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ENGINEERING MECHANICS

ENGINEERING MECHANICS. Part Ⅱ Mechanics of Materials or Strength of Materials Mechanics of deformable bodies Theoretical Mechanics------particles or rigid bodies Mechanics of Materials-----deformable solid bodies. Introduction.

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ENGINEERING MECHANICS

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  1. ENGINEERING MECHANICS Part Ⅱ Mechanics of Materials or Strength of Materials Mechanics of deformable bodies Theoretical Mechanics------particles or rigid bodies Mechanics of Materials-----deformable solid bodies Kylinsoft

  2. Introduction Mechanics of Materials is a branch of applied mechanics, it deals with the behaviors of deformablesolid bodies subjected to various types of loading. Behaviors: stress, strain, deformation Abilities: (a)Strength —the ability to prevent failure (b) Rigidity — the ability to resist deformation (c) Stability — the ability to keep the original equilibrium state (d) Toughness — the ability to resist fracture Kylinsoft

  3. Introduction Solid bodies(1D) and various types of loading axially loaded members in tension or compression shafts in torsion beams in bending columns in compression The principle of superposition The resultant response in a system due to several forces is the algebraic sum of their effects when separately applied only if each effect is linearly related to the force causing it. Kylinsoft

  4. Tension and Compression Torsion Bending Kylinsoft

  5. Torsion: Loaded by pairs of forces——torque, twisting couples, or twisting moments. There will be a rotation about the longitudinal axis of one end of the bar with respect to the other. Kylinsoft

  6. Bending Forces acting transverse to the axis of structural member

  7. Chapter 4 inner forces 4.2 Method of Section for Internal Force The resistance forces that set up within a body to balance the effect of the externally applied forces. Make an imaginary cut at cross section perpendicular to longitudinal axis. Using for forces representing action of the moved part upon remained one, the forces are continuously distributed over the cross section. Kylinsoft

  8. N A P P P P L 4.3 Tension and compression Prismatic bar loaded by axial force at the centroid of ends (Axially loaded members). s Resultant of forces N For equilibrium：S x=0 N - P=0 Axial internal force ： N=P Kylinsoft

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  11. A C P2=15kN B P3=10kN x P1=5kN 1 1 A N1 P1=5kN 2 2 B P3 =10kN N2 N 5kN x 10kN Axial Force Diagram Example determine the axial forces and draw the axial-force diagram. N1-P1=0 N1=P1=5kN N2 = -10kN Kylinsoft

  12. Example P1=15kN, P2=10kN, P3=5kN, determine the axial forces on the cross sections 1-1, 2-2, and draw the axial-force diagram. Solution: X=0 N1 +P1=0 N1=-P1=-15kN X=0 N2 +P1 -P2=0 N2=-5kN N(kN) x (-) (-) -5 -10 Kylinsoft

  13. T 4.4 diagram of torque4.4.1 Transmission of power by circular shafts Motor driven shaft Angular speed ω(rad/s) Transmitted torque T(N·m) Φ—angular rotation Work: Power: f—frequency of revolution, Hz=s-1 n—number of revolutions per minute (rpm) If N is expresses in kilowatt or horsepower, 1 hp=735.5 W Kylinsoft

  14. n m m n n m x T n n m T n 4.4.2 Diagram of Torsion Moment Equilibrium Smx = 0 Method of Section for Internal Force T - m = 0 T = m Sign convention for torque ——right hand law. Kylinsoft

  15. 10 20 15 a b c d 10 20 15 I b c d T3 T1 T2 20 15 II c d T 15 5kN-m III d x 5kN-m 15kN-m Determine the torsion moment of the shaft SMx = 0 15-20+10+T1 = 0 T1 = -5kN-m 15-20+T2 = 0 T2 = 5kN-m 15+T3 = 0 T3 = -15kN-m Kylinsoft

  16. Kylinsoft

  17. A B q P 4.5 Shear Forces and Bending Moments 4.5.1 Beams Beam: the structural member to resist forces acting transverse to its axis Simple beam Simply supported beam Kylinsoft

  18. P P2 P1 A B A B Beam with an overhang Cantilever beam Kylinsoft

  19. Plane Bending • having symmetric cross sections • Loads act in the symmetric plane • the bending deflections occur in this plane Kylinsoft

  20. Plane bending and 3D bending Kylinsoft

  21. Type of Loads Concentrated Load P Distributed Load q Couple M Kylinsoft

  22. Reactions in various types of supports FAx Simple Beam FBy FAy FAx Beam with an Overhang FAy FBy FAx Cantilever Beam MA FAy How to find the reactions? Kylinsoft

  23. P B m n A x P M M V M V A x V o V M V M V B V V V M M V M M M 4.5.2 Shear Forces and Bending Moments Method of section SY=0 P - V =0 V = P SMo=0 M - Px =0 M = Px Sign Convention Kylinsoft

  24. P a M x V RA 2P M 2P P V a a I I RB B RA RB A x l Find the Bending Moment and Shear Force on cross section of I-I Bending Moment and Shear Force Reactions RA, RB S MB=0 RAl - P(l-a) - 2Pa = 0 RA + RB - 3P = 0 S Y=0 RA - P - V = 0 S Y=0 RA x - P(x-a) - M = 0 S M0=0 Kylinsoft

  25. C P a b B A C x RB RA l M(x) A V(x) x RA Shear-Force and Bending-Moment Diagrams Concentrated Loads Reactions Equations for shear force and bending moment (0 < x < a) RA - V(x) = 0 SY = 0 S M= 0 RA x - M(x) = 0 Kylinsoft

  26. C P a b B A C x l P a M(x) A x V(x) RA (a < x < l) RA - V(x) - P = 0 SY = 0 S M = 0 Rax - P(x-a) - M(x) = 0 Kylinsoft

  27. P a b B A C x l V M (0 < x < a) (a < x < l) Kylinsoft

  28. b a3 a2 P3 P1 P2 a1 RA b RB RB A B P3 M(x) B x l V(x) x l V P1 P2 RA P3 M RB M2 M3 M1 A beam loaded by several concentrated forces Reactions Ra and RB (0 < x < a1) V(x) = RA M(x)=RAx (a1< x < a2) V(x) = RA- P1 M(x)=RA x -P1(x-a1) (a2< x < a3) V(x) = RA- P1 – P2 M(x)= RA x -P1(x-a1)- P2(x-a2) (a3< x < l) V(x) = - RB M(x)=RB (l - x) Kylinsoft

  29. q A B x l q A RB RA M(x) x RA V V(x) M Distributed Loads RA - V(x) - qx = 0 SY=0 S M = 0 (0 < x < l) Kylinsoft

  30. q B A x l M(x) B l-x V(x) V ql M Cantilever beam loaded by uniformly distributed forces SY=0 V(x) - q (l-x) = 0 V(x) = q (l-x) S M = 0 Kylinsoft

  31. q=3kN/m，m=3kNm q m B C A D x 2m 2m 4m RB RA M V 8.5kN + 3.5kN 6kN 4.83m + 6.04kNm 7kNm 4kNm 6kNm Beam with an overhang RA=14.5kNRB=3.5kN (0<x<2m) V(x) = - qx M(x) = - qx2 / 2 (2m<x<6m) V(x) = RA- qx=14.5 - 3x M(x) =RA(x-2) - qx2 / 2 = 14.5(x-2)-1.5x2 x = 4.83m M (x = 4.83m)=14.5(4.83-2)-1.5×4.832 = 6.04kNm (6m<x<8m) V(x) = - RB= -3.5 kN M(x) = RB(8-x) = 3.5(8-x) Kylinsoft

  32. y x o q(x) x dx M(x)+dM M(x) V(x)+dV V(x) dx C q(x) 4.5.3 Relationships between Loads, Shear Forces and Bending Moments SY = 0 SM= 0 Kylinsoft

  33. 10kN-m F D E A B C RA 1.5m 1.5m 1.5m RB 20kN 10kN-m C RA B V M 1.5m RA V M 1.5m Shear force and bending moment diagrams Reactions RA= 8.9kN RB= 11.1kN Shear force and bending moment at special cross sections Special cross sections of beam are where loads occur abrupt change Kylinsoft

  34. 2m 2m 4m RB M V 8.5kN + 3.5kN 6kN 4.83m V V + 6.04kNm 7kNm 4kNm M M 6kNm Techniques of how to draw shear force and bending moment diagram q=3 m=3 B C A D x q is positive when it acts downward q=0, V=c, M=f(x) q=c, V=f(x), M=f(x2 ) q=f(x), V =f(x2 ) , M= f(x3 ) V has a sudden increase where P acts upward. M has a sudden increase where m acts clockwise. RA Kylinsoft

  35. P Pa A C B P a a Pa A C B RB RA MA a a V P RA Pa M M Pa Examples RA= P MA= 0 RA= 0 RB= P V P Kylinsoft

  36. 10kN-m F D E A B C RA 1.5m 1.5m 1.5m RB 20kN 10kN-m RA V M C V RA V M 1.5m 11.1 x M 8.9 3.35 13.35 x 16.65 Examplesimple beam RA= 8.9kN RB= 11.1kN A:V= -8.9kN M=0 B:V= -8.9kN M= -13.35kN-m C:V= -8.9kN M= -3.35kN-m D:V= -8.9kN M= -16.65kN-m E:V= 11.11kN M= -16.65kM-m F:V= 11.11kN M= 0 Kylinsoft

  37. M x V 7kN m=3.6kN-m P=3kN q=10kN/m x A D B 3kN C 5kN 0.6m 1.25kN-m 0.6m 1.2m 2.4kN-m RA RB 1.2kN-m 1.8kN-m example RA=10kN RB=5kN Kylinsoft

  38. qa q q a qa a a 2qa qa qa V 0. 5qa2 M x x 0. 5qa2 example Find mistakes in shear force and bending moment diagram Correct Kylinsoft

  39. q C B A L/2 L/2 3qL/8 M V RB=qL/8 RA=3qL/8 3L/8 qL/8 9qL2/128 qL2/16 Example Kylinsoft

  40. V V Kylinsoft

  41. V Kylinsoft

  42. How to get the inner force in following bent shaft • Establish coordinate; • Find reactions; • Treat loads; • Find inner force components in different segments. Kylinsoft

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