MEC 0011 Statics Lecture 6 Prof. Sanghee Kim Fall_ 2012

5.4 Two- and Three-Force Members Two-Force Members - When forces are applied at only two points on a member, the member is called a two-force member - two force on a member  a. same magnitude b. opposite direction c. same line of action

Three-Force Members - When subjected to three forces, the forces are concurrent or parallel line of action of F1 and F2 intersect at O line of actin of F3 must also pass through point O

Example 5.13 The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A.

Solution Free Body Diagrams BD is a two-force member Lever ABC is a three-force member Equations of Equilibrium

Compare the force exerted on the toe and heel of a 120-lb woman when she is wearing regular shoes and stiletto heels. Assume all her weight is placed on one foot and the reactions occur at points A and B as shown. Exercise

Solution - Stiletto heal shoe - Regular shoe

5.5 Free-Body Diagrams Single pin Single bearing Single hinge Others (properly aligned) Force + couple moment Force Single pin Single bearing Single hinge

Example 5.14

5.6 Equations of Equilibrium Vector Equations of Equilibrium For two conditions for equilibrium of a rigid body in vector form, ∑F = 0 ∑M O = 0 Scalar Equations of Equilibrium If all external forces and couple moments are expressed in Cartesian vector form ∑F = ∑F x i + ∑F y j + ∑F z k = 0 ∑M O = ∑M x i + ∑M y j + ∑M z k = 0 external forces couple moments

5.7 Constraints for a Rigid Body Redundant Constraints More support than needed for equilibrium Statically indeterminate: more unknown loadings than equations of equilibrium Deformation conditions at the point of supportsMechanics of materials 5 unknowns  3 equilibrium eqs 8 unknowns  6 equilibrium eqs

Improper Constraints Instability caused by the improper constraining by the supports When all reactive forces are concurrent at this point, the body is improperly constrained

Example 5.15 The homogenous plate has a mass of 100kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket joint at N, and a cord at C, determine the components of reactions at the supports.

Free Body Diagrams - Five unknown reactions acting on the plate - Each reaction assumed to act in a positive coordinate direction Equations of Equilibrium Solution

Equations of Equilibrium Components of force at B can be eliminated if x ’, y ’ and z ’ axes are used Solving, above equations A z = 790N B z = -217N T C = 707N