Responses of structural materials Mechanical ... courses/0_fundamentals of... · Department of Mechanics, Materials and ... lateral support against buckling. ... FUNDAMENTALS OF STRUCTURES

FUNDAMENTALS OF STRUCTURES 2018 STRUCTURAL MATERIALS page 1

Budapest University of Technology and Economics

Department of Mechanics Materials and Structures English courses General course 2018 Fundamentals of Structures BMEEPSTG201

Responses of structural materials Mechanical characteristics testing strength

evaluation


Introduction

Structural materials are used to construct the loadbearing structural part of building constructions

Loadbearing structures of buildings should safely support loads acting on

the structure (without damage rupture loss of stability falling over or down) and transmit loads to the subsoil under the building

Their most important characteristic is having adequate strength

Strength is the greatest force per unit area that the material can bear

without damage rupture

Tensile strength compression strength and shear strength can be

distinguished according to the way of actuation the force is acting on to the cross-section of the investigated member


Normal stress and normal strength

Normal stress σ (Nmm2) force per unit area acting perpendicular to the plane

under consideration It can be compression or tensile stress

Normal strength f (Nmm2) capacity force per unit area acting perpendicular

to the plane under consideration It can be compression or tensile strength


Shear stress and shear strength

Shear stress τ (Nmm2) force per unit area acting in the plane under

consideration Shear strength τR (Nmm

2) capacity force per unit area acting in the plane

under consideration


Stresses and deformations

Tension Compression

Compression stress provokes shortening

Tensile stress provokes elongation

Torsion

Shear stress ndash due to torsion for example ndash provokes distortion


Stress-deformation diagrams

visualize the better the mechanical behaviour of the structural materials

Strain is the specific deformation ε (mmm=o) = 119897

119871 or deformation

(that is elongation or shortening) per unit length

Stress-strain diagrams σ=f(ε)


General requirements of structural materials

structural requirements non-structural requirements

strength space limitation

tensile strength (ft) thermal insulation

compression strength (fc) sound insulation shear strength (fτ) economy prize

deformability density specific weight (γ)

elastic elastic-plastic plastic brittle behaviour ultimate deformation

fire resistance

durability resistance to corrosion re-modelling possibilities


Mechanical characteristics of some structural materials

Materials Strength Deformability fd γ ratio

compression tension at σ=fd at rupture (kNm2)(kNm

3)=

Nmm2

mmm=o =m

Brick walls ceramic brick+lime mortar

02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320

Timber (parallel to fibre direction)

15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500

high strength steel 1400-1800 1400 1800 7-9 15 17200

Reinforced concrete see concrete see concrete 1200

1kNcm2 = 10 Nmm

2

Max column length that the material can support its weight In compression continuous

lateral support against buckling


Numerical example

Gmanasymp08 kN= 800 N

Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2

Compare this stress You can feel with material strengths indicated above


1 Uniaxial stress-strain relationships of structural materials

Definitions

Stress = σ = force per unit area = A

F (Nmm

2)

Stress at rupture= strength (compression or tensile strength according to

the direction of F)

Strain = ε = specific deformation =length

ncontractioor alongation=

l

l

Perfectly plastic behaviour the material deforms under compression or tension without stress increment


Characteristic mechanical behaviours in uniaxial stress state

elastic rigid-plastic elastic-plastic nonlinear

Hooks law describes the linear elastic mechanical behaviour

E

where E modulus of elasticity (Nmm2)

Definition of E the value of the stress to be applied to double (or ndash in case of compression ndash to halve) the length of the member (ε=1)


Characteristic σ-ε relationships of some important structural materials

timber steel concrete Idealized stress-strain relationships

timber steel concrete


Test specimens used for uniaxial tests



Static bending test of a timber specimen Tension test of steel specimen


Mechanical behaviour of the tested materials

Timber Strength perpendicular to grains is different (very low) Linear elastic

behaviour Strength is similar to compression strength of concrete

Steel

After linear elastic behaviour yield produces great plastic deformation The same strength in compression or tension The strength is approximately

10 times greater than that of timber


Concrete

Compression and tensile strength is very different After short linear

behaviour non-linear σ-ε curve can be observed Ultimate deformation is

much smaller than that of steel Brickwork

Negligible tensile strength The strength of brickwork is much smaller than

that of ceramic bricks Compression strength is at about by one order of magnitude smaller than that of concrete


2 Strength as probability variable

21 Statistical strength data no of occurrence relative occurrence

nom

A=1 An

fnom

strength f (Nmm2) strength

Strength distribution diagram Density function of strength

(histogram) distribution

The probability of not exceeding fnom= nn A

A

A


22 Statistical evaluation of test results (Statistical strength evaluation9

Characteristics of the strength distribution

Mean value fm=n

fi n number of test data (min 10 if

qualification is based on unknown scatter)

Scatter s=1n

)ff( 2mi

Threshold value (characteristic or nominal value)

fk= fnom = fm ndash ts

where t is the so called Student factor which for n=10 and 5 risk is t=179

(5 threshold value the probability of the occurrence of a strength value smaller than fk is 5)


Relation of characteristic and design value

relative occurrence

strength distribution curve

An

fd fk f =fm strength (Nmm2)

design (d) characteristic (k) and mean (m) strengths

Design value fd= m

kf

The safety factor γm for some important structural materials

115 for reinforcement (steel in reinforced concrete)

13 for timber 15 for concrete

The probability of fd not being exceeded is 01


Example test data evaluation of a timber test specimen subjected to axial compression in direction of the grains

Specimen data 32x32x5 cm timber No Cross-section Rupture load Rupture strength

A Nu fu=NuA fui- fm (fui- fm)2

mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829


Mean value of the rupture strength

673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2

The characteristic compression strength

fk= fm- ts= 3267 ndash 179295middot = 2739 Nmm2

for n= 10 and 5 risc the Student factor t=179 Design compression strength

fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2

the safety factor of timber γm= 13


TESTING OF PREFABRICATED REINFORCED

CONCRETE BEAMS IN2005 IN THE LABORATORY

OF THE DEPARTMENT OF STRENGTH OF

MATERIALS AND STRUCTURES










Introduction

Structural materials are used to construct the loadbearing structural part of building constructions

Loadbearing structures of buildings should safely support loads acting on

the structure (without damage rupture loss of stability falling over or down) and transmit loads to the subsoil under the building

Their most important characteristic is having adequate strength

Strength is the greatest force per unit area that the material can bear

without damage rupture

Tensile strength compression strength and shear strength can be

distinguished according to the way of actuation the force is acting on to the cross-section of the investigated member












under consideration



Tension Compression



Torsion

















fire resistance






3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2


























under consideration



Tension Compression



Torsion

















fire resistance






3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2




















under consideration



Tension Compression



Torsion

















fire resistance






3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2

















Tension Compression



Torsion

















fire resistance






3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2






























fire resistance






3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2























fire resistance






3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2



















3)=

Nmm2

mmm=o =m


02-1 0 08 4 120

Concrete 10-50 07-4 1 25-35 320


15-30 15-30 2 4 1900

Steel mild 180-280 180-280 1-2 25 2500



1kNcm2 = 10 Nmm

2




Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2
















Numerical example


Apalm 100x100 =10 000 mm2

soil 1010000

800

A

G

soil

man Nmm2=001kNcm

2




Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2

















Definitions


F (Nmm

2)


the direction of F)



l

l






E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2



















E
















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2




























Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2
























Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2





















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2



















Steel




Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2
















Concrete









nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2


















nom

A=1 An

fnom





A

A




Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2


















Mean value fm=n



Scatter s=1n

)ff( 2mi







relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2

















relative occurrence


An



Design value fd= m

kf









mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2



















mmxmm kN Nmm2 Nmm

2 N

2mm

4

1 32 x 32 350 3418 151 228

2 32 x 32 312 3047 -22 484

3 32 x 32 284 2773 -494 2440

4 32 x 32 375 3662 395 1560

5 32 x 32 348 3398 131 172

6 32 x 32 334 3262 -005 0031

7 32 x 32 372 3633 366 1340

8 32 x 32 294 2871 -396 1568

9 32 x 32 337 3291 024 0058

10 32 x 32 340 3320 053 0281

fm= nfuii =3267510=3267 Nmm2 2

mui )ff( 7829



673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2

















673210

75326m

n

ff u Nmm

2

The scatter

s= 1n

)ff( 2mi

=

9

2978 295 Nmm

2




fd= m

kf

=

31

3927 2107 Nmm

2= 2107 kNcm

2
































































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Responses of structural materials Mechanical ... courses/0_fundamentals of... · Department of Mechanics, Materials and ... lateral support against buckling. ... FUNDAMENTALS OF STRUCTURES