Structural Curriculum for Construction Management and Architecture Students 1 Prepared by: Ajay Shanker, Ph.D., P.E. Associate Professor Rinker School of Construction Management University of Florida 1.5. General Concepts of Beam Design

Design Parameters: ► Introduction to beam design Resources: ► AISC Steel Construction Manual Learning Concepts and Objectives: ► Understand the main steps for beam design Activities: ► Learning beam design through AISC Steel Construction Manual 2

3 M u ≤ ɸ b M n M u = the required strength M n = the nominal flexural strength (the flexural capacity of the beam) ɸ b = the resistance factor = 0.9 for flexure 1.5. General Concepts of Beam Design

4 Limit States to determine M n 1.Yielding: yielding is the upper limit for all shapes. 2.Local buckling: buckling of elements before they are able to reach yield. 3.Lateral-torsional buckling: a combination of lateral buckling and twist General Concepts of Beam Design

Yielding occurs when stresses exceed the yield stress of the member General Concepts of Beam Design 1. Yielding ☞ Figure Stress-strain diagram

General Concepts of Beam Design X X Section X- X d b Factored Uniform Load, w u Span, L 1. Yielding Figure (a). Simply supported beam with uniform loading. Figure (b). Section cut through beam.

Section X-X General Concepts of Beam Design FyFy Compressio n Tensio n b d/2 PNA 1. Yielding FyFy C = F y A C T = F y A T d/4 Figure Loading on section cut of beam.

General Concepts of Beam Design 1. Yielding Y i = The distance from the centroid of the Area, A i, to the plastic neutral axis. M n = F y Z x F y = Yield strength of the steel Z x = Plastic Section Modulus = the moment of the area about the PNA

9 ► How do we define the plastic neutral axis, PNA? ► Thus, the area below the PNA must equal the area above the PNA General Concepts of Beam Design Plastic Section Modulus, Z x ACAC b C = F y A C T = F y A T PNA ATAT Figure Balances forces on section cut of beam.

General Concepts of Beam Design A1 Y1 Y2 d/2 d/4 b Plastic Section Modulus, Z x PNA For a rectangle the PNA is in the middle. A2 Figure Calculating plastic section modulus.

11 PNA Z x =A 1 Y 1 + A 2 Y 2 =12x x0.5 =6 + 6 =12 in 3 Z x =A 1 Y 1 + A 2 Y 2 =12x1 + 12x1 = = 24 in General Concepts of Beam Design Plastic Section Modulus, Z x 1” 12 in A 1 =12 in 2 A 2 =12 in 2 Y 2 =1/2 in Y 1 =1/2 in 2” 6 in Y 2 =1” Y 1 =1” Section 1Section 2 Figure (a) and (b). Calculating plastic section modulus.

12 Z x =A 1 Y 1 + A 2 Y 2 =12x x1.5 = = 36 in 3 Z x =A 1 Y 1 + A 2 Y 2 =12x3 + 12x3 = =72 in General Concepts of Beam Design Plastic Section Modulus, Z x 4 in PNA 3 in A 1 =12 in 2 A 2 =12 in 2 Y 2 =1.5 in Y 1 =1.5 in Y 1 =3 in Y 2 =3 in 6 in Section 3 Section 4 Figure (a) and (b). Calculating plastic section modulus.

13 Z x =A 1 Y 1 + A 2 Y 2 + A 3 Y 3 + A 4 Y 4 =8x x2 + 4x2 + 8x4.5 = =88 in General Concepts of Beam Design Plastic Section Modulus, Z x 8 in PNA 10 in 1 in Y 1 =4.5 in Y 4 =4.5 in Y 2 =2 in Y 3 =2 in A1A1 A4A4 A2A2 A3A3 Total Area = A 1 + A 2 + A 3 + A 4 = = 24 in 2 Section 5 Figure Calculating plastic section modulus.

General Concepts of Beam Design Plastic Section Modulus, Z x PNA 10 in 1 in 1/2 in Y 1 =4.5 in Y 2 =2 in Y 3 =2 in Y 4 =4.5 in Z x =A 1 Y 1 + A 2 Y 2 + A 3 Y 3 + A 4 Y 4 =10x x2 + 2x2 + 10x4.5 = =98 in 3 Total Area = A 1 + A 2 + A 3 + A 4 = = 24 in 2 A1A1 A4A4 A2A2 A3A3 Section 6 Figure Calculating plastic section modulus.

General Concepts of Beam Design PNA 10 in 18 in 1 in 1/4 in Y 1 =8.5 in Y 2 =4 in Y 3 =4 in Y 4 =8.5 in Total Area = A 1 + A 2 + A 3 + A 4 = = 24 in 2 A 1 =10 in 2 A4A4 A2A2 A3A3 Z x = A 1 Y 1 + A 2 Y 2 + A 3 Y 3 + A 4 Y 4 = 10x x4 + 2x4 + 10x8.5 = = 186 in 3 Section 7 Figure Calculating plastic section modulus.

16 Z x = 186 in 3 Z x = 98 in 3 Z x = 88 in 3 Z x = 24 in 3 Z x = 12 in 3 Z x = 36 in 3 Z x = 72 in All seven shapes have same area of 24 in 2 but have increasing Z x and moment capacity. Shape 7 has (186/12 = 15.5 times) more moment capacity than Shape 1. Figure Comparing section modulus for different shapes of same area.

17 ShapeWt/ftZx, in 3 W8x W10x W12x W14x W16x W18x W21x W24x Plastic Section Modulus of Different W- Shape of Approximately Same Weight

General Concepts of Beam Design M M Compression, leads to possible local buckling Tension, no buckling A A M M Flange Local Buckling (FLB) A A M M Web Local Buckling (WLB) A A 2. Local Buckling ☞ Figure Local buckling of beam section under compression.

19 ► If a shape is capable of reaching the plastic moment without local buckling it is said to be a compact shape. ► Most standard shapes are compact. All current W, S, M, C and MC shapes except W21x48, W14x99, W14x90, W12x65, W10X12, W8x31, W8x10, W6x15, W6x9, W6x8.5, and M4x6 have compact flanges for F y = 50 ksi (345 MPa); all current ASTM A6 W, S, M, HP, C and MC shapes have compact webs at F y < 65 ksi (450 MPa) General Concepts of Beam Design 2. Local Buckling AISC Specification Chapter F2 User note states:

20 ► Yielding is the upper limit on strength ► However, lateral-torsional buckling, based on beam unbraced length, may control strength 1.5. General Concepts of Beam Design 3. Lateral Torsional Buckling ☞

21 ► The compression portion of the bending member tries to behave like a column but can’t ► Tension region resists buckling down ► Tension region also resists buckling laterally ► Thus, the shape twists as it buckles laterally 1.5. General Concepts of Beam Design 3. Lateral Torsional Buckling

22 ► To control lateral-torsional buckling the beam must be properly braced. ► Intermediate points along the span may be braced against translation and twisting. ► The distance between braced points is referred to as the unbraced length, L b General Concepts of Beam Design 3. Lateral Torsional Buckling ☞

General Concepts of Beam Design Beam Lateral Bracing Examples Cross beam acts as a lateral brace since it will prevent lateral displacement of the girder’s compression flange. Compression Flange, top Tension Flange, bot. Continuous concrete floor slab provides continuous bracing for the compression flange, L b =0, no LTB. Concrete Slab Lateral displacement of the bottom compression flange is prevented by the diagonal members (typically angles). Compression Flange, top Tension Flange, bot. Tension Flange, top Compression Flange, bot. Figure Beam Lateral Bracing Examples.

24 ► If L b ≤ L p, the limit state of yielding controls bending strength: ► If L b > L r the limit state of elastic lateral- torsional buckling controls. ► If L p < L b < L r the limit state of inelastic lateral-torsional buckling controls General Concepts of Beam Design 3. Lateral Torsional Buckling

25 ☞ 1.5. General Concepts of Beam Design Lateral bracing points from secondary beams Figure Partial floor framing plan

LATERAL SUPPORT OF STEEL BEAMS 47 kip-ft 28.5 kip-ft CASE II W10 X 12 CASE III CASE I CASE II CASE III CASE I

CASE I if lateral brace is spaced ’ CASE II if lateral brace is spaced 2.75’-8’ CASE III if lateral brace is spaced more than 8’ W10X12 CASE I Design Moment Strength = = 47 kip-ft BRACING DISTANCE BRACES W10X12

CASE II Design Moment Strength reduces as increases AT kip-ft Linear variation in & W10X12 LATERAL BRACES

AT Design Moment Strength reduces as increases Should be avoided for load bearing floor beams. kip-ft CASE III

Top flange embedded in concrete ADDITIONAL CASE I CONSTRUCTION DETAILS x x Y Y X-X Y-Y CONCRETE W-SHAPE STEEL STUDS

Use 50 ksi and select a shape for a typical floor beam AB. Assume that the floor slab provides continuous lateral support. The maximum permissible live load deflection is L/180. The service dead loads consist of a 5-inch-thick reinforced- concrete floor slab (normal weight concrete), a partition load of 20 psf, and 10 psf to account for a suspended ceiling and mechanical equipment. The service live load is 60 psf.

F y = 50ksi Case 1 LIVE LOAD = 60psf DEAD LOADS 1) 5” Slab 2) Partition = 20psf 3) Ceiling, HVAC = 10psf GIRDER (PRIMARY BEAMS) not to exceed Figure 1

DESIGN LOAD (kips/ft) on AB = w u x TRIBUTORY AREA LENGTH OF BEAM x = kips/ft 12” of Slab = 150 psf 6” of Slab = 75 psf 1” of Slab = 12.5 psf 5” of Slab = 62.5 psf (12.5 psf for every inch of concrete thickness) *5 x 12.5 = 62.5

w u = kips/ft MuMu = == ft-kips CASE 1

139.7 STRENGTH OF W14 X 26 = ft-kips STRENGTH OF W16 X 26 = ft-kips Page # 3-127

Page # 1-22

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Page # 1-20

SELECT W16 X 26 SHAPEAREA W 14 x W16 x Page # 1-21

EXTRA SELF WEIGHT MOMENT = = 3.3 ft-kips MOMENT STRENGTH APPLIED MOMENT W16 X 26 is OK == = 2” = Page # 3-211

w = = 360 lb/ft = kips/ft = 0.03 kips/in = = 0.75” 2” OK = ; = in = kips/in

Typical Copes for a shear connection of a large girder to column web. Note that duct holes have to be strengthened by plates. Also, holes are at third point where shear & moment are not maximum.

Cantilever construction for projected balcony. If shear studs are noticed on beams and column then those members have to be encased in concrete for increasing fire resistance of steel.

Details of web opening in steel girders for HVAC ducts.