Bracing

Ahmed Elgamal (Draft, 2010)

Bracing for Earthquake Resistant Design

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September 18, 2002 (2010 update) Ahmed Elgamal

Rigid Roof Idealization and Column Stiffness Relative to the columns, the roof structural system might be quite rigid, resulting in the classical deformed shape shown in the Figure. In this deformed configuration, the column stiffness is dictated by a state of fixed-fixed boundary conditions.

u

h

In order to define the lateral column stiffness (k) for this fixed-fixed state, we start with the beam Equation of equilibrium and the appropriate boundary conditions (see next page for derivations):

z

EI w’’’’ = 0 w(0) = w’(0) = w’(h) = 0, and w(h) = u Resulting in:

w = ( (3z2/h2) - (2z3/h3) ) u

F

u

h

With shear force Q = - EI w’’’ , the shear force F at h becomes: F = -EIw’’’(h) = (12EI / h3) u and k(column) is therefore,

k = (12EI /

w k

F=ku

h3) 2

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Bending beam Equation EI w’’’’ = 0 (case of zero pressure acting along the beam length) EI w’’’ = c1 EI w’’ = c1 z + c2 EI w’ = c1 z2/2 + c2 z + c3 (slope) EI w = c1 z3/6 + c2 z2/2 + c3 z + c4 (displacement) w (0) = 0 results in c4 = 0 w’(0) = 0 results in c3 = 0 w’(L) = 0 results in c2 = - ( c1/ h ) w (L) = u results in c1 = - ( 12 EI / h3 ) u Therefore w = ( (3z2/h2) – ( 2z3/h3) ) u Note: Moment (M) = - EI w’’ Shear force = M’ = - EI w’’’ 3

30’

Ahmed Elgamal

Bracing 1) Bending stiffness of 4 I-beams in the NS and EW directions 2) Axial stiffness of 4 slender rod braces in the EW direction

h = 12’

Building is supported laterally by:

Mass Take weight (w)of roof as 30 lb/ft2 and calculate the mass (m) m

East-West Direction (EW) North-South direction (NS)

w 30  30  20   46.63 lb-sec2/in = 0.04663 kip-sec2/in g 386.4

Note: Acceleration of gravity (g) = 386.4 in/sec2

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First version: September 18, 2002 (2010 update)

Ahmed Elgamal

30’

Es = 29,000 ksi (Steel Young’s Modulus)

Rigid roof

Steel I-Beam columns (Section a-a)

x y (W8x24 Steel I-beam) h = 12’

a

Ix = 82.8 in4 Iy = 18.3 in4

a

a

a

a a

a

Note that Iy is much smaller than Ix

x y

Brace (1 in diameter circular bar) Cross sectional area A = 0.785

a

East-West Direction (EW)

in2

North-South direction (NS)

In NS-direction





12 29x10 3 82.8  12EI  k NS  4 3 x   4  38.58 kips in 12x12 3  h 

September 18, 2002 (2010 update)

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Ahmed Elgamal

NS-direction Equation of motion: muNS  uNSg   cu NS  k NSuNS  0



where c  ζccr  ζ 2 km



and z Maybe ≈ 1.2-1.5% for steel

or uNS  uNSg   2z NS NS u NS  NSNSuNS  0 where n(NS)= sqrt (38.58 / 0.04663) = 28.76 radians/sec Note:

fn(NS) = 28.76 / (2 x 3.1428) = 4.58 Hz (cycles/sec)

and

Tn(NS) = 1/ 4.58 = 0.22 seconds

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3



In EW-direction



 12EI y  12 29x10 3 18.3 k EW  4 3   4  8.52 3 h   12x12  

kips in

As can be seen, lateral stiffness of the columns in the EW direction is much lower than that in the NS direction. The braces will change this situation dramatically. Thus, we will rely on brace stiffness since column stiffness is relatively small and not intended for lateral support (only to carry vertical load). Laterally,

fs = kbrace u ,

or kbrace = fs / u

From geometry,

fs = p cos q and

resulting in

kbrace = (p/d) cos2 q

u = (d / cos q

p brace

u q

fs

In the brace, axial stress is related to axial strain by (p/A) = Es (d/L)

(A is brace cross-sectional area)

so that (p/d) = (AEs/L) , and therefore

kbrace = ( AEs / L )

cos2

q

q

Ahmed Elgamal

From the building geometry

20

cosθ 

12  20 2

2

 0.8575



L = sqrt (202 + 122) = 23.3 ft



0.785 29 10 3 0.85752  59.8 kips in 23.3 12 Braces are slender in this case, and therefore only provide added stiffness when subjected to tensile force (the braces sag or buckle when in compression. As such, only two braces will be providing lateral stiffness at any given time. As such, and

k brace 

k EW (bracing)  2  59.8  119.6 kips in Since the stiffness due to bracing is much larger than that due to the 4 columns, we will only rely on the bracing for stiffness in this direction:





muEW  uEWg   cu EW  kEW uEW  0 or, uEW  uEWg  2z EW EW u EW  EWEW uEW  0

where n(EW)= sqrt (119.6 / 0.04663) = 50.64 radians/sec Note:

fn(EW) = 50.64 / (2 x 3.1428) = 8.06 Hz (cycles/sec)

and

Tn(EW) = 1/ 8.06 = 0.12 seconds

now,

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Question: Will all 4 braces in the frame be effective at the same time? Euler Buckling Load http://en.wikipedia.org/wiki/Buckling :  2 EI , Lk = k L = Effective Buckling Length

Ncr 

L2k

4 πr 4 π0.5" I  0.049 in 4 4 4 1 L k  23.3ft   23.3ft  279.6 in 1

I

N cr 

 2 29000ksi0.049in 4 

279.6in (279.6in )

 0.179 kips  179lbs

Answer: When the brace experiences compression, buckling load is minimal. This is why, we used only two braces at a time in our calculations (of the 4 installed braces) 9

Buckling Restrained Braces (BRBs)

http://jiano.typepad.com/photos/un categorized/brb_02.jpg

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Bracing Systems (a) Diagonal, b) Chevron, and c) V-Braced

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Ahmed Elgamal

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Ahmed Elgamal

Testing

http://peer.berkeley.edu/~yang/NEESZipper/Summary.html

http://nees.buffalo.edu/projects/zipperframes/CollaboratoryResearchGT-UCB-UCB-UB-USF.pdf

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Energy Dissipation Devices

Viscous Dampers

From Publication by Alessandro Martelli and Massimo Forni

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