Post on 04-Apr-2018
7/31/2019 Basic Structural Design Consideration
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By . Professor,
Department of
Civil
EngineeringBUET
Consultant
7/31/2019 Basic Structural Design Consideration
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BUILDING DESIGN
ARCHITECTURAL DESIGN STRUCTURAL DESIGN
AESTHETICS
STRENGTH
FUNCTIONALITY
SERVICEABILITY
STABILITY DEFLECTION
DURABILITY VIBRATION
ECONOMIC
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DEAD LOAD
WIND LOAD
LIVE LOAD EARTHQUAKE LOAD
OTHER LOAD
SOIL PRESSURE
RAIN LOAD
SNOW LOAD
TEMPERATURE LOAD
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LOAD COMBINATION
ACCORDING TO BNBC THERE ARE 26 LOAD COMBINATION
1.4 DL
1.4DL+1.7LL
1.05DL+1.275WLY
1.0 DL‐1.27 WLY 1.3305DL‐1.4025EQLY+1.275LL
1.3305DL+1.4025EQLY+1.275LL
1.05DL+1.275LL+1.275WLX
‐
0.9DL+1.3WLX 1.3305DL+1.4025EQLX
1.3305DL‐1.4025EQLX . . .
1.05DL+1.275LL+1.275WLY
0.9 ‐1.3
0.9DL+1.3WLY
1.3305DL+1.4025EQLY
1.3305DL‐1.4025EQLY
1.05DL+1.275LL‐1.275WLY 0.9DL‐1.3WLY
1.3305DL+1.4025EQLX+1.275LL1.05DL+1.275WLX
0. 14 +1.430.614DL‐1.43EQLX
1.05DL‐1.275WLX 1.3305DL‐1.4025EQLX+1.275LL
. .
0.614DL‐1.43EQLY
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LOAD TRANSFER MECHANISM
XIALLY
FLEXURAL
SHEAR + FLEXURE
PURE SHEAR
AXIAL + FLEXURE
TORSION
TORSION + SHEAR
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MATERIALS PROPERTIES
STRESS‐STRAIN PROPERTIES
RUPTURE/ FAILURE PROPERTIES
FATIGUE
BEARING CAPACITY
CONSOLIDATION
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STRESS‐STRAIN PROPERTIES
Concrete Stress
‐Strain
Curve
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Stress‐Strain Curve For Steel
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Creep and Stress Rupture Properties
reep ropert es
Creep is a time‐dependent deformation of a material
while under an a liedload that is below its yield
strength. It is most often
occurs at elevated ,
materials creep at room
temperature. Creep
terminates in rupture if
steps are not taken tobring to a halt.
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of cyclic loading in the elastic regime. Failure is the end result of a process involving the initiation and growth of a crack, usually at the site of a stress
concentration on the surface.
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AXIALLY LOADED MEMBER
Short Reinforced Concrete Compression y Short ‐ slenderness does not need to be considered –
column will not buckle
y Only axial loadCross‐sectional Areas:
s
=
A c = Area of concrete A g = Total area
F = stress in steel
Fc = stress in concrete
From Equilibrium:P = A cf c + A sf s
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Short Concrete Columns
For ductile failure – must assure that steel
crushes.Strain in steel at yield ~0.002ε = 0.002 corresponds to max. stress in concrete.Concrete crushes at a strain ~ 0.003
= ’ s y c c
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Reinforcement Ratio
y = A A
y ACI 318 limits on ρ for columns: 0.01≤ρ≤0.08 (practical ρmax = 0.06)
y Substitute ρ=A s/A g and A g=A s+A c into equilibrium
equation:P = A g[ρf y +f’c(1‐ ρ)]
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Short Concrete Columns
P = A g[ρf y +f’c(1‐ ρ)]Safet Factors
y Resistance factor, Ф = 0.65 (tied), Ф = 0.70 (spiral)
y ’ c . , ,
y Stray moment factor for columns, K1
y = 1
y K1=0.85 for spiral reinforcement
ФPn = Ф K1 A g[ρf y +0.85f’c(1‐ ρ)]
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Short Column Design Equation= ’ ‐ 1
, u n
⎤⎡ −uP '1
⎥⎦⎢⎣−c
gc y K f f .
)'85.0( 1φ
[ ])1('85.01
ρ ρ φ −+≥c y
ug
f f K A
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Transverse Reinforcement
Used to resist bulge of concrete and buckling of steel
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LATERAL TIES
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FLEXURAL MEMBER
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Data:y
Rectangular Beam Design
y Material properties – f’c, fyy All section dimensions – b and h
Required:-
1. Calculate the dead load and find Mu
2. d = h – cover – stirrup – db /2 (one layer)
and find As
4. Use As to find a
5. Use a to find As (repeat…)
6. Choose bars for As and check ρ max & min
7. Check Mu<φ Mn (final condition)
8. Desi n shear reinforcement stirru s
9. Check deflection, crack control, steel
development length.
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PURE SHEAR
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TORSION
TORSIONAL SPECIMEN
TORSIONAL STRESS DISTRIBUTION