A Simply supported beam with a concentrated load at mid-span: Loading Stages

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1 A Simply supported beam with a concentrated load at mid-span: Loading Stages P L/2 L PL/4 MOMNT F b < 1 lastic F b = 2 lastic F b = 3 lastoplastic 4 F b = Plastic hinge Plastic Dr. M.. Haque, P.. (LRFD: Bending Members) Page 1 of 12

2 1 Bending Stress and Stain: LASTIC Strain < Yield Strain Є < Є y NA Stress < Yield Stress f b < f b = (M c) / I = M/S y f y f y = (M y) / I Dr. M.. Haque, P.. (LRFD: Bending Members) Page 2 of 12

3 2 Bending Stress and Stain: YILD Strain = Yield Strain Є = Є y NA Stress = Yield Stress f b = f b = (M c) / I = M/S y f y f y = (M y) / I Dr. M.. Haque, P.. (LRFD: Bending Members) Page 3 of 12

4 3 Bending Stress and Stain: lasto-plastic (Partially Plastic) Strain > Yield Strain Є > Є y NA Stress = Yield Stress f b = Dr. M.. Haque, P.. (LRFD: Bending Members) Page 4 of 12

5 4 Bending Stress and Stain: Plastic (Fully Plastic) Strain > Yield Strain Є > Є y NA Stress = Yield Stress f b = Plastic N.A. Plastic Moment: M p = Z Where Z = Plastic Section Modulus Dr. M.. Haque, P.. (LRFD: Bending Members) Page 5 of 12

6 4 F b = C=A c a Plastic hinge T=A t Plastic N.A. Plastic Stage From equilibrium of forces: C = T A c = A t A c = A t Plastic Moment, M p = (A c ) a = (A t ) a = (A/2) a = Z Where A = Total Area; A c = Compression Area; A t = Tension Area a = distance between the centroids of the two half-areas Z = (A/2) a = Plastic section modulus XAMPL 1: Compute the plastic moment, M p for a W16 X 57 of A992 Steel. Solution: From AISC Table, for W16 X 57, A= 16.8 in 2 ; Z x = 105 in 3 For A992 Steel, = 50 ksi M p = Z = (50)(105) = 5250 k-in = 5250/12=437.5 k-ft. Dr. M.. Haque, P.. (LRFD: Bending Members) Page 6 of 12

7 AISC Classification of Shapes as (1) Compact (2) Non-compact, and (3) Slender depending on width to thickness ratios. (See AISC Table B 4.1) Flange Criterion: If the flange is continuously connected to the web, and b f 2 t f the shape is COMPACT 0.38 The shape is NONCOMPACT 0.38 b f t f The shape is SLNDR b f 2 t f 1.0 Dr. M.. Haque, P.. (LRFD: Bending Members) Page 7 of 12

8 Web Criterion: The shape is COMPACT h t w 3.76 The shape is NONCOMPACT 3.76 h 5.70 t w The shape is SLNDR h t w 5.70 NOTS: The web criterion is met by all standard I- and C- shapes listed in the AISC Manual, so only flange criterion needs to be checked. Dr. M.. Haque, P.. (LRFD: Bending Members) Page 8 of 12

9 XAMPL 2: Check compactness for W16 X 57 A992 Steel. A 992 Steel, = 50 ksi AISC Table W16 X 57 b f /(2t f ) = 4.98; 0.38 (29000/50) = 9.15 > 4.98, Hence, the flange is COMPACT h/t w = d/t w = 33.0; 3.76 (29000/50) = > 33.0, Hence, considering web, section is COMPACT [NOT for all shapes in the AISC Manual, section is compact considering web. Therefore, W16 X 57 is COMPACT. Dr. M.. Haque, P.. (LRFD: Bending Members) Page 9 of 12

10 LRFD: Bending Member M u φ M n Where Φ = Resistance factor = 0.90 (LRFD) M u = Factored Moment φ M n = Design Moment Strength LATRALLY SUPPORTD COMPACT BAMS Where M n = M p M p = Z <= 1.5 M y The limit of 1.5 M y is to prevent excessive working load deformations, and is satisfied when Z <= 1.5 S or Z/S <=1.5 NOT: For C-, I and H shapes bent about the strong axis (x-x), Z/S will always be <= 1.5. For I and H shapes bent about the weak axis (y-y), Z/S will never be <= 1.5. Dr. M.. Haque, P.. (LRFD: Bending Members) Page 10 of 12

11 XAMPL 3: The beam shown in figure is a W16 X 57 of A992 steel. It supports a reinforced concrete slab that provides a continuous lateral support of the compression flange. Is the beam adequate? Service DL= 450 lb/ft LL = 550 lb/ft 30 ft SOLUTION: Wu = 1.2( ) + 1.6(0.55) = k/ft Mu = wu L 2 / (30) 2 / 8 = k-ft. Check compactness for W16 X 57 A992 Steel. A 992 Steel, Fy = 50 ksi AISC Table W16 X 57 bf /(2tf) = 4.98; 0.38 SQRT( 29000/50) = 9.15 > 4.98 Hence, the flange is COMPACT h/tw = d/tw = 33.0; 3.76 SQRT(29000/50) = > 33.0, Hence, considering web, section is COMPACT [NOT for all shapes in the AISC Manual, section is compact considering web. Therefore, W16 X 57 is COMPACT. Because the beam is compact and laterally supported, Mn = Mp = FyZx = 50(105) = 5250 k-in/12 = k-ft. Check for Mp <= My Zx /Sx = 105/92.2 = < 1.5 OK Φ Mn = 0.9 (437.5 ) = k-ft > k-ft OK The design moment is greater than the factored load moment. So W16 X 57 IS ADQUAT Dr. M.. Haque, P.. (LRFD: Bending Members) Page 11 of 12

12 XAMPL 4: Select the least-weight wide-flange A992 steel beam section. It supports a reinforced concrete slab that provides a continuous lateral support of the compression flange. Service DL= 450 lb/ft LL = 550 lb/ft SOLUTION: Since, at this point beam W section s weight is unknown, calculate Mu without the beam self weight Wu = 1.2(0.45) + 1.6(0.55) = 1.42 k/ft Mu = wu L 2 /8 = 1.42 (30) 2 / 8 = k-ft. Mn = Mp = Fy Z φmn = φ Fy Z Considering Mu Z req = Mu /( φ Fy) = (k-ft) X12/ (0.9 x 50) = 42.6 in 3 From AISC Table, W 12 X 30, Zx= 43.1 Check: φmn = φ Fy Z = 0.9(50)(43.1) = k-in = k-ft Wu = (0.03)= k/ft Mu = (30) 2 /8 = k-ft > NG Select W12 X 35, Zx=51.2 Check: φmn = φ Fy Z = 0.9(50)(51.2) = 2304 k-in = 192 k-ft Wu = (0.035)= k/ft Mu = (30) 2 /8 = k-ft < 192 OK ANSWR: W 12 X ft Dr. M.. Haque, P.. (LRFD: Bending Members) Page 12 of 12

AISC LRFD Beam Design in the RAM Structural System

AISC LRFD Beam Design in the RAM Structural System Model: Verification11_3 Typical Floor Beam #10 W21x44 (10,3,10) AISC 360-05 LRFD Beam Design in the RAM Structural System Floor Loads: Slab Self-weight: Concrete above flute + concrete in flute + metal