General Flashcards
Fourier Transform
This is a way to describe an oscillating function in terms of angle and intensity by changing the function to be a function of angular frequency (rad/sec) or wavenumber (k=seconds to complete a wave).
It uses eulers identity to switch the imaginary parts of the frequency into cos and sine terms. This is then converted into polar coordinates in the intensity phase space.
Use the table to convert regular functions in fourier equations that are used to analyze a frequency.
In the manual the period is given so 2 pi over the angular frequency give syou seconds and the functions for several common signal inputs.
Even and odd functions
f(-x)=f(x) means the function is even (x2 cos)
f(-x)=-f(x) means that it is odd (x3 sin)
Orthogonal functions
This is when the integral of f(x)*g(x)=0
Taylor Series
Taylor Series represent using a series of bisected functions to model an original function f(x). They converge when |x-a|
Fourier transform of a cosine wave
This will yield one line because each and every interval is on the same frequency which means your peak will show that one frequency.
Fourier transform of a series of signals with t(duration)=tau
This will yield several bumps because thetime domain shows 2 intensities within one 360 degree rotation.
Integers, rational, and irrational numbers
Integers are whole numbers
Rational numbers are numbers that can be written as a ratio of two integers
irrational numbers cannot be expressed as a ratio of two integers. They include numbers that repeat or do not terminate Ex: (3)^.5/2
This is irrational because 3^.5 has an operator
Real numbers: include all of the above
factorial
n!=(n)*(n-1)*…(2)*(1)
Complex Numbers (Add, Subtract, multiply, divide)
Add/subtract: combine like terms.
Mult.: FOIL
Divide: Multiply by the conjugate of the denominator
Using conjugates
if A=a+ib and A#=a-ib then A*A#=a2 + b2
Sets
A set is a collection of objects or entities. If X belongs to a set S then X ∈ S where S={n0,n1…n} or a set could be defined by a function S={X:X3>27 }
Joining Sets
∈
C ∪ D = {X:X∈C or X∈D) This is the union of sets C and D
C ∩ D = {X:X∈C and X∈D) This is the intersection of sets C and D
And if all of a set (set A) are part of a larger set (set B) then AcB where the c is the inclusion symbol
Ø
This is the null set symbol. It means that there are not any elements within a set.
Exponent Laws
n√a=a(1/n)
am/an= am-n am*an= am+n
(am)n=am*n
(a*b)m= am*bm
Simplify n √am
n √am= a(1/n)*m = a (m/n)
Simplify n √ab
n √ab= a(1/n) * b(1/n)
How to take the root of a function with a divisor
Multiply the denominator through and divide the whole term by the diviseor with the root operator used.
Rule for binomial expansion
if we have a binomial given by (A+X)n then the expansion is given by B*An-m * Xm
Where: m=term # starting from 0 to n+1
B=(The exponent*coefficient of the term m-1)/m
Ex: Term 3 of (A+X)7 is given by A7+7A6X+(7*6/2)A5X2…
Determinants
Determinants are the values of matrices. The order of a determinant is given by the sqaure root of the number of elements which are the numbers inside of the matrix.
determinant nomenclature
the position of any value is given by aij where i= row # and j = collumn #
Finding determinants of second order matrix
If the determinant is second order then the determinant is given by:
Product of primary diagonal-Product of secondary diagonal
Where: Primary diagonal are the values given by a11 and a22 and the secondary diagonal is given by the values a21 and a12
Minor of a matrix
This is the reduced matrix that can be used to find the determinant of higher order matrices. The order is given by crossing out the values that intersect at the value of interest and finding the determinant of the resulting matrix multiplied by the value of interest.
Sign of determinant parts
If the leading coefficient is in the position ij where i+j is
even then the coefficient is positive
odd then the coefficient is negative
Solving Linear Equations with two unknowns
when given a set of equations like
a1x+b1y=c1 and a2x+b2y=c2
Multiply the the first equation by -a2 and the second by a1 so that when added the first terms cancel. This enables you to have only one unknown to be solved for.
y=(-a2c1+a2c2)/(a1b2-a2b1) and x=(c1b2-c2b1)/(a1b2-a2b1)
Solving linear equations with 3 unknowns
If we have three equations we can create three matrices; C=A*b where A=3x3 coefficient matrix, b = 3x1 matrix of xyz (or any three unknowns) and C= 3x1 matrix of constants
Then b= A-1*C
Scalar Product
a dot b= ax*bx+ay*by+az*bz=|a|*|b|*cos(alpha)
Where: a dot b represents the magnitude of the shared a and b vector space
Cross Product
axb is the cross product of a and b and represents the magnitude of the distance between the tips of a nd b in the direction that is normal to the plane created by a and b.
axb is given by the determinate of a and b
axb=
|ax ay az|
|bx by bz|
i j k |
How to solve third order algebraic equations when one root is given
When one root is given you can divide the third order equation by the root to make it a quadratic which is then solved via the quadratic formula.
Long division for equations
To do long division of an equation you always eliminate the higher order term.
Basic logarithm function
The answer to a log is the exponent
logbn=a -> ba=n and n=blogb^n
logb1=?
=0 because b0=1
logbb=?
=1 because b1=b
expand: logbm*n
logbm*n= logbm+logbn
expand: lobb(m/n)
logb(m/n)=logbm-logbn
logb(1/n)=?
logb(1/n)=-logbn because logbn-1=-1*logbn
logbm(1/n)=?
logbm(1/n)=(1/n)*logbm
radians to degrees formula
radians= degrees * (pi/180) AKA there are 180o in pi radians
30o to radians
30o=pi/6 radians
pi/3 radians to degrees
60o
45o to radians
pi/4
275o to radians
5pi/4 radians
3pi/2 to degrees
270o
330o to radians
-pi/6 or 11pi/6
5pi/3 to degrees
300o
120o to radians
2pi/3 radians
rpm to rad/sec
rpm*(360/rev)*(pi rad/180)*(min/60 sec)=rad/sec
sin, cos, tan of 17pi/6
17pi/6= 12pi/6+5pi/6
sin(5pi/6)=1/2
cos(5pi/6)= -3.5/2
tan(5pi/6)= -1/3.5
sin, cos, tan 5pi/3
sin (5pi/3)= -3.5/2
cos(5pi/3)= 1/2
tan(5pi/3) = -3.5
polar coordinate transform
(x, y) = (r, θ)
where: r=(x2 + y2) and θ=tan-1(y/x)
x=rcosθ and y = rsinθ
Law of cosines
for a triangle with angles A,B,C opposite to sides a,b,c
c2 = a2 + b2 - 2abcos(C)
Law of sines
a/sin(A)=b/sin(B)=c/sin(C)
Using law of cosines
Law of cosines can be used to find:
Distances between objects starting at one point and travelling in different directions.
diagonals of a parallelagram
General Equation for a line
Ax+By+C=0
m=-A/B and
b=-C/B
slope intercept line equation
y=mx+b where:
m=(y2-y1)/(x2-x1)
two point line formula
y-y1/x-x1=(y2-y1)/(x2-x1)
Slope relationship between perpendicular lines
m1=-1/m2
angle between two lines where m1>m2
tan(α)=(m1-m2)/(1+m1*m2)
general equation for conic sections
ANY second order equation
Ax2 + Bxy + Cy2 + Dx + Ey + F
criterion for ellipse, parabola, and hyperbola
ellipses occur when (B2-4AC)<0
parabolas occur when (B2-4AC)=0
Hyperbolas occur when (B2-4AC)>0
ellipse equation
(x-h)2/a2 + (y-k)2/b2 = 1
where C=(h,k) and a,b are the radius in the x and y directions
Limits
Limits are the value a function approaches at some independent variable value.
If, the function exists at the limit then it is continuous.
L’Hospital’s Rule
Suppose we have two functions, f(x) and F(x) where both go to zero as x approaches A. The limit for f(x)/F(x) is given by f’(x)/F ‘ (x)
Formal definition of a derivative
The derivative of a function, y=f(x) is the instantaneous rate of change or
dy/dx=lim (Δx->0) (Δy)/(Δx) and can be proven by subbing y+Δy for y and x+Δx for x within a function given by y=f(x) and then solving for Δy/Δx and taking the limit as Δx approaches 0
mulitplication rule for derivatives
d/dx u*v= u dv/dx + v du/dx
power rule for derivatives
y=n xm
y’ = (m*n)xm-1
d/dx un = ? where u = f(x)
d/dx un = nun-1 du/dx via the chain rule
d/dx (vu) where both v and u =f(x)
d/dx(vu) = u*vu-1 dv/dx + vu lnv du/dx
d/dx (u/v)=?
d/dx (u/v) = [v(du/dx) - u(dv/dx)] / v2
d/dx (logcu) = ? where u=f(x) and c=constant
d/dx (logcu) = 1/u logce * du/dx
How to treat derivatives
Derivatives act as differentials so they can be inversed or have parts cancel (eg (dy/dt)/(dx/dt)=dy/dx
Implicit differentiation
if f(x,y)=c then to find dy/dx
- ) differentiate f(x,y) where d/dx(xy)= y + x(dy/dx)
- ) isolate dy/dx
partial derivatives
These are written with the funky d symbol (∂) or f ‘x and are formally defined by
if z=f(x,y) then ∂f/∂x = lim Δx->0 [f(x+Δx,y) - f(x,y)]/ (Δx)
where: if the variable is not in the denominator it is treated like a constant
chain rule for partial derivatives
if z=f(x,y) and x=f(t) ; y =f(t) then ∂z/∂t = [(∂z/∂x * ∂x/∂t) + (∂z/∂y * ∂y/∂t)]
Meaning of f ‘ and f ‘’ being greater than zero
if f ‘ >0 then the slope of f at whatever x is positive so f(x+dx)>f(x)
if f ‘’ >0 then the slope of f at x is increasing therefore f ‘ (x+ dx)> f ‘ (x)
Procedure for finding maxima/minima
- ) find the first derivative
- ) Find the values of x so that the first derivative (f ‘ )=0 . These are the function’s critcal values
- ) find the second derivative f ‘’
- ) Substitute the critical values from 2 into f ‘’
if f ‘’ (x)>0 then at x there is a local minima
if f ‘’ (x) <0 then at x there is a local maxima
Solving for local maxima/minima when f’‘=0
To evaluate if f(x) is a local maxima or minima when f ‘’ is 0 you must find f ‘ (x+1) and f ‘ (x-1) and use those values to evaluate what f(x) is
Integration
This is the process of finding the function where the function’s derivative is given
d( ∫ f(x) dx) = f(x) dx
∫=the function whose differential is
anti-derivative: Power rule
∫ axb dx = [axb+1/(b+1)] + c
definite integral definition
∫ab f(x) dx = F(b)-F(a)
Integration by Parts
When the integrand can be split into two functions; one that represents a u where u=f(x) and another that represents dv where dv=g(x)dx and g(x) is easily integrated.
Under these conditions: ∫ udv = uv - ∫ vdu
Common functions for integration by parts
Generally allow ln to be u (du=dx/x) and x, ex, or other easily integrated functions be dv
How to use dy in a double integral
When using dy the slice is taken perpendicular to the y axis.
- ) with f(x) solve for x so that way x=f(y)
- ) determine the bounds and integrate ∫ f(y) dy
Formula for generating a solid around the x axis
V= ∫ab pi y2 dx
where y= f(x)
Laplace Transformation
These involve the transformation of derivatives where ∫0inf e -st F(t) dt = f(s) and can be written f(s) = L {F(t)} = L{F} = F(s)
∫ e x n =?
∫ e x n dn = (1/x) e x n + c
Finding Fr for coplanar and parallel forces
Fr= ΣF
dr = ΣFi*di / ΣFi where d is the perpendicular distance to some point O
Fr for concurrent force systems?
Fr = Σ Fi where each F is a vector so vector addition is used. This is a system where all the forces intersect at one point.
Non Concurrent force systems: Fr and dr = ?
In a nonconcurrent system the forces do not intersect at a common point and the body is not static. A moment around the center of gravity is given by
Σ Fi * di = Fr * dr
where the direction of Fr is determined by vector addition
Couple
A couple is a torque that is produced by two parallel but opposite forces. The resulting moment is given by M= F*d where d is the total distance between the parallel forces and F is the force of one of the forces
The magnitude of the couple is not a function of location. AKA the couple has the same magnitude regardless the center of the moment.
Solving static structures: Method of Joints
The idea of this method is that each joint must be in equilibrium therefore if we isolate each and every joint and solve for the internal forces then each will be in equilibrium. Procedure:
- ) Draw FBD with everything internal in tension
- ) Solve for the reaction forces (use logic not tension)
- ) choose a joint where 2 unknowns exist and solve.
- ) rinse and repeat
Equations of Equilibrium
This says that for a static body:
ΣFx = ΣFy = ΣFz = ΣM0 = 0
When solving a structure we can assume that the internal reactions cancel each other out and only treat the system by analyzing the external forces with these equations.
Method of Sections
This “cuts” the structure and analyzes the internal forces at the cut. Procedure:
- ) Draw FBD
- ) solve for Rxns
- ) “cut” the structure and draw internal forces in tension. Do NOT cut more than three members at once.
- ) ΣMo = 0 where o is the intersection of two of the forces which leaves the third unknown and the external forces which makes solving easy.
- ) ΣFx = ΣFy =0 solve for the other two unknowns
What is the tension in a cable/rope equal to?
Tension in a cable or a rope is constant throughout the cable/rope. The direction does NOT matter.
Finding moments in 3-space
- ) break forces into their three components
- ) ΣMx = Fz*dz + Fy * dy It is the same as 2-space but everything that is not in the direction of the axis is considered to be causing a moment about the said axis.
Friction
Friction is the contact resistance between bodies in contact with one another and is proportional to the normal force. It acts parallel to the interface across the surface and is summated at the point where the normal force acts.
Fmax= N * μs
Fmotion=N * μk where μs >μk
Total force of two interacting solids
R = N + Ff where Ff = μ * N and the angle of R is given by tan ϕ = Ff /N= μ and ϕ is the angle between Ff and N not to the horizontal
Belt Friction
This is the idea that a belt going over a solid has unequal tensions where the greater tension is in the direction of impending motion.
T2max = T1 *e^(μs*β) where β is the angle of contact in radians
Centroids solving for CG
x̄ = [Σ W*x]/W and similar with the y bar
Where: x represents the distance of a homogenous chunk of the object from any defined x axis and W is the weight of that chunk.
Centroid of areas
x̄ = ∫ x dA/∫dA
centroid of lines
This is fundamentally similar to the centroid of an area but uses a line integral given by x̄ = ∫ x dL/∫dL where L refers to the line coordinates
This is most common for arcs where dL=rdθ and x=rcosθ and y=rsinθ
Theorem of Pappus
This says that when you rotate a line about an axis 360o that the resulting surface area is given by A=L 2 pi X where X is equal to the cg or centroid
Theorem of Guldinus
This states that the volume of an area revolved around an axis is given by V = 2*pi*x*D where x is the centroid or cg of the area
Strain and Elongation
When a material is put into a stress state it experiences deformation by a total amount given by δ . and the strain is given by δ/L = ε
Twisting or angular deformation is given by γ = δ/L = tan ϕ where ϕ is the angle from vertical to the deformation
Modulus of elasticity
E= σ/ε = (P/A)/(δ/L) and represents the per unit stress over per unit strain within the elastic region of deformation
Poissons ratio
This the ratio of lateral to longitudinal strain
Coefficient of thermal expansion
δT = L α (ΔT)
Solving statically indeterminate structures
These are structures where the number of unknowns exceeds the number of equations of equilibrium so we need to use the material properties of the material to solve.
- ) Draw FBD for a selected member
- ) ID known and unknowns. Write all equations to constrain via equilibrium then for material properties
- ) Use material properties to solve for an unknown. Plug into Equations of Equlibrium.
Twisting
When a shaft is placed under some torque T it experiences a shear stress τ where dF= τ dA and the internal resisting moment is given by τ r dA = dMT
finding shear from a torque
J= ∫ r2 dA = the polar moment of inertia across the cross sectional area
Jmax= pi*d4/32
and τ = Tr/J
angle of twist
This is the angle of rotation caused by a torque.
It is related to the shear force by the shear modulus G.
τ / γ =G where γ is the shear strain
G= TL/θJ where J is from 0 to R
Shear strain from twisting
γ = C θ /L where θ is in radians
The angle of twist for compound shafts
This is the summation of individual twists. J does not depend on the material but G is material dependent
Beam analysis
“walk the beam” These are usually subject to several loads and moments.
- ) solve reactions
- ) “walk the beam” to find moment and stress diagrams
Sign convention for beam analysis
Shear is positive when walking the beam left to right and the shear points upwards.
Moments are positive when the forces on either side of the point make the beam bend like a smiley face.
Shear diagrams at x=0 and x=L
The shear is ALWAYS 0 at x=0 and x=L because otherwise the beam would be mobile and not static
Relating moments to shear
∫AB dM = ∫AB V dl where V=V(l) and is determined by the shear diagram
This means that the change in the bending moment from A to B = the area under the shear curve from A to B
Flexure Formula
This is the formula for understanding the tension/compression within a bent beam. It says that in a uniform beam that σy/y = constant and that the total moment Mt = σmax/ymax ∫ y2 DA = Iy σmax/ymax
Position of the neutral axis
This influence the moment about the x axis and represents the line of the beam where σ=0 It is given by the centroid of the cross section.
second moment of area formula
Ix = ∫-l/2l/2 y2 dA where the bounds are measured from the centroid
Parallel Axis Theorem
This says that the moment of inertia about any axis parallel to the neutral axis is equal to Ib = INA + A*D2 where D is the distance from the NA to the reference axis
General procedure for moment of inertia of composites
- ) Divide each cross-section into several chunks with known Icg
- ) located the neutral axis via finding the centroid (C = ∫ y dA/∫dA)
- ) Calculate INA about each centroid.
- ) Use the parallel axis theorem to relate each component to the overall cross-sectional I where holes are treated as negative values and the others summate.
What are the units of moment of inertia?
L4 when solving for anything with moment of inertia try to use inches and centimeters
Units of moments and torque
L*F
Units of shear in shear diagrams
This is simply put in terms of pounds. It represents the total shear throughout the cross section
Units of G/E
This is in F/L2 because strain is unitless (L/L)
Types of supports that create counter moments
Fixed connection to walls create opposite moments. These also will have shear at their open end.
Dynamics
This is the study of motion of particles and bodies. Particles are point objects where it is assumed to be concentrated at one point. Bodies are systems of particles that form an object where the relative motion of different parts of the object must be considered.
Newton’s Three Laws of Motion
- ) A particle acted upon by a balanced force system has no acceleration. AKA an object in motion stays in motion unless acted upon by an external force
- ) F=ma
- ) Action and reaction forces are always equal and opposite
Kinematic equations
Assuming that a is a constant
S= Vot + at2/2
Vf = Vo + at
Vf2 = Vo2 + 2a*s
How to use the kinematic equations
When a body is under constant acceleration and the problem is trying to describe the motion of a particle or body in terms of time and position identify the known variables (to, tf, Vo, Vf, a) and the unknown variables.
Use the kinematic equations to solve.
Relating s, V, and a
given s = f(t) then s ‘ = V(t) and s ‘’ = a(t)
Also s = ∫ V(t) = ∫∫ a(t)
Change of position with derivatives
ds = (VdV)/f(V)
Position as a function of force when F=f(t)
x = (1/m) ∫∫f(t) dt dt + C1*t + C2
Angular Velocity
ω = θ/t where θ is in radians and the change in position is given by rθ
Relating velocity and angular velocity
V = r * ω
Angular acceleration
This is the change of angular velocity given by d2 θ/dt2 = α
Also α*dθ = ω*dω
Do the kinematic equations apply to circular motion?
Yes. Replace V with ω; a with alpha; S with theta
flight of a particle given its initial conditions
y:
centripital and centrifugal acceleration
Centripital acceleration is an = V ω = rω2 = V2/r and is directed towards the center of rotation
Centrifugal acceleration is at = dV/dt = rα + 2ω dr/dt and is directed in the direction tangent to the path of the object at that moment
centripetal and centrifugal forces
the centripetal F; Fn = WV2/gr = W*an/g
The centrifugal F; Ft= (W/g)*dV/dt = (W/g) * at
Ideal bank angle
tanθ = V2/gR + tanθf for sliding down the curve and -μ for sliding up
where: θf = friction angle = tan-1 μ
θ = ideal bank angle where an object travelling at V through a curve with R will neither slide up or down.
Forces of a rotation that is not about the object’s center of mass
ΣFn = (W/g)*ro* ω2
ΣFt = (W/g)*ro * α
ΣMz = α Iz
where: ro=the distance from the center of rotation to the center of mass of the object in rotation
Iz = Io + m*ro2
alpha= rotational acceleration
ro ω2 = an and ro * α=at
Radius of Gyration
k is a distance and it can regarded as the distance from the axis of inertia to the center of gravity for an object that is rotating around a center that is not outside of the object.
k = (I/m).5
Rotational dynamic equilibrium
This is a state where V is constant and an object is rotating.
The first step is to draw FBD and find the resultant force on the said object.
In this state (W/g)ro α d -Izα = 0 where d is the distance from the CG to the point of rotation
Fundamental work energy relationship
The work done on a translating object where the force is in the direction of motion is equal to the change in kinetic energy of that object. This is given by
F*S=(W/2g)(Vf2-Vi2)
Work energy with variable force
Work = kS2/2 where k is the spring constant and S is the spring’s displacement.
Spring constants in parallel
if springs are in parallel then kf= Σ ki
Spring constants in a series
For springs in a series (1/kf) = Σ(1/ki)
Frequency of oscillation: springs
ω = +-[2k/m].5 and f (1/s) = +-(1/2pi)[2k/m].5
Period of oscillating springs
T = 1/f = 2pi[m/2k].5
Conservative Systems
These are systems that lack friction or other dissipative forces. Therefore
KE + PE = K + U = c
and (d/dt)(KE+PE) = 0
Solving translating systems
- ) FBD
- ) ID known forces, unknown forces, and what you are attempting to solve for
- ) ΣF = ma = external forces
Oscillations: Steps to solve via energy
- ) FBD
- ) if the system is conservative then
PE = W*h + kx2/2 = Wg(L-Lcosθ) + (k/2)(bsinθ)2
KE = (1/2)mV2= (1/2) m (ωL)2 where ω= dθ/dt
- ) d/dt(PE+KE)=0 and for small θ sinθ=θ and cos=1
- ) Simply and (1/2pi)*(coefficient of θ) = f
Momentum
Momentum is p=m*V and because dV/dt=a then F=dp/dt where p is a vector
Impulse
This is a force that acts on an object for a short duration of time such that F(t)dt = impulse force.
∫t0t F(t) = (V-V0)m = ∫vv mdV
Force vs. time curves
If force = F = f(t) then (Σ ∫F(t))/m = dV over the period
Conservation of Momentum
IF there are no external forces acting on a system then the internal components of a system have the property where m1 * V1 = m2 * V2
It generally works when dt~0 but there is a dV
Conservation of Momentum: applications
conservation of momentum should be used when there is an impulse and a change in mass/velocity that occurs over dt~0
Ex: a bullet hitting a pendulum
How to solve a bullet hitting a pendulum
- ) apply the conservation of momentum to find v
- ) assuming it is a conservative system then KE1 + PE1 = KE2 + PE2
Impacts
These are when two objects collide and it is assumed that the time over which the collision takes place ~0. Therefore conservation of momentum and impulse physics applies.
Elastic impact
This assumes that there is ~0 loss of energy due to deformation and impact is over a time interval =0
Coefficient of restitution
this is a constant that is used to adjust for deformation upon impact of otherwise elastic objects such that
e = (relative V before)/(relative V after) = separation V/approach V where the relative nature of the velocity is described as the absolute value of their convergence/divergence
Coefficient of restitution when collision is non-parallel
When collisions are non parallel but both are spheres you can break the problem down into the x and y componenets where the y component does not change unless an external force acts upon it. Therefore the coefficient of restitution only applies to the x velocities.
Solving Vi and Vf with e and nonparallel collisions
- ) Use conservation of momentum in the x and y directions. Vy does not change.
- ) e = abs(V2/V1) (summate for each object)
- ) separate e so that one of the two terms is the unknown in the conservation of momentum.
- ) You should now have two equations and two unknowns. Solve
- ) Use the resulting vs initial velocities to find changes in momentum, KE, or other
Solving plane surfaces below water
Use pressure prisms and remember that the Fr (fluid) does NOT go through the CG of the plane.
Solving manometer/barometer problems
- ) Choose an A and B point
- ) move from point A to B where when moving down you add P= γh
- ) Remeber that equal heights are of equal pressures
Center of pressure
When an object is submerged the center of pressure measured from the fluid surface is given by Xc = ∫ (X2 dA) / ∫(X dA)
or through using the parallel axis theorem: Xc-Xo=Io/AXo where Xo is the CG measured from the free surface
Tension in a thin-walled pressure cylinder
There are generally two stresses acting on the walls. There is the hoop stress σ1 and the longitudinal stress, σ3.
Note: at the boundary of the pressure and the wall there is a pressure that acts opposite to the internal pressure. On the outer edge there is no pressure.
Hoop stress formula
If the cylinder’s wall is >(1/10)*diameter then σ = pr/t where t is the thickness, p is the pressure, and r is the radius
Thick walled cylinders
These are cylinders where the wall is thicker than (1/10)*diameter so that the hoop stress within the material varies with radial position.
f= [(ri2 + ro2)/(ro2-ri2)] = inner σ/internal p = σ/p
Buoyancy force
This is the force of the displaced fluid. It acts through the center of gravity of the displaced fluid in the up direction
Steps to solve: dams and gates
- ) FBD
- ) find all relevant weights, hydrostatic forces, and buoyant forces and the places that they are acting on the other object.
- ) summate forces to find reactions
- ) summate moments to find positions
Bernoulli’s theorem
if a fluid is both inviscid (frictionless) and incompressible
then E= (P/γ) + z + V2/2g
Energy Equation
The energy equation is the equivalent to bernoullis when there is head loss or energy is added to the system. To use identify to place where the system has less energy to add loss. Head loss and power is in terms of feet of energy.
(P1-P2)/γ + (z1-z2) + (V12 - V22)/2g = HL - Hp
Head loss due to friction
Frictional head loss is proportional to the length of the pipe. Inversely proportional to the diameter to some power because SA = f(d2). Varies as a function of the roughness of the pipe and flow turbulence. Varies with velocity like solids but is independent of pressure.
hf = f (L/d) (V2/2g) where L is the pipe length, f is the friction factor, V is the average velocity ( ∫ V dA/ ∫dA)
Orifice coefficients
This is a result of the vena contracta effect where the change in momentum causes the discharge of flow to have a smaller diameter than the outlet. The smallest diameter in comparison to the orifice diameter is related by the coefficient of contraction.
How to use the coefficient of contraction
Cc = smallest area of jet/ area of orifice
Q for fluids
Q is the volumetric flowrate given in ft3/s
velocity from a orifice
V = Cv [2gh].5 where Cv is the velocity coefficient (usually .98) and h is the distance from the free surface to the discharge.
This only works if there is not a turbine or other thing being moved by the flow.
Minor losses
These are losses due to changes in the geometry of the flow (conservation of momentum requires external forces)
HL = Σ ksys (V2/2g)
Viscosity
The viscosity of a fluid is defined by its internal resistance to flow.
dynamic viscosity
This is the ratio of applied shear to the change in velocity throughout a fluid.
μ = τ /(dV/dy) where y is the vertical distance from the stationary surface to the point of interest within the fluid.
Kinematic viscosity
This is ν = μ/ρ and has units of L2/s. It represents the viscosity of a fluid without relating it to the density of the fluid.
Newtonian fluids
These are common fluids where viscosity is independent of shear.
non-newtonian fluids
for a non-Newtonian fluid the viscosity depends on the rate of applied shear such that it becomes more viscous as the shear increases
Reynolds number (pipes + open channels)
Pipes: Re = ρVd/μ = Vd/ν
Open channels: VL/ν where L is the characteristic length
The reynolds number is related to whether a flow is considered turbulent or laminar which dictates whether the flow lines of the fluid move in parallel or at random.
pump head for a pump moving a material to a resivor with a higher z
Work/ γ * Q = (z2-z1)+(P2-P1)/γ + (V22-V12)/2g +hl
Where: hl = f(l/D)(V2/2g) where V is the V exiting the pump and throughout the pipe of length l and diameter d
hl can also be minor losses with are equal to the summation of the loss coefficients times V2/2g
What is Q
Q is the volumetric flow rate it is given in cubic feet/sec
How to solve multi-pipe problems
- ) draw diagram and ID end states for P, V, Z, l, and D + unknowns
- ) write energy equation from 1->2 , 1-> 3… where hL = f1(l1/d1)(V12/2g) + f2(l2/d2)(V22/2g) + minor losses. This is added to the second state of the fluid.
- ) Conservation of matter says that Qin = Qout
- ) Solve
Pitot Tube
This is a L-shaped tube that measures the stagnation pressure of a moving flow. If we consider h to be the difference in heights of a fluid in a pitot tube and a static pressure tube (vertical tube) then Vflow= [2gh].5