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Reg 162 note dr norizal

1. 1. 1 | P a g eLECTURE 1Youngs modulus,Also known as the tensile modulus or elastic modulus is a measure ofthe stiffness of an elastic material and is a quantity used to characterize materials.It is defined as the ratio of the stress along an axis over the strain along that axis inthe range of stress in which Hookes law holds. In solid mechanics, the slope ofthe stress-strain curve at any point is called the tangent modulus. The tangentmodulus of the initial, linear portion of a stress-strain curve is called Youngsmodulus. It can be experimentally determined from the slope of a curve createdduring tensile tests conducted on a sample of the material. In anisotropic materials,Youngs modulus may have different values depending on the direction of theapplied force with respect to the materials structure.Elastic propertiesFor the description of the elastic properties of linear objects like wires, rods,columns which are either stretched or compressed, a convenient parameter is theratio of the stress to the strain, a parameter called the Youngs modulus of thematerial. Youngs modulus can be used to predict the elongation or compression ofan object as long as the stress is less than the yield strength of the material.
2. 2. 2 | P a g eBulk Elastic PropertiesThe bulk elastic properties of a material determine how much it will compressunder a given amount of external pressure. The ratio of the change in pressure tothe fractional volume compression is called the bulk modulus of the material.A representative value for thebulk modulus for steel isand that for water isThe reciprocal of the bulkmodulus is called thecompressibility of the substance.The amount of compression ofsolids and liquids is seen to bevery small.MaterialDensity(kg/m3)YoungsModulus109N/m2Ultimate Strength(or “UltimateStress”) Su106N/m2Yield Strength(or “YieldStress”) Sy106N/m2Steels7860 200 400 250Aluminum 2710 70 110 95Glass 2190 65 50b...Concrete 2320 30 40b...Wood 525 13 50b...Bone 1900 9b170b...Polystyrene 1050 3 48 ...
3. 3. 3 | P a g eLECTURE 2CENTROIDS AND MOMENT INERTIA Centroids axis in all body and structure The centroid of a two dimensional surface is a point that corresponds to the center ofgravity of a very thin homogeneous plate of the same area and shape.Formula centroid gravityArea centroid gravity= 0 ̅ = r= bh ̅ =0= h ̅ =̅=b̅ =h ̅b
4. 4. 4 | P a g eExample 1 :150mm30mm70mm30mmFind the centroid gravity??Find the y-axis :Y1 = (70 + ) = 85Y2 = ( ) = 35Use formula(150*30)(85) + (30*70) (35) = [(150*30) + (30*70)] YCYC = 69.1MMd1=15.9d=34.1 85MM35 69.1A1A2Calculation must from bottomAIYI + A2Y2 = (A1+A2) YCA2A1YC
5. 5. 5 | P a g eMOMENT INERTIA (I) Also known as the Second Moment of the Area is a term used to describe the capacity of across-section to resist bending. It is a mathematical property of a section concerned with a surface area and how that areais distributed about the reference axis. The reference axis is usually a centroid axis It is used to determine the state of stress in a section. It is used to calculate the resistance to bending. It can be used to determine the amount of deflection in a beam.FORMULASecond Moment of the AreaIxx = Sum (A)(y2)In which:Ixx = the moment of inertia around the x axisA = the area of the plane of the objecty = the distance between the centroid of the object and the x axisIx:Iy:
6. 6. 6 | P a g eExample 2:5 cm10 cm 10 cm5 cmI xx for a 5 (10 ) ^ 3 / 12 = 416.6 cm4I xx for b 10 ( 5)^ 3 / 12 = 104 cm4ab
7. 7. 7 | P a g eTransfer formula• There are many built-up sections in which the component parts are not symmetricallydistributed about the centroid axis.• To determine the moment of inertia of such a section is to find the moment of inertia ofthe component parts about their own centroid axis and then apply the transfer formula.• The transfer formula transfers the moment of inertia of a section or area from its owncentroid axis to another parallel axis. It is known from calculus to be:FORMULAOr ( expand from above)[ ] [ ]
8. 8. 8 | P a g eExample 3 :15015030d1=15.9d=34.1 85MM35 69.1 7030Find the moment inertia.Use formula : [ ] [ ]: [ ] []= 4774546A2A1YCA1A2
9. 9. 9 | P a g eLECTURE 3BENDING STRESFormulaUnit: N/mm^2WHERE: Z= I/Y Y= center gravity to participate section M= bending momentF= =
10. 10. 10 | P a g eExample 4:100mmWhere M=15KNm250mmFind the bending stress.Y= d/2 = 250/2 = 125mmI ===130208333.3 mm^4Use formula : Z= I/YZ= 130208333.3/125=1041666.667mm^3Then use formula stress :f= m/zF= (15*10^6)/(1041666.67)=14.40N/MM^2CHANGE M TOMM.
11. 11. 11 | P a g eBENDING FORMULA in Combined StressZ is termed as Modulus of section = I/yI – moment inertia; y – distance from Neutral axis to stress levelM – Applied moment (from loads)Combined Stress – bending and compression
12. 12. 12 | P a g eEXAMPLE 5 :1000 mm100mm xxyyColumn / wall sectionI xx = 1000 (100)^3 /12 = 83.33 e 6 mm4Zxx = 83.33e6 / 50 = 1.66 e 6 mm3Axial Load P = 50 kNApplied Moment = 5 kNmTotal Stress on section = P/A +- M/Z50e3 / 1e5 + 5 e 6 / 1.66 e6 = 0.5 + 3 = 5.3 N/mm2Or 0.5 – 3 = -2.5 N/mm2 tensile stress in wallCompressive stress 5.3N/mm2+ M/Z =Tensile stress -2.5N/mm2Compressive(P)0.5N/mm2 3N/mm2POSITIVE – COMPRESIONNEGATIVE – TENSILESTRESS
13. 13. 13 | P a g eSHEAR STRESSESThe formula for shear stress in beams is a as belowWhere :V - shear force at that sectionQ – first moment of areaI – moment of inertiab- breadth of web
14. 14. 14 | P a g eExample:for rectangular section :b- 50 mm ( breadth )d- 100 mm ( depth ) 50mmI = 50 (100) ^3 /12 = 4.16e6 mm4 x xV = 5 kNQ = Ay = 50 x 50 = 2500 x 50/2= 62.5e3 mm3T ( shear stress at XX ) = 5e3 . ( 62.5e3 ) /( 4.16e6). 50 = 1.5N/mm2Average shear stress = V/A = 5000 / 50 x 100 = 1 N/mm2Maximum shear stress in timber beam = 0.9 to 1.45 parallel to grainSo above beam may fail if it is made from timberIf concrete beam grade 20 then the allowable shear is 0.26 N/mm2 so this beammay too
15. 15. 15 | P a g eKOMPOSIT SECTIONSSometimes sections of beams or columns can be made of 2 or more different materials such asa sandwich section or compbined beam and concrete deck as shown below.Both materials may be bolted or glued but they are supposed to act together / combined as awhole unit.In sandwich section the internal core takes the shear and the outer skin takes the bending .Composite sectionIf the section is acting as a whole , then the strain across section is still linear .Outer skinInner core
16. 16. 16 | P a g eEc (compressive strain)Neutral axisEt( tensile strain)Because a stronger materials such as steel has a higher E value compared to concrete , a higherload is taken by the steel and the stress is not uniform as in a homogenous section.
17. 17. 17 | P a g eWe have to convert the section into a “transformed section” so that we can analyze the stressin the section.Skin stressCore stressStress in steelIn timberSteelTimberE steelE timber
18. 18. 18 | P a g eThe Modula ratio = 203.4 / 12.1 = 16.8So in a section shown below the steel plate can be converted into a timber plate by;B16.8 BtimberSteel
19. 19. 19 | P a g eExample :Determine the maximum bending stress to the timber and steel if the applied moment on theabove section is 5.65 kNm.E ( steel) = 203.4 kN/mm2E ( timber) = 12.1 kN/mm2M ( Modula ratio ) = 203.4/12.1 = 16.8Transformation B = 75 x 16.8 = 1260 mm1260mm6.25mm137.5mm
20. 20. 20 | P a g eStress at top of section = 8.15 N/mm2Actual Steel Stress at this top section = 8.15 x 16.8 = 136.92 N/mm2Maximum bending stress ( at timber section ) where y = 137.5 / 2 = 68.75mm
21. 21. 21 | P a g eIf the sandwich composite section is made of outer skin cement board and inner corelightweight concrete of density 1000 kg/m3 , find out the maximum bending stress in thecement board and stress at the LWC section.Timberstresssteel
22. 22. 22 | P a g e