Deformation of Concrete

Deformation of Concrete

Fall 2010

Dept of Architecture

Seoul National University

7 th week

Special cases of uncracked

sections

Chap. 3. Special Cases of uncracked

sections and calculation of displacement

1. Introduction 2. Prestress loss

3. Effect of presence of non-prestressed steel 4. Reinforced concrete section w/o prestress 5. Approximate equations

6. Graphs

7. Multi-stage prestressing

8. Displacement

3.1 introduction

1. One type of concrete

2. Reinforcing bars and prestressing steel

are located in one layer

3.2 prestress loss

Total reinforcement area

st ns ps

A  A  A

Modulus of elasticity

st ns ps

E  E  E

c ns ps

P P P

    

ps ps ps

P A 

  

Equilibrium

ns ns ns

P A 

  

Compatibility condition

ns s

E

st

  

  

Creep + Shrinkage + due to

 P

_c

   

     

2

0 0

0

0 0

, 1

, ,

ps pr cst

ns c c st

cs

st st c c c c

t t t P P y

E E E t t t E t t A I

   

      

  

      

 

       

 

0 0 0 0

2 2 0

, / ,

1 1

,

cst st st c cs st st pr ps

c

st st st

c c

t t t A E E t t t E A A

P A E y

A E t t r

          

  

 

   

 

pr r pr

  

  

Stress change in concrete

c c st

c c

P P y

A I y

  

  

1

2

c st

c

c c

P y y

A r

   

    

 

Since ²

c c c

I  A r



o



ns

E

st

y

st

  

    



_o



ps

E

st

y

st pr

   

      

3.2.1

  and  

       

o 0 0 o 0

0

, ,

,

c cs

c c

t t t t t P

E t t A

    

   

   

^{0 0}

 

0

, ,

c st

c c

t t t P y

E t t I

   

  

,

115 MPa 1700 MPa

ps

f

ptk



_

  

  

3 4

2

2 2

0

42.588*10 m 0.357 m

0.1193 m 1400 kN

390 kN-m 30 GPa

c c c

c

I A r P M E t















Intrinsic relaxation

Characteristic tensile strength

 

1 o

2

1 1

ref ref st

A B N I B N

M Py

B I M B A

E E AI B





 

        

 

           

   

 

3 4

2

2 2

0

42.588*10 m 0.357 m

0.1193 m 1400 kN

390 kN-m 30 GPa

c c c

c

I A r P M E t

 











 

o

2 0

1 1400

390 1400 * 0.454

t ref

I B

B A

E AI B



 _

 

     

        

 

 

0 6.533 MPa

cps t

   Strain distribution right after pressing

       

 

0 0 0 0

2 2 0

, / ,

1 1

,

cst st st c cs st st pr ps

c

st st st

c c

t t t A E E t t t E A A

P A E y

A E t t r

          

  

 

   

 

Assume the relaxation factor



_r

 0.7

0.7( 115) 80.5

pr r pr

  

     

 

, 0 8.824 GPa E t tc 

 

t t, 0 3.0

 

 

2

2 2

0

0.357 m 0.1193 m

30 GPa

c c

c

A r E t







In the absence of non-prestressed steel

c ps

P P

  

3 6

242.7 *10

Pa=216.7MPa 1120 *10

ps c

ps

ps ps

P P

A A

  

_

     

To find an improved estimate the relaxation reduction factor

Relaxation reduction coefficient



_r

pr r pr

  

  

Reduced relaxation

0

216.7 115

0.081 1250

ps pr

p

 



     

          



r is a function of

 and 

p0 ptk

1250 0.706 1770

f

    

r

0.8

 

0.8( 115) 92 MPa

pr r pr

  

  

   

       

 

0 0 0 0

2 2 0

, / ,

1 1

,

cst st st c cs st st pr ps

c

st st st

c c

t t t A E E t t t E A A

P

A E y

A E t t r

      

  

 

   

 

ps c

ps

ps ps

P P

A A

  

   

1

2

c st

c

c c

P y y

A r

   

    

 

       

o 0 0 o 0

0

, ,

,

c cs

c c

t t t t t P

E t t A

    

   

   

^{0 0}

 

0

, ,

c st

c c

t t t P y

E t t I

   

  

3.4 RC w/o prestress

3.5 Approximate equation

3.6 Graph

3.8 Displacement

• 3.8.1 unit load theory

Real Strain and Curvature Distribution

Virtual Axial and Moment Distribution due to unit virtual force at coordinate j



O



N

uj

M

uj

j

1 

O

N dx

uj

 M dx

uj

     

Displacement System Equilibrium System

3.8.2 Elastic weight: Conjugate Beam

 Load 



Shear

Moment

1 2

3 4

 

1

1

S  2 s b a 

 

3

1

S  2 s b c  S

2

 sa

S

4

 sc

1 2

1 3

1 3

1 2 1

2 2 3

2 3

1 2

1 1 2

1 1

2 3

Q  S  S

3 3 4

1 1

2 3

Q  S  S

2 1 2 3 4

1 2 1 2

2 3 2 3

Q  S  S  S  S



²



1 0

1

s x

Q Ax Bx C dx

s

 





    

 

2

2

3 1

s

s

Q Ax Bx C x dx

s

 





    



²



²



²



2 0

2

s s

s

x x

Q Ax Bx C dx Ax Bx C dx

s s

   

          

   

 

y  Ax

2

 Bx C 

2

2 3 3

, ,

2 2

a b c a b c

A B C a

s s

    

  

Assume parabolic variation of strain at O and curvature.

Note 2s=l



1 2 3



¹2

3

7 / 48 / 8 / 48

, , / 24 5 /12 / 24

/ 48 / 24 7 / 48

l l l

Q Q Q l l l

l l l







    

  

      

D

1

D

2

D

4

D

3

Apply a unit moment at the left support

 1

 1/ 2

 

¹

2 2

3

1 0, 0.5, 1

Q

D Q

Q

   

     

   

Moment distribution by a unit load

D

1

D

2

D

4

D

3

Apply a unit moment at the left support

1 1/ 2

3

 

¹2

3

1 1, 0.5, 0 Q

D Q

Q

   

   

   

Apply a unit load at the center

1 4l

1

4 2

3

1 1, , 0

4 l Q

D Q

Q

   

 

         

 

1 1

2 2 2

3 3

7 / 48 1/ 8 1/ 48

1 8 8

0, , 1 1/ 24 5 /12 1/ 24 0, ,

2 24 48

1/ 48 1/ 8 7 / 48

D l l

 

 

 

      

   

    

               

 

^{1 1}

1 2 2

3 3

7 / 48 1/ 8 1/ 48

8 16 8

1,1,1 1/ 24 5 /12 1/ 24 , ,

48 24 48 1/ 48 1/ 8 7 / 48

O O

O O

O O

D l l

 

 

 

      

     

 

                        

Axial deformation



O

Axial force distribution due to a unit load

1.0