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L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 1
Wall Rigidity and
Out-of-Plane Walls
CENG 232 Lecture 7
Shear Wall Rigidity (Text 3.8.2)
For buildings with rigid diaphragms, lateral loads are distributed based on relative stiffness of walls
Wall rigidity is the shear required to cause a unit deflection of the wall:
R = V
Shear Wall Rigidity (Text 3.8.2)
Deformations are due to: Flexural deformation Shear deformation Foundation rotation or uplift
Assuming the shear wall is a deep cantilever beam:
Youngs Modulus:
Shear Modulus:
=VH 33EmI
+1.2VHAEv
+ F2HL
Em = 900 f mEv = 0.4Em
Foundation rotation is usually ignored
Shear Wall Rigidity (Text 3.8.2)
For distribution of EQ loads, only the relative rigidity is needed.
For walls with the same thickness, relative rigidity is:
The force resisted by each wall is:
Ri =1
4 HL
3+ 3 HL
i
Vi =RiRi
i=1
n
V
L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 2
Shear Wall Rigidity (Text 3.8.2)
If the top of the wall is restrained from rotation (fixed-fixed):
For walls with the same thickness, relative rigidity is:
=VH 312EmI
+1.2VHAEv
Ri =1
HL
3+ 3 HL
i Load to each shear wall can be calculated
based on relative stiffness.
Determining relative stiffness can be more difficult for walls with openings like this
The walls of this school have architectural CMU veneer over loadbearing CMU.
L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 3
Varying wall stiffness in a CMU building
Rigidity of Walls with Openings (Text 3.8.2)
Modified stiffness for walls with openings:
wall = solid wall solid strip + piers
Example 3.8.3: Rigidity of Walls with Openings
Wall broken into 6 segments
Relative rigidity found for each
30 kip force is distributed according to relative rigidity
Example 3.8.3: Rigidity of Walls with Openings
Solution:
L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 4
Out-of-Plane Loads on Walls (Text 3.9)
Shear walls loaded out-of-plane are not part of the lateral-load resisting system, and need only support their own loads.
Because wall deflections are comparable to the wall thickness, P- effects must be considered.
Christchurch, New Zealand, 2011. An example of out-of-plane failure of an unreinforced masonry wall.
Christchurch, New Zealand, 2011. Failure of an reinforced masonry wall.
Out-of-Plane Loads on Walls (Text 3.9.1)
Out-of-plane EQ loads different for structural and non-structural walls.
Structural walls: part of lateral load resisting system
SDS = design spectral response acceleration in short period range
I = importance factor
Wp = weight of wall
Fp = 0.4SDS I Wp
Non-structural: see text section 3.9.1 for loads
Fp min = 0.1 Wp
L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 5
Walls can be designed to span vertically, or horizontally between vertical supports:
Out-of-Plane Loads on Walls
Effective wall height for design is based on supports:
This wall is being braced while under construction to reduce its effective height.
Out-of-Plane Analysis of Slender Walls (Text 5.5.1)
Out-of-plane wall design MSJC requirements valid if one of the following is satisfied:
Pu = factored axial load at location of max moment
Ag = gross cross-sectional area
H = effective height of wall
t = wall thickness
Pu 0.05 f m Ag0.05 f m Ag < Pu 0.20 f m Ag;
H t 30
L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 6
Out-of-Plane Analysis of Slender Walls (Text 5.5.1)
Equations based on pinned-pinned walls. Use equivalent span for non-pinned walls.
Wall moments due to: Lateral loads (w) Eccentricity of floor loads (Pf) of wall weight x parabolic
shape deflection (Pw x 2/3)
(assumes parabolic deflected shape, with top of wall c.g. at 2/3 of )
Out-of-Plane Analysis of Slender Walls (Text 5.5.1)
Moment at the middle of the wall is:
Note that the wall moment is based on the wall deflection (), which is based on the wall moment.
(an iterative process)
M = wH2
8 +Pf e2 + Pf + Pw( )
Cracked Section Properties:
Cracked moment:
Cracked moment of inertia:
Mcr = Sn fr See Table 1.6.2 for fr
Icr = nAse d c( )2 +bc33
n = modular ratio
Ase =P + As fy
fy
Out-of-Plane Analysis: Closed form solution for wall deflection
(Text 5.5.1)
Substitute wall moment equation into deflection equation to get closed-form (no iteration) solution for deflection:
=
wH 28 +
Pf e2
Mcr 1
IcrIg
48EmIg5H 2 Pw + Pf( )
=
wH 28 +
Pf e2
48EmIg5H 2 Pw + Pf( )
Uncracked: Cracked:
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L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 7
Design of Slender Walls (Text 5.5.2)
After wall displacement and corresponding moment are determined, the section can be designed.
Design requirements: Out-of-plane moments including P- effects: Horizontal deflection at mid-height under service
(unfactored) loads, s, must be less than 0.007H.
Mn Mu
Design of Slender Walls (Text 5.5.2)
One layer of reinforcing:
Mn = As fy + Pu( ) d a2
a = As fy + Pu0.8 f mb
Design of Slender Walls (Text 5.5.2)
Two layers of reinforcing:
Mn = As fy + Pu( ) d a2
Pu d
t2
a = As fy + Pu0.8 f mb
Design of Slender Walls: Effective Width
Maximum width of compression block is the smallest of: 1. Center to center of bar spacing 2. 6t (t = wall thickness) 3. 72 inches
For partially grouted walls, compression zone cannot include the ungrouted part of the wall. Calculate T-beam shape compression block.
L7 - Rigi(ty, Out-of-Plane Walls
CENG 232 - Winter 2015 8
Design of Slender Walls: Max Reinforcement
Strain in the reinforcing steel must be at least 1.5 times the yield strain when the masonry reaches its strain limit.
(see text 5.5.2 for derivation)
max =As,maxbd
max = 0.64 f m
fy0.0025
1.5 fy Es + 0.0025
Pbdfy
where = 1.5, mu = 0.0025
Example: Design of Slender Walls