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Department of Civil and Environment Engineering

Department of Civil and Environment Engineering

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Department of Civil and Environment Engineering

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  1. Department of Civil and Environment Engineering CGN 4980/CGN 6939 FE/Graduate Seminar Review Notes Fall 2005

  2. TOPICS • Forces • Moments • Equilibrium • Trusses and Frames • Plane Areas • Friction

  3. Types of Forces • Concurrent coplanar • Non-Concurrent coplanar • Concurrent non-coplanar • Non-Concurrent non-coplanar • External forces • Internal forces

  4. Representation of Forces A vector is represented graphically by an arrow which defines the magnitude direction and sense

  5. Vector Operations

  6. Resultant of Coplanar forces Cartesian Vector Notation Component vectors F1 = Fxi + Fyj F1 = F1xi + F1yj F2 = -F2xi + F2yj F3 = -F3xi - F3yj

  7. Resultant - Vector notation Vector Resultant FR= F1 + F2+ F3 = (FRx)i + (FRy)j FR= F1 + F2+ F3 = (FRx)i + (FRy)j Resultant - Scalar notation FRx=  Fx FRy=  Fy

  8. Rectangular Components of a Vector A= Ax + Ay+ Az Unit vector uA = A => A = AuA A Cartesian vector representation A= Axi + Ayj+ Azk Magnitude of a cartesian vector

  9. Direction of a Cartesian Vector Unit vector Important relations A = AuA = Axi + Ayj+ Azk

  10. Position Vector Position vector r is defined by r = xi + yj+ zk rA+ r = rB Direction Vector Direction vector rAB is defined by rAB= (xB – xA)i + (yB – yA)j + (zB – zA)k

  11. Dot Product A.B = AB cos  0 1 Commutative law A.B = B.A i.i = (1) (1) cos 0 o = 0 i.j = (1) (1) cos 90 o = 0 2 Multiplication by a scalar a(A.B) = (aA).B =A.(aB) = a(B.A) i.i = 1 j.j = 1 k.k = 1 i.j = 0 j.k = 0 k.j = 0 3 Distributive law A.(B+D) = A.B + (A.D) A.B = AxBx + AyBy + AzBz

  12. Dot Product Applications 1) Angle formed between two vectors or lines A//= |A| cos u A// = (A.u)u A.B A = A + A A= A - A 2) Component of a vector // or perpendicular to a line Projection of A along a line is the dot product of A and the unit vector uwhich defines the direction of the line

  13. Dot Product Applications z F= (300j) N B 3m A y 2m 6m x Determine the magnitude of components of the force F, parallel and perpendicular to member AB The magnitude of the component of F along AB is equal to the dot product of F and the unit vector uB, which defines the direction of AB.

  14. Cross Product C = A x B = AB sin  Magnitude of C C = AB sin  Direction of C C = A x B =( AB sin  )uc 1 Commutative law 2 Multiplication by scalar A x B B x A a(A x B) = (aA) x B =A x (aB) = (B x A) a A x B = - B x A 3 Distributive law A x (B + D) = (A x B) + (A x D)

  15. Cartesian Vector Formation i x j = k i x k = -j i x i = 0 j x k = i j x i = -k j x j = 0 k x i = j k x j = -i k x k = 0 AxB = (Ax i + Ay j + Az k) x (Bx i + By j + Bz k) A x B NB : j element has a –ve sign

  16. Equilibrium of Forces 1 Two dimensional Forces  F = 0  Fxi +  Fyj = 0  Fx = 0  Fy = 0 => 2 Three dimensional Forces •  Fx = 0 • Fy = 0 • Fz = 0  F = 0  Fxi +  Fyj +  Fzk = 0 =>

  17. M0 = rF sin = F( r sin ) = Fd Moment of a Forces -Vector M0 = r x F 1) Magnitude 2) Direction determined by right hand rule Cartesian Vector Formulation M0 = rBxF M0= rCxF M0= rCxF Transmissibility M0 = (ryFz – rzFy)i – (rxFz – rzFx)j + (rxFy – ryFx)k

  18. Moment of a Forces -Vector Resultant Moment of Forces -Vector MRo = ( r x F) Resultant Moment of Forces -Scalar M0 = Fd MR0 =Fd

  19. Moment of a Force About a Specific Axis 1 Scalar analysis Ma = F da da is the perpendicular or shortest distance from the force line of action to the axis 2 Vector analysis Mb = ub . (r x F) Mb Mb= Mbub

  20. Moment of a Couple - A Couple is a pair of equal and opposite parallel forces - Two Couples producing the same moment are equivalent 1 Scalar analysis 2 Vector analysis M= F d M= r x F

  21. Equilibrium of a Rigid Body • F = 0 • M= 0 Equilibrium in 2D

  22. Equilibrium in 3D • F (x,y,z)= 0 • M0(x,y,z) = 0

  23. Structural Analysis Assumptions All loads are applied at the joints Members are joined together by smooth pins

  24. Structural Analysis Method of Joints Draw free body diagram Establish sense of known forces Orient x an y axes Apply equilibrium equation Start from one simple joint and proceed to others

  25. Structural Analysis Zero Force Members If two members form a truss joint and no external load or support reaction is applied to the joint, the member must be zero-force member

  26. Structural Analysis Zero Force Members If three members form a truss joint for which two of the members are co-linear, the third member is a zero-force member provided no external force or support reaction is applied to that joint,

  27. Structural Analysis Zero Force Members (a) a (b) (d) (c) (e)

  28. Structural Analysis Method of Sections Determine external reaction Cut members through sections where force is to be determined Draw free body diagram Determine external reaction Apply equilibrium equations

  29. Friction Impending Motion (static) F s= sN s= sN Motion (P > Fk kinetics) F k= kN

  30. Friction

  31. Friction

  32. Friction

  33. Friction

  34. Sample Problem The rod has a weight W and rests against the floor and wal for which the coefficients of static friction are mA and mB, respectively. Determine the smallest value of q for which the rod will not move.

  35. Impending Motion at All Points FB NB W slipping must occur at A & B L sin q FA NA Equilibrium Eqs.

  36. Center of Gravity Similar equations for line and volumes

  37. Example Find the x and y coordinates of the centroid Find the Centroid

  38. 3 2 1

  39. Example Find the x and y coordinates of the centroid 1 2 3

  40. Moment of Inertia Moment of Inertia Polar Moment of Inertia

  41. Moment of Inertia Parallel axis theory Polar Moment of Inertia Radius of Gyration