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Structural AnalysisDoc Bs Pointers

Dr. David M. Bayer, PEProfessor Emeritus Retired

Last Update: August 31, 2006

Notes:1. These notes are intended to be a resource to the current sections of Structural Analysis at UNC Charlotte. The course instructo

instructions and presentations have precedence over anything presented here.

2. The comments that follow will often be dependent on the fact that we are working with two-dimensional structures.You should make every effort to UNDERSTAND these comments, not to memorize them (that wont work).

UNDERSTANDING

Successful students in this class will obtain a basic understanding of structural analysis of determinateand simple indeterminate structures.

The more you understand freebody diagrams, the better you will perform in this class! Be aware of thisas the course progresses and youll benefit.

Determinate StructuresA structure is determinate if the equations of statics are sufficient to permit a complete analysis(reactions, shears, moments etc).E = modulus of elasticity of the material (steel E = 29,000 ksi); Slope of initial part of stress-straincurveI = Moment of Inertia; property of cross section dimensions (rectangle I = Base*Height3/12)E and I are not required to solve for reactions, shears, moments and axial forces.

Indeterminate Structures:A structure is indeterminate if it cannot be analyzed by the equations of statics alone.E and I are required to solve for reactions, shears, moments and axial forces.

BUT you dont know what the values of I are until you ANALZE the structure and then DESIGN it.Thus, the design of such structures is trial and error make assumptions for the I values, analyze thestructure, design it compare the designed I values to your assumptions make any necessary changesand then analyze it again design it again etc.

Deflections and Rotations:E and I are required to solve for rotations and deflections.

Design (sizing) Structural Members:Cross-section properties are required to solve for tensile, compressive and flexural stresses.These properties include the principal axes moments of inertia (Ix and Iy), cross-section area (A) and

various dimensions.

Linear elastic structures:We have to assume all structures presented in this class will act linearly elastically. For this to be true,the sections used for each member must be designed (sized) to assure linear elasticity that will be donin the follow-up design courses (steel, concrete, timber, masonry etc). The stress-strain diagram forsuch a structure is a straight line.

Page 1 Pointers by Dr. David M. Bayer, Univ of NC at Charlotte Printed: 07/24/12

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Principle of Superposition (Leet page 188):If a structure behaves in a linearly elastic manner, the force of displacement at a particular pointproduced by a set of loads acting simultaneously can be evaluated by adding (superimposing) the forces odisplacements at the particular point produced by each load of the set acting individually.In other words, the response of a linear, elastic structure is the same if all loads are appliedsimultaneously or if the effects of the individual loads are combined.

Expected Competencies at the conclusion of the course:Be able to identify and model beams, trusses and frames.Be able to identify degrees of indeterminacy in structures.Be able to determine support reactions and internal forces in statically determinate structures using themethod of statics.Be able to determine displacements in statically determinate structures using the method of virtual woras well as the method of conjugate beams.Be able to determine support reactions and internal forces in statically indeterminate beams and framesby the slope deflection method and the moment distribution method followed with basic statics.

Thoughts from the old school (suggestions for now from Doc B)

NEVER use a scale as a straightedge. Ask me why. Take off your hat when indoors and especially in the classroom. Basic courtesy. Raise your hand and be recognized before asking a question. Courtesy is a simple expectation from instructor to a student, and vice-versa. Speak loud enough for everyone in the room to hear.

Fifty Three Rules for this class

1. Go to the bathroom before class (especially on quiz days)2. I expect you to UNDERSTAND the procedures in this class NOT just how to do them, but to

UNDERSTAND what you are doing and being able to SHOW me that.3. Staple will be in the upper left corner at an angle of 45 degrees sloping lower left to upper right

4. Do Not write in the left margin of the paper (DNWIM)5. Each homework problem must have a complete statement of the GIVEN and REQUIRED along with

appropriate figures and standard dimensions.6. On each and every freebody used, show all forces required for equilibrium (SAFRFE)7. Define all variables used (UV)8. Show UNITS on all intermediate answers and final answers, all loads and all dimensions (U)9. When you write an equilibrium equation show your sign convention, freebody used (superscript) and

direction or point at which moments are being summed.10. Speak LOUDLY when asking/answering questions in class.11. Equations continued on successive lines should CLEARLY INDICATE that continuation!12. Shear at a concentrated load must be defined by the shear LEFT of that load and the shear RIGHT

of that load.13. My EXPECTATIONS on the final exam are MORE than any other graded material!14. If you dont have enough time to complete all the work required, do as much as you can, BEING SURE

to do it correctly!15. still in progress

BASICS questions, comments, and general course requirements

Why are you pursuing the BSCE degree?Seriously, have you really thought about the responsibilities involved?Do you understand the practice of civil engineering involves very directlythe lives of the public .. people


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your family and mine?With that simple motivator, I would hope you would pursue your courses appropriately ethically!

CODE OF ETHICS

Are you familiar with any Code of Ethics for engineers and engineering? If not, Id strongly encourageyou to visit the ASCE webpages and read the ASCE Code of Ethics there are others and they all havsimilar statements.LINKS:http://www.asce.org/inside/codeofethics.cfmhttp://www.onlineethics.org/

MOST IMPORTANT BASIC CONCEPTS for this course? your final course grade will be directly related to the percentage of HW problems YOU work and

YOU understand! [i.e. HW average very closely approximates your course average] Two basic equations used throughout the course MB

AB= 0 = and Fxwfb= 0 =

(pay attention to the instructions on these two equations and follow them ALWAYS) SAFRFE whenEVER you draw a freebody, Show All Forces Required For Equilibrium! STUBS this concept is critical and if you will pay attention, will save you time and points SCALE not paying attention to the scale of shear and moment diagrams. Scale helps! NOT having enough motivation to want to learn

Biggest mistake made by many students taking this class? NOT taking detailed notes (probably most important thing to do) NOT studying their notes after class reworking them or adding more detail while it is new NOT paying attention in class NOT doing virtually ANY studying outside of class NOT really trying to understand what they are doing NOT doing assigned work homework reading/studying material NOT attending class on a regular basis

SIMPLE STUFF: your name or initials should be included on EVERY PAGE submitted for consideration if you KNOW you have an error on work submitted, if you will TELL me that you know including enoughinformation to convince me that you indeed understood something was wrong, they youll receive morecredit that you would otherwise. if your HOMEWORK grade is to count 15% of your final course grade, and the exam 30-50%, then youHW grade could be more important than a regular quiz grade! Plus, I have absolute proof throughexperience that your HW grade is more often than not, very close to your course grade! This shows thahomework is critical in this class! So why not accept that, and work your own homework problems?

If Dr. Bayer grades your homework problems, the grade WILL count up to 15% towards your course

average. If Dr. Bayer corrects your paper, it will not count towards your grade. If a grader is used,then Dr. Bayer might select one or more problems for a HW set to grade, leaving the others to thegrader. The homework graded by other than Dr. Bayer will not count toward your course average.However, it might effect your final grade borderline grades might be decided by what I describe asHW effort.

If you receive a notation of DIM, G/R, USE, or UFC on any homework problem or quiz, then your gradewill be reduced on all future work where you do not follow the guidelines represented.DIM use proper dimensioning technique (extension lines, dimension lines, arrowheads etc)G/R a complete GIVEN and REQUIRED with the SOLUTION line is required

http://www.asce.org/inside/codeofethics.cfmhttp://www.onlineethics.org/http://www.asce.org/inside/codeofethics.cfmhttp://www.onlineethics.org/

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USE use a straightedge (after receiving this note, you are expected to use a st-edge on all work)UFC use a French curve (after receiving this note, you are expected to use a fr-curve on all work)

NOTE TAKING/STUDYING: Critical! Im absolutely convinced that either my students do not take vergood notes, or, they never USE them after taking them! Your notes are virtually wastedif you neverstudy them except for the day or so (or less) before a major test! I make a concerted effort to give yotime to not only copy the material off the board, but to write the comments Im making in class (Iliterally look to see how many pencils are moving and when they mostly stop I continue).

Each homework problem submitted must have a complete statement of the problem (Given and Requiredfollowed by your Solution. Problems submitted without a complete Given/Required will be acceptedinitially WITH at least a 15% reduction in grade; After a couple of HW assignments, they will NOT beaccepted at all (and you will not be allowed to complete them and resubmit).

Homework Problems WILL have a CLEAR separation between the GIVEN/REQUIRED and theSOLUTION. Use a horizontal line with the word solution as shown (can be a solid line):--------SOLUTION----------------------------------------------------------------------------------------------

When you write an equilibrium equation write as follows and include sign conventionabove the :

MBII

= 0 = II

indicates the freebody being used, B indicates the point summing moments about.Fx

wfb= 0 = wfb indicates whole freebody and x indicates directionThe superscript is not required if it is OBVIOUS to which freebody it applies.ALWAYS show the = 0 or you tempt a NO EQUATION error.

On homework problems, use a STRAIGHTEDGE when drawing figures. Draw figures at LEAST roughly tscale (I recommend a 6 scale broken into tenthsof an inch). Freehand figures WILL be accepted IFAND ONLY IF you submit a sample to Dr. Bayer for approval otherwise, freehand drawings will not beaccepted.

CR cant read: If I cannot read something that you submit, and I show it to you and you cannot read

it either, then that assignment will be given the grade of zero.UV undefined variable: If you write an equation with an unknown, that unknown should be defined(normally by showing a force/moment on a freebody and naming it).MUD magnitude, units, direction: ALL reactions calculated should include all parts of MUD.MUDL magnitude, units, direction, location: Typically used for maximum deflection problems.CLOUD show answers, when reasonable, in a CLOUD as shown in class.STUBS show stubs on fb diagrams with length and thickness! Show internal forces ON that cut face!

Be careful with round off rounding off a number from 106.666666 to 107 can cause fairly significantdifferences in other calculations using the rounded off number! You can write the answer as 106.6indicating a continuing decimal AND when you use that number in your calculator input a lot of sixe

to better approximate the exact number. REM: these are NOT necessarily significant digits, but willgive an indication of correctness that might not otherwise be obtained.

BASICS Structural Analysis introduction

LIMIT STATE a limit state is a condition at which a structure of some part of a structure fails tofulfill its intended function.Two types of Limit States:(1) STRENGTH limit state are based on the safety or load carrying capacity of structures includingplastic strengths, buckling, fatigue, overturning etc


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(2) SERVICEABILITY limit state refer to the performance of structures under normal service (actualoads and are concerned with the uses and/or occupancy of structures including excessive deflections,slipping, vibrations and cracking.

ELASTIC STRUCTURE: The assumption we must make in structural analysis classes is that the structuris elastic. This means that the stresses in the members a limited to a point well below the elastic limit othe material used.

SUPERPOSITION: If a structure is elastic, the effects of forces on the structure can be determined

individually and added together to get the effect of all the forces acting.

SIMPLE SPAN: A beam with a pin at one end and a roller at the other.CANTILEVER: A beam fixed at one end with no other supportsCANTILEVERED SIMPLE SPAN: A simple span with one or both ends extending past the supportsPROPPED CANTILEVER: A beam fixed at one end and a roller at the other.

The number of unknown forces at a roller reaction= one. That force is always perpendicular to thesurface on which the roller rides.USE: Show the force perpendicular to the surface and give it a name (i.e. R). It will often be helpful tobreak that force into two components. BUT, when doing so, make the components a function of R (i.e.0.6R and 0.8R). DO NOT give the components new names like Rx and Ry! If you use sine or cosine oa given angle, find the VALUE of that trig function rather than writing sin 30 etc

The number of unknown forces at apin reaction= one. That force is also in an unknown direction.Therefore there are two unknowns at a pin reaction! Rather than use the force and an angle as thoseunknowns, components are typically used (i.e. Axand Ay instead of A and ).Warning: The slope of the ground surface is totally meaningless for a pin reaction! That is not true forROLLER reactions, but it IS true for PIN reactions. Be careful!

A fixed endreaction is exactly like the pin reaction with one major exception. There can be NO rotatio

at the fixed end, resulting in the addition of a moment reaction.

A linkis a structurethat is connected to other structures by an internal hinge at two locations, or, by aninternal hinge and a pin support; there are no forces applied at any points in the structure except at thetwo internal hinges (or internal hinge and pin support) where the link is connected to other structures.Thus, the members of a link are considered weightless! Therefore, the link becomes a two-forcemember, with the line of action of the forces on the line connecting the two internal hinges (or internalhinge and pin support) where the link is connected to other structures. The structureis composed of onor more structural members.The recognitionof a link will often make the solution of reactions easier. At the same time, if the link isnotrecognized, the structure is often still solvable for reactions but not as easily.

The most common link is that in a traditional truss where all members are connected with frictionless pin(internal hinges or pin support), all forces are applied at the internal hinges. If a link is composed of asingle straight member, then that member is a pure tension or a pure compression member!

When you cut a structureto create a freebody, you must introduce the internal forces that the part ofthe structure being discarded was placing on the structure you are retaining for the new freebody. Intwo dimensional analyses, you must always consider the following internal forces: shear force, axial forcand a moment. The shear and axial force are components of a single internal force at an unknown angle athe cut section.


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USE: it is often easier to use a horizontal and vertical component of the single internal force instead ofthe shear and axial force as unknowns.WARNING: Many of the serious mistakes made will be the failure to follow this requirement!

Stubs: This is a concept introduced by Dr. Bayer and will not be found in a textbook. Many texts willinstruct you to cut a structure through an internal hinge. A problem that arises is what do you do wiany forces applied to that internal hinge. In Dr. Bayers classes you will NEVER cut the structurethrough an internal hinge. Instead, cut the structure just to one side or the other of the internal hingThis creates a short stub to the side you made the cut. That stub has no length but is shown with ashort length for clarification of where the cut was made. Then, appropriate internal forces are placed othat cut face of the stub! When you show that stub, give it a short length (~ 1/8) and a small thicknesstoo (two dimensions)!

An internal hingeis a connection of one part of a structure to another allowing only the transmission of shear force and an axial force from one side to the other. There can be no internal moment at an internhinge! Thus, when you cut the structure at an internal hinge, you introduce only two unknowns.Comment: show internal hinges using an open circle (o) do NOT show it as a solid filled circle ()!

Equation of Condition: The use of internal hinges in a structure allows a special case to be used in

calculating reactions. By definition, there can be no internal moment at an internal hinge. Thus, when yocut the structure at an internal hinge, you introduce only a shear force and an axial force. The equationof condition is simply the moment at the internal hinge equals zero, or, MB=0 where B is the location ofthe internal hinge.ERRORin Kassimalis text: On page 53 in Kassimalis second edition, he states that the equation ofconditions are MB

AB=0 and MBBC=0. That is in error the equations of condition in his case are simply

MBAB=0 and MB

BC=0 no summation!

You can solve for only three unknown reactions using the standard equilibrium equations. Somedeterminate structures have more than three unknown reaction components but will also include someinternal hinges. If there are no links in a given structure, you will ALWAYS have to cut the structure at

EACH internal hinge creating new freebodies in order to solve for reactions! Those internal hingesrepresent equations of conditiongiving additional equations beyond the standard equilibrium equations.

Given a system of forces, you can ALWAYS eliminate any two of those forces from an equilibriumequation. Heres the only two ways: if the 2 forces intersect, sum moments ( M=0) about their point of intersection. if the 2 forces are ||, then sum forces ( F=0) perpendicular to their direction.

When trying to solve any given freebody, you have only TWO EQUATIONS available: M=0 or F=0. Yocan use them multiple times, but the only equations you can write will be one of those two! Before you

write either THINK about what you are trying to solve for, then determine which equation will give yothe result you are looking for!

For a two-dimensional structure, the necessary and sufficient conditionsfor equilibrium are: Fx = 0 Fy = 0 Mz = 0There are two alternative sets of equations that can be used, but each has a specific contingency:

(1) Fq = 0 MA = 0 MB = 0(2) MA = 0 MB = 0 MC = 0

You should know what the contingencies are, and more importantly, WHYthey are contingencies!


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When you write an equilibrium equation in this class, and it has more than ONE unknown, then you havevery likely written the wrong equation at that point in your calculations! You should THINK about anyequation that you are about to write and determine how many unknowns will be in it. Concentrate on whayou are trying to solve for! The most common place in this course to have to solve simultaneous equationis when solving for reactions on a three-hinged archand when solving Slope Deflection equations(indeterminate method).

A three-hinged archis a structure composed of two rigid portions connected to each other by a singleinternal hinge, while each portion is connected to ground with a pin-support.General Solution for reactions:1. Using the whole freebody, write the equation M=0 about one of the pin-reactions to give an equation

with two unknowns (the reactions at the other pin reaction).

2. Construct a freebody by cutting the structure at the internal hinge and including the reactions that

are in the first equation.

3. Write the equation M=0 about the internal hinge resulting in another equation with the same

unknowns as in the first equation.

4. You now have two equations with two unknowns. Solve the equations simultaneously.

5. Use those two reactions on the whole freebody to solve

Given a determinate structure with a single internal hinge, your thought process should be: is the internal hinge part of a LINK? If so, use it and continue if the internal hinge is NOT part of a link (or you dont recognize it), then you MUST cut the structureat the internal hinge and use the left or right-hand freebody to solve for reactions.

Give a determinate structure with two or more internal hinges if there is no reaction between the two hinges, then the first freebody to be used in most (all?) casesis the freebody from one internal hinge to the other.

TOOLBOXF = 0

M = 0 Equation of condition Support: roller Support: pin Support: pin with a link to show direction Support: fixed end Links Scissors to cut structure into freebodies Mathematical relationships between load, shear and moment

PROCEDURES FOR TOOLBOX How to remove two unknowns from an equilibrium equation How to recognize a link How to handle a three hinged arch How to handle a beam with two adjacent internal hinges How to handle a beam with two adjacent internal hinges and a support between them How to recognize when extra freebodies will be required How to recognize determinate and indeterminate structures

TRUSSES Eyeball Method


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Truss bar forces (axial loads in the members) using the EYEBALL METHOD (joints): to be solvable by the eyeball method, at least one of the summation of forces equations must have onlyone unknown, and, the other may have only one additional unknown.

If both have components of two unknowns, it can be solved by simultaneous equations, OR, possibly by thEyeball Method if you move to another joint(s) and later come back to the joint in question. be sure you check summation of forces vertically and horizontally before you leave a joint. be sure to transfer the arrowheads to the other ends of a member in the correct direction.

ZERO MEMBERS given a freebody of a joint from a truss, if all of the forces have the same line ofaction EXCEPT one then that one must be a zero member.

FLEXURAL MEMBERS IN A TRUSS (cookbook): solve for the reactions of the structure (does not always have to be done, but will not have a check ifnot done). remove each flexural member to a freebody of its own. draw a freebody of what is left of the structure (no flexural members). solve for the shears on the flexural freebodies; transfer those shears to the other freebody.Begin solving for bar forces on the freebody of what is left.

when you get to a joint with an axial force from one of the flexural freebodies, solve for that forcethen transfer it to the flexural freebody,then solve for the axial load at the opposite end of that freebody,then transfer that axial load back to the freebody of what is left and keep solving.

SHEAR, MOMENT, AXIAL and ROUGH DEFLECTED SHAPE DIAGRAMS

The intensity of the load diagram at a point = the slope of the shear diagram at that same point.

The value of the shear diagram at a point = the slope of the moment diagram at that same point.

The area between two points on the load diagram = CHANGEin the value of the shear between the same

two points. The area between two points on the shear diagram = CHANGEin the value of the moment between the

same two points.

THREE WAYS to determine the value of the moment at a given point on a beam, use the load diagram and areas of the shear diagram use a freebody cut at the point you want the moment and write an equilibrium equation use the definition of moment

Short Version of above four relationships for use while drawing those diagrams:VALUE ABOVE = SLOPE BELOW

AREA ABOVE = CHANGE BELOWNOTE: you can apply those to the load, shear, and momentrelationships but do NOTtry to extend thosto the RDS diagram!

ZERO SLOPE (zs): Mark known points of zero slope on both the shearand the momentdiagrams.If the value above = zero, then the slope below = zero.

Two Points and a Zero Slope refers to drawing the shape of a shear or moment diagram between two points


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(1) when you know the slope at one of the points is zero (horizontal tangent line to curve)(2) when you know the curve is a smooth curve (diagram above it has no discontinuities).

To determine where a shear or a moment diagram cross at a value of zero:Determine the change in the value of the diagram itself between two points.Equate that change to the area of the diagram above between those same two points.That area will be written as a function of the distance you are looking for (as an unknown).

We normally will draw shear and moment diagrams from left to right.

If you draw from right to left, the sign conventions used are reversed.

When moving from left to right, the SHEAR diagram ALWAYS moves in the direction of the load.

A MOMENT diagram can have a vertical jump ONLY at the location of an applied or reaction moment.

At an internal hinge, the value of the moment = zero. But the SHEAR does NOT have to be zero and infact is normally NOT zero (zero would be a special case of loading).

Drawing the V and M diagrams to scale both horizontally and vertically CAN help you to get the propershape of the diagram. Drawing out of scale CAN cause errors!

The RDS (rough deflected shape) is the shape of the centerline of the structure AFTER the loads havebeen applied. Draw these with a SINGLE LINE (that represents the centerline).

The RDS (rough deflected shape) will have to go thru support points and will have a smiling curvature ifthe moment is positive and a frowning curvature when the moment is negative.If the moment is zero then there is no curvature (unless it was present before loading).Try playing with a coat hanger wire to play with deflected shapes.

If the RDS diagram has a reverse curvature (inflexion point) be sure to PROJECT that inflexion point(PIP) from the moment diagram to the RDS diagram and put a slash mark at that point on the RDS.

For the RDS diagram, be sure to show all supports. If there is no given deflection of a support, then thRDS diagram must go thru the support points (CNLS = cannot leave support).

RDS diagram: Internal hinges (not a support) can deflect (i.e. deflection at an internal hinge is notnecessarily zero, so, unless you KNOW it is zero, do not SHOW it as zero).

The RDS at a fixed end can only be one of three possibilities.All three options have the SAME tangent at the fixed end.

If there are no applied or reaction moments on a beam, then the shear diagram areas will algebraicallyadd to zero. If there is an applied moment or reaction moment on a beam, then the shear diagram areas willalgebraically add up to the value of the moment. If there is more than one moment, then the shear areas will add up to the algebraic sum of the appliedmoments.


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The SHEAR at any section of a beam is equal in magnitude but opposite in direction to the algebraic sumof all forces perpendicular to the longitudinal axis at the section in question, taken to either side of thatsection.

The MOMENT at any section of a beam is equal in magnitude but opposite in direction to the algebraicsum of the moments of all forces to either one side or the other of the cross section (includingapplied/reactive moments) taken about the centroid of the cross section.

The AXIAL load at any section of a beam is equal in magnitude but opposite in direction to the algebraic

sum of all forces parallel to the longitudinal axis at the section in question, taken to either side of thatsection.

THE PICTURES BELOW SUMMARIZES A LOT OF THE INFORMATION ABOVE:

[These are available at my website as PDF files or JPEGS]


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ADVICE ON SOLVING FOR REACTIONS, DRAWING V-M-A-RDS DIAGRAMS

This is an email I wrote to one student in the class (Fall 2005) it might be helpful here:

... after the brief time I watched you thinking through the problem after class yesterday ... and

after thinking about it before I got up this morning (my thinking time) ... I came up with one word that you need to

consider and work at:


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FOCUSA note sent to a struggling student in the Fall of 2005

It seemed you were jumping around ... jumping ahead in the solution before you got the FIRST things you neede

... almost hyper-active. I think we need to get you geared down ... calmer. I'm not sure how. You seemed to

want to start writing equations BEFORE you had an appropriate freebody! Don't do that. Draw the freebody

FIRST ... and only after giving thought to what freebody you need.

I vaguely remember writing something on your first Quiz ... it might have even be close to what I'm thinking now.

DEVELOP A STRATEGY and WORK to STICK TO IT.

==============================

If a problem asks for V-M-A-RDS diagrams ... there are a series of things you need to do ... and you cannot jump

forward without completing each step (at least not very successfully):

1. Hole in the paper ... one-by-one, SHOW the reactions at each support, NAME them, and ASSUME a

direction for each (you said you didn't like to do that up front ... do it ... assume a direction by showing anarrowhead).

2. Look for LINKS associated with PIN reactions.

3. CONSIDER your procedure:

(a) If you only have THREE unknown reactions, you can use the three equilibrium equations to solve for

reactions.

(b) if you have more than THREE unknown reactions you are going to need an equation(s) of condition or

something similar (like parallel chords with no diagonal in a truss).

4. If you will need to draw additional freebodies to "make use" of the equations of condition, THINKabout which

freebody you will need by "imagining it" and how many unknowns it would have. THEN draw the freebody ...

show the forces at the CUTS ... ASSUME a direction ... NAME them. Solve the freebody for what you need

and move to other freebodies as needed.

AFTER YOU HAVE REACTIONS,

TO DRAW V-M-A-RDS DIAGRAMS:

1. You simply have to KNOW ... not JUST have memorized ... you have to KNOW the relationships between loa

and shear .... and shear and moment. I find several students can "recite" the relationships, but do NOT really

KNOW what they mean. They are really very simple and we use them OVER and OVER in drawing the V and M

diagrams!

2. Project lines perpendicular to the beam itself at all important points ... supports, internal hinges, breaks in

loadings ... project down through ALL the diagrams.

3. Try to draw the shear diagram going from LEFT to RIGHT. Then, the shear ALWAYS moves in the direction o

the load.

(a) Plot the values at the important points.


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(b) Sketch in the "curves" joining the points.

4. To draw the moment diagram ...

(a) first plot the values that you already know ... points of zero moment and applied/reaction moments

(b) IMPORTANT: determine specifically what additional values you need to complete the M diagram

(c) For each of those additional values, determine whether to calculate it using AREAS of the V-diagram OR

a freebody cut at the location the moment value is needed.

5. To draw the axial load diagram ... remember, tension is positive ... where you have axial loads applied, therewill be a jump/change in the value ... but it will be constant over much of the beam.

6. To draw the rough deflected shape:

(a) show the supports

(b) project inflection points from the moment diagram (points where the moment changes sign)

(c) determine the curvature over sections of the beam like I've shown in class ... drawing a "smile" or a "frown"

over the span where that curvature will occur.

(d) show that curvature, matching the support constraints.

(c) show a 'slash' where the inflection points occur

Well, this got to be much longer than I anticipated ... I hope it will help some. I hope to see you during office

hours to work on your progress. No one wants you to be successful anymore than I, unless it is yourself (and I

hope that is the case).

Doc B

CONJUGATE BEAM

The Conjugate Beam Method is used to determine values of rotation and deflection at points on a givenbeam. Well call the given beam, the Real Beam (RB). The Conjugate Beam will be referred to by CB.The differential equation of the elastic curve (deflected shape) is: d2y/dx2 = M/(EI)

Integrating that equation once yields the equation for the slope of the beam: dy/dx = slope (rotation)Integrating one more time yields the equation for the deflection of the beam: y = deflection

A mathematical model will be made using the real beam and its moment diagram. The real beam will beconverted to a CONJUGATE beam by replacing real support conditions with their conjugate. This beawill be loaded with the moment diagram divided by EI [the M/EI diagram].The shear diagram of the conjugate beam = rotation of the real beam! VCB = ROTRBThe moment diagram of the conjugate beam = deflection of the real beam! MCB= DEFLRBThe relationships between the conjugate beam load diagram, the VCB and MCB are exactly the samerelationships we use for typical shear and moment diagrams.

Units for the conjugate beam are NOT the same as a real beam. Remember, this is a mathematical mode for the conjugate beam using typical loading/dimensions for the real beam: load = (k-ft) / (EI units) THINK about it (M) / (EI) VCB = ROTRB = (k-ft

2) / (EI units) where CB = Conjugate Beam and RB = Real Beam MCB = DEFLRB = (k-ft

3) / (EI units)... EI units typically = (k/in2) (in4) = k-in2 When plugging into the values of the VCB and MCB to get actual values, you need to understand what theunits are and make appropriate CONVERSIONS to get rotations in radians and deflection in inches Your answers for actual values of rotations and/or deflections should indicate a DIRECTION (cw or


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ccw up/down left/right etc) and should have units indicated! MUDL = Magnitude, Units, Direction, Location

Using the original beam (not the loading), construct its CONJUGATE BEAM.Load the conjugate beam with the M-diagram of the real beam divided by EI (at each point).Construct the shear diagram of the conjugate beam that = ROTATION or SLOPE of the real beam.Construct the moment diagram of the conjugate beam that = DEFLECTION of the real beam.

There is LOGIC behind the conjugate supports.

REAL BEAM SUPPORT Rotation? Deflection? CONJUGATE SUPPORT Shear? Moment?

Pin at end of beam Yes - No Roller Yes - No

Roller at end of beam Yes - No Pin Yes - No

Internal pin or roller Yes - No Internal hinge Yes - No

Internal hinge Yes - Yes Internal pin or roller Yes - Yes

Fixed end No - No Free end No - No

Free end Yes - Yes Fixed end Yes - Yes

USEFUL HINT: Since the conjugate beams moment diagram IS the deflection of the beam, then it

should be obvious that the RDS diagram SHOULD LOOK very much like the MCB! You should be able touse that hint to help you draw the moment diagram for the conjugate beam!

When you have a VARIABLE I, it is important that you NOT construct the loading for the conjugate beausing more than one I value. The Is will be constant over fixed lengths of the beam and this will allowyou to write all the variable Is as a function of a relative I that you define.For example, if I1 = 400 in

4, I2 = 600 in4, I3 = 1200 in

4, and I4 = 1600 in4.

LET I = 200 in4

THEN I1 = 2II2 = 3II3 = 6II4 = 8I

Now use those values i.e divide M by 2EI in span one, 3EI in span two, etc. Then when you substitutevalues of E and I for final answers when required, use the I = 200 in4 when needed for rotations ordeflections of the real beam.

When determining the direction of the rotation at a point on the real beam, remember that the RB (MCis the deflected shape of the real beam. Thus, you can SEE the rotation on that diagram!

Conjugate beams do NOT have to be structurally STABLE, but will be in equilibrium due to the appliedloading (M/EI).

Conjugate beams will ALWAYS be determinate.

INFLUENCE LINES FOR DETERMINATE STRUCTURES

AnINFLUENCE LINEis a diagram whose ordinates show the magnitude of some function of a structure

as a unit load moves across it. Each ordinate gives the value of the function when the unit load is at that point. The functions represented by an influence line include reactions, shears, moments, forces and


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deflections.

Influence lines are primarily used to determine where to place live loads to cause maximum or minimumvalues of the function in question. They may also be used to calculate the values of the functions for aspecific set of loads.

Influence lines for determinate structures are straight lines or a series of straight lines.

Heinrich Muller-Breslaus principle:The deflected shape of a structure represents to some scale the influence line for a function (such asreaction, shear or moment) if the function in question is allowed to act through a unit displacement.

When drawing the deflected shapes using Heinrich Muller-Breslaus principle, there are a few things toremember that will help you ILs for REACTIONS:

When drawing an influence line for a reaction, the vertical jump of the diagram is 1K at the reactionpoint.

ILs for SHEARS:

When drawing an influence line for the shear at a point (not at an internal hinge) the slope to the left of that point and the slope to the right of that point MUST be the same(because the moment has not been released). the total vertical jump at that point must be equal to 1K.

When drawing an influence line for the shear at an internal hinge the slope to the left of that point and the slope to the right of that point are NOT required to besame (because there is no moment at an internal hinge). the total vertical jump at that point must be equal to 1K.

When drawing an influence line for the shear just left (or right) of a support, the side of the cutadjacent to the support cannot move vertically. So, the other side will move vertically 1k.

ILs for MOMENTS:

When drawing an influence line for a moment at a point, the moment is released, but NOT the shear athat point. Think of it as inserting an internal hinge in the beam at that point. There can be noseparation (jump) in the diagram at the point in question. In allowing the moment to act thru a unit displacement, remembering that the influence line for adeterminate structure is a series of straight lines, it is often necessary to move the point in questionupwards. If the point in question cannot be moved upward, then a rotation on each side should beattempted.

When drawing an influence line for the moment at a support location, the point cannot be movedvertically because the reaction is not released.

The values of the influence lines drawn using Heinrich Muller-Breslaus principle for Reaction or shear ILs: can be determined by geometry

Moment ILs: cannot be determined by geometry for the first value. After an initial value isdetermined, the other points can be determined by geometry.

Values for influence lines can always be determined by appropriate load diagrams using basic statics tosolve for the function under consideration.


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For our class, the sign conventions for various functions are as follows for horizontal beams: for reactions, UP is positive for shear use the sign convention that we use for shear diagrams: up on the left, down on the right for moment use the sign convention that we use for moment diagrams: SMILE is positive

To determine the value of the function given its influence line and a set of loads for concentrated loads (downward applied loads is positive), multiply the load by the signed value of thinfluence line at the point of the applied load. for a uniform load, multiply the load (k/ft) by the area of the influence line over the length of theuniform load.

INFLUENCE LINES FOR INDETERMINATE STRUCTURES

Influence lines for indeterminate structures are NOTstraight lines or a series of straight lines.

Heinrich Muller Breslaus principle STILL applies deflected shapes will have curvature!

VIRTUAL WORK (Beams and Frames) [Kassimalis Text]

The three terms to the right of the equals mark are, respectfully work done by the internal virtualforces acting thru a deflection in the real structure.(1) due to elongation or shortening of the member due to axial loads (bar forces or truss action)(2) due to bending (deflections) of the member (flexure or beam action). And(3) due to changes in temperature (elongation or shortening).

The term to the left includes ALL virtual external forces the applied force and all reactions times

the deflection in the real system at that point and in the direction of the virtual forces.

Equating the left side to the right simply says the external work = internal work.

Normally, there are no support settlements/movements and thus the only unknown will be the deflectionat the point of your applied load (normally 1k), in the direction of your applied load. The problemstatement steers you to where to apply the 1k load, and in what direction at the point the deflectionis desired, and in the direction it is desired.

This method can handle support deflections, but those deflections must generally be given data. Thereactions of the virtual system act thru the deflections of each support causing work. If the forceis in the direction of the support movement, the work is positive and if not negative.

When a given problem involves two or more of the terms on the right side, or, if there are supportmovements to consider, Doc B prefers to solve the problem in pieces and then add the pieces togethealgebraically to get a final answer. One can certainly write one equation including all the appropriateterms, but it can quickly become cumbersome.

TOTAL = SUPPORT + TRUSS + BEAM


[ ] +

+

= LTF

EI

MdxM

AE

FLFF

vvvv)(

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Truss Action is easily calculated using a tabular format. If your last column is for the term FvFL/A, andyou use typical units, you can sum that column and your units will be k2ft/in2 to get your answer,divide by E (ksi) and multiply by the conversion factor 12 in/ft to give answer in inches.

Data for truss action is oftengivenin tabular format with room left to add columns to the right of thegiven data. Use that for TRUSS action calculations. I would start a NEW table for BEAM action becausyou really dont need much of the other column data (see below).

Beam Action is also simplified by the use of a tabular format using the following column headings:Section M Mv MMv Range and simplifying each sections MMv term PRIOR to putting it into thevirtual work equation integral. The range tells you the upper and lower limits on the integral.

Be sure to CLEARLY indicate the origin for x used in writing your moment equations.Use subscripts if you are using more than one origin (x1, x2, x3 etc).

The number of sections you will need for the beam action integrals will depend on BOTH the real systeand the virtual system. You must be able to write an equation for M and for Mv that is valid over theentire range specified. If you have an applied concentrated load in either system, that will be a start stop point on a given range!

When writing the equation for M or Mv in a given segment, imagine cutting the structure at ANY pointwithin that segment and solving for the moment at that point by fb or definition, but remembering thatthe point location is not a constant, but is dependent on the value of x. The equation must be valid forany value of x in the range specified.

VIRTUAL WORK (Beams and Frames) [Leet and Uangs Text]

[page 354 eq 10.24 -> page 358 eq 10.27 + page 367 approx eq 10.31]

Work of external virtual forces (Q forces) thru defl in real system (P system)

EQUALSwork of internal virtual forces thru related change in real system member lengths

[ ]IE

dxMMLLT

EA

LFFQ P

x

xQfabr

PQP

+

++

=

2

1

][

Truss Action Temp Fabr errors Beam Action

Q = virtual forces (including reactions)

Normally this is a single unit load and associated structural reactions.

P = deflection of the real structure at the point of application of the Q-force,

and in its same direction.

FQ = bar forces in Q-system (virtual-system)

FP = bar forces in P-system (real system)

L = length of the member

A = cross section area of the member

E = modulus of elasticity of the member

= coefficient of thermal expansion

T = change in temperature

L = change in length of the member (usually a fabrication error)MQ = equation for moment at a point in the Q-system (virtual system)

MP = equation for moment at a point in the P-system (real system)

I = moment of inertia about the bending axis of the cross section


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The four terms to the right of the equals mark are, respectfully work done by the internal virtual forcesacting thru a deflection in the real structure.(1) due to elongation or shortening of the member due to axial loads (bar forces or truss action)(2) due to changes in temperature (elongation or shortening).(3) due to changes in lengths due to fabrication errors. And(4) due to bending (deflections) of the member (flexure or beam action).

The term to the left includes ALL virtual external forces the applied force and all reactions times

the deflection in the real system at that point and in the direction of the virtual forces.

Equating the left side to the right simply says the external work = internal work.

Normally, there are no support settlements/movements and thus the only unknown will be the deflectionat the point of your applied load (normally 1k), in the direction of your applied load. The problemstatement steers you to where to apply the 1k load, and in what direction at the point the deflectionis desired, and in the direction it is desired.

This method can handle support deflections, but those deflections must generally be given data. Thereactions of the virtual system act thru the deflections of each support causing work. If the force

is in the direction of the support movement, the work is positive and if not negative.

When a given problem involves two or more of the terms on the right side, or, if there are supportmovements to consider, Doc B prefers to solve the problem in pieces and then add the pieces togethealgebraically to get a final answer. One can certainly write one equation including all the appropriateterms, but it can quickly become cumbersome.

TOTAL = SUPPORT + TRUSS + BEAM

Truss Action is easily calculated using a tabular format. If your last column in that tabular format is fothe term FQFPL/A, and you use typical units, you can sum that column and your units will be k

2ft/in2 t

get your answer, divide by E (ksi) and multiply by the conversion factor 12 in/ft to give answer in inches.

Data for truss action is oftengivenin tabular format with room left to add columns to the right of thegiven data. Use that for TRUSS action calculations. I would start a NEW table for BEAM action becausyou really dont need much of the other column data (see below).

Beam Action is also simplified by the use of a tabular format using the following column headings:Section MQ MP MQMP Range and simplifying each sections MQMP term PRIOR to putting it into thevirtual work equation integral. The range tells you the upper and lower limits on the integral.

Be sure to CLEARLY indicate the origin for x used in writing your moment equations.Use subscripts if you are using more than one origin (x1, x2, x3 etc).

The number of sections you will need for the beam action integrals will depend on BOTH the real systeand the virtual system. You must be able to write an equation for MQ and for MP that is valid over theentire range specified. If you have an applied concentrated load in either system, that will be a start stop point on a given range!

When writing the equation for MQ or MP in a given segment, imagine cutting the structure at ANY pointwithin that segment and solving for the moment at that point by freebody or definition, but rememberin


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that the point location is not a constant, but is dependent on the value of x. The equation must be validfor any value of x in the range specified.

SLOPE DEFLECTION and MOMENT DISTRIBUTION VOCABULARY

It is critically important to understand some basic terms and not confuse them during your work in both SDand MD. Following is some of the vocabulary that is important to know:

Fixed-End Moments, FEMs (FEMAB, FEMBA etc) are moments used in the solution process. Each span isconsidered fixed at both supports at the ends of the span, and end moments are calculated due to the

applied loads on that span. FEMs are one of the four cases we added together to come up with the SDequations the other three are: rotation at left end, rotation at right end, and relative displacement ofone end relative to the other. Of those four, the FEMs are the only one that takes into considerationthe loading on the member.

Member End Moments (MAB, MBA etc) are what the procedure is setup to solve for! These are the actuamoments shown at the end of a freebody of the member (cut inside of supports, showing no supports).

Joint Rotations (A, B, C etc) are the actual rotations of the structural member at a joint after theloads are applied to the structure. The angle is measured from a tangent at the joint BEFORE loading toa tangent at that same joint AFTER loading.

Chord Rotations (=/L) are used in the solution process. The chord is drawn from the support at oneend of a span to the support at the other end of the span. The rotation of that chord is measured from

the chord prior to any loading/settlement, to the chord AFTER loading/settlement that is the chordrotation, . If the structure has NO sidesway, then the chord rotation will ONLY be caused by supporsettlement(s). [NOTE: tan =/L, but for small deflection theory, the tangent of the angle is equal tothe angle itself in radians.]

Joint Equations are the equations that force compatibility between the SD equations. A joint freebodyis constructed and moments are summed about that joint. The summation of all the end moments thatframe into that joint must equal zero [ MB

= 0 = MB1 + MB2 + MB3 + + MBi ]. The number of moments inthat equation will be equal to the number of members framing into that joint.Typical equation: MBA + MBC = 0 where two beams frame into joint B.

Simple End Span with or without a cantilever: An end span is simply the last span at that point in thestructure it is simple if the exterior support of the span is a pin or a roller. It is NOT simple if the

exterior support is a fixed end.

SLOPE DEFLECTION (Beams)

SLOPE DEFLECTION EQUATIONS (two sets):Set One:

MLR = 2E1K1 [2L + R -31] + FEMLRMRL = 2E1K1 [ L + 2R -31] + FEMRL

Set Two: (simplified equation for use with a span where the moment at one end is equal to ZERO)MIX = 3EK [I -1] + FEMIX - 0.5 [FEMXI]

[obviously, joint x has M = 0 MXI is not in the eq]. I = interior joint, X = exterior joint

NOTE:Never EVER mix the two sets (specifically do not use one equation from Set One with the single equation from Set Tw Use Set Onefor all interior spans (as opposed to end spans, or end spans with a cantilever.

Use Set Onefor all end spans with the exterior end FIXED. The rotation at the fixed end = zero (FIX = 0).Use Set Twofor all simple end spans.Use Set Twofor all simple end spans with a cantilever that has no loading on it.Never EVER write SD equations for a cantilever!


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JOINT EQUATIONS:

The joint equation for ALL interior joints will include the member end moment for each member thatframes into that joint. If two members frame into joint K, the equation is MKJ + MKL = 0 If three members frame into joint K, the equation is MKJ + MKL +MKI= 0For an end span with a loaded cantilever (i.e. moment at the end support is not zero), the jointequation should come from a freebody of the cantilever cut on the side of the joint away from thecantilever. Always draw the freebody, and always recognize the direction of the moment exposed JOINT end moment. Be consistent with our basic sign convention direction opposite to that of membe

end moments. For a fixed end, you will never use a joint equation (because you get no compatibility equation).

The simple end span equation is used when you KNOW the moment at one end of a span equals ZERO.Failure to use the simplified equation when you can, will result in one more equation to be solved for onemore unknown AFTER the SD equations have been combined with the JOINT equations.

To derive the simple end span equation, set the member end moment at the simple end equal to zero andadd the two equations such that you eliminate the rotation () at the simple end. This will eliminate onof the unknowns to be solved for and thus the number of equations to be solved.

SIGN CONVENTION (Slope Deflection and Moment Distribution):For Leet & Uangs Texbook: Clockwise member end momentsand chord rotations are positive.

NOTE: anyjoint end moment would be ccw positive to be consistent with the basic SD equations.For Kassimalis Textbook: Counter-clockwise member end momentsand chord rotations are positive.

NOTE: anyjoint end moment would be cw positive to be consistent with the basic SD equations.

COOKBOOK:

THINK SHOW the chord rotation on a sketch of the whole structure (if one exists). Solve for FEMs, Ks (or EKs), and s

Use actual value of EK when you have a -term! EK will typically have units of kip-ft. You can use relative value of EK when all s = zero. [Let K = ***, Then K1 =(K) etc] Write a set of equations (or equation) for each span; But, NEVER for a cantilever.

Use the pair of standard equations, OR, the simplified equation for a simple end span, but never evermix the two sets of equations for the same span.

Write the joint equations for each interior support joint. Substitute the right side of the SD equations into the joint equations eliminating the M ij terms. Solve the simplified set of equations for the unknowns (normally EKs, but sometimes the actual

value of the s). Back substitute the EKs or s back into the SD equations gives the member end moments

Even if you know what a member end moment is, substitute into its SD equation to be sure you GET

that value an easy check on your work. CHECK YOUR WORK after you solve for the member end moments are the joint equations valid? Show freebodies of each span, with the member end moments and solve for the shears at each end.

Keep FIXED ENDS a part of the member freebodies (do not cut at the fixed end). Draw the shear, moment, and RDS diagrams.

ANOTHER COOKBOOK

1. Write Slope Defl Equations First calc FEM's, EK's, 's and decide on set of 2 equations or set of 1 equatio

for each span; NOT cantilevers For 's, show the chord rotations, and solve for associated term2. Write Joint Equations One for all interior joints; if loaded cantilever, use fb as described in


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pointers/class

3. Combine those equations Resulting equations will contain only 's or EK's

4. Solve for 's or EK's

5. Back Sub into SD equations See equation solving recommendations in the Pointers File

solving for mbr end mom's

6. Place MEM's on fb diagrams Must cut structure to expose specific end moment .. MBC cut on C-side of B

7. Continue with basic statics to solve for shears, reactions, V-M-diagrams later to DESIGN with

NOTE: You will NOT be given a problem that REQUIRES you to solve more than two unknowns manually.

However, if you do NOT use the Simplified SD equation appropriately, you might end up with more!

The actual number of equations you will need to solve will always be smaller than the number of SDequationsandjoint equations. You will end up solving a set of equations for rotations (s or EKs) andthen back-substitute those values into the SD equations to solve for the member end moments.

SOLVING SIMULTANEOUS EQUATIONS. When solving two equations in two unknowns it is best tomultiply one of the equations by a constant such that when you add it to the other equation, one of thetwo unknowns is eliminated.Many students want to solve two equations in two unknowns by first solving one equation for one of thetwo unknowns as a function of the other then substitute that into the other equation to solve for one

unknown. That method is NOTrecommended for the method of slope deflection!

If is not equal to zero, or stated differently if you are given a support deflection or have to solve foa support deflection, then the actual value of EK is required. Thus you would solve for the actual jointrotations (s).

SLOPE DEFLECTION (Frames with single degree of sidesway)

Side-sway is the relative movement, or displacement, of the frame in a direction which does not haverestricted movement due to end conditions. (ie. The top of a frame consisting of two vertical columns ana single beam connecting the tops together.)

A frame will undergo side-sway if its joints are not restrained against translation. One exception to thisis a symmetric frame with a symmetric load. The -term in the SD equations was introduced specificallyto account for such sidesway.

You must assume a deflection of a joint in the direction of the sidesway movement and relate other jointmovements to your assumption neglecting elongation or shortening of members. For a simple frame wione degree of sidesway, those movements will be equal. These joint movements will result in non-zero terms in some members (draw the chord rotations to clarify those terms).

Side-sway can occur in either horizontal or vertical directions. It can be introduced into a frame by anyof the following:

A force acting on the frame. Unsymmetrical loading on the frame.

Unsymmetrical frame. Varying E and/or I and/or lengths.

To solve, determine K or EK values, s, and FEMs, Write corresponding SD Equations for each member anoted before. Be sure to pay attention to the direction of the FEMs and the s.

Determine the appropriate Joint Equations. Write the shear equation (see below). The shears are in the direction of the side-sway.

Note: You should have as many SD equations, joint equations plus the shear equation as you do unknowns.

SHEAR EQUATION:


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The shear equation, which is another equilibrium equation, comes from determination of the shear inthe frame. It is based on the condition that the sum of all forces, producing and resisting side-swayacting on the free body of the entire frame must be zero.

In the case of a single beam being supported at its ends by columns with a horizontal load applied atthe one of the joints, the shear equation would be:

o V1 + V2 + Q = 0

o V1 comes from the summation of moments about the end of a freebody of the first column.

o V2 comes from the summation of moments about the end of a freebody of the second column.

o Q is the value of the applied load at the joint.

Substitute the appropriate SD equations into the joint and shear equations and solve simultaneousequations for s and .

Simultaneous equations are best solved with the assistance of a calculator (ie. TI-89) or a computerprogram (ie. MathCAD).

Back substitute s and into SD Equations and solve for member-end-moments.Remember to check yourself by substituting these values into the joint and shear equations.

MOMENT DISTRIBUTION (Beams)

Basic recording details:

the ONLY numbers you write down in the MD distributionprocedure should be the fixed-end moments(FEMs), distributed moments, carry-over moments and, the last should be the TOTALS of each columin the procedure. each carry over moment should be clearly indicated using an arrow from the distributed moment to th

place moment is carried over to! [Dist Mom carry over moment] after a joint is balanced, a single line should be drawn UNDER the distributed moment that balancedthe joint. after the procedure is completed, draw a double line under each column of numbers indicating you havecompleted the distribution and that the number below that double line is the sum of the column ofnumbers (that sum should be in a cloud and thus clearly marked as the final member end moments).

when recording the various numbers, it is cleaner and easier for YOU to follow if you never record online ABOVE where you are currently working i.e. dont back up. The distributed moments and the carrover moments can be on the same line. Or, the carry over moments can be on the line BELOW thedistributed moments. But, the carry over moment should never be recorded on lines above thedistributed moment. All this is for simplicity, accuracy, and understanding of the procedure itself.

The CARRY OVER FACTOR for a span with a constant E and I, will always be = .The carry over factor is derived using a propped cantilever with a moment applied to the roller end.Using slope deflection, you can show that the fixed end reaction moment equals one-half of the appliedmoment applied. Distributing a moment to the end of a span is equivalent to the propped cantilever andone-half of that distributed moment carries overto the other end of the span.

The UNBALANCED MOMENT is the sum of ALL the moments around a particular support.The unbalanced moment is distributed at its joint according to the distribution factors.The amount of moment to be distributed is equal to the unbalanced moment with opposite sign.

The DISTRIBUTION FACTORS for a given joint indicate the % of the moment to be distributed to eacmember framing into that joint.The distribution factor for one member (Dij) is equal to its stiffness (Kij) divided by the summation of thstiffness of all members framing into THAT joint (Ki): Dij = Kij/Ki


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Dij = distribution factor at joint i for the member ij framing into the joint.

The sum of the distribution factors at each support must be equal to 1 (i.e. 100% of the distributedmoment has got to go somewhere).

The distribution factors for a support (joint) at a beginning point on a structure:(A) if the support is a simple support (roller or pin) with or without a cantilever, the interior side of thesupport has a DF of 1. Why? If the joint is at the end of the member, there is no structure to take anyof the distributed moment. If there is a cantilever, the cantilever has no stiffness in that it cannotresist any moment at the support!

(B) if it is a FIXED end, then the wall has an infinite stiffness compared to the member and will takeALL the moment distributed at that joint thus the member has a DF of zero and the fixed end has aDF of 1 most textbooks do NOT show the 1 on their moment distribution procedures.The fixed end is what our class calls the MOMENT MONSTER Always feed the Moment Monster when it is feeding time (give it the carry over it deserves) NEVER EVER take ANY moment away from the Moment Monster it does NOT like that!

MODIFIED STIFFNESS PROCEDURE:For? Modified Stiffness for simple end spans with or without a loaded/unloaded cantilever

modify the stiffness (K) only of the simple end span(s) by (i.e. use K) calculate the distribution factors [using the modified stiffness value(s)] balance the simple end span(s) perform the required carryover(s) continue with the regular MD procedure, BUT NEVER CARRY OVER BACK TO THE SIMPLE END SUPPORT(S)WHY ? The actual stiffness of interior spans is 4EI/L and for simple end spans is 3EI/L. The is in recognition of thoseactual stiffness values.

Specificsfor modified stiffness of a simple end span with a loaded cantilever:Follow the procedure described above with these specifics:

show the fixed end moments for the end span as normally shown determine the member end moment on the cantilevered side of the simple support using the MD signconvention:

For Leet & Uangs Texbook: Clockwise member end momentsand chord rotations are positive.For Kassimalis Textbook: Counter-clockwise member end momentsand chord rotations are

positive. The unbalanced moment is the sum of the cantilever member end moment and the fixed-end moment othe interior span. That sum (moment), distributed with opposite sign, is all taken by the interior side ofthe simple end support. perform the required carry over continue

If you do NOTuse the modified stiffness for simple end spans, then carry over moments to the simpleend support are required! You will be ok with this procedure, but will be wasting a lot of TIME!

How to handle a SUPPORT DEFLECTION using Moment Distribution: Support deflection is handled simply by including the Fixed-End Moments induced by the supportdeflection with the Fixed-End Moments due to any applied loading. The fixed-end moments will have thesame sign (both positive, or, both negative). The equations for the Fixed End Moment values are includewith the other FEM equations.


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To USE the final member end moments to solve for reactions/shears, you MUST choose freebodies thatexpose the member end moments that were determined by the MD procedure.I recommend using freebodies of EACH SPAN.When a cantilever is involved, show the freebody of the end span WITH the cantilever.For joints at the end of the beam (roller, pin, or fixed), show that support and cut the freebody to theend spans side of the interior support. Those moments that were initially determined by basic statics, do NOT need to be exposed as isillustrated by the moment at a roller support adjacent to a cantilever.

MOMENT DISTRIBUTION (Frames with single degree of sidesway)

Frames that are subject to side-sway are typically unbraced. These unbraced frames generate chordrotations in members that must be accounted for in the design of the members.

Frames with side-sway are solved in a multi-step process using moment distribution. The first step is to create a non-sway frame and complete a normal moment distribution:

Determine the member end moments by the method of moment distribution described above. Determine the direction of the side-sway and introduce a fictitious joint or end condition that will

make it a non-sway frame. Solve for the new reaction at the fictitious joint that you introduced to create a non-sway frame. Note: There should be only one new reaction that you are trying to solve for and it should be in the

direction of the side-sway. I would recommend going ahead and solving for all reactions to ensure that the structure has

balancing forces. Better to catch an error now.

The second step is to create a new structure. This structure is the same as the original structure except with none of the original loads. Add a force, F, at the same joint at which you placed your fictitious joint before and in the direction

of the assumed side-sway.

The FEMs will be introduced by the sidesway of the structure resulting in chord rotations for somemembers. The FEMs can be calculated using the Slope Deflection equations where all but the terms are zero.

To determine FEMs, you can either assume a value for the moments with the appropriate sign andratio to the other FEMs, or you can solve for them by assuming a = 1 inch and using the equation:

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LEIFEM =

Pay attention to sign which indicates direction. Determine member end moments by the method of moment distribution described above. Using appropriate freebodies, solve for the force, F. Again, I would recommend solving for all reactions to ensure that the structure has balancing forces

The third step is to determine the Correction Factor The correction factor, CF, is the needed adjustment to account for our assumption of the amount of

side-sway in the frame. The CF is determined by the reactions you determined from the first two steps. Here you will need

divide the reaction at the fictitious joint in step one by the force, F, in step two. To check your CF, multiply your force, F, by the CF and you should get the RXN you calculated in the

first step.

The fourth step is to determine the actual member end moments accounting for the effect of side-sway


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To determine the actual member end moments you will need to add the determined member endmoment due to non-sway from the first step to the product of the member-end-moment due to side-sway from the second step and the correction factor for each member end moment.

That is for the member end moment at AB, MAB = MABNon-Sway + (CF * MABSide-sway)

The last step is solve for reactions and draw V,M, RDS of the structure.

This method can be applied in the analysis of multistory frames where joints may have three or moremembers connecting together.

MOMENT DISTRIBUTION versus SLOPE DEFLECTION

When should you select Moment Distribution? when Slope Deflection?

One big problem with Slope Deflection is the requirement to solve simultaneous equations. Once the SDequations have been substituted into the joint equations the result is a set of equations that must besolved. For beams, the number of equations is typically equal to the number of unknown joint rotations ithe SD equations. If there are too many equations to realistically solve by hand, then MD will likely bethe preferred method.

For beams both SD and MD can handle support settlement (=/L). However, it is literally built-in to thSD equations and that might give SD an advantage. In order to use MD with support settlement, the SDequations must be used to calculate the Fixed-End Moments, then MD can be used as normal.

We have not covered beams with internal hinges in either SD or MD. Thus, if a structure HAS an internhinge, the problem is NOT intended to be solved by SD or MD!

QUIZ and FINAL EXAM POINTERS

Take the time to review briefly each problem and make an effort to work them

There will be determinate and indeterminate structures on the final exam and it is your responsibility torecognize that and apply an appropriate procedure!

Items typically overlooked that WILL have points deducted include: PIP/SLASH show the projection line for reverse curvature points on RDS diagram and use the slash SAFRFE 3d STUBS UV dnc U clearly mark your answers (cloud?) the use of at least a rough SCALE on V-M diagrams (that can resolve problems with correct curvatureof the diagrams themselves!

Be sure you do what the problem asks for.

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