Lecture 22Exam 3 ReviewIntroduction to Engineering Analysis

4-9-18

Test 3

• Wednesday 4/11/2018o 8:00 – 9:50 am

• Covers lectures 14 - 21

• Sec-01 DARRIN 308• Sec-02 Darrin 318 – This is not a typo• Sec-03 DARRIN 308• Sec-04 DARRIN 308• Sec-05 DARRIN 308

Test 3 Rooms

Test 3Four problems

1. Rigid Body Equilibrium (3D) 25%2. Comp. Areas and/or Distr. Loads 25%3. Trusses (MoJ/MoS) 25%4. Linear Algebra 25%

Test 3• Closed book and closed notes. • For maximum credit, show all set-ups and all details

necessary (see syllabus).• You can use only a hand-held calculator for calculations

(no smartphone, laptop, etc.). • Cell phones, audio devices, headphones and laptops must

be off and stored away. Any student found deviating from this policy will be asked to leave the exam room, given a grade of zero, and no retest will be allowed.

ALAC Review Session

• Location : DCC 330• Time : 8 PM until 10 PM• Date : April 9th, 2018

Students entitled to extra time should remain in test room until escorted by a TA to a different room (DCC 239) to finish their test.

Test 3

Grade-Challenging Sessions

• Monday 4/16/18 in DARRIN 236• Tuesday 4/17/18 in DARRIN 236• 6:00 – 8:00 pm

Makeup Test 3• Only students with a valid reason and a

note from the office of the Dean of Students will be allowed to take it:o A valid reason is approved/provided only by

the office of the dean of students.o “My alarm did not work” or “ I thought that the test

starts at 9:00” etc. are not valid reasons.

• If you miss the test without a valid reason your grade will be zero.

• No makeup for the makeup test

Makeup Test 3

• Wednesday 4/18/2018• 5:00 – 6:50 pm• Room: DCC 236 – Change from

previous exams

3-D Force Balance Diagrams

SUPPORT REACTIONS IN 2-D

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

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

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

SUPPORT REACTIONS IN 3-D (Table 5-2)

As a general rule, if a support prevents translation of a body in a given direction, then a reaction force acting in the opposite direction is developed on the body. Similarly, if rotation is prevented, a couple moment is exerted on the body by the support.

A few examples are shown above. Other support reactions are given in your text book (Table 5-2).

EQUATIONS OF EQUILIBRIUM (Section 5.6)

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 (E-of-E). 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. Any forces occurring at the point where moments are taken do not appear in the moment equation since they pass through the point.

EXAMPLE I

a) Use the established x, y and z axes.b) Draw a FBD of the rod.c) Write the forces using scalar equations.d) Apply scalar equations of equilibrium to solve for the

unknown forces.

Given: The rod, supported by thrust bearing at A and cable BC, is subjected to an 80 lb force.

Find: Reactions at the thrust bearing A and cable BC.

Plan:

EXAMPLE I (continued)FBD of the rod:

Applying scalar equations of equilibrium in appropriate order, we get

∑ F X = AX = 0; AX = 0

∑ F Z = AZ + FBC – 80 = 0;

∑ M Y = – 80 ( 1.5 ) + FBC ( 3.0 ) = 0;

Solving the last two equations: FBC = 40 lb, AZ = 40 lb

EXAMPLE I (continued)FBD of the rod

M X = ( MA) X + 40 (6) – 80 (6) = 0 ; (MA ) X= 240 lb ft CCW

∑ M Z = ( MA) Z = 0 ; (MA ) Z= 0

= 40 lb

Now write scalar moment equations about what point? Point A!

ExampleBar AC rests against a smooth surface at end C and is supported at end A with a ball-and-socket joint. The cable at B is attached midway between the ends of the bar. Determine the reactions at supports A and C and the tension in the cable at B.

Az

AyAx

C

TBD

FBE

Z

yx

We need a FBD of this system. We take the pen looking piece extending from A to C, and identify the forces acting on it

Az

AyAx

C

TBD

FBE

Z

yx

Next, let us write an expression that captures the force experienced in the string extending from B to E:

�⃑�𝐹𝐵𝐵𝐵𝐵 = 8000.4 ̂𝚤𝚤 − 0.4 ̂𝚥𝚥 − 0.6�𝑘𝑘

0.4 2 + −0.4 2 + −0.6 2N

Following the rules for this course with regard to precision, this simplifies to

�⃑�𝐹𝐵𝐵𝐵𝐵 = 388.1 ̂𝚤𝚤 + −388.1 ̂𝚥𝚥 + −582.1 �𝑘𝑘 N

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 × �⃑�𝐹𝐵𝐵𝐵𝐵

What is a position vector that brings us from A to B?

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 = 0.4 ̂𝚤𝚤 + 0.8 ̂𝚥𝚥 + (−0.6)�𝑘𝑘

And we already wrote a vector for �⃑�𝐹𝐵𝐵𝐵𝐵 :

�⃑�𝐹𝐵𝐵𝐵𝐵 = 388.1 ̂𝚤𝚤 + −388.1 ̂𝚥𝚥 + −582.1 �𝑘𝑘 N

So, execute the cross product:

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 × �⃑�𝐹𝐵𝐵𝐵𝐵 =̂𝚤𝚤 ̂𝚥𝚥 �𝑘𝑘

0.4 0.8 −0.6388.1 −388.1 −582.1

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 × �⃑�𝐹𝐵𝐵𝐵𝐵 =̂𝚤𝚤 ̂𝚥𝚥 �𝑘𝑘

0.4 0.8 −0.6388.1 −388.1 −582.1

This becomes:

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 × �⃑�𝐹𝐵𝐵𝐵𝐵 = −698.5 ̂𝚤𝚤 + (−465.7)�𝑘𝑘

We now have three expressions for the moments created about point A:

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 × �⃑�𝐹𝐵𝐵𝐵𝐵 = −698.5 ̂𝚤𝚤 + (−465.7)�𝑘𝑘

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐵𝐵 × 𝑇𝑇𝐵𝐵𝐵𝐵 = (−.4457𝑇𝑇𝐵𝐵𝐵𝐵) ̂𝚥𝚥 + (0.5542𝑇𝑇𝐵𝐵𝐵𝐵)�𝑘𝑘

𝑟𝑟𝐴𝐴 𝑡𝑡𝑡𝑡 𝐶𝐶 × 𝐶𝐶 = 1.6𝐶𝐶𝑧𝑧 ̂𝚤𝚤 + (−0.8𝐶𝐶𝑧𝑧) ̂𝚥𝚥

Now, add all the ̂𝚤𝚤 pieces, or components:

−698.5 + 1.6𝐶𝐶𝑧𝑧 = 0 = �𝑀𝑀𝐴𝐴𝐴𝐴

From this line, we find that 𝐶𝐶𝑧𝑧 is equal to 437 N. (but use 436.6 for further calculations)

Az

AyAx

C

TBD

FBE

Z

yx

The last step involves finding the reaction force(s) at “A”: Use the other forces already written and add up the components for each direction:

�𝐹𝐹𝐴𝐴𝐴𝐴 = 388.1 + −0.3714 𝑇𝑇𝐵𝐵𝐵𝐵 + 𝐴𝐴𝐴𝐴

�𝐹𝐹𝐴𝐴𝐴𝐴 = −388.1 + 0.7428 𝑇𝑇𝐵𝐵𝐵𝐵 + 𝐴𝐴𝐴𝐴

�𝐹𝐹𝐴𝐴𝑧𝑧 = −582.1 + −0.5571 𝑇𝑇𝐵𝐵𝐵𝐵 + 𝐶𝐶𝑧𝑧 + 𝐴𝐴𝑧𝑧

Plug in 𝑇𝑇𝐵𝐵𝐵𝐵 and 𝐶𝐶𝑧𝑧 and find 𝐴𝐴𝐴𝐴 = −97.1 N𝐴𝐴𝐴𝐴 = −194 N𝐴𝐴𝑧𝑧 = 582 N

Centroids and Distributed Loads

CONCEPT OF CENTROID

The centroid coincides with the center of mass or the center of gravity only if the material of the body is homogenous (density or specific weight is constant throughout the body).

If an object has an axis of symmetry, then the centroid of object lies on that axis.

In some cases, the centroid may not be located on the object.

The centroid, C, is a point defining the geometric center of an object.

DISTRIBUTED LOADING

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

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.

EXAMPLESWe consider today rectangular and triangular loading diagrams whose centroids are well defined and shown on the table posted on LMS. That table also includes centroids for other shapes of lines, areas & volumes.

Look at table posted on LMS. You should find the rectangle and triangle cases. Finding the area of a rectangle and its centroid is easy!

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

EXAMPLES

The rectangular load: FR = 400 × 10 = 4,000 lb and = 5 ft.x

The triangular loading:FR = (0.5) (600) (6) = 1,800 N and = 6 – (1/3) 6 = 4 m. Please note that the centroid of a right triangle is at a distance one third the width of the triangle as measured from its base.

x

Now let’s complete the calculations to find the concentrated loads (which is a common name for the resultant of the distributed load).

SIMPLE TRUSSES (Section 6.1)

If a truss, along with the imposed load, lies in a single plane (as shown at the top right), then it is called a planar truss.

A truss is a structure composed of slender members joined together at their end points.

A simple truss is a planar truss which begins with a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equationM = 2 J – 3.

ANALYSIS & DESIGN ASSUMPTIONS When designing the members and joints of a truss, first it is necessary to determine the forces in each truss member. This is called the force analysis of a truss. When doing this, two assumptions are made:

1. All loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members.

2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases where the joints are formed by bolting the ends together.

With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling.

ZERO-FORCE MEMBERS (Section 6.3)

You can easily prove these results by applying the equations of equilibrium to joints D and A.

If a joint has only two non-collinear members and there is no external load or support reaction at that joint, then those two members are zero-force members. In this example members DE, DC, AF, and AB are zero force members.

Zero-force members can be removed (as shown in the figure) when analyzing the truss.

ZERO – FORCE MEMBERS (continued)

Again, this can easily be proven. One can also remove the zero-force member, as shown, on the left, for analyzing the truss further.

Please note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various different loading conditions.

If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member, e.g., DA.

THE METHOD OF JOINTS (Section 6.2)

When using the method of joints to solve for the forces in truss members, the equilibrium of a joint (pin) is considered. All forces acting at the joint are shown in a FBD. This includes all external forces (including support reactions) as well as the forces acting in the members. Equations of equilibrium (∑ FX= 0 and ∑ FY = 0) are used to solve for the unknown forces acting at the joints.

A free-body diagram of Joint B

STEPS FOR ANALYSIS

1. If the truss’s support reactions are not given, draw a FBD of the entire truss and determine the support reactions (typically using scalar equations of equilibrium).

2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling on the pin) unless you can determine by inspection that the forces are compression loads.

3. Apply the scalar equations of equilibrium, ∑ FX = 0 and ∑FY = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression).

4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined.

THE METHOD OF SECTIONS

In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss.

Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude as the forces at the joint. This result is based on the equilibrium principle and Newton’s third law.

STEPS FOR ANALYSIS

1. Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general).

2. Decide which side of the cut truss will be easier to work with (minimize the number of external reactions).

3. If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E.

STEPS FOR ANALYSIS (continued)

4. Draw the FBD of the selected part of the cut truss. You need to indicate the unknown forces at the cut members. Initially, you may assume all the members are in tension, as done when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per the assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)

5. Apply the scalar equations of equilibrium (E-of-E) to the selected cut section of the truss to solve for the unknown member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut!

STEPS FOR ANALYSIS (continued)

EXAMPLE

a) Take a cut through members KJ, KD and CD.

b) Work with the left part of the cut section. Why?

c) Determine the support reactions at A. What are they?

d) Apply the E-of-E to find the forces in KJ, KD and CD.

Given: Loads as shown on the truss.

Find: The force in members KJ, KD, and CD.

Plan:

EXAMPLE (continued)

Analyzing the entire truss for the reactions at A, we getΣ FX = AX = 0. A moment equation about G to find AY results in:

∑MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = 56.7 kN

Now take moments about point D. Why do this?

+ MD = – 56.7 (9) + 20 (6) + 30 (3) – FKJ (4) = 0FKJ = − 75.1 kN or 75.1 kN ( C )

56.7kN

EXAMPLE (continued)

Now use the x and y-directions equations of equilibrium.

↑ + Σ FY = 56.7 – 20 – 30 – (4/5) FKD = 0;

FKD = 8.38 kN (T)

→ + Σ FX = (– 75.1) + (3/5) ( 8.38 ) + FCD = 0;

FCD = 70.1 kN (T)

56.7 kN

Determinant of a 3x3 matrix

If A is a 3x3 matrix

a11 a12 a13

a21 a22 a23

a31 a32 a33

A=

Then we define

det(A)=a22 a23

a32 a33

= a11

a11 a12 a13

a21 a22 a23

a31 a32 a33

a21 a23

a31 a33

-a12a21 a22

a31 a32

+a13

det(A)=

1 5 -3

1 0 2

3 -1 2

Example 2

0 2

-1 2= 1 1 2

3 2-5 1 0

3 -1-3

=1[0*2 - (-1*2)] –5 [1*2-2*3] - 3[-1*1-0*3]=25

Example 1: Find the inverse of1 2 3

2 5 3

1 0 8A=

1 2 3

2 5 3

1 0 8

1 0 0

0 1 0

0 0 1

* –2 * –1

1 2 3

0 1 -3

0 -2 5

1 0 0

-2 1 0

-1 0 1* 2

1 2 3

0 1 -3

0 0 -1

1 0 0

-2 1 0

-5 2 1* -1

1 2 3

0 1 -3

0 0 1

1 0 0

-2 1 0

5 -2 -1* 3

1 2 3

0 1 0

0 0 1

1 0 0

13 -5 -3

5 -2 -1* -3

1 2 0

0 1 0

0 0 1

-14 6 3

13 -5 -3

5 -2 -1* -2

1 0 0

0 1 0

0 0 1

-40 16 9

13 -5 -3

5 -2 -1

Thus:

A-1=

-40 16 9

13 -5 -3

5 -2 -1

Example: consider the system of linear equations:

x1+2x2+3x3=5

2x1+5x2+3x3=3

x1 +8x3=17

We can write the this system as AX=B

1 2 3

2 5 3

1 0 8A=

x1

x2

x3

X=

5

3

17B=

In example 1 we found that the inverse of A-1 is :

A-1=

-40 16 9

13 -5 -3

5 -2 -1

By theorem 1 the solution of the system is:

X=A-1B=

-40 16 9

13 -5 -3

5 -2 -1

5

3

17

1

-1

2=

or: x1=1, x2=-1, x3=2

Good Luck!

See you bright and early in the DCC at 8 AM Wednesday Morning