Mechanics For Engineers: Statics
Mechanics for Engineers: StaticsApplicationThe tension in the cable supportingthis person can be found using theconcepts in this chapter 2013 The McGraw-Hill Companies, Inc. All rights reserved.2-1
Scalar triple product A B (Axi Ayi Azi ) (Bxi Byi Bzi) A B (AyBz - AzBy)i (AzBx - AxBz)j (AxBy - AyBx)k𝑖𝑗𝑘 𝐴 𝐵 𝐴𝑥 𝐴𝑦 𝐴𝑧𝐵𝑥 𝐵𝑦 𝐵𝑧𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐴 𝐵. 𝐶 𝐵𝑥 𝐵𝑦 𝐵𝑧𝐶𝑥 𝐶𝑦 𝐶𝑧
Mechanics for Engineers: StaticsScalar Product of Two Vectors The scalar product or dot product betweentwo vectors P and Q is defined as P Q PQ cos scalar result Scalar products:- are commutative,- are distributive,- are not associative, P Q Q P P Q1 Q2 P Q1 P Q2 P Q S undefined Scalar products with Cartesian unit components, P Q Px i Py j Pz k Qx i Q y j Qz k i i 1 j j 1 k k 1 i j 0 j k 0 k i 0 P Q Px Qx Py Q y Pz Qz P P Px2 Py2 Pz2 P 2 2013 The McGraw-Hill Companies, Inc. All rights reserved.3-3
Mechanics for Engineers: StaticsScalar Product of Two Vectors: Applications Angle between two vectors: P Q PQ cos Px Qx Py Q y Pz Qzcos Px Qx Py Q y Pz QzPQ Projection of a vector on a given axis:POL P cos projection of P along OL P Q PQ cos P Q P cos POLQ For an axis defined by a unit vector: POL P Px cos x Py cos y Pz cos z 2013 The McGraw-Hill Companies, Inc. All rights reserved.3-4
Mechanics for Engineers: StaticsMixed Triple Product of Three Vectors Mixed triple product of three vectors, S P Q scalar result The six mixed triple products formed from S, P, andQ have equal magnitudes but not the same sign, S P Q P Q S Q S P S Q P P S Q Q P S Evaluating the mixed triple product, S P Q S x Py Qz Pz Q y S y Pz Qx Px Qz S z Px Q y Py Qx 2013 The McGraw-Hill Companies, Inc. All rights reserved.SxSySz PxPyPzQxQyQz 3-5
Mechanics for Engineers: StaticsResultant of Two Forces force: action of one body on another;characterized by its point of application,magnitude, line of action, and sense. Experimental evidence shows that thecombined effect of two forces may berepresented by a single resultant force. The resultant is equivalent to the diagonal ofa parallelogram which contains the twoforces in adjacent legs. Force is a vector quantity. 2013 The McGraw-Hill Companies, Inc. All rights reserved.2-6
Mechanics for Engineers: StaticsAddition of Vectors Trapezoid rule for vector addition Triangle rule for vector addition Law of cosines,CBCBR 2 P 2 Q 2 2 PQ cos B R P Q Law of sines,sin A sin B sin C PRQ Vector addition is commutative, P Q Q P Vector subtraction 2013 The McGraw-Hill Companies, Inc. All rights reserved.2-7
Mechanics for Engineers: StaticsResultant of Several Concurrent Forces Concurrent forces: set of forces which allpass through the same point.A set of concurrent forces applied to aparticle may be replaced by a singleresultant force which is the vector sum of theapplied forces. Vector force components: two or more forcevectors which, together, have the same effectas a single force vector. 2013 The McGraw-Hill Companies, Inc. All rights reserved.2-8
Mechanics for Engineers: StaticsSample ProblemSOLUTION:A barge is pulled by twotugboats. If the resultant ofthe forces exerted by thetugboats is 5000 lbf directedalong the axis of the barge,determine the tension in eachof the ropes for a 45o.Discuss with a neighbor howyou would solve this problem. 2013 The McGraw-Hill Companies, Inc. All rights reserved. Find a graphical solution by applyingthe Parallelogram Rule for vectoraddition. The parallelogram has sidesin the directions of the two ropes and adiagonal in the direction of the bargeaxis and length proportional to 5000 lbf. Find a trigonometric solution byapplying the Triangle Rule for vectoraddition. With the magnitude anddirection of the resultant known andthe directions of the other two sidesparallel to the ropes given, apply theLaw of Sines to find the rope tensions.2-9
Mechanics for Engineers: StaticsSample Problem Graphical solution - Parallelogram Rulewith known resultant direction andmagnitude, known directions for sides.T1 3700 lbf T2 2600 lbf Trigonometric solution - Triangle Rulewith Law of SinesT1T25000 lbf sin 45 sin 30 sin105 T1 3660 lbf T2 2590 lbf 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 10
Mechanics for Engineers: StaticsWhat if ? At what value of a would the tension in rope2 be a minimum?Hint: Use the triangle rule and think abouthow changing a changes the magnitude of T2.After considering this, discuss your ideas witha neighbor. The minimum tension in rope 2 occurs whenT1 and T2 are perpendicular. 2013 The McGraw-Hill Companies, Inc. All rights reserved.T2 5000 lbf sin 30 T2 2500 lbfT1 5000 lbf cos 30 T1 4330 lbfa 90 30 a 60 2 - 11
Mechanics for Engineers: StaticsRectangular Components of a Force: Unit Vectors It’s possible to resolve a force vector into perpendicularcomponents so that the resulting parallelogram is arectangle. Fx and Fy are referred to as rectangularvector components and F Fx Fy Define perpendicular unit vectors i and j which areparallel to the x and y axes. Vector components may be expressed as products ofthe unit vectors with the scalar magnitudes of thevector components. F Fx i Fy j Fx and Fy are referred to as the scalar components of F 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 12
Mechanics for Engineers: StaticsAddition of Forces by Summing Components To find the resultant of 3 (or more) concurrentforces, R P Q S Resolve each force into rectangular components,then add the components in each direction: Rx i R y j Px i Py j Qx i Q y j S x i S y j Px Qx S x i Py Q y S y j The scalar components of the resultant vector areequal to the sum of the corresponding scalarcomponents of the given forces.R y Py Q y S yRx Px Qx S x Fx Fy To find the resultant magnitude and direction,22 1 R yR Rx R y tanRx 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 13
Mechanics for Engineers: StaticsSample ProblemSOLUTION: Resolve each force into rectangularcomponents. Determine the components of theresultant by adding the correspondingforce components in the x and ydirections.Four forces act on bolt A as shown.Determine the resultant of the forceon the bolt. 2013 The McGraw-Hill Companies, Inc. All rights reserved. Calculate the magnitude and directionof the resultant.2 - 14
Mechanics for Engineers: StaticsSample ProblemSOLUTION: Resolve each force into rectangular components.force magrF1 150rF280rF3 110rF4 100x compy comp 129.9 75.0 27.40 75.2 110.0 96.6 25.9Rx 199.1 R y 14.3 Determine the components of the resultant byadding the corresponding force components. Calculate the magnitude and direction.R 199.12 14.3214.3 Ntan a 199.1 N 2013 The McGraw-Hill Companies, Inc. All rights reserved.R 199.6Na 4.1 2 - 15
MOMENT 𝑀 𝐹𝑑 𝑴 𝒓 𝑭 Mo r R r P r Q(a)
MomentofaForceAboutaPoint A force vector is defined by its magnitude anddirection. Its effect on the rigid body also dependson it point of application. The moment of F about O is defined asMO r F The moment vector MO is perpendicular to theplane containing O and the force F. Magnitude of MO measures the tendency of the forceto cause rotation of the body about an axis along MO.M O rF sin FdThe sense of the moment may be determined by theright-hand rule. Any force F’ that has the same magnitude anddirection as F, is equivalent if it also has the same lineof action and therefore, produces the same moment.3 - 17
Moment of a Force About a Point Two-dimensional structures have length and breadth butnegligible depth and are subjected to forces contained inthe plane of the structure. The plane of the structure contains the point O and theforce F. MO, the moment of the force about O isperpendicular to the plane. If the force tends to rotate the structure counterclockwise,the sense of the moment vector is out of the plane of thestructure and the magnitude of the moment is positive. If the force tends to rotate the structure clockwise, thesense of the moment vector is into the plane of thestructure and the magnitude of the moment is negative.
Varignon’s Theorem The moment about a given point O of theresultant of several concurrent forces is equalto the sum of the moments of the variousmoments about the same point O. r F1 F2 r F1 r F2 Varigon’s Theorem makes it possible toreplace the direct determination of themoment of a force F by the moments of twoor more component forces of F.3 - 19
Mechanics for Engineers: StaticsExpressing a Vector in 3-D SpaceIf angles with some of the axes are given: The vector is Fcontainedin the plane OBAC. Resolve F into horizontaland vertical components.Fy F cos yFh F sin y 2013 The McGraw-Hill Companies, Inc. All rights reserved. Resolve Fhinto rectangularcomponentsFx Fh cos F sin y cos F y Fh sin F sin y sin 2 - 20
Mechanics for Engineers: StaticsExpressing a Vector in 3-D SpaceIf the direction cosines are given: 2013 The McGraw-Hill Companies, Inc. All rights reserved. With the angles between F and the axes,Fx F cos x Fy F cos y Fz F cos z F Fx i Fy j Fz k F cos x i cos y j cos z k F cos x i cos y j cos z k is a unit vector along the line of action of Fand cos x , cos y , and cos z are the directioncosines for F2 - 21
Mechanics for Engineers: StaticsExpressing a Vector in 3-D SpaceIf two points on the line of action are given:Direction of the force is defined bythe location of two points,M x1 , y1 , z1 and N x2 , y2 , z 2 rd vector joiningM and Nrrr d xi dy j dz kd x x 2 x1 d y y 2 y1 d z z 2 z1rrF F r 1rrr d x i d y j d z kdFdyFdFdxFx Fy Fz zddd 2013 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 22
Mechanics for Engineers: StaticsSample ProblemSOLUTION: Based on the relative locations of thepoints A and B, determine the unitvector pointing from A towards B. Apply the unit vector to determine thecomponents of the force acting on A.The tension in the guy wire is 2500 N.Determine: Noting that the components of the unitvector are the direction cosines for thevector, calculate the correspondingangles.a) components Fx, Fy, Fz of the force acting onthe bolt at A,b) the angles x, y, z defining the direction ofthe force (the direction cosines) 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 23
Mechanics for Engineers: StaticsSample ProblemSOLUTION: Determine the unit vector pointing from Atowards B.rrrAB 40m i 80m j 30m kAB 40m 80m 30m 222 94.3 m 2013 The McGraw-Hill Companies, Inc. All rights reserved.r 40 r 80 r 30 r i j k 94.3 94.3 94.3 rrr 0.424 i 0.848 j 0.318k Determine the components of the force.rrF F rrr 2500 N 0.424 i 0.848 j 0.318krrr 1060N i 2120 N j 795 N k 2 - 24
Mechanics for Engineers: StaticsSample Problem Noting that the components of the unit vector arethe direction cosines for the vector, calculate thecorresponding angles. cos x i cos y j cos z k 0.424 i 0.848 j 0.318k x 115 .1 y 32.0 z 71.5 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 25
Mechanics for Engineers: StaticsWhat if ?SOLUTION: Since the force in the guy wire must bethe same throughout its length, the forceat B (and acting toward A) must be thesame magnitude but opposite indirection to the force at A.FBAFABrrFBA FABrrr 1060N i 2120 N j 795 N k 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 26
Mechanics for Engineers: StaticsRectangular Components of the Moment of a ForceThe moment of F about O, M O r F , r xi yj zk F Fx i Fy j Fz k M O M xi M y j M z k i x jy kzFxFyFz yFz zFy i zFx xFz j xFy yFx k 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 27
Mechanics for Engineers: StaticsRectangular Components of the Moment of a ForceThe moment of F about B, M B rA / B F rA / B rA rB x A xB i y A y B j z A z B k F Fx i Fy j Fz k i M B x A xB Fx 2013 The McGraw-Hill Companies, Inc. All rights reserved. j y A yB k z A z B FyFz3 - 28
Mechanics for Engineers: StaticsRectangular Components of the Moment of a ForceFor two-dimensional structures, M O xFy yFz kMO M Z xFy yFz M O x A x B Fy y A y B Fz kMO M Z x A x B Fy y A y B Fz 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 29
Mechanics for Engineers: StaticsSample ProblemSOLUTION:The moment MA of the force F exertedby the wire is obtained by evaluatingthe vector product, M A rC A FThe rectangular plate is supported by the bracketsat A and B and by a wire CD. Knowing that thetension in the wire is 200 N, determine the momentaboutA of the forceexertedthewire at C. 2013The McGraw-HillCompanies,Inc. Allbyrightsreserved.3 - 30
Mechanics for Engineers: StaticsSample ProblemSOLUTION: M A rC A F rC A rC rA 0.3 m i 0.08 m j F F 200 N rC DrC D 0.3 m i 0.24 m j 0.32 m k 200 N 0.5 m 120N i 96 N j 128 N k i M A 0.3j0k0.08 120 96 128 M 7.68 N m i 28.8 N m j 28.8 N m k 2013 The McGraw-Hill Companies, Inc. All rights reserved. A3 - 31
Mechanics for Engineers: StaticsMoment of a Force About a Given Axis Moment MO of a force F applied at the point A about apoint O, MO r F Scalar moment MOL about an axis OL is the projection of themoment vector MO onto the axis,𝑀𝑂𝐿 𝜆 𝑀𝑂 𝜆 𝑟 𝐹 Moments of F about the coordinate axes,M x yFz zFyM y zFx xFzM xF yFzy All rights reserved.x 2013 The McGraw-Hill Companies,Inc.3 - 32
Mechanics for Engineers: StaticsMoment of a Force About a Given Axis Moment of a force about an arbitrary axis, M BL M B rA B F rA B rA rB The result is independent of the point B along the given axis. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 33
Mechanics for Engineers: StaticsSample ProblemA cube is acted on by a force P as shown.Determine the moment of Pa)b)c)d) 2013 The McGraw-Hill Companies, Inc. All rights reserved.about Aabout the edge AB andabout the diagonal AG of the cube.Determine the perpendicular distance betweenAG and FC.3 - 34
Mechanics for Engineers: StaticsSample Problem Moment of P about A, M A rF A P rF A ai a j a i j P P 2 i 2 j P 2 i j M A a i j P 2 i j M A aP 2 i j k Moment of P about AB,M AB i MA i aP 2 i j k M AB aP 2 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 35
Mechanics for Engineers: StaticsSample Problem Moment of P about the diagonal AG, M AG M A rA G ai aj ak1 i j k rA Ga 33 aP MA i j k 21 aP M AG i j k i j k 32aP 1 1 1 6M AG 2013 The McGraw-Hill Companies, Inc. All rights reserved.aP 63 - 36
Mechanics for Engineers: StaticsSample Problem 3.5 Perpendicular distance between AG and FC, P 1 P 0 1 1 P j k i j k 236 0Therefore, P is perpendicular to AG.M AGaP Pd6ad 6 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 37
COUPLE M F(a d) – Fa or M Fd M rA F rB (-F) (rA – rB) x F M r F
Mechanics for Engineers: StaticsMoment of a Couple Two forces F and -F having the same magnitude, parallel linesof action, and opposite sense are said to form a couple. Moment of the couple, M rA F rB F rA rB F r FM rF sin Fd The moment vector of the couple is independent ofthe choice of the origin of the coordinate axes, i.e.,it is a free vector that can be applied at any pointwith the same effect. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 39
Mechanics for Engineers: StaticsMoment of a CoupleTwo couples will have equal moments if F1d1 F2 d 2 the two couples lie in parallel planes, and the two couples have the same senseor the tendency to cause rotation inthe same direction. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 40
Mechanics for Engineers: StaticsEquivalent Couples 2013 The McGraw-Hill Companies, Inc. All rights reserved.2 - 41
Mechanics for Engineers: StaticsAddition of Couples Consider two intersecting planes P1 and P2with each containing a couple M 1 r F1 in plane P1 M 2 r F2 in plane P2 Resultants of the vectors also form a couple M r R r F1 F2 By Varigon’s theorem M r F1 r F2 M1 M 2 Sum of two couples is also a couple that is equal to thevector sum of the two couples 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 42
Mechanics for Engineers: StaticsCouples Can Be Represented by Vectors A couple can be represented by a vector with magnitude and direction equal to themoment of the couple. Couple vectors obey the law of addition of vectors. Couple vectors are free vectors, i.e., the point of application is not significant. Couple vectors may be resolved into component vectors. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 43
Mechanics for Engineers: StaticsResolution of a Force Into a Force at O and a Couple Force vector F can not be simply moved to O without modifying its action on the body. Attaching equal and opposite force vectors at O produces no net effect on the body. The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couplesystem. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 44
Mechanics for Engineers: StaticsResolution of a Force Into a Force at O and a Couple Moving F from A to a different point O’ requires the addition of adifferent couple vector MO’ M O' r F The moments of F about O and O’ are related, M O ' r ' F r s F r F s F MO s F Moving the force-couple system from O to O’ requires theaddition of the moment of the force at O about O’. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 45
Mechanics for Engineers: StaticsSample Problem 3.6SOLUTION: Attach equal and opposite 20 lb forces inthe x direction at A, thereby producing 3couples for which the moment componentsare easily computed. Alternatively, compute the sum of themoments of the four forces about anarbitrary single point. The point D is agood choice as only two of the forces willproduce non-zero moment contributions.Determine the components of the singlecoupleequivalentto theshown. 2013The McGraw-HillCompanies,Inc. couplesAll rights reserved.3 - 46
Mechanics for Engineers: StaticsSample Problem 3.6 Attach equal and opposite 20 lb forces in the xdirection at A The three couples may be represented by three couplevectors,M x 30 lb 18 in. 540 lb in.M y 20 lb 12 in. 240lb in.M z 20 lb 9 in. 180 lb in. M 540 lb in. i 240lb in. j 180 lb in. k 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 47
Mechanics for Engineers: StaticsSample Problem 3.6 Alternatively, compute the sum of themoments of the four forces about D. Only the forces at C and E contribute tothe moment about D. M M D 18 in. j 30 lb k 9 in. j 12 in. k 20 lb i M 540 lb in. i 240lb in. j 180 lb in. k 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 48
Mechanics for Engineers: StaticsSystem of Forces: Reduction to a Force and Couple A system of forces may be replaced by a collection offorce-couple systems acting a given point O The force and couple vectors may be combined into aresultant force vector and a resultant couple vector, R R FM O r F The force-couple system at O may be moved to O’with the addition of the moment of R about O’ , R R M O' M O s R Two systems of forces are equivalent if they can bereduced to the same force-couple system. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 49
Mechanics for Engineers: StaticsFurther Reduction of a System of Forces If the resultant force and couple at O are mutuallyperpendicular, they can be replaced by a single force actingalong a new line of action. The resultant force-couple system for a system of forceswill be mutually perpendicular if:1) the forces are concurrent,2) the forces are coplanar, or3) the forces are parallel. 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 50
Mechanics for Engineers: StaticsFurther Reduction of a System of Forces System of coplanar forcesis reduced to a force-couple system R and M OR that ismutually perpendicular. System can be reduced to a single forceby moving the line of action of R untilits moment about O becomes M OR In terms of rectangular coordinates,xRy yRx M OR 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 51
Mechanics for Engineers: StaticsSample ProblemSOLUTION:a) Compute the resultant force for theforces shown and the resultantcouple for the moments of theforces about A.For the beam, reduce the system offorces shown to (a) an equivalentforce-couple system at A, (b) anequivalent force couple system at B,and (c) a single force or resultant.Note: Since the support reactions arenot included, the given system willnot maintain the beam in equilibrium. 2013 The McGraw-Hill Companies, Inc. All rights reserved.b) Find an equivalent force-couplesystem at B based on the forcecouple system at A.c) Determine the point of applicationfor the resultant force such that itsmoment about A is equal to theresultant couple at A.3 - 52
Mechanics for Engineers: StaticsSample ProblemSOLUTION:a) Compute the resultant force and theresultant couple at A. R F 150 N j 600 N j 100 N j 250 N j R 600 N j R M A r F 1.6 i 600 j 2.8 i 100 j 4.8 i 250 j RM A 1880 N m k 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 53
Mechanics for Engineers: StaticsSample Problemb) Find an equivalent force-couple system at Bbased on the force-couple system at A.The force is unchanged by the movement of theforce-couple system from A to B. R 600 N jThe couple at B is equal to the moment about Bof the force-couple system found at A. R R M B M A rB A R 1880 N m k 4.8 m i 600 N j 1880 N m k 2880 N m k RM B 1000 N m k 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 54
Mechanics for Engineers: StaticsSample Problem 3.10SOLUTION: Determine the relative position vectorsfor the points of application of thecable forces with respect to A. Resolve the forces into rectangularcomponents. Compute the equivalent force, R FThree cables are attached to thebracket as shown. Replace theforces with an equivalent forcecouple system at A. 2013 The McGraw-Hill Companies, Inc. All rights reserved. Compute the equivalent couple, R M A r F 3 - 55
Mechanics for Engineers: StaticsSample Problem Resolve the forces into rectangularcomponents. FB 700 N rE B 75 i 150 j 50k rE B175 0.429 i 0.857 j 0.289k FB 300 i 600 j 200k N SOLUTION: Determine the relative positionvectors with respect to A. rB A 0.075 i 0.050k m rC A 0.075 i 0.050k m rD A 0.100 i 0.100 j m 2013 The McGraw-Hill Companies, Inc. All rights reserved. FC 1000 N cos 45 i cos 45 j 707 i 707 j N FD 1200 N cos 60 i cos 30 j 600 i 1039 j N 3 - 56
Mechanics for Engineers: StaticsSample Problem Compute the equivalent force, R F 300 707 600 i 600 1039 j 200 707 k R 1607i 439 j 507k N Compute the equivalent couple, R M A r F ijk rB A F B 0.07500.050 30i 45k300 600 200 ijk rC A F c 0.075 0 0.050 17.68 j707 0 707 ijk rD A F D 0.100 0.100 0 163.9k60010390 R M A 30 i 17.68 j 118 .9k 2013 The McGraw-Hill Companies, Inc. All rights reserved.3 - 57
Mechanics for Engineers: Statics Rectangular Components of a Force: Unit Vectors 2 - 12 Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. & F x and F y are referred to as the scalar components of F F x i F y j & F & It’s possible to resolve a force vector into .File Size: 2MBPage Count: 57
Engineering Mechanics Statics 8th Edition meriam Solutions Manual Author: meriam Subject: Engineering Mechanics Statics 8th Edition meriam Solutions ManualInstant Download Keywords: Engineering Mechanics Statics
Bruksanvisning för bilstereo . Bruksanvisning for bilstereo . Instrukcja obsługi samochodowego odtwarzacza stereo . Operating Instructions for Car Stereo . 610-104 . SV . Bruksanvisning i original
Statics (or vector mechanics) is the branch of mechanics that is concerned with the analysis of loads (or forces and moments) on physical systems in static equilibrium. Systems that are in static equilibrium are either at rest or the system's center of mass moves at a constant velocity. Problems involving statics use trigonometry to find a .
Shear-Force and Bending Moment Diagrams dx dM V dx dV w 9 MEM202 Engineering Mechanics - Statics MEM Shear-Force and Bending Moment Diagrams One Concentrated Load Several Concentrated Loads. 10 MEM202 Engineering Mechanics - Statics MEM Shear-Force and Bending Moment Diagrams Uniform load over the entire span Uniform load over part of File Size: 1001KBPage Count: 14
philosophiae naturalis principia mathematica. isaac newton. [1687] syllabus 7 e14 - applied mechanics: statics syllabus 8 e14 - applied mechanics: statics.
10 tips och tricks för att lyckas med ert sap-projekt 20 SAPSANYTT 2/2015 De flesta projektledare känner säkert till Cobb’s paradox. Martin Cobb verkade som CIO för sekretariatet för Treasury Board of Canada 1995 då han ställde frågan
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Hotell För hotell anges de tre klasserna A/B, C och D. Det betyder att den "normala" standarden C är acceptabel men att motiven för en högre standard är starka. Ljudklass C motsvarar de tidigare normkraven för hotell, ljudklass A/B motsvarar kraven för moderna hotell med hög standard och ljudklass D kan användas vid
