AIA St. Louis January 4 th, 2017 Phillip Shinn, PE Jacobs Engineering Washington University in St. Louis, Sam Fox School of Design and Visual Arts

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1 AIA St. Louis January 4 th, 2017 Phillip Shinn, PE Jacobs Engineering Washington University in St. Louis, Sam Fox School of Design and Visual Arts Owners are NOT Architects (but they ARE the Owners) Steel Beams: Using C b Values Upcoming AISC Steel Code Change

2 Structures I - Fall 2016 At Washington University: 35 of 52 Architecture Students are young women: 67%

3 Owners are NOT Architects But they ARE the Owners

4 Shirk Center Illinois Wesleyan University Bloomington, Illinois Hastings & Chivetta, Architects Theiss Engineers, Structural

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9 Heman Park Recreation Center Addition University City, Missouri Jacobs Engineering Architects and Structural

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16 Kemper Art Museum Washington University in St. Louis Maki & Associates, Architects Jacobs Engineering, Structural

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20 Steel Beam Using C b values

21 Using C b Values The Architect asks the structural engineer to design a roof beam to be as small and as unobtrusive as possible.

22 Using C b Values Roof beam: DL=20 psf, LL=20 psf span = 45 feet spacing = 14 feet assume 80 plf for beam weight For deflection (ASD loads) SIMPLE SPAN w(beam) DL+LL = 14(20+20) + 80 = 640 plf w (beam) LL = 14(20) = 280 plf I x (req d) for L/240 = 905 in 4 I x (req d) for L/360 = 594 in 4 For deflection, Use W21x48, I x =969 in 4 ; Z x =107 in 3 For Strength LRFD SIMPLE SPAN w u (beam) = 1.2(14(20)+80)+1.6(14(20)) = 880 plf M u = w u L 2 /8 = 222,750 ft-lbs Use Fy=50 ksi Z x (req d) = 59.4 in 3 No Good For Strength LRFD, choose W18x35, Z x =66.5 in 3 ; I x =513 in 4

23 Using C b Values Deflection controls. To reduce deflection, use fixed ends at the beams supports. This reduces the deflection to 1/5 th of its simple span value. Now, I x (req d) for L/240 = 181 in 4 I x (req d) for L/360 = 119 in 4 For deflection, choose W14x22, I x =199 in 4 ; Z x = 33.2 in 3 This reduces the moment to 2/3 s of its simple span value. Now, M u = w u L 2 /12 = 148,500 ft-lbs Z x (req d) = 39.6 in 3 For Strength LRFD, choose W14x26, Z x = 40.2 in 3 ; I x =246 in 4 Use W14x26?? THIS ASSUMES THE BEAM IS FULLY BRACED WRONG! (?)

24 Using C b Values Try to use a W14x26 with Fixed End Moment connections. Need to be concerned about STRENGTH (for beams that s shear and moment strength) STABILITY (is the beam braced?) SERVICEABILITY (for beams that s deflection) Moment strength (beam braced) and deflection have been accounted for. What about Shear and Stability? And what about (gasp!) Cost?

25 Using C b Values What about Cost? Well, put your money where your mouth, or your design, is. What about Shear? Shear rarely controls steel beam design only on short span, heavily loaded beams. What about Stability? Aye, there s the rub! Stability. Right now we have a W14x26 beam that s been designed as though it was completely braced. Is it? Is it braced, that is?

26 Using C b Values The top flange is braced by the metal deck welded to the top of the beam. More importantly, the compression flange is braced, if the compression is, indeed, at the top. What if the compression flange is at the bottom? Remember, compression is our enemy we need to brace the beam so the compression flange doesn t buckle. More about that later.

27 Using C b Values 0 0 Our FIRST design, W21x48, a simple span beam, allowed rotation at the ends, so the end moments were zero. In this case the moment is Positive along the entire length of the beam. Compression is on the side of the graph the TOP, in this case. A positive moment has compression on top of the beam. Compression is our enemy brace the top flange, where the compression is.

28 Using C b Values Deck at the top braces the beam. 0 0 Inflection Point (M=0) Nothing braces the bottom of the beam. Nothing braces the bottom of the beam. Our second design, W14x26 (BRACED), a fixed-end beam, had the end moments as negative, with compression on the bottom. In this case the compression is on the bottom at the end portions of the beam, and compression is at the top for the center portion of the beam. A positive moment has compression on top of the beam, and a negative moment has compression at the bottom of the beam.

29 Using C b Values So, we should (or could) brace the bottom flange of the beam: Well, maybe not.

30 Using C b Values So, The beam is unbraced what happens? Beams (members in bending or flexure) will reach their maximum capacity in one of three ways: 1) Unbraced beams: Failure by Elastic Lateral-Torsional Buckling 2) Partially braced: Failure by Inelastic Lateral-Torsional Buckling 3) Fully braced beams: Failure by complete inelastic material yielding.

31 Using C b Values 1) Unbraced beams: Failure by Elastic Lateral-Torsional Buckling Elastic: will return to its previous shape/orientation when load is removed looks fine. Buckles before any part of the section yields Fails at low stress we don t get all we paid for Lateral: compression flange moves sideways. Torsional: entire section twists. This is a stability failure, not a strength failure.

32 Using C b Values 2) Partially braced: Failure by Inelastic Lateral-Torsional Buckling Inelastic: will NOT return to previous shape/orientation when load is removed it s BENT Buckles after some part of the section yields Fails at a yield stress occurs, but only a part of the section has yielded. We STILL don t get all we paid for. Lateral: compression flange moves sideways. Torsional: entire section twists. This is mostly a stability failure, but strength is involved.

33 Using C b Values 3) Fully braced beams: Failure by complete inelastic material yielding. Complete: the entire section has yielded. Completely Inelastic: REALLY BENT. Develops the maximum capacity of the section. We get our money s worth. Best bag for the buck.

34 Using C b Values What does it mean to get our money s worth with braced beams? We can use a smaller, lighter, and more economical beam. That alone has value it helps us stay within budget, and saves money for non-structural uses. These values apply to non-aesthetic uses primarily non-exposed structures. And, At the same time, we have a greater choice of sections to use, and we can either use shallower or more slender shapes, which may be enhance the aesthetic value for exposed structures. So we want to brace our beams. Don t we?

35 Using C b Values What constitutes BRACING? WHAT do we brace? And how do we do it? With beams, we need to stop the twist : Assuming vertical loads (gravity): Grab the top & bottom of the beam, & stop the beam from twisting.

36 Using C b Values What constitutes BRACING? WHAT do we brace? And how do we do it? With beams, we need to stop the twist : Assuming vertical loads (gravity): Grab the top & bottom of the beam, & stop the beam from twisting. { moving sideways is OK just don t twist}

37 Using C b Values What constitutes BRACING? Bracing has to have a certain STRONG: +/- 1% of compression force. Bracing has to have a certain STIFFNESS often overlooked. Long braces tend to NOT have stiffness, especially in compression. Try to keep braces short and fat. To get more stiffness, keep braces as short as possible (L), and then large enough (A) to maintain their stiffness. Look at L(length) and A(area) D=movement=(NON-stiffness) D= PL AE

38 Using C b Values What constitutes BRACING? Bracing can be torsional composite steel beams with a concrete slab can have the BOTTOM FLANGE braced torsionally by the concrete slab at the TOP of the beam.

39 Using C b Values With metal deck AND with compression on the BOTTOM, the bottom moves sideways and causes the beam to twist. Lateral-Torsional Buckling will occur there s very little to stop the twist. There s nothing to stop the twist.

40 Using C b Values With metal deck AND with compression on the BOTTOM, the bottom moves sideways and causes the beam to twist. Lateral-Torsional Buckling will occur there s very little to stop the twist a stiffener plate will NOT brace the beam by itself. There s nothing to stop the twist.

41 C b uses and values It looks like our beam is UNBRACED, and we will have to design in that way. Here are the equations we need to go through:

42 C b uses and values And these:

43 C b uses and values C b ONLY applies when the beam is UNBRACED. There is NO C b for braced beams.

44 C b uses and values

45 C b uses and values Forget the formulas (EXCEPT, perhaps for C b ) All we need is the moment to be resisted, and the length between braced points on the beam huh? I thought we were unbraced! First of all, our LRFD maximum moment, M u = w u L 2 /12 = ft-kips The beam is braced at each bearing end, at each connection we have to be sure of that. So, the distance between braced points would be from one end of the beam to the other, which equals the span, so the distance between braces would be 45 feet.

46 C b uses and values Forget the formulas (EXCEPT, perhaps for C b ) Assuming the beam is Braced at 22-6 at midspan. This does NOT check For deflection. Only for moment strength and stability.

47 C b uses and values Forget the formulas (EXCEPT, perhaps for C b ) Assuming the beam is Braced at 22-6 at midspan. This does NOT check For deflection. Only for moment strength and stability. W10x49 OK Mcapacity = 176 ft-k OK HERE w/ fixity, We needed I x = 181 in 4

48 Using C b Values Deck at the top braces the beam feet 26 feet 0 0 Inflection Point (M=0) Nothing braces the bottom of the beam. Nothing braces the bottom of the beam. Is the inflection point a braced point? Is the beam braced where the inflection point occurs? Can the beam twist at that location? NO, NO, NO, NO, NO. The inflection point is NOT a braced point. Our unbraced length is 45 feet.

49 C b uses and values We need a moment capacity (LRFD, ΦMn = ft-kips) and an I x (req d) for L/240 = 181 in 4 USING C b = 1.0 W10x68 ΦMn = ft-kips Ix = 394 in 4 OK W12x65 ΦMn = ft-kips Ix = 533 in 4 OK W14x68 ΦMn = ft-kips Ix = 723 in 4 OK W16x67 ΦMn = ft-kips Ix = 954 in 4 OK W18x76 ΦMn = ft-kips Ix = 1,330 in 4 OK W21x83 ΦMn = ft-kips Ix = 1,830 in 4 OK

50 C b uses and values What C b should we use?

51 C b uses and values What C b should we use? For our beam, M max = 148.5; M A =M=18.56; M B =74.25 ft-kips C b = < 3.0 maximum (in the Commentary) Using the absolute values of the moments

52 C b uses and values What C b should we use? Check the Commentary:

53 C b uses and values What C b should we use? For top of the beam loading, multiply by C b = x (0.714 for top loading) = 1.70 Using the absolute values of the moments

54 C b uses and values We need a moment capacity (LRFD, ΦMn = ft-kips) and an I x (req d) for L/240 = 181 in 4 USING C b = x 2.38 = 1.70 W10x39 ΦMn = ft-kips Ix = 209 in 4 OK W12x40 ΦMn = ft-kips Ix = 310 in 4 OK W14x43 ΦMn = ft-kips Ix = 428 in 4 OK W16x50 ΦMn = ft-kips Ix = 659 in 4 OK W18x50 ΦMn = ft-kips Ix = 800 in 4 OK W21x57 ΦMn = ft-kips Ix = 1,170 in 4 OK

55 C b uses and values What C b should we use? For our beam, M 1 = 148.5; M 0 =148.5; M CL =74.25 ft-kips C b = 3.0 This requires that the top flange be braced!! Which it is. Does this make sense?

56 C b uses and values We need a moment capacity (LRFD, ΦMn = ft-kips) and an I x (req d) for L/240 = 181 in 4 USING C b = 3.0 W10x39 ΦMn = ft-kips Ix = 209 in 4 OK W12x30 ΦMn = ft-kips Ix = 238 in 4 OK W14x30 ΦMn = ft-kips Ix = 291 in 4 OK W16x31 ΦMn = ft-kips Ix = 375 in 4 OK W18x35 ΦMn = ft-kips Ix = 510 in 4 OK W21x44 ΦMn = ft-kips Ix = 843 in 4 OK

57 C b uses and values Beginning design: Next design: Next design: W21x48 Simple Span W14x26 Fixed Ends, assumed fully braced W10x49 Fixed Ends, braced at mid-span Next design: W12x65 Fixed Ends, unbraced, C b =1.00 Next design: W10x39 Fixed Ends, unbraced, C b =1.70 Next design: W12x30 Fixed Ends, unbraced, C b =3.00 W14x30 Fixed Ends, unbraced, C b =3.00

58 C b uses and values Column/Beam analogy: Why does a beam want to buckle? And what is the C b doing? These are columns, all carrying 100 kips at the base. Which one will want to buckle first? The first column, with a constant axial load, is the worst case.

59 C b uses and values Column/Beam analogy: What is the worst case for a beam? It is a constant compression force throughout the length of the beam. We accomplish this by placing concentrated moments at each end of the beam as shown. THIS IS = C b 1.0

60 C b uses and values Our actual Fixed-End Beam Forces. This is NOT C b = 1.0

61 C b uses and values With uplift (wind, probably) on a beam braced on top, all compression is on the bottom. The rods will buckle, and the top chord member is on its own! Use a C b = 2.00

62 C b uses and values

63 C b uses and values

64 C b uses and values

65 Using C b Values Roof beam: Check the W14x26 beam with WIND UPLIFT Uplift Wind = LL=10 psf net uplift span = 45 feet spacing = 14 feet For Strength LRFD SIMPLE SPAN (Conservative) w u (beam) = 1.6(14(10)) = 224 plf M u = w u L 2 /8 = 56,700 ft-lbs Use Fy=50 ksi Unbraced beam design, L u =45 feet, C b =2.0 ΦM n = 58,940 ft-pounds OK

66 AESS Architecturally Exposed Structural Steel

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73 AESS Architecturally Exposed Structural Steel The new (2016?) AISC Specification (the Steel Code) Will require that the connection design is NOT to be left up to the contractor. The SEOR shall design all AESS connections. The Architect and the engineer will need to determine what type of connections are appropriate and desirable. double angle / single plate / moment / welded / bolted / welded and bolted / the sky s the limit BUT the control of the connections appearance needs to be in the hands of the designers.

74 C b uses and values QUESTIONS?