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## Department of Civil and Environment Engineering

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**Department of Civil and Environment Engineering**CGN 4980/CGN 6939 FE/Graduate Seminar Review Notes Fall 2005**TOPICS**• Forces • Moments • Equilibrium • Trusses and Frames • Plane Areas • Friction**Types of Forces**• Concurrent coplanar • Non-Concurrent coplanar • Concurrent non-coplanar • Non-Concurrent non-coplanar • External forces • Internal forces**Representation of Forces**A vector is represented graphically by an arrow which defines the magnitude direction and sense**Resultant of Coplanar forces**Cartesian Vector Notation Component vectors F1 = Fxi + Fyj F1 = F1xi + F1yj F2 = -F2xi + F2yj F3 = -F3xi - F3yj**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**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**Direction of a Cartesian Vector**Unit vector Important relations A = AuA = Axi + Ayj+ Azk**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**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**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**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.**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)**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**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 =>**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**Moment of a Forces -Vector**Resultant Moment of Forces -Vector MRo = ( r x F) Resultant Moment of Forces -Scalar M0 = Fd MR0 =Fd**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**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**Equilibrium of a Rigid Body**• F = 0 • M= 0 Equilibrium in 2D**Equilibrium in 3D**• F (x,y,z)= 0 • M0(x,y,z) = 0**Structural Analysis**Assumptions All loads are applied at the joints Members are joined together by smooth pins**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**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**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,**Structural Analysis**Zero Force Members (a) a (b) (d) (c) (e)**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**Friction**Impending Motion (static) F s= sN s= sN Motion (P > Fk kinetics) F k= kN**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.**Impending Motion at All Points**FB NB W slipping must occur at A & B L sin q FA NA Equilibrium Eqs.**Center of Gravity**Similar equations for line and volumes**Example**Find the x and y coordinates of the centroid Find the Centroid**3**2 1**Example**Find the x and y coordinates of the centroid 1 2 3**Moment of Inertia**Moment of Inertia Polar Moment of Inertia**Moment of Inertia**Parallel axis theory Polar Moment of Inertia Radius of Gyration