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Statics FE Review 032712.Pptx FE Stacs Review h0p://www.coe.utah.edu/current-undergrad/fee.php Scroll down to: Stacs Review - Slides Torch Ellio [email protected] (801) 587-9016 MCE room 2016 (through 2000B door) Posi'on and Unit Vectors Y If you wanted to express a 100 lb force that was in the direction from A to B in (-2,6,3). A vector form, then you would find eAB and multiply it by 100 lb. rAB O x rAB = 7i - 7j – k (units of length) B .(5,-1,2) e = (7i-7j–k)/sqrt[99] (unitless) AB z Then: F = F eAB F = 100 lb {(7i-7j–k)/sqrt[99]} Another Way to Define a Unit Vectors Y Direction Cosines The θ values are the angles from the F coordinate axes to the vector F. The θY cosines of these θ values are the O θx x θz coefficients of the unit vector in the direction of F. z If cos(θx) = 0.500, cos(θy) = 0.643, and cos(θz) = - 0.580, then: F = F (0.500i + 0.643j – 0.580k) TrigonometrY hYpotenuse opposite right triangle θ adjacent • Sin θ = opposite/hYpotenuse • Cos θ = adjacent/hYpotenuse • Tan θ = opposite/adjacent TrigonometrY Con'nued A c B anY triangle b C a Σangles = a + b + c = 180o Law of sines: (sin a)/A = (sin b)/B = (sin c)/C Law of cosines: C2 = A2 + B2 – 2AB(cos c) Products of Vectors – Dot Product U . V = UVcos(θ) for 0 <= θ <= 180ο ( This is a scalar.) U In Cartesian coordinates: θ . U V = UxVx + UYVY + UzVz V In Cartesian coordinates to find θ? UVcos(θ) = UxVx + UYVY + UzVz -1 θ = cos [(UxVx + UYVY + UzVz)/UV] Products of Vectors – Dot Product UVcos(θ) = UxVx + UYVY + UzVz -1 θ = cos [(UxVx + UYVY + UzVz)/UV] However, usuallY we use vectors so that we do not have to deal with the angle between the vectors. Products of Vectors – Dot Product U . V = UVcos(θ) for 0 <= θ <= 180ο How could You find the U θ Projec'on of vector U, in 90o The direc'on of vector V? V Ucos(θ) eV = V/V à eV = 1 . U eV = (U)(1)cos(θ) = Ucos(θ)= the answer The projec'on of U in the direc'on of V is U do0ed with the unit vector in the direc'on of V. Products of Vectors – Cross Product Also called the vector product V V U x V = UVsin(θ)e e θ for 0 <= θ <= 180o U U Note: U x V = - V x U + If θ = 0 U x V = 0 i j If θ = 90o U x V = UVe In Cartesian coordinates: k i x i = 0 i x j = k j x i = - k etc. Products of Vectors – Cross Product If U = 2i + 3j – k and V = -i + 2j –k, then find U x V. i j k i j U x V = 2 3 -1 2 3 -1 2 -1 -1 2 Mul'plY and use opposite same sign. 3k 2i 2j -3i j 4k Then add all six values together, but note that i can add to i, j to j, and k to k, but i cannot add to j, etc. U x V = - i + 3j + 7k units Products of Vectors Mixed Triple Product U x V . W W V U x V = UVsin(θ)e U Projecon of UVsinθ UVsin(θ)e φW V on the θ direcon U of W Products of Vectors Mixed Triple Product Note: No i j k Otherwise, same procedure Ux UY Uz . U x V W = Vx VY Vz Wx WY Wz Pulley Problem (frictionless pins) T T ΣFy = 0 = 2T – T1 – mg T1 = 2T – mg m T1 T1 T ΣFy = 0 = 2T1 – mg mg m – mAg = 2(2T – mg) T1 – mg – mAg mg m mAg A T = (3m + mA)g/4 Moment of a Force 2-dimensional, in the plane of the screen y . z x O. F The moment of F D 90o about point O is the perpendicular distance from point O to the line of ac'on of F 'mes the magnitude |F| of F, where posi've is counter clockwise (CCW). Moment of a Force z 3-dimensional y e x O. r θ F Then: D 90o r x F = rFsin(θ)e = rsin(θ)Fe = (D)(F)e = M Note that since M = DF à D = M/F The magnitude of the force F is 100N. Determine the moment of F about point O and about the x-axis. y MO = rOB x F B C rOB 360mm O x F 500mm D A 600mm z . You should determine: MO = (-20.9i + 25.1k) N m ?? j term?? . Then find Mo x axis Mo x axis = -20.9 N m ?? vector form?? Couples B F causes translation and, in r0B F general, rotation. -F rBA Let –F be: 0 • Equal in magnitude to F r A 0A • Opposite direction of F A arbitrary • Not collinear with F. point Then F and –F form a plane, and cause rotation, but no translation. _____________________________________________________________________________________________________________________ Pick an arbitrary point in 3-D space and draw position vectors from that point to a point on the line of action of F (point A), and a point on the line of action of -F (point B). Draw position vector rBA. Then: r0B + rBA = r0A Or: rBA = r0A - r0B _____________________________________________________________________________________________________________________ M0 = (r0A x F) + (r0B x (-F)) = (r0A - r0B) x F M0 = rBA x F The forces F and -F form a couple. The moment of a couple is the same about every point in space. Therefore, the moment of a couple is a free vector. Also, two couples that have the same moment are equivalent. Equivalent SYstems 1 2 A . F F F A . -F 3 4 F F Mc F A . A . -F equivalent to 1 Review For stac equilibrium: ΣF = 0 and ΣM = 0 In 3-D You have 6 independent equaons: 3 force and 3 moment in 2-D You have 3 independent equaons: 2 force and 1 moment or 1 force and 2 moment (omen) or 0 force and 3 moment (omen) Reactions (2-D) Rough surface or F F pinned support A B Ax B Ay Smooth surface or roller support F F Fixed, built-in, or MA Ax cantilever support Ay Rope or cable (tension) Spring (tension or compression) 2-D Beam Solution 10 N 0 80 4 m 2 m 10 N 0 MA Ax 80 4 m Ay ΣFx = 0 = Ax – 10cos(80) Ax = 1.74 N ΣFy = 0 = Ay – 10sin(80) Ay = 9.85 N ΣΜA = 0 = MA – 10sin(80)(4) . MA = 39.4 N m 3-D Beam Solution – Given T find MA y Is this bar properly constrained and statically determinate? B Rigidly held x Yes and yes A T z Options: y Sum forces in a direction May B Sum moments about a point Ay x Sum moments about an axis Az A T Ax MAx Maz Which option is best here? ΣMA = 0 = MAxi + MAyj + MAzk + (rAB X T) Then group the i, j, and k components and solve. 3-D Beam Solution – Given W, find T y Is this bar properly constrained and statically determinate? B x Yes (?) and no rope with tension T A W Options: Ball and socket supports Sum forces in a direction Axis Sum moments about a point Sum moments about an axis Which option is best here? Sum moments about the axis. Special Case àTwo-force Bodies If a bodY in equilibrium has forces at two and FB onlY two locaons and B no moments, then the A forces are equal in magnitude, opposite in FA direc'on, and along the line of ac'on between the two points. Special Case àThree-force Bodies If a bodY in equilibrium has F forces at three and onlY C C three locaons and no moments, then the forces B FB are either concurrent or A parallel. UsuallY, You can use the trigonometric or FA graphical techniques of Chapter 2. Trusses The truss model requires that all members be two-force members. This further requires the assumptions that: • the connections between the members are frictionless pins. • all forces and reactions are applied at connection points. As a result, the forces in prismatic members are along the members. Work truss problems efficiently. First look at the physics of the problem to see: • if you can solve for the forces in any members by inspection. • if you need to find the reactions. • if there is symmetry in loading and geometry that can be used. If the problem is not solved directly from the physics, then, • use the method of joints to solve for the unknowns if they are near a known force that can be used in the solution. • use the method of sections to solve for the unknowns if they are not near a known force that can be used in the solution. Handle Frames and machines 2 ft 3 ft Pin 1 m 3 m 3 m 5 ft 1 m C 100 lb B D ½ m 2 m 4 ft 700 lb A E Note 1 m 3 m 3 m ΣMA = 0 = – (4.5)(700) + (7)(Ey) Ey = 450 N 1 m C ΣFy = 0 = Ay – 700 + 450 B D ½ m Ay = 250 N 2 m ΣFx = 0 = Ax + Ex Can’t solve yet. 700 lb Ax A Ex E Ay Ey Frames Cx Cy 700 N 1 m 3 m 3 m 700 N E 1 m C x 450 N B D ½ m M yields E = 150 N so 2 m Σ C x 700 N A E 1 m 3 m 3 m ΣMA = 0 = – (4.5)(700) + (7)(Ey) Ey = 450 N 1 m C ΣFy = 0 = Ay – 700 + 450 B D ½ m Ay = 250 N 2 m ΣFx = 0 = Ax + Ex Can’t solve yet.
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