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Torch Ellio ellio@eng.utah.edu (801) 587-9016 MCE room 2016 (through 2000B door) Posion 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 Connued

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 Projecon of vector U, in 90o The direcon of vector V? V Ucos(θ) eV = V/V à eV = 1 . U eV = (U)(1)cos(θ) = Ucos(θ)= the answer The projecon of U in the direcon of V is U doed with the unit vector in the direcon 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

Mulply 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 acon of F mes the magnitude |F| of F, where posive 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 (oen) or 0 force and 3 moment (oen)

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 800 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 direcon, and along the line of acon 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. 700 N

Ax A Ex E Ax = 150 N Ay Ey Centroids

Area Moments of Inera Moment of inertia about the x axis: 2 Ix = A y dA Note y not x in the integral. Moment of inertia about the y axis: 2 4 Iy = A x dA Units are length Polar moment of inertia: y 2 J0 = A r dA 2 2 = A (x + y ) dA dA

= Ix + Iy y r Product of Inertia: x x

Ixy = A xy dA Radius of Gyraon The radius of gyraon (k) is the distance at which the enre area would need to be located to give the same moment of inera as the actual area.

2 2 2 2 Ix = A y dA = A kx dA = = kx A dA = kx A

Therefore, kx = Ix/A Units are length.

Used in buckling computaons, and found in tables of standard material shapes. Find the moment of inera of a rectangle about its base.

Moment of inertia about the x axis: h a+b h I = y2dA = y2dxdy = y2[(a+b)-a]dy x A 0 a 0 = b[h3 – 0]/3 = 1/3(bh3) y

dy h dx x a b

Parallel axis theorem The moment of inera of a figure about any axis is equal to the y y’ moment of inera of the figure A about the parallel axis through the x’ x centroid of the figure plus the area d mes the distance between the axes L squared. 2 IL = Ix’ + Ad You MUST move to and from the centroidal axis only!

Find the moment of inera of a rectangle about its centroidal axis x’.

We found the moment of inertia about the x axis: 3 2 2 Ix = 1/3(bh ) But, Ix = Ix’ + Ad àIx’ = Ix - Ad 3 2 3 Therefore Ix’ = 1/3 (bh ) – bh(h/2) = 1/12 (bh ) y

h x’ d = h/2 x a b

Composite Structures Find the moment of inera and radius of gyraon of the figure about its base. •Select the parts to use. 1” 1”

•Find the moment of inera of each part about 6” the base. I labeled the base as the x axis. 4” 10” •Add the moments of inera to get the total, which is the moment of inera about the base. 6” x •Find the radius of gyraon by dividing the moment of inera by the area, and taking the square root. Composite Structures Find the moment of inera and radius of gyraon of the figure about its base. A way: Subtract the red rectangle from the blue. 1” 1”

3 3 4 Ix =1/3 (bh ) = 1/3 [(6”)(10”) ] = 2000 in 6” 3 2 Ix =1/12 (bh ) + Ad 7” 4” 10” = 1/12 [(4”)(6”)3] + (4”)(6”)(7”)2 = 1248 in4 4 4 4 Ixtotal = 2000 in – 1248 in = 752 in 6” x

A = (6”)(10”) – (4”)(6”) = 36 in2

4 2 1/2 kx = [(752 in )/(36 in )] = 4.57 in The homogeneous slender bar has x mass m, cross-section area A, and x sin length l. Use integration to find the θ (θ) L mass moment of inertia of the bar about the axis L through its centroid.

m = ρAl dm = ρAdx -l/2 <= x <= l/2

r2 = x2 sin2θ

Integration from -l/2 to l/2 yields: I = (1/12) ρA sin2θ l3

Then substitute m for ρAl yielding: I = (1/12) sin2θ ml2

P Py Px

f W N

Static (3 Eq.) Impending Motion (3 Eq. + fmax = µsN)

Dynamic (only f = µkN unless constant velocity à (3 Eq. + f = µkN) f

45o

Px 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 fmax = µsN. If f <= fmax, 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

f1

µs1 P W1 N1

µs2 N1 Not W1 f1

f2 W2 N2 Type Problems (continued)

4. f > fmax Dynamics problem , but f = µkN.

5. Dynamic, but with constant velocity à You can solve it just like a static

equilibrium problem, and you may use f = µkN. P Py 6. Various things could happen. Px Ask à What could happen? Look at the physics of the problem. Develop a plan f What is the limiting case? W

N wall

Given W , W , θ, and the coefficients of static A B F A friction between all surfaces, what is the largest B force F for which the boxes will not slip? θ y If you are given a mass, then don’t forget to W convert it to a weight. A x

f2 f2 = µsN2

WB B f1 = µsN1 θ “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 µs = 0.1 and µk = 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?

µsβ T2 = T1 e where: T2 > T1 You must decide this. µ could be either µs or µk β is in radians, and often > 2π Note also that, µsβ = ln(T2/T1) ?

For (a) what are T1 = F For (b) 40 lb T2 = 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 A

1.5 ft Water 1.5 ft

1 ft 1 ft

1.5 ft 1.5 ft 1 ft

B Bx Width of dam (into figure) 10 ft By (3)(62.4) Weight density of water 62.4 lb/ft3 Water Pressure

Ignoring atmospheric pressure and the weight of the dam, what are the reactions at A and B A Fw = (1)(10)(1.5)(62.4) = 936 lb 1.5 ft FP = (1/2)[(3)(62.4)](10)(3) = 2808 lb Fw 1 ft ΣMB = 0 = - 3A + (0.5)(936) + (1)(2808) A = 1092 lb FP 1.5 ft 1 ft ΣFx = 0 = 1092 – 2808 + Bx Bx Bx = 1716 lb By (3)(62.4) Σfy = 0 = - 936 + By B = 936 lb y y

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. + ΣMA = 0 = 10 – (10)(5) + Ey (8) Ey = 5 N + Σfy = Ay – 10 + 5 Ay = 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. + ΣMA = 0 = 10 – (10)(5) + Ey (8) Ey = 5 N + Σfy = Ay – 10 + 5 Ay = 5 N

For 0 < x < 2m: Signs by

Vx = 5 N beam + ΣMx = 0 = - 5x + Mx convention . Mx = 5x (N m)

Draw the shear force and bending moment diagrams.

For 0 < x < 2m:

Vx = 5 N . Mx = 5x (N m)

For 2m < x < 4m:

Vx = 5 N + ΣMx = 0 = - 5x + 10 + Mx . Mx = 5x – 10 (N m)

For 4m < x < 6m:

Find Vx & Mx Draw the shear force and bending moment diagrams.

For 0 < x < 2m:

Vx = 5 N . Mx = 5x (N m)

For 2m < x < 4m:

Vx = 5 N + ΣMx = 0 = - 5x + 10 + Mx . Mx = 5x – 10 (N m)

For 4m < x < 6m:

Find Vx & Mx + Σfy = 0 = 5 – (x – 4)(5) – Vx Vx = 25 – 5x (N) + ΣMx = 0 = - 5x + 10 + Mx + (x - 4)(5)(x – 4)/2 force = area x distance Note that the loading is not the same as it would 2 . Mx = - (5/2) x + 25x – 50 (N m) 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: + Σfy = 0 = 5 – 10 – Vx Vx = -5 N + ΣMx = 0 = - 5x + 10 + 10(x – 5) + Mx . Mx = - 5x + 40 (N m)

For 0 < x < 2m:

Vx = 5 N . Mx = 5x (N m)

For 2m < x < 4m:

Vx = 5 N . Mx = 5x – 10 (N m)

For 4m < x < 6m:

Vx = 25 – 5x (N) 2 . Mx = - (5/2) x + 25x – 50 (N m)

For 6 < x < 8: Vx = -5 N . Mx = - 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 Mx = 10 N m . For x = 5m Mx = 12.5 N m . For x = 6m Mx = 10 N m