Destructive Tests of Materials

In order to determine the mechanical properties of materials certain tests are carried-out. Depending on the nature of tests, these are classified as follows:

  • Non-destructive tests (NDT) in which the materials or their specimens are not broken.
  • Destructive tests in which the specimens of materials are fractured or deformed.

Destructive Tests of Materials

Generally specimens are tested on machines in the laboratory under static, dynamic or impact loadings. Most common among mechanical destructive tests are the following:

  1. Tensile test
  2. Compression test
  3. Shear and Bending tests
  4. Torsion test
  5. Hardness tests
  6. Impact tests
  7. Fatigue test
  8. Creep test

Now, we shall discuss them in details separately.

Tensile Test

This test is performed on machines such as Hounsfield tensometer, Universal testing machine (UTM), Instron or other material testing system (MTS).

Hounsfield tensometer: A typical Hounsfield tensometer consists of several load-scales. Its loading capacity ranges between 0 to 31 kgf on the smallest load-scale, and 0 to 2000 kgf on the highest load-scale. Least count of these scales is 1 kgf and 10 kgf respectively.

It is used for testing of thin sections and low tensile strength materials such as leather, rubber, thin wire, nylon, plastics and plywood etc.

Universal testing machine: The UTM’s have large load capacity, generally ranging from 0 – 4 tonne and 0 – 40 tonne. Materials with large sections and high tensile strength such as steel and aluminum bars are tested on them.

Tensometer is generally mechanically operated while UTM’s are hydraulically operated. They can be controlled by computers or can be operated manually.

Tensile testing standards and specimens: While conducting the experiment, a specimen is prepared in accordance to a standard like ASTM (American Society of Testing Materials), ASM (American Society of Metals, Chinese Standards or Indian Standards.

We shall follow Indian standards (IS). Different tensile test specimens shown in Figures 1(a – d) are in conformity with IS. Their gap ends may be cylindrical, pinholed, threaded or shouldered. Their cross-section may be either circular or rectangular. 

destructive tests of materials

Gauge length: The gauge length lis taken equal to

l = 5.65√A ……for a circular section specimen.

l = 4√A …………for a rectangular strip.

Where, A is the cross-sectional area.

According to other standards, these relations are as follows:

ASTM ——— l = 4.51√A

BS ————- l = 4.0√A

 DIN ———–l = 11.2√A

The test performance is judged for gauge length and not the full length L of specimen. The two ends of larger cross-section are gripped by the machine jaws. The fillet is provided so that there is no stress concentration.

Procedure of Tensile Testing

The tensile load P is applied gradually. For each incremental value of P, the elongation δl of the gauge length is observed by precision instruments such as a dial gauge. The observation of change in length beyond yield point may be done by a precision scale. Fluctuation of the live pointer on dial of the machine indicates occurrence of yield point. The specimen, finally, fractures with a thud sound.

The two broken pieces of the specimen become hot after breaking. It is due to the conversion of internal friction into heat. The internal friction is caused due to the slipping of planes of atoms during loading.

When the two pieces cool down, we measure the gauge length and also the diameter at the neck of fractured ends. The fractured ends are generally of cup and cone type.

Observations and calculation: Calculation of the stress and strain, Young’s modulus, yield stress, ultimate strength and breaking stress is done using the relations. The tress-strain curves are plotted. UTM’s are generally equipped with an automatic stress-strain curve plotting system.

We may obtain percent reduction in cross-sectional area Ar, and percent elongation (%δl) by

%Ar = (Ao – Ab)/Ao

% δl = (lblo)/lo

Where Ao, and lo, are the original cross-sectional area and gauge length of the specimen, and Ab, lb are corresponding dimensions on breaking.

Compression Test

The uniaxial compression test is also carried-out on a Universal Testing Machine (UTM) which has been already discussed.  In this test, the specimen is prepared in the form of a cube, or in cylindrical form.

Limitations. The compression test suffers from the problems of end restraint and lateral instability. The end restraint is caused by the lateral expansion called buldging as shown in Figure 2(c). Its effect is to produce transverse compression which is undesired. Its detrimental effect may be minimized by taking a high ratio of depth of the cylinder h and its diameter d.

Contrary to this, the ratio is kept small to avoid lateral instability due to buckling action. However the most suitable value is taken as (h/d) < 2. The shape and size of compression specimen has an effect on the compressive strength. This strength decreases with an increase in (h/d) ratio.

compression test of materials

Failure mode. The procedure of experimentation and observation of data are almost similar to that of the tension test described above. Most compression tests are performed on brittle materials. Generally they fail in shear.

Shear develops along a diagonal plane. The maximum shear stress develops on a plane inclined at 45° from the direction of compressive load. However, the spread of shear cracks may be deviated from this plane due to the effects of end restraint. It can be noticed in cast iron and concrete. The directionality of grains in wood also affects the shear failure along the 45° shear plane.

Shear and Bending (or Flexure) Tests

Three-point and four-point beam bending test: The shear and flexural tests are performed by three-point and four-point beam bending tests shown schematically in Figures 3(a – b).

shear and bending tests of materials

Simply supported beam of material to be tested is subjected to a concentrated load P. The experiment may be conducted on UTM and Tensometer using suitable attachments, or on a machine specially meant for it.

The load P is increased gradually in increments and the central deflection δ is recorded. This helps in plotting the load-deflection characteristic of the material.

Modulus of rupture σr for brittle materials may also be determined from bending test. In this case, load P is applied until rupture (breaking) of the beam occurs. The σ, may be approximated by beam flexure formula as

σr = My/I

Where I is the area moment of inertia of the beam’s cross-section and y is half of the depth of beam. Here maximum resisting moment (equal to maximum bending moment) M corresponds to elastic condition which is not the case at rupture. Thus the relation is approximate.

Calculation for the result: Variation of shear force diagram (SFD) and bending moment diagram (BMD) in the two cases are shown in the Figures 3(a – b). Case of pure bending exists in ‘a’ region of the beam, Figure 3(b), as SF = 0 and BM = constant in this region.

The flexural strength i.e. the bending stress σ at the surface of specimen and the maximum shear stress t at the midplane for a rectangular section beam may be determined by

σ = 6M/bH2 and,

σ = 3F/2bh

Where M is the maximum bending moment at failure; b and h are the width and depth of the beam respectively, and F is the maximum shear force in the beam.

Span over depth ratio (I/A) of 16 is normally an acceptable value in a three-point bending test.

Plot of Likely Stress-Strain Diagram in Pure Bending

The stress-strain diagram for a material in pure bending is shown in Figures (a – b) below. The beams having rectangular, circular, I cross-section, or any other shape having two axes of symmetry will exhibit a curve as illustrated in Figure (a).

tensile test of materials

This curve is identical in tension and compression, and the neutral axis coincides with the horizontal axis of symmetry.

For some materials, the stress-strain diagram differs in tension and compression. In this case, the neutral axis does not coincide with the horizontal axis of symmetry. Such curves may also be seen in beams of T-section.

Torsion Test

The ability of a material to resist twisting moment (torque) is determined by a torsion test. In this test, the specimen of a solid circular section or a hollow tube (torsion tube) is taken.

It is held in a Torsion Testing Machine at one end, and at the other end the torque T, is given. This torque causes twist in the specimen. The twist is shown by deformed dotted lines which were rectangular grids before application of torque.

The torque is increased gradually and the corresponding angle of twist θ is measured. The angle of twist is measured by a precision angular instrument or by a graduated dial for different values of torque.

torsion test of materials

The shear strain y can be calculated by

y = s/lu = rθ

Where s is the unit shearing displacement in unit length lu of a rectangular grid Here r is mean radius of thin hollow tube or outer radius of a solid shaft and θ the angle of twist per unit length. The torque is read on the dial of the machine.

Torque Vs twist diagram:  From the observed data, the torque versus twist diagram is plotted whose likely shape is shown in Figure 5(c). Its variation is linear up to point A which implies that the elastic limit is from O to A.

Calculation to determine torsional shear stress: The effect of torque is to create shear stress in the specimen. The torsional shear strength t at the outer periphery of the solid circular specimen of diameter d is obtained from

t = 16T/πd3

And for hollow circular specimen by

t = 16TDo/ (πD4o – D4i)

Where Do, and Di, are outer and inner diameters respectively.

Plot between shear stress and shear strain: A plot between shear stress and shear strain is shown in Figure 5(d). The elastic limit is up to A, and material obeys Hooke’s law in OA range. Slope of this curve yields

shear stress/shear strain = G, the shear modulus.

Impact Tests

In manufacturing locomotive wheels, coins, connecting rods etc., the components are subjected to impact loads (or shock loads). These loads are applied suddenly. The stress induced in components is many times more than the stress produced by gradual loading. Hence materials should be able to sustain such loads.

Therefore, impact tests are performed to assess shock absorbing capability of materials subjected to suddenly applied shock loads. These capabilities are expressed as (i) rupture energy, (ii) modulus of rupture, and (iii) notch impact strength.

impact tests of materials

Machine set-up and test procedure: In this test, a notched specimen (described later in this article) placed as a beam is given an impact, and the energy required to propagate the crack resulting in failure of specimen is measured.

The impact test is performed on load Impact Testing Machine shown in Figure (6). The specimen is broken by a swinging hammer of length R and weight W. This hammer is initially placed at certain height A; in OA position. It thus possesses

potential energy Ui, equal to

Ui = WHi = WR(1 – cos α)

Where α is the angle of hammer before strike. In a typical machine, this energy (capacity) is 30 kgf-m.

This hammer falls, strikes the specimen and breaks it when released from the initial position. The hammer then reaches to OB position and attains a height of hf. Thus the energy after rupture Uf; becomes

Uf = Whf = WR(1 – cos β)

Where, β is the angle of hammer after strike.

Result: The energy U required to rupture the specimen may be calculated as under:

U = Ui – Uf = Whi – Whf

= WR(cos β – cos α)

This value may be directly read on the graduated scale of machine’s dial. This rupture or fracture energy is a measure of toughness of the material. The angles α and β may be read directly on an indicator in some machines.

Notch impact strength: The notch impact strength Is is now determined from the following relation:

Is = U/Ae

Where, Ae is the effective cross-sectional area of the specimen below the notch before test.

Modulus of rupture: Once the rupture energy U is known, the modulus of rupture is obtained from the following relation:

Ur = U/Ve

Where, Ve is the effective volume of specimen. It is found from

Ve = Aete

Where te is effective thickness of specimen below the notch.

Types of Impact Tests and Specimens

Two types of notch impact tests are commonly conducted. These are

  • Charpy test, and
  • Izod test.

In both the tests, the standard specimen is in the form of a notched beam. In Charpy test, the specimen is placed as simply supported beam while in Izod test as cantilever beam.

Standard sizes of these specimens are shown in related figures. The specimens have standard V-shaped notch of 45°. Notches of U-shape and keyhole-shape are also common. The notch is located on the tension side of the specimen during impact loading. Depth of the notch is generally taken as t/5 to t/3 where t is the thickness of the specimen.

Purpose of keeping a notch and notch sensitivity: Purpose of keeping a notch on the tension side is double fold. First purpose is that the stress raises to a peak value at the base of the notch due to elastic stress concentration.

Second is that the yield stress is raised due to elastic and plastic actions. The two effects combine together and break the specimen due to brittle fracture more readily in presence of a sharp notch than in an un-notched specimen.

The nature of a ductile material to behave as brittle material in the presence of a notch is called notch sensitivity. The energy of rupture decreases when the sharpness of V-notch increases. This is due to the increase in stress concentration.

The salient features and comparison between Charpy and Izod tests are as under.

.

 ConsiderationCharpy testIzod test
1Placement of specimeneasier on machine as specimen is centrally locatedneeds adjustment in location of specimen
2Temperature suitabilitybetter for low temperature testssuitable for room and high temperature tests
3Type of beamsimply supportedcantilever
4Strike of hammeron the opposite side of the notchon the same side of the notch
5Length of the specimensmallerlonger
6Cross-section of the specimenequalequal
7Falling height of the hammerfrom top most heightfrom certain predetermined height
8Angle of strike (α)α > 90°, normally 160°α < 90°, normally 90°

Leave a Comment

Your email address will not be published. Required fields are marked *