Rechenübung zu Allgemeine Technische Mechanik 1

Robert Werner

Inhalt:

• Einführung in die Vektorrechnung (Grundlagen)

• Theorie zur Statik (Teil 1)

• Theorie zur Kinematik und Kinetik (Teil 2)

Einführung in die Vektorrechnung (Grundlagen)

3-D CARTESIAN VECTOR TERMINOLOGY

Consider a box with sides Ax, Ay, and Az meters long.

The vectorA can be defined asA = (Ax i + Ay j + Az k) m

The projection of the vectorA in the x-y plane is A´. The magnitude of this projection, A´, is found by using the same approach as a 2-D vector: A´ = (Ax

2 + Ay2)1/2 .

The magnitude of the position vectorA can now be obtained as

A = ((A´)2 + Az2) ½ = (Ax

2 + Ay2 + Az

2) ½

A UNIT VECTOR

Characteristics of a unit vector:a) Its magnitude is 1.

b) It is dimensionless.

c) It points in the same direction as the original vector (A).

The unit vectors in the Cartesian axis system are i, j, and k. They are unit vectors along the positive x, y, and zaxes respectively.

For a vector A with a magnitude of A, an unit vector is defined as UA = A / A .

The direction or orientation of vectorA is defined by the angles α, β, and γ.These angles are measured between the vector and the positive x, y and z axes, respectively. Their range of values are from 0° to 180°

Using trigonometry, “direction cosines” are found using the formulas

These angles are not independent. They must satisfy the following equation.

cos ²α + cos ²β + cos ²γ = 1

This result can be derived from the definition of a coordinate direction angles and the unit vector. Recall, the formula for finding the unit vector of any position vector:

or written another way, u A = cosα i + cosβ j + cosγ k .

TERMS (continued) DEFINITION

The dot product of vectors A and B is defined as A•B = A B cosθ.

Angle θ is the smallest angle between the two vectors and is always in a range of 0º to 180º.

Dot Product Characteristics:1. The result of the dot product is a scalar (a positive or

negative number).

2. The units of the dot product will be the product of the units of the A and B vectors.

DOT PRODUCT DEFINITON (continued)

Examples: i • j = 0

i • i = 1

A • B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k)

= Ax Bx + AyBy + AzBz

USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS

For the given two vectors in the Cartesian form, one can find the angle bya) Finding the dot product, A • B = (AxBx + AyBy + AzBz ),

b) Finding the magnitudes (A & B) of the vectors A & B, and

c) Using the definition of dot product and solving for θ, i.e.,

θ = cos-1 [(A • B)/(A B)], where 0º ≤ θ ≤ 180º .

DETERMINING THE PROJECTION OF A VECTOR

You can determine the components of a vector parallel and perpendicular to a line using the dot product.

Steps:

1. Find the unit vector, Uaa along line aa

2. Find the scalar projection of A along line aa by

A|| = A • U = AxUx + AyUy + Az Uz

3. If needed, the projection can be written as a vector, A|| , by using the unit vector Uaa and the magnitude found in step 2.

A|| = A|| Uaa

4. The scalar and vector forms of the perpendicular component can easily be obtained by A⊥ = (A 2 - A||

2) ½ and A ⊥⊥⊥⊥ = A – A||

(rearranging the vector sum of A = A⊥⊥⊥⊥ + A|| )

DETERMINING THE PROJECTION OF A VECTOR (continued)

MOMENT IN 2-D

The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque).

MOMENT IN 2-D (continued)

In the 2-D case, the magnitudeof the moment is MO = F d

As shown, d is theperpendiculardistance from point O to the line of actionof the force.

In 2-D, the directionof MO is either clockwise or

counter-clockwise depending on the tendency for rotation.

MOMENT IN 2-D (continued)

For example, MO = F d and the direction is counter-clockwise.

Often it is easier to determine MO by using the components of Fas shown.

Using this approach, MO = (Fy a) – (Fx b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinnedat O and deciding which way the body would rotate because of the force.

Fa

b

dO

abO

F

Fx

Fy

MOMENT IN 3-D

Moments in 3-D can be calculated using scalar (2-D) approach but it can be difficult and time consuming. Thus, it is often easier to use a mathematical approach called the vector cross product.

Using the vector cross product,MO = r ×××× F .

Here r is the position vector from point O to any point on the line of action of F.

CROSS PRODUCT

In general, the cross product of two vectors A and B results in another vector C , i.e., C = A ×××× B. The magnitude and direction of the resulting vector can be written as

C = A ×××× B = A Bsin θ UC

HereUC is the unit vector perpendicular to both A and Bvectors as shown (or to the plane containing theA and B vectors).

CROSS PRODUCT

The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product.

For example: i ×××× j = k

Note that a vector crossed into itself is zero, e.g., i ×××× i = 0

CROSS PRODUCT (continued)

Of even more utility, the cross product can be written as

Each component can be determined using 2 × 2 determinants.

MOMENT IN 3-D (continued)

So, using the cross product, a moment can be expressed as

By expanding the above equation using 2 × 2 determinants, we get:

MO = (ry Fz - rz Fy) i - (rx Fz - rz Fx ) j + (rx Fy - ry Fx ) k

The physical meaning of the above equation becomes evident by considering the force components separately and using a 2-D formulation.

VECTOR ANALYSIS

Our goal is to find the moment of F(the tendency to rotate the body) about the axis a’-a.

First compute the moment of F about any arbitrary point Othat lies on the a’-a axis using the cross product.

MO = r ×××× FNow, find the component of MO along the axis a’-a using the dot product.

Ma = ua • MO

VECTOR ANALYSIS (continued)

Ma can also be obtained as

The above equation is also called the triple scalar product.

In the this equation,

ua represents the unit vector along the axis a’-a axis,

r is the position vector from any point on the a’-a axis to any point A on the line of action of the force, and

F is the force vector.

MOMENT OF A COUPLE

A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d.

The moment of a couple is defined as

MO = F d (using a scalar analysis) or as

MO = r ×××× F (using a vector analysis).

Here r is any position vector from the line of action of –Fto the line of action ofF.

MOMENT OF A COUPLE (continued)

The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F d

Moments due to couples can be added using the same rules as adding any vectors.

Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body.

Theorie zur Statik (Teil 1)

THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS

For analyzing an actual physical system, first we need to create an idealized model.Then we need to draw a free-body diagram showing all the external (active and reactive) forces.Finally, we need to apply the equations of equilibriumto solve for any unknowns.

STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS

1. If not given, establish a suitable x - ycoordinate system.

2. Draw a free body diagram (FBD) of the object under analysis.

3. Apply the three equations of equilibrium (EofE) to solve for the unknowns.

IMPORTANT NOTES

1. If we have more unknowns than the number of independent equations, then we have a statically indeterminate situation.We cannot solve these problems using just statics.

2. The order in which we apply equationsmay affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force, then solving ∑ Fx = 0 first allows us to find the horizontal unknown quickly.

3. If the answerfor an unknown comes out asnegative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.

PROCEDURE FOR DRAWING A FREEBODY DIAGRAM

1. Draw an outlined shape.Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape.

2. Show all the external forces and couple moments.These typically include: a) applied loads, b) support reactions, and, c) the weight of the body.

Idealized model Free body diagram

PROCEDURE FOR DRAWING A FREE BODY DIAGRAM (continued)

3. Label loads and dimensions:All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA, etc.. Indicate any necessary dimensions.

Idealized model Free body diagram

CONDITIONS FOR RIGID-BODY EQUILIBRIUM

In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces).

For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.

∑ F = 0 and ∑ MO = 0

Forces on a rigid body

Forces on a particle

EQUATIONS OF EQUILIBRIUM

A body is subjected to a system of forces that lie in the x-yplane. When in equilibrium, the net force and net moment acting on the body are zero. This 2-D condition can be represented by the three scalar equations:

∑ Fx = 0 ∑ Fy = 0 ∑ MO = 0

Where point O is any arbitrary point.

Please notethat these equations are the ones most commonlyusedfor solving 2-D equilibrium problems. There are two other sets of equilibrium equations that are rarely used. For your reference, they are described in the textbook.

x

y

F1

F2

F3

F4

O

EQULIBRIUM EQUATIONS IN 3-D

As stated earlier, when a body is in equilibrium, the net force and the net moment equal zero, i.e., ∑ F = 0 and ∑ MO = 0 .

These two vector equations can be written as six scalar equations of equilibrium (EofE).These are

∑ Fx = ∑ Fy = ∑ Fz = 0

∑ Mx = ∑ My = ∑ Mz = 0

The moment equations can be determined about any point. Usually, choosing the point where the maximum number of unknown forces are present simplifies the solution.Those forces do not appear in the moment equation since they pass through the point. Thus, they do not appear in the equation.

CONSTRAINTS FOR A RIGID BODY

Redundant Constraints: When a body has more supports than necessary to hold it in equilibrium, it becomes statically indeterminate.A problem that is statically indeterminatehas more unknowns than equations of equilibrium.

Are statically indeterminate structures used in practice? Why or why not?

IMPROPER CONSTRAINTS

In some cases, there may be as many unknown reactions as there are equations of equilibrium. However, if the supports are not properly constrained, the body may become unstable for some loading cases.

Here, we have 6 unknowns but there is nothing restricting rotation about the x axis.

CG / CM FOR A SYSTEM OF PARTICLES

Consider a system of n particles as shown in the figure. The net or the resultant weight is given as WR = ∑W.

Similarly, we can sum moments about the x and z-axes to find the coordinates of G.

By replacing the W with a M in these equations, the coordinates of the center of mass can be found.

Summing the moments about the y-axis, we get

x WR = x1W1 + x2W2 + ……….. + xnWn

where x1 represents x coordinate of W1, etc..

CG / CM / CENTROID OF A BODY

A rigid body can be considered as made up of an infinite number of particles. Hence, using the same principles as in the previous slide, we get the coordinates of G by simply replacing the discrete summation sign ( ∑ ) by the continuous summation sign ( ∫ ) and Wby dW.

Similarly, the coordinates of the center of mass and the centroidof volume, area, or length can be obtained by replacing W by m, V, A, or L, respectively.

DISTRIBUTED LOADING

In many situations a surface area of a body is subjected to a distributed load. Such forces are caused by winds, fluids, or the weight of items on the body’s surface.

We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body.

In such cases, w is a function of xand has units of force per length.

MAGNITUDE OF RESULTANT FORCE

Consider an element of length dx.

The force magnitude dF acting on it is given as

dF = w(x) dx

The net forceon the beam is given by

+ ↓ FR = ∫L dF = ∫L w(x) dx = A

Here A is the area under the loading curve w(x).

LOCATION OF THE RESULTANT FORCE

The force dF will produce a moment of (x)(dF) about point O.

The total moment about point O is given as

+ MRO = ∫L x dF = ∫L x w(x) dx

Assuming that FR acts at , it will produce the moment about point O as

+ MRO = ( ) (FR) = ∫L w(x) dx

x

xx

Comparing the last two equations, we get

LOCATION OF THE RESULTANT FORCE (continued)

You will learn later that FR acts through a point “C,” which is called the geometric center or centroid of the area under the loading curve w(x).

Zur richtigen Auslegung der Streben dieser Brücke müssen die in ihnen wirkenden Kräfte bestimmt werden. Ein Fachwerk ist eine Konstruktion aus schlanken, miteinander verbundenen Teilen. Die Verbindung der Holzstreben oder Metallstäbe erfolgt über Schrauben oder Schweißverbindungen mit einer Platte, dem sogenannten Knotenblech (a). Man kann auch einen Bolzen oder einen Stift durch jedes Teil stecken (b).

Ein typisches räumliches Fachwerk, das ein Dach trägt. Kugelgelenke verbinden die Stäbe.

CHARACTERISTICS OF DRY FRICTION

Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body.

Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion.

For the body shown in the figure to be in equilibrium, the following must be true: F = P, N = W, and Wx= Ph.

CHARACTERISTICS OF FRICTION (continued)

To study the characteristics of the friction force F, let us assume that tipping does not occur (i.e., “h” is small or “a” is large). Then we gradually increase the magnitude of the force P. Typically, experiments show that the friction force F varies with P, as shown in the left figure above.

FRICTION CHARACERISTICS (continued)

The maximum friction forceis attained just before the block begins to move (a situation that is called “impending motion”). The value of the force is found using Fs = µs N, where µs is called the coefficient of static friction. The value of µsdepends on the materials in contact.

Once the block begins to move, the frictional force typically drops and is given by Fk = µk N. The value of µk

(coefficient of kinetic friction) is less than µs .

DETERMING µµµµs EXPERIMENTALLY

A block with weight w is placed on an inclined plane. The plane is slowly tilted until the block just begins to slip.

The inclination, θs, is noted. Analysis of the block just before it begins to move gives (using Fs = µs N):

+ ∑ Fy = N – Wcosθs = 0

+ ∑ Fx = µs N – Wsin θs = 0

Using these two equations, we get:

µs = (Wsin θs ) / (Wcosθs ) = tan θsThis simple experiment allows us to find the µs between two materials in contact.

PROCEDURE FOR ANALYSIS

Steps for solving equilibrium problems involving dry friction:

1. Draw the necessary free body diagrams. Make sure that you show the friction force in the correct direction(it always opposes the motion or impending motion).

2. Determine the number of unknowns. Do not assumeF = µs N unless the impending motion condition is given.

3. Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns.

Theorie zur Kinematik und Kinetik – Dynamik (Teil 2)

MoI – DEFINITION

For the differential area dA, shown in the figure:

d Jx = y2 dA ,d Jy = x2 dA , and,d JO = r2 dA , where JO is the polar

moment of inertia about the pole O or zaxis.

The moments of inertia for the entire area are obtained by integration.

Jx = ∫A y2 dA ; Jy = ∫A x2 dA

JO = ∫A r2 dA = ∫A ( x2 + y2 ) dA = Jx + Jy

The MoI is also referred to as the second moment of an areaand has units of length to the fourth power (m4).

RADIUS OF GYRATION OF AN AREA

For a given area A and its MoI, Jx , imagine that the entire area is located at distance kx from the x axis.

2Then, Jx = kx A or . This kx

is called the radius of gyration of the area about the x axis. Similarly;

and

The radius of gyration has units of length and gives an indication of the spread of the area from the axes. This characteristic isimportant when designing columns.

A

kx

x

y

O O /k J m=y y /k J m=

x x /k J m=

PARALLEL-AXIS THEOREM FOR AN AREA

This theorem relates the moment of inertia (MoI) of an area about an axis passing through the area’s centroidto the MoI of the area about a corresponding parallel axis. This theorem has many practical applications, especially when working with composite areas.

Consider an area with centroidC. The x' and y' axes pass through C. The MoI about the x-axis, which is parallel to, and distance dy

from the x' axis, is found by using the parallel-axis theorem.

PARALLEL-AXIS THEOREM (continued)

Using the definition of the centroid:

y‘ = (∫A y' dA) / (∫A dA) . Now

since C is at the origin of the x' – y' axes,

y' = 0 , and hence ∫A y' dA = 0 .

Thus Jx = Jx’ + A dy2

Jx = ∫A y2 dA = ∫A (y' + dy)2 dA

= ∫A y' 2 dA + 2 dy ∫A y' dA + dy2 ∫A dA

y = y' + dy

Similarly, Jy = Jy’ + A dx2 and

JO = JC + A d2

STEPS FOR ANALYSIS

1. Divide the given areainto its simpler shaped parts.

2. Locate the centroid of each part and indicate the perpendicular distancefrom each centroid to the desired reference axis.

4. The MoI of the entire area about the reference axis is determined by performing an algebraic summationof the individual MoIs obtained in Step 3. (Please note that MoI of a hole is subtracted).

3. Determine the MoI of each “simpler” shaped part about the desired reference axis using the parallel-axis theorem ( Jx = Jx’ + A dy

2 ) .

CONCEPT OF THE MMI

Consider a rigid body with a center of mass at G. It is free to rotate about the zaxis, which passes through G. Now, if we apply a torque M about the z axis to the body, the body begins to rotate with an angular acceleration α .

M and α are related by the equation M = I α . In this equation, I is the mass moment of inertia (MMI)about the z axis.

The MMI of a body is a property that measures the resistance of the body to angular acceleration. This is similar to the role of mass in the equation F = m a. The MMI is often used when analyzing rotational motion (done in dynamics).

G·

M

DEFINITION OF THE MMI

Consider a rigid body and the arbitrary axis p shown in the figure. The MMI about the p axis is defined as I = ∫m r2 dm, where r, the “moment arm,” is the perpendicular distance from the axis to the arbitrary element dm.

The MMI is always a positive quantity and has a unit of kg ·m2

or slug · ft2.

p

RELATED CONCEPTS

Parallel-Axis Theorem:Just as with the MoI for an area, the parallel-axis theorem can be used to find the MMI about a parallel axis p’ that is a distance d from the axis through the body’s center of mass G. The formula is Ip’ = IG + m d2 (where m is the mass of the body).

Theradius of gyrationis similarly defined as

Finally, the MMI can be obtained by integration or by the method for composite bodies. The latter method is easier for many practical shapes.

·G

p’d

m

/k I m=