This document discusses compression testing and summarizes: 1. It describes the barrel shape of compressed specimens and types of failure under compression. 2. It outlines limitations of compression tests and precautions needed for the tests. 3. It provides information on specimen size, shape, and dimensions for different test purposes and defines terms like elastic limit stress, ultimate compressive strength, and modulus.

1. Compression
• The reason for barrel shape:
1
2
• Types of failure:
Brittle material
(Cast iron)
Semiductile materials
(Brass)
Ductile materials
(Steel)
Shear failure
55-60º(
Shear failure
50º
failure
(θ = 45º+ Φ /2)
θΦ
• Limitations of compression test:
1load axial
2unstable loading
3
4
• Precautions
1-
2-)Lо = (2-10) dо(Buckling
1
P
Δ
L
55 - 60°
Δ
L
P
50°
P
Δ
L

2. Pp.l
Aо
Pmax
Aо
Aо
Lо
½ Pp.l
X ∆Lp.l
Aо
Lо
⅔ Pmax
X ∆Lmax
L
D
L
D
3
43-14(°
• The use of spherical bearing blocks
1
2
•
UseSizeSpecimen
For bearing materialsL = 0.9DShort
Determination of compressive strengthL = (2- 3)Dmedium
Drawing the P- ΔL and determining the
Young's modulus
L = (8-10)DLong
1.10
2.1.5Strength
)L/D > 2) to provide suitable middle region free from end restarints
σp.lElastic (proportional) limit stress
σmaxUltimate compressive strength
M.RModulus of resilience
M.T.
Modulus of toughness
2
Spherical block
Specimen
Machine head

3. ∆Lf
Lо
ε1
σ1
2
φ
% Contraction
tan βSecant modulus at stress σ1
• Prove that (θ = 45º+ )
θ
Friction force = x
=μF
μtan Φ
F= (Pcos θ) (tan Φ)
τ = force/area
Area = A/ cos θ
τ = (cos θ/A)[Psin θ - Pcos θ tan Φ]
τ = (P/A)[ sin θ cos θ - cos 2θ tan Φ]
τ = (P/A)[ ½ sin 2θ - cos 2θ tan Φ]
θτ
dτ/dθ = 0
cos 2θ +2 sin θ cos θ tan Φ = 0
3
N
μ N
θ
θ
P
P
P sin θ
P cosθ tanφ
P cos θ

4. 2
φ
cos 2θ + sin 2θ tan Φ = 0
cos 2θ / sin 2θ + tan Φ = 0
-tan Φ = cot 2θ
-tan Φ = tan (90 - 2θ)
θ = 45º+
Φ
( ) Determine the relative displacement of points A and D of a steel rod of
variable cross section area shown in Fig. when subjected to 4 concentrated
load forces P1, P2, P3, and P4 . (Es = 200 GPa). Cross sectional area of AB and
CD = 0.001 m2
, and BC = 0.002 m2
ΔL =
A1 = 0.001 m2
A2 = 0.002 m2
A3 = 0.001 m2
Tension Compression Tension
4
P1
= 100 KN
2m
P 4
NK 001 =
P 3
NK 001 =
P 2
NK 001 =
A B C
D
1m 5.1m
L P
A E
P 1
L 1
A E 1
P 2
L 2
A E 2
P 3
L 3
A E 3
01 x 2 x 001 3
00201 x 9
100.0 x
05101 x 1 x 3
01 x 002 9
200.0 x
01 x 5.1 x 05 3
00201 x 9
100.0 x
NK001
2m
NK 05NK 051NK 051NK001
1m
5.1m
NK 05

5. ΔL1 =
ΔL1 =
ΔL2 =
ΔL2 =
ΔL3 =
ΔL3 =
ΔL1 = 0.001 m ΔL2 = 0.000375 ΔL3 = 0.000375
ΔLtot = ΔL1 + ΔL3 – ΔL2 = 0.001 m (Tension)
( ) A brass bar having cross-sectional area of 10 cm2
is subjected to axial forces
as shown in Fig. 1. Find the total change in length of the bar. Eb = 1.05 x 106
Kg/cm2
ΔL =
Tension Compression Compression
5
5000 Kg
60 cm
1000 Kg2000 Kg8000 Kg
100
mc
021mc
L P
A E
P 1
L 1
A E 1
P 2
L 2
A E 2
P 3
L 3
A E 3
06 x 0005
50.101 x 6
01 x
0003001 x
50.101 x 6
01 x
021 x 0001
50.101 x 6
01 x
gK 0005
06mc
gK 0001gK 0003gK 0003
gK0005
001
mc
021mc
gK 0001

6. ΔL1 =
ΔL1 =
ΔL2 =
ΔL2 =
ΔL3 =
ΔL3 =
ΔL1 = 0.0286 cm ΔL2 = 0.0286 cm ΔL3 = 0.0114 cm
ΔLtot = ΔL1 + ΔL3 – ΔL2 = 0.0114 cm
( ) A member formed by connecting a steel
bar to an aluminum bar is shown in
the Fig. (3). Assuming that the bars are
presented from buckling sidewise,
calculate the magnitude of force P that
will cause the total length of the
member to decrease 0.25 mm. The
values of elastic modulus for steel and
aluminum are 210 KN/mm2
and 70
KN/mm2
respectively.
ΔL =
ΔLtot = ΔLs + ΔLAl = 0.25
6
P L
E A
300 P
21005 x 05 x
P 083
07001 x 001 x
rab munimulA
001001 x
rab leetS
0505 x
P
083
mm
mm 003

7. + = 0.12
P = 224.4 kg
( ) A bar ABC shown in the Fig. (2)
consists of two parts AB and BC, each part
being 1 meter long and having cross-section
area of 2 cm2
and 3 cm2
respectively. The
bar is suspended from A and there is a
rigid horizontal support at 2.001 m from A.
A force of 10 tons acting vertically
downwards is applied at B. Determine (i)
the reaction produced by the rigid horizontal support, (ii) the stresses in parts
AB and BC of the bar. Take E = 2 x 106
kg/cm2
(i) ΔL =
ΔLtot = ΔLAB - ΔLBC = 0.001
+ = 0.1
7
A
A1
= 2 cm2
10
not
1m
m 1
1mm
A 2
mc 3 = 2
B
C
L P
A E
001 x (R – 01)
201 x 6
2 x
001 x R
201 x 6
3 x
noisserpmoC
R
R
noisneT
01-
R
01-
R
PBA
A о
6.3 – 01
2
PCB
A о
6.3
3

8. R = 3.6 kg
(ii)
σAB = = = 3.2 kg/cm2
σBC = = = 1.2 kg/cm2
( ) A reinforced concrete column of cross-section is
shown in the figure. The section is reinforced by four
steel bars of 20 mm diameter each. If the column carry a
load of 80 ton, find the load carried by the steel and
concrete, the contraction in the column is 0.5 mm. Find
also the stress on the steel bars and on the concrete. Es =
206 GPa, Ec = 14 GPa
Ptot = 80 ton = PS + PC ΔL = 0.5 mm
ΔL =
ΔLtot = ΔLS = ΔLC
AS = 4 x (π/4)(20)2
= 1257 mm2
AC = (400 x 600) – 1257 = 238743 mm2
8
40
mc
06
mc
P S
A S
P s
L
60201 x 3
7521 x
P c
L
4101 x 3
347832 x
L P
A E
7.5
7521
P C
A C
3.47
347832

9. Pmax
Aо
250
177
0.05x 100
25
∆Lf
Lо
%
455
0.0075/25
σ
ε
0.5 = =
PS L = 129.5 x 106
PC L = 1671.2 x 106
PS = 0.07749 PC
PC = 74.3 ton
PS = 5.7 ton
σS = = = 4.5 x 10-3
ton / mm2
σC = = = 3.11 x 10-4
ton / mm2
( ) A concrete cylinder 15 cm in diameter with a gauge length of 25 cm is used
to obtain the stress-strain diagram in compression. If the specimen failed at
a load of 250 KN and a total contraction of 0.05 cm, determine the following:
(i) The ultimate strength (ii) The percentage contraction
(iii) The secant modulus at a stress of 455 N/cm2
and contraction of
0.0075cm
D = 15 cm L = 25 cm Pmax = 250 KN ΔLmax = 0.05 cm
Ao = (π/4) (15)2
= 177 cm2
i) Ultimate strength σmax = = = 1.41 KN/cm2
ii) % Contraction = = = 0.2%
iii) tan α = = = 1516.7 KN/ /cm2
9

10. ( ) The following data were obtained in the compression test of a cast iron
cylindrical specimen 35 mm diameter and 250 mm height. The load (P) and
the corresponding decrease in height (h) were as follows:
P (KN) 0 27 68 102 136 203 272
h(mm) 0 0.056 0.14 0.208 0.28 0.457 0.71
Draw the ordinary stress-strain diagram and determine the following:
(i) The crushing stress (ii) The initial tangent modulus
(iii) The secant modulus at a stress of 2100 kg/cm2
(iv) Modulus of toughness
Lо = 250 mm Aо = π/4 (35)2
= 962 mm2
σ (N/mm2
) 0 28 71 106 141.4 211 282.7
ε x 10-4
0 2.24 5.6 8.3 11.2 18.3 28.4
100
50
100
150
200
250
300
0 5 10 15 20 25 30
σ
ε

11. Pmax
Aо
272
962
εp.l
σp.l
∆Lp.l
x Aо
Pp.l
x Lо
0.056 x 962
27x 250
210
18.75 x 10-4
σ
ε
i) The crushing stress σmax = = = 0.283 KN/mm2
ii) Modulus of elasticity E = = = = 125.3 KN/mm2
iii) The secant modulus
σ = 2100 kg/cm2
= 210 N/mm2
From drawing (ε) = 18.75 x 10-4
tan α = = = 11.2 x 104
N/ /mm2
( ) A compression test was carried out on a cast iron cylindrical specimen of 3.4
cm diameter and 25 cm length. The loads and corresponding decrease in
height were as follow:
Load (KN) 0 27 68 102 136 203 272
Decrease in height (mm) 0 0.056 0.14 0.208 0.28 0.457 0.71
Draw the ordinary stress-strain diagram and determine the following:
(i) The fracture stress (ii) The modulus of elasticity
(iii) The secant modulus at a stress of 21 KN/cm2
Lо = 25 cm Aо = π/4 (3.4)2
= 9.1 cm2
σ (KN/cm2
) 0
2.9
7
7.47 11.2 14.94 22.3 29.89
ε x 10-3
0
2.2
4
5.6 8.3 11.2 18.3 28.4
11
σ
ε
0
5
10
15
20
25
30
35
0 5 10 15 20 25 30

12. Pmax
Aо
272
9.1
εp.l
σp.l
∆Lp.l
x Aо
Pp.l
x Lо
0.056 x 9.1
27x 25
21
17.25 x 10-3
σ
ε
0
5
10
15
20
25
30
0 5 10 15 20 25 30 35 40
σ
ε
i) The fracture stress σmax = = = 29.9 KN/cm2
ii) Modulus of elasticity E = = = = 1324.6 KN/cm2
iii) The secant modulus
σ = 2100 kg/cm2
= 21 KN/cm2
From drawing (ε) = 17.25 x 10-3
tan α = = = 1217.4 KN/cm2
( ) A compressive test was carried out on a brass cylinder of 10 cm2
cross-
section area and gauge length of 200 mm. The load and corresponding
decrease in height were as follow:
Load (Kg) 0 2700 6750 10100 12500 17500 20000 27000
Decrease in
height (mm)
0 0.055 0.1375 0.205 0.255 0.365 0.445 0.700
Draw the stress-strain diagram and find:
(a) Ultimate compressive strength
(b) Modulus of resilience
(c) Elastic compressive strength
(d) The modulus of elasticity
(e) Modulus of toughness
(f) Contraction %
(g) Secant modulus at a stress of 2400 Kg/cm2
Lо = 200 mm Aо = 1000 mm2
σ (Kg/mm2
) 0 2.7 6.75 10.1 12.5 17.5 20 27
ε x 10-4
0
2.7
5
6.875 10.25 12.75 18.25 22.25 35
12

13. 2400
29 x 10-4
σ
ε
0.7x 100
200
∆Lf
Lо
%
Aо
x Lо
⅔ Pmax
x ∆Lmax
1000 x 200
⅔ x 27000 x 0.7
Pmax
Aо
27000
1000
εp.l
σp.l
∆Lp.l
x Aо
Pp.l
x Lо
0.365 x 1000
17500x 200
Pp.l
Aо
17500
1000
Aо
x Lо
½Pp.l
x ∆Lp.l
1000 x 200
½ x 17500 x 0.365
(a) Ultimate compressive strength σmax = = = 27 Kg/mm2
(b) Modulus of resilience (M.R.) = = = 0.016 Kg/mm2
(c) Elastic compressive strength σP.l = = = 17.5 Kg/mm2
(d) Modulus of elasticity E = = = = 9.6 ton/mm2
(e) Modulus of toughness = = = 0.063 Kg/mm2
(f) % Contraction = = = 0.35%
(g) The secant modulus
σ = 2400 kg/cm2
= 24 kg/mm2
From drawing (ε) = 29 x 10-4
tan α = = = 8.3 x 105
Kg/cm2
13

14. ( ) A compression test was carried out on a concrete cylinder of 20 cm in diameter
and gage length of 25 cm. The following data load and were recorded:
Load, ton 0 1.8 5.4 9 12.6 14.4 18 21.5 23.4 25.2 26 27
∆L, mm x 10-2
0 0.66 2.05 4 6.6 8.3 15.4 23.4 29.7 40 47 59.5
Draw the stress-strain diagram and find:
a- Ultimate compressive strength
b- The initial modulus of elasticity
c- The modulus of toughness
d- Secant modulus at a stress of 45.5 Kg/cm2
e- Contraction %
f- Explain the fracture characteristic of test specimen
Lо = 25 cm Aо = (π/4) (Do)2
= 314 cm2
σ (Kg/cm2
)
0 5.7 17.2
28.
7 40.1 45.9 57.3 68.5 74.5 80.3
82.
8 86
ε x 10-3
0 2.64 8.2 16 26.4 33.2 61.6 93.6
118.
8 160 188 238
14
0
10
20
30
40
50
60
70
80
90
100
0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255
σ
ε

15. Pmax
Aо
27000
314
εp.l
σp.l
∆Lp.l
x Aо
Pp.l
x Lо
2.64
5.7x 1000
45.5
34.5 x 10-3
σ
ε
59.5x 100
250x 100
∆Lf
Lо
%
Aо
x Lо
⅔ Pmax
x ∆Lmax
314 x 25 x 1000
⅔ x 27000 x 59.5
(a) Ultimate compressive strength σmax = = = 86 kg/cm2
(b) Modulus of elasticity E = = = = 2159 kg/cm2
(c) Modulus of toughness = = = 0.136 kg/cm2
(e) % Contraction = = = 0.238 %
(d) The secant modulus σ = 45.5 kg/cm2
From drawing (ε) = 34.5 x 10-3
tan α = = = 1319 Kg/cm2
( ) A compression test was carried out on a brass cylinder of 1000 mm2
cross
section area and gauge length of 200 mm.
Based on the shown stress-strain
diagram calculate the following:
1- The proportional limit stress
2- The modulus of elasticity
3- The modulus of resilience
4- The modulus of toughness
5- Secant modulus at stress 210 MPa
6- Ultimate compressive strength
7- Contraction %
8-Explain the fracture characteristic of
test specimen
15
55 - 60°
0
50
100
150
200
250
300
0 0.002 0.004
Stress,MPa
Strain