracture failure. The nominal strength for block shear is the lesser of the following two cases because only that will cause the final separation of the block from the member. Rn = lesser of 1) 0.6Fu Anv + Ubs Fu Ant 2) 0.6Fy Agv + Ubs Fu Ant Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Nominal tension rupture strength = Ubs Fu Ant Nominal shear rupture strength = 0.6 Fu Anv Shear yielding strength = 0.6 Fy Agv 0.6Fy yield shear strength = τy 0.6Fu ultimate shear strength = τu Ø = 0.75 (LRFD) Agv = gross area subjected to shear Anv = net area in shear Ant = net area in tension Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Ubs = tensile rupture strength reduction factor (subscript ‘bs’ stands for block shear) = 1.0 when tensile stress is uniform, such as in all tensile members and gusset plates and single row beam end connections = 0.5 when tensile stress is not uniform such as for multiple row beam end connections Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Note: For welded connections, welded length provided longitudinally multiplied with the thickness of the connected leg becomes the area in shear. Area in tension is considered as the transverse area of the connected leg alone. The requirements for pin-connected members are separately given in AISC-D5. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi

Example 2.4: Using LRFD procedure, investigate the shear rupture failure mode for the angle L102 x 102 x 6.4 attached with three 20mm diameter rivets to a 10mm gusset plate, as shown in the figure The material is A36 steel. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 24

Solution: Capacity of section: Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 26

Block Shear Failure Along Path a-b-c: Agv = ( )(6.4) = 1216 mm2 Anv = (2.5)(20+3)(6.4) = 848 mm2 Ant = (0.5)(20+3)(6.4) = mm2 Ubs = 1 Ø Rn = lesser of 1) 0.6Fu Anv + Ubs Fu Ant = 0.75/1000 [0.6 x 400 x x 400 x 169.6] = kN Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 27

2) 0.6Fy Agv + Ubs Fu Ant = 0.75/1000 [0.6 x 250 x x 400 x 169.6] = kN Ø Rn = kN Ø Rn ≤ Øt Tn ( kN ≤ kN) Hence, block shear failure is the governing limit state and factored capacity of the member is reduced from kN to kN due to it. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 28

How can we increase the block shear failure strength?
Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 29

Strength of Threaded Parts A certain reduction in strength of bolts occurs due to threads, compared with solid rods, because of the reduction in net area of the rod. However, for convenience, this reduction is included in the strength and then full area of rod is considered to calculate the load carrying capacity. The maximum slenderness ratio of threaded rods should preferably be 2000 (corresponding to L/D ratio of 500). Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 30

Strength of Threaded Parts Nominal tensile strength, Rn = Fn Ab LRFD design tensile strength = ØRn, Ø=0.75 where Ab = nominal unthreaded body area of bolt gross area or area corresponding to outer diameter of bolt Fn = nominal tensile strength = 310 MPa for A307 bolts Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 31

= 620 MPa for A325M bolts = 780 MPa for A490M bolts The tensile strength of connecting elements is taken as Rn = Fu Ae, where for bolted splice plates, Ae = An ≤ 0.85Ag. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 32

Design Procedure/Design Flowchart Known Data Service or working loads, TD, TL, and TW, etc. and length of member, L Find factored tension (Tu) in LRFD method using load combinations. For example, Tu=1.2TD+1.6TL for gravity loads alone Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 33

Find Areq as the bigger out of that required for yielding in the gross section and fracture in the net section. where any reasonably assumed value of U may be considered and R is the assumed ratio of An with respect to Ag. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 34

Find Minimum Connected Leg Width (is always required for riveted connections, but also preferred for the welded case, bmin in any case should be greater than or equal to 50mm) where d = diameter of rivet which may be assumed to be 15mm if not known. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 35

Selection of Trial Section: It depends on the following four criteria: Asel ≥ Areq. Selection should be of minimum weight and smaller size. Connected leg width ≥ bmin. Compatibility of connections with other members is to be provided. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 36

Check Tensile Capacity: Find actual values of U and An if rivet pattern and diameter of rivets are known from connection design. (LRFD) If anyone of the above conditions is not satisfied, revise the section. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 37

Calculate rx, ry, and rz for built-up sections or directly note these values from tables for hot rolled sections. Find rmin = smallest of rx, ry, and rz Check Max. Preferable Slenderness Ratio: Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 38

Check Fatigue Strength: If loading cycles > 20,000 increase the section accordingly Design Lacing: Decide spacing of stay plates or arrangements and sizes of lacing in case of built-up sections. For finding spacing of stay plates, maximum slenderness ratio of individual elements may be equated to the max. allowed slenderness ratio that is 300. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 39

End Tie Plate Size: Minimum length = 2/3 s Minimum thickness = s / 50 where s is the distance between the lines of welds or fasteners on two components of built- up section. The longitudinal spacing of welds or fasteners at tie plates should not exceed 150mm. Design the Connections: Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 40

Check Block Shear Failure: The block shear strength must be checked at the connection, if the connection details are available. Write the final selection very clearly Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 41

Analysis of a Tension Member The term analysis means to calculate the maximum tensile load that may be applied on a particular member whose dimensions and sizes are already known. The same flow chart as given above may be used for the analysis if the given section is taken as trial section, leaving the initial steps to arrive at this trial section. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 42

Example 2.5: Calculate the factored load capacity of a double channel section member of A36 steel according to AISC LRFD Specification. The member is 5m long and consists of Cs 200 x 20.5, with flanges turned out and with clear gap of 100mm. Assume that there can be as many as two 15mm rivets at any one cross-section (one in each flange). U ≈ 0.80. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 43

Solution: Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 44

Solution: Loading cycles are assumed to be less than 20,000 and hence no reduction in strength due to fatigue is considered. The factored tensile capacity is kN, but the connecting width criterion is not satisfied. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 47

Example 2.6: Select a W-section to resist a dead tensile load of kN and a service tensile live load of kN using A36 steel and AISC LRFD Specification. The member is to be 9m long and is to be connected through its flanges only. Assume that there can be as many as four 20mm rivets at any one cross-section (two in each flange). Fasteners per line are at least three and bf of the W-section may be assumed to be lesser than 2/3 d for the initial calculation of shear lag factor. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 48

Solution: TD = 1020 kN ; TL = 680 kN ; L = 9m Tu = 1.2 TD TL = 2312 kN Areq= larger of Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 49

bmin = 3.25d + 26 = 3.25 (20) + 26 ≈ 91 mm (the web does not have bolts) Approximate minimum flange width required = 91 x 2 = 182 mm Options for selection of section: W200 x 86 A = 11,000 mm2 W310 x 86 A = 11,000 mm2 W410 x 85 A = 10,800 mm2 Weight is relatively lesser for this Section but the depth is excessively Large. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 50

Trial Section: W200 x 86 A = 11,000 mm2 bf = 209 mm tf = 20.6 mm rx = 92.7 mm ry = 53.3 mm tw = 13 mm Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 51

Capacity check: In the absence of the detailed connection detail, the AISC commentary and the table in Design Aids gives the efficiency factor as U = 0.9 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 52

rmin = smaller of rx and ry = 53.3 mm Loading cycles are assumed lesser than 20,000, if not given. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 53

Design connections: Block shear cannot be checked until the connection design is available. Final Selection: W200 x 86 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 54

General Considerations for Selection of Sections The type of connections used for the structure often affects the choice of member type. It is very difficult to apply bolts/rivets between some steel sections and the required gusset plates, while the same sections may easily be welded to the gusset plates. For example, plate-members are to be welded to other members in most the cases when the two plates are lying perpendicular to each other. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 55

The designer should select members such that connections to other members in the structure are easy. More parts of the section as for as possible are connected at the end to improve joint efficiency and to obtain a compact arrangement. Most commonly, W-sections have gusset plates on both sides of the section connected with the flanges. Filler plates are to be used if depths of the joining sections are different. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 56

Gusset plates present within the two angles connect double angles. Sometimes zero-force members are required for internal stability of frame, for minor loads like fans, false ceiling, etc., for future changes in loading and for temperature effects. These may also be used to reduce effective lengths of other members. Section is selected for these members keeping in view the following: Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 57

Preferably slenderness ratio equal to limiting maximum value for compression members is maintained, which is equal to 200. Using this criterion if the size becomes excessive, maximum slenderness ratio of tension members may be provided. However, if still the section is excessively bigger, a section comparable with other truss members may be used. 2. Connected legs should have a minimum width for proper connection. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 58

If the zero force member is a top or bottom chord member, continue the same section as present in the adjoining panel. Member under Stress Reversal The maximum factored tensile and compressive forces acting at different time instants due to different load combinations may be represented by the following notation: Tu = magnitude of ultimate tensile factored force. Pu = magnitude of factored compressive force. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 59

There are four possibilities of design based on the relative magnitude of Tu and Pu, as explained in the following cases: CASE 1 The tensile force is less than the compressive force and the connections are to be made using welding, then the tensile force may be neglected. Similarly, if the tensile force in case of riveted/bolted connections is less than 75% of the compressive force, it may be neglected. The member is designed as a pure compression member. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 60

CASE 2 The compressive force is less than 10% of the tensile force and the maximum slenderness ratio is 300, the compressive force may be neglected and the member is designed as a pure tension member. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 61

CASE 3 Tu > ( L2 ) Pu where L = length of member in meters. The member may be designed for a tension of Tu. However, during the capacity check, it is made sure that the compression capacity ØcPn is greater than or equal to Pu. It is better to keep the slenderness ratio up to 200 for these members. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 62

CASE 4 If the conditions of Case-1 through Case-3 are not satisfied, the section is to be designed for Pu as a compression member. It is checked later that ØtTn is greater than or equal to Tu. Note: The factored force may be replaced with the service force in case of allowable stress design. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 63

Example 2.7: Design member of a roof truss using LRFD procedure, carrying a factored compressive force of 450 kN and a factored tensile force of 840 kN; L = 6 m. Built-up section consisting of two channels back- to-back with a total width of 300 mm is to be used. Check the member under stress reversal. Welded connections are to be used. Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 64

Solution: Pu = 450 kN Tu = 840 kN ( L2 ) Pu = (1.54)(450) = 693 kN Tu > ( L2 ) Pu 840 kN > 693 kN ► Design first as a tension member and then check for Pu. (CASE-3) Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 65

For welded connections, bmin = 50 mm for welded connections Options Available: 1. C 150 x MC 310 x 15.8 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 66

Trial Section: 2Cs 150 x 15.6 A = 1990 mm2 d = 152 mm bf = 51 mm Ix = 633 x 104 mm4 Iy = 36.0 x 104 mm4 x = 12.7 mm rx = 56.4 mm ry = 13.4 mm c.g. x Fig Location of centroid for a channel section Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 67

Check for bmin Both ‘d’ and bf > bmin (OK) Capacity Check: Approximate rx & ry: (using Design Aids) rx =0.36 h=0.36(152)=55mm ry =0.60 b=0.60(300-2 x 51)=118.8mm 51 152 300 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 68

Exact rx & ry: (preferable and must for final trial) rx =56.4mm as for a single channel section Iy =2 x 36.0 x x 1990( )2 =5038 x 104 mm4 rmin = 56.4 mm 51 152 300 Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 69

Design of Lacing: Check for Compressive Strength: These parts will be completed after doing the next chapter. Loading cycles are assumed to be less than 20,000 Design Connections Nov-18 Steel Structures by Dr. Zahid Ahmad Siddiqi 70

Assignment: PROBLEMS of Chapter 02
Submission Time: After 14 Days

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