FE Sta'cs Review h0p://www.coe.utah.edu/current- undergrad/fee.php Scroll down to: Sta'cs Review - Slides Torch Ellio0 ellio0@eng.utah.edu (801) 587-9016 MCE room 2016 (through 2000B door)
Posi'on and Unit Vectors If you wanted to express a 100 lb force that was in the direction from A to B in vector form, then you would find e AB and multiply it by 100 lb. (- 2,6,3). A Y r AB r AB = 7i - 7j k (units of length) e AB = (7i-7j k)/sqrt[99] (unitless) z O x B.(5,- 1,2) Then: F = F e AB F = 100 lb {(7i-7j k)/sqrt[99]}
Another Way to Define a Unit Vectors Direction Cosines The θ values are the angles from the coordinate axes to the vector F. The cosines of these θ values are the coefficients of the unit vector in the direction of F. If cos(θ x ) = 0.500, cos(θ y ) = 0.643, and cos(θ z ) = - 0.580, then: z Y F θ y O θ x x θ z F = F (0.500i + 0.643j 0.580k)
Trigonometry hypotenuse θ adjacent opposite right triangle Sin θ = opposite/hypotenuse Cos θ = adjacent/hypotenuse Tan θ = opposite/adjacent
Trigonometry Con'nued A b C a c B any triangle Σangles = a + b + c = 180 o Law of sines: (sin a)/a = (sin b)/b = (sin c)/c Law of cosines: C 2 = A 2 + B 2 2AB(cos c)
Products of Vectors Dot Product U. V = UVcos(θ) for 0 <= θ <= 180 ο ( This is a scalar.) U In Cartesian coordinates: θ U. V = U x V x + U y V y + U z V z In Cartesian coordinates to find θ? V UVcos(θ) = U x V x + U y V y + U z V z θ = cos - 1 [(U x V x + U y V y + U z V z )/UV]
Products of Vectors Dot Product UVcos(θ) = U x V x + U y V y + U z V z θ = cos - 1 [(U x V x + U y V y + U z V z )/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 <= 180 ο How could you find the Projec'on of vector U, in The direc'on of vector V? e V = V/V à e V = 1 U. e V = (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. U θ Ucos(θ) 90 o V
Products of Vectors Cross Product Also called the vector product V V U x V = UVsin(θ)e e θ for 0 <= θ <= 180 o U U Note: U x V = - V x U + If θ = 0 U x V = 0 i j If θ = 90 o 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 Projec'on UVsin(θ)e φ W θ U of UVsinθ V on the direc'on of W
Products of Vectors Mixed Triple Product Note: No i j k Otherwise, same procedure U x U y U z U x V. W = V x V y V z W x W y W z
Pulley Problem (frictionless pins) T T ΣF y = 0 = 2T T 1 mg T 1 = 2T mg T 1 T 1 ΣF y = 0 = 2T 1 mg mg m A g = 2(2T mg) T 1 mg m A g mg m A g T = (3m + m A )g/4 m m A m T
Moment of a Force 2- dimensional, in the plane of the screen y. z x O. F The moment of F D 90 o 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 x e O. r θ F Then: D 90 o 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. M O = r OB x F y B C r OB 360mm O x F 500mm D A 600mm z You should determine: M O = (- 20.9i + 25.1k) N. m?? j term?? Then find M o x axis M o x axis = - 20.9 N. m?? vector form??
0 arbitrary point r 0B r 0A -F B r BA A A F Couples F causes translation and, in general, rotation. Let F be: Equal in magnitude to F Opposite direction of F Not collinear with F. 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 r BA. Then: r 0B + r BA = r 0A Or: r BA = r 0A - r 0B M 0 = (r 0A x F) + (r 0B x (-F)) = (r 0A - r 0B ) x F M 0 = r BA 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 A. F F M c F A. - F equivalent to 1
Review For sta'c equilibrium: ΣF = 0 and ΣM = 0 In 3- D you have 6 independent equa'ons: 3 force and 3 moment in 2- D you have 3 independent equa'ons: 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 A x B A y Smooth surface or roller support F F Fixed, built-in, or M A A x cantilever support Rope or cable (tension) A y Spring (tension or compression)
2-D Beam Solution 10 N 80 0 4 m 2 m 10 N M A A x 80 0 4 m ΣF x = 0 = A x 10cos(80) A x = 1.74 N ΣF y = 0 = A y 10sin(80) A y = 9.85 N ΣΜ A = 0 = M A 10sin(80)(4) M A = 39.4 N. m A y
3-D Beam Solution Given T find M A y Is this bar properly constrained and statically determinate? B Rigidly held Yes and yes A T z Options: y Sum forces in a direction M ay B Sum moments about a point A y x Sum moments about an axis A z A T A x M Ax M az Which option is best here? ΣM A = 0 = M Ax i + M Ay j + M Az k + (r AB X T) Then group the i, j, and k components and solve. x
3-D Beam Solution Given W, find T y Is this bar properly constrained and statically determinate? Yes (?) and no rope with tension T A W Options: Ball and socket supports Sum forces in a direction Sum moments about a point Sum moments about an axis Which option is best here? Sum moments about the axis. Axis B x
Special Case à Two- force Bodies If a body in equilibrium has forces at two and only two loca'ons and no moments, then the forces are equal in magnitude, opposite in direc'on, and along the line of ac'on between the two points. F A A B F B
Special Case à Three- force Bodies If a body in equilibrium has forces at three and only three loca'ons and no moments, then the forces are either concurrent or parallel. Usually, you can use the trigonometric or graphical techniques of Chapter 2. F A F C C A B F B
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.
Frames and machines 1 m 3 m 3 m Handle 2 ft 3 ft 5 ft Pin 1 m C B D ½ m 2 m 700 lb A E Note 100 lb 4 ft 1 m 3 m 3 m 1 m C B D ½ m 2 m 700 lb A x A E x E A y E y ΣM A = 0 = (4.5)(700) + (7)(E y ) E y = 450 N ΣF y = 0 = A y 700 + 450 A y = 250 N ΣF x = 0 = A x + E x Can t solve yet.
Frames C x 1 m 3 m 3 m C y 700 N 700 N 1 m C B D ½ m 2 m 700 N A E ΣM C yields E x = 150 N E x 450 N so 1 m 3 m 3 m 1 m C B D ½ m 2 m 700 N A x A E x E A y E y ΣM A = 0 = (4.5)(700) + (7)(E y ) E y = 450 N ΣF y = 0 = A y 700 + 450 A y = 250 N ΣF x = 0 = A x + E x Can t solve yet. A x = 150 N
Centroids
Area Moments of Iner'a Moment of inertia about the x axis: I x = A y 2 da Note y not x in the integral. Moment of inertia about the y axis: I y = A x 2 da Units are length 4 Polar moment of inertia: y J 0 = A r 2 da = A (x 2 + y 2 ) da da = I x + I y y r Product of Inertia: x x I xy = A xy da
Radius of Gyra'on The radius of gyra'on (k) is the distance at which the en're area would need to be located to give the same moment of iner'a as the actual area. I x = A y 2 da = A k x 2 da = = k x 2 A da = k x 2 A Therefore, k x = I x /A Units are length. Used in buckling computa'ons, and found in tables of standard material shapes.
Find the moment of iner'a of a rectangle about its base. Moment of inertia about the x axis: I x = A y 2 h a+b h da = y2 0 dxdy = y2 0 [(a+b)-a]dy y = b[h 3 0]/3 = 1/3(bh 3 ) a a dy dx b h x
Parallel axis theorem The moment of iner'a of a figure about any axis is equal to the moment of iner'a of the figure about the parallel axis through the centroid of the figure plus the area 'mes the distance between the axes squared. y y A d x x L I L = I x + Ad 2 You MUST move to and from the centroidal axis only!
Find the moment of iner'a of a rectangle about its centroidal axis x. We found the moment of inertia about the x axis: y I x = 1/3(bh 3 ) But, I x = I x + Ad 2 à I x = I x - Ad 2 Therefore I x = 1/3 (bh 3 ) bh(h/2) 2 = 1/12 (bh 3 ) a b h x d = h/2 x
Composite Structures Find the moment of iner'a and radius of gyra'on of the figure about its base. Select the parts to use. 1 1 Find the moment of iner'a of each part about 6 the base. I labeled the base as the x axis. 4 10 Add the moments of iner'a to get the total, which is the moment of iner'a about the base. 6 x Find the radius of gyra'on by dividing the moment of iner'a by the area, and taking the square root.
Composite Structures Find the moment of iner'a and radius of gyra'on of the figure about its base. A way: Subtract the red rectangle from the blue. 1 1 I x =1/3 (bh 3 ) = 1/3 [(6 )(10 ) 3 ] = 2000 in 4 6 I x =1/12 (bh 3 ) + Ad 2 7 4 10 = 1/12 [(4 )(6 ) 3 ] + (4 )(6 )(7 ) 2 = 1248 in 4 I xtotal = 2000 in 4 1248 in 4 = 752 in 4 6 x A = (6 )(10 ) (4 )(6 ) = 36 in 2 k x = [(752 in 4 )/(36 in 2 )] 1/2 = 4.57 in
The homogeneous slender bar has mass m, cross-section area A, and length l. Use integration to find the mass moment of inertia of the bar about the axis L through its centroid. θ x sin(θ) x L m = ρal dm = ρadx -l/2 <= x <= l/2 r 2 = x 2 sin 2 θ Integration from -l/2 to l/2 yields: I = (1/12) ρa sin 2 θ l 3 Then substitute m for ρal yielding: I = (1/12) sin 2 θ ml 2
P P y P x W N f Static (3 Eq.) Impending Motion (3 Eq. + f max = µ s N) f Dynamic (only f = µ k N unless constant velocity à (3 Eq. + f = µ k N) 45 o P x
Type Problems P 1. You don t know where you are on the curve. Assume static equilibrium. Solve equilibrium equations to find f. P Solve f max = µ s N. If f <= f max, then the assumption is good. 2. Impending motion, and you know where. f You have 4 equations, so just solve it. N W 3. Impending motion, but you don t know where. Ask à what can happen? P f 1 µ s 1 P W 1 N 1 µ s 2 N 1 Not W 1 f 1 W 2 N 2 f 2
Type Problems (continued) 4. f > f max Dynamics problem, but f = µ k N. 5. Dynamic, but with constant velocity à You can solve it just like a static equilibrium problem, and you may use f = µ k N. P P y 6. Various things could happen. P x Ask à What could happen? Look at the physics of the problem. Develop a plan What is the limiting case? W N f
wall Given W A, W B, θ, and the coefficients of static friction between all surfaces, what is the largest force F for which the boxes will not slip? If you are given a mass, then don t forget to convert it to a weight. F θ A W A B y x f 2 f 2 = µ s N 2 WB B θ f 1 = µ s N 1
Belt friction The box weighs 40 lb. The rope is wrapped 2 ¼ turns around the fixed wooden post. The coefficients of friction between the rope and post are µ s = 0.1 and µ k = 0.08. (a) What minimum force is needed to support the stationary box? (b) What force must be exerted to F raise the box at a constant rate? T 2 = T 1 e µ s β where: T 2 > T 1 You must decide this. µ could be either µ s or µ k β is in radians, and often > 2π Note also that, µ s β = ln(t 2 /T 1 )? For (a) what are T 1 = F For (b) 40 lb T 2 = 40 lb F µ = 0.1 0.08 β = 4.5π 4.5π?
Water Pressure Ignoring atmospheric pressure and the weight of the dam, what are the reactions at A and B A 1.5 ft Water A 1.5 ft 1 ft 1.5 ft B Width of dam (into figure) 10 ft Weight density of water 62.4 lb/ft 3 1 ft 1.5 ft 1 ft B x B y (3)(62.4)
Water Pressure Ignoring atmospheric pressure and the weight of the dam, what are the reactions at A and B F A w = (1)(10)(1.5)(62.4) = 936 lb F 1.5 ft P = (1/2)[(3)(62.4)](10)(3) = 2808 lb ΣM B = 0 = - 3A + (0.5)(936) + (1)(2808) A = 1092 lb ΣF x = 0 = 1092 2808 + B x B x = 1716 lb Σf y = 0 = - 936 + B y B y = 936 lb y F w 1 ft F P 1.5 ft 1 ft B x B y (3)(62.4) Sign convention? x
Beams We will discuss an example of a beam with point forces, a point couple, and a constant distributed load. We will look at segments of the problem, but you need to think of it as a continuous solution with the figures aligned and the text and figures progressing down the page in an organized manner. The total solution is shown to the right.
Draw the shear force and bending moment diagrams. To find the reactions, you may model the distributed load as a point load. ΣM A + = 0 = 10 (10)(5) + E y (8) E y = 5 N Σf y + = A y 10 + 5 A y = 5 N However, you must go back to the distributed load while completing the problem. To do otherwise would cause errors that might be unsafe.
Draw the shear force and bending moment diagrams. To find the reactions, you may model the distributed load as a point load. ΣM A + = 0 = 10 (10)(5) + E y (8) E y = 5 N Σf y + = A y 10 + 5 A y = 5 N For 0 < x < 2m: Signs by V x = 5 N beam ΣM + x = 0 = - 5x + M x convention M x = 5x (N. m)
Draw the shear force and bending moment diagrams. For 0 < x < 2m: V x = 5 N M x = 5x (N. m) For 2m < x < 4m: V x = 5 N ΣM x + = 0 = - 5x + 10 + M x M x = 5x 10 (N. m) For 4m < x < 6m: Find V x & M x
Draw the shear force and bending moment diagrams. For 0 < x < 2m: V x = 5 N M x = 5x (N. m) For 2m < x < 4m: V x = 5 N ΣM x + = 0 = - 5x + 10 + M x M x = 5x 10 (N. m) For 4m < x < 6m: Find V x & M x Σf y + = 0 = 5 (x 4)(5) V x V x = 25 5x (N) ΣM x + = 0 = - 5x + 10 + M x + (x - 4)(5)(x 4)/2 force = area x distance M x = - (5/2) x 2 + 25x 50 (N. m) Note that the loading is not the same as it would be, if the point load through the centroid, that was used to find the reactions, were used here.
When you are past the distributed load, then you can model it as a point load through the centroid. For 6 < x < 8: Σf y + = 0 = 5 10 V x V x = -5 N ΣM x + = 0 = - 5x + 10 + 10(x 5) + M x M x = - 5x + 40 (N. m)
For 0 < x < 2m: V x = 5 N M x = 5x (N. m) For 2m < x < 4m: V x = 5 N M x = 5x 10 (N. m) For 4m < x < 6m: V x = 25 5x (N) M x = - (5/2) x 2 + 25x 50 (N. m) For 6 < x < 8: V x = -5 N M x = - 5x + 40 (N. m) Note that all sections of the plots are Constant or linear f(x) except 4 < x < 6. For it you can pick values. For x = 4m M x = 10 N. m For x = 5m M x = 12.5 N. m For x = 6m M x = 10 N. m