RCH 614 Note Set 5 S2012abn Moments & Supports Notation: = perpenicular istance to a force from a point = name for force vectors or magnitue of a force, as is P, Q, R x = force component in the x irection y = force component in the y irection D = free boy iagram L = beam span length M = moment ue to a force R x = resultant component in the x irection R y = resultant component in the y irection w = name for istribute loa W = name for total force ue to istribute loa x = horizontal istance x = x axis irection y = y axis irection = angle, in a trig equation, ex. sin, that is measure between the x axis an tail of a vector Moment of a orce bout an xis Two forces of the same size an irection acting at ifferent points are not equivalent. They may cause the same translation, but they cause ifferent rotation. DEINITION: Moment moment is the tenency of a force to make a boy rotate about an axis. It is measure by, where is the istance perpenicular to the line of action of the force an through the axis of rotation. C M (about ) C M (force at C) not equivalent or the same force, the bigger the lever arm (or moment arm), the bigger the moment magnitue, i.e. M 1 M 2 1
RCH 614 Note Set 5 S2012abn Units: SI: Nm, KNm Engr. English: lb-ft, kip-ft Sign conventions: Moments have magnitue an rotational irection: OUR TEXT: positive - negative CW + CCW - MOST OTHER TEXTS, INCLUDING PHYSICS TEXTS: positive - negative CCW + CW - Moments can be ae as scalar quantities when there is a sign convention. - + Repositioning a force along its line of action results in the same moment about any axis. = M = M = - M = M = force is completely efine (except for its exact position on the line of action) by x, y, an M about (size an irection). 2
RCH 614 Note Set 5 S2012abn The sign of the moment is etermine by which sie of the axis the force is on. positive or negative Varignon s Theorem: The moment of a force about any axis is equal to the sum of moments of the components about that axis. Q P. 2 1. M P Q 1 2 Proof 1: Resolve into components along line an perpenicular to it (90)... from to line = 0 from to = = = = sin cos M = 0 cos cos cos Proof 2: Resolve P an Q into P & P, an Q & Q. Q. P. Q P. P. Q.. P from to line = 0 M by P = P M by Q = Q M = P Q + an we know from Proof 1, an by efinition: P + Q.=. We know an from above, so again M = = Q 3
RCH 614 Note Set 5 S2012abn y choosing component irections such that = 0 to one of the components, we can simplify many problems. EQUILIRIUM is the state where the resultant of the forces on a particle or a rigi is zero. There will be no rotation or translation. The forces are referre to as balance. NEWTON S IRST LW: If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant spee in a straight line (if originally in motion). R H 0 Ry V ND M 0 x x y 0 Moment Couples Moment Couple: Two forces with equal magnitue, parallel lines of action an opposite sense ten to make our boy rotate even though the sum of forces is 0. The sum of the moment of the forces about any axis is not zero. 1 M 2 1 M 2 M ( 1 2) M : moment of the couple (CW) M oes not epen on where is. M epens on the perpenicular istance between the line of action of the parallel forces. M for a couple (efine by an ) is a constant. n the sense (+/-) is obtaine by observation. Just as there are equivalent moments (other values of an that result in M) there are equivalent couples. The magnitue is the same for ifferent values of an resulting or ifferent values of an resulting. 300 N 100 mm M 300 N 200 N M M 200 N 120 N 250 mm 120 N 150 mm 4
RCH 614 Note Set 5 S2012abn Equivalent orce Systems Two systems of forces are equivalent if we can transform one of them into the other with: 1.) replacing two forces on a point by their resultant 2.) resolving a force into two components 3.) canceling two equal an opposite forces on a point 4.) attaching two equal an opposite forces to a point 5.) moving a force along its line of action 6.) replacing a force an moment on a point with a force on another (specific) point 7.) replacing a force on point with a force an moment on another (specific) point * base on the parallelogram rule an the principle of transmissibility The size an irection are important for a moment. The location on a boy oesn t matter because couples with the same moment will have the same effect on the rigi boy. ition of Couples Couples can be ae as scalars. Two couples can be replace by a single couple with the magnitue of the algebraic sum of the two couples. Resolution of a orce into a orce an a Couple The equivalent action of a force on a boy can be reprouce by that force an a force couple: If we rather have acting at which isn t in the line of action, we can instea a an at with no change of action by. Now it becomes a couple of separate by an the force move to. The size is =M - The couple can be represente by a moment symbol: ny force can be replace by itself at another point an the moment equal to the force multiplie by the istance between the original line of action an new line of action. 5 M=
RCH 614 Note Set 5 S2012abn Resolution of a orce into a orce an a Moment Principle: ny force acting on a rigi boy (say the one at ) may be move to any given point, provie that a couple M is ae: the moment M of the couple must equal the moment of (in its original position at ) about. - M= IN REVERSE: force acting at an a couple M may be combine into a single resultant force acting at (a istance away) where the moment of about is equal to M. Resultant of Two Parallel orces Gravity loas act in one irection, so we may have parallel forces on our structural elements. We know how to fin the resultant force, but the location of the resultant must provie the equivalent total moment from each iniviual force. R C a b D C x D R M C a b R x x a b R Equilibrium for a Rigi oy REE ODY DIGRM STEPS; 1. Determine the free boy of interest. (What boy is in equilibrium?) 2. Detach the boy from the groun an all other boies ( free it). 3. Inicate all external forces an moments which inclue: - action on the free boy by the supports & connections - action on the free boy by other boies - the weigh effect (=force) of the free boy itself (force ue to gravity) 6
RCH 614 Note Set 5 S2012abn 4. ll forces an moments shoul be clearly marke with magnitues an irection. The sense of forces an moments shoul be those acting on the boy not by the boy. 5. Dimensions/angles shoul be inclue for moment computations an force component computations. 6. Inicate the unknown angles, istances, forces or moments, such as those reactions or constraining forces where the boy is supporte or connecte. Reactions can be categorize by the type of connections or supports. reaction is a force with known line of action, or a force of unknown irection, or a moment. The line of action of the force or irection of the moment is irectly relate to the motion that is prevente. The line of action shoul be inicate on the D. The sense of irection is etermine by the type of support. (Cables are in tension, etc ) If the sense isn t obvious, assume a sense. When the reaction value comes out positive, the assumption was correct. When the reaction value comes out negative, the assumption was opposite the actual sense. DON T CHNGE THE RROWS ON YOUR D OR SIGNS IN YOUR EQUTIONS. With the 3 equations of equilibrium, there can be no more than 3 unknowns. COUNT THE NUMER O UNKNOWN RECTIONS. 7
RCH 614 Note Set 5 S2012abn Reactions an Support Connections Structural nalysis, 4 th e., R.C. Hibbeler 8
RCH 614 Note Set 5 S2012abn Loas, Support Conitions & Reactions for eams Types of orces Concentrate single loa at one point Distribute loaing sprea over a istance or area P 1 P 2 concentrate uniformly istribute istribute Types of supports: Statically eterminate (number of unknowns number of equilibrium equations) L L L simply supporte (most common) overhang cantilever Statically ineterminate: (nee more equations from somewhere L restraine, ex. L continuous (most common case when L 1 =L 2 ) L Distribute Loas Distribute loas may be replace by concentrate loas acting through the balance/center of the istribution or loa area: THIS IS N EQUIVLENT ORCE SYSTEM. w is the symbol use to escribe the loa per unit length. W is the symbol use to escribe the total loa. w x W x W w 0 w x 2 x W 2 W/2 w w x W W/2 2w x/2 x/2 2x/3 x/3 9 x/2 x/6 x/3
RCH 614 Note Set 5 S2012abn Example 1 (pg 31) Verify that the beam reactions satisfy rotational equilibrium for the rigi boy. Check the summation of moments at points, & C. Example 2 (pg 32) 10
RCH 614 Note Set 5 S2012abn Example 3 (pg 34) 8. *Note: The figure has been change to show w rather than W. * Example 4 (horizontal) on the bar at the free en. 4' 11
RCH 614 Note Set 5 S2012abn Example 5 12