Concrete Laboratory Preparation 2010-11

Calculation of design and predictedcracking and ultimate loads

Dr Iman Hajirasouliha

Structures & Materials 3

H23SM3

Outline:

Introduction to the laboratory exercise

Bending behaviour

Measurement of strains and deflections

Calculation of predicted loads

Jobs for you to do BEFORE the laboratory

Laboratory 2 - 4 pm on Tuesdays :

Group A 8 February 2011

Group B 15 February 2011

Group C 22 February 2011

Group D 1 March 2011

Others join group A

Report hand-in 2 weeks later.

Feedback Tuesday 3 May 2011

Location : ground floor L2

La a

P

Four-point Bending Test

Load spreader beam

La a

P

Nominal beam dimension 8” deep x 4” wide(203 x 102 mm)

Effective span approx. L = 1.370 m

La a

P

2 no. H16or H10 bars

25 mmcover tomain bars(not thelinks!)

Nominal beam dimension 8” deep x 4” wide(203 x 102 mm)

Effective span approx. L = 1.370 m

1. Mid-span deflection d

Structural response using 2 methods :

1. Mid-span deflection d

Structural response using 2 methods :

Plot P vs d

Dial Deflection Gauge

2. Curvature

M E s

I R y= =

Divide through by E …

R

(Radius of Curvature )

M 1 eE I R y

= =

1/R = M / EI = strain gradient

Curvature = strain gradient

Measurement of strains through the beam

using DEMEC gauge

DEMEC pips (5 pairs willbe glued to your beams)

Centre-line of beam

Measurement of strains through the beam

using DEMEC gauge

DEMEC pips (5 pairs willbe glued to your beams)

Centre-line of beam

Will be demonstrated in the laboratory

DEMEC gauge

Centre-line of beam

200 mm

0.0034

DEMEC factor: 0.001 = 4 x 10-6 strain

Measure the change in the DEMECreading, converting this to strain

DEMEC pips moveapart = tension

DEMEC pips movecloser = compression

Measure the change in the DEMECreading, converting this to strain

DEMEC pips moveapart = tension

DEMEC pips movecloser = compression

Measure the change in the DEMECreading, converting this to strain

using the instrument’s factor

DEMEC pips moveapart = tension

DEMEC pips movecloser = compression

Plot strains e along y-axis of beam for all5 pairs of DEMEC

X

X

e

y

XX

X

X

X

e

y

XX

X

Draw best straight line through 5 values

X

X

e

y

XX

X

Draw best straight line through 5 values

Slope = e /y = M / EI = Pa / 2EI

Plot P v e /y

P

d or e/y

So now you have graphs of load vsdeflection and strain gradient …

Theoretical Predictions

1.“Design” using EC2 with nominalvalues and safety factors

2.“Fundamental” using actual test data

Applied load P orbending moment

Basic load vs deformation behaviour for RC beam

Actual behaviour in a test

Deflection or e/y

Applied load P orbending moment

Basic load vs deformation behaviour for RC beam

Actual behaviour in a test

Theoreticalprediction

Deflection or e/y

Basic load vs deformation behaviour for RC beam

Why different ?

Deflection or e/y

No cracking

Stiffness based onuncracked cross section Iu

Centroidaxis

Centroid isnot the

same asneutral axis

at ULS !

Mo

me

nt

Deflection or e/y

During crackingCentroid

axis

Mo

me

nt

Deflection or e/y

Mo

me

nt

After cracking

Stiffness based on crackedcross section Ic

Centroidaxis

Deflection or e/y

After cracking

Centroidaxis

MRd

Mc

Cracking moment of resistanceMc = fctm Zbottom fctm = 2.90 N/mm2

Deflection or e/y

Now do the flexural stiffness andstrength calculations …

Young’s modulus of steel / concrete = m

Today only m = 200 / 33.3 = 6.0

As mAs

xx

modular ratio

Direct deflection calculation

Say 100

Say 170

Uncracked 2nd MoA

Say200

xu

xu =

Today only b = 100 mm

h = 200 mm

Direct deflection calculation

Say 100

Say 170

Uncracked 2nd MoA

Say200

xu

(m-1)As = 5.0 x 157= 785 mm2

xu =

Why ?

Direct deflection calculation

Say 100

Say 170

Uncracked 2nd MoA

Say200

xu

(m-1)As = 5.0 x 157= 785 mm2

xu =

(m-1) because the steel replacesthe area of concrete it occupies

Direct deflection calculation

Say 100

Say 170

Uncracked 2nd MoA

Say200

xu

xu =100 x 2002/2 + 785 x 170

From thetop

100 x 200 + 785

xu = 102.6 mm

(m-1)As = 5.0 x 157= 785 mm2

Direct deflection calculation

Uncracked 2nd MoA

102.6

xu = 102.6 mm

Iu = 100 x 2003/12

+ 100 x 200 x (102.6 – 200/2)2

+ 785 x (170 – 102.6)2

= 70.4 x 106 mm4

Say 170

Say 100

Say200

(m-1)As = 5.0 x 157= 785 mm2

Direct deflection calculation

Uncracked 2nd MoA

102.6

xu = 102.6 mm

Iu = 100 x 2003/12

+ 100 x 200 x (102.6 – 200/2)2

+ 785 x (170 – 102.6)2

= 70.4 x 106 mm4

and ZB = 70.4 / 97.4 = 0.722 x 106 mm3

97.4

ZB section modulus at the bottom of the beam

Then, just before cracking occurs

Mc ≤ fctm ZB

Mc ≤ 0.722 x 106 x 2.90 x 10-6 = 2.09 kNm

From EC2: fctm= 0.3 fck2/3

fctm is flexural tensile cracking strength

La a

Find Pc

La a

Find PcLoad rig = 38 kg

Concrete = 25 kN/m3

La a

Pc

Self weight w = 0.2 x 0.1 x 25 = 0.5 kN/m

P/2 P/2

F = 0.38 kN

2.09 kNm here

Load rig = 38 kg

Concrete = 25 kN/m3

La a

Pc

Self weight w = 0.2 x 0.1 x 25 = 0.5 kN/m

P/2 P/2

Mc = Pc a/2 + etc etc

F = 0.38 kN

1.37 m0.432

Pc

Self weight w = 0.2 x 0.1 x 25 = 0.5 kN/m

P/2 P/2

Prove yourself that

Pc = 8.75 kN

F = 0.38 kN

0.432

First crackCentroid

axis

P

Pc = 8.75 kN

Deflection or e/y

Anticipated cracking load

? kN

Direct deflection calculation

170

Cracked 2nd MoAxc

xc =100 xc

2/2 + 942 x 170

From thetop

100 xc + 942

mAs = 6.0 x 157 =942 mm2

Say 100

Direct deflection calculation

170

Cracked 2nd MoAxc

xc =100 xc

2/2 + 942 x 170

From thetop

100 xc + 942

mAs = 6.0 x 157 =942 mm2

Say 100

Solving the quadratic

xc = 48 mm

Direct deflection calculation

170

Cracked 2nd MoAxc

xc =100 xc

2/2 + 942 x 550

From thetop

100 xc + 942

mAs = 6.0 x 157 =942 mm2

Say 100

Solving the quadratic

xc = 48 mm

Ic = 100 x 483/3

+ 942 x (170 – 48)2

= 17.7 x 106 mm4

8.75

?

P

Deflection or e/y

z

Fs

Fc

Ultimate moment of resistance :

Fs = 0.87 x 500 x 157 = 68295 N

Fc = 0.567 fck x 100 x 0.8X

fck = 30 N/mm2

z

Fs

Fc

Ultimate moment of resistance :

Fs = 0.87 x 500 x 157 = 68295 N

Fc = 0.567 fck x 100 x 0.8X

X = 50.2 mm < 0.6d

(Note how close this is to xc = 48 mm)

z = 170 – 0.4 x 50.2 = 150 mm

MRd = 68295 x 150 x 10-6 = 10.24 kNm

x

Ultimate moment of resistance :

Fs = 0.87 x 500 x 157 = 68295 N

Fc = 0.567 fck x 100 x 0.8X

X = 50.2 mm < 0.6d

(Note how close this is to xc = 48 mm)

z = 170 – 0.4 x 50.2 = 150 mm

MRd = 68295 x 150 x 10-6 = 10.24 kNm

PRd = 46.5 kN

8.75

46.5

P

Deflection or e/y

If mean partial safety factor fordead and live load is about 1.4

Anticipated failure load

?

Mark crack patterns and loads like this -

Before your lab class :

Repeat these exercises for thedimensions given in the handout, for H10

and H16 bars

including the design gradients of the

load v deflection and e/y graphs

Bring your hard hats

Steel toe cap boots (we have extra pairs)

Wear grubby clothes

Bring calculations and graphs showinganticipated design beam loads, etc.

Camera

Hand-in also includes your Mix Design Sheet

End