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Application of Strut and Tie
Models to PrestressedMembers Anchorages
Presented by: Antonio SerbiUniversity of South Florida
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Content:
Problem DefinitionStress Distribution TheoryB & D RegionsStrut & Tie Model Theory
S & T Model Application to Anchorage ZoneQ & A
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Stress Distribution Theory:
St. Venant's Principle - states that " the localizedeffects caused by any load acting on the body willdissipate or smooth out within regions that aresufficiently away from the location of the load
"
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Stress Distribution Theory:
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B and D regions :
D region B region
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Strut & Tie Theory :
Concept - Use of uniaxially stressed trussmembers to model stressflow
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Strut & Tie Theory :
Elements of a Strut & Tie Model:
Nodes - Concrete
Struts - Concrete Ties - Steel
P
P/2
P/2
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Strut & Tie Theory :
Geometry of a Strut & Tie Model:
P
P/2
P/2
F
T=P/2 tan (F
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Strut & Tie Theory :
Important Considerations:
Equilibrium must be maintained
Tension in concrete is neglected
Forces in struts and ties are uni-axialExternal forces applied ONLY at nodesPrestressing is treated as a loadDetailing for adequate anchorage (detailing)
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Strut & Tie Theory :
P
P/2
P/2
Stress - Strain Compatibility RelationConcrete StressMild Steel StressPT Bars...
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Strut & Tie Theory :
Constructing the Model:Sketch Force FlowDetermine Truss GeometryDetermine Forces
Node SizeStrut SizeTie Location
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Strut & Tie Theory :Sketch Force Flow:
Join St. Venants stressareas with Bernoullistress areas.(For ALL Cases)
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Strut & Tie Theory :Truss Geometry:
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Strut & Tie Theory :Truss Geometry:
Use Direct Load Paths
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Strut & Tie Theory :Size Nodes:
CCC NodeStress = 0.85 ffc
CCT NodeStress = 0.75 ffc
f= 0.70 (Concrete Bearing)
CTT NodeStress = 0.60 ffc
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Strut & Tie Theory :Validate Model:
P
P/2
P/2
F T=P/2 tan (F )
C=(P/2)
/cos(F
)
ITERATIVEPROCESS
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Anchorage Zone :Design Example (Collins & Mitchell Ex 9.7):
DATA:1/2 in. diameter, Lo-Lax strandsfpu = 270 ksi0.75 fpu at jackingfc = 5000 psi
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Anchorage Zone :Design Example, cont. (Collins & Mitchell Ex 9.7):
1. Find PP = 4 x 0.153 x 270 = 165 kip
2. Check Bearing Pressure Under PlatesA = 7 x 7 - p x 22/4 = 45.9 in2
fb= 165/45.9 = 3.60 ksiAllowable Nodal Stress = 0.85 f fc
= 0.85 x 0.9 x 5= 3.83 ksi > 3.60 ; OK
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Anchorage Zone :Design Example, cont. (Collins & Mitchell Ex 9.7):
3. Bernoulli Stresses and Forcesfc= 4 strands x 165 kip / 966 in
2= 0.683 ksi
A of flanges = 40 x 7 = 280 in2A of web = 7 x 58 = 406 in2
Force in flanges = 240 x 0.683= 191 kipForce in web = 406 x 0.683= 277 kip
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Anchorage Zone :
(c) Plan
Forces:(kip)
CD = 126.7DE = 173.9
GH = 39.4
Design Example, cont. (Collins & Mitchell Ex 9.7):
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Anchorage Zone :Design Example, cont. (Collins & Mitchell Ex 9.7):
6. Detail Reinforcement#4 closed stirrups can resist:Fs = 0.9 x 60 x (2 x 0.20) = 21.6 kip
Member DE (web), T = 173.9 kip therefore173.9/21.6 = 8.1, use 9 closed #4 stirrups
Member GH (flange), T = 39.4 kip therefore39.4/ 10.8 = 3.6, use 4 (single leg) #4
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Anchorage Zone :Design Example, cont. (Collins & Mitchell Ex 9.7):
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Anchorage Zone :Design Example, cont. (Collins & Mitchell Ex 9.7):
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