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ES 201 Statics

Review for Exam #2 Chapter 4-5

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Homework #11

5–2. Draw the free-body diagram of member AB, which issupported by a roller at A and a pin at B. Explain thesignificance of each force on the diagram.

NA – Force on roller ABx, By – Forces of pin at B

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Homework #11

5–7. Draw the free-body diagram of the “spanner wrench” subjected to the 20-lb force. The support at A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of each force on the diagram.

NB – Force on Wrench at BAx, Ay – Forces of pin at A

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Homework #12

5–22. The articulated crane boom has a weight of 125 lb andcenter of gravity at G. If it supports a load of 600 lb, determine the force acting at the pin A and the force in the hydraulic cylinder BC when the boom is in the position shown.

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Homework #12

5–34. Determine the horizontal and vertical componentsof reaction at the pin A and the normal force at the smoothpeg B on the member.

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MOMENT OF A FORCE - SCALAR FORMULATION

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

Then 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.

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

Fa

b

dO

abO

F

F x

F y

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CROSS PRODUCT (Section 4.2)

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 B sin uC

As shown, uC is the unit vector perpendicular to both A and B vectors (or to the plane containing the A and B vectors).

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CROSS PRODUCT (continued)

Cartesian Vector Formulation:

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

= Ax Bx (i i ) + Ax By (i j ) + Ax Bz (i k)

+ Ay Bx (j i ) + Ay By (j j ) + Ay Bz (j k)

+ AzBx (k i ) + Az By (k j ) + AzBz (k k)

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

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CROSS PRODUCT (continued)

Also, the cross product can be written as a determinant.

Each component can be determined using 2 2 determinants.

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MOMENT OF A FORCE – VECTOR FORMULATION (Section 4.3)

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.

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MOMENT OF A FORCE – VECTOR FORMULATION (continued)

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

By expanding the above equation using 2 2 determinants,

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

If a body is acted upon by several forces, then the resultant moment of the forces about point O can be determined by vector addition of the moment of each force: MRO = ∑( r x F).

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VECTOR ANALYSIS

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

MO = r F

Now, find the component of MO along the axis a’- a using the

dot product.

Ma’-a = ua • MO

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

(Section 4.5)

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VECTOR ANALYSIS (continued)

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.

Ma’- a can also be obtained as

The above equation is also called the triple scalar product.

MOMENT OF A COUPLE

(Section 4.6)

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 –F to the line of action of F.

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

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MAGNITUDE OF RESULTANT FORCE

The net force on the beam is given by

+ FR = L dF = L w(x) dx = A

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

Consider an element of length dx.

The force magnitude dF acting on it is given as

dF = w(x) dx

(Section 4.9)

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LOCATION OF THE RESULTANT FORCE

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

The force dF will produce a moment

of (x)(dF) about point O.

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LOCATION OF THE RESULTANT FORCE

(continued)

Comparing the last two equations, we get

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

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EXAMPLE

Until you learn more about centroids, we will consider only rectangular and triangular loading diagrams whose centroids are well defined and shown on the inside back cover of your textbook.

Note that triangle presents a bit of a challenge but still is pretty straightforward.

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CONDITIONS FOR RIGID-BODY EQUILIBRIUM (Section 5.1)

Forces on a particle

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 (no translation)

and MO = 0 (no rotation)Forces on a rigid body

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FREE-BODY DIAGRAMS (Section 5.2)

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

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

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

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SUPPORT REACTIONS IN 2-D

As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction.

Similarly, if rotation is prevented, a moment is exerted on the body in the opposite direction.

A few example sets of diagrams are shown above. Other support reactions are given in your textbook (Table 5-1).

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EQUATIONS OF EQUILIBRIUM (Section 5.3)

Fx = 0 Fy = 0 MO = 0

where point O is any arbitrary point.

Please note that these equations are the ones most commonly used for solving 2-D equilibrium problems.

A body is subjected to a system of forces that lie in the x-y plane. 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:

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TWO-FORCE MEMBERS

(Section 5.4)

If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B.

The solution to some equilibrium problems can be simplified if we recognize members that are subjected to forces at only two points (e.g., at points A and B).

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STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS

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

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

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

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1) Resolve the 100 N force along the x and y axes.

2) Determine MO using a scalar analysis for the two force components and add those two moments together.

EXAMPLE #1

Given: A 100 N force is applied to the frame.

Find: The moment of the force at point O.

Plan:

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EXAMPLE

Given: A force is applied to the tool to open a gas valve.

Find: The magnitude of the moment of this force about the z axis of the value.

Plan:

1) Use Mz = u • (r F).

2) Note that u = 1 k.

3) The vector r is the position vector from A to B.

4) Force F is already given in Cartesian vector form.

A

B

A

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EXAMPLE

Given: The loading on the bottom of the slab has an equivalent resultant force is equal but opposite to the resultant of the distributed loading acting on the top of the plate.

Find: The intensities of w1 and w2 .

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GROUP PROBLEM SOLVING

Given:Spring CD remains in the horizontal position at all times due to the roller at D. If the spring is unstretched when θ=0° and the bracket achieves its equilibrium position when θ=30°.

Find: the stiffness k of the spring and the horizontal and vertical components of reaction at pin A.

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No Homework !Prepare for Exam #2