W2L2 Problems 11, 13
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1 W2L2 roblems 11, 13 Bulk Stress and Bulk Modulus (French pg This considers changes in the total volume associated with a uniform stress in the form of a pressure change think of a piston pushing down an air column initial volume, initial pressure, (before the piston pushes down Now increase uniform stress b b pushing down the piston (note that units are alread N/m 2 Strain is the fractional volume change B or B bulk stress bulk modulus bulk strain Note that this is negative because (b convention compression is considered to be created b a positive 1
2 Calculating B for a gas: this depends on the tpe of environment the gas is subjected to: isothermal or adiabatic 1. ISOTHERMAL: This assumes temperature stas constant as ou squeeze (note: this means that ou are somehow removing heat because the gas will normall heat up as ou squeeze it For constant temp Bole s Law: const. In differential form: + 0 Gives: This is just B from before! So for isothermal (constant temp: B (isothermal Gas at atmospheric pressure: B 10 5 N/m 2 (a (for solids 5 or 6 orders of magnitude higher see table 2
3 2. ADIABATIC: Gas heats up when ou compress it. If ou don t remove the heat, then we have adiabatic conditions. Heating greater pressure (so gas will be stiffer higher B For adiabatic conditions the relationship is: take logs differentiate wrt rearrange γ ln + γ ln 1 d d γ + 0 d d const. const. γ -B B γ (adiabatic tpicall 1.4 for a diatomic gas Comparing: isothermal B adiabatic B γ Adiabatic conditions make the gas stiffer. As ou compress it, the increased temperature works against ou b increasing pressure back! 3
4 Shear Stress and Shear Modulus (French pg. 55 Consider the following block of material: A shear force F is applied to the surface as shown * Area A F (acting over A A F l shear angle α l Get deformation in shear Deformation is characterized b a shear angle α, which is called the shear strain small α: α l shear stress F A A top surface area (not crosssectional shear stress F nα A shear modulus shear strain Note that for this block, in order to maintain translational and rotational equilibrium ou must also appl a horizontal force of F and a CCW torque of lf. 4
5 problem 5
6 Twisting or torsion produces a shear stress Twisting is one of the most common forms of shearing in a material. Figure (a below is similar to the one on the previous page, and figure (b shows how the angles translate to a twisting geometr. Later on we will consider how we can use this twisting geomet to model twisting behaviour and oscillations. 6
7 Summar of Elastic Constants oung s modulus n shear modulus B bulk modulus σ oisson s ratio τ ε F A α tensile (linear stress tensile strain shear stress shear strain bulk stress volume strain All stresses in a ( or Ma is more common All strains are dimensionless (N/m 2 (N/mm 2 Relationship Between Elastic Constants All of these parameters are related there can onl be 2 independent parameters, then these define the rest (ie. if ou know two ou can figure out the rest The two most commonl used are: derive now B 3(1 2σ No shear (n in this derive in tutorial (No, actuall we don ttoo mess 2n`(1 + σ No bulk (B in this 7
8 Deriving the First Relationship: E. Consider that a uniform bulk stress arises from the application of 3 perpendicular (and equal tensile stresses τ, τ, τ z. B 3(1 2σ The following was derived in tutorial Q10: τ z / ε + ε + ε Z τ 1 [ τ + τ + τ 2σ ( τ + τ + τ ] τ Add up ε s in all 3 directions to get / 1 2σ 3 (1 2σ Rearrange ( τ + τ + τ z z From last lecture: [ τ σ ( τ τ ] ε 1 + z 8 z ( τ τ 1 [ τ ] σ z 1 [ τ σ ( τ τ ] ε + ε z z + -3 if all stresses are equal. This is negative since the τ are in tension which is considered a negative pressure B 3(1 2σ
9 Constrained Modulus oung s modulus holds provided the sides of an object are free to move according to the oisson s ratio effect What if oisson s ratio epansion/contraction is constrained? e. rubber cube in a steel hole of same size τ z rigid soft rigid Because it cannot epand freel it is much stiffer than it would be if it were not constrained higher modulus Equation for a constrained modulus: (to be derived as Tutorial question #11 Constrained Modulus: M (1 σ (1 2σ (1 + σ 9
10 E. In a material with a oisson ratio of σ 0.33, how much stiffer is the material if its sides are constrained (not allowed to move in or out? This problem is just to give ou a feeling for how much constraint can change the inherent stiffness of a material. Using the equation on the previous page: M (1 σ (1 2σ (1 + σ lug in σ 0.33 Get M 1.48 So b constraining the movement of the sides (so the can t move in or out ou have effectivel made the material 1.5 stiffer than before!! 10
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