Unit-1 Mechanical Properties
Unit-1 Mechanical Properties
Unit-1 Mechanical Properties
Structure
1.1 1.2 Introduction
Objectives
True Stress and True Strain Compression Test Universal Testing Machine Notched-bar Impact Tests Temperature Effects Hardness
1.8.1 1.8.2 1.8.3 1.8.4 1.8.5 1.8.6 1.8.7 1.8.8 Scratch Hardness Indentation Hardness Brinell Hardness Rockwell Hardness Vickers Hardness Knoop Hardness Other Methods of Hardness Measurement Relationship between Hardness and Other Properties
1.9
Fatigue
1.9.1 1.9.2 1.9.3 1.9.4 1.9.5 Fatigue Curve Fatigue Mechanism Statistical Nature of Fatigue Determination of Fatigue Strength Factors Affecting Fatigue Behaviour
1.1 INTRODUCTION
Undoubtedly, the most commonly performed test on engineering materials is the static tension test. From this test many mechanical properties of material are evaluated and understanding about its behaviour is developed. 5
Materials Applications
At the outset the distinction between static and dynamic load must be understood carefully. A static load is the one which changes very slowly with respect to time during its application. Once applied, the static load remains unchanged. Very common static tests which are performed for evaluation of mechanical properties are axial tension, compression loading, flexural loading or torsional loading. Axial load is applied along the axis of specimen. Flexural loading is achieved by application of load perpendicular to the axis while torsional loading result from eccentric transverse forces which do not cause bending or change in length. Direct shearing test is also performed sometimes but is not a very popular test. In all these the load is slowly increased from zero till failure occurs; rate of loading being such that definition of static load is not violated. Hardness is a very important mechanical property which is also determined under static load by indentation. Hardness and tensile strength are related. Fatigue is behaviour of material under load or stress which keeps changing with time and failure depends upon stress level and number of stress reversal. Creep is yet another behaviour under statically applied load over a long period of time. Temperature plays an important role in deciding this behaviour.
Objectives
After studying this unit, you should be able to understand the effect of force on solids qualitatively as well as quantitatively, differentiate between the true stress and true strain, analyse the compression test, know universal testing machine, understand tensile strengths and effect of strain rate, know different impact tests, explain temperature effects, understand hardness and fatigue, and identify material problems at elevated temperatures.
extension begins to increase can be observed from the specimen and as the extension begins to increase at a faster rate (it can be judged from experience) the extensometers if attached to specimen may be removed. With further increase in the load the extension increases at increased rate and record of extension may be obtained from displacement between the cross-heads. At some level the load becomes stationary and the specimen either fails or begins to reduce in cross-section rapidly. The latter phenomenon is known as necking and is followed by reduction in load accompanied by further extension until the specimen fractures into two pieces.
Fixed Upper Cross Head Grip or Holder Tension Specimen
Screw Column Grip or Holder Moving Cross Head Space for Compression Specimen Fixed Lower Cross Head
The results of tension test can be entirely presented in form of a stress-strain diagram in which stress is plotted as ordinate and strain as abscissa. It may, however, be emphasized here that in the definition of stress; it is assumed that area of cross-section will not change. It must have become clear by this time that during a tension test area of cross-section and length both change considerably. Therefore, there is a need to reconsider the definition of stress. The stress is defined as ratio of load to original area of cross-section (denoted by A0). It will be defined as engineering stress. The prefix engineering is often dropped and whenever term stress is used, it is understood that it is engineering stress. The ratio of load to actual or current area of cross-section is defined as true stress. Likewise, the ratio correspondingly the true strain is the sum of strains over small ranges of load upto the current load. Engineering stress () and strain () will be
= =
P A0 L L0
. . . (1.1)
. . . (1.2)
where, P is load at any moment. A0 the original area of cross-section, L, the change in length when load changed from zero to P and L0 is the original length. Figure 1.2 and 1.3 show typical stress-strain diagrams for common engineering materials. Diagram for ductile materials are depicted in Figure 1.2, while Figure 1.3 shows the same for brittle materials. These two sets have the same strain scale and it can be seen that a ductile materials shows considerable deformation before it fails while a brittle material shows little deformation. To distinguish between these two types of material, materials with strain less than 5% at fracture are regarded as brittle and those having strains greater than 5% at fracture are called ductile. The type A ductile materials (Figure 1.2(a)) include steels, aluminium alloys, copper alloys etc., while type B ductile materials are mild steel and structural steels. The difference between type A and type B ductile materials must be noted carefully. In both cases the stress and strain vary linearly upto strain a. This region of deformation is elastic in the sense that if load is removed at any point before reaching point a, the specimen will regain its original length and area of cross-section. Beyond point a, ductile material of type A changes the relationship which is more linear. During this deformation, the strain changes at a faster rate than stress, although this rate tends to decrease. Type B of
Materials Applications
ductile material shows distinctly different between behaviour than that shown in Figure 1.2(a). From a slightly higher point b (Figure 1.2(b)), the stress drops suddenly to point c, remains approximately constant over a range, then follows a pattern similar to that of type A.
Yield Strength d e c Stress Stress a o - Plastic Strain Offset c Lower Yield Point Elastic Upper d Yield Limits a b Point
o o Strain
o Strain
(a) Type A
(b) Type B
Stress
o Strain
Stress o Strain
yp el lp
(a) Type A
(b) Type B
The sudden drop in stress from b to c, increase in strain from c to c while stress remains approximately constant is known as yielding. The point b from which the stress drops to point c, is known as upper yield point while point c is known as lower yield point. The deformation at approximately constant stress is termed as yield deformation. Once a ductile material has exceeded the elastic deformation it enters into plastic deformation range. During this deformation, the stress reaches maximum value at d, where necking in the specimen begins. The stress at point d (the engineering stress) is termed as Ultimate Tensile Strength which is a very important property. If loading is continued, the specimen fails eventually at e and stress reduces from point d to e. the stress at point e is called fracture stress. The deformation between point d and e in Figure 1.2 is plastic in the sense that if the specimen is unloaded at any point between d and e the specimen will not regain its original size. Figure 1.4 shows the range of elastic and plastic deformations. The unloading of a specimen from any point f between a and d is known in Figure 1.5. Unloading follows a curve which is somewhat parallel to original elastic line. When specimen is fully unloaded to point g, some strain still remains in it. This residual strain is the plastic component of strain by p, a part of strain between f and g has been recovered and this is elastic part of strain at point f and denoted by e. Thus, strain at point f,
a d e
f = fp + fe
. . . (1.3)
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o Elastic Plastic
f a h
g p e
On reloading the specimen from point g, curve follows a path very close to unloading curve and eventually joints the original curve at h. The area between loading and unloading curve (shown hatched in Figure 1.5) represents the loss of energy due to unloading-cycle and is termed hysteresis. Type B brittle materials (Figure 1.3(b)) do not show linear relationship between stress and strain. Concrete is the example. The maximum stress sustained by such material is defined as ultimate tensile strength. Type A brittle materials, like cast iron, will also have ultimate tensile strength and they show linear relationship between and .
SAQ 1
(a) (b) (c) (d) (e) (f) (g) (h) (i) What is meant by static load? What tests are performed under static load? Sketch a tension test specimen. Define stress and strain for a tension test specimen whose original area of cross-section is A0 and original length is L0. Why should you call the stress defined in (d) as engineering stress? What is true stress? How do you measure elongation of a tension specimen? Distinguish between a ductile and a brittle material. Give example of ductile and brittle materials. What are upper and lower yield point and ultimate tensile strength? Distinguish between elastic and plastic deformation of a tension test specimen.
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= E i.e. E =
. . . (1.4)
The constant of proportionality E is called Modulus of Elasticity or Youngs Modulus. Apparently the slope of elastic part of stress-strain diagram (shown as oa in Figures 1.2 and 1.3) is the modulus of elasticity. Modulus of elasticity is a constant for a given class of material and variation in small alloying contents does not cause change in E. For example, wide range of carbon steel has E = 210 103 MPa. If stress-strain curve is established accurately the slope of elastic part is E of the material. However, it may be noted that this is not a reliable method because tension tests require such loads that may cause small deformations in loading system, part of which may look like deformation of the specimen.
50 d c e a d e b
Figure 1.6 : (a) With Strain Scale Expanded Showing Yield Deformation and (b) With Strain Scale Shortened to Show Ultimate and Fracture Point alongwith Yield Point
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The yielding behaviour of mild and structural steel is shown in Figure 1.6(a) on an expanded scale of abscissa or strain axis. This figure exphasises the fact that total deformation during yielding is about ten times the deformation upto elastic limit. The yielding is shown to end at point e beyond which the plastic deformation continues as is shown in Figure 1.6(b). The point c is the upper yield point and d the lower yield point. The difference between upper and lower yield point exists because movement of the ends of the specimen as produced by two machine cross-heads, does not proceed as fast as the
yield deformation of material. This fact may be confirmed by allowing the loading crosshead to move very fast, which is the condition of the impact loading. In such a case only one yield point (i.e. the upper yield point) will exist. For materials which show upper and lower yield points, the latter is taken as the yield strength because it is less affected by variables. Most ductile materials do not yield and show curve of Figure 1.2(a) or as in Figure 1.7. For such materials the end of elastic behaviour is defined as attaining a certain plastic strain. The method of determination of this strength value is to measure a certain offset strain from origin as in Figure 1.7 represented by oA. From point A a line parallel to elastic strength line part is drawn to meet the curve at d. the stress corresponding to point d is defined as the yielding strength or proof strength of the material. The values of offset strain have been standarised for different materials. The yield strength determined by the offset method is always described as yield strength for an offset. Table 1.1 describes the values of offset strains commonly used.
d
o o
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(a)
(b)
(c)
(d)
Figure 1.8 : Sequence of Deformation of a Cylindrical Specimen with Reference to Figure 1.6 (a) Just Before Point, a, (b) Just Before Point, f, (c) Just After Point, f and (d) At Point g
Figure 1.9 shows how the plastic deformation is distributed along the length of the specimen. Different gauge points have been shown marked on the specimen. It can be seen that by choosing any pair of gauge points, equidistant from the centre, one may arrive at a value of percent elongation which will differ from that obtained by choosing a different pair of gauge points. This example shows that percent elongation is not a unique quantity for a material but may vary with gauge length. For this reason gauge length is standardised so that comparison of ductility between two materials may be made consistently. For cylindrical specimen gauge length of five times the diameter is chosen as a standard. But in any case percent elongation is described with gauge length on which it has been calculated.
Elongation
The value of percent elongation may be obtained from diagram as the strain at the time of fracture multiplied by 100. More practically, the two pieces of specimen after fracture may be placed together touching at fracture surface and the distance between gauge points noted to measure final length Lf. Then, % elongation =
L f Lo Lo 100
. . . (1.5)
The % reduction in area of cross-section is also used as a measure of ductility. It is defined as : % reduction in area =
Ao A f Ao 100
. . . (1.6)
where A and L represent the area of cross-section and length respectively, f and o are respectively the suffixes to denote final and original values. It may be understood by above description that both % elongation and % reduction in area would define the local ductility of the material. Maximum uniform strain, which is the strain just before necking begins or the strain corresponding to ultimate stress may sometimes be used to indicate the ductility of a material. Ductility is an important property of material which governs its ability to be deformed in such processes as drawing, forging and extrusion. Adequate ductility ensures that the materials during these processes will not fracture. All associated property by virtue of which sheets can be rolled from material is termed malleability.
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SAQ 2
(a) (b) (c) Distinguish between proportional limit and elastic limit. Which one is higher in a stress-strain diagram obtained from a tension test? Use sketch to distinguish between upper and lower yield points. Also show the phenomenon of yielding. Yield strength which is the stress at lower yield point in a tension test is used to define elastic strength. How do we define elastic strength in those materials which do not show yielding? What are the measures of ductility? Name the manufacturing processes which are performed upon ductile materials. What is a ductility associated property and in which process it is useful? On a uniform cross-section bar two gauge marks A and B are made at a distances of 200 mm. The length between A and B is divided into 8 equal intervals, each equal to 25 mm. The bar is put under tensile load in the machine and pulled until fracture occurs. The two broken pieces are put together and intervals measured. The measure distances before test and after fracture are shown in figure below.
A 1 2 3 4 5 6 7 B
25 mm all intervals
27.1
27.2
28.0
30.0
32.2
27.5
27.0
26.8
All mm
Figure
Calculate the percent elongation for 50 mm length between 3 and 5. Calculate the percent elongation for 100 mm length between 2 and 6. Calculate the percent elongation for 150 mm length between 1 and 7. Calculate the percent elongation for 200 mm length between A and B. Plot % elongation against gauge length and conclude that gauge length has to be standardised.
1.2.4 Toughness
When the load acts upon the specimen, certain work is done. This work it stored in the specimen as its strain energy. If load P acts upon the specimen and causes certain displacement, dL the work done,
U =
P dL
. . . (1.7)
U = A0 L0
The integral
P dL = A0 L0
. . . (1.8)
d is nothing but the area under stress-strain curve. It means that the
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work done upon the specimen per unit volume during stretching, which is stored in the specimen as strain energy, can be measured by area under the curve. The maximum
Materials Applications
elastic energy per unit volume that material can absorb without attaining plastic state is known as modulus of resilience, while the energy stored in the specimen at any stress within elastic limit may be referred to as resilience. The toughness of a material is understood to be its ability to absorb energy during entire elastic and plastic deformation. The modulus of toughness is measured as area under entire stress-strain curve and is the energy absorbed by material of the specimen per unit volume upto fracture. From Figure 1.10 it can be seen that modulus of toughness will depend upon both the ultimate tensile strength and strain at fracture and material which is very ductile will exhibit a higher modulus of toughness as is the case with mild steel. On the other hand, modulus of resilience depends upon yield strength (or proportional limit or elastic limit) and hence a material with higher yield strength will have higher modulus or resilience. Higher toughness is a desirable property in materials used for gears, chains, hooks, freight car coupling etc. Higher resilience is desirable in springs.
High Carbon Steel Mild Steel Mild Steel High Carbon Steel
Figure 1.10 : Curves for High Toughness (Mild Steel) and Low Toughness (High Carbon Steel) Materials
For ductile materials like mild steel (Figure 1.10) the modulus of toughness may be calculated by approximate formula : Modulus of toughness = u f =
Y + u f 2
. . . (1.9)
where Y and u are respectively yield and ultimate tensile strengths, and f is the strain as fracture. Modulus of reliance is the area of triangle whose height is Y and base is the strain 1 corresponding to Y, i.e. Y. Hence modulus of resilience is Y Y . 2 Since
Y =
Modulus of resilience =
Y E
2 Y 2E
. . . (1.10)
Strain Hardening Referring to Figure 1.5 in Section 1.2 it was stated that on reloading from g the new curve joins the original curve at h, where point h is higher than point a. In other words it can be said that due to the initial loading of the specimen upto point f its yield strength has increased to h for if it were not so, the reloading curve must follow a path similar to original curve showing a yield strength equal to the height of point a. This rise in yield strength due to loading in plastic region is termed as strain hardening. If the material does not strain harden during plastic deformation it will extend at constant stress level. The materials that extend at constant stress level are called ideally plastic. Very soft materials behave in this manner. The explanation to strain hardening is found in defects in crystalline materials at atomic level. Dislocations are such defects. Strain hardening is a practical method of increasing elastic strength. Wires that make helical springs are strain hardened. 14
y yy = P/A
P/2A
P/2A P/2A x
y x P 45
0
(a)
(b)
(c)
Figure 1.11 : (a) A Tension Specimen; (b) State of Stress at any Pont with respect to x-y axes when yaxis is Parallel to Axis of the Specimen; and (c) State of Stress with respect to x y axes when an Angle 45o with Axis of Specimen, the Maximum Shearing Stress equal to P/2A Acts Parallel to x and y Axes at any Point
On the other hand a ductile material is stronger against tension than against shearing stresses. In a tension specimen, the maximum shearing stress acts upon a plane that is inclined at an angle 45o with the axis as shown in Figure 1.11(c). However, due to formation of neck the state of stress changes considerably. Still the final fracture of a ductile material shows a characteristics cup-and-cone fracture surface. The cup and cone match at an angle 45o with the axis, showing that the final fracture has occurred along these planes all around the periphery. If a highly polished specimen is tested in tension then on the surface at yield point fine lines will appear. These lines are generally inclined beginning of sliding of atomic planes under the effect of maximum shearing stress (Figure 1.11(c)).
Direction of Load
Cup
Cone
45
45
(a)
Figure 1.12 : (a) Cup and Cone Fracture of a Ductile Material in Tension; and (b) Flat Fracture Surface with Granular Appearances in Case of Brittle Material
Direction of Load
(b)
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Materials Applications
Example 1.1 A steel specimen shows upper yield point at 210 MPa and lower yield point at 200 MPa. If modulus of elasticity, E, for steel is 210 103 MPa. Calculate modulus of resilience. Solution Modulus of resilience =
2 Y 2E
Where, Y is yield strength. Y corresponds to lower yield point. Hence, Y = 200 MPa
Modulus of resilience =
N-m/m3
or
Example 1.2 A steel specimen of 10 mm diameter and 50 mm gauge length was tested in tension and following observations were recorded. Load at upper yield point = 20600 N Load at lower yield point = 19650 N Maximum load = 35550 N Gauge length after fracture = 62.43 N Calculate modulus of resilience and modulus of toughness. Also calculate % elongation. E = 210 103 N/mm 2 . Solution Area of cross-section of specimen, A0 =
2 d 4
A0 =
Yield strength,
Modulus of resilience =
2 Y 2E
(250)2 2 210 10
3
N-mm/mm3
...
= 148.8 10 3 N-mm/mm3
(i) 16
(Compare with Example 1.1 to note that for higher yield strength modulus of resilience is larger). Modulus of toughness =
= 87.8 N-mm/mm3
% elongation = 25%
SAQ 3
(a) (b) (c) Define modulus of resilience and modulus of toughness. What is strain hardening? Is it used in practice? A steel wire having Y = 190 MPa is required to have a modulus of resilience of 140 10 6 N-m/mm3. The yield strength can be increased by strain hardening. What should be the % increase in yield strength. E = 210 103 N/mm2. Why should a ductile material fracture in cup and cone configuration whereas brittle material fails on cross-section of a bar tested in tension? What are Luders bands? Are these bonds shown by cast iron? Give reasons. Why should modulus of resilience of spring material be high? For a given material modulus of resilience can be increased either by increasing Y or decreasing E. Which one is possible?
P A
where, A is the current area of cross-section corresponding to load P during the tension test. True strain is defined as summation of incremental strain over the entire load range from zero to P. If the initial length L0 changes to length L during load change of 0 to P, true strain
L0
dL L = ln L L0
Because of the final form of the true strain, it is often called logarithmic strain. Since,
L = L0 + L
17
Materials Applications
L . L0
Stress
Strain
Figure 1.13 : Comparison of Engineering Stress-Strain Diagram with True Stress-Strain Diagram
It is well known fact that volume of material remains constant during plastic deformation, which means
A0 L0 = AL or
With this fact it can be shown that
A0 L = A L0
. . . (1.14)
= ln
A L = ln 0 L0 A
. . . (1.15)
Properties in Terms of True Stress and Strain Under some circumstance, particularly when the material is in plastic range of deformation, certain material properties are defined in terms of true stress and strain. True stress plotted against true stain is best represented as in Figure 1.14(a) with properly choosing strain scale because it is in plastic range. In most cases the relationship is non-linear and the slope of the curve at any given strain is called modulus of strain hardening at that strain. Several experimental observations suggest that and satisfy following relation
= k n
. . . (1.16) . . . (1.17)
or
ln = ln k + n ln
where, k and n for a given material can be regarded as its properties, k is called strength coefficient while n is strain hardening exponent. As is obvious from Eq. (1.17) the plot between ln and ln is a straight line whose slope is equal to n and intercept on ln axis is ln k. 18
The constants k and n can be determined experimentally if diameter of specimen corresponding to load P is recoded. A special gauge will have to be used for measuring least diameter along gauge length with increasing load.
True Stress,
1 True Strain ,
In
Strength Coefficient, k In
(a)
(b)
Figure 1.14 : True Stress Plotted as Function of True Strain (a) Linear Scale; and (b) In Scale
One may refer to Figure 1.2 and once more may note that at point d the stress is maximum and even when strain keeps increasing stress keeps reducing after point d. in an experiment wherein a ductile material is being tested perceptible reduction in diameter is observed after ultimate point, d. The condition of beginning of reduction of diameter in the localised region is termed necking and it is the neck where diameter is always minimum. This phenomenon is also called instability and cannot be reversed. However, if load is removed very fast further necking may be stopped. If the load is maintained then necking will continue until fracture. It is possible to establish the condition of beginning of necking in terms of strain hardening exponent. Starting from the fact that tangent to load elongation curve at highest point will be horizontal, it can be written that dP = 0 or, d ( A) = 0 where, is true stress and A is the area of cross-section of tension specimen at maximum load.
or
dA + A d = 0
d dA = A
. . . (1.18)
Also since volume of specimen of length L and cross-section area A remains constant, i.e. AL = constant. d (AL) = 0 or, i.e. L dA + A dL = 0
dA dL = = d A L
. . . (1.19)
d = d
or,
d = d
d ( k n ) = k n d
. . . (1.20)
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i.e. or,
n k n 1 = k n
= n
. . . (1.21)
Thus it is seen that when true strain becomes equal to strain hardening exponent n, necking will begin and fracture will ultimately occur. Metal working process like drawing, rolling, extrusion etc. may be thus performed safely it at no stage the true strain is allowed to reach the value of strain hardening exponent. The strain hardening exponent can be determined from tension test. Such determination will be exemplified in solved example. Example 1.3 A copper specimen of 64 mm gauge length and 12.80 mm dia. was tested in tension. Following two diameters were recorded in the plastic range of deformation. Load = 25.75 kN, d1 = 12.176 mm Load = 24.25 kN, d2 = 8.581 mm Calculate strength coefficient and strain hardening exponent. Solution Original area of cross-section,
A0 =
2 d0 = (12.8)2 4 4
A0 = 128.6144 mm 2
2 d1 = (12.176)2 4 4
A1 = 116.3802 mm 2
2 d 2 = (8.581)2 4 4
A2 = 57.8023 mm 2
A0 A
= ln 1
= 0.0999 1
and
2 = ln
2 = 0.7998
Thus two pairs of values are obtained. Use these values of and in Eq. (1.15), i.e. ln = ln k + n ln
and
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Subtract first equation from second 6.039 5.399 = n ( 0.2234) n ( 2.3036) i.e. or, (i) Using this value in one of above equations 5.399 = ln k + 0.3077 ( 2.3036) (ii) True stress-true strain relationship for copper is
= 449.35 0.3077
0.64 = 2.0802 n
n=
...
...
(iii)
Example 1.4
At what load the specimen of copper of last example will begin to neck in a tension test? What will be its ultimate tensile strength?
Solution
The condition for necking is u = n = 0.3077 Hence using this value of u , the true stress at necking
0.3077 u = 449.35 (0.3077)
= 449.35 0.696
2 u = 312.67 N/mm
u = ln
A0 Au
128.6144 Au
0.3077 = ln
128.6144 = 1.36 Au
Au = 94.55 mm2
Pu = u Au = 312.67 94.55 N
= 230 N/mm2.
SAQ 4
(a) Distinguish between true stress and engineering stress. Distinguish between true strain and engineering strain. 21
Materials Applications
Define strength coefficient and strain hardening exponent. How can these properties be determined? What should be limiting strain in process of wire drawing? A steel specimen with gauge length of 62.5 mm and diameter of 12.5 mm was tested under tension. Thee gauge length at maximum load of 72.5 kN was 71.55 mm and gauge length at fracture load of 66.6 kN was 80.5 mm. Find specimen diameter at maximum load and fracture load. [Use A0 L0 = AL, dia. at maximum load = 11.68 mm, dia. at fracture = 11.01 mm].
(e)
Find true stress and true strain at maximum load and fracture in (d). [ u = 0.1345, f = 0.25]
Compression Specimen
22
At atomic level, unlike tensile stress case, no definite maximum exists in stress vs atomic bond curve and thus there is not limit to repulsive force that can be build up between atoms. Thus, the compression will not become the cause of failure and will always result
in elastic strain. It is the shearing stress component which causes the slip of atomic planes and results in bulging of the specimen, and plastic deformation. With this consideration the stress-strain diagram within elastic limit will be the extension of that in tension with the difference that the elastic behaviour in compression ends a higher values of stress and strain as shown in Figure 1.16(a).
+
Compression Stress - +
Tension
Strain
(a)
(b)
Figure 1.16 : (a) Elastic Stress-Strain Relationship in Tension and Compression and (b) Behaviour of Rubber in Tension and Compression
Long chain ploymeric materials like rubber, cork or wood show distinctly different behaviour in tension and compression. In tension the coiled molecules are first uncoiled and then strained whereby such materials show low stiffness followed by higher stiffness in tension. On the other hand compression may tend to cause further coiling in the initial stage followed by elastic compression of molecules showing slight stiffening effect in immediate decrease of stiffness. As the cells are compacted with increasing compressive load, the stiffness will further increase.
Shear Plane
Even in brittle materials some lateral bulging will tend to occur. This lateral deformation is greatly restricted at flat specimen ends due to friction between the platens and the specimen ends. This friction reduces for longer specimen. If the compression specimen is too long, it may have a tendency to buckle. To minimise the influences of friction and buckling the length to diameter ratio of compression specimen has to be chosen very carefully. For determination of compression strength this ratio is between two and three.
23
Materials Applications
However, for determination of elastic properties the ratio of length to diameter may be chosen between eight and ten. If this ratio is less than 1.5 the fracture plane might intersect the end which is undesirable. It may be noted that the brittle material in which the atomic bonds are not re-established will ultimately fracture due to slip which may occur along one or several planes. The later failure is fragmentation. While the cracks, pores and holes play very important role in reducing tensile strength the same do not become active under compressive stress. They rather close and net compressive strength is higher than tensile strength particularly in case of brittle materials. Figure 1.18 shows this fact for cast iron, and Table 1.2 describes the same for several materials.
140 0 Stress (MPa) 140 280 420 560 700 840 Compression 0.06 0.04 0.02 0.0 0.01 Tension
Figure 1.18 : Stress-Strain Diagram for Cast Iron in Tension and Compression
[Note : Concrete 1.3 mix by volume, water cement ratio 0.64 (by volume) cured for 28 days.]
Compression test is terminated if excessive deformation occurs and specimen does not fracture. The hydrostatic compression is the loading in which every element is subjected to equal compression stresses in all three directions. This state of stress which is obtainable when a body is submerged in water does not cause any distortion but produces only volume reduction. In such a test the specimen my not fail at all because at atomic level such state of stress will tend to push atoms into each other and no slip would occur, hence no plastic deformation.
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circular concentric grooves and created to hold lubricant, particularly in specimens of ductile materials. The problem of bulging is minimal in concrete and hence l/d = 2 is often used for this material. In yet another method the ends are made conical hollow to receive similar loading plate as shown in Figure 1.19(b). The angle of cone is equal to angle of friction between the materials of platen and specimen, and it is not readily known. It is important that compression force must be applied such that rate of straining is constant. Most materials required to be tested in compression are viscoelastic in nature and do not reach equilibrium readily. Prescribed strain rates are, therefore, maintained.
R =
(a)
(b)
Figure 1.19 : (a) Self Aligning Platen; and (b) Conical Hollow End of Compression Specimen
SAQ 5
(a) (b) (c) (d) (e) Sketch how a mild steel specimen will change shape under increasing compressive load. Why does cast iron show higher compressive strength than tensile strength? The stiffness of rubber first increases in tension and then decreases whereas it increases in compression. Give reason. Why should a compression test specimen, which is compressed between two rigid platens, bulge around mild section? Why should axis of machine coincide with the axis of compression test specimen? What methods are adopted in place of flat pattern to ensure coaxiality? The ratio of length to diameter in cast iron compression specimen is 3 whereas it is 2 in concrete specimen. Give reason. A concrete specimen with l/d ratio less than 1.5 tested in compression. Sketch the failure plane. Why this specimen is not recommended? Sketch the same plane if l/d = 3. In which case l/d between 8 and 10 is preferable in compression specimen.
(f) (g)
(h)
Materials Applications
screw type varies in size and capacity over wide range of a few newtons to a few tones. Hydraulic machines are generally of high capacity. These machines have limited speeds of movement of cross-heads so that loads applied are within the definition of static. The strains rates greater than 10 6 mm/mm.s are usually not available on universal testing machines. Modern computer operated machines now-a-days make wide range of strain rates available and they can be even used for cycling stresses. Instron and MTS machines are examples. Whatever by the type of machine general features remain same, a universal machine applies axial load which can act as tension, compression, shear or bending load upon the specimen depending upon positioning of the specimen. The application of load is effected by relative movement between two cross-heads. Generally machines have two cross-heads rigidly connected with each other and a third cross-head is positioned between them. Each machine has proper specimen holding and supporting devices to apply desired type of load upon the specimen. The load applied upon the specimen has to be balanced and measured and as such each machine should have this device. Figure 1.20 shows essential features of a mechanical type of universal testing machine. The upper and lower cross-heads are rigidly connected with middle cross-head can be moved down by a screw and gear arrangement driven by an electric motor. While moving down the middle or movable cross-head can apply a tension load on a specimen gripped between the moving cross-head and upper cross-head. Alternatively, the middle cross-head can apply a compressive load on a specimen placed on lower cross-head. The fixed cross-heads are supported over a compound lever system at whose end a poise or balance weight can be shifted to achieve balance of the lever and thus measure the applied force. The balance weight is often replaced by a swinging pendulum whose rotation is transmitted to an indicator through gear train for direct indication of load. In a hydraulic universal testing machine (Figure 1.21) the rigid frame containing upper and lower cross-head is attached to an accurately finished piston which exactly matches with the hydraulic cylinder. The middle cross-head, whose position is adjustable remains stationary when the load is applied upon the specimen. An electric motor rotates a hydraulic pump that forces the oil into the cylinder whereby the rigid assembly of lower and upper cross-head is lifted up. During this movement a tensile load will act upon the specimen held between upper and middle cross-head and a compressive force will act upon a specimen placed on the lower cross-head. The hydraulic pressure on the bottom of the hydraulic cylinder is transmitted to a hydraulic capsule which is connected to a Bourdon tube gauge which can be calibrated to read the force directly. The middle cross-head can be moved up or down by separate positioning screws and during its positioning movement the hydraulic pump is not operated. Both types of machine often incorporate the mechanical graphical devices to plot the load displacement curves. Modern machines incorporate load cell and displacement gauges whose electrical signals respectively proportional to load and displacement can be plotted automatically on X-Y plotter when the test is proceeding.
Fixed Cross Head Tension Specimen Balance Weight Movable Cross Head Screw Fixed Cross Head Levers
26
Gears
Motor
Mechanical Properties and Their Determination Figure 1.20 : Screw-Gear Type Universal Testing Machine
Upper Cross Head Columns Space for tension Specimen Middle Cross Head Space for Compression Specimen Bourdon Tube Gauge Scale
Motor
SAQ 6
(a) What is the difference between a universal and special purpose machine? Name a few special purpose machines. [Special purpose testing machines : Torsion testing machines, Compression testing machine, Harness tester, Impact testing machine.] Show a compression test specimen placed in a universal testing machine. What are the difference between a conventional universal testing machine and modern computer controlled testing machine? What devices are used for recording load and elongation in universal testing machines? A tension test specimen records 40,000 N force over a period of min. During this time the initial gauge length of 100 mm changes to 120 mm. If the mechanical efficiency of the machine is 80%, calculate the power consumed by driving motor. (16.6 W).
Static tensile tests are often performed at very low strain rate [10 6 mm/mm (sec)] and variations permissible in conventional testing machines are generally so limited that they do not influence the static properties. But machines capable of varying strain rate by as much as 100,000 time have been developed and tensile tests in such machines have revealed that tensile properties are quite susceptible to changes under increasing strain rates. In general it is observed that yield strength and ultimate tensile strength increase with increasing strain rates, percentage elongation seems to first increase with strain rate but then decreases of remains constant. One important influence of high strain rate is that for such material as low-carbon steels which do not show yield point in normal tensile tests, yield point appears. Further, yield strength out of all static properties is not affected by strain rate. Figures 1.22 and 1.23 show how these properties are influenced by strain rate. The effect of strain rate on mechanical properties become more pronounced at elevated temperatures. 27
Materials Applications
The load applied at strain rate higher than 10 mm/mm (sec) is classified as impact load, and under impact loads ductility of a material has tendency to reduce.
630 560 490
Stress (MPa) 100 / Sec 0.5 / Sec Strain Rate 300 / Sec
30
40
50
Figure 1.22 : Stress-Strain Diagram for Mild Steel at different Strain Rates
80 Y / u 60 u 40 Y % Elongation 20
-2 3
10
1 10
100
10
A number of accidental failures of ships during Second World War drew attention of engineers and metallurgist on the problem of brittle fracture. The said ships had all failed showing brittle fracture wherein the crack propagated suddenly without any plastic deformation and producing loud report due to sudden release of strain energy. As a result fracture tendency in material like steel develops due to (a) (b) (c) a triaxial state of stress, low temperature, and high strain rate or rapid rate of loading.
It has been recognised that most brittle failures in service conditions take place due to combined effects of triaxial state of stress, such as existing at a notch and low temperature. These effects are, however, accentuated, by high strain rate, hence the laboratory tests combining all these three conditions are performed to determine tendency of a material to fail in brittle manner. Such tests known as impact tests are performed in impact testing machine one such machine is shown in Figure 1.24.
Scale Swinging Pendulum h2 Specimen
h1
28
The Charpy impact test is performed on a square cross-section specimen, having a notch on one side in the central cross-section. The specimen is placed in charpy impact testing machine shown in Figure 1.24 such that swinging pendulum strikes the specimen at the central cross-section but on the opposite side of the notch. The specimen is simply supported on the platform of the machine as illustrated in Figure 1.25(a). The commonly used impact specimen is 10 mm square crosssection beam of 55 mm length, supported over 40 mm span and having a V-notch in the central plane whose included angle is 45o, depth is 2 mm and root radius is 0.25 mm.
Impact Load 2 mm Impact Load 55 mm 22 mm 45
0
10 mm 40 mm
(a)
(b)
The charpy test machine shown in Figure 1.24 is rigid and strong structure of two columns on a heavy base. The columns carry a heavy swinging pendulum at their top which swings on a frictionless pin, and support platform at the bottom for the
29
Materials Applications
specimen. When the pendulum is in its vertical position its striking edge is level with the central cross-section of the specimen. The mass of the striker (or hammer) is concentrated in vertical plane. A circular disc scale mounted centric with the pin of the pendulum reads its position but is often calibrated in terms of the potential energy of the pendulum. To perform the experiment the specimen is first placed in position by the help of a centring device, the pendulum is raised to and held at its extreme position and then allowed to fall. At its extreme position, h1 from the specimen, the pendulum has potential energy Wh1 where W is the weight of the pendulum. In this position the arm of the disc scale is in its extreme position, reading total amount of energy stored in the hammer. The speed of approximately released from its raised position and strikes the specimen at a speed of approximately 4.8 m/sec. Under this condition the specimen is forced to bend with a rate of 103 mm/mm (sec) resulting into instantaneous fracture or plastic deformation. The pendulum continues its swing after fracturing or deforming the specimen and rises to height of h2. Thereafter, the pendulum swings back to oscillate pointer which does not swing back with the pendulum. As already pointed out the disc scale is calibrated to read energy and the position of the dead pointer reads the difference of potential energies of the pendulum in two extreme positions corresponding to h1 and h2. Naturally this difference of the energy is that which was absorbed in fracturing the specimen. Thus if Uf denotes the impact roughness, Uf = W (h1 h2) . . . (1.17) It may, however, be recognised that some energy loss may occur in the bearing of the pendulum and also due to resistance of air offered to swinging pendulum. These two losses may be determined by simple experiment in which the pendulum is allowed to fall from is extreme position without placing the specimen in its path. The other extreme position to which pendulum rises will be slightly lower than its initial position. The difference of the potential energies in two positions, indicated by the position of the dead pointer, is the loss of energy to bearing friction and air resistance. This energy loss (UL) has to be subtracted from Uf to obtain corrected value of impact toughness, i.e. Uf = Wh1 Wh2 UL
Izod Impact Test
. . . (1.18)
Izod impact test specimen, shown in Figure 1.25(b), is either circular or square in cross-section and contains a V-notch at one end. The specimen is clamped at notched end in vertical plane as cantilever beam. The swinging hammer strikes the free end at a distance of 22 mm from the notch. The general construction of Izod impact testing machine is same as that of Charpy machine except the hammer. The mass of hammer in Charpy machine is distributed in a vertical plane whereas in Izod machine it is distributed in horizontal plane. Now-a-days same hammer is being used in machine with attachable strikers for charpy and izod specimens.
SAQ 7
(a) (b) (c) (d) (e) How does strain rate influence yield strength, ultimate tensile strength and percent elongation? Distinguish between shapes of hammers used in Izod and Charpy impact tests. Distinguish between types of loading and specimens for Charpy and Izod impact tests. What are the factors that make material behave in brittle manner? A hammer weighing 50 N at the end of a swinging arm of length 800 mm is lifted to a height of 1500 mm from the level of Charpy test specimen. With what speed the hammer will strike the specimen.
30
1 mv 2 and calculate 2
v = 2 g h = 2 9.81
If the height of specimen from ground level is known, then the PE of hammer in lower most position may be subtracted from KE.] (f) The hammer of (e) is released and no specimen is placed in its path. The hammer rises to a height of 1450 mm. How much energy is lost in friction and air resistance? [2.5 N-m]
20
40
Temperature ( C)
Presence of notch provided one condition for brittle fracture. The qualitative behaviour of material can be understood by accepting the fact that the material has distinct properties as yield strength and fracture strength. Both of these properties decrease with increasing temperature, reducing strain rate and triaxility. However, the reduction rate of fracture strength is much less than that of yield strength, and hence at certain point these curves
31
Materials Applications
intersect as shown in Figure 1.27. Thus the yield strength which is initially higher than fracture strength becomes lower after the point of intersection. The specimen loaded on right hand side of the ordinate CA will first reach yield stress and will plastically deform. Contrarily the specimen loaded on left hand side of CA will first reach fracture stress and will thus fail in a brittle manner without showing any plastic deformation. The line CA can be called the line of transition from brittle to ductile. In practice CA becomes a band in which change from brittle to ductile behaviour takes place very fast and its width depends upon the nature of material.
Yield
Strength
Fracture C
Figure 1.27 : Various of Yield and Fracture Strengths with Temperature, Triaxiality and Strain Rate
After each impact test, the fracture parts must be explained for details. A brittle fracture, as below temperature Tr exhibits a granular surface which is characteristics of cleavage fracture and absence of any shear deformation. As the temperature rises the part of the notched section, the region near the surface, tends to fracture in shear mode whereby energy absorption in fracture increases. This shear deformation results in transverse contraction of notch section and the percent contraction is sometime reported to indicate the ductility of the material. The fracture surface on which failure occurs due to shear is marked by fibrous appearance and considerably transverse contraction in the notched section. Figure 1.28 illustrates the fracture surface. (a) (b) (c) Granular surface, Mixed granular and fibrous appearance with some side contraction, and Mixed fracture with considerably fibrous appearance and side contraction, 2d is the total transverse contraction at the notch.
Fibrous Appearance d d
Granular Surface
(a)
(b)
In addition to temperature, there are several other factor that affect the impact toughness. One important factor is the machine itself. Each machine will absorb some energy when the specimen is struck by pendulum and this amount though very small will vary from machine to machine and thus the impact toughness will be affected. The notch root radius is starndarised and any deviation from it will influence the result. A sharper radius will tend to reduce the impact toughness while a coarser radius will increase the amount of energy absorbed during fracture.
Metallurgical Factors
32
Two alloying elements, namely carbon and manganese are highly effective in altering the impact toughness of steel. Each 0.1% increase in carbon percentage raises the ductility transition temperature by about 14oC. Each 0.1% addition of manganese in steel reduces the transition temperature by about 5oC. For a satisfactory notch impact toughness a 3 : 1 ratio for manganese percentage to that of carbon is often suggested. Phorphorous also raises the transition temperature, and steel with high phosphorous content is not preferred at low temperature. Nickel generally has beneficial effect on notch impact property while chromium has little effect. Silicon tends to increase the transition temperature if in excess of 0.25%. Nitrogen has detrimental effect on notch toughness, though it is hard to determine its effect because of interaction with other elements. Molybdenum increases the ductility transition temperature almost as rapidly as carbon. The presence of oxygen in steel is highly dangerous as it raises the transition temperature in the higher temperature range. Deoxidation practice as followed in production of killed and semi-killed steels is highly beneficial in improving impact properties. Grain size has pronounced effect on transition temperature. In general the transition temperature reduces with decreasing grain size. This effect is observed both in mild steel and higher alloyed steel. Since in thick hot rolled plates it is not possible to obtain uniform fine grain size throughout the thickness, the transition temperature will be appreciably higher. The loss of fracture energy in specimen of thicker section is also due to geometrical factors, such as finding a weaker point is easier in the large section than in smaller section of the same material.
SAQ 8
(a) (b) (c) (d) Impact test is normally performed on steel. Give reasons. Define transition temperature and explain its usefulness in selecting material for ships. Explain how yield strength and fracture strength are affected by decreasing temperature and increasing strain rate and triaxiality of stress. A steel has transition temperature at 0oC. The transition range is over 10oC. Show the appearance of impact fracture surface of charpy specimens tested respectively at 5oC and 5oC. The alloying elements in steel affect impact strength. What are the effects of C, Mn, P and Si? What is killed steel? Why this treatment is given on steel? Which of the following will decrease the transition temperature? Is decreasing of transition temperature advantageous? Explain. (i) (ii) (iii) (iv) (v) Carbon addition Mn addition Mo addition Deoxidation Decreasing grain size.
1.8 HARDNESS
The technological use of material has made the term hardness very common but it still remains a poorly defined material property. Even a layman can differentiate between a harder and softer material, but it may be difficult to designate a material by its
33
Materials Applications
quantitative hardness because it is difficult to correlate the property with stress or strain. In general hardness may be understood as a property which provides resistance to permanent plastic deformation. Such a deformation may leave a permanent impression on the material surface. For engineering purpose the property of hardness is defined as resistance to indention or scratching. Based upon this two methods of hardness measurement, viz. (a) (b) are used. Ability of material to resist abrasion, cutting or penetration is also attributed to hardness. scratch hardness, and indentation hardness
34
Specimen
Figure 1.29 : The Specimen and Ball Indenter for Brinell Hardness
If,
P = Applied load, kgf, D = diameter of the ball, mm, and d = diameter of the indentation, mm.
then,
BHN =
P D (D 2 D d )
2 2
. . . (1.19)
Although, from the definition and Eq. (1.19) it is obvious BHN has units of kgf/mm2, this does not convey and physical meaning because the load P is not uniformly distributed over the indented surface. The BHN value obtained for a material is fairly reproducible provided the load P and diameter D are same for each test. To obtain the same BHN at any other load the ratio d/D must be constant which requires that P/D must be constant. Thus for P = 3000 kgf, D = 10 mm, P/D = 30. Hence if a load of 2000 kgf is to be used the ball dia must be 8.15 mm, and for a 1 mm diameter indenter a load of 30 kgf must be used. The calculated value of BHN may be slightly in error because of the assumption that the diameter of indented surface is same as that of indenter which is not exactly true. After the removal of the load, some elastic recovery takes place whereby the final diameter of the surface becomes different than ball diameter. For harder materials elastic recovery is greater than for softer materials. An anomaly may arise in measurement of diameter of indentation diameter, d, due to localized deformation of metal in the region of indentation. Figure 1.30(a) shows the cross-section through indentation in a cold-worked metal. Because such a metal has little capacity to strain harden the metal has a tendency to pile up along the edge of impression. This behaviour is known as pilling up or ridging. Although the measured d is larger than actual diameter of indentation, yet the hardness is calculated on the basis of measured d as shown in Figure 1.30(a) because the ridge also carries part of the load. Figure 1.30(b) shows the sinking in of an annealed metal. The depression around the edge of metal will apparently increase the diameter of the impression. To arrive at the true value of d, the indenter is often coated by a dye before making the indentation. The diameter then can be measured on the coloured ring.
d
(a)
(b)
Figure 1.30 : Cross-section through Brinell Indentation on Two different Materials Showing (a) Ridging; and (b) Sinking In
While making indentation on a plate specimen care must be exercised to avoid any interference between the indentation and the edge of specimen and the interference between two indentations. For this reason it is advisable that the centre of the indentation should be at least 1.5 D away from any edge. Same distance should be maintained between two neighbouring indentations. Similarly the thickness of the plate must be at least equal to the diameter of indenter so that the plastically deformed zone below the indentation does not interfere with the back surface. 35
Materials Applications
120
(a)
Rockwell Scale B F G E
Load
Hardness
(b) Figure 1.31 : (a) Brale Indenter (Diamond Cone); and (b) The Spherical Ball Indenter of Steel, d is the Depth of Penetration
One combination of load and indenter will not be able to produce a wide range of hardness. Therefore, loads with three indenters are used in different combinations to provide wide range of measure hardness of several materials. Figure 1.31 shows the different scales of Rockwell hardness which is often expressed as R with a suffix A, B or C to indicate the scale. The scale (RC) is very commonly used for steel. The Rockwell hardness is many times preferred over Brinell hardness because this test can be performed on very small parts in finished or unfinished condition. Since the indentation is very small, the finished surface is not spoilt and unfinished surface does not affect the indentation. Also since no measurement of indentation and consequent calculation are needed, the method is direct, fast and free from personal error. However, several precautions regarding specimen and indentation have to the observed. The thickness of the plate specimen must be at least ten times the depth of indentation. This will avoid any effect of indentation to pass through the thickness. The distance between two adjacent impressions should at least be three times the size of indentation. The hardness read from indenting the curved surface should be corrected for curvature. The surface on which indentation is made must be clean and smooth and it should be well seated upon a clean platform. The rate of load application is controlled in Rockwell hardness testing machine and it is achieved by a dashpot adjustment corresponding to a standard rate of loading.
SAQ 9
(a) 36 What do you understand by hardness? Arrange following substances from hardest to softest.
Diamond, steel, copper, gypsum, corundum, calcite. (b) (c) (d) What is Mhos hardness scale? Why is it difficult to measure hardness of steel on Mohs scale? How is hardness measured? What is Brinell hardness? If for a ball of 10 mm diameter a load of 3000 kgf is applied on the ball for measuring BHN, what force will have to be applied upon a 5 mm ball? Is Rockwell hardness measured by indenting the surface? A diamond cone indenter under a load of 100 kgf penetrates through 0.15 mm in steel A but 0.1 in steel B. Is A harder than B? Explain. Describe the procedure for finding Rockwell hardness. What limitations are placed on making ball indentation in a 12 mm thick steel plate in Brinell hardness measurement? Explain ridging and sinking in. Compare Rockwell hardness with Brinell hardness measurement in respect of a steel plate 5 mm thick.
(e)
1.854 P d2
. . . (1.20)
where, P is the indenting load in kgf. d is the average of two diagonal and is the angle between opposite faces of diamond come which is 136o.
135
o
Diamond Indenter
Indentation d d
Figure 1.32 : Diamond Indenter and its Square Indentation in Vickers Hardness Test
The impression of the diamond cone is square under any load, hence the impressions are geometrically similar irrespective of load. This provides a great convenience of a continuous scale from very soft material to very hard material without changing the load or indenter. Generally for a given load hardness from 5 to 1500 VHN can be measured with Vickers indenter, and it makes VHN independent of load except at very high loads. However, loads between 1 kgf to 120 kgf are commonly used for VHN measurement for very soft to very hard metals. For the reason of independence of VHN of load, this method has widely been accepted as a research tool.
37
Materials Applications
However, Vickers hardness measurement cannot be made on poorly finished surfaces and an element of personal error is introduced because of accurate measurement of the diagonal of indentation. In addition, the anomalies of ridging in case of cold worked metals and of sinking in in case of annealed metals are also observed in Vickers impression as in case of Brinell indentation. The barrel shape of indentation causes an apparent increase in hardness due to ridging. Pin-cushin shape of indentation causes an apparent decrease in VHN due to sinking in as shown in Figure 1.33 along with perfect Vickers indentation.
(a)
(b)
(c)
Figure 1.33 : (a) Perfect Indentation of a Diamond Pyramid; (b) Barrrel Shaped Impression due to Ridging; and (c) Pin Cushion Shaped Impression due to Sinking In
Knoop hardness measurement is such a method in which a diamond indenter having pyramidal faces with very large included angle is used (Figure 1.34). This indenter is pressed against a metallographically finished surface to produce an indentation of the diamond shape (Figure 1.34). The longer diagonal of the impression is approximately seven times the shorter diagonal. The depth of indentation is about one thirteenth of the longer diagonal. The Knoop Hardness Number (KHN) is the measure of the ratio of load to projected area of impression. Thus, KHN = P d C
2
. . . (1.21)
where d is the length of longer diagonal and C is a factor of diagonal pyramid, supplied by the manufacturer. The indentation load is very small, often as low as 30 gms. The small impression is measured under microscope and also since the area over which the hardness measurement is made is very small, KHN is sometimes called microhardness. It is highly essential that the surface of sample must have metallographic finish as well as the indenting load controlled carefully. Any error in measurement of d can cause error in hardness. The error in locating ends of the impression becomes larger at lower load and hence the hardness has a tendency to increase with reduction in load and corrections for such error are available.
38
Yet another method of dynamic hardness measurement was developed by Martel. He used a pyramidal indenter upon which a ram falls to produce an indentation of sample surface under dynamic loading condition. The Martel Hardness Number (MHN) is defined as :
MHN =
where, W = weight of falling arm, h = height of fall, and V = volume of indentation.
Wh V
. . . (1.22)
For paving materials the hardness measurement method has been developed in which wear and abrasion of samples is measured. A number of preweighed pieces of concentrate aggregate of bricks are placed in drum which is subsequently rotated for a fixed number of cycles. Due to tumbling in the drum the sample pieces wear and lose weight. The percentage loss of weight is used as a measure of wear or abrasion hardness. This is known as Rattler method.
80 60 50 40 30 20 10 0 Sceleroscope no. 70
Vickers
Figure 1.35 : Approximate Relationship between Tensile Strength and different Hardness Numbers
The dependence of both strength and hardness on the binding forces between atoms will certainly correlate the strength and hardness, a fact which was pointed out in Section 1.21. However, because the mechanism of indentation is quite complex and involves such factors as triaxial stress state, friction, strand hardening and creep, it is not theoretically possible to correlate hardness with properties of strength, friction and creep etc. However, attempts have been made to produce closely approximated relations and one such between hardness and ultimate tensile strength is that ultimate tensile strength equals 3.4 BHN in N/mm2. Figure 1.35 describes the relationship between hardness on different scales and ultimate tensile strength of structural steels. Similar relationship exist for different materials but they will be different than the ones shown in Figure 1.35. The techniques used for improving the material strength, such as heat treatment, mechanical treatments or alloying also result in the increase in hardness. Therefore, hardness check of specimens is the quick and easy method of ascertaining the success of such treatment. The hardness test is most widely used for all manufacturing operations as a measure of uniformity and quality.
SAQ 10
(a) Explain how Brinell hardness and Vickers hardness are similar. What is the difference between two methods? 39
Materials Applications
(b)
What are the difficulties in using Vickers hardness measurement on cold worked and annealed specimens? Why should a barrel shape of indentation in case of cold worked metal result in apparent increase in hardness? Which method of hardness measurement is recommended for a region having hardness gradient? Which method of hardness measurement requires surface to be finished metallographic? In a given material which is perfectly homogeneous at microscopic level Knoop hardness is measured at several points in a narrow zone with force decreasing from 100 gf to 30 gf. Will the results be closely equal on there is possibility of variation in hardness? Which method uses loss of weight as measure of hardness? For which materials this method is used. Arrange materials A, B and C in order increasing hardness.
Material Hardness A 80 RB B 300 BHN C 50 RC
(f) (g)
(h)
What should be the hardness of material in RC which is required to have a tensile strength of 1054 MPa?
1.9 FATIGUE
In most service conditions the parts of structures and elements of machines are subjected to loads that change with respect to time. The aerodynamic loads on aircrafts, the earth resistance to earth moving machines, the wind load on buildings and direct load on rail road rolling stock etc. all change with time and so do the stresses, caused by these loads. The important point to be noted is that material behaves differently under these variable loads than under static load. The behaviour of material under variable loads is referred to as fatigue. In recent past several failures of structures and machinery have been attributed to fatigue and this recognition of fatigue problem has given a great impetus to researches in the area because of huge loss of life and capital resulting from such failures. However, the present day understanding of fatigue phenomenon has reached a stage where exploitation of material without endangering human life and compromising economical gains has become already possible. The main reason for fatigue behaviour to have assumed such significance is the weakness of material under variable load. A stress level at which failure under static condition will not take place is not safe if the stress repeats with respect to time. Theoretically there is no stress level for most materials which will be indefinitely safe. However, practically some safe stress level can be defined and determined for some specified number of cycles. Yet another characteristics of fatigue which makes it an important design consideration, is that fatigue takes place in brittle manner. So there is not plastic deformation giving a warning for impending failure.
40
is also termed fluctuating. About 90% of data are collected under fully reversed stress cycle in which the mean stress = 0 and stress ratio
R=
where,
min = 1 max
Time
Time
(a)
(b)
Figure 1.36 : (a) Random Stress Variation; and (b) Sinusoidal Stress Variation (Fluctuating Stress Cycle)
The stresses that are applied upon the fatigue test specimens can be axial, bending torsion or their combination. In each case a particular type of machine is required. The simplest type of machine, which is largely used, is rotating bending type. In this machine, as shown in Figure 1.37, a dead weight is applied upon the specimen through bearing while the specimen rotates. Generally the specimen is loaded as a cantilever (Figure 1.37) or under four points bending. A fatigue testing machine must have a counter to record number of cycles and microswitch to stop motor when specimen fails.
Micro Switch Specimen Loading Bearing Motor Holder Revolution Counter
Load
The fatigue test consists in choosing a stress level and running the specimen at that level until its fails. The number of cycles after which the specimen fails is known as the life of specimen at that stress level and recorded against it. Thus, a number of observations for stress levels and corresponding number of cycles to failure are obtained. Generally several specimens at each stress level are tested. For obtaining fatigue curves the stress levels are plotted as ordinates and logarithms of number of cycles are plotted as abscissa. For plotting these observations semi-log papers are used. A typical plot of fatigue curve is shown in Figure 1.38(a). It can be seen that fatigue curve tends to flatten for large number of cycles. The highest stress level is so chosen that number of cycles of failure is around 104. The fatigue curve is seldom investigated for cycles less than this. Further, the fatigue tests are rarely continued
41
Materials Applications
beyond 108 cycles because such long endurances are not common in service and time required for them will be very long. An important property of great principal importance is deduced from fatigue curve. The fatigue strength of material (Figure 1.38(a)) is defined as the maximum alternating stress which a material will withstand for given number of cycles. This number of cycles is mostly taken at 107 cycles except in some special case. Nevertheless, whenever fatigue strength is described the number of cycles of which it has been determined must also be mentioned.
Log N
Log 10
Log N
(b)
There are some materials for which failure does not occur once the specimen has endured nearly 106 cycles. The - log N curve is then composed of two straight lines, one in the finite region an inclined line, and other a horizontal line representing non-failures. The stress at which the curve becomes horizontal is known as fatigue limit or endurance limit. Figure 1.38(b) illustrates such a curve and is obtained for such materials as mild steel. It may be noted here that mild steel is the material which exhibits yield point in tension test.
cycle. The total number of cycles to failure as recorded has been found to coincide with the number of conchoidal marks. Figure 1.39 shows the characteristics fatigue fracture surface. The proportionate area of any one zone is indicative of level of stress. That is if smooth zone is much larger than rough zone it indicates that the stress level was low and vice-versa. If there are several points of crack initiation, it indicates that some geometrical stress concentration on the surface was present. In rotating bending the conchoidal rings tend to turn and centre on a different diameter. Figure 1.39(b) shows this tendency of crack to propagate in a direction opposite to that of rotation.
Smooth Zone
(a)
(b)
SAQ 11
(a) (b) Draw fatigue curve and define fatigue strength.
(c) (d)
43
Materials Applications
(e)
How does the fatigue fracture surface verify the concept of progressive fatigue crack growth?
A 1 e = + 0 + d0 N 2
A = i ni N = ni
. . . (1.22)
where i denotes the level of stress as 0, 1, 2 etc. 0 is the stress level which is lower and at which a non-failure is obtained. It can be seen from Table 1.3 that 3 is the stress level at which no failure is obtained. Apparently 3 is the lowest stress, which is equal to 0.
1 2 3
Stress
10
12
14
16
18
Specimen Numbers
ni = N = 9
i ni = A = 12
12 1 e = 3 + d 0 + 9 2
If, for example,
3 = 280 N/mm 2 and d 0 = 5 N/mm 2 e = 289 N/mm 2
Practical applications place some stress concentration on the surface. Corrosion and high temperature damage surface. Good surface finish and surface mechanical and heat treatment. (like shot peening, surface rolling, carburizing, nitriding, cyaniding) help obtain improved fatigue properties. Vanadium is the only alloying element that improves fatigue strength of steel.
Example 1.5
For determining the fatigue strength of medium carbon steel 15 specimens, identical in all respect were tested in cantilever rotating beam fatigue testing machine. The ultimate tensile strength of steel was known to be 550 MPa and test began at 275 MPa with a fixed step of 5 MPa. Determine the fatigue strength if results obtained are described below.
Sl. No. 1 2 3 4 5 6 7 8 9 Stress (MPa) 275 280 285 280 285 290 285 290 285 Survival/Failure S S F S S F S F F
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Materials Applications
10 11 12 13 14 15
S F F S F S
Solution
As a first step plot survival (o) and failure (x) pattern form visual depiction. From the plot it is seen that the lowest level at which survival is obtained is o = 275 MPa. Calculations are made by the help of the table.
290 285 Stress 280 275
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Failure Survival
Specimen Numbers
Survival ni 3 3 2 0
i ni 0 3 4 0
N = ni = 8
A = i ni = 7
A 1 e = + 0 + d0 N 2
Substituting
0 = 275 MPa, d 0 = 5 MPa, A = 7, N = 8
7 1 e = 275 + 5 + 8 2
= 281.875 MPa. Say 282 MPa (Ans.)
SAQ 12
(a) For determining fatigue strength of highly homogenized fine grained steel whose ultimate tensile strength was 950 MPa, fifteen identical specimens were tested. The results obtained are shown as survival failure pattern in Figure 1.42. Calculate fatigue strength.
485 480 475
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Specimen Numbers
Describe staircase method of determining fatigue strength. Enumerate all factors that affect fatigue behaviour. Why is surface of a shaft very important to controlling fatigue strength? Mention the factors that reduce fatigue strength. Mention the factors that improve fatigue strength.
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Lever P
Materials Applications
the actual complicated arrangement for measurement of deformation of creep specimen may be like one shown in Figure 1.43(b).
Figure 1.43 : (a) Schematic of Creep Testing Equipment; and (b) An Arrangement for Measuring Deformation of Specimen in a Furnace
The creep curves are plotted between strain and time for given stress and temperature. One typical curve is shown in Figure 1.44 for constant stress and temperature T. To start with the specimen is loaded at point O which corresponds to zero strain and zero time. Immediately after application of stress the specimen suffers from elastic strain upto point A. As the time passes the specimen continues to extend with decreasing rate upto point B and with a constant rate from B to C. From point C the rate of increase of creep strain increases upto point D where the fracture or rupture takes place.
I First Stage Strain () II Second Stage E F B 0 Time (t) C Elastic Recovery Permanent Plastic Strain III Third Stage D , T
Although the description of creep curve appears to be quite simple, the collection of data involves complicated control, measurement and recording system. The important point to note is that the total time required in the creep curve may be very large, ranging from a few months to as long as 10 years. The instruments are supported to be working throughout his time. Generally a battery of machines is required to make simultaneous studies at different temperatures and stress levels. A creep curve is conveniently divided into three stages. The stage of primary creep extends from point A to B, this covers a small part of total period. The creep during this stage is often referred to as transition creep and rate of creep decreases because of strain hardening of material. From B to C, it can be seen that rate of creep is constant. This is known as stage II or creep and is identified by its constant rate. During this stage minimum constant creep rate is attained. This steady state creep is the result of balance between two effects of strain hardening and softening due to temperature. In service conditions most machine parts which are required to have infinite life work in secondary creep zone. The final or tertiary creep from point C is marked with increasing creep rate ultimately culminating into rupture or separation of specimen into two pieces. This is termed stage III of creep. It may be pointed out there that during tertiary creep the necking also takes place which increases effective stress and thus rate of strain also increases. 48
If creep load is removed at any point like E during stage II then immediate elastic recovery takes place as shown from E to F. This is followed by a gradual strain recovery to permanent strain which does not recover at all (Figure 1.44).
T3
3 2
Log t r
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Materials Applications
In the range of temperature and for stress levels that are used in practice creep is a very slow process and for an appreciable deformation to take place it may be a very long time. For collecting information on creep of a new alloy it may generally take very long time (about 10,000 hours or 11 years). There is no way by which these tests can be accelerated and technological development cannot wait for assimilation of data on a new material which has been developed say before a year or two. Efforts were concentrated on means to develop methods which could predict long-term creep behaviour by making use of short-term test. Such methods depended upon extrapolation of data. One such parameter which is commonly used is described here.
Larson-Miller Parameter
This parameter, denoted by m, is expressed as m = T (ln t C ) or, and . . . (1.23) . . . (1.24) . . . (1.25)
m ln t = + C T
ln = k1 m + c1
In above equations, T is temperature in K, t is time for rupture or a specified deformation in hours, is stress in N/mm2 or MPa. C, k1 and c1 are constants which are determined from experiment. The advantages of the parameter is that by performing quick rupture test or creep test at high temperature and high stress Eq. (1.25) can be established. Then this equation can be extrapolated for any combination of , t; , T or T, t in which t can be large, can be small and T can be large or small. This parameter can determine T for a given combination of and t or for a given combination of T and t or t for a given combination of and T. This parametric method is a good design tool.
Example 1.6
From creep tests on 713 C alloy the constant C in Larson-Miller parameter is determined as 85.75. In rupture test a specimen of this material fails after 500 hours at 34 MPa and 1373 K while another specimen at a stress level 136 MPa and at temperature of 1308 K fails after 3 hours. Calculate stress to cause failure after 30,000 hours at 1173 K.
Solution
= 11.36 104
k1 =
1.387 10 4 1.26
= 1.1 10 4
(i)
ln = 1.1 10 4 m + 17.409
m3 = 1173 (ln 30,000 + 85.75)
...
which means that a stress of 150.2 MPa will cause specimen to rupture at 1173 K after 30,000 hours. Note that performing test for 30,000 hours is highly time consuming and this result has been obtained by performing two tests which required 500 + 3 = 503 hours.
Example 1.7
Larson-Miller parameters for an alloy at tests levels 10 MPa and 30 MPa are determined respectively as 18 104 and 16 104. Find the parameter for 20 MPa.
Solution
Write equation
ln = k1 m + c1
By subtraction
2 k1 104 = 1.1
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Materials Applications
For
k1 = 0.55 10 4
c1 = 2.3 + 18 0.55 = 12.21
ln = 0.55 10 4 m + 12.21
= 20 MPa
m=
Example 1.8
For material of Example 1.7 calculate life in hours at temperature of 900oC, and at = 20 MPa.
Solution
Larson-Miller parameter is given by m = T (ln t + 85.75) Use But T = 900 + 273 = 1173 K
m = 16.75 104
ln t =
Stress relaxation is creep related phenomenon. It is loss of stress in a part under constant strain over a given period of time at elevated temperature. For examples bolts are tightened to a fixed strain and stress to keep the cover of a pressure vessel leak proof. If the pressure vessel is used at high temperature the stress in bolts will reduce whereas the bolts will not lose their extension.
SAQ 13
(a) (b) Describe how one can find Larson-Miller parameter by performing creep rupture tests. If Larson-Miller parameters for a material is known at a stress and linear relationship between this parameters and stress is known, how can you determine life of the material at a given temperature?
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1.11 SUMMARY
Mechanical properties from the basis for calculations of dimensions of machine or structure of part. Conveniently, these properties are determined by experiments in laboratory. Tensile tests perform universal testing machine help define such properties as ultimate tensile strength, yield strength % elongation, modulus of resilience and modulus of toughness. These properties have been discussed. The idea of static loading, rate of straining, elastic and plastic deformations and concepts of true stress and strains have been introduced. The tendency of material to become brittle and factors that may cause such tendency are important considerations before engineers. Impact tests on notched specimen for determination of such tendency are performed. The material in practice is many times required to resist scratching and wearing. Hardness of materials, which can be measured on several scales through indentation are described. Different types of indenters and the impressions they leave on surface and factors by which they are influenced are discussed. The stresses applied on machine parts and structures very with time. The materials fail at low stress level under varying stresses and behaviour is called fatigue. The highest stress level at which material will not fail for 107 cycles is defined as fatigue strength. A host of factors affect fatigue strength. Determination of fatigue strength from experiments is a tedious exercise. This is illustrated through exercise. The material deforms continuously under constant stress/load over a long period of time. This behaviour, accentuated with high temperature, called creep is discussed. It has been discussed how this behaviour is considered for calculating life time or temperature out of three variables. Brief description of materials to be used for high temperature has also been given.
Creep
Extrusion
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Materials Applications
Fatigue
: Failure, at relatively low stress levels, of structures that are subjected to fluctuating and cyclic stresses. : The maximum stress level that a material can sustain, without failing, for some specified number of cycles. : The measure of a materials resistance to deformation by surface indentation. : One of two tests (see also Charpy test) that may be used to measure the impact energy of a standard notched specimen. An impact blow is imparted to the specimen by a weighted pendulum. : The point on a stress-strain curve at which the proportionality between stress and strain ceases, that is, the point at which plastic deformation begins. : A stress that persists in a material that is free of external forces or temperature gradients. : Failure that is accompanied by significant plastic deformation; often associated with creep failure. : A force applied so as to cause or tend to cause two adjacent parts of the same body to slide relative to each other, in a direction parallel to their common plane. : The tangent of the shear angle, that results from an applied shear load. : The instantaneous applied shear load divided by the original cross-sectional area across which it is applied. : The concentration or amplification of an applied stress at the tip of a notch or small crack. : The maximum engineering stress, in tension, that may be sustained without fracture. Often termed ultimate (tensile) strength. : A measure of the amount of energy absorbed by a material as it fractures. Toughness is indicated by the total area under the materials tensile stress-strain curve. : The natural logarithm of the ratio of the instantaneous length of a deformed specimen to its original length. : The instantaneous applied load divided by the instantaneous cross-sectional area of specimen. : The onset of plastic deformation. : The stress required to produce a very slight yet specified amount of plastic strain; a strain offset of 0.002 is commonly used.
Fatigue Strength
Proportional Limit
Toughness
True Strain
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