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4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Lecture-07
Introduction to Earthquake Resistant Design of Reinforced
Concrete Structures
By: Prof Dr. Qaisar Ali
Civil Engineering Department
NWFP UET Peshawardrqaisarali@nwfpuet.edu.pk
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Topics
Introduction
Effects of Earthquake
Deformation due to Earthquake
Horizontal and Vertical Earthquake Shaking
How Architectural Features Affect Buildings DuringEarthquakes
Earthquake Design Philosophy
Seismic Loading Criteria
Static & Dynamic Lateral Force Procedures
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4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Topics
Gravity vs. Earthquake Loading in RC Buildings
ACI Special Provisions for Seismic Design
ACI Provisions for SMRF
ACI Provisions for IMRF
Miscellaneous Considerations
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction
Earth’s Interior
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction
Earthquake results from the sudden movement ofthe tectonic plates in the earth’s crust.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction
Effect of Earthquake
The movement, taking place at the fault lines, causesenergy release which is transmitted through the earth inthe form of waves. These waves reach the structurecausing shaking.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction
Seismic Events around the globe
Mostly takes place at boundaries of Tectonic plates
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Dots represents an earthquake
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction
Types of Waves Generated Due to Earthquake
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Body Waves Surface Waves
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Displacement due to Earthquake
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Horizontal and Vertical Shaking
Earthquake causes shaking of the ground in all three directions.
The structures designed for gravity loading (DL+LL) will benormally safe against vertical component of ground shaking.
The vertical acceleration during ground shaking either adds to orsubtracts from the acceleration due to gravity.
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Horizontal and Vertical Shaking
The structures are normally designed for horizontal shakingto minimize the effect of damages due to earthquakes.
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake Types with respect to Depth of Focus
Shallow
Depth of focus varies between 0 and 70 km.
Deep
Depth of focus varies between 70 and 700 km.
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake characteristics with respect to distancefrom epicenter
Introduction
0.05 ≤ T ≤ 0.320 Hz ≥ f ≥ 3.33 Hz
Low period & high frequency field
0.3 ≤ T ≤ 1.0 sec3.33 Hz ≥ f ≥ 1 Hz
1.0 ≤ T ≤ 10 sec
25 km
Large period & low frequency field
Moderate period & low frequency field
Epicenter
1 Hz ≥ f ≥ 0.1 Hz
Near Field: 0 to 25 km
Intermediate Field: 25 to 50 km
Far Field: Beyond 50 km
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Resonance risk for structures w.r.t near, intermediateand far field earthquakes
The natural time period of a structure is its important characteristicto predict behavior during an earthquake of certain time period(Resonance phenomenon).
For a particular structure, the natural time period is a function ofmass and stiffness {T = 2π√(M/K)}
“T” can be roughly estimated from: T = 0.1 × number of stories
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Resonance risk for structures w.r.t near, intermediateand far field earthquakes
Introduction
Low rise Structure
(upto 3 stories)
Epicenter
Medium rise Structure
(upto 5 stories)High rise Structure
(Above 5 stories)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake Recording
Seismograph
Using multiple seismographsaround the world, accuratelocation of the epicenter of theearthquake, as well as itsmagnitude or size can bedetermined.
Working of seismograph shownin figure.
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake Recording
Richter Scale
In 1935, Charles Richter (US)developed this scale.
The Richter scale is logarithmic,So, a magnitude 5 Richtermeasurement is ten timesgreater than a magnitude 4;while it is 10 x 10, or 100 timesgreater than a magnitude 3measurement.
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake Recording
Some of the famous
earthquake records
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake Occurrence
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Seismic Zones
Introduction
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
How Architectural Features Affect Buildings During Earthquakes?
Importance of Architectural Features
The behavior of a building duringearthquakes depend critically on its overallshape, size and geometry, in addition tohow the earthquake forces are carried to theground.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
How Architectural Features Affect Buildings During Earthquakes?
Importance of Architectural Features
At the planning stage, architects and structural engineers mustwork together to ensure that the unfavorable features are avoidedand a good building configuration is chosen.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
How Architectural Features Affect Buildings During Earthquakes?
Other Undesirable Scenarios
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Soft Storey
How Architectural Features Affect Buildings During Earthquakes?
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Earthquake Design Philosophy
Performance level
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Building Code of Pakistan
In Pakistan, the design criteria for earthquake loading are basedon design procedures presented in chapter 5, division II ofBuilding Code of Pakistan, seismic provision 2007 (BCP, SP2007), which have been adopted from chapter 16, division II ofUBC-97 (Uniform Building Code), volume 2.
Seismic Loading Criteria
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Lateral Force Determination Procedures
The total design seismic force imposed by an earthquake onthe structure at its base is referred to as base shear “V” in theUBC-97.
The design seismic force can be determined based on:
Dynamic lateral force procedure [sec. 1631, UBC-97 or sec. 5.31, BCP-2007].
Static lateral force procedure [sec. 1630.2, UBC-97 or Sec. 5.30.2, BCP 2007], and/or
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Dynamic Lateral Force Procedure
UBC-97 section 1631 include information on dynamic lateral forceprocedures that involve the use of:
Time history analysis.
Response spectrum analysis.
The details of these methods are presented in sections 1631.5and 1631.6 of the UBC-97.
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Dynamic Lateral Force Procedure
Time History Analysis (THA)
T
Ground acceleration T
LateralDisplacement
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Dynamic Lateral Force Procedure
Response Spectrum Analysis (RSA)
T
a (ft/sec2)
T
Response
a (ft/sec2)
T
a (ft/sec2)
T
Response
Response
Ts1 = 0.3 sec
Ts2 = 1.0 sec
Ts3 = 2.0 sec
D1
D2
D3
(Ts1,D1)
(Ts2,D2)
(Ts3,D3)
Structural Time period
Peak Response
T
T
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
UBC-97 Response Spectrum Curve(Acceleration vs. Time period)
Dynamic Lateral Force Procedure
Response Spectrum Analysis (RSA)
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
=
Seismic Loading Criteria
Static Lateral Force Procedure
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
The total design base shear (V) in a given direction can bedetermined from the following formula:
V = (CνI/RT) W
Where,
Cν = Seismic coefficient (Table 16-R of UBC-97).
I = Seismic importance factor (Table 16-K of UBC-97 )
R = numerical coefficient representative of inherent over strength andglobal ductility capacity of lateral force-resisting systems (Table 16-Nor 16-P).
W = the total seismic dead load defined in Section 1630.1.1.
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
The total design base need not exceed [ V = (2.5CaI/R) W ]
Where, Ca = Seismic coefficient (Table 16-Q of UBC-97)
The total design base shear shall not be less than [ V = 0.11CaIW ]
In addition for seismic zone 4, the total base shear shall also notbe less [ V = (0.8ZNνI/R) W ]
Where, Nν = near source factor (Table 16-T of UBC-97);
Z = Seismic zone factor (Table 16-I of UBC-97)
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 1: Find Site Specific details.
Step 2: Determine Seismic Coefficients
Step 3: Determine Seismic Importance factor “I”
Step 4: Determine “R” factor
Step 5: Determine structure’s time period
Step 6:Determine base shear “V” and apply code maximum andminimum.
Step 7: Determine vertical distribution of “V”.
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 1: Find Site Specific details.
Following list of data needs to be obtained:
Seismic Zone
Soil type
Past earthquake magnitude (required only for highest seismic zone).
Closest distance to known seismic source (required only for highest seismiczone).
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 1: Site Specific details.
i. Seismic Zone Source: BCP SP-2007
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 1: Find Site Specific details.
ii. Soil Type
As per UBC code, if soil type is not known, type SD shall be taken.
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 1: Find Site Specific details.
iii. Past Earthquake magnitude: This is required only for seismic zone 4to decide about seismic source type so that certain additional coefficientscan be determined.
iv. Distance to known seismic zone is also required to determineadditional coefficients for zone 4.
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 2: Determination of Seismic Coefficients.
Cv:
Nv (required only for zone 4):
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 2: Determination of Seismic Coefficients.
Ca:
Na (required only for zone 4):
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 3: Determination of Seismic Importance Factor.
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 4: Determination of “R” Factor.
R factor basically reduces base shear “V” to make the systemeconomical. However the structure will suffer some damage as explainedin the earthquake design philosophy.
R factor depends on overall structural response of the structure underlateral loading.
For structures exhibiting good performance, R factor will be high.
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 4: Determination of “R” Factor.
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 5: Determination of structure’s time period.
Structural Period (By Method A, UBC 97): For all buildings, the value Tmay be approximated from the following formula:
T = Ct (hn)3/4
Where,
Ct = 0.035 (0.0853) for steel moment-resisting frames.
Ct = 0.030 (0.0731) for reinforced concrete moment-resisting frames andeccentrically braced frames.
Ct = 0.020 (0.0488) for all other buildings.
hn = Actual height (feet or meters) of the building above the base to the nth level.
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 6: Determination of Base Shear (V).
Calculate base shear meeting the following criteria:
0.11CaIW ≤ V = (CνI/RT) W ≤ (2.5CaI/R) W
For zone 4, the total base shear shall also not be less than:
V = (0.8ZNνI/R) W
Seismic Loading Criteria
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Static Lateral Force Procedure
Steps for Calculation of “V”:
Step 7: Vertical Distribution of V to storeys.
The joint force at a particular level x of the structure is given as:
Fx = (V – Ft)ωxhx/∑ωihi (UBC sec. 1630.5)
{ i ranges from 1 to n, where n = number of stories }
Ft = Additional force that is applied to the top level (i.e., the roof) inaddition to the Fx force at that level.
Ft = 0.07TV {for T > 0.7 sec}
Ft = 0 {for T ≤ 0.7 sec}
Seismic Loading Criteria
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example: Calculate the base shear and storey forces of a five storeybuilding given in the figure. The structure is constructed on stiff soilwhich comes under soil type SD of table 16-J of UBC-97. Thestructure is located in zone 3.
Static Lateral Force Procedure
60′-0″
12′-0″
800 kip
800 kip
800 kip
800 kip
700 kip
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Selection of Ca and Cv.
Base Shear (V) = {CvI/RT}W
Static Lateral Force Procedure
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Selection of I.
Base Shear (V) = {CvI/RT}W
Therefore, I = 1.00
Static Lateral Force Procedure
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Selection of R.
Base Shear (V) = {CvI/RT}W
Therefore, R = 8.5 (for SMRF)
Static Lateral Force Procedure
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Calculation of T and W
Base Shear (V) = {CvI/RT}W
T = Ct (hn)3/4
= 0.030 × (60)3/4
= 0.646 sec.
W = w1 + w2 + w3 + w4 + w5
= 4 × 800 + 700
= 3900 kip
Static Lateral Force Procedure
hn = 60′-0″
12′-0″
w1 = 800 kip
w2= 800 kip
w3 = 800 kip
w4 = 800 kip
w5 = 700 kip
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Therefore,
V = {CvI/RT}W
= {0.54 × 1.00/ (8.5 × 0.646)} × 3900 = 383 kips
The total design base need not exceed the following:
V = (2.5CaI/R) W
= {(2.5 × 0.36 × 1.00)/ (8.5)} × 3900 = 413 kips
The total design base shear shall not be less than the following:
V = 0.11CaIW
= 0.11 × 0.36 × 1.00 × 3900 = 154.44 kips
Therefore, V = 383 kip
Static Lateral Force Procedure
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Vertical distribution of base shear to stories
The joint force at a level x of the structure is given as:
Fx = (V – Ft)ωxhx/∑ωihi
{ i ranges from 1 to n, where n = number of stories }
Ft = Additional force that is applied to the top level (i.e., the roof) in addition to theFx force at that level.
Ft = 0.07TV {for T > 0.7 sec}
Ft = 0 {for T ≤ 0.7 sec}
Static Lateral Force Procedure
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Vertical distribution of base shear to stories
Fx = (V – Ft)ωxhx/∑ωihi
∑ωihi = 800×12 + 800×24 + 800×36 + 800×48 + 700×60 = 138000 kip
Therefore for the case under consideration, Force for storey 1 is:
F1 = (383 – 0) × 800 × 12/ {(138000)} = 27 kip
Storey forces for other stories are given in table below.
Static Lateral Force Procedure
Table Storey shears.Level
x hx (ft) wx(kip) wxhx (ft-kip) wxhx /(Σwihi) Fx (kip)
5 60 700 42000 0.304 1174 48 800 38400 0.278 1073 36 800 28800 0.209 802 24 800 19200 0.139 531 12 800 9600 0.070 27
Σwihi = 138000 Check ΣFx =V = 383 kip OK
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution: Vertical distribution of base shear to stories
Static Lateral Force Procedure
Table: Storey shears.Level
x hx (ft) wx(kip) wxhx (ft-kip) wxhx /(Σwihi) Fx (kip)
5 60 700 42000 0.304 1174 48 800 38400 0.278 1073 36 800 28800 0.209 802 24 800 19200 0.139 531 12 800 9600 0.070 27
Σwihi = 138000 Check ΣFx =V = 383 kip OK
800 kip
800 kip
800 kip
800 kip
700 kip
383 kips
117 kips
107 kips
80 kips
53 kips
27 kips
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Reversal of stresses takes placeduring earthquake shaking, fig.
Earthquake shaking reversestension and compression inmembers
Reinforcement is required onboth faces of members.
Gravity vs. Earthquake Loading in Reinforced Concrete Building
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
The principal goal of the Special Provisions is to ensure adequatetoughness under inelastic displacement reversals brought on byearthquake loading.
The provisions accomplish this goal by requiring the designer toprovide for concrete confinement and inelastic rotation capacity.
No special requirements are placed on structures subjected to low orno seismic risk.
Structural systems designed for high and moderate seismic risk arereferred to as Special and Intermediate respectively.
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ACI Special Provisions for Seismic Design
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Based on moment resisting capacity, there are three types of RCframes,
SMRF (Special Moment Resisting Frame),
IMRF (Intermediate Moment Resisting Frame),
OMRF (Ordinary Moment Resisting Frame).
Some general requirements will be presented first, which arecommon to all frames. Specific requirements for each type of frameare presented later on.
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ACI Special Provisions for Seismic Design
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General Requirements
Concrete in members resisting earthquake induced forces
Min f’c = 3000 Psi (cylinder strength) for all types
No maximum limit on ordinary concrete
5000 psi is maximum limit for light weight
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ACI Special Provisions for Seismic Design
4/25/2012
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General Requirements
Reinforcement in members resisting earthquake induced forces
Grade 60, conforming to ASTM A 706 (low alloy steel)
Grade 40 or 60, conforming to ASTM A 615 (billet steel) provided that
Fy (actual) – Fy (specified) ≤ +18 Ksi
Actual Ultimate / Actual Fy ≥ 1.25
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ACI Special Provisions for Seismic Design
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General Requirements
Hoops, Ties and Cross Ties
Confinement for concrete is provided by transverse reinforcementconsisting of stirrups. hoops, and crossties.
To ensure adequate anchorage, a seismic hook (shown in figure) is usedon stirrups, hoops and crossties .
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ACI Special Provisions for Seismic Design
(Seismic Hook)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General Requirements
Hoops, ties and Crossties: Advantages
Closely spaced horizontal closed ties in column help in three ways:
i. they carry the horizontal shear forces induced by earthquakes, and therebyresist diagonal shear cracks,
ii. they hold together the vertical bar and prevent them from excessivelybending outwards (in technical terms, this bending phenomenon is calledbuckling), and
iii. they contain the concrete in the column. The ends of the ties must be bentat 135° hooks. Such hook ends prevent opening of hoops andconsequently buckling of concrete and buckling of vertical bars.
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ACI Special Provisions for Seismic Design
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provisions for Flexural Members
These provision applies to flexural members with:
Factored axial compressive force ≤ Agf’c/10.
Note: These provisions generally apply to beams because axial load on beamsis generally 0 or less than Agfc′/10.
However they are also applicable to columns subjected to axial load lessthan Agfc′/10.
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ACI Provisions for Special Moment Resisting Frames (SMRF)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provisions for Flexural Members
1. Size: The members must have:
a. clear span-to-effective-depth ratio of at least 4, (Ln/d ≥ 4)
e.g., for Ln = 15 ft, d = 16″, Ln/d = 15 × 12/16 = 11.25 > 4, O.K.
b. a width-to-depth ratio of at least 0.3, b/d ≥ 0.3
e.g., for width (b) = 12″ and depth (h) = 18″, b/h = 12/18 = 0.67 > 0.3, O.K.
c. A web width of not less 10 inches.
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ACI Provisions for Special Moment Resisting Frames (SMRF)
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provisions for Flexural Members
2. Flexural Reinforcement
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ACI Provisions for Special Moment Resisting Frames (SMRF)
Asl−
Asl+ ≥ (Asl−)/2
As− or As+ (at all section) ≥ (maximum of As at either joint)/4
Asr−
Asr+ ≥ (Asr−)/2
ρmin = 3√f’c/fy, 200/fy (at critical sections)
ρmax = 0.025 (at critical sections)
Min. 2 bars continuous at all sections
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provisions for Flexural Members
3. Transverse Reinforcement
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ACI Provisions for Special Moment Resisting Frames (SMRF)
s ≤
d/48 × smallest longitudinal bar diameter24 × hoop bar diameter12”≤ 2”
≥ 2h ≥ 2hs ≤ d/2
Column
Column
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provisions for Flexural Members
4. Lap Splice
68
ACI Provisions for Special Moment Resisting Frames (SMRF)
Lap splice length =1.3 ld = 1.3× 0.05 (fy/ √fc′)db
50 db for fc′ 3 and fy 40 ksi
70 db for fc′ 3 and fy 60 ksi
Spacing of stirrups Least of d/4 or 4 inches
Lapping prohibited in regions where longitudinal bars can yield in tension
Lapping of Longitudinal bars
≥ 2h ≥ 2h
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
69
Mechanical Splice of Longitudinal Reinforcement
Mechanical Splices shall conform to 21.2.6.
Section 21.2.6 says that welded splice shall conform to 12.14.3.2which states “A full mechanical splice shall develop in tension orcompression, as required, at least 125 % of the specified yieldstrength (fy) of the bar.
ACI Provisions for Special Moment Resisting Frames (SMRF)
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
70
Welded Splice of Longitudinal Reinforcement
Welded Splices shall conform to 21.2.7.
Section 7.3.6 says that welded splice shall conform to 12.14.3.4which states “ A full welded splice shall develop at least 125 % ofthe specified yield strength (fy) of the bar.
ACI Provisions for Special Moment Resisting Frames (SMRF)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Frame Members Subjected to Bending and Axial Load
The provision applies to members with:
Factored axial compressive force > Agf’c/10
71
ACI Provisions for Special Moment Resisting Frames (SMRF)
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Frame Members Subjected to Bending and Axial Load
1. Size
a) Each side at least 12 inches
b) Shorter to longer side ratio ≥ 0.4.
i.e. 12/12, 12/18, 12/24 OK; but 12/36 not O.K
72
ACI Provisions for Special Moment Resisting Frames (SMRF)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Frame MembersSubjected to Bending and AxialLoad
2. Longitudinal Reinforcement
73
ACI Provisions for Special Moment Resisting Frames (SMRF)
0.01 ≤ ρg ≤ 0.06
Clear span, hc
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for FrameMembers Subjectedto Bending andAxial Load
3. Trans. Rein.
74
ACI Provisions for Special Moment Resisting Frames (SMRF)
h2
h1
s ≤ 0.25 × (smaller of h1 or h2)
6 × long. bar dia.so
s ≤ 6 × long. bar dia.6”
lo
lo ≥ Larger of h1 or h2
Clear span/618”
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Frame Members Subjected to Bending and Axial Load
3. Trans. Rein.
75
ACI Provisions for Special Moment Resisting Frames (SMRF)
4” ≤ so = 4 + [(14 – hx)/3] ≤ 6”
hx = max. value of hx on all column faces
hx ≤ 14”
hx hx hx
hx
hx
6db ≥ 3” 6db extension
Alternate90-deg hooks
Provide add.trans. reinf. if thickness > 4”
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Frame Members Subjectedto Bending and Axial Load
4. Lap Splice
76
ACI Provisions for Special Moment Resisting Frames (SMRF)
Tension lap splicewithin center half ofmember length
Lap splice length =1.3 ld = 1.3 × 0.05 (fy/ √fc′)db
50 db for fc′ 3 and fy 40 ksi
70 db for fc′ 3 and fy 60 ksi
Spacing of ties in lap splice not more than smaller of d/4 or 4″
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Joints of Special Moment Frame
ACI Provisions for Special Moment Resisting Frames (SMRF)
Beam
Column
Beam Column Joint
Column ties (with 135o) hook continued through joint(ACI 21.5.2)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Joints of Special Moment Frame
Successful seismic design of frames require that the structures beproportioned so that hinges occur at locations that leastcompromise strength. For this, “weak Beam-strong column”approach is used.
After design, the member capacities are calculated based ondesigned section.
At joint, Column flexural capacity ≥ 1.20 x Beam flexural capacity
78
ACI Provisions for Special Moment Resisting Frames (SMRF)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
79
Joints of Special Moment Frame
Minimum Flexural Strength of Column at Joint
M+nc,t
M-nc,b
M+nb,r
M-nb,l
M+nc,b + M-
nc,t ≥ 6(M+nb,l + M-
nb,r)/5
M-nc,t
M+nc,b
M+nb,l M-
nb,r
M-nc,b + M+
nc,t ≥ 6(M-nb,l + M+
nb,r)/5
ACI Provisions for Special Moment Resisting Frames (SMRF)
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Joints of Special Moment Frame
To prevent beam column joint cracking, ACI Code 21.5.1requires that the column dimension parallel to the beamreinforcement must be at least 20 times the diameter of thelargest longitudinal bar.
80
ACI Provisions for Special Moment Resisting Frames (SMRF)
Beam longitudinal reinforcement with diameter (db)
20×db
Beam
Column
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Joints of Special Moment Frame
Beam longitudinal reinforcement that is terminated within acolumn. must be extended to the far face of the column core.The development length (ldh) of bars with 90° hooks must be notless than 8db, 6 inch, Or fydb/ (65 √ fc′).
81
ACI Provisions for Special Moment Resisting Frames (SMRF)
Beam longitudinal reinforcement
Beam
Column
ldh
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Flexural Members
1. Size: No special requirement (Just as ordinary beamrequirement).
2. Flexural steel: Less stringent requirement as discussed next.
3. Transverse steel: Same as for SMRF.
4. Lap: No special requirement (Just as ordinary beamrequirement).
82
ACI Provisions for Intermediate Moment Resisting Frames (IMRF)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provisions for Flexural Members
2. Flexural Reinforcement
83
ACI Provisions for Intermediate Moment Resisting Frames (IMRF)
Asl−
Asl+ ≥ (Asl−)/3
As− or As+ (at all section) ≥ (maximum of As at either joint)/5
Asr−
Asr+ ≥ (Asr−)/3
ρmin = 3√f’c/fy, 200/fy (at critical sections)
εt ≥ 0.004
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Columns
1. Size: No special requirement (Just as ordinary columnrequirement)
2. Flexural steel: No special requirement (Just as ordinary columnrequirement)
3. Transverse steel: Less Stringent requirement as given next.
4. Lap: No special requirement (Just as ordinary columnrequirement)
84
ACI Provisions for Intermediate Moment Resisting Frames (IMRF)
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43
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Provision for Columns
85
ACI Provisions for Intermediate Moment Resisting Frames (IMRF)
h2
h1
≤ so/2
Trans. reinf. based on Mn and factored tributary gravity load
so ≤
8 × smallest long. bar dia.24 × tie bar dia.0.5 × min. (h1 or h2)12”
lo
lo ≥ Larger of h1 or h2
Clear span/618”
s ≤ d/2 (As per ACI 11.5.4)
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
IMRF are not allowed in regions of high seismic risk, however,SMRF are allowed in regions of moderate seismic risk.
Unlike regions of high seismic risk, two way slab system withoutbeams are allowed in regions of moderate seismic risk.
In regions of low or no seismic risk ordinary moment resistingframes OMRF are allowed but IMRF and SMRF may also beprovided.
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Miscellaneous Considerations
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 1: Detail the selected frame of E-W interior frame of thegiven structure as per SMRF requirements. The structure isalready designed for the following seismic zone data.
Seismic zone: 4
Magnitude of earthquake ≥ 7.0
Slip rate ≥ 5.0
Closest distance to known seismic source > 15 km.
Soil type: SD (stiff).
Concrete compressive strength = 3 ksi,
Steel yield strength = 40 ksi
Modulus of elasticity of concrete = 3000 ksi.
87
Design Example
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Given 3D structure:
88
Design Example
20 ft 20 ft 20 ft 20 ft
15 ft
15 ft
15 ft
10.5 ft
10.5 ft
10.5 ft (floor to floor)
SDL = 40 psfLL = 60 psf
SDL = 40 psfLL = 60 psf
SDL = 40 psfLL = 60 psf
fc′ = 3 ksify = 40 ksi
Slab-Beam Frame Structure
Beams: 15″ × 18″ (d =15 inch)Columns: 15″ square
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design Example
Selected portion of E-W interior Frame:
89
20 ft 20 ft 20 ft 20 ft
Portion of frame considered
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design Example
Reinforcement from Design:
90
For the beam in the given frame, we haveAsmin = 0.005 x15 x 15 = 1.125 in2
Asmax = 0.025x15 x 15 = 5.625 in2
Negative reinforcement at the ends and positive reinforcement at the mid span must be greater than Asmin and reinforcement at all locations must be less than the Asmax
This is OK for the values on the given beam.For the column, we haveAsmin = 0.01 x15 x 15 = 2.25 in2
This is also OK.20 ft 20 ft 20 ft 20 ft
1.61 1.611.02 1.021.21
2.25 2.25
Reinforcement in in2
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design Example
Calculation of number of bars:
91
20 ft 20 ft 20 ft 20 ft
No. of bars = As/Ab
Use No. 5 bar,Negative reinforcement at joint:Left joint:1.61/0.31= 5.19 (take 6 bars in 2 layers)Right joint:1.61/0.31= 5.19 (take 6 bars in 2 layers)Positive bars (mid span):1.21/0.31 = 3.9 (take 4 bars in 1 layer)Positive bars (at joint):1.01/0.31 = 3.29 (take 4 bars in 1 layer)Column reinforcement:2.25/ 0.31 = 7.25 (take 8 bars for even distribution of bars at all faces of column)
6 bars 6 bars4 bars 4 bars4 bars
8 bars 8 bars
No. of #5 bars
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
SMRF Requirements Checklist
Provisions for Beams
Sizes
ln/d = 20 × 12/15 = 16 > 4 (ACI 21.3.1.2 satisfied)
Width/ depth = 15/18= 0.833 > 0.3 (ACI 21.3.1.3 satisfied)
Width = 15″ > 10″, O.K.
Therefore 12″ wide and 18″ beams are OK
Design Example
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
SMRF Requirements Checklist
Provisions for Beams
Flexural Reinforcement
Design Example
As+ (at joints) ≥ ½ As− (at joints)4 #5 bars ≥ ½ (6 #5 bars)
As (any section) ≥ ¼ Max. As at joints2 #5 bars ≥ ¼ (6 #5 bars)OK OK
Asl− = 6 #5 Asr− = 6 #5
Asmid+ = 4 #5Asl+ = 4 #5 Asr+ = 4 #5
93
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
SMRF Requirements Checklist
Provisions for Beams
Transverse Reinforcement
Design Example
s ≤
d/4 = /4 = 3.75″8 × smallest long. bar dia.= 8 × 5/8= 5″24 × hoop bar diameter = 24 × 3/8= 9″12”≤ 2”
2h = 36″s ≤ d/2 = 15/2 = 7.5″
Column
2h = 36″
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
SMRF Requirements Checklist
Provisions for Beams
Lap Splice: If required then,
Not to be provided within joints. Not to be provided within 2h region from face ofthe support.
Spacing of hoops within lap = least of d/4 or 4″ c/c = 3.75″ c/c
Lap splice length =1.3 ld = 1.3×0.05 (fy/ √fc′)db ≈ 30″ = 2.5′
50 db = 50 × (5/8) = 31.25″ ≈ 2.5′ for fc′ 3 and fy 40 ksi
Design Example
2h=36″ 2h=36″
95
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
SMRF Requirements Checklist
Provisions for Columns
Size: All columns are 15″ square, therefore the size is more than the
least required for SMRF (i.e., 12″).
Flexural Reinforcement: All columns are reinforced with 8 #5 barswhich gives ρg = 0.011, within the specified range 0.01 ≤ ρg ≤ 0.06.
Design Example
96
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSMRF RequirementsChecklist
Provision for Columns
Transverse Reinforcement:
lo = max (larger columndimension, hc/6, 18″) = 18″
Spacing of ties in lo region isleast of = smaller columndimension/4, 6 × long bardia = 3.75″
Spacing in the remaining
region will be least of 6 ×
long bar dia or 6″ = 3.75″
hc = 8.5′
lo
lo
15” × 15” column
8, #5 bars
97
#3 @ 3.75” c/cthroughout height
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSMRF Requirements Checklist
Provisions for Columns
Lap Splice:
Tension lap splice within centerhalf of member length.
Spacing of ties in lap splice notmore than smaller of d/4 or 4″.Continue 3.75 inch spacing.
Lap length = 1.3 × 0.05 (fy/ √fc′ )db=30″ ≈ 2.5′
And from 50db = 50×(5/8) =
31.25″
hc = 8.5′
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSMRF Requirements Checklist
Provision for Joints
To prevent beam column joint cracking, ACI Code 21.5.1 requires thatthe column dimension parallel to the beam reinforcement must be atleast 20 times the diameter of the largest longitudinal bar.
20 × 5/8 = 12.5″
Column dimension parallel to beam long bar = 15″, OK
99
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSMRF Requirements Checklist
Provision for Joints
6 #5 bars
6 #5 bars2″
ColumnBeam
Joint
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example 2: A two dimensional frame of the building is shown infigure 01. All beams in the frame are 12″ wide and 18″ deep. Allcolumns are 12″ square. It is required only for frame BFGC to:
a. Provide suitable number of bars in beams and columns for which flexuralreinforcement is shown in the figure (Use only #5 bars as main reinforcementand #3 bars as transverse reinforcement).
b. Provide suitable spacing for stirrups and ties when shear reinforcement asper design is 0.23 in2/ft in all parts of the frame.
c. Satisfy all SMRF requirements for beams and columns.
d. Present appropriately proportioned structural details of beam and column.Also draw the beam to column joint detail at detail X as shown in fig. 01.Take fc′ = 3 ksi and fy = 40 ksi.
Design Example
101
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Given Building:
Design Example
Figure 01: RC Building Frame
3.2 in2
2.4 in2
1.2 in2
2.6 in2 2.6 in2 2.4 in2
1.2 in21.8 in2
3.0 in2 3.0 in2
Detail X
3.2 in2 3.2 in2 3.2 in2
F G
E H
B CA D
J K
1.44 in2 1.44 in2
1.0 in21.2 in2 1.2 in2
10 ft ln =15 ft 10 ft
10 ft
10 ft
1.2 in2 1.2 in2 1.0 in21.0 in2 1.0 in2 1.0 in2
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (a)
As given in problem, # of bars using #5 bars are as below:
Design Example
2.6 in2 2.6 in2
1.8 in2
3.0 in2 3.0 in2
3.2 in2 3.2 in2
F G
E H
B CA D
J K
10 ft 15 ft 10 ft
1.2 in2 1.2 in2
# of bars = As/Ab
Negative reinforcement at joint:2.6/0.31= 8.38 (take 9 bars in 2 layers)3.0/0.31= 9.67 (take 10 bars in 2 layers)Take 10 bars at joint for beam FG.Positive bars (mid span):1.8/0.31 = 5.80 (take 6 bars in 2 layers)Positive bars (at joint):1.2/0.31 = 3.87 (take 4 bars in 1 layer)Column reinforcement:3.2/ 0.31 = 10.32 (take 12 bars for even distribution of bars at all faces of column)
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (a)
Design Example
(4+2) #5(5+5) #5 (5+5) #5
12 #5 12 #5
F G
E H
B CA D
J K
10 ft 15 ft 10 ft
4 #5 4 #5
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (b)As given in problem, Shear reinforcement spacing with #3 bar is:
Design Example
0.23 in2/ft
F G
E H
B CA D
J K
10 ft 15 ft 10 ft
Spacing= (Ab/As) × 12With #3, 2 leg vertical stirrups (Ab = 0.22 in2)Stirrup spacing for beams:sb = (0.22/0.23) × 12 = 11.47″ c/cAs per ACI minimum reinforcement and maximum spacing requirement, spacing should not exceed:i. Avfy/50bw = 14″ c/cii. d/2 = 16.5/2 = 8.25″ c/ciii. 24″ c/civ. Avfy/{0.75√(fc′)bw = 17.85″ c/cFinally sb = 8.25″ c/cTie spacing for column:s = (0.22/0.23) × 12 = 11.47″ c/cSpacing for tie bars according to ACI 7.10.5.1 is minimum of:16 × dia of main bar =16 × 5/4 =10″ c/c48 × dia of tie bar = 48 × (3/8) =18″ c/cLeast column dimension =12″ c/cFinally sc = 10″ c/c
0.23 in2/ft
0.23 in2/ft
105
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (b)As given in problem, Shear reinforcement spacing with #3 bar is:
Design Example
F G
E H
B CA D
J K
10 ft 15 ft 10 ft
#3, 2 legged @ 8.25″ c/c
#3, ties @ 10″ c/c #3, ties @ 10″ c/c
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (c)
SMRF Requirements Checklist
Provisions for Beams
Sizes
ln/d = 15 × 12/16.5 = 10.9 > 4 (ACI 21.3.1.2 satisfied)
Width/ depth = 12/18 = 0.66 > 0.3 (ACI 21.3.1.3 satisfied)
Width = 12″ > 10″, O.K.
Therefore 12″ wide and 18″ beams are OK.
Design Example
107
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (c): SMRF Requirements Checklist
Provisions for Beams
Flexural Reinforcement
Design Example
Asl− = (5+5) #5Asr− = (5+5) #5
Asmid+ = (4+2) #5Asl+ = 4 #5 Asr+ = 4 #5
As (at all critical sections) ≥ Asmin
As Asmin = 4 #5 bars
As (at any section) ≤ Asmax
Asmax = 0.025bd = 16 #5 barsOK
As+ (at joints) ≥ ½ As− (at joints)4 #5 bars ≥ ½ {(5+5) #5 bars}
As (any section) ≥ ¼ Max. As at jointsProvide at least 3 bar
N.G
Provide at least 5 bars at joint. O.K
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (c): SMRF Requirements Checklist
Provisions for Beams
Recommended Flexural Reinforcement
Design Example
109
Asl− = (5+5) #5 Asr− = (5+5) #5
Asmid+ = (4+2) #5Asl+ = (4+2) #5 Asr+ = (4+2) #5
3 #5
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (c): SMRF Requirements Checklist
Provisions for Beams
Transverse Reinforcement
Design Example
s ≤
d/4 = 16.5/4 = 4.125″ ≈ 4″8 × smallest long. bar dia.= 8 × 5/8= 5″24 × hoop bar diameter = 24 × 3/8= 9″12”≤ 2”
2h = 36″s ≤ d/2 = 16.5/2 = 8.25″
Column
2h = 36″
Note: Spacing calculated in part (b) (i.e., sb = 8.25″)of this problem does not govern for 2h regions.However it governs elsewhere.
110
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (c): SMRF Requirements Checklist
Provisions for Beams
Lap Splice: If required then,
Not to be provided within joints. Not to be provided within 2h region from face ofthe support.
Spacing of hoops within lap = least of d/4 or 4″ c/c = 4″ c/c
Lap splice length =1.3 ld = 1.3×0.05 (fy/ √fc′)db ≈ 30″ = 2.5′
50 db = 50 × (5/8) = 31.25″ ≈ 2.5′ for fc′ 3 and fy 40 ksi
Design Example
2h=36″ 2h=36″
111
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Solution of Part (c): SMRF Requirements Checklist
Provisions for Columns
Size: All columns are 12″ square, which is equal to least required for
SMRF (i.e., 12″).
Flexural Reinforcement: All columns are reinforced with 12 #5 barswhich gives ρg = 0.025, within the specified range 0.01 ≤ ρg ≤ 0.06.
Design Example
112
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (c): SMRF Requirements Checklist
Provision for Columns
Transverse Reinforcement:
lo = max (larger column dimension,hc/6, 18″) = 18.5″
Spacing of ties in lo region is least of =smaller column dimension/4, 6 × longbar dia = 3″
Spacing in the remaining region will be
least of 6 × long bar dia or 6″ = 3.75″
hc = 9.25′
lo
lo
12” × 12” column
12, #5 bars
G.L
113
Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (c): SMRF Requirements Checklist
Provisions for Columns
Lap Splice:
Tension lap splice within centerhalf of member length.
Spacing of ties in lap splice notmore than smaller of d/4 or 4″
Lap length = 1.3 × 0.05 (fy/ √fc′ )db=30″ ≈ 2.5′
And from 50db = 50×(5/8) =
31.25″
hc = 9.25′
lo
lo
12” × 12” column
G.L
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (c): SMRF Requirements Checklist
Provision for Joints
To prevent beam column joint cracking, ACI Code 21.5.1 requires thatthe column dimension parallel to the beam reinforcement must be atleast 20 times the diameter of the largest longitudinal bar.
20 × 5/8 = 12.5″
Column dimension parallel to beam long bar = 12″, Almost OK.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (c): SMRF Requirements Checklist
Provision for Joints
(5+5) #5 bars
(4+2) #5 bars2″
ColumnBeam
Joint
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (d):
Presentation of appropriately proportioned structural details ofbeam, column and selected joint.
In this part, structural drawings showing all calculated SMRF detailshave been asked to draw. This has already been done in part (c) alongwith each member (beam and column) SMRF checks.
However in the examination, it is better to draw all drawings here in part(d) at the same place to avoid waste of time.
Part (d) is shown next.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (d):
Structural Details for beam (Long section):
s = 4″
≤ 2”
2h = 36″s = 8.25″
Column
2h = 36″
(5+5) #5(5+5) #5
(4+2) #5(4+2) #5 (4+2) #5
Note: Lap splice is provided only if required.
A
A
B
B
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (d):
Structural Details for Beam (Cross sections):
12″
18″
(5+5) #5
(4+2) #5
Section A-A
12″
18″
3 #5
(4+2) #5
Section B-B
#3, 2 legged @ 4″ c/c
#3, 2 legged @ 8.25″ c/c
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (d):
Structural Details for Column:
lo = 18.5″
12” × 12” column
12, #5 bars
G.Llo = 18.5″
#3 ties @ 3″ c/c (throughout)
Lap length = 2.5′
12 #5
#3ties @ 3″ c/c
12″
12″
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Design ExampleSolution of Part (d):
Structural Details for Joints:
(5+5) #5 bars
(4+2) #5 bars2″
ColumnBeam
Joint
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
ReferencesACI 318UBC-97BCP SP-2007Earthquake tips from IITK.
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Prof. Dr. Qaisar Ali CE 404 Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
The End
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