STATICS - Weebly

STATICSFORCEA force is a vector quantity. It is defined when its (1)magnitude, (2) point of application, and (3) direction areknown.RESULTANT (TWO DIMENSIONS)The resultant, F, of n forces with components Fx,i and Fy,ihas the magnitude of1 2F d ! Fx, i n d ! Fy, i n H2n2ni 1i 1The resultant direction with respect to the x-axis usingfourquadrant angle functions isi arctan e ! Fy, ini 1ni 1The vector form of a force isF Fx i Fy jcosx Fx /F; cosy Fy /F; cosz zFz /FSeparating a force into components when the geometry offorce is known and R x 2 y 2 z 2Fx (x/R)F;Fy (y/R)F;Fz (z/R)FMOMENTS (COUPLES)A system of two forces that are equal in magnitude, oppositein direction, and parallel to each other is called a couple. Amoment M is defined as the cross product of the radius vectorr and the force F from a point to the line of action of the force.M rF;Mx yFz – zFy,My zFx – xFz, andMz xFy – yFx.SYSTEMS OF FORCESF FnM (rn Fn)Equilibrium RequirementsFn 0Mn 0rc mn rn rc mnrn / mn, wherethe mass of each particle making up the system,the radius vector to each particle from a selectedreference point, andthe radius vector to the center of the total mass fromthe selected reference point.The moment of area (Ma) is defined asMay xnanMax ynanMaz znan! Fx, i oRESOLUTION OF A FORCEFx F cos x; Fy F cos y; Fz F cosCENTROIDS OF MASSES, AREAS, LENGTHS, ANDVOLUMESFormulas for centroids, moments of inertia, and first momentof areas are presented in the MATHEMATICS section forcontinuous functions. The following discrete formulas are fordefined regular masses, areas, lengths, and volumes:The centroid of area is defined asxac May /Ayac Max /Awith respect to center ofzac Maz /Athe coordinate systemwhere A anThe centroid of a line is defined asxlc ( xnln)/L, where L ylc ( ynln)/Lzlc ( znln)/LlnThe centroid of volume is defined asxvc ( xnvn)/V, where V vnyvc ( ynvn)/Vzvc ( znvn)/VMOMENT OF INERTIAThe moment of inertia, or the second moment of area, isdefined asIy x2 dAIx y2 dAThe polar moment of inertia J of an area about a point is equalto the sum of the moments of inertia of the area about any twoperpendicular axes in the area and passing through the samepoint.Iz J Iy Ix (x2 y2) dA rp2A, whererp the radius of gyration (see the DYNAMICS section andthe next page of this section).STATICS49

Moment of Inertia Parallel Axis TheoremThe moment of inertia of an area about any axis is defined asthe moment of inertia of the area about a parallel centroidalaxis plus a term equal to the area multiplied by the square ofthe perpendicular distance d from the centroidal axis to theaxis in question.I lx Ixc dx2 AI ly Iyc d 2y A, wheredx, dy distance between the two axes in question,Ixc , Iyc the moment of inertia about the centroidal axis, andI lx, I ly the moment of inertia about the new axis.Radius of GyrationThe radius of gyration rp, rx, ry is the distance from areference axis at which all of the area can be considered to beconcentrated to produce the moment of inertia.rx Ix A; ry Iy A; rp J AProduct of InertiaThe product of inertia (Ixy, etc.) is defined as:Ixy xydA, with respect to the xy-coordinate system,Ixz xzdA, with respect to the xz-coordinate system, andIyz yzdA, with respect to the yz-coordinate system.The parallel-axis theorem also applies:I'xy Ixc yc dxdy A for the xy-coordinate system, etc.wheredx x-axis distance between the two axes in question, anddy y-axis distance between the two axes in question.FRICTIONThe largest frictional force is called the limiting friction.Any further increase in applied forces will cause motion.FsN, whereF friction force,s coefficient of static friction, andN normal force between surfaces in contact.SCREW THREAD (also see MECHANICALENGINEERING section)For a screw-jack, square thread,M Pr tan (), where is for screw tightening,– is for screw loosening,M external moment applied to axis of screw,P load on jack applied along and on the line of the axis,r the mean thread radius, the pitch angle of the thread, and tan the appropriate coefficient of friction.50STATICSBELT FRICTIONF1 F2 e , whereF1 force being applied in the direction of impendingmotion,F2 force applied to resist impending motion, coefficient of static friction, and the total angle of contact between the surfaces expressedin radians.STATICALLY DETERMINATE TRUSSPlane TrussA plane truss is a rigid framework satisfying the followingconditions:1. The members of the truss lie in the same plane.2. The members are connected at their ends by frictionlesspins.3. All of the external loads lie in the plane of the truss andare applied at the joints only.4. The truss reactions and member forces can be determinedusing the equations of equilibrium.F 0; M 05. A truss is statically indeterminate if the reactions andmember forces cannot be solved with the equations ofequilibrium.Plane Truss: Method of JointsThe method consists of solving for the forces in the membersby writing the two equilibrium equations for each joint of thetruss.FV 0 and FH 0, whereFH horizontal forces and member components andFV vertical forces and member components.Plane Truss: Method of SectionsThe method consists of drawing a free-body diagram of aportion of the truss in such a way that the unknown trussmember force is exposed as an external force.CONCURRENT FORCESA concurrent-force system is one in which the lines of actionof the applied forces all meet at one point. Atwo-force body in static equilibrium has two applied forcesthat are equal in magnitude, opposite in direction, andcollinear. A three-force body in static equilibrium has threeapplied forces whose lines of action meet at a point. As aconsequence, if the direction and magnitude of two of thethree forces are known, the direction and magnitude of thethird can be determined.

STATICS51CbCxxxbxhxhhabhaxhCbCaCbbC b3h/4Iy b3h/12Iy3[ ()]2h 3 (3a b )12Ix 2[3]3() a 2b 2 sinθ cosθ 62I y ab sinθ (b a cosθ )Ix32I xc a 3b sin 3θ 12xc (b a cos θ)/2))] 12)] 36() [ab sinθ (b a cos θ )] 12 (a b sin θ ) 32h 3 a 2 4ab b 236(a b )I xc (J bh b 2 h 2 12[ (I x bh 3 3I y b3h 3I yc b 3 h 12I xc b h 3 12I y bh b ab a2I x bh 12I ycyc (a sin θ)/2[ (Iyc bh b ab a2I xc bh 3 36 bh3/12IxI yc b 3h/36Ix c bh 3 /36 bh3/12IxI yc b 3h/362Area Moment of InertiaIx c bh 3 /36A ab sin θA h(a b ) 2h(2a b )yc 3(a b )yc h/2xc b/2A bhyc h/3xc (a b)/3A bh/2yc h/3xc b/3A bh/2yc h/3xc 2b/3A bh/2Area & Centroid(2) 18()rx2ry2) (b a cosθ )2 3 (ab cosθ ) 62 (a sinθ ) 3()ry2c b 2 a 2 cos 2 θ 12rx2c (a sinθ )2 12(h 2 a 2 4ab b 218(a b )2(h3a b)rx2 6(a b )rx2c rp2 b 2 h 2 12(rx2 h 2 3ry2 b 2 3ry2c b 2 12) h 6 b 2 ab a 2 62 b ab a2rx2c h 2 12ry2crx2ry2rx2c h 2 18rx2 h 2 6ry2 b 2 6ry2c b 2 18rx2c h 2 18rx2 h 2 6ry2 b 2 2ry2c b 2 18rx2c h 2 18(Radius of Gyration)2Housner, George W. & Donald E. Hudson, Applied Mechanics Dynamics, D. Van Nostrand Company, Inc., Princeton, NJ, 1959. Table reprinted by permission of G.W. Housner & D.E. Hudson.yyyyyyFigure[] bh 2 (2a b ) 24() Abh 4 b 2 h 2 4I xc yc a 3b sin 2 θ cos θ 12I xy] [Ah(2a b )] 12I xc y c 0I xy[ bh 2 (2a b ) 72I xc yc [Ah(2a b )] 36I xy Abh 12 b 2 h 2 24 Abh 4 b 2 h 2 8I xc yc Abh 36 b 2 h 2 72IxyIxc yc Abh 36 b 2 h 2 72Product of Inertia

52STATICSyyaaabaC2aCCCCxxxxx2[]sin 3θ2a3 θ sinθ cosθyc 0xc A a2 θ sin 2θ2A a 2θ2a sinθxc 3 θyc 0yc 4a/(3π)xc aA πa /2yc axc a4))Iy Ix ]4Aa 22sin 3 θ cosθ1 4θ sin θ cosθ2sin 3θ cosθAa 21 3θ 3sinθ cosθ4[[Iy a4(θ sinθ cosθ)/4Ix a (θ – sinθ cosθ)/44I x πa 4 8I y 5π a 4 8I ycI xc()a 4 9π 2 64 72π πa 4 8(5πaπbIx Iy πa 2b 2 44J π a4 b4 2I xc I y c π a 4 b 4 4(J πa 4 2yc aA π (a2 – b2)I x I y 5π a 4 4I xc I y c π a 4 4Area Moment of Inertiaxc aA πa2Area & Centroid]) b ) 4 (ry2)2)2ry2 rx2 a22sin 3θ cosθ1 4θ sinθ cosθ[[]2sin 3θ cosθa21 3θ 3sinθ cosθ4a 2 (θ sinθ cosθ )4θa 2 (θ sinθ cosθ )2ry 4θrx2 ry2 5a 2 4rx2 a 2 4(a 2 9π 2 6436π 222ryc a 4rx2c rp2 a 2 b 2 2rx2( (5a a 22rx2c ry2c a 2 b 2 4rp2rx2 ry2 5a 2 4rx2c ry2c a 2 4](Radius of Gyration)2Housner, George W. & Donald E. Hudson, Applied Mechanics Dynamics, D. Van Nostrand Company, Inc., Princeton, NJ, 1959. Table reprinted by permission of G.W. Housner & D.E. Hudson.yyyFigure(I xc y c 0I xy 0I xc y c 0I xy 0I xy 2a 4 3I xc y c 0 πa 2 a 2 b 2I xy Aa 2I xc y c 0I xy Aa 2I xc y c 0)Product of Inertia

STATICS53bCxhy (h/b 1/n)x1/nbxbbbCy (h/b n)xnaCaCxhxn 1bn 2n 1h2(n 2)n 1b2n 1nbhn 1yc xc A h n 1yc 2 2n 1xc A bh (n 1)yc 3b/8xc 3a/5A 2ab/3yc 0xc 3a/5A 4ab/3Area & Centroidhb 3n 3nIx bh 33(n 3)nIy b3h3n 1Iy bh 3Ix 3(3n 1)Iy 2ba /73Ix 2ab3/15I y 4 a 3b 7I yc 16a b 1753I xc I x 4ab 3 15Area Moment of Inertia 12a 1752n 1 2h3(n 1)n 1 2b3n 1ry2 n 1 2bn 3h 2 (n 1)3(3n 1)rx2 ry2 rx2 ry2 3a 2 7rx2 b 2 5ry2 3a 2 7ry2crx2c rx2 b 2 5(Radius of Gyration)2Housner, George W. & Donald E. Hudson, Applied Mechanics Dynamics, D. Van Nostrand Company, Inc., Princeton, NJ, 1959. Table reprinted by permission of G.W. Housner & D.E. Hudson.yyyyFigureIxy Aab/4 a2b2Ixy 0I xc y c 0Product of Inertia

MOMENT OF INERTIA The moment of inertia, or the second moment of area, is defined as I y x2 dA I x y2 dA The polar moment of inertia J of an area about a point is equal to the sum of the moments of inertia of the area about any two point. I z J I y I x (x 2 y2) dA