1. CONCEPTS OF FORCE-STRESS and DEFORMATION- STRAIN

2. For a body in equilibrium (not in motion) subjected to some external forces P i, there are internal forces developed within the body. P 1, P 2...:External forces F : Internal forces P1P1 P2P2 P3P3 P4P4

3. Internal forces can be shown on an imaginary cut section. P3P3 P4P4

4. P1P1 P2P2 P3P3 P4P4 P3P3 P4P4   Since the body was initially in equilibrium, half of this body should also be in equilibrium.   The internal forces are derived from the equilibrium equations and can be defined as the forces to bring the body to equilibrium.   Remember force is a vectorical quantity which has a magnitude and direction.

5. P 1 +P 2 +P 3 +P 4 = 0 P 1 +P 2 +P 3 +P 4 = 0  For the original body → ΣM = 0 ΣM = 0 P 1 +P 2 +F = 0 P 1 +P 2 +F = 0  For half of the body → ΣM = 0 ΣM = 0  F can be solved from the above set of equilibrium equations of the half plane.

6. STRESS

7.  If you look at that cut section little bit closer; Force acting on an infinite small area can be shown; ΔsΔs ΔFΔF  That force is called the STRESS.  In other words stress is the force intensity (force per unit area) acting on a material. or Stress = Area Force A F σ = 

8. Stress Shear ( τ ) : acts parallel to the area. Normal (σ) : acts perpendicular to the area For example; if the cut section is perpendicular to x- axis z y x ΔFΔF ΔFyΔFy ΔFxΔFx ΔFzΔFz ΔAΔA ΔFxΔFx σ x = ΔAΔA ΔFyΔFy τ y = ΔAΔA ΔFzΔFz τ z =

9.  However, stresses are always represented in tensorial (!not vectorical!) notation.  The plane it is acting on is also presented.  Therefore, if you take an infinitesmall volume element you can show all of the stress components  The first subscript indicates the plane perpendicular to the axis and the second subscript indicates the direction of the stress component. Stress Tensor

10.  In tensorial notation the stress components are assembled in a matrix. S =  For equilibrium it can be shown that : τ ij = τ ji for i ≠ j τ xy = τ yx τ xz = τ zx τ yz = τ zy  This symmetry reduces the shear stress components to three.

11.  Stresses can be grouped in several ways.  Stress Static: A constant and continuous load causes a static stress. Dynamic: Loads having different magnitudes and at different times cause dynamic stresses.  Stress Biaxial tension or compression Uniaxial tension or compression Triaxial compression Pure shear

12. W Spring is in uniaxial tension W Column is in uniaxial compression Baloon Membrane forces (biaxial tension) Hydrostatic pressure (triaxial compression)

13. Simple tension: cable Note:  = M/A c R here. Common States of Stress A o = cross sectional area (when unloaded) FF o   F A o   F s A  M M A o 2R2R F s A c Torsion (a form of shear): drive shaft Ski lift (photo courtesy P.M. Anderson)

14. (photo courtesy P.M. Anderson) Canyon Bridge, Los Alamos, NM o   F A Simple compression: Note: compressive structure member (  < 0 here). (photo courtesy P.M. Anderson) A o Balanced Rock, Arches National Park Common States of Stress

15. Bi-axial tension: Hydrostatic compression: Pressurized tank   < 0 h (photo courtesy P.M. Anderson) (photo courtesy P.M. Anderson) Fish under water  z > 0   Common States of Stress

16. DEFORMATION

17.  Deformation: is the change in the shape or dimension of a material. In other words when the relative position of points within a body changes “deformation” takes place. a) Elongation: occurs under tensile stresses. b) Shortening: under compressive stresses c) Rotation: due to shear stresses P A Δ2Δ2 B’ B A’ Δ1Δ1 Total elongation of the rod is Δ 2 (cm, mm, length) Elongation between AB is (Δ 2 -Δ 1 )

18. STRAIN

19.  Strain: represents the deformation of materials per unit length and is unitless (cm/cm, mm/mm) Strain = Deformation Original length l-l0l-l0 =ε = ΔlΔl l0l0 l0l0 εl=εl= ΔlΔl l0l0 ε d = ΔdΔd d0d0 l0l0 l d d0d0 P P (+) Tensile (elongation) (-) Shortening

20.  When pure shear acts on an element, the element deforms into a rhombic shape.  For convenience the element is rotated by an angle γ/2 and represented as shown. Ξ τ γ/2 y x A’ A γ y x γ DC B B’  For small angles γ = tanγ → γ = AA’ AD (radians)

21.  A pure shear strain is produced in torsion. γ B A A’ L θ: Angle of twist of radial line AB to position A’B r: radius of cross-sectional area AA’ = rθ θ r γ = AB AA’ A’ A γ = L θrθr

22.  Stress has units: N/m 2 or kgf/cm 2 or psi Engineering Stress Shear stress,  : Area, A F t F t F s F F F s  = F s A o Tensile stress,  : original area before loading Area, A F t F t  = F t A o 2 f m N =

23. Tensile strain: Lateral strain: Shear strain: Strain is always dimensionless. Engineering Strain  90º 90º -  y xx   =  x/y = tan   L o    L  L w o Adapted from Fig. 6.1 (a) and (c), Callister 7e.  /2  L L o w o

24. Stress-Strain Testing Typical tensile test machine Adapted from Fig. 6.3, Callister 7e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.) specimen extensometer Typical tensile specimen Adapted from Fig. 6.2, Callister 7e. gauge length

25. Adapted from Fig. 6.11, Callister 7e. yy Typical response of a metal F = fracture or ultimate strength Necking TS engineering stress engineering strain Maximum stress on engineering stress-strain curve.