Lec05 STRUCTURE SYSTEMS to resist EARTHQUAKE (Earthquake Engineering هندسة الزلازل & Assc.Prof Nasser El-Shafey)

STRUCTURE SYSTEMS
to resist
EARTHQUAKE
Prepared by
Assc . Prof. Dr. Nasser. El-Shafey

Non-Structural Damage
All items, which are not part of structural system, considered
"NON-STRUCTURAL", and include such elements as:
Exterior cladding and curtain walls.
Parapet walls.
Partitions, doors, windows.
Suspended ceilings.
Mechanical, Plumbing, Electrical and Communications
equipment.
Elevators.
Furniture and equipment.
These items must be stabilized with bracing to prevent their
damage or total destruction.

Lateral Load Resisting Elements
Vertical Elements
Walls – Bearing walls / Shear Walls only
Moment-Resisting Frames (limited ductility)
“Dual” System [Frames (carry 25% of
lateral loads) + shear Wall)]
Tube System
Bundled-Tube System
• Floors and roofs can be used as rigid horizontal planes, to
transfer lateral forces to vertical resisting elements such as
walls or frames.
• Foundation – various types

buildings have Shear Walls , Moment-Resistant Frames or
Braced Frames generally have:
 Equal Floor Heights
 Symmetrical or Unsymmetrical Plans
 Uniform Sections and Elevations
 Maximum Torsional Resistance
Shear walls : are capable of transferring lateral forces from
floors and roofs to the foundation
Braced Frames:: frames that transfer lateral loads from floors
and roofs to foundations. used where shear walls impractical
Moment-Resistant Frames: Column/beam joints are designed
to take both shear and bending, joints carefully designed to be
stiff to allow some deformation for energy dissipation

1.Determine W, γI and λ
2.Determine location of the building(zone) and get ag
3.Calculate the fundamental period T1
4.Specify soil type and city in which building located,
determine the type of response spectrum (Type 1 or
Type 2) and get S, TA, TB, TC.
5.Get the value of Sd (T).
6.Substitute in the equation of Fb
Summary of procedure for Base Shear calculation

Fb= Sd (T). λ .W/g λ = 0.85 if T ≤ 2Tc
λ = 1.0 If T > 2Tc
Ordinary frames; flexural walls-R.C. R= 5.0
Moment resisting frames; R.C. with adequate ductility. R= 7.0
High occupancy buildings: schools, assembly halls, etc. γ=1.20
Ordinary buildings. γ=1.00
T = Ct H3/4 C t = 0.075 for RC framed structures
0.05 for shear wall structures
= 1 for R.c

Structural design load (W) (code 8-7-1-7)
Building weight above foundation = Σ D.L+ (Factor) Σ L.L
W= D.L.+ 0.25 L.L for residential buildings.
W= D.L.+ 0.5 L.L. for common buildings, malls, schools
W= D.L.+ L.L. for silos, tanks, stores, libraries, garages
Sub Soil Class S TB TC TD
A 1 0.05 0.25 1.2
B 1.35 0.05 0.25 1.2
C 1.5 0.1 0.25 1.2
D 1.8 0.1 0.3 1.2
Type 1 Response Spectrum

points to
considered while
locating shear
walls

Shear Walls
Wall with Shear
Deformation
Wall with Flexural
Deformation
Wall with both Shear and
Flexural Deformation
• Large width-to-thickness ratio; else like a column
• Height-to-width

14m
0.4m 0.4m
3.6m
4m
Area = 860,000 mm2
Shear Area = 540,000 mm2
(= 0.15m x 3.6m)
Inertia = 1.867 x 1012 mm4
E = 25,500 MPa
G = 10,500 MPa
Stiffness due to point load at thetop
Example
Wall Section
0.4m
W
0.15m thick

Moment Resisting Frame
• Components
– Beams
– Columns
– Joints
• Joints: Most frames have joints where the angle between
connecting members is maintained, i.e., rigid joints.
Approximate Analysis of:
allows to get a simple estimate of member sizes and to check
the magnitude of computer analysis results

Frame with rigid joints and with very flexiblebeams.
BMD
BMD
Frame with rigid joints and with infinitely rigid beams
/2
ph/2l

Shears on Different Columns Lateral Forces Lateral Shears
If the storey shear at the top level is 120 kN say, then the shear force
on an internal column in 40kN, and on an external column is 20kN.
120kN
20kN20kN
40kN 40kN 120kN
Exterior Columns Assumed to Carry One Half Shears of
Internal Columns

Shears on Different Members
20kN
120kN
40kN 40kN
Top right beam shear is found by
considering a free body. The beam
axial force is first computed from .
horizontal equilibrium as 20 kN.
Then, by taking moments about
column mid-height, the beam shear
is 20kN x 0.3*3.6m / (0.5x7.2m)= 6kN.
20kN
20kN
6kN
0.3 x 3.6m
0.5 x 7.2m
6kN
The beam moment demand is therefore
0.5 x 7.2m * 6 KN = 21.6 KN.m due to earthquake
loads. This can be combined with gravity loads
for design. Seismic axial forces in columns are
generally small in theinternal columns A similar
process used to obtain all moments, shears and
axial forces throughout the frame
20kN

The wall structure deflected shape is
 Straight line for point load at top
 Approximately a quarter cycle of sine function in case of earthquake force
Frame
Deformation:
Cantilever beam
(flexural beam; ignoring shear deformation) Zero Slope :: Small inter-storey
displacement
::Large inter-storey
displacement
Zero Slope:: Small inter-storey
displacement
Large inter-storey
displacement
Wall-Frame Systems
Building has walls and frames which shear lateralloads
• Extreme 1 ::Walls too rigid compared to frames, Frames deform as per walls
• Extreme 2 ::Frames too rigid, Walls deform as per frames
• Walls and frames comparable:: Interaction through floor
diaphragm

Wall-Frame Interaction
Rigid Frame
“Shear Mode”
Deformation
Shear Wall
Bending Mode Deformation
Combine Deformations
Interacting
Forcestension
Combine
compression
Building has walls and frames which shear lateralloads
• Extreme 1 ::Walls too rigid compared to frames, Frames deform as per walls
• Extreme 2 ::Frames too rigid, Walls deform as per frames
• Walls and frames comparable:: Interaction through floor
diaphragm

Wall-Frame Interaction
• Walls :: flexural deformations
• Frames :: deformations are like shear beam
P
This can be considered
in analysis
Buildings must be designed
to carry interaction forces

Tube Systems
Shear lag
A Compression Columns B
Plan
Tension Columns
Force Plan
A B
1
2
Variation in axial
force
in columns
Bundled Tube
Other Systems

Lec05 STRUCTURE SYSTEMS to resist EARTHQUAKE (Earthquake Engineering هندسة الزلازل & Assc.Prof Nasser El-Shafey)

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