FE/Graduate Seminar Review Notes 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

A FORCE is caused by the action of one body on the another body A FORCE is caused by the action of one body on the another body. Forces are described by: Its point of application. Its magnitude. Its direction or two points along its line of action

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

Vector Operations

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 = A uA 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 Direction Vector Position vector r is defined by r = xi + yj+ zk rA + r = rB Position vector r is a fixed vector which locates a point in space relative to another point. Direction Vector Direction vector rAB is defined by rAB = (xB – xA)i + (yB – yA)j + (zB – zA)k

2 Multiplication by a scalar 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 (to determine component of forces parallel and perpendicular to a line ) 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 u which defines the direction of the line

Dot Product Applications F= (300j) N A B z 6m 2m 3m x y 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.

2 Multiplication by scalar A x B B x A 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 =>

Moment of a Forces -Vector M0 = r x F 1) Magnitude M0 = rF sin = F( r sin ) = Fd 2) Direction determined by right hand rule Cartesian Vector Formulation M0 = rB x F M0 = rC x F 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 = Mb ub

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 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 Apply equilibrium equations

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

Friction

Friction

Friction

Friction

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.

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

Similar equations for line and volumes Center of Gravity Similar equations for line and volumes

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

1 2 3

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

Polar Moment of Inertia

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