MEC201 Mechanics of Solids

Deformation, Strain and Material Properties Fluid Mechanics is the study of the motion of the fluids, the forces that result from such motion, and the forces required to cause such motion. We are surrounded by fluids all around us. The motions of air in our atmosphere and water in our ocean and river determine the weather, so essential for life on earth. In fact we are concerned with motion of fluids in a more intimate manner as well. The flow of blood in our veins and arteries, of the other bodily fluids, of air that we breathe are all necessary for maintaining life.

Hooke Law δ P P/A δ/L L Stress= E×Strain P

The Fundamental Strategy of Deformable-Body Mechanics Deformation depends on loading, material and geometry Strain depends on stress AND material. NOT on geometry Stress depends on loading and geometry

The Fundamental Strategy of Deformable-Body Mechanics Load Stress Equilibrium Geometry Macro Strain Material Property Micro Deformation Geometry Micro Macro

Tug of War Cross-section: 6 cm2 Section AB BC CD DE EF Tension, N 250 500 800 550 300 Stress, kPa 416.7 833.3 1333.3 916.7 500.0 Strain 4.16×10- 3 8.35×10- 3 13.33×10- 3 9.20×10- 3 5.00×10- 3 Length, m 1.5 2.0 Elongation, m 6.24×10- 3 16.70×10- 3 20.02×10- 3 13.78×10- 3 10.00×10- 3

Measuring the height of Kutub x dx W = ρAxg = T σ = T/A = ρxg ε = σ/E = ρxg/E dδ = εdx = ρgxdx/E W(x) T

Measuring the height of Kutub or For a Nylon wire the density ~ 0.8X103 kg/m3, and E ~ 400 MPa, we get δ ~ 52 mm For steel density is 7.6X103 kg/m3, and E is 200 GPa, we get δ ~ 1 mm

Deflection in a Truss RA,y = 20 kN RA,x = − 20 kN RB,y = 20 kN 1 m A C B RA,y = 20 kN RA,x = − 20 kN RB,y = 20 kN C 20kN A B TAC = 28.8 kN TBC = − 20 kN Member Force kN Length m Area m2 Stress MPa Strain Elongation AC 28.3 1.41 1.77×10−4 160.1 7.6×10−4 1.07×10−3 BC − 20 1 −113.2 −5.4×10−4 −0.54×10−3

Deflection in a Truss X-displacement of C ~ shortening of BC =−0.54 mm y X-displacement of C ~ shortening of BC =−0.54 mm y-displacement of C ~ EF + FC1 = CD/cos45o + FG(=EC) ~ 1.25 mm

Statically Indeterminate Problems Reaction at middle support (and hence, at all supports depends on the bending of plank.

A Simple Example 2.6 m 1.3 m F1 F2 150 kN 1 m

Statically Indeterminate Problems Consideration of static equilibrium and determination of loads Consideration of relations between loads and deformations, (first converting loads to stresses, then transforming stresses to strain using the properties of the material, and then converting strains to deformations), Considerations of the conditions of geometric compatibility

Statically Indeterminate Structure Taking moments about the pivot point, 2R1 + 2R2 – P = 0 P Geom. Comp. δ1 = δ2 R1L1/E1A1 = R2L2/E2A2

Statically Indeterminate Structure R2 - F - R1 = 0; R1L – Fx = 0 h + δ1 = 2(h - δ2) Geom. Comp. h F R1 = kδ1 R2 = kδ2 h L x

Statically Indeterminate Structure P P = R1+ R2 R1 R2 R1 = (E1A1/L1)δ1 R2 = (E2A2/L2)δ2 Geom. Comp. δ1 = δ2

Stress-Strain Relationship σxx εxx = σxx/E σxx

Poisson Ratio σxx εyy = - ν εxx ν is Poisson ratio σxx

Stress-Strain Relationship Let us consider εxx. σxx produces an εxx = σxx /E σyy produces an εyy = σyy /E, which through Poisson ratio gives εxx = -νεyy = - νσyy/E. Similarly for σzz .

Generalized Hooke’s Law Shear stresses do not cause any normal strain εxx = σxx/E – νσyy/E - ν σzz/E = [σxx – ν(σyy + σzz)]/E Similarly for εyy and εzz

Geometric compatibility: An Example F σyy = −F/A, σzz = 0 Geometric compatibility: εxx = 0 y x εxx = [σxx – ν(σyy + σzz)]/E 0 = [σxx – ν(σyy + 0)]/E, → σxx =νσyy = − ν F/A εyy = [σyy – ν(σxx + σzz)]/E = [−F/A + ν F/A]/E = −(1− ν)F/AE

Shear Strain θ2 θ θ1 Shear strain γ is π/2 − θ Shear strain is also seen as: θ1 − θ2 (with clockwise angles as positive θ

Shear Strain θ2 θ1 θ1 = 0.013/0.2 = 0. 065 θ2 = 0.012/0.2 = 0. 06 Coordinates after deformation (in mm) are: A(0,0), B(0.194, 0.013), and D(−0.012, 0.196). x y D C B A θ2 θ1 θ1 = 0.013/0.2 = 0. 065 θ2 = 0.012/0.2 = 0. 06 γxy = 0.65 − 0.60 = −0.05 radians

Shear Stress θ Shear strain γ is related to shear stress τ by γxy = τxy/G, where G is shear modulus θ It can be shown that γxy does not depend on other components of stress.

Shear Modulus Material G, GPa Aluminium 25 Steel 80 Glass 26-32 Soft Rubber 0.003- 0.001

Vibration Isolator Shear stress τ = Wall 8,000 N Rubber blocks 10 cm × 10 cm with 12 cm height Shear stress τ = 4,000 N/ (0.1 m)(0.12 m) = 3.33×105 Pa Shear strain γ = τ/G 3.33×105 Pa/1 GPa = 3.33×10−4 Vertical deflection of load = 3.33×10−4×0.10 m

Elastic Properties We have so far introduced three elastic properties of materials. Material E, GPa G, GPa ν Aluminium 70 25 0.33(1/3) Steel 200 80 0.27(1/4) Glass 50-80 26-32 0.21-.27 Soft Rubber 0.0008-0.004 0.003- 0.001 0.50

Thermal Strains There is no thermal shear strain Material α ×10-6/oC Steel ~10 Aluminium ~20

Generalized Hooke’s Law εxx = [σxx – ν(σyy + σzz)]/E + αΔT εyy = [σyy – ν(σzz + σxx)]/E+ αΔT εzz = [σzz – ν(σxx + σyy)]/E+ αΔT γxy = τxy/G, γyz = τyz/G, and γzx = τzx/G

An Example σyy Steel: εx = 0.6×10−4; εy = 0.3×10−4 σxx Find σxx and σyy. E = 200 GPa, ν = 0.3

Another Example Aluminium rod, rigid supports. Temperature raised by ΔT. What are the stresses? εxx = 0 = [σxx/E + αΔT]

An Example p Tank is flush when empty. Find end forces when pressure is p p Due to p: σzz = pr/ 2t, σθθ = pr/ t z If end forces F, axial stress due to it is F/2πrt εzz = [(pr/ 2t − F/2πrt) −νpr/t ]/E Equate it to 0 and determine F

Stress-Strain Relationship A material property. Tensile Test Machine, UTM

Extensometer

Stress-Strain Curve: Elasticity

Failure Modes Necking Brittle Failure Ductile Failure σ (= F/Ao) ε (=∆L/Lo) Necking Brittle Failure Ductile Failure

Plastic Deformation, Yield Strength ε (= ΔL/L0) σ (= F/A0) Y Yield stress, σY 0.02% Permanent set

Strain Hardening ε (= ΔL/L0) σ (= F/A0) Y Ultimate stress Y1 B

Stress-Strain in Brittle Materials

Idealized Stress-Strain Curves ε σ (a) Rigid ε σ (b) Perfectly elastic ε σ (c) Elastic-Plastic ε σ (d) Perfectly plastic ε σ (e) Elastic- Plastic (strain hardening) Increase in yield strength

Pre-Stressing A Section AA A

Pre-stressing (a) Tendon being stresses during casting. Tension in tendon, no stress in concrete. (b) After casting, the force is released and the structure shrinks. (d) FBD of concrete. The residual force in the tendon is trying to compress the concrete, which is in tension now. (c) FBD of tendon. The concrete does not let the tendon shrink as much as it would on its own. This results in residual tension in the tendon.

Pre-stressing: A simple example A concrete beam of cross-sectional area 5 cm×5 cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20 kN. The external tension in steel released after the concrete is set. What is the residual compressive stress in the concrete? T = 20 kN →σ = 255 Mpa →ε = 1.21×10- 3 →δ = 2.42 mm

Pre-stressing: A simple example 2.42 mm δs δc F δs + δc = 2.42 mm

Failure Under Compression Buckling

Buckling of a Bridge

Elastic Buckling