The document describes the static compression test. It discusses: 1) The test involves applying a compressive load opposite to tension. 2) Specimen geometry is important - a height to diameter ratio of 2-10 is recommended. The ends must be smooth, flat, and perpendicular to the axis. 3) Properties measured include compressive strength, elastic modulus, yield strength. Brittle materials fail by shearing, ductile materials compress without failing.

1. STATIC COMPRESSION
TEST

2. THE STATIC COMPRESSION TEST
The compression test is the opposite of the tension test with
respect to direction of applied stress.
LIMITATIONS OF COMPRESSION TEST
1- The difficulty of applying a truly concentric or axial load.
2- Friction between the heads of the testing machine or bearing plates
and the end surfaces of the specimen results in formation of lateral
expansion of the specimen which gives the specimen its barrel shape.
3- The relatively unstable character of this type of loading contrasted
with tensile loading.
4- The relatively large cross section area of the compression test
specimen required to obtain a proper degree of stability of the tested
piece, that needs machines with high capacity.
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3. THE STATIC COMPRESSION TEST
3
Sec A-A
Sec B-B
A A
B B

4. THE STATIC COMPRESSION TEST
1. TEST SPECIMENS.
The general requirements of compression test specimens
Square, rectangular and circular cross sections can be used for uniform
stressing of the compression specimen, circular section is to be preferred
over other shapes.
 With respect to relative dimensions
• A height to diameter ratio of 10 is suggested as a practical upper limit
and 2 is suggested as a lower limit.
• A ratio of length to diameter of 2 or more is commonly employed,
although the height to diameter ratio used varies for different
materials.
 With respect to shape of cross section
The selection of the actual size depends on:
1- Type of material
2- Type of measurements
3- Testing machines available
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5. With respect to geometry
THE STATIC COMPRESSION TEST
The ends to which load is applied must be smooth, parallel, flat and perpendicular to the
axis of the specimen.
2. APPARATUS
Because these specimens are rather large, testing machines of relatively high
capacity are needed.
3. Requirements in specimen positioning
1- The loaded faces of the specimen should normal to its geometric axis.
2- The loaded faces of the machine must bear uniformly against the specimen.
3- The resultant load must coincide with the geometric axis of specimen.
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6.  Usually one end of the specimen should bear
on a spherical seated block. The purpose is
to overcome the effect of a small lack of
parallelism between the heads of the machine
and the end faces of the specimen.
 It is desirable that the spherical seated
bearing block be at the upper end of the test
specimen. it is important for the center of the
spherical surface of this block to lie in the flat
face that bears on the specimen and for the
specimen itself to be carefully centered with
respect to the canter of the spherical surface.
SPHERICAL SEATING BLOCK
Fig 6-4 Arrangements of the specimen and a
spherical seated block
THE STATIC COMPRESSION TEST
3. PROCEDURES
In commercial tests, the only property determined is the compressive strength.
4. MECHANICAL PROPERTIES IN COMPRESSION
various mechanical properties may be defined from the initial portion
in the same manner as for tension, such as the modulus of elasticity, E,
the proportional limit stress, spL, and the yield strength, sy. 6

7. Stiffness for Non-linear material behaviour
1- For Non-linear material behaviour, the stiffness of the material can
be estimated by the initial tangent modulus which is the slope of
the tangent to the stress-strain curve at the origin.
2- The stiffness of these types of materials can be also compared at any
stress level by the tangent modulus. This tangent modulus is the
slope of the tangent to the stress-strain curve at the point
corresponding to this stress level.
3- This comparison can be also made by the secant modulus which is
the slope of the line OB on the curve, sB given.
4- The modulus of elasticity can be expressed by the chord modulus
which is the slope of line connecting two specified points on the
curve as points C, D.
THE STATIC COMPRESSION TEST
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8. THE STATIC COMPRESSION TEST
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1
2
3 4

9.  Brittle Materials (as cast iron)
The brittle materials will then fail by shearing on definite angle (55-60)
 Semi-ductile material (as brass)
If the material is semi-ductile such as brass, it will continue to contract
under load and it will fail by shearing on a definite angle (50).
 Ductile Metals (as mild steel)
If the material is mild steel, it will contract and flatten without failure.
THE STATIC COMPRESSION TEST
5. FAILURE OF MATERIALS IN COMPRESSION
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Contraction,

10. THE STATIC COMPRESSION TEST
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11. THE STATIC COMPRESSION TEST
11
At the instant of failure (rupture) there is a tendency for one surface to
slide over the other. The particular plane or surface along which rupture
takes place is therefore a function of the internal friction of the material
as well as the magnitude of the shearing resistance of the material (shear
stress), see Fig .6-8 therefore the angle of the plane of rupture  will be
 = 45 +  /2., where  is the angle of friction of the material and it is
function of the grain size.

12. 6. EFFECT OF VARIABLES
1. Size and Shape
Almost all compression specimens are
right regular prisms or cylinders. As the
L/D increased the measured strength
decreased due to buckling effect.
As the D or size of specimen increase the
measured strength decreased due to
increasing of internal defects and
impurities.
2. Test Speed
The more rapid the rate the higher the
indicted strength.
THE STATIC COMPRESSION TEST
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13. Example
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 follows:
THE STATIC COMPRESSION TEST
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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
s (Kg/mm2 ) 0 2.7 6.75 10.1 12.5 17.5 20 27
 x 10-4 0 2.75 6.875 10.25 12.75 18.25 22.25 35
Solution
(a) Ultimate compressive strength (d) Modulus of elasticity
(b) Modulus of resilience (e) Modulus of toughness
(c) Elastic compressive strength (f) Contraction %
(g) Secant modulus at a stress of 2400 Kg/cm2
s = 2400 kg/cm2 = 24 kg/mm2 From drawing () = 29 x 10-4
E= tan α = 24/ 29 x 10-4 = 8.3 x 105 Kg/cm2

14. 14
(a) Ultimate compressive strength smax = Pmax/ Aо = 27000/10 = 2700 Kg/cm2
(b) Modulus of resilience (M.R.) = ( ½ P P.L × ∆LP.L ) / (Aо × Lо)
= ( 1/2 × 12500× 0.0255 ) / (10 × 20) = 0.79 Kg/cm2
(c) Elastic compressive strength sP.l = P P.L / Aо = 12500/ 10 = 1250 Kg/cm2
(d) Modulus of elasticity E = s P.L / ε P.L = ( P P.L × Lо ) / (Aо × ∆LP.L )
= ( 12500× 20 ) / (10 × 0.0255) = 9.8 ×105 Kg/cm2
(e) Modulus of toughness (M.T) = T/V = (2/3 P max ∆Lmax )/ (Aо × Lо)
= 2/3 ( 27000 × 0.7 ) / (10 × 20 )= 63 Kg/cm2
(f) % Contraction = ∆Lmax / Lо % = (0.7 × 100) /200 = 0.35 %
0
5
10
15
20
25
30
0 5 10 15 20 25 30 35 40
s

Lо = 200 mm = 20 cm
Aо = 10 cm2

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