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  1. 1. PN. IRINAH BINTI ABDULLAH 013 - 7443474
  2. 2. JOHOR BAHRU POLYTECHNIC DEPARTMENT OFMECHANICAL ENGINEERINGKAJIDAYA BAHAN 1 / J3009CONTINUOUS ASSESSMENT (CA)a. Quiz (2) = 20%b. Assignments (2) = 30%c. Case Study (1) = 20%d. Test (2) = 30%FINAL ASSESSMENT (FA)a. Final Exam = 50%
  3. 3. REFERENCES• 1. Ahmad Zafri Bin Zainudin & Yazid Bin Yahya (1998) Mekanik Pepejal 1. Universiti Teknologi Malaysia.• 2. Ahmad Zafri Bin Zainudin (1998) Mekanik Bahan 1. Universiti Teknologi Malaysia.• 3. Beer & Johnston (1992) Mechanic Of Materials. McGraw Hill.• 4. G.H.Ryder (1983) Kekuatan Bahan, Edisi Bahasa Malaysia. UPM.• 5. I.Granet (1980) Strength Of Materials For Engineering Technology. Reston.
  6. 6. Unit 1
  7. 7. example 1 : Figure 1.0 shows a steel bar has a rectangular cross section measuring 30mm x 25mm. 40kN tensile force is imposed on the bar. Calculate the resulting stress. 40kN 40kN Figure 1.0
  8. 8. Example 2 : Figure 1.1 shows a block cylinder 30cm in length, have a diameter of 10cm. This block is a compression load of 70kN and shrink 0.02cm. Find a compressive stress and compressive strain. 70kN 70kN figure 1.1
  9. 9. example 3 : A glass bar that has a 2m long and cross section of 25mm x 20mm. The bar is forced by 52kN load. Determine the elongation of the glass given modulus of elasticity 60GPa.
  10. 10. practice1. A brick that has a cross section of 15mm x 20mm and length 30cm as shown in Figure 1.2 bear Abu weight of 68kg. Given that the Young modulus 30GPa. Calculate stresses and length changes that occur on the bricks. figure 1.2
  11. 11. Unit 2
  12. 12. 2.0 Hooke’s Law elastic material is a material that changes shape easily if the burden imposed against him. It expenses or not subject to the condition and would return to real condition.
  13. 13. ∆P Kawasan anjal. Tegasan berkadarterus dengan terikan Graf 2.1
  14. 14. Max. forceForce, P Elongation,∆ L yeild Strain hardeningA = Yeild point D = Lower yield pointB = Elastic point E = Maximum forceC = Upper yield point F = Rupture
  15. 15. i. Yield stress= force on point C cross sectional area PY σy = AO ii. Max. stress = max.force on point E cross sectional area Pm σm = Ao iii. Percent of elongation= change on length original length X 100% ∆L %L = ×100% L
  16. 16. iv. Percent of area = change on area original area X 100% ∆A %A = ×100% Ao
  17. 17. example 1 :The tensile tests were conducted on asubstance, the data acquired is recorded asfollows: Gauge length = 260mm Final gauge length = 295mm Original diameter = 30mm End of Diameter = 19mmForce , kN 20 60 100 140 160 170 172 176 178elongation x 10-3 50 160 260 360 410 440 470 550 720Force , kN 180 190 220 240 257 261 242 229 180elongation x 10-3 760 900 1460 1990 3120 4500 5800 5850 760
  18. 18. Draw a graph of load versus elongation and determine:a.Youngs modulusb. The yield stressc. The maximum stressd. Percentage of areae. Percent of elongation
  19. 19. Example 2 :The tensile tests were conducted on a rod with a diameterof 6.4mm and length of 30mm, the determinant of thematerial are as follows:Tensile strength = 500MN/m2Yield stress = 210MN/m2% elongation = 20%% reduction in area = 60%calculate:a. maximum loadb. Minimum load at yield levels Ac. minimum lengthd. maximum diameter
  20. 20. UNIT 3
  21. 21. 3.1 Safety Factor Formula, n σy σm σ allowed = σ allowed = n n Py Pmn= n= P work Pwork
  22. 22. 3.2 safety margin n -1Note: the ultimate tensile or yield stress must be two times then allowable stress ,and FK should be larger than 1
  23. 23. example : A hollow cylinder used to support a compressive load of 13kN. Ultimate stress is 350kN/m2, the cylinder is 1 cm thick and safety factor is 3.0. calculate:a. allowable stressb. cross-sectional areac. outside diameter of the cylinder
  24. 24. 3.6 Strain EnergyForce Stain energy elongation
  25. 25. Example :A bar is loaded by a force as in Figure 3.1.calculate the strain energy stored in the bar.Given E = 200GN/m2. 200kN ∅40mm 0mm 52 200kN figure 3.1
  26. 26. 3.7 Poisson Ratio G A B F H E D CA tensile force P is act on square ABCD and it will changeshape into EFGH, the action of this force, the strain thatoccurs towards x is known as LONGITUDINAL strain (ε x),while the square is also a shortage in the y-direction widthmeasurement known as LATERAL strain (ε y).
  27. 27. LONGITUDINAL LATERAL strain strain ∆L σ ∆dεx = = εy = L E d ∆L d ∆d L
  28. 28. example :A load of 45kN is act on figure 3.2 below. GivenE = 200GN/m2, Ʋ = 0.4 calculate:a. stressb. longitudinal strainc. lateral strain 250mm 30mm 45kN 45kN 50mm figure3.2
  29. 29. 3.8 Shear stress 3.8.1 shear stress is the frictional between two surfaces abed and EFGH as shown in Figure 3.3Shear force,P d c a b e f Shear force,P g h Figure 3.3
  30. 30. Example 1 :Figure 3.4 shows a punch of diameter 20 mm is used to drill a hole on the plate of 8mm thick. Force 120kN is loaded at punch, determine the shear stresses that occur on the plate. 120kN punch D=20mm 8mm Shear act on the surface Figure 3.4
  31. 31. Example 2 : Three plates are connected using two ribet as figure 3.5. If shear stress is 35N/m2. calculate the diameter of ribet. 8kN Figure 3.5
  32. 32. 3.8.2 Shear strain will lead to distortions on onesurface or structural blocks as in Figure 3.4 x Shear force,P γ y Figure 3.4
  33. 33. Example : V=120kN db=25mm γh=30mm h b a=50mm Rajah 3.5 Figure 3.5 above shows a pure copper plates the elastic material located above its subjected to horizontal forces V. Determine the shear stress, shear strain and horizontal displacement of the plate. Given the rigidity modulus G = 45GPa.