Modulus of Elasticity and
Poisson Ratio of Concrete
Benjie Cho and Mulugojam Alemu
Undergraduate Civil Engineering
Univ.. of Southern California
Objective

Find the Modulus of Elasticity of Concrete
 Find Poisson’s Ratio of Concrete
Introduction

Compressive Strength of Concrete
– Standardardized test of concrete
 American
Society for Testing and Materials
(ASTM)
– Specifications include
 Correct mix
 Properly affixing strain gauges
 Properly capping the cylinder with sulfur
Procedure

Select (5) 4in diameter by 12in length
concrete with 28-day age with 4 ksi design
strength.
 Cap all the bearing surfaces with sulfur to
make the end flat.
 Mark area where strain gauges will be
attached.
 Clean area with chemical cleaners and
sand paper.
Procedure





Affix strain gauges on their designated area,
one vertical and one horizontal, using glue.
Attach wire to strain gauges by soldering.
Test voltage of the strain gauges.
Place and center samples on the Satec
Universal Testing Machine.
Connect the wires to the P3500 strain
indicator, which in turn is connected to a
computer for recording.
Procedure

Set gauge factor to 2.055 +/- .5% and zero
the strain readings.
 Begin loading the specimens and record
values of strain for given loads.
 Continue testing until failure of the
specimen.
Organizing the data

Compute the stress by dividing the load
by the cross-sectional area.
 Graph the strength against the vertical
and horizontal strains of each specimen.
 Find values for the Young’s Modulus and
Poisson’s ratio from the data.
 Calculate the theoretical values the
Young’s Modulus and Poisson’s ratio.
Calculation procedures

Young’s Modulus
– E=(s1-s2)/(e2-.000005)
 s1=The
stress corresponding to the longitudinal
strain of 50 micro strain.
 s2=The stress corresponding to .4f ‘c.
 E2=The longitudinal strain corresponding to s2.
 Based on ASTM C 469

Poisson’s Ratio
 n=(Lateral Strain)/(Longitudinal Strain)
Example
S 2 = 2 0 8 8 .9 3 9 6 p si
E = (2 0 8 8 .9 3 9 6 -1 5 0 .6 3 )/ (9 6 2 -5 0 ) = 2 .0 6 E
S 1 = 1 5 0 .6 3 p si
e   m icro stra in
S tre s s (p s i)
C y lin d er #1
6000
4000
H o rizo nt
2000
E
0
-1 0 0 0
V er tic al s tr ain
0.4f c '
0
1000
2000
3000
S tra in (1 0 E -6 )
4000
5000
Cylinder #2
C ylin d er #2
6000
5000
Horizontal S train
S tre s s (p s i)
V ertic al s train
4000
0.4fc'
3000
2000
1000
0
-500
0
500
1000
S train (10E -6)
1500
2000
2500
Cylinder #3
C ylinder #3
6000
5000
Horizontal Strain
S t re s s ( p s i)
4000
0.4 f 'c
3000
2000
Vertical Strain
1000
0
-500
-1000
0
500
1000
Strain (10E-6)
1500
2000
2500
Cylinder #4
C ylin d er #4
S tr e s s (p s i)
5000
4000
3000
2000
1000
0
-500
Horiz ontal s train
V ertic al s train
0.4fc '
0.4
500
V ertical
1500
Str ain (10E-6)
2500
3500
Cylinder #5
C ylin d er #5
5000
4500
4000
Ho rizo n ta l Stra in
S t re s s ( p s i)
3500
3000
2500
0 .4 fc'
2000
1500
'c
Ve rtica l Stra in
1000
500
0
-1 0 0 0
-5 0 0
0
500
1000
Str a in (1 0 E-6 )
1500
2000
2500
3000
Selected Values of Stress
and Strain
C ylin d e r # 1
S tre s s (p s i)
S tra in (la t.)
S tra in (lo n g .)
P o is s o n 's
1 0 0 6 .7 6 4
6
440
0 .0 1 3 8 2 5
2 0 0 4 .2 1 7
-1 1
947
0 .0 1 3 6 4 0
3 0 1 0 .2 6 5
-2 0
1462
0 .1 3 6 8 0 0
4 0 0 7 .7 1 9
-1 0 0
2014
0 .0 4 9 4 6 0
5 2 2 1 .2 1 4
-2 2 2
3367
0 .0 6 5 9 3 0
Avg.
P e a k S tre s s (f 'c ) T h e o re tic a l E
P o is s o n
6
0 .0 5 5 9 3 1
5 2 2 1 .2 1 4 4 .1 1 8 7 x 1 0
E x p e rim e n ta l E
2 .0 6 1 7 x 1 0
6
C ylin d e r # 2
S tre s s (p s i)
S tra in (la t.)
S tra in (lo n g .)
P o is s o n 's
2 0 0 0 .9 7 1
-1 0 7 .8
4 3 8 .3
0 .2 4 5 9 5 0
1 0 0 7 .3 6 9
-3 9
1 3 4 .4
0 .2 9 0 1 7 9
3 0 1 1 .6 8 2
-1 6 7 .9
8 0 0 .8
0 .1 3 6 8 0 0
4 0 0 5 .2 0 5
-1 9 1 .4
1 1 8 8 .3
0 .0 4 9 4 6 0
5 0 7 7 .3 4 9
-8 .6
1 9 3 8 .9
0 .0 6 5 9 3 0
Avg.
P e a k S tre s s (f 'c ) T h e o re tic a l E
P o is s o n
6
0 .1 5 7 6 6 3 8
5 0 7 7 .3 4 9 4 .0 6 1 5 x 1 0
E x p e rim e n ta l E
3 .3 6 5 0 6 x 1 0
6
Selected Values of Stress
and Strain
C ylin d e r # 3
S tre s s (p s i)
S tra in (la t.)
S tra in (lo n g .)
P o is s o n 's
1 0 0 2 .1 1 7
-5 0
3 0 3 .1
0 .1 6 4 9 6
2 0 1 1 .4 7 5
-1 0 7
7 0 5 .0
0 .1 5 1 6 7
3 0 0 5 .7 1 4
-1 6 2 .5
1 0 8 .6
0 .1 4 6 5 8
4 0 0 3 .9 3 1
-1 9 0 .6
1 5 3 8 .3
0 .1 2 3 9 0
5 2 1 8 .1 2 1
-1 9 9 .2
2 1 9 6 .1
0 .0 9 0 7 1
Avg.
P e a k S tre s s (f 'c ) T h e o re tic a l E
P o is s o n
6
0 .1 3 5 5 6 4
5 2 1 8 .1 2 1 4 .1 1 7 4 8 x 1 0
E x p e rim e n ta l E
2 .8 4 6 9 7 x 1 0
6
C ylin d e r # 4
S tre s s (p s i)
S tra in (la t.)
S tra in (lo n g .)
P o is s o n 's
1 0 0 2 .1 1 7
-6 4 .8
4 1 4 .1
0 .1 5 6 4 8
2 0 0 7 .5 7 4
-1 4 5 .3
8 8 0 .5
0 .1 6 5 0 2
3 0 0 5 .7 1 4
-2 4 6 .1
1 4 0 3 .1
0 .1 7 5 4 0
4 4 6 0 .3 8 6
-1 8 5 .9
2 4 0 5 .5
0 .0 7 7 2 8
Avg.
P e a k S tre s s (f 'c ) T h e o re tic a l E
P o is s o n
6
0 .1 1 4 8 3 6
4 4 6 0 .3 8 6 3 .8 0 6 8 x 1 0
E x p e rim e n ta l E
2 .2 9 5 3 2 2 x 1 0
6
Data
C ylin d e r # 5
S tre s s (p s i)
S tra in (la t.)
S tra in (lo n g .)
P o is s o n 's
1 0 0 0 .2 0 7
-6 0 .2
4 2 1 .1
0 .1 4 2 9 6
2 0 1 0 .2 8 1
-1 1 6 .4
825
0 .1 4 1 0 9
3 0 0 3 .8 0 4
-1 2 5 .8
1 2 4 4 .6
0 .1 0 1 0 8
4 6 1 6 .3 5 8
-7 0 9 .4
2 4 7 0 .3
0 .2 8 7 1 7
Avg.
P e a k S tre s s (f 'c ) T h e o re tic a l E
P o is s o n
6
0 .1 3 4 4 6 0
4 6 1 6 .3 5 8 3 .8 7 2 8 x 1 0
E x p e rim e n ta l E
2 .4 3 4 1 4 6 x 1 0
6
Conclusion

All specimens performed under the
theorectical values of E.
– Average E= 2.60E6 psi
 Average Poisson Ratio= .119691
Conclusion

Concrete performed to design
specifications.
– E= 1.5 - 5 ksi
 n= .1

Xiao, Yan. Experimental Analysis of Engineering Materials.
University of Southern California lecture notes 2002.