FE Statics Review Sanford Meek Department of Mechanical Engineering Kenn 226 (801)581-8562 meek@mech.utah.edu
Statics! forces = 0!moments = 0 2
Vectors Scalars have magnitude only Vectors have magnitude and direction 3
Vectors Vector addition and subtraction! A +! B =! R A B!!! R! B = A R A B R A -B 4
Vectors Resultants Law of sines sin! A = sin " B = sin# C Law of cosines C β A α γ B C 2 = A 2 + B 2! 2ABcos! 5
Vectors Cartesian coordinates The only system that is generally used in statics x, y, z axes i, j, k unit vectors 6
Vectors! A = A ˆi + A ˆj + A kˆ x y z ˆk  ĵ A = maga! = A + A + A 2 2 2 x y z î! ˆ A! Ax " ˆ! Ay " ˆ! Az " A = = ˆ # $ i + # $ j+ # $ k A % A & % A & % A & 7
Direction cosines The components of the UNIT vector,! ˆ A! Ax " ˆ! Ay " ˆ! Az " A = = ˆ # $ i + # $ j+ # $ k A % A & % A & % A & are the cosines of the angles made by the vector with the coordinate axes. cos! = cos " = cos# = Ax A A y A Az A 2 2 2 cos! cos " cos # 1 + + = 8
Position Vectors! r = x î + y ĵ+ z ˆk! A A A A r = x î + y ĵ+ z ˆk B B B B Final point (head of arrow) initial point (tail of arrow)! r = ( x! x )î + ( y AB B B! y ) A ĵ+ ( z B! z ) ˆk A 9
Force Vectors Simply the force magnitude times the unit position vector!! ˆ rab F = FF = Frˆ AB = F rab! r = (! AB 5')ˆi + (! 3'cos20 )ˆj + (3'sin 20 + 6') kˆ 10
Dot Product Use: find the component of a vector along a certain direction component of force along a particular axis!! AiB!! AiB = AB cos! = A B + A B + A B x x y y z z A!! 11
Dot Product Use: find the component of a vector along a certain direction component of force along a particular axis The result is a scalar.!! AiB!! AiB = AB cos! = A B + A B + A B x x y y z z A!! 12
What part of A is aligned with the line? Dot Product What part is perpendicular?! A! = A! " A! a =perpendicular part uˆ a! A a = A cos! uˆ a! A cos! = A i uˆ Projected component (scalar): a 13
Moment (Torque) Only the perpendicular part of force gives torque 14
Cross Product Remember Right Hand Rule IEEE logo 15
Cross Product In terms of i, j, k i x j = k j x k = i k x i = j j x i = -k k x j = -i i x k = -j i x i = 0 j x j = 0 k x k = 0 i j k i positive i j k i negative 16
Moment of a Force Primary use of the cross product in statics posibon vector from O to any point on the line of acbon of F 17
Moment About an Axis Triple scalar product In this application, axis a is the y-axis, making unit axis vector u a simply base vector j. 18
Resultant Moments 19
Resultant Moments 20
Equal and opposite forces acting at a distance Force Couples answers.com No net force 21
Force Couples These all have the same net moment! Q: Why is this a useful observabon? A: It speeds up computabons. Look for couples! 22
Force Couples r! d! r = vector connecting any two points on lines of action d! F = perpendicular distance between forces! = The force at the tip of r M!! = r! F! M = Fd 23
Equivalent Force-Moment Systems PRINCIPLE OF TRANSMISSIBILITY P P F d F For both of these, the equivalent load at P is the same You can slide any force along its line of action without affecting the equivalent load! M=Fd F 24
Equivalent Force-Moment Systems An equivalent shifted force always exists for a set of coplanar forces. The net force for coplanar forces is itself in the plane. The moment for coplanar forces is perpendicular to the plane. d = ( ) M R O F For these, you can always find a location to place the net force such that the moment about O is unchanged. 25
Distributed Loads Demand the moment to be the same: 26
Equilibrium! forces = 0!moments = 0 No acceleration at rest or constant velocity 27
Equilibrium Free Body Diagram Define coordinate system Write all known and unknown forces and moments in terms of the coordinate system Sum all forces to zero Sum all moments to zero 28
Support Reactions zero horizontal resistance. has vertical resistance. zero moment resistance. has horizontal resistance. has vertical resistance. has moment resistance. has horizontal resistance. has vertical resistance. zero moment resistance. 29
Two-Force Members A two- force member is a body or part that is subjected to exactly two forces, and no moments. The equilibrium equabons then imply that those two forces must be equal in magnitude, opposite in direcbon, and be collinear (i.e., act along the same line of ac4on.) 30
Trusses A structure composed of two-force members tension or compression Loads are applied only at the joints Joints are frictionless pins Member weight is insignificant Usually, one support is a roller to allow expansion and contraction from temperature changes 31
Trusses Method of Joints FBD at each joint (pin joints) T C Method of joints: apply force balance at pins. Since force balance has only two equabons (ΣF x =0 and ΣF x =0), pick joints that have only two unknown forces! Force balance gives: T As part of your checks, first guess tension T or compression C. 32
Trusses Method of Joints FBD at each joint (pin joints) Joint (pin) FBD has the applied loads and link loads. Link (spar or element) FBD has only axial load! Links must be two- force members. (thus equal and opposite forces) 33
Trusses Method of sections make a cut in the structure 34
Trusses Method of sections: Make cuts with no more than 3 unknowns. Apply 3 statics equations (ΣF x =0, ΣF y =0, ΣM=0) to get the 3 unknowns! 35
Trusses Steps for solving truss problems Find the reactions by looking at the truss as a single rigid body Use the method of joints if the unknown forces to be solved are near or at the joint with a known (applied) force. Use the method of sections if the unknown forces to be solved are not at or near the known (applied) forces. 36
Frames Not necessarily two-force members loads can be applied anywhere on themembers 37
Frames Internal forces They don t appear in this FBD: 38
Internal Forces Make a virtual cut in a member and analyze the internal forces 39
Internal Forces Sign conventions 40
Internal Forces Load, shear, and moment relationships Shear is the integral of the load Moment is the integral of the shear Beam slope is the integral of the moment Beam deflection is the integral of the slope 41
Friction configuration matters! 42
Friction To be in equilibrium, friction is however large it needs to be treated as a pin support at an unknown location N F = = W P ( F! µ N ) moment balance: 43
Friction Impeding motion Frictional limit: At some point, P will reach a peak value, F s, large enough to start dragging the block. F s = µ s N 44
Friction F = µ s N F = µ k N 45
Friction F! µ s N ( F = µ N only at impending motion) s Methodology: assume that fricbon is a pin support located at an unknown posibon. Find these three unknowns (horizontal and verbcal pin reacbons, as well as pin locabon) by enforcing three equabons of equilibrium. Validate no slip assumpbon by confirming that F < µ s N. 46
Friction To slip, both ends must slip, which implies that BOTH ends will be at impending slip. Seing F A = µ A N A and F B = µ B N B, the FBD has THREE unknowns: N A, N B, and angle θ. Use the THREE equilibrium equabons to solve for them! Answer: 47
Wedges 48
Screws Screws are wedges around a cylinder How many threads? Lead = Number of threads x Pitch Pitch = 1/(number of threads per inch) 49
Screws Upward Motion Self locking 50
Screws Downward Motion Non- self- locking FricBon angle is smaller than the thread (wedge) angle The moment must resist the mobon Self- locking FricBon angle is bigger than the thread (wedge) angle The moment must start the mobon 51
Belts Tension is different at each end because of fricbon! (Contrast with a pully) 52
Belts dt " = µ d! T 53
Belts Tension increases in the direction of impending motion T T 2 1! = e µ (! 2 "! 1) 2! 1 If fricbonless (µ=0), then T 2 = T 1 (like pulley). With fricbon, however, T 2 > T 1 (consistent with limit µ ). Usually written as: where β is the angle of contact 54
Moment of Inertia about an axis polar moment of inerba
Moment of Inertia Parallel axis theorem Use this if you know Moment of Inertia about the primed axes passing thru the centroid, but you seek it with respect to some different set of parallel axes. d 2 = dx 2 + dy 2
Moment of Inertia Radius of Gyration The equivalent moment of inertia of a body that is a point mass at a distance (the radius of gyration) from the axis. Ix = Akx 2 kx = I x A Iy = Aky 2 ky = I y A I o = Ak o 2 ko = I o A
Moment of Inertia Moment of inertia for Composite Areas: Add them together (subtract for holes). Make sure that all the moments of inertia for all parts are found about the same axis. Use tables and parallel axis theorem.
Centroids - Lines
Centroids - Areas
Centroids Composite Bodies