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by: MJ F.

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# Study Guide for Statics Exam 1 CE 301

Marketplace > Kansas > Civil Engineering > CE 301 > Study Guide for Statics Exam 1
MJ F.
KU

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## About this Document

Study guide hits all topics covered in class with definitions and examples, problems are review problems as given in class.
COURSE
Statics and Dynamics
PROF.
Lyon, Robert/Carter, Nicole
TYPE
Study Guide
PAGES
6
WORDS
CONCEPTS
Statics & Dynamics, review, forces, Exam 1, Engineering, statics
KARMA
50 ?

## Popular in Civil Engineering

This 6 page Study Guide was uploaded by MJ F. on Monday September 5, 2016. The Study Guide belongs to CE 301 at Kansas taught by Lyon, Robert/Carter, Nicole in Fall 2016. Since its upload, it has received 10 views. For similar materials see Statics and Dynamics in Civil Engineering at Kansas.

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Date Created: 09/05/16
1 Statics Exam 1: Study Guide Fall 2016  Important Terms: a) Vocab and fun words  Scalar – signed value (i.e. and integer) without a direction Ex: 1, ­2   Vector – length and magnitude (think calc) Ex: < 2, 3, ­4>                                    Note: It is good to review vector operations!  Free Body Diagram (FBD) – Structure or object outline or representation with all  active force draw and labeled as signed vectors on the object  Moment of a Force – the tendency of an object to rotate, also known as torque  Equation: M= F*D  Positive – counter clockwise  Negative ­ clockwise b) Theorems  Varignon’s Theorem – the moment of the force sum is equal to the sum of all the  force moments   Equations: a) Equilibrium Equations  Sum of the forces in x: Fx = Ax + Bx + Cx…   Sum of the forces in y: Fy= Ay + By + Cy…  Sum of the moments at a point: Ma = B*Da + C*Da +E*Da… Note: If the sum of the forces equals 0 then there is no translation, if the sum of the  moments equals 0 then there is no rotation. b) Distributed Load Equations ∫ ×⋅ⅆ A A  Equivalent force locati:   ∫ ⅆA    This equation is more often used as the  A area of applied force divided by the equilibrium distance from the origin point. This is also simply the x value of the centroid location of the distributed force’s area.              Ex: For a square the distance would be half of the base. A triangle would be 2/3 of the base (1/3 from the high end).  2 20N D= 2m    |­­­­­­­4m­­­­­­­­­| ∫ ×⋅ⅆ A  Equivalent force magnitude:  A    This equation is most often seen as  simply the area of the distributed load with the applied force magnitude subbed in for  height. Ex: Triangle with a base of 3m and a applied for a 30N would have the  equation:     Fr = ½ *3m*30N = 45Nm c) Joint Types (Examples can be found in table 5­1 of our textbook.):  Smooth pin or hinge: has both and x and y force component. Detonated by a trapezoid shape with a curved top.  Roller: Has only a single force component in the direction of the direct force due to  contact between the joint and the member. Detonated by a trapezoidal shape with  curved top and a series of circles (i.e. wheels) on the bottom, or as simple a circle (i.e. wheel).   Fixed: Has both an x and y component along with a moment in the direction opposing the pull of the member or truss. Detonated by a wall support or plates screwed from  the member into a wall.    Problem Types and Solution Procedures: a) Two Force Members:  Only have forces acting on the member at 2 locations. Ex:   Ay      Ax                     Bx      By                  AB is a 2 force member  Design AssumptionsC  (1) All pins and pinned joints within a truss or structure are frictionless (2) Loads are only applied at the joints (This is a requirement for a 2 force load, and  if any member only has forces applied at the joints it must be a 2 force load  member).  If the member in question is straight it is under an axial load and is in either tension or compression.  3 b) Method of Joints:   If the truss is in equilibrium then the joints are in equilibrium.  All forces at the joint pass through the pin. This means all the forces intersect and  there is no moment. Procedure: (1) Determine the external forces and reactions. (2) Choose a joint with only 2 member forces and solve for the unknowns with the  equations of equilibrium (Sum of the x and Sum of the y). (3) Select a new 2 force joint and repeat steps 1 & 2 until all unknowns are solved  for. Note: A member’s reaction to an applied force (either tension or compression)  matches the type of the applied force (either tension of compression). Think   Newton’s 3    Law.  Ex: a tension force is applied to a beam the beam will apply a  tension for in the opposite direction.  c)   Method of Sections: This is a more concise and sometime quicker way of solving for the unknowns within a truss or structure. Real world application: damage analysis to a structure. Procedure: (1) Cut the truss into 2 pieces. Cutting through the members with forces you want to  solve for.  Note: Since we only have 3 equilibrium equations we can only cut through 3  members at a time. (2) Draw each half of the truss as a FBD and compensate for the forces applied by the other half as external forces.  (3) Choose a cut member to analyze and solve for the unknowns with the 3  equilibrium equations.  Pro Tip: Assume all forces are in tension (i.e. your variables are positive) in your  FBD, this will all you to keep track of the signs of the forces easier.

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