1. 1/1 SOE 1032 SOLID MECHANICS Website www.ex.ac.uk~TWDavies/solid_mechanics www.ex.ac.uk~TWDavies/solid_mechanics course organisation,lecture notes, tutorial problems,deadlines Course book:J M Gere, Mechanics of Materials, Nelson Thornes, 2003, £29.

2. 1/2 TUTORIALS Three groups, A, B, C Three sessions Monday, Tuesday and Friday weeks 8, 9, 10, 11

3. 1/3 LABORATORY SESSIONS ONE DEMONSTRATION Week 10 Write up deadline Friday week 11 Hand in to 307 and get it date stamped.

4. 1/4 LEARNING TRIANGLE MEYOU BOOK

5. 1/5 DEFORMATIONS to be studied Static objects Extension or compression of a rod under an axial load Twisting of a rod by applied torque Bending of a beam subjected to point loads, uniformly distributed loads and bending moments

6. 1/6 BASIC SCIENCE USED Newton’s 3 rd Law, equilibrium (STATICS) Auxiliary relationships based on material properties – e.g. Hooke’s Law

7. 1/7 POSSIBLE MATERIALS NATURAL MANUFACTURED

8. 1/8 NATURAL MATERIALS WOOD STONE

9. 1/9 MANUFACTURED MATERIALS METALS (examples used in this course) PLASTICS CONCRETE AND BRICK CERAMICS AND GLASS

10. 1/10 SCOPE OF COURSE STRESS AND STRAIN IN SIMPLE SYSTEMS DEFORMATIONS IN TENSION, COMPRESSION & TORSION STRESS & BENDING IN BEAMS MOHR’S CIRCLE i.e. essential parts of Chapters 1 to 5 and 7 (see reading list).

11. 1/11 NORMAL STRESS Axial force per unit X-sectional area P/A =  N/m 2 or Pa (like pressure) Tensile stress (positive) Compressive stress (negative) eg m=100 kg held by rod of A = 1 cm 2 g = 10ms -2  =P/A = mg/A = 1000/10 -4 = 10 MPa

12. 1/12 SHEAR STRESS TANGENTIAL FORCE PER UNIT AREA P/A =  N/m 2 or Pa

13. 1/13 NORMAL STRAIN Change in length  caused by normal stress  =  /L (dimensionless) Tensile strain Compressive strain eg  = 2 mm, L = 2 m, then  = 1 mm/m or  = 0.1%

14. 1/14 UNIAXIAL STRESS Conditions are that: Deformation is uniform throughout the volume (prismatic bar) which requires that: Loads act through the centroid Material is homogeneous See Section 1.2 in book

15. 1/15 LINE OF ACTION OF AXIAL FORCES FOR UNIFORM STRESS DISTRIBUTION Prismatic bar of arbitrary cross-section A Axial forces P producing uniformly distributed stresses  = P/A

16. 1/16 BALANCE THE MOMENTS

17. 1/17 Mechanical Properties Strength compression tension shear Elasticity, plasticity, ductility,creep Stiffness, flexibilty Used to relate deformation to applied force

18. 1/18 Mechanical Testing Tensile test machine

19. 1/19 Tensile test for mild steel

20. 1/20 Nominal and true SS Nominal stress based on initial area True stress based on necked area Nominal strain based on initial length True strain based on current length Use nominal values when operating within elastic limit

21. 1/21 Test data – linear scale

22. 1/22 TENSILE TEST SPECIMEN

23. 1/23 BRITTLE MATERIAL

24. 1/24 CAST IRON Compression Tension

25. 1/24 STONE Compression Tension Strain Stress 5-200MPa

26. 1/26 WOOD Not an isotropic or homogeneous material Stronger and stiffer along the grain Stronger in tension than compression Fibres buckle in compression Very high strength/weight ratio Very high stiffness/weight ratio

27. 1/27 COMPRESSION TEST

28. 1/28 Compression test - concrete

29. 1/29 ELASTICITY

30. 1/30 PLASTICITY

31. 1/31 DUCTILITY

32. 1/32 CREEP

33. 1/33 LINEAR ELASTICITY STRAIGHT LINE PORTION OF STRESS- STRAIN CURVE  = E.  (Hooke’s Law) E is the modulus of elasticity or Young’s Modulus and is the slope of the curve For stiff materials E is high (steel 200GPa) For plastics E is low (1 to 10 GPa)

34. 1/34 POISSON’S RATIO Tensile stretching of a bar results in lateral contraction or strain (and v v) For homogeneous materials axial strain is proportional to lateral strain Poisson’s Ratio = - (lateral/axial strain) = - (  ’ /  ) For a bar in tension  is positive and  ’ is negative, and v.v. for compression.

35. 1/35 Poisson (1781-1840) Normal values 0.25-0.35 for metals Concrete about 0.2 Cork about 0 (makes it a good stopper) Auxetic materials have NEGATIVE

36. 1/36 Axial and lateral deformation L  B B. .  =  /L

37. 1/37 Volume change V 1 =L*B 2 (original volume) V 2 =L(1+  )*(B(1-  * )) 2 (final volume) V 2 =L B 2 (1+  -2  -2  2 +  2 2 +  3 2 ) V 2  L B 2 (1+  -2  )

38. 1/38 DILATION  V=L B 2  (1 -2 ) (change in volume)  V/V 1 =  (1 -2 ) or (1 -2 )  /E = DILATION Max value of is 0.5 Since is 1/4 to 1/3 then dilation is  /3 to  /2

39. 1/39 SUMMARY Stress - Force/Area Strain – (Extension or compression)/Length For some materials and subcritical loads strain is proportional to stress (Hooke) Change in length proportional to change in width (Poisson). Shrinkage/expansion. Characteristics determined by experiment