hospital building Eisenhower Medical Center (EMC),. Rancho Mirage, CA, USA, designed and analyzed for. Floor Vibrations due to Human Activity (8.7.2009 to.
PRACTICAL TRAINING REPORT
Presented by
K.Senthil Kumar, 90108512013 M.E.,Structural Engg, ACCET, Karaikudi.
Firm Profile • Francis Santiago and Associates is a partnership firm started in Chennai, India in April 2007. • The founding members of the firm Francis Santiago, PE (CA) - The Managing partner. Dr.M.L.Santiago, PhD - The Partner in the firm. • Francis Santiago and Associates provides Structural consultancy services. Their Clients are based in California. Now, planning to expand their market in India and Middle East.
Task assigned during Training My training comprised of two parts, 1. Floor Vibration due to Human Activity - A new hospital building Eisenhower Medical Center (EMC), Rancho Mirage, CA, USA, designed and analyzed for Floor Vibrations due to Human Activity (8.7.2009 to 23.7.2009). 2. Entry Canopy Design for a Hospital - A Cantilevered Steel Framed Entry Canopy designed for a Hospital named as Sequoia Hospital, CA, USA as per Indian Code (24.7.2009 to 7.8.2009).
1. Floor Vibration due to Human Activity • Basic Concept of Floor vibrations due to human activity was studied. • American Institute of Steel Construction (AISC) Steel Design Guide Series 11 Floor Vibration due to Human activity was studied. • AISC design guide 11 was used as the reference material to analyze the hospital.
Eisenhower Medical Center (EMC), Rancho Mirage, CA, USA. • A new hospital building Eisenhower Medical Center (EMC), Rancho Mirage, CA, USA, a five story, 200000 Sq ft Hospital. • Steel Special Moment Frames and Composite steel framing are used as the main structural system.
• In EMC hospital, a bay of a floor was taken and solved for floor vibration due to walking excitation by using Design guide 11.
• I learnt software RAM STEEL. • This software was used for floor vibration analysis due to human activity. • EMC Hospital was modeled using RAM STEEL. • A bay of a floor was taken and analyzed.
• Hand calculation and RAM STEEL results were compared.
Eisenhower Medical Center (EMC)
Design for Walking Excitation Bay: 1st floor, B-C, 6-5 Beam Properties: W 18 x 40 A = 11.8 in2 Ixx = 612 in4 d = 17.9 in
Girder Properties: W 24 x 68 A = 20.1 in2 Ixx = 1830 in4 d = 23.7 in
Deck Properties: Concrete Wc = 115 pcf fc = 3 ksi Total depth = 7 in Deck height = 3 in Load: Dead load = 4 psf Live load = 11 psf Slab + Deck weight = 55 psf
Beam Mode Properties : With an effective concrete slab width of 120 in = 10 ft < 0.4 Lj = 0.4 x 30 = 12 ft. Considering only the concrete above the steel form deck, and using a dynamic modulus of elasticity of 1.35 Ec, the transformed moment of inertia is: Ec = W1.5 x √fc = (115)1.5 x √3 = 2133.505 ksi. n = modular ratio = Es/ 1.35Ec = 29000/(1.35 x 2133.505) = 10.06
y = [11.8 (3+18/12) – (120/10.06) x 4 x (4/2)] / [11.8 + (12/10.06) x 4] = 2.786 in below from top of form deck
Ij = [612+11.8 x (3+18/2 – 2.786)2 + (120/10.06) x (43/12) + (120/10.06) x4x (2.786+4/2)2 ] = 2770.32 in4 For each beam, the uniform distributed loading is, wj = 10 ( 11+55+4+40/10) = 740 plf
The corresponding deflection is, Δj = 5 wjLj4 /384 Es x Ij = (5 x 740x 304x1728) / (384 x 29x106 x 2770.32) = 0.167 in The beam mode fundamental frequency is, fj = 0.18√(g/Δj) = 0.18 x √(386/0.167) = 8.65 Hz
Using an average concrete thickness of 4+3/2 = 5.5 in , the transformed moment of inertia per unit width in the slab direction is, Ds = 12 de3 / 12 n = 12 x 5.53 / 12 x 10.06 = 16.53 in4/ft
The transformed moment of inertia per unit width in the beam direction is (beam spacing is 10 ft) Dj = Ij / S = 2770.32 / 10 = 277.032 in4 / ft
The Effective beam panel width with Cj = 2 (interior panel) Bj = Cj (Ds/Dj) ¼ Lj = 2 x (16.53/277.032) ¼ x 30 = 29.65 ft. which must be less than 2/3 times the floor width. The actual floor width is at least 3 times the girder span, 3x 30 = 90 ft. 2/3 x 90 = 60 ft > 29.65 ft Therefore, the effective beam panel width is 29.65 ft
The weight of the beam panel is calculated from equation, factor of 1.5 to account for continuity Wj = 1.5 x (wj/S) x Bj x Lj =1.5 x (740/10) x 29.65 x 30 = 98.73 kip.
Girder Mode Properties: With an effective concrete slab width of 0.4 Lg = 0.4 x 30 x 12 = 144 in < Lj Considering the concrete in the deck in ribs, the tansformed moment of inertia is: Average concrete depth = 4 + 3/2 = 5.5 in. y = [20.1 (1.5+23.7/2) – (144/10.06) x 5.5 x (5.5/2)] / [20.1 + (144/10.06) x 5.5] = 0.524 in below effective slab
Ig = 1830+20.1x(1.5+23.7/2 - 0.524)2 +(144/10.06)x(5.53/12)+(144/10.06)x5.5x0.524+5.5/2)2 = 6178.33 in4 For each girder, the uniform distributed loading is, wg = Lj(wj/S) + girder weight per unit length = 30(740/10) + 68 =2288 plf The corresponding deflection is, Δg = 5 wgLg4 /384 Es x Ig = (5 x 2288x 304x1728) / (384 x 29x106 x 6178.33) = 0.232 in
The girder mode fundamental frequency is, fg = 0.18√(g/Δg) = 0.18 x √(386/0.232) = 7.34 Hz Dg = Ig / Lj = 6178.33/30 = 205.94 in4/ft The Effective girder panel width with Cg = 1.8 for girders supporting beams connected to the girder web. Bg = Cg (Dj/Dg) ¼ Lg = 1.8 x (277/205) ¼ x 30 = 58.15 ft, which must be less than 2/3 times the floor width.
The actual floor width is at least 3 times the joist span, 3x 30 = 90 ft. 2/3 x 90 = 60 ft > 58.15 ft. Therefore, the effective girder panel width is 58.15 ft Wg = (wg/Lj) x Bg x Lg = (2288/30) x 58.15 x 30 = 133 kip.
Combined Mode Properties: In this case , the girder span (30 ft) is greater than the joist panel width (29.65 ft) ,therefore girder deflection Δg is not reduced. The Floor fundamental frequency is, fn = 0.18 √(g /Δj+ Δg) = 0.18 √(386 /0.167+ 0.232) = 5.59 Hz
The Equivalent panel mode panel weight is, W = [(Δj /Δj+ Δg) x Wj + (Δg /Δj+ Δg) x Wg] = [(0.167 /0.167+ 0.232) x 98.73+ (0.232 /0.167+0.232) x 133] = 118.65 kip For Occupancy without full height partitions, β = 0.03 βW = 0.03 x 118.65 = 3.55 kips ap / g = Po x e (-0.35 fn) / βW From table 11 of AISC design guide series 11, Po = 65 lb = 0.29 KN – a constant force representing the excitation
ap / g = Po e (-0.35 fn) / βW = 65 e (-0.35 x 5.59) / 3550 = 0.0025 ap = 0.25 % g < 0.5 % g The floor is therefore judged satisfactory. Fundamental frequency of the system is less than 9 Hz, the minimum stiffness requirement of 5.7 kip per in (1 KN /mm) does not apply.
Comparison of Hand Calc and RAM steel software package Values: Parameter Member
Beam Panel
RAM Steel
W 18 x 40
W 18 x 40
740
737.10
Effective I , in4
2770.32
2516.80
Frequecny , Hz
8.65
8.24
PanelWidth ,ft
29.65
30.38
Panel weight, K
98.73
100.80
Uniform load, plf
Parameter Member
Girder Panel
RAM Steel
W 24 x 68
W 24 x 68
2288
2279.3
Effective I , in4
6178.33
6159.8
Frequecny , Hz
7.34
7.38
PanelWidth ,ft
58.15
56.82
133
129.5
Uniform load , Plf
Panel weight, K
Parameter
Combined System (Floor)
RAM Steel
Frequency , Hz
5.59
5.50
Panel weight, K
118.65
116.7
0.25
0.27
ap / g
2.MAIN ENTRY CANOPY • A Cantilevered Steel Framed Entry Canopy designed for a Hospital named as Sequoia Hospital, CA, USA. • Entry Canopy drawings was studied. IS 800:2007, IS 1893:2002 part-II, National Building Code of California (NBC 2005) was studied. • Entry Canopy was designed for dead load, live load and Earthquake loading. • I learnt Software ETAB . Using ETAB software entry Canopy model was done.
• From that time period was calculated, which is used for response spectra analysis. Comparison of response spectra was done. • Entry Canopy beam, column, was designed as per IS 800:2007 and footing as per IS 2911 part 1:1979.
• Loading Calculation works done manually as well as using software ENERCALC.
• Loading: Dead Load: 7/8” Cement Plaster of Metal Lathing Metal Framing Structural Pan Deck 5/8” Thick Gypsum Board Décor roofing Miscellaneous
= = = = = =
Faux Wood Lattice Miscellaneous
= 4.0 psf = 1.0 psf
Live Load (IS 875 Part-2 –Table-2)
= 15 Psf
10 psf 1.5 psf 3.0 psf 2.5 psf 2.5 psf 10 psf
Response Spectra 3
2.5
2
1.5
1
0.5
0 0
0.5
1
1.5
2
2.5
• Seismic Load: The total design lateral force, VB = AH x W (IS 1893 Part -1,2002) AH = Z/2 x I/R x Sa/g I = 1.5
Z = 0.4
R = 1.5
From E-Tabs Model, T = 0.265 Sec Sa/g = 2.5 where T is in between 0.1 to 0.4 AH = Z/2 x I/R x Sa/g = 0.4/2 x 1.5/1.5 x 2.5 = 0.5
Beam Design • HSS 3 X 3 X 3/16 (BEAM):
• Step 1: wDL = 5 Psf x 5’2”+ 6.83 Plf = 32.66 Plf Factored Load = 1.5 x 32.66 = 49 Plf • Step 2: Maximum Bending Moment, M max = wL2 /8 = 7.35 k-in Maximum Shear Force, Vmax = wL/2 = 0.245 kip • Step 3: Required Section Modulus (ZP) = M/fy x γmo fy = 46ksi , γmo = Partial Safety Factor = 1.10 (Table 5 IS 800:2007) ZP = 7.35 x 1.1 / 46 = 0.176 in3
• Step 4: Sectional Properties Area, A = 1.89 in2 Ixx = 2.46 in4 ZP = 1.97 in3 Section Classification:
ε = √( 250/fy ) = (250/320)1/2 = 0.88 b/t f = 3/0.1875 = 16 < 29.3 ε = 25.78 Hence, the Section is classified as Plastic Section.
• Step 5 : Adequacy of Section Modulus Required Plastic Section Modulus , ZP reqd = 0.176 in3 < ZP provd= 1.97 in3 , Hence O.K. • Step 6: Design Shear Strength of Section Design Shear Strength, Vd = fy x Av / γmo x √3 Av =A x h / (b+h) = 1.89/2 = 0.945 in2
(IS 800:2007 ,8.4.1.1)
Vd = 46 x 0.945 /1.1 x √3 = 22.81 kip > 0.245 kip Also, 0.6 Vd =13.86 kip > 0.245 kip Hence O.K.
(IS 800:2007,8.2.1.2)
• Step 7: Resistance to Shear Buckling d/tw = 3/0.1875 =16 < 67 ε = 67 x 0.88 =58.9 (IS 800:2007 ,8.4.2) Hence O.K
• Step 8: Check for design capacity of beam From section ,8.2.2 in IS 800:2007, for Rectangular / Tubular Hollow sections or Solid bars, resistance to lateral torsional buckling need not to be checked separately (Member may be treated as laterally supported). The Factored Design Shear Strength, 0.6 Vd = 13.86 kip > Vmax = 0.245 kip Therefore, Md = βb x ZP x fy / γmo
βb = 1.0 for Plastic and Compact sections. Md = 1 x 1.97 x 46 / 1.1 = 82.38 k-in Maximum Bending Moment, Mmax = 7.35 k-in Md=82.38 k-in > Mmax = 7.35 k-in Hence O.K • Step 9: Check for deflection Δ = 5 wL4 /384 EI =(5 x 49x 1204) / (12 x 384 x 29000x 2.46 x 1000) = 0.154 in Δ = 0.154 in (L/780) < (L/240) , Hence O.K
2. HSS 12 X 4 X 3/16 (BEAM)
3. HSS 14 X 4 X 3/16 (BEAM)
4. HSS 12 X 4 X 1/4 (BEAM)
5. HSS 12 X 4 X 1/4 (BEAM)
6. HSS 12 X 4 X 1/4 (BEAM)
7. HSS 16 X 12 X 5/8 (COLUMN)
For Seismic analysis in the long direction ,a rigid diaphragm analysis is done.
Computation of Centre of Mass
• Step 1: Axial Load =P1 + P2 = 3.71 + 22.814 = 26.52 kip Moment about Z- axis , MZ = TEQ x 19.5’+(0.3 VD + VT) x 14.5’ - CEQ x 14.5’ = 145.9 k-ft
Moment about Y-axis, My = Vw x (19.52/2) + VD x 14.5’ = 81.95 k-ft
• Step 2: Sectional Properties A = 30.3 in2 Izz = 1090 in4 ZP = 165 in3 Ze = 136 in3 r z = 6 in
t = 0.581 in Iyy = 700 in4 ZP = 135 in3 Ze= 117 in3 r y = 6 in
• Step 3: Section Classification ε = √( 250/fy ) = √ (250/320) = 0.88 d/t w = 16/0.581 = 27.54 < 84 ε = 74 Hence, the Section is classified as Plastic Section.
• Step 4 : Bending and Axial Compression Members subjected to combined axial compression and biaxial bending shall satisfy the following interaction relationships
1.(P/Pdy )+ Ky x Cmy (My /Mdy )+ KLT x (Mz / Mdz) ≤ 1.0 2.(P/Pdz )+ 0.6 x Ky x Cmy (My /Mdy )+ Kz x Cmz (Mz / Mdz) ≤ 1.0
Pd = Ae x fcd
fcd = ( fy / γmo) / ɸ + [ɸ2 - λ2]0.5 = (χ x fy) / γmo
λ
= √ (fy /fcc)
ɸ
= 0.5 [1+α (λ-0.2) + λ2]
fcc = π2 E / (KL/r )2 Pdz = 28.018 x 30.3 = 849 kip Pdy = 20.99 x 30.3 = 636 kip
Moment Capacity about Z-Z axis : Mdz = βb x ZP x fy / γmo βb = 1.0 for Plastic and Compact sections. ZP = 165 in3 Mdz = 1 x 165 x 46 / 1.1 = 6900 k-in For Cantilever , (1.5 x Ze x fy ) / γmo = (1.5 x 136x 46 ) / 1.1 = 8531 k-in (1.5 x Ze x fy ) / γmo > Mdz Therefore, Mdz = 6900 k-in
Moment Capacity about Y-Y axis : Mdy = βb x ZP x fy / γmo = 5645 k-in • Ky = 1 + ( λ y - 0.2 ) ny ≤ 1+0.8 ny Ky = 1 + ( 1.25- 0.2 ) 0.0417 = 1.044 1+0.8 ny = 1+0.033 = 1.033 Therefore, Ky =1.033 • Kz = 1 + ( λ z - 0.2 ) nz ≤ 1+0.8 nz = 1.0246
KLT =[1- (0.1 x λ LT x ny) / (C mLT – 0.25) ] ≥ [1- (0.1 x ny) / (C mLT – 0.25) ] λ LT =√(fy/fcr,b) fcr,b can be calculated from table 14 in IS 800:2007 KLT =[1- (0.1 x 0.866 x 0.0467) / (0.6 – 0.25) ] ≥ [1- (0.1 x 0.0467) / (0.6 – 0.25) ] =0.988 > 0.9866 Therefore, KLT = 0.988
Check for Axial Compression and Biaxial Bending 1. (P/Pdy )+ Ky x Cmy x (My /Mdy )+ KLT (Mz / Mdz) ≤ 1.0 =0.4015 ≤ 1.0 Hence O.K 2. (P/Pdz )+ 0.6 x Ky x Cmy (My /Mdy )+ Kz x Cmz x (Mz / Mdz) ≤ 1.0 =0.252 ≤ 1.0 Hence O.K.
• Step 6: Check for Deflection (From Enercalc Software) Δ max = 1.604 in (only gravity) Δ max = 5.71 in (includes seismic load) In IS 1893(Part 1) :2002 ,Section 7.11.1 there shall be no drift limit for single storey building which has been designed to accommodate storey drift. Hence O.K
8.Design of Foundation: Providing Pile foundation with pile cap. Step 1: Classification of Pile Stiffness factor (R or T) is used to predict whether the pile long or short. Factor R is used only for Clay soil. For Sands, T is used. T = (EI/K) 1/5 Assume, Diameter of Pile = 2.46 ft = 0.75 m Height of Pile = 9.84 ft = 3 m
Ec =293.99 psi = 20.67 kg/cm2 I = π d4 / 64 = π x 29.5264 / 64 = 37287.92 in4 From Geotechnical Report, the site contains dense sand, hence K = 1.245 kg/ cm3 (from IS: 2911(Part I/Sec 2) – 1979 T = (EI/K) 1/5 T = (293.99 x 37287.92/17.71)1/5 = 14.39 in = 0.365 m L =3 m > 4T, Hence the Pile is classified as Long and Elastic.
• Depth of Piles: 4T value is considered sufficient to develop the maximum resistance in the soil. Depth beyond 4T value will not have much benefit. Therefore this length is sufficient. Step 3: Determine Maximum Bending Moment e, is the distance of force acting from ground level. e = 17 ft (Distance from G.L. to resultant force acting) e / T = 17/1.19 = 14.28 For Fixed head pile (having pile cap), corresponding Zf / T value is 1.32.
Depth of Fixity, Zf = 1.32 x 1.19 =1.57 ft
• Step 3: Determine Maximum Bending Moment Total Length = e + Zf = 17 + 1.57 = 18.57 ft Total Lateral force = 5.4 - 5.47 + 4.95 + 5.57 + 0.053 x19.5 = 11.48 kip M max = (11.48 x 19.08) / 2 = 1314 k-in Maximum Axial load, Pu = 26.52 kip Assuming 20 mm bars with 40 mm cover, d’ = 50 + 12.5 = 62.5 d’/D =62.5 /750 = 0.083 (from SP16 , section 3.2)
Using Chart, d’/D = 0.1
Pu / (fck x D2) = 120.545 x 103 / (20 x7502) = 0.0107 Mu / (fck x D3) = 15.12 x 106 / (20 x 7503) = 0.0074 Use Fe 415 bars, chart 55 from SP 16,
p/fck = 0.01 p = 0.01 x 20 = 0.2 Ast = ( 0.2/100) x (π x 2.462) / 4 = 0.0095 ft2
Base plate Design • Load combination used – 1.2 D.L+0.5 L.L+1.2 EL (Strength design) – IS 800:2007 Table 4 Mu = 288.4 k-ft Tu = Mu /d = (288.4 x 12) /19 = 182 kip. Critical section from face of column is 3” Mu (Plate) = 182 x 3 = 546 k-in
t = √(6Mu / σbs) = 1.673” thick base.
Anchor Bolt Design size of bolt = 1.25” dia Length of bolt = 18” Controlling strength = 248.5 kip > 182 kip (Tu) Hence O.K
Anchor Bolt Design size of bolt = 1.25” dia , Length of bolt = 18” Controlling strength = 248.5 kip > 182 kip Hence O.K
Conclusion • Floor Vibration due to human activity – new concept. • Entry Canopy design based on IS 800:2007. • RAM Steel , E Tab ,ENERCALC - softwares new to me. • In this occasion I would like to thank Francis Santiago & Associates .
THANK U
