SAFE & ECONOMIC DESIGN OF R/C STRUCTURES by Use of …

SAFE & ECONOMIC DESIGN

OF R/C STRUCTURES

by

Use of the Proper Bending Moments

Prof. Dr.-Ing. habil. Piotr Noakowski Technische Universität Dortmund, Exponent Industrial Structures Düsseldorf

JUBILEE SCIENTIFIC CONFERENCE “PRACTICAL APPLICATIONS OF INNOVATIVE SOLUTIONS RESULTING FROM SCIENTIFIC RESEARCH”

THE OBJECTIVE

How to make the r/c structures

safer and more economic by accepting

the stiffness drop due to cracking?


THE THESIS

Determination of the bending moment

by neglecting cracking

and then using these moments for the design

is a very wrong approach!


Continuous Deformation Theory - CDT

CONTENT

Considered Structure 5 Figures

Phenomenon of Redistribution 4 Figures

Continuous Deformation Theory 4 Figures

Structure Modeling 2 Figures

Structure Analysis 5 Figures

Findings 1 Figures

----------------------------------------------------------

21 Figures

ANALYZED STRUCTURE

Front View

ANALYZED STRUCTURE

Building body


0.2

2

ANALYZED STRUCTURE

Slab-Column System 3

.14

Slab

Column


ANALYZED STRUCTURE

Bending Moments + Tension Forces

Tension due to

constraint shrinkage

Staircase


q

ΔT

Mq

MΔT

PHENOMENON of REDISTRIBUTION

Beam loaded by q and DT

q

DT

Mq

MΔT


Moment drop caused by a total cut

Mq

MΔT

q

DT


Moment drop caused by a partial cut


Cracks causing moment redistribution

k1

km

k2

km

M

N

Computation node

CONTINUOUS DEFORMATION THEORY

Computation node


not cracked section

cracked section

computation node

k

Mcr


Deformation Low M - k

EI(Ec)

EI(r, N)

first cracks

steel yielding


e w

Bond Stresses t

t

w

Steel Strains e

Compatibility Se = d ●

Equilibrium St Cs = s As ● ● t = f(d) Bond law

Steel Displacements d

a


General relations

(1) Bond equation

[t(y)/(A fcm2/3)]1/N = 4/(ds Es)

t(y) dy dy

(2) Bond distribution

t(y) = k yp

(3) Integration

[k yp/(A fcm2/3)]1/N = 4/(ds Es) k/[(p+1) (p+2)] y(p+2)

(4) Solving for p

p/N = p+2 p = 2 N/(1-N)

(5) Solving for k

[k/(A fcm2/3)]1/N = 4/(ds Es) k/[(p+1) (p+2)]

k = [2 (1-N)2/(1+N) (A fcm2/3)1/N/(ds Es)]

N/(1-N)


Bond Stress Distribution

STRUCTURE MODELING

Grid System, 92 nods, 139 bars

STRUCTURE MODELING

Slab System, 633 elements


N

10

1.0

0.5

ρ Case 1 2 3 4

ΔT [K] 15 15 15 15

q [kN/m] 20 40 20 20

N [kN] 0 0 600 0

ρ [%] 0.4 0.4 0.4 1.2

N

10

ρ

STRUCTURE ANALYSIS

Behavior of a beam loaded by DT, q and N

DT N q


-149

-176

-349

-1.2

-2.1

-250

STRUCTURE ANALYSIS

Case 1: Standard

ΔT = 15 K

q = 20 kN/m

N = 0

ρ = 0.4 %

Mq [kNm]

MDT [kNm]

d [mm]

-40%

-50%

+75%


-425

-116

-349

-1.9

-6.3

-499

0

Mq [kNm]

MDT [kNm]

d [mm]

STRUCTURE ANALYSIS

Case 2: Load

ΔT = 15 K

q = 40 kN/m

N = 0

ρ = 0.4 %

-15%

-67%

+232%

-213

-66

-349

-1.2

-5.6

-250

0

STRUCTURE ANALYSIS

Case 3: Tension

ΔT = 15 K

q = 20 kN/m

N = 600 kN

ρ = 0.4 %

Mq [kNm]

MDT [kNm]

d [mm]

-19%

-81%

+366%

-187

-262

-409

1.1

1.6

-250

0

ΔT = 15 K

q = 20 kN/m

N = 0

ρ = 1.2%

STRUCTURE ANALYSIS

Case 4: Reinforcement

Mq [kNm]

MDT [kNm]

d [mm]

-25%

-36%

+45%


FINDINGS

Cracking is an important phenomenon since

..the design moments M are decreased but

..the deflections d are increased.

Cracking is also a “mysterious” phenomenon since

..the moments M depend on q, N and r but

..the deflections d do not depend on Ec but on r & N

Finally, the determination of the design moments

without considering the stiffness drop due to cracking

is a joke!


Thank you very much

TV Tower Düsseldorf

QUESTIONS?

SAFE & ECONOMIC DESIGN OF R/C STRUCTURES by Use of …

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