Study Guide for Statics Exam 1
Study Guide for Statics Exam 1 CE 301
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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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