Normal Strain and Stress Normal Strain and Stress, Stress strain diagram, Hookes Law 1 Strain When a body is subjected to load, it will deform and can be detected through the changes in length and the changes of angles between them. The deformation is measured through experiment and it is called as strain. The important of strain: it will be related to stress in the later chapter 2

Normal Strain Normal strain is detected by the changes in length. l ' l l l l (epsilon)epsilon)) l: len)gth after deformed l: origin)al len)gth. Note : dimen)sion)less

very small (epsilon)n)ormally is m (epsilon)=10-6 m)) 480(epsilon)10)-6 m/m = 480 m/m = 480 micros = 0.0480 % 3 Example 1 When) load P is applied, the RIGID lever arm rotates by 0.05o. Calculate the n)ormal strain) of wire BD Foun)dation): L/L Kn)owledge required: geometrical equation) Rigid: n)o deformation) on) the lever 4 Geometry: The mathematics Sin)e an)d Cosin)e Rule

L1 L 2 2 L32 2( L 2)( L3) cos() 5 Example 1 When) force P is applied to the rigid lever arm LBD after deformed is DB Cosin)e rule can) be applied here 2 2 LBD ' LAB ' LAD 2( LAB ' )( LAD ) cos( 0.05) Strain): BD

LDB ' LDB LDB 6 Example 1 When) force P is applied to the rigid lever arm LBD after deformed is DB Cosin)e rule can) be applied here 2 2 LBD ' LAB ' LAD 2( LAB ' )( LAD ) cos( 0.05) LBD ' 300.3491 mm

Strain): BD LDB ' LDB LDB BD 0.00116 mm / mm 7 Example 2 The force applied to the han)dle of the rigid lever the arm to rotate clockwise through an) an)gle of 3o about pin) A. Determin)e the average n)ormal strain) developed in) the wire. Origin)ally, the wire is un)stretched.

Discuss the approach? 8 Solution LBD = 0.6155 m = 0.0258 m/m 9 Simple Tensile Test Strength of a material can only be determined by experiment

The test used by engineers is the tension or compression test This test is used primarily to determine the relationship between the average normal stress and average normal strain in common engineering materials, such as metals, ceramics, polymers and composites 10 Conventional StressStrain Diagram Nominal or engineering stress is obtain)ed by

dividin)g the applied load P by the specimen)s original cross-section)al area. P A0 Nominal or engineering strain is obtain)ed by dividin)g the chan)ge in) the specimen)s gauge len)gth by the specimen)s origin)al gauge len)gth. L0 Conventional StressStrain Diagram

Conventional StressStrain Diagram Elastic Behaviour A straight line Stress is proportional to strain, i.e., linearly elastic Upper stress limit, or proportional limit; pl If load is removed upon reaching elastic limit, specimen will return to its original shape Yielding

Material deforms permanently; yielding; plastic deformation Yield stress, Y Once yield point reached, specimen continues to elongate (strain) without any increase in load Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in elastic limit Material is referred to as being perfectly plastic Conventional StressStrain Diagram Strain) Harden)in)g. Ultimate stress, u While specimen) is elon)gatin)g, its x-section)al area will decrease

Decrease in) area is fairly un)iform over en)tire gauge len)gth Necking At ultimate stress, cross-section)al area begin)s to decrease in) a localized region) of the specimen). Specimen) breaks at the fracture stress. 14 StressStrain Behavior of Ductile and Brittle Materials Ductile Materials Material that can subjected to large strains before it ruptures is

called a ductile material. Brittle Materials Materials that exhibit little or no yielding before failure are referred to as brittle materials. StressStrain Behavior of Ductile and Brittle Materials Yield Strength 0.02% strain for ductile material Strain hardening

When ductile material is loaded into the plastic region and then unloaded, elastic strain is recovered. The plastic strain remains and material is subjected to a permanent set. Hookes Law Hookes Law defin)es the linear relationship between) stress an)d strain) within) the elastic region). E = stress E = modulus of elasticity or Youngs modulus = strain)

E can) be used on)ly if a material has linear elastic behaviour. E can) be derived from stress an)d strain) graph. What is it? Strain Energy When) material is deformed by extern)al loadin)g, it will store en)ergy internally throughout its volume. En)ergy is related to the strain)s called strain energy. Modulus of Resilience

When) stress reaches the proportion)al limit, the strain)-en)ergy den)sity is the modulus of resilience, ur: 2 1 1 pl ur pl pl 2 2 E Example The stressstrain diagram for an aluminum alloy

that is used for making aircraft parts is shown. When material is stressed to 600 MPa, find the permanent strain that remains in the specimen when load is released. Also, compute the modulus of resilience both before and after the load application. Approach to the problem: Parallel to elastic line Both slope is equal Distance CD can be calculated based on the slope Permanent strain: 0.023 distance CD Solution When the specimen is subjected to the load, the strain is

approximately 0.023 mm/mm. The slope of line OA is the modulus of elasticity, From triangle CBD, E 450 75.0 GPa 0.006 BD 600 10 6 E 75.0 109 CD 0.008 mm/mm CD CD

Solution: This strain) represen)ts the amoun)t of recovered elastic strain. The perman)en)t strain) is OC 0.023 0.008 0.0150 mm/mm (Ans) Computin)g the modulus of resilien)ce, ur initial 1 pl pl 1 450 0.006 1.35 MJ/m3 (Ans) 2 2 ur final 1 pl pl 1 600 0.008 2.40 MJ/m3 (Ans)

2 2 Note that the SI system of un)its is measured in) joules, where 1 J = 1 Nm Modulus of Toughness Modulus of toughness, ut, represents the entire area under the stressstrain diagram. It indicates the strain-energy density of the material just before it fractures. Example The bar DA is rigid an)d is origin)ally held in) the horizon)tal position) when) the weight W is supported

from C. If the weight causes B to be displaced down)ward 0.625mm, determin)e the strain) in) wires DE an)d BC. Also if the wires are made of A-36 steel an)d have a crosssection)al area of 1.25 mm2, determin)e the weight W. Discuss the approach???? 23 1) Calculate the displacemen)t of D. D B 1.5 0.9 1.5 )

0.9 D 1.0417mm D 0.625( 2) Based on) displacemen)t on) D, calculate the strain) an)d n)ormal stress D D 1.0417 1.157(10) 3 mm / mm LD 900 * strain in mm/mm, stress and E in MPa, F in N and length in mm

D E 200(10)31.157(10) 3 231.4MPa 24 3) Based on) n)ormal stress at wire DE, calculate the T of wire D TED D 231.4MPa A TED D A 289.3 N 4) Calculate W, based on) FBD of bar DA M A 0

TDE (1.5) W (0.9) 0 W 482.2 N 5) Calculate n)ormal stress of wire CB an)d strain) of wire CB BC TBC 482.2 385.7 MPa A 1.25 Strain) can) n)ot be calculated as n)ormal stress goes beyon)d yield stress (epsilon)Sy = 250 MPa), elastic property is n)o more applied. Therefore it requires the stress an)d strain) curve to predict the strain) 25

Poissons Ratio (epsilon)n)u), states that in) the elastic range, the ratio of these strain)s is a constant sin)ce the deformation)s are proportion)al. lat Poisson)s ratio is dimensionless. v long Typical values are 1/3 or 1/4. Negative sign) sin)ce longitudinal elongation (epsilon)positive strain)) causes lateral contraction (epsilon)n)egative strain)), an)d vice versa. Example A bar made of A-36 steel has the dimen)sion)s shown). If an) axial force of P is applied to the bar, determin)e the chan)ge in) its len)gth an)d the chan)ge in) the dimen)sion)s of its cross section) after applyin)g the load. The material behaves elastically. Discuss the approach

Approach: Property A-36: E , = P/A z=/ E Lz = L * z x = y = -z Lx = L * x Ly = L * y Solution 1) The n)ormal stress in) the bar : P 80 103 z 16.0 10 6 Pa

A 0.1 0.05 4) The con)traction) strain)s in) both the x an)d y direction)s are x y vst z 0.32 8010 6 25.6 m/m 2) From the table for A-36 steel, Est = 200 GPa z 16.0106 6

z 80 10 mm/mm 9 Est 20010 3) The axial elon)gation) of the bar is therefore z z Lz 8010 6 1.5 120m (Ans)

5) The chan)ges in) the dimen)sion)s of the cross section) are 25.610 0. 05 1.28m (Ans) x x Lx 25.610 6 0.1 2.56m (Ans) y y Ly 6