1 Learning outcomes Equilibrium body & Free Body Diagram - PowerPoint PPT Presentation

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Learning outcomes • Equilibrium body & Free Body Diagram • Elastic & Plastic Deformation • Engineering Stress & Strain • This subject also involves the deformations and stability of a body when subjected to external forces.

Introduction • Mechanics of materials is a study of the relationship between the external loads on a body and the intensity of the internal loads within the body. • This subject also involves the deformations and stability of a body when subjected to external forces.

Equilibrium of a Deformable Body External Forces 1. Surface/ Linear/ Concentrated Forces - caused by direct contact of other body’s surface 2. Body Forces - other body exerts a force without contact

Equilibrium of a Deformable Body Internal Resultant Loadings  Objective of free body diagram (FBD) is to determine the resultant force and moment acting within a body.  In general, there are 4 different types of resultant loadings: a) Normal force, N b) Shear force, V c) Torsional moment or torque, T d) Bending moment, M Free-body diagrams = diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. A free-body diagram is a special example of the vector diagrams.

Equilibrium of a Deformable Body Equations of Equilibrium  Equilibrium of a body requires a balance of forces and a balance of moments     F 0 M 0 O  For a body with x , y , z coordinate system with origin O,       F 0 , F 0 , F 0 x y z       M 0 , M 0 , M 0 x y z  Best way to account for these forces is to draw the body’s free -body diagram (FBD).

Elastic Deformation 1. Initial 2. Small load 3. Unload bonds stretch return to initial d F F Elastic means reversible! Linear- elastic Non-Linear- elastic d 7

Plastic Deformation (Metals) 1. Initial 2. Larger load 3. Unload bonds p lanes stretch still & planes sheared shear d plastic d elastic + plastic F F linear linear d elastic elastic d plastic Plastic means permanent ! 8

Engineering Stress • Normal/Tensile stress, s : • Shear stress, t : Ft Ft F Area, A Fs Area, A Fs Ft t = F s Ft F s = F t lb N A o = f or 2 2 A o in m original area  Stress has units: before loading N/m 2 or lb f /in 2 9

Stress Normal Stress σ  Force per unit area acting normal to Δ A  F z s  lim  z   A A 0 Shear Stress τ  Force per unit area acting tangent to Δ A  F t  x lim  zx   A A 0  F t  y lim  zy   A A 0

Common States of Stress • Simple tension: cable F F Ao = cross sectional area (when unloaded) s  F s s A o Ski lift (photo courtesy • Torsion (a form of shear): drive shaft P.M. Anderson) M Fs A o A c t  Fs A o M 2 R Note: t = M / A c R here. 11

OTHER COMMON STRESS STATES • Simple compression: A o Canyon Bridge, Los Alamos, NM (photo courtesy P.M. Anderson) Note: compressive structure member s  F ( s < 0 here). A Balanced Rock, Arches o National Park (photo courtesy P.M. Anderson) 12

OTHER COMMON STRESS STATES • Bi-axial tension: • Hydrostatic compression: (photo courtesy Pressurized tank Fish under water (photo courtesy P.M. Anderson) s q > 0 P.M. Anderson) s < 0 s z > 0 h 13

Engineering Strain (terikan) • Tensile strain: • Lateral strain: d /2 -d e  d e L  L L o w o L o w o d L /2 • Shear strain: q g =  x / y = tan q  x 90º - q y Strain is always 90º dimensionless. 14

Strain Energy  Strain-energy density is strain energy per unit volume of material Strain σ e ∆ U = u = ∆ V 2 If material behavior is linear elastic, Hooke ’ s law applies, ( ) σ σ σ 2 = u = e 2 2 E E 15

Strain Energy  When material is deformed by external loading, energy is stored internally throughout its volume  Internal energy is also referred to as strain energy  Stress develops a force, STRAIN ENERGY  F = σ  A = σ (  x  y ) 16

Shear Stress-Strain Graph REVIEW  Hooke ’ s law for shear Tensile Strength (TS), M Fracture point, f Proportional G is shear modulus of elasticity or modulus Limit of rigidity G can be measured as slope of line on τ - γ diagram, G = τ pl / γ pl The three material constants E , ν , and G is related by τ = G γ E G = 2(1 + ν ) 17

Shear Stress-strain  Use thin-tube specimens and subject it to torsional loading  Record measurements of applied torque and resulting angle of twist 18

Shear Stress-strain  Material will exhibit linear-elastic behavior till its proportional limit, τ pl  Strain-hardening continues till it reaches ultimate shear stress, τ u  Material loses shear strength till it fractures, at stress of τ f 19

Stress-Strain Testing • Typical tensile test • Typical tensile machine specimen specimen extensometer gauge length 20

SE Stress – Strain Behavior Table 1 - Room-Temperature Elastic and Shear Modulus, and Poisson’s Ratio for Various Metal Alloys

STRESS – STRAIN BEHAVIOR • Modulus of Elasticity, E : (also known as Young's modulus) • Hooke's Law: s = E e s F E e Linear- F elastic simple tension test 24

Stress-Strain Behavior There are some material such as gray cast iron, concrete and many polymers which this elastic portion of the stress-strain curve is not linear Tangent modulus = E (Modulus of elastic) Schematic stress – strain diagram showing non-linear elastic behavior, and how secant and tangent moduli are determined.

Stress-Strain Behavior  Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metal Force versus inter-atomic separation for weakly and strongly bonded atoms. The magnitude of the modulus of elasticity is proportional to the slope of each curve at the equilibrium inter-atomic separation r o . 26

EXERCISE 1 A piece of copper originally 305 mm long is pulled in tension with a stress of 276MPa. If the deformation is entirely elastic, what will be the resultant elongation?

EXERCISE 2 A piece of aluminum originally long is half of 600mm pulled in tension with a stress of 275MPa. If the deformation is entirely elastic, what will be the resultant elongation? Ans: 1.195x10 -3

Poisson’s Ratio  When body subjected to axial tensile force, it elongates and contracts laterally  Similarly, it will contract and its sides expand laterally when subjected to an axial compressive force 29

Poisson’s Ratio  Strains of the bar are: Early 1800s, S.D. Poisson realized that within elastic range, ration of the two strains is a constant value, since both are proportional.  ν is unique for homogenous and isotropic material  Why negative sign? Longitudinal elongation cause lateral contraction (-ve strain) and vice versa  Lateral strain is the same in all lateral (radial) directions  Poisson ’ s ratio is dimensionless, 0 ≤ ν ≤ 0.5 30

El Elas asti tic c Proper erties ties Of Mat ater erials als Poisson’s ratio

Exercise III From the tensile stress-strain behavior for the brass specimen shown in the figure, determine the following: a) The modulus of elasticity b) The yield strength at a strain offset of 0.002 c) The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm d) The change in length of a specimen originally 250 mm long that is subjected to a tensile stress of 345MPa

s Yield Strength, y • Stress at which noticeable plastic deformation has occurred. when e = 0.002 @ 0.2% s tensile stress, Elastic Plastic s y s y = yield strength P (yielding point) Note: for 2 inch sample e = 0.002 =  z / z 0.002 Strain   z = 0.004 in Offset Method engineering strain, e e = 0.002 A typical s-e behavior for metal showing elastic/plastic deformation, proportional limit (P) & s Y determined using 0.002 strain offset value. 34

Yield Strength EXAMPLE 3.1 (SOLN) At 0.2% strain, extrapolate line (dashed) parallel to OA till it intersects stress-strain curve at A’ σ YS = 469 MPa 35

Yield Strength : Comparison Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 2000 Steel (4140) qt 10 00 Yield strength, s y (MPa) Ti (5Al-2.5Sn) a W(pure) 700 in ceramic matrix and epoxy matrix composites, since 600 since in tension, fracture usually occurs before yield. Cu (71500) cw in tension, fracture usually occurs before yield. 500 Mo (pure) Steel (4140) a 400 Steel (1020) cd , 300 Room T values Al (6061) ag Hard to measure, Hard to measure Steel (1020) hr 200 ¨ Ti (pure) a Ta (pure) Cu (71500) hr 100 dry 70 PC 60 Nylon 6,6 Al (6061) a PET 50 humid PVC 40 PP 30 HDPE 20 LDPE Tin (pure) 10 36

Tensile Strength, TS • Maximum stress on engineering stress -strain curve. TS F = fracture or s y engineering ultimate strength stress Neck – acts Typical response of a metal as stress concentrator strain engineering strain • Metals: occurs when noticeable necking starts. • Polymers: occurs when polymer backbone chains are aligned and about to break. 37