NGINRING COUNCIL CRTIFICAT LVL NGINRING SCINC C03 TUTORIAL 4 - STRSS AND STRAIN This tutorial ma be skipped if ou are alread familiar with the topic You should judge our progress b completing the self assessment eercises. These ma be sent for marking at a cost (see home page). On completion of this tutorial ou should be able to do the following. Define the elastic properties of materials. Define and use direct stress. Define and use direct strain. Define and use Modulus of lasticit. Define and use Poisson s Ratio. Define and calculate lateral strains. Sole problems inoling two dimensional stress sstems. tend the work to 3 dimensional stress sstems. Define and calculate olumetric strain. Define and use Bulk Modulus. Derie the relationships between the elastic constants. D.J.Dunn freestud.co.uk
INDX. MATRIAL TYPS. DIRCT STRSS 3. DIRCT STRAIN 4. MODULUS OF LASTICITY 5. POISSON S RATIO. THR DIMNSIONAL STRSS AND STRAIN. VOLUMTRIC STRAIN. BULK MODULUS.3 RLATIONSHIP BTWN TH LASTIC CONSTANTS
. MATRIAL TYPS An elastic material will spring back to its original shape and size when the forces causing it to deform are remoed. lastic materials hae seeral constants which ou either alread know or will learn soon. These are Modulus of lasticit Modulus of Rigidit Bulk Modulus Poissons' Ratio G K ν Materials ma be further classified as follows. ISOTROPIC MATRIAL In this tpe of material the elastic constants are the same in all directions so if a specimen is cut from a bulk material, the direction in which it is cut has no affect on the alues. This applies to most metals with no pronounced grain structure. ORTHOTROPIC MATRIAL In this tpe of material, the elastic constants hae different alues in the, and z directions so the results obtained in a test depend upon the direction in which the specimen was cut from the bulk material. This applies to materials with grain structures such as wood or rolled metals. NON-ISOTROPIC MATRIAL In this tpe of material, the elastic constants are unpredictable and the results from an two tests are neer the same. This applies to materials such as glass and other ceramics.. DIRCT STRSS When a force F acts directl on an area A as shown in figure, the resulting direct stress is the force per unit area and is gien as F/A. F is the force normal to the area in Newtons A is the area in m (sigma) is the direct stress in N/m or Pascals. Since Pa is a small unit kpa, MPa and GPa are commonl used. If the force pulls on the area so that the material is stretched then it is a tensile force and stress. This is regarded as positie. If the force pushes on the surface so that the material is compressed, then the force and stress is compressie and negatie. Figure 3
3. DIRCT STRAIN Consider a piece of material of length L as shown in figure. The direct stress produces a change in length L. The direct strain produced is (epsilon) defined as L/L The units of change in length and original length must be the same and the strain has no units. Strains are normall er small so often to indicate a strain of 0- we use the name micro strain and write it as µ. For eample we would write a strain of 7 0- as 7 µ. Tensile strain is positie and compressie strain is negatie. 4. MODULUS OF LASTICITY Man materials are elastic up to a point. This means that if the are deformed in an wa, the will spring back to their original shape and size when the force is released. It has been established that so long as the material remains elastic, the stress and strain are related b the simple formula / is called the MODULUS OF LASTICITY. The units are the same as those of stress. WORKD XAMPL No. A metal bar which is part of a frame is 50 mm diameter and 300 mm long. It has a tensile force acting on it of 40 kn which tends to stretch it. The modulus of elasticit is 05 GPa. Calculate the stress and strain in the bar and the amount it stretches. SOLUTION F 40 03 N. A πd/4 π 50/4 3 mm F/A (40 03)/(3 0-) 0.37 0 N/m 0.37 MPa / 05 0 N/m / (0.37 0)/(05 0).4 0- or.4 µ L/L L L.4 0-300 mm 0.08 mm 4
5. POISSON'S RATIO Consider a piece of material in dimensions as shown in figure. The stress in the direction is and there is no stress in the direction. When it is stretched in the direction, it causes the material to get thinner in all the other directions at right angles to it. This means that a negatie strain is produced in the direction. For elastic materials it is found that the applied strain () is alwas directl proportional to the induced strain () such that ν ν (Nu)is an elastic constant called Poisson s' ratio. The strain produced in the direction is - ν Figure If stress is applied in direction then the resulting strain in the direction would similarl be - ν Now consider that the material has an applied stress in both the and directions (figure 3). The resulting strain in an one direction is the sum of the strains due to the direct force and the induced strain from the other direct force. Hence ν ν ( ν )...(A) Figure 3 Similarl ( ν )...(B) The modulus must be the same in both directions and such a material is not onl elastic but ISOTROPIC. WORKD XAMPL No. A material has stresses of MPa in the direction and 3 MPa in the direction. Gien the elastic constants 05 GPa and ν 0.7, calculate the strains in both direction. SOLUTION ( ν ) { 0 0.7 3 0 } 050 5.8µ ( ν ) { 3 0 0.7 0 } µ 050 5
WORKD XAMPL No.3 A material has stresses of - MPa in the direction and 3 MPa in the direction. Gien the elastic constants 05 GPa and ν 0.7, calculate the strains in both direction. SOLUTION ν ν ( ) {- 0 0.7 3 0 } 050 3.7µ ( ) { 3 0 0.7 (- 0 )} 7.3µ 050 Note that we do not hae to confine ourseles to the and directions and that the formula works for an two stresses at 0 o to each other. In general we use and with corresponding strains and. 5. CONVRTING STRAINS INTO STRSSS We hae alread deried ν ν Rearrange to make the subject. Substitute this into the first formula ( + ν ) If we do the same but make ν Rearrange ν + ν ( + ν ) the subject of the formula we get ν ( + ν ) SLF ASSSSMNT XRCIS No.. Sole the strains in both directions for the case below. 80 GPa ν 0.3-3 MPa 5 MPa (Answers 3.78 µ and -5 µ). Sole the stresses in both directions for the cases below. 00 GPa ν 0. -4. µ -8.05 µ (Answers -.5 MPa and - MPa) 3. 05 GPa ν 0.5 0.3 µ.5 µ (Answers 00 kpa and 500 kpa)
. THR DIMNSIONAL STRSS AND STRAIN quations A and B were deried for a dimensional sstem. Suppose a material to be stressed in mutuall perpendicular directions, and z. The strain in an one of these directions is reduced b the effect of the strain in the other two directions and the three strains are ( ν νz ) [ ν( + z )]...(C) ( ν νz ) [ ν( + z )]...(D) z ( z ν ν ) [ z ν( + )]...(). VOLUMTRIC STRAIN Now consider a cube of material is stressed in the direction b a compressie pressure as shown in figure 4. The change in olume is L L. Now consider that the cube is strained b an equal amount in the and z directions also. With er little error the total change in olume is 3L L. The original olume is L3. When a solid object is subjected to a pressure p such that the olume is reduced, the olumetric strain is change in olume/original olume. Figure 4 In the case of a cube this becomes 3L L/ L3 3 L/ L 3...(F) is the equal strain in all three directions. It follows that when a material is compressed b a pressure which b definition must be equal in all directions, the olumetric strain is three times the linear strain in an direction.. BULK MODULUS It has been established that the olumetric strain in an elastic material is directl proportional to the stress such that / K K is called the Bulk Modulus. This is another of the elastic constants of a material..3 RLATIONSHIP BTWN TH LASTIC CONSTANTS When the material is compressed b a pressure p the stress is equal to -p because it is compressie. The bulk modulus is then p K...(G) From equation F we hae 3 From equations C, D and the strains in all direction being equal and the stresses being equal to -p, we hae ( p + νp) 3 The olumetric strain is 3 ( p + νp)...(h) 7
Combining equations G and H we hae p K...(I) 3 3 ( ) ( ν) p + νp This shows the relationship between, K and ν. It ma be shown that the relationship of the shear modulus G to the other elastic constants is gien b G...(J) + ν ( ) You would need to stud three dimensional stress sstems in depth in order to derie this equation. WORKD XAMPL No.4 A solid piece of metal of olume 8000 mm3 is compressed b a pressure of 80MPa. Determine the bulk modulus gien that the elastic modulus is 7 GPa and Poissons' ratio ν is 0.34. Determine the change in olume. SOLUTION 7 0 K 73. GPa 3( ν) 3(- 0.34) K - p V V p -80 0 K 73. 0 V V 0.0008 8000 0.0008 8.5 mm 3 SLF ASSSSMNT XRCIS No.. Find the bulk modulus K and shear modulus G for aluminium gien that the elastic modulus is 7 GPa and Poissons' ratio ν is 0.34. (Answers 73. GPa and.4gpa).. A cube is stressed in 3 mutuall perpendicular direction, and z. The stresses in these directions are 50 kpa 80 kpa z -00 kpa Determine the strain in the direction. ν is 0.34 and is 7 GPa. (Answer 800 0-) 8