using matlab in engineering static

Using MATLAB in Statics

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A Guide to Using MATLAB inECE210 Statics

Dhaval Shah, Robert Hinks and S. D. RajanDepartment of Civil Engineering

Arizona State UniversityTempe, AZ 85287-5306

August 2000


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TABLE OF CONTENTS

Chapter Page

1 Introduction 3

2 Sample Problems and Further Exercises 42.1 Forces on a Particle 5

2.1.1 Problem 2.1.1 Forces on a barge 62.1.2 Problem 2.1.2 Crate on an inclined plane 82.1.3 Problem 2.1.3 Equilibrium of a suspended lamp 10

2.2 Rigid Bodies and Equivalent Force Systems 132.2.1 Problem 2.2.1 Automobile towing bar 142.2.2 Problem 2.2.2 Calculation of a torque 172.2.3 Problem 2.2.3 Calculation of resultant force and moment 18

2.3 Distributed Forces: Centroids and Center of Gravity 212.3.1 Problem 2.3.1 Centroid of a quadrant of ellipse 222.3.2 Problem 2.3.2 Centroid of a jetway 242.3.3 Problem 2.3.3 Equivalent force in an automobile 26

2.4 Equilibrium of Rigid Bodies 282.4.1 Problem 2.4.1 Overhanging roof 292.4.2 Problem 2.4.2 Tow truck mechanism 312.4.3 Problem 2.4.3 Folding platform 342.4.4 Problem 2.4.4 Power-line pole 36

2.5 Truss Analysis 392.5.1 Problem 2.5.1 Cantilevered truss analysis 402.5.2 Problem 2.5.2 Truss analysis of a jack support stand 422.5.3 Problem 2.5.3 Space frames supporting window

washing scaffold 452.5.4 Problem 2.5.4 Lever system 492.5.5 Problem 2.5.5 Plow mechanism 51

2.6 Beam and Frame Analysis 532.6.1 Problem 2.6.1 Airplane wing acting as a cantilevered beam 542.6.2 Problem 2.6.2 Computation of tension in cable segments 56

2.7 Friction 582.7.1 Problem 2.7.1 Block on an inclined plane 592.7.2 Problem 2.7.2 Leveling of refrigerator using a wedge 61


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Chapter 1Introduction

There is a class of engineering problems that requires numerical solution either because finding exact solutionsis tedious, or because exact solutions cannot be found. Computer programs are indispensable in findingnumerical solutions. A wide choice of options is available including traditional high-level programminglanguages such as FORTRAN, C, C++ etc. MATLAB provides a different option. It is an integratedenvironment where numerical computations can be carried out either using MATLAB commands, or using auser-written program. Excellent references are available to study, understand and use the capabilities ofMATLAB. The reader is encouraged to use these references. An excellent list is available at the following website: http://www.mathworks.com/support/books/index.php3.

The Student Edition of MATLAB has several manuals and tutorials. One of them MATLAB: The Language ofTechnical Computing (using_ml.pdf) provides a good starting point. The developers of MATLAB, Mathworks,Inc. have an excellent web site (http://www.mathworks.com) where additional resources can be found. In addition,there are several books available that combine numerical analysis and MATLAB.

In this guide we will take a look at solving problems in statics using custom MATLAB programs. In otherwords, we will develop custom computer programs using the MATLAB programming language. Theseprograms are developed usually using the MATLAB editor, and the computer program is saved in one or more.m files. All the problems are taken from the current textbook for ECE210 Engineering Mechanics: Statics byR.W.Soutas-Little and D.J.Inman, a Prentice-Hall publication (ISBN: 0-13-769001-0).

The first step in this process is to understand the (statics) theory necessary to solve the problem. Second, onemust write an algorithm (a sequence of steps) that outlines how the problem input can be used along with thetheory to generate the required output. Third, these steps must be translated into correct MATLAB statements.Inherent in these steps is the possibility of making logical errors (errors that are based on flawed theory) orsyntax errors (errors in translating the algorithmic steps into MATLAB statements). Each one of us must findways of debugging these errors.

As a starting point, become familiar with the MATLAB environment. Run a few simple examples that arecommand driven. Later, develop a few simple programs in MATLAB to solve very simple problems whosesolution can be easily checked by hand. Once you have mastered this, you are ready to understand the examplesin this guide and solve the computer-based problems in the ECE210 text.


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Chapter 2Sample Problems and Further Exercises

The problems in this chapter follow the sequence in a standard statics text, and include problems from thefollowing topics.

(1) Forces on a Particle(2) Rigid Bodies and Equivalent Force Systems(3) Distributed Forces: Centroids and Centers of Gravity(4) Equilibrium of Rigid Bodies(5) Truss Analysis(6) Beam and Frame Analysis(7) Friction

Each of these topics has two or more example problems that have been solved using MATLAB. Every problemis usually divided into two parts – (1) the MATLAB solution to the problem, and (2) suggested furtherexercises. In (2), additional questions are posed and you are required to modify the .m file to find the numerical(or, even symbolic) solution.

Note that the MATLAB solution is not unique. The computer statements can usually be structured in a varietyof ways. However, the answers in almost all cases, are unique.


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2.1 Forces on a Particle

Three problems are solved in this section.

This Guide’s ProblemID

Corresponding Problem from ECE210Text

Problem Title

Problem 2.1.1 Problem 2.11 on page 39 Forces on a bargeProblem 2.1.2 Problem 2.106 on page 90 Crate on an inclined planeProblem 2.1.3 Problem 2.109 on page 91 Equilibrium of a suspended lamp


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Problem 2.1.1 Forces on a barge

A barge is pulled by two tugboats as shown in the Figure P2.11. To move thebarge along in the water properly, the tugboats must exert a resultant force of5000 lb along the direction of motion of the barge. First determine the tension ineach rope if the position of tug b is such that β = 45. Second, suppose that tug b

can move anywhere such that 0 90≤ ≤β o . Determine the angle β at whichthe angle in the Tug B’s rope is maximum, and again, at which the tension hasminimum value, still maintaining the resultant of 5000 lb along the x-direction.

MATLAB File

% 1 2 3 4 5 6 7% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 2-11 on pg. 39 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with finding out the force "Tb" in the% rope between the tug B and the barge when the angle Beta% = 45 and the resultant force is 5000 lb. The second part of% the problem deals with finding out the maximum and minimum% force Tb while 'Beta' varies between 0 and 90 degrees.%% CONCEPT KEYWORDS: Free-body diagrams, Equilibrium equations%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% Tb Force in rope connecting barge to tug B To be computed% Ta Force in rope connecting barge to tug A To be computed% Beta Angle between horizontal and rope% connecting barge to tug B Given% Alpha Angle between horizontal and rope% connecting barge to tug A Given% FResultant Resultant force Given% AInc Angular increment to be used with angle Beta To be defined% TbMax Max. force in rope to tug B To be computed% AngleMax Corresponding value of angle beta To be computed% TbMin Min. force in rope to tug B To be computed% AngleMin Corresponding value of angle beta To be computed% ----------------------------------------------------------------------------% Set default values FResultant = 5000; Alpha = 30; Beta = 45; AInc = 10;%% Convert angles to radians AngleToRadians = pi/180; BetaR = Beta * AngleToRadians;


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AlphaR = Alpha * AngleToRadians;%% ===== Part 1 of the problem ==================================================% Write the two equations of equilibrium and eliminate Ta. b = (sin(BetaR) * cos(AlphaR) / sin(AlphaR)) + cos(BetaR);% Compute the force in rope from barge to tug B Tb=FResultant/b;% Display the result fprintf('The force in the rope connecting barge to tug B is Tb=%g lb\n', Tb);%% ===== Part 2 of the problem ==================================================% The angle beta will be varied between 0 and 90 degrees in increments of AInc% degrees.% We will store these values in a vector called AnglesV% We will store the corresponding value of the force Tb in vector TbV TbMin = 2*FResultant; TbMax = 0;% Now loop through all possible values of beta I = 0; for Angle = 0:AInc:90 I = I + 1; AnglesV(I) = Angle; BetaR = AnglesV(I)*AngleToRadians; b = (sin(BetaR) * cos(AlphaR) / sin(AlphaR)) + cos(BetaR);% Compute the force in rope to tug B TbV(I)= FResultant/b;% Track the maximum value if TbV(I) > TbMax TbMax = TbV(I); AngleMax = AnglesV(I); end% Track the minimum value if TbV(I) < TbMin TbMin = TbV(I); AngleMin = AnglesV(I); end end

% Now we will plot the force in rope Tb versus the angle beta plot (AnglesV, TbV); grid; xlabel ('Beta (degrees)'); ylabel ('Tb (lb)');

% Display the result fprintf ('The minimum force in the rope is Tbmin = %g lb. at angle %g\n',... TbMin, AngleMin); fprintf ('The maximum force in the rope is Tbmax = %g lb. at angle %g\n',... TbMax, AngleMax);

% ==============================================================================% FURTHER EXERCISES% (1) Get a better answer for part 2 by changing the value of AInc.% What is your conclusion?% (2) Do the orientation of the maximum and minimum forces depend% on the resultant force?


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Problem 2.1.2 Crate on an inclined plane

A 100-kg crate is held in equilibrium as it rests on africtionless inclined plane making an angle θ with thehorizontal (see Figure P2.106). (a) Compute the tension in the

cable T and the normal force at the surface for θ = 30o . (b)Compute the tension in the cable T and the normal force at the

surface for θ at 5o increments, increasing from 0 to 90. Isthere some value of θ for which T = N?

MATLAB File

% -----------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 2-106 on pg. 90 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with finding out the tension in the cable% that holds the crate resting on an inclined surface at an% angle Theta with respect to the horizontal.% The second part of the problem deals with finding out the% tension in the cable for different values of Theta.

% CONCEPT KEYWORDS: Free-body diagrams, Equilibrium equations%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% T Force in cable connecting crate to wall To be computed% Theta Angle between inclined plane and horizontal Given% M Mass of crate Given% W Weight of crate To be computed% AInc Angular increment to be used with angle theta To be defined% N Normal Component of weight of crate% (90deg. to inclined plane) To be computed% g Acceleration due to gravity = 9.81 m/s^2 Given

% -------------------------------------------------------------------------------

% Set default values: Theta = 30; M = 100; g = 9.81; AngleToRadians = pi/180; W = M * g; ThetaR = Theta*AngleToRadians; AInc = 1;


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% Using the equilibrium equations along the axis of the inclined plane: T = W * sin(ThetaR); % equation (A)% Display results fprintf('The force in the cable is T = %g N\n', T);% Normal reaction N = W * cos(ThetaR); % equation (B) fprintf('The normal force N = %g N\n', N);

%=====================part b===================================================% Theta is now stored in a vector (ThetaV) with Theta varying from 0 to 90 degrees% in increments of AInc.% ThetaV is now a vector of 18 elements (with Ainc = 5) from 0 to 90.

ThetaV = (0:AInc:90);

% Reuse Eqns. (A) and (B) and store the force values in vector TV and NV

TV = W * sin(ThetaV * AngleToRadians); NV = W * cos(ThetaV * AngleToRadians);

% Find for what value of Theta, T = N AngleEquals = -1; TOL = 0.1; I = 0; for Angle = 0:AInc:90 I = I + 1;% check whether the T and N are almost equal if (abs(TV(I)-NV(I)) <= TOL) AngleEquals = Angle; break; end% check whether the (T-N) has changed sign if ((Angle ~= 0 & Angle ~= 90) & (TV(I-1) <= NV(I-1) & TV(I) >= NV(I))) AngleEquals = Angle; break; end end fprintf('The normal force N = T when theta=%g\n', AngleEquals);

% The plot function plots the different values of T against its corresponding% value of Theta varying from 0 to 90 degrees. plot (ThetaV, TV); grid; Xlabel ('Theta (degrees)'); Ylabel ('T (N)');

% ==============================================================================% FURTHER EXERCISES% (1) If the cable connecting the crate to the wall is replaced by a spring system% whose stiffness is known (say k=1000 N/m), what would be the displacement% in the spring?% (2) How would you solve the problem if the cable is not parallel to the inclined% plane but is at an angle of 20 degrees clockwise from the plane, i.e. the% point of attachment of the cable to the wall is 20 degrees higher than the% point of attachment of the cable to the crate?


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Problem 2.1.3 Equilibrium of a suspended lamp

A 25-lb light fixture is suspended from a ceiling by four cordsattached at a point, with the geometry given in Figure P2.109.(a) Assuming that gravity acts along the z-axis, calculate thetension in the four cords. The dimensions are given in feet. (b)Investigate the effect of the tensions in the cords of movingthe lamps to different heights by changing the z-coordinate ofthe point where all the four cords meet from 5 ft to 10 ft andthen to 1 ft.

MATLAB File

% 1 2 3 4 5 6 7

% ------------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 2-110 on pg. 91 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the tension in the 4 cords% that hold the lamp hanging vertically along the "z axis" from% the ceiling as a function of the connection point with respect% to the ceiling.

% CONCEPT KEYWORDS: Free-body diagrams, Equilibrium equations%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% O Point O is the origin Given% Ax,Ay,Az The x, y, z coordinates of point A Given% Bx,By,Bz The x, y, z coordinates of point B Given% Cx,Cy,Cz They are the x, y, z coordinates of point C Given% T1,T2,T3,T4 Forces in cords AC,OC,BC,CL respectively To be computed% Lzac,Lxac,Lyac Direction cosines of cord AC To be computed% Lzoc,Lxoc,Lyoc Direction cosines of cord OC To be computed% Lzbc,Lxbc,Lybc Direction cosines of cord BC To be computed% Lac Length of cord AC To be computed% Loc Length of cord OC To be computed% Lbc Length of cord BC To be computed% W Weight of lamp Given


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% HInc Height increment with Cz To be defined% T1Max, T2Max% T3Max Max. force in cord T1, T2 and T3 To be computed

% ---------------------------------------------------------------------------------% Set default values W = 25; HInc = 1; Ax = 0; Ay = 4; Az = 0; Bx = 4; By = 0; Bz = 0; Cx = 2; Cy = 1; Cz = 5; Ox = 0; Oy = 0; Oz = 0;% ============================ part (a) ===========================================% Calculate the length of the cords Lac = sqrt((Cx-Ax)^2 + (Cy-Ay)^2 + (Cz-Az)^2);

Lbc = sqrt((Cx-Bx)^2 + (Cy-By)^2 + (Cz-Bz)^2); Loc = sqrt((Cx-Ox)^2 + (Cy-Oy)^2 + (Cz-Oz)^2);

% Calculate the direction cosines of cords Lzac = (Cz-Az)/Lac; Lxac = (Cx-Ax)/Lac; Lyac = (Cy-Ay)/Lac;

Lzbc = (Cz-Bz)/Lbc; Lxbc = (Cx-Bx)/Lbc; Lybc = (Cy-By)/Lbc;Lzoc = (Cz-Oz)/Loc; Lxoc = (Cx-Ox)/Loc; Lyoc = (Cy-Oy)/Loc;

% Form the linear equations for the 3 unknowns T1, T2, T3% T1lxac + T2lxoc + T3lxbc = 0.% T1lyac + T2lyoc + T3lybc = 0.% T1lzac + T2lzoc + T3lzbc = 25.

% Solving the above equations using [A][x]=[B] A = [Lxac Lxoc Lxbc; Lyac Lyoc Lybc; Lzac Lzoc Lzbc];

B = [0;0;W];x = inv(A)*B;

% Display resultsfprintf ('The force in the cord AC is T1 = %g lb\n', x(1));fprintf ('The force in the cord OC is T2 = %g lb\n', x(2));fprintf ('The force in the cord BC is T3 = %g lb\n', x(3));

% Calculate force in chord CL (T4)T4 = W;fprintf('The force in the cord CL is T4 = %g lb\n', T4);

%=========================part (b) ==============================================% Now loop through all possible values of Beta

I = 0; T1Max = 0; T2Max = 0; T3Max = 0;for Height = 1.0:HInc:10 I = I + 1;

% We define the variable for the z coordinate of point C as a vector CzV CzV(I) = Height; Lac = sqrt((Cx-Ax)^2 + (Cy-Ay)^2 + (CzV(I)-Az)^2);

Lbc = sqrt((Cx-Bx)^2 + (Cy-By)^2 + (CzV(I)-Bz)^2); Loc = sqrt((Cx-Ox)^2 + (Cy-Oy)^2 + (CzV(I)-Oz)^2);

% Calculate the direction cosines of cords Lzac = (CzV(I)-Az)/Lac; Lxac = (Cx-Ax)/Lac; Lyac = (Cy-Ay)/Lac; Lzbc = (CzV(I)-Bz)/Lbc; Lxbc = (Cx-Bx)/Lbc; Lybc = (Cy-By)/Lbc; Lzoc = (CzV(I)-Oz)/Loc; Lxoc = (Cx-Ox)/Loc; Lyoc = (Cy-Oy)/Loc;


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A = [Lxac Lxoc Lxbc; Lyac Lyoc Lybc; Lzac Lzoc Lzbc]; B = [0;0;W]; X = inv(A)*B;

% The tension in the cords are now stored in the vector for every values of Cz% in the vector CzV

T1V(I) = X(1); T2V(I) = X(2); T3V(I) = X(3);if (T1V(I) > T1Max) T1Max = T1V(I); end;if (T2V(I) > T2Max) T2Max = T2V(I); end;if (T3V(I) > T3Max) T3Max = T3V(I); end;

endfprintf ('The max. force in cord AC is T1 = %g lb\n', T1Max);fprintf ('The max. force in cord OC is T2 = %g lb\n', T2Max);fprintf ('The max. force in cord BC is T3 = %g lb\n', T3Max);

% Plot the forces in the cords T1, T2, T3 for different heights at which the% lamp hangs.

plot (CzV,T1V,'r'), hold on, text (CzV(2),T1V(2),'T1'),plot (CzV,T2V,'b'), hold on, text (CzV(2),T2V(2),'T2'),plot (CzV,T3V,'g'), text (CzV(2),T3V(2),'T3'), hold on,grid onxlabel ('Cz (ft)')ylabel ('Forces (lb)')

% ==============================================================================% FURTHER EXERCISES% (1) What conclusion(s) can you draw from the graph?% (2) Does the change in length of one cord affect the forces in all the cords?% (3) When is the system in unstable equilibrium? Is this physically possible?


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2.2 Rigid Bodies and Equivalent Force Systems




Problem Title

Problem 2.2.1 Problem 3.24 on page 130 Automobile towing barProblem 2.2.2 Problem 3.43 on page 133 Calculation of a torqueProblem 2.2.3 Problem 3.76 on page 166 Calculation of resultant force and

moment


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Problem 2.2.1 Automobile towing bar

Consider the automobile towing bar illustrated in FigureP3.24. A 4000-N force is applied to the trailer hitch asillustrated and causes the force to change through the angletheta as the lead car pulls on the bar AB. Calculate themoments about the bumper connection point B as a functionof the angle θ . Calculate the value of Mb at θ = 0 45, , and

90° .

MATLAB File

% 1 2 3 4 5 6 7

% -----------------------------------------------------------------------------

% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 3-24 on pg. 130 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with finding out the moment caused by a% 4000 N force applied to the trailer hitch as a function% of the angle of application, Theta, of the 4000 N force.%% CONCEPT KEYWORDS: Rigid Bodies, Equivalent Force Systems.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% P Force of 4000N applied at an angle Theta Given% Px Component of force P along X axis To be computed% Py Component of force P along X axis To be computed% C Point of application of force P Given% B Point about which the moment of force P is% required Given% M Moment of Force P about point B To be computed% Lbc Distance between points B and C Given% Alpha Angle that Lbc makes with the Y axis Given% Theta Angle at which the force is applied


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% with respect to the X axis Given% Lx Lever arm of the component of force P% along X axis To be computed% Ly Lever arm of the component of force P% along Y axis To be computed% AInc Angular increment to be used with angle Beta To be defined% ------------------------------------------------------------------------------------

% Set default values P = 4000;

Alpha = 60;Lbc = 1;AInc = 1;AngleToRadians = pi/180;Theta = 45;

%=============================part1==========================================% Given

ThetaR = Theta*AngleToRadians;Px = P*cos(ThetaR);Py = P*sin(ThetaR);

% Calculate the lever arm of the force P componentsAlphaR = Alpha*AngleToRadians;Lx = Lbc * sin(AlphaR);Ly = Lbc * cos(AlphaR);

% Calculate the moment about point B at theta = 45 degreesM = Px * Ly - Py * Lx;

fprintf ('The moment about point B at theta 45 degrees is %g N-m\n',M);

%====================part 2=================================================% Now loop through all possible values of Theta I = 0; Index0=1; Index45=1; Index90=1;

for Angle = 0:AInc:90% ThetaV is a vector holding values from 0 to 90 degrees, the MV is a% moment vector of the same size as Theta and so are the components% Px and Py of the force P. I = I + 1; ThetaV(I) = Angle;

Px = P*cos(ThetaV(I) * AngleToRadians); Py = P*sin(ThetaV(I) * AngleToRadians); MV(I) = Px * Ly - Py * Lx; if (Angle == 0) Index0 = I; end; if (Angle == 45) Index45 = I; end; if (Angle == 90) Index90 = I; end;

end

% Plot the moment about point B for different values of thetaplot (ThetaV,MV)

text(ThetaV(Index0),MV(Index0),'M@0'), text (ThetaV(Index90),MV(Index90),'M@90'), text (ThetaV(Index45),MV(Index45),'M@45'),

grid onxlabel ('Theta (degrees)');ylabel ('Moment (N-m)');

% ==============================================================================% FURTHER EXERCISES


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% (1) What would be the forces acting at point A and B if the force P% acts at Theta = 0 degrees?% (2) At what angle should the force be placed with the positive X% axis such that the moment about point A is 0?


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Problem 2.2.2 Calculation of a torque

Given that M i j ko = + +20 10 5$ $ $ and F i j k= − −10 10 5$ $ $ kN, where M Fo. = 0 , calculate the value of the moment arm r that

will produce the given moment by application of the given force.

MATLAB File

% 1 2 3 4 5 6 7% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 3-42 on pg. 133 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing a moment vector 'r'% required to produce a known moment by a known force.%% CONCEPT KEYWORDS: Rigid Bodies, Varignon's Theorem%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% Mx,My,Mz Moment components Given% Fx,Fy,Fz Force components Given% Rx,Ry,Rz Components of relative position vector To be computed% ----------------------------------------------------------------------------%% Set default values% Define M = (Mx)i + (My)j + (Mz)k% Define F = (Fx)i + (Fy)j + (Fz)k

M(1) = 20; M(2) = 10; M(3) = 5; F(1) = 10; F(2) = -10; F(3) = -20;

% Let R = (Rx)i + (Ry)j + (Rz)k% Varignon's theorem states M = R X F% Using the process discussed in pages 124-126,% F X M% R = -----% F . F%

Numerator = cross (F, M);Denominator = Dot (F, F);

R = Numerator/Denominator;

% Display results fprintf ('The position vector R = (%g)i + (%g)j + (%g)k\n',... R(1),R(2),R(3));

% ==============================================================================% FURTHER EXERCISES% (1) What magnitude of force is required to produce a moment vector of% M = 20i - 30k lb.ft for a position vector r = 3i + 2k.% (2) Given a force vector F = 2i + 3j, find out what would be the force% vector along a position vector r = 8j -3k.%


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Problem 2.2.3 Calculation of a resultant force and moment

Four forces act on a cross sectional piece of a truck frame (seeFigure P3.76). Calculate the resultant force and moment atpoint A, and find a point B somewhere along the X-axis wherethe resultant alone will represent the effect of all the fourforces. Here, let F kN1 10= ,F kN2 10= , F kN3 3= , andF kN4 4= .

MATLAB File

% 1 2 3 4 5 6 7

% ---------------------------------------------------------------------------

% AUTHOR: Dhaval Shah

% LAST UPDATE: August 4, 2000

%% REFERENCE: Problem 3-76 on pg. 166 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing a resultant force and moment% at point A for a given force system.% The next part of the problem is to find a point B on the x% axis where the resultant alone will represent all the forces.%%% CONCEPT KEYWORDS: Rigid Bodies: Varignon's Theorem%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% O Point O is the origin Given% A() The x, y coordinates of point A Given% B() The x, y coordinates of point B To be computed


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% C() The x, y coordinates of point C (F2) Given% D() The x, y coordinates of point D (F4) Given% E() The x, y coordinates of point E (F3) Given% F() Resultant Force vector about point A Given% M() Moment vector about point A Given% Rda() Relative position vector AD Intermediate qty.% Rea() Relative position vector EA Intermediate qty.% Rca() Relative position vector CA Intermediate qty.% Roa() Relative position vector OA Intermediate qty.

% ----------------------------------------------------------------------------

% (X, Y, Z) Coordinates of points A, D, E, F, O A(1) = 0.3; A(2) = 0.3; A(3) = 0; C(1) = 0.6; C(2) = 0; C(3) = 0; D(1) = 0; D(2) = 0.6; D(3) = 0; E(1) = 0.6; E(2) = 0.6; E(3) = 0; O(1) = 0; O(2) = 0; O(3) = 0;

% F1, F2, F3, F4 as vectors

F1(1) = 0; F1(2) = 10; F1(3) = 0; F2(1) = 0; F2(2) = 10; F2(3) = 0; F3(1) = -3; F3(2) = 0; F3(3) = 0; F4(1) = 4; F4(2) = 0; F4(3) = 0;

% Define Rd/a, Re/a, Rc/a, Ro/a D = [D(1); D(2); D(3)]; E = [E(1); E(2); E(3)]; C = [C(1); C(2); C(3)]; A = [A(1); A(2); A(3)]; O = [O(1); O(2); O(3)];

Rda = D - A; Rea = E - A; Rca = C - A; Roa = O - A;

% ================ calculation of moment=================================% Moment of the above force vectors about point A involves cross-products% M = Rd/a X F4 + Re/a X F3 + Rc/a X F2 + Ro/a X F1 M = cross (Rda, F4) + cross (Rea, F3) + cross (Rca, F2) + ...

cross (Roa, F1); fprintf ('Vector M = [(%g)i + (%g)j + (%g)k] kN-m\n', M(1), M(2), M(3));

%================= calculation of resultant force F ======================== F = F1 + F2 + F3 + F4; fprintf ('Vector F = [(%g)i + (%g)j + (%g)k] kN\n', F(1), F(2), F(3));

%============================ part b ======================================% Given that point B lies on the X axis. B(2) = 0; B(3) = 0; BINC = 0.01; TOL = 0.1;

% We can find Bx, i.e. B(1) by computing the moment about B using% M = Rd/b X F4 + Re/b X F3 + Rc/b X F2 + Ro/b X F1% and then setting the z component of M to zero. This will give us an% equation with Bx as the unknown. Solving ...


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for BX = 0:BINC:0.6 B(1) = BX;

Rdb = D - B; Reb = E - B; Rcb = C - B;

Rob = O - B; M = cross (Rdb, F4) + cross (Reb, F3) + cross (Rcb, F2) + ... cross (Rob, F1);

if (abs(M(3)) < TOL) fprintf ('The coordinates of point B are (%g, %g) m\n', B(1), B(2)); break; end end

% ==============================================================================% FURTHER EXERCISES% (1) Calculate the resultant force and moment about point O.% (2) Compute a position vector from 'O' to 'A'. Use the resultant% force vector obtained above to compute a force vector along the% given position vector, Ro/a.


21

2.3 Distributed Forces: Centroids and Center of Gravity




Problem Title

Problem 2.3.1 Problem 4.15 on page 190 Centroid of a quadrant of an ellipseProblem 2.3.2 Problem 4.51 on page 205 Centroid of a jetwayProblem 2.3.3 Problem 4.60 on page 209 Equivalent force in an automobile


22

Problem 2.3.1 Centroid of a quadrant of ellipse

Calculate the centroid of the area found by the quadrant of anellipse (see Figure P4.15). The equation of an ellipse is

xa

yb

FHG

IKJ + F

HGIKJ =

2 2

1

%0 1 2 3 4 5 6 7% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 4-15 on pg. 190 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the centroid of the area% formed by a quadrant of an ellipse whose equation is:% (x/a)^2 + (y/b)^2 = 1.

% CONCEPT KEYWORDS: Centroids of Areas by First Moment.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% O Point O is the origin Given% a length on the major axis (X) Given% b length on the minor axis (Y) Given% A Area of the ellipse To be computed% Xc X coord. of centroid of the ellipse To be computed% Yc Y coord. of centroid of the ellipse To be computed

% ----------------------------------------------------------------------------

% To calculate the centroid of the ellipse, we first need to calculate% the area(A) as follows. Consider a vertical strip of infinitesimally% small thickness 'dx' at any 'x' having height of 'y'. Let the area of% this strip be dA = y.dx. As this strip moves along the x axis from '0'% to 'a', we need to integrate the above equation dA from '0' to 'a'.

% Defining the symbols to be used for integration. syms a b theta

% We now change from the cartesian coordinate system to the polar coordinates x = a*cos(theta); y = b*sin(theta);

% Taking the derivative of the above equations to obtain the area A dx = diff(x,theta);


23

dy = diff(y,theta);

% A = integral (xdy) dA = x*dy; A = int(dA,0,pi/2);

% To obtain the centroid of the given ellipse: Yc = [integral (ydA)]/A% Xc = [integral (xdA1)]/A1 dA1 = y*dx; A1 = int(dA1,0,pi/2); cgy = y*dA; cgx = x*dA1; Yc = int(cgy,0,pi/2)/A; Xc = int(cgx,0,pi/2)/A1;

% Display the results fprintf('The Area formed by the quadrant of the ellipse is: ');pretty(A) fprintf ('The X coordinate of the centroid is: '); pretty(Xc) fprintf ('The Y coordinate of the centroid is: '); pretty(Yc)

% FURTHER EXERCISES% (1) How would you modify the above program to obtain the area, and the centroid% of the semi ellipse.% (2) If we form a rectangle of dimension (a X b), then write a code to calculate% the area obtaind by deducting the quadrant of the ellipse from the rectangle.% (3) Solve the above problem for a special case when a = b.


24

Problem 2.3.2 Centroid of a jetway

The jetway leading to an aircraft is a telescopic device formedby a series of interconnected rectangular parallelepiped. Thedesigner of this device needs to know where the center ofgravity is at each position as the jetway telescopes out alongthe X-axis. The length of each rectangular parallelepiped is20 ft , and the walls are 3in thick. The first section is

20 10 10ft ft ft× × . Using the centreline coordinate systemillustrated in Figure P4.51, compute the position of thecentroid as the section extends from zero to 15 ft past theedge of the first section.

MATLAB File

% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 4-51 on pg. 205 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: To obtain the centre of gravity of the jetway leading to an% aircraft at each position as it telescopes out along the% x-axis. Given the length of each rectangular parallelepiped% is 20ft.and the walls are 3in. thick. The first section is% 20ft X 10ft X 10ft

% CONCEPT KEYWORDS: Centroids, Composite bodies.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% L1, B1, H1 Length, breadth, height of the% first section Given% T Wall Thickness Given% B2, H2 Breadth and height of telescopic% section To be computed% Lt Length of the telescopic section Given% V1 Volume of first section To be computed% V2 Volume of telescoping section To be computed% X1, Y1, Z1 X,Y,Z coordinates of the centroid of % the first section Given% X2, Y2, Z2 X,Y,Z coordinates of the centroid of % telescoping section To be computed% X, Y, Z X, Y, Z coordinates of the centroid% jetway To be computed% ----------------------------------------------------------------------------


25

% Set default values: Units feet. L1 = 20; B1 = 10; H1 = 10; T = 3/12; AInc = 1; % calculate the volume of the first section: V1 = V1outer - V1inner V1outer = L1*B1*H1; V1inner = L1*(B1-T)*(H1-T); V1 = V1outer - V1inner;% Define the X, Y, Z centroid of the first section X1 = L1/2; Y1 = 0; Z1 = 0;% Compute the breadth and the length of the telescopic section. B2 = B1 - T; H2 = H1 - T;% Now run through the loop for different values of the length of the jetway. I = 0; for Lt = 0:AInc:15 I = I + 1;% Volume of the (telescopic) second section

L2 = Lt; V2outer = L2*B2*H2;

V2inner = L2*(B2-T)*(H2-T); V2 = V2outer - V2inner; X2 = 20 + (Lt/2);% Compute the X, Y, Z coordinate of the centroid of the entire jetway X(I) = (V1*X1 + V2*X2)/ (V1 + V2); Lj(I) = Lt; end

% Plot the X coordinate of the centroid of the entire section against the% different values of the lengths of the jetway. plot (Lj, X) grid Xlabel ('Length (ft)') Ylabel ('Centroid (X) (ft)')

% ==============================================================================% FURTHER EXERCISES% (1) Would the centroid of the entire section change, if the breadth and the height% of the parallelopiped is changed?% (2) If one of the walls of the parallelopiped is kept open such that the jetway is% is open on one of its sides, what would be the new centroid of the entire% section for Lj = 10ft?


26

Problem 2.3.3 Equivalent force in an automobile

A side-impact air bag for an automobile may be modeled as a distributed load by w x x lb ft( ) ( cos( ))= +100 1 . Compute theequivalent force and point of action for this system. Use Figure P4-60.

MATLAB File

% ---------------------------------------------------------------------------



%

% REFERENCE: Problem 4-60 on pg. 209 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: To compute the equivalent force and its point of action% for a system subjected to a distributed load by% w(x) = 100(1 + x*cos(x))%% CONCEPT KEYWORDS: Centroid of Area, First Moment.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% W Distributed load Given% A Area under the load curve To be computed% Xc X coord. of centroid of the distributed% load To be computed% ----------------------------------------------------------------------------

% Set absolute values:


27

a = 2; wc = 100;

% To calculate the centroid of the distributed load we need to calculate% the area, A. Consider a horizontal strip of infinitesimally small% thickness 'dx' at any 'x' having length of 'w(x)'. Let the area of this% strip be dA = dA = w(x)dx. As this strip moves along the x axis, we need% to integrate with the limits of integration from '0' to 'a'.

% Numerically integrate using Trapezoidal RuleA = 0; xinc = 0.01;

for x = 0:xinc:a; x1 = x + xinc; A = A + 0.5 * xinc * wc * ((1 + x*cos(x)) + (1 + x1*cos(x1))); end% Display the results.

fprintf ('The equivalent force in %g lb\n', A);

% To obtain the centroid of the given load, Xc = [integral (xdA)]/AXc = 0; xinc = 0.01;

for x = 0:xinc:a; x1 = x + xinc; Xc = Xc + 0.5 * xinc * wc * (x*(1 + x*cos(x)) + ... x1*(1 + x1*cos(x1))); end Xc = Xc/A;% Display the results fprintf ('The X coordinate of the centroid is %g ft\n', Xc);

% ==============================================================================% FURTHER EXERCISES% (1) What would be the desirable features of the distributed load?


28

2.4 Equilibrium of Rigid Bodies

Four problems are solved in this section.



Problem Title

Problem 2.4.1 Problem 5.31 on page 250 Overhanging roofProblem 2.4.2 Problem 5.35 on page 252 Tow truck mechanismProblem 2.4.3 Problem 5.73 on page 278 Folding platformProblem 2.4.4 Problem 5.80 on page 281 Power-line pole


29

Problem 2.4.1 Overhanging roof

An overhanging roof for an outdoor café consists of a beam 5m long of mass 100 kg (See Figure P5.31). Consider designingthe support system. That is, choose where to attach the cable(pick l ) so that the tension in the cable and the reaction forceat the pin are the smallest. Compute values of the reactionforces for l = 0 5115 5. , , . ,...., .

MATLAB File

% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 5-31 on pg. 250 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with finding out an appropriate length on% the beam of mass 100kg to attach the cable so that the tension% in the cable and the reaction force at the pin are the smallest.% The cable is connected between point A on the wall and point B% on the beam.%% CONCEPT KEYWORDS: Equilibrium, Rigid bodies.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% T Force in the cable To be computed% L Length at which the cable needs to be attached To be computed% Theta Angle between horizontal and the cable Given% h Height between the pin connection on the wall% and point A (where cable is anchored). Given% M Mass of the beam Given% AInc Increment to be used with length l To be defined% TMin Min. force in cable To be computed% MinStress Min. stress in cable To be computed

% ----------------------------------------------------------------------------% Set default values h = 1; M = 100; AInc = 0.5; g = 9.81; W = M * g; % Weight of the beam.

% ===== Part 1 of the problem ================================================


30

% Resolving the tension in the cable into its 2 components.% Th = T*cos(ThetaR);% Tv = T*sin(ThetaR);% Taking moments about the hinge to obtain the value of Tv:% Tv X L = M X 2.5% Tv = 2.5M/L% Now using the equilibrium equations along X and Y direction: J = 1; % array increment

for I = 0.5:AInc:5% Length 'l' varying from 0.5 to 5 and stored in an array L L(J) = I; ThetaV(J) = atan(h/L(J)); Tv(J) = 2.5*W/(L(J)); % Vertical component of the tension Rv(J) = W - Tv(J); % Vertical component of reaction% Compute the tension in the cable

T(J) = Tv(J) / sin(ThetaV(J)); % Tension in the cable. Th(J) = T(J)*cos(ThetaV(J)); % Hor. component of the tension Rh(J) = Th(J); % Horizontal component of raction R(J) = sqrt((Rv(J))^2 + (Rh(J))^2); % Reaction at the hinge J = J + 1; end

% Now we will plot the force in cable 'T' and reaction 'R' versus the length 'L'plot (L, T),hold on, plot(L, R),

text (L(2),T(2),'Tension'), text(L(5),R(5),'Reaction') grid, xlabel ('Length (m)') ylabel ('T (N)')

% Obtain the minimum values of the tension and the reaction forces TMin = min(T); RMin = min(R); for I = 1:1:J-1 if T(I) == TMin LMin = L(I); end if R(I) == RMin LRMin = L(I); end end

% Display the resultfprintf ('The minimum tension in the cable is Tmin=')fprintf ('%g N at %g m from the pin\n', TMin, LMin);

fprintf ('The minimum reaction in the cable is Rmin=') fprintf ('%g N at %g m from the pin\n', RMin, LRMin);

% ==============================================================================% FURTHER EXERCISES% (1) Compute the minimum tension in the cable by varying the height 'h'% and keeping the length 'L' at 'LMin' obtained above.% (2) If the beam cross section varies such that for first 2.5m from the% hinge, the cross-sectional area is twice as much as the rest of the% beam, calculate the minimum tension in the beam for M = 100 kg, L = 4m.


31

Problem 2.4.2 Tow truck mechanism

Consider the tow truck mechanism of Figure P5.35, and calculate the value of the reactionforces and the tension in the cable for values of θ from 0 to 60° in increments of 5° .Incrementing θ in this manner corresponds to changing the position of the motor used tohoist an object (usually an automobile) relative to the truck frame. Such information isuseful in the design of the hoist system. Is there a preferred position θ , where themagnitude of the reaction force at A is a minimum?

MATLAB File

% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 5-35 on pg. 251 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the value of the reaction% forces and the tension in the cable for different values of% 'Theta' from 0 to 60 degrees in increments of 5 degrees.% Calculate the value of 'Theta' at which the reaction is minimum.%% CONCEPT KEYWORDS: Equilibrium, Rigid bodies.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% T Force in the supporting cable To be computed% Tac Force in the hoist cable To be computed% Lac Length at bar AC Given% Theta Angle between hoist cable and the lift bar% AC Given% Beta Angle between supporting cable and the% lift bar AC Given% Alpha Angle between lift bar AC and horizontal Given% Rp, Rt Reaction components at A To be computed% AInc Increment to be used with angle Theta To be defined% TMin Min. force in cable To be computed

% ----------------------------------------------------------------------------% Set default values


32

W = 1500; AInc = 5; Lac = 8; Beta = 20; Alpha = 60; AngleToRadians = pi/180; BetaR = Beta * AngleToRadians; AlphaR = Alpha * AngleToRadians;

% ===== Part 1 of the problem ================================================% Force in the hoist cable is equal to the supported weight

Tac = W;% Resolving the tension in the cable into its 2 components.% Tacp is the component perpendicular to the lift bar and% Tact is along the bar.% Tt = Tac*cos(Theta);% Tp = Tac*sin(Theta);% Taking moments about point A, we can obtain T.% T*sin(Beta) * (2/3)Lac = Lac*W*cos(Alpha) - Lac*Tp (eqn. (A))

% Now using the equilibrium equations along X and Y direction:% Angle Theta is varying from 5 to 60 and is stored in array ThetaV. J = 0; % array increment

for I = 0:AInc:60 J = J + 1; ThetaX = I; ThetaR(J) = ThetaX * AngleToRadians; Tacp = Tac*sin(ThetaR(J)); Tact = Tac*cos(ThetaR(J));% Using the equation(A) as shown above, we can compute the value of T

TX = (W*cos(AlphaR)*Lac - Tacp*Lac)/(sin(BetaR)*(2/3)*Lac);% Ensure that cable is always in tension if (TX < 0) J = J-1; break; end; T(J) = TX; ThetaV(J) = ThetaX;% Compute the two components of T (along and perpendicular to the bar)

Tt = T(J)* cos(BetaR); Tp = T(J)* sin(BetaR);

% Compute the two components of reaction R at A Rp(J) = W*cos(AlphaR) - Tacp - Tp; Rt(J) = W*sin(AlphaR) + Tact + Tt;

% Compute the resultant R(J) = sqrt((Rp(J))^2 + (Rt(J))^2); end% Now we will plot the force in cable 'T' and reaction 'R' versus the angle 'Theta'

plot (ThetaV, T),hold on, plot(ThetaV, R), text (ThetaV(2),T(2),'Tension'), text(ThetaV(5),R(5),'Reaction'), grid xlabel ('Theta (degrees)') ylabel ('T (lb)') TMin = min(T); RMin = min(R);

% Compute the angles at which the reaction and the tension in the cable are minimum


33

for I = 1:1:J if T(I) == TMin AngleMinT = ThetaV(I); end if R(I) == RMin AngleMinR = ThetaV(I); end end

% Display the resultfprintf ('The minimum reaction at point A is Rmin=')fprintf ('%g N when Theta = %g degrees\n', RMin, AngleMinR);fprintf ('The minimum force in the cable is Tmin=');fprintf ('%g N when Theta=%g degrees.\n', TMin, AngleMinT);

% ==============================================================================% FURTHER EXERCISES% (1) Compute the minimum tension in the cable by varying the length of lift bar Ac% from 2ft to 10ft.% (2) If the supporting cable was connected at point C at one end and maintaining its% other end at hinge, compute the maximum tension in the cable for different values% 'Theta' varying from 20 to 60 degrees.


34

Problem 2.4.3 Folding platform

A folding platform is used to hold parts, as well as toconserve floor space when not in use. The platform issupported by a hinge at C, which is assumed to supportnegligible moments, a leg at B, modeled as a frictionlesssupport, and a removable pin at A, modeled as athrustless bearing, again with negligible moments. If theplatform is loaded as illustrated in Figure P5.73,compute the reaction forces at A, B, and C. Ignore thethickness of the platform.

MATLAB File

% ---------------------------------------------------------------------------


% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 5-73 on pg. 278 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the value for the reaction% forces at points A, B, and C when a vertical force P% is applied at point D.%% CONCEPT KEYWORDS: Equilibrium, Rigid bodies.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- --------- ------% P Applied vertical force (at D) Given% O Point O is the origin Given% A,B,C,D X,Y,Z coordinates of points A,B,C,D Given


35

% X Reactions at A,B,C To be computed%==================================================================================

% Set default values P(1) = 0; P(2) = -3000; P(3) = 0; A(1) = 0; A(2) = 0; A(3) = 1.2; B(1) = 1; B(2) = 0; B(3) = 1.2; C(1) = 0.5; C(2) = 0; C(3) = 0; D(1) = 0.5; D(2) = 0; D(3) = 0.4; Ra(1) = 0; Ra(2) = 1; Ra(3) = 1; Rb(1) = 0; Rb(2) = 1; Rb(3) = 0; Rc(1) = 1; Rc(2) = 1; Rc(3) = 1;

% There are six unknowns (Ray, Raz, Rby, Rcx, Rcy, Rcz)% and six equations of equilibrium that we can write as% S X = Q

for i=1:1:6 Q(i) = 0; for j=1:1:6 S(i,j) = 0; end; end;% Force equilibrium equations

S(1,4) = 1; S(2,1) = 1; S(2,3) = 1; S(2,5) = 1; S(3,2) = 1; S(3,6) = 1;% Take moments about the hinge C S(4,1) = -1.2; S(4,3) = -1.2; S(5,2) = 0.5; S(6,1) = -0.5; S(6,3) = 0.5;

Rdc = D - C; M = cross (Rdc, P);% Generate the right-hand side vector

Q = [-P(1); -P(2); -P(3); -M(1); -M(2); -M(3)];% Solve

X = inv(S)*Q;% Display the results

fprintf ('The (y,z) reactions at point A are ');fprintf ('%g N and %g N\n', X(1), X(2));

fprintf ('The (y) reaction at point B is '); fprintf ('%g N\n', X(3));

fprintf ('The (x,y,z) reactions at point C are ');fprintf ('%g N, %g N and %g N\n', X(4), X(5), X(6));


36

Problem 2.4.4 Power-line pole

In Figure P5.80, power-line pole is held in place by aball-and-socket connection at A, two cables at C and B,and a rod at D. Ignoring the mass of the pole, computethe reaction forces at A, B, C, and D for the loadindicated.

MATLAB File

%0 1 2 3 4 5 6 7% ------------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 5-80 on pg. 281 in Engineering Mechanics Statics by Soutas.%

% PROBLEM OVERVIEW: This problem deals with computing the tensions T1, T2, R% in order to compute the reaction forces at A, D, C and B.

% CONCEPT KEYWORDS: Free-body diagrams, Equilibrium equations%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% O Point O is the origin Given% Ax,Ay,Az The x, y, z coordinates of point A Given% Bx,By,Bz The x, y, z coordinates of point B Given% Cx,Cy,Cz The x, y, z coordinates of point C Given% Dx,Dy,Dz The x, y, z coordinates of point D Given% Ex,Ey,Ez The x, y, z coordinates of point E Given% Fx,Fy,Fz The x, y, z coordinates of point F Given


37

% Gx,Gy,Gz The x, y, z coordinates of point G Given% T1,T2,R Forces in cords DC,BE,DF respectively To be computed% Lzcg,Lxcg,Lycg Direction cosines of chord GC To be computed% Lzdf,Lxdf,Lydf Direction cosines of chord DF To be computed% Lzbe,Lxbe,Lybe Direction cosines of chord BE To be computed% Lgc Length of chord GC To be computed% Ldf Length of chord DF To be computed% Lbe Length of chord BE To be computed% F Force applied Given% Rxa Reaction at point A along X axis To be computed% Rya Reaction at point A along Y axis To be computed % Rza Reaction at point A along Z axis To be computed

% ---------------------------------------------------------------------------------% Set default values FFx = -30; FFy =0; FFz = 0;

F = FFx; Ax = 0; Ay = 0; Az = 0; Bx = 0; By = 1; Bz = 10; Cx = 0; Cy = -1; Cz = 10; Dx = 0; Dy = 0; Dz = 7; Ex = 3; Ey = 3; Ez = 0;

Fx = -2; Fy = 2; Fz = 0;Gx = 3; Gy = 0; Gz = 0;

% ============================ part (a) ===========================================% Calculate the length of the cords

Lgc = sqrt((Cx-Gx)^2 + (Cy-Gy)^2 + (Cz-Gz)^2);Lbe = sqrt((Ex-Bx)^2 + (Ey-By)^2 + (Ez-Bz)^2);

Ldf = sqrt((Dx-Fx)^2 + (Dy-Fy)^2 + (Dz-Fz)^2);

% Calculate the direction cosines of cordsLzcg = (-Cz+Gz)/Lgc; Lxcg = (-Cx+Gx)/Lgc; Lycg = (-Cy+Gy)/Lgc;Lzbe = (Ez-Bz)/Lbe; Lxbe = (Ex-Bx)/Lbe; Lybe = (Ey-By)/Lbe;Lzdf = (Dz-Fz)/Ldf; Lxdf = (Dx-Fx)/Ldf; Lydf = (Dy-Fy)/Ldf;

% Form the equilibrium equations.% T1V = T1((Lxcg)i + (Lycg)j + (Lzcg)k))% T2V = T2((Lxbe)i + (Lybe)j + (Lzbe)k))% RV = R((Lxfd)i + (Lyfd)j + (Lzfd)k))% MaV = Rc/a X T1V + Rb/a X T2V + Rd/a X R% Compute components of tension vectors T1V, T2V, R:

T1x = Lxcg; T1y = Lycg; T1z = Lzcg;T2x = Lxbe; T2y = Lybe; T2z = Lzbe;Rx = Lxdf; Ry = Lydf; Rz =Lzdf;

% Compute relative position vectorRcax = (Ax-Cx); Rcay = (Ay-Cy); Rcaz = (Az-Cz);Rbax = (Ax-Bx); Rbay = (Ay-By); Rbaz = (Az-Bz);Rdax = (Ax-Dx); Rday = (Ay-Dy); Rdaz = (Az-Dz);

% Define arrays for all position vectors and tensions T1, T2, RRca = [Rcax; Rcay; Rcaz]; Rba = [Rbax; Rbay; Rbaz];

Rda = [Rdax; Rday; Rdaz]; T1V = [T1x; T1y; T1z]; T2V = [T2x; T2y; T2z]; RV = [Rx; Ry; Rz];% Compute components of MaV :M1

M1 = cross(Rca,T1V);M2 = cross(Rba,T2V);


38

M3 = cross(Rda,RV);% Compute the moments due to the given force F:

Rx = 0; Ry = 0; Rz = 10; R = [Rx; Ry; Rz]; F = [Fx; Fy; Fz];

Mf = cross(R,F);% Solve the linear equations [A][x] = [B]% A = [ Ax Ay Az T1 T2 T3 ]

a = [1 0 0 Lxcg Lxbe Lxdf]; b = [0 1 0 Lycg Lybe Lydf]; c = [0 0 1 Lzcg Lzbe Lzdf]; d = [0 0 0 M1(1) M2(1) M2(1)]; e = [0 0 0 M1(2) M2(2) M3(2)]; f = [0 0 0 M1(3) M2(3) M3(3)];

A = [d;e;f;a;b;c]; B = [-Mf(1);-Mf(2);-Mf(3);-FFx;0;0]; X = inv(A)*B;

% Display resultsfprintf('The reaction at A in x direction is Xa = %g\n',X(1));fprintf('The reaction at A in x direction is Ya = %g\n',X(2));fprintf('The reaction at A in x direction is Za = %g\n',X(3));

% FURTHER EXERCISES% (1) Solve the above problem for F =50j + 20k.% (2) If the connection at A is not a ball and socket type and if% Ma = 25i + 45j + 50k for F = 60i + 30K, calculate the tension in the% cables T1, T2, R.


39

2.5 Truss Analysis

Five problems are solved in this section.



Problem Title

Problem 2.5.1 Problem 6.11 on page 301 Cantilevered truss analysisProblem 2.5.2 Problem 6.20 on page 302 Truss analysis of a joack support standProblem 2.5.3 Problem 6.60 on page 316 Space frames supporting window-

washing scaffoldProblem 2.5.4 Problem 6.85 on page 329 Lever systemProblem 2.5.5 Problem 6.95 on page 333 Plow mechanism


40

Problem 2.5.1 Cantilevered truss analysis

Consider the cantilevered truss as sketched in FigureP6.9, and calculate the forces in each member if a windload of approximately 1000 N is added at joint D. Thatis, let L N L N1 21000 500= =, , and L N3 1000= .State which members are in tension and which are incompression, and determine the reaction forces as well.

MATLAB File

% ------------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 6-11 on pg. 301 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the internal forces in the% members and the reactions in the truss subjected to a wind load% of 1000KN at joint D. Given L2 = 500N and L1 = 1000N are the% forces acting joints C and E respectively.

% CONCEPT KEYWORDS: Free-body diagrams, Equilibrium equations%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% L1,L2,L3 Forces applied at joints C, E, D% respectively To be computed% Lac,Lab,Lcd,Lce Lengths of different truss members To be computed% Fac,Fab,Fcd,Fce Forces in different truss members To be computed% Fbc,Fdb,Fde Forces in different truss members To be computed% Theta1 Angle between CE and DE To be computed% Theta2 Angle between AC and BC To be computed% Theta3 Angle between BC and BD To be computed% Rxa Reaction at point A along X axis To be computed% Rya Reaction at point A along Y axis To be computed % Rxb Reaction at point B along X axis To be computed % Ryb Reaction at point B along Y axis To be computed

% ---------------------------------------------------------------------------------% Set default values:

L1 = 1000; L2 = 500; L3 = 1000;


41

Lac = 1; Lce = 2; Lab = 1.5; Lcd = 1;

% Compute the angles :Theta1 = atan(Lcd/Lce);

Theta2 = atan(Lab/Lac);

% Consider the equilibrium of joint E:Fed = L2/sin(Theta1);

Fce = Fed*cos(Theta1);

% Consider the equilibrium of joint D:Fdb = (Fed*cos(Theta1)+L3)/cos(Theta1);

Fcd = Fdb*sin(Theta1) - Fed*sin(Theta1);

% Consider the equilibrium of joint C:Fcb = (L1 - Fcd)/sin(Theta2);

Fca = Fcb*cos(Theta2) + Fce;

% Consider the equilibrium of joint B:Fba = Fcb*sin(Theta2) + Fdb*sin(Theta1);

% Compute the support reactionsRax = Fca;

Ray = Fba; Rbx = Fdb*cos(Theta1) + Fcb*cos(Theta2); Rby = 0;

% Display results:fprintf('The forces in members are: Fec = %gN(T)\nFed = %gN(C)\n',...

Fce, Fed); fprintf('Fdc = %gN(C)\nFdb = %gN(C)\nFbc = %gN(C)\nFba = %gN(T)\n',... Fcd, Fdb, Fcb, Fba);

fprintf('Fca = %gN(T)\n',Fca);

% ==============================================================================% FURTHER EXERCISES% (1) Compute the force in member CD if the force L3 = 1000N.% (2) Vary the length of AB from 1 to 10ft and calculate the corresponding force in% member CD.


42

Problem 2.5.2 Truss analysis of a jack support stand

Suppose that you are asked to design the stand illustrated inFigure P6.19 by choosing the angle θ that gives that gives thesmallest value of force in each member. A constraint on yoursolution is that the stand must be 0.15 m high (that is, DB = 0.15m) and that the width of the stand (as measured by the line AC)must be less than 0.4 m. The maximum load the stand must holdis 1500 N.

MATLAB File

% ------------------------------------------------------------------------------

% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 6-20 on pg. 302 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the smallest value of force% in each member by choosing the angle 'Theta'.% Given that the stand BD must be 0.15m high and the width of the% stand AC must be less than 0.4 m. The maximum load on the stand% is 1500N.

% CONCEPT KEYWORDS: Free-body diagrams, Analysis of structures%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% F Applied load at joint B Given% Lad, Ldc, Lbd Lengths of different truss members Given% Fad, Fab, Fbc, Fdc Forces in different truss members To be computed% Theta Angle between AD and AB To be computed% Ha Reaction at point A along X axis To be computed% Va Reaction at point A along Y axis To be computed % Vc Reaction at point C along Y axis To be computed

% ---------------------------------------------------------------------------------% Set default values:

F = 1500; Lac = 0.2; Lcd = 0.1; Lad = 0.1; Lbd = 0.15; AInc = 0.01;

RadiansToDegrees = 180/pi;


43

% Run through the loop for different values of Lad.% We will assume that the stand is symmetric.

J = 0;for I = 0.01:AInc:0.3

J = J + 1; Lad(J) = I; Lcd(J) = Lad(J);% Compute Theta for differnt values of Lad Beta(J) = atan(Lbd/Lcd(J)); Theta(J) = atan(Lbd/Lad(J)); end% Compute the different forces in each member and reactions% by using equilibrium equations

FTotalMin = 100*F;for I = 1:1:J

% Compute vertical reaction at C Vc = (F*Lad(I))/(Lad(I) + Lcd(I));

% Compute vertical reaction at A Va = F - Vc;

Fab(I) = Va/sin(Theta(I));% Apply equilibrium equations at A

Fbc(I) = Vc/sin(Beta(I));% Apply equilibrium equations at B

Fdc(I) = Fbc(I)*cos(Beta(I));% Apply equilibrium equations at C

Fad(I) = Fab(I)*cos(Theta(I));% Apply equilibrium equations at D

Fdb(I) = 0; FTotal = abs(Fab(I)) + abs(Fbc(I)) + ... abs(Fdc(I)) + abs(Fad(I)) + abs(Fdb(I)); if (FTotal < FTotalMin) FTotalMin = FTotal; LMin = Lad(I); ThetaMin = RadiansToDegrees*Theta(I); end% Convert from radians to degrees ThetaD(I) = RadiansToDegrees*Theta(I); end fprintf ('Min force is for Lad=%g m or Theta=%g degrees\n',... LMin, ThetaMin);

% Plot different forces in each member against Theta plot (ThetaD,-Fab),text(ThetaD(2),-Fab(2),'Fab') hold on,plot(ThetaD,-Fbc),text(ThetaD(4),-Fbc(4),'Fbc') hold on,plot(ThetaD,Fdc),text(ThetaD(5),Fdc(5),'Fdc') hold on,plot(ThetaD,Fdb),text(ThetaD(10),Fdb(10),'Fdb') hold on,plot(ThetaD,Fad),text(ThetaD(10),Fad(7),'Fad') grid on xlabel ('Theta (degrees)'); ylabel ('Member Force (N)');

% ==============================================================================% FURTHER EXERCISES% (1) Keeping the angle 'Theta' constant at 25 degrees, vary the height of% member DB between 0.1 m and 0.3 m and plot the forces in each member% against the height DB.% (2) Reformulate the design problem statement to obtain a more meaningful


44

% design result.


45

Problem 2.5.3 Space frames supporting window-washing scaffold

A window-washing scaffold is supported by two space frames,one of which is illustrated in Figure P6.60. The support at A isa ball and socket, and those at B and C are frictionlesssupports; C also has a pin restraint in the x-direction, to keepthe frame from rotating. Joint D is centered 1 m above theintersection of BC and AE, which coincides with the midpointof line BC, which is 1 m long. (a) How many unknowns andhow many equations does this system represent if solved bythe method of joints? (b) Compute the reaction forces at pointsand the forces in each member.

MATLAB File

% ---------------------------------------------------------------------------% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 6-60 on pg. 316 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing a the reaction forces% and the force in each member of a window-washing% scaffold supported by 2 space trusses.%%% CONCEPT KEYWORDS: Analysis of Structures, Space Truss%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% O Point O is the origin Given% Ax,Ay,Az The x, y, z coordinates of point A Given% Bx,By,Bz The x, y, z coordinates of point B Given% Dx,Dy,Dz The x, y, z coordinates of point D Given% Ex,Ey,Ez The x, y, z coordinates of point E Given% Cx,Cy,Cz The x, y, z coordinates of point F Given% Fab,Fac,Fad Force vectors in members AB, AC, AD To be computed% Feb,Fec,Fed Force vectors in members EB, EC, ED To be computed% Fbc,Fbd,Fcd Force vectors in members BC, BD, CD To be computed% Nab,Nad,Nac Unit vectors along AB, AD, AC To be computed% Neb,Ned,Nec Unit vectors along EB, ED, EC To be computed% Nbc,Nbd,Ncd Unit vectors along BC, BD, CD To be computed% RBz Reaction at B along Z axis To be computed% RCx Reaction at C along X axis To be computed% RCz Reaction at C along Z axis To be computed% RAx,RAy,RAz Reactions at A along X, Y, Z axis To be computed

% ----------------------------------------------------------------------------


46

% Set default valuesF = 10;

Ax = 0; Ay = 0; Az = 0; Dx = 0.5; Dy = 0; Dz = 1; Ex = 2; Ey = 0; Ez =0; Cx = 0.5; Cy = 0.5; Cz = 0; Bx = 0.5; By = -0.5; Bz = 0;% Define M = (Mx)i + (My)j + (Mz)k% Define F = (Fx)i + (Fy)j + (Fz)k% Define R = (Rx)i + (Ry)j + (Rz)k% Rd/a has 3 components : Rdax, Rday, Rdaz:

Rdax = (Dx - Ax); Rday = (Dy - Ay); Rdaz = (Dz - Az);% Unit vector along Da:

Lda = sqrt(Rdax^2 + Rday^2 + Rdaz^2);Ndax = Rdax/Lda; Nday = Rday/Lda; Ndaz = Rdaz/Lda;

% Rb/a has 3 components : Rbax, Rbay, Rbaz:Rbax = (Bx - Ax); Rbay = (By - Ay); Rbaz = (Bz - Az);

% Unit vector along BA:Lba = sqrt(Rbax^2 + Rbay^2 + Rbaz^2);Nbax = Rbax/Lba; Nbay = Rbay/Lba; Nbaz = Rbaz/Lba;

% Rc/a has 3 components : Reax, Reay, Reaz:Rcax = (Cx - Ax); Rcay = (Cy - Ay); Rcaz = (Cz - Az);

% Unit vector along CA:Lca = sqrt(Rcax^2 + Rcay^2 + Rcaz^2);Ncax = Rcax/Lca; Ncay = Rcay/Lca; Ncaz = Rcaz/Lca;Rebx = (Ex - Bx); Reby = (Ey - By); Rebz = (Ez - Bz);

% Unit vector along EB:Leb = sqrt(Rebx^2 + Reby^2 + Rebz^2);Nebx = Rebx/Leb; Neby = Reby/Leb; Nebz = Rebz/Leb;

% Rb/a has 3 components : Rbax, Rbay, Rbaz:Recx = (Ex - Cx); Recy = (Ey - Cy); Recz = (Ez - Cz);

% Unit vector along BA:Lec = sqrt(Recx^2 + Recy^2 + Recz^2);Necx = Recx/Lec; Necy = Recy/Lec; Necz = Recz/Lec;

% Rc/a has 3 components : Reax, Reay, Reaz:Redx = (Ex - Dx); Redy = (Ey - Dy); Redz = (Ez - Dz);

% Unit vector along CA:Led = sqrt(Redx^2 + Redy^2 + Redz^2);Nedx = Redx/Led; Nedy = Redy/Led; Nedz = Redz/Led;Rbcx = (Bx - Cx); Rbcy = (By - Cy); Rbcz = (Bz - Cz);

% Unit vector along Da:Lbc = sqrt(Rbcx^2 + Rbcy^2 + Rbcz^2);Nbcx = Rbcx/Lbc; Nbcy = Rbcy/Lbc; Nbcz = Rbcz/Lbc;

% Rb/a has 3 components : Rbax, Rbay, Rbaz:Rbdx = (Bx - Dx); Rbdy = (By - Dy); Rbdz = (Bz - Dz);

% Unit vector along BA:Lbd = sqrt(Rbdx^2 + Rbdy^2 + Rbdz^2);Nbdx = Rbdx/Lbd; Nbdy = Rbdy/Lbd; Nbdz = Rbdz/Lbd;

% Rc/a has 3 components : Reax, Reay, Reaz:Rcdx = (Cx - Dx); Rcdy = (Cy - Dy); Rcdz = (Cz - Dz);

% Unit vector along CA:Lcd = sqrt(Rcdx^2 + Rcdy^2 + Rcdz^2);Ncdx = Rcdx/Lcd; Ncdy = Rcdy/Lcd; Ncdz = Rcdz/Lcd;

% Form the linear equations to determine the force in each member% and the reactions.% For joint A:% Ndax(AD) + Nbax(AB) + Ncax(AC) + Ax = 0;% Nday(AD) + Nbay(AB) + Ncay(AC) + Ay = 0;


47

% Ndaz(AD) + Nbaz(AB) + Ncaz(AC) + Az = 0;% For joint B:% -Nbax(BA) + Nbex(BE) + -Nbcx(BC) + -Nbdx(BD) = 0;% -Nbay(BA) + Nbey(BE) + -Nbcy(BC) + -Nbdy(BD) = 0;% -Nbaz(BA) + Nbez(BE) + -Nbcz(BC) + -Nbdz(BD) + Bz = 0;% For joint C:% -Ncax(CA) + Ncex(CE) + Nbcx(CB) + -Ncdx(CD) + Cx = 0;% -Ncay(CA) + Ncey(CE) + -Nbcy(CB) + -Ncdy(CD) = 0;% -Ncaz(CA) + Ncez(CE) + -Nbcz(CB) + -Ncdz(BD) + Cz = 0;% For joint E:% -Nedx(ED) + -Nebx(EB) + -Necx(EC) = 0;% -Nedy(ED) + -Neby(EB) + -Necy(EC) = 0;% -Nedz(ED) + -Nebz(EB) + -Necz(EC) + -10 = 0;% For joint D:% -Ndax(DA) + Ndbx(BE) + Ndcx(BC) + -Nedx(BD) = 0;% -Nday(DA) + Ndbz(BE) + Ndcy(BC) + -Nedy(BD) = 0;% -Ndaz(DA) + Ndbz(BE) + Ndcz(BC) + -Nedz(BD) = 0;

% Solve for: [A][X] = [B]% A = [ Fda Fba Fca Fed Fec Feb Fbc Fbd Fcd Ax Ay Az Cx Cz Bz]

a = [ Ndax Nbax Ncax 0 0 0 0 0 0 1 0 0 0 0 0 ];b = [ Nday Nbay Ncay 0 0 0 0 0 0 0 1 0 0 0 0 ];c = [ Ndaz Nbaz Ncaz 0 0 0 0 0 0 0 0 1 0 0 0 ];d = [ 0 -Nbax 0 0 0 Nebx -Nbcx -Nbdx 0 0 0 0 0 0 0 ];e = [ 0 -Nbay 0 0 0 Neby -Nbcy -Nbdy 0 0 0 0 0 0 0 ];f = [ 0 -Nbaz 0 0 0 Nebz -Nbcz -Nbdz 0 0 0 0 0 0 1 ];g = [ 0 0 -Ncax 0 Necx 0 Nbcx 0 -Ncdx 0 0 0 1 0 0 ];h = [ 0 0 -Ncay 0 Necy 0 Nbcy 0 -Ncdy 0 0 0 0 0 0 ];i = [ 0 0 -Ncaz 0 Necz 0 Nbcz 0 -Ncdz 0 0 0 0 1 0 ];j = [ 0 0 0 -Nedx -Necx -Nebx 0 0 0 0 0 0 0 0 0 ];k = [ 0 0 0 -Nedy -Necy -Neby 0 0 0 0 0 0 0 0 0 ];l = [ 0 0 0 -Nedz -Necz -Nebz 0 0 0 0 0 0 0 0 0 ];m = [-Ndax 0 0 Nedx 0 0 0 Nbdx Ncdx 0 0 0 0 0 0 ];n = [-Nday 0 0 Nedy 0 0 0 Nbdy Ncdy 0 0 0 0 0 0 ];o = [-Ndaz 0 0 Nedz 0 0 0 Nbdz Ncdz 0 0 0 0 0 0 ];

A = [ a; b; c; d; e; f; g; h; i; j; k; l; m ; n; o];B = [ 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 10; 0; 0; 0];x = inv(A)*B;

% Display resultsfprintf ('Fda = %g kN\n',x(1));fprintf ('Fba = %g kN\n',x(2));fprintf ('Fca = %g kN\n',x(3));fprintf ('Fed = %g kN\n',x(4));fprintf ('Fec = %g kN\n',x(5));fprintf ('Fed = %g kN\n',x(6));fprintf ('Fbc = %g kN\n',x(7));fprintf ('Fbd = %g kN\n',x(8));fprintf ('Fcd = %g kN\n',x(9));fprintf ('RAx = %g kN\n',x(10));fprintf ('RAy = %g kN\n',x(11));fprintf ('RAz = %g kN\n',x(12));fprintf ('RCx = %g kN\n',x(13));fprintf ('RCz = %g kN\n',x(14));fprintf ('RBz = %g kN\n',x(15));


48

% ==============================================================================% FURTHER EXERCISES% (1) Solve the above problem, if C has a pin restraint in the y-direction% instead of x-direction.% (2) Is the structure statically determinate if C has a pin restraint in both% x and y directions?


49

Problem 2.5.4 Lever system

Compute the forces at A, B, and C, as well as F, for the lever system of Figure P6.84 analytically in terms of the four lengths a, b, c,and d.

MATLAB File

% ------------------------------------------------------------------------------



%

% REFERENCE: Problem 6-85 on pg. 329 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the forces at A, B, and C,% as well as F, for the lever system in terms of P, a, b, c,% and d. Point 'D' is the point where 'P' is applied and 'E'% is the point where 'F' is applied.

% CONCEPT KEYWORDS: Analysis of Structures: Free Body Diagrams%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% F Unknown force applied at E To be computed% Lad, Lac, Lbc, Lbe Distances on the beam Given% P Force applied at 'D' Given% Vb Reaction at point B along X axis To be computed% Va Reaction at point A along Y axis To be computed % Vc Reaction at point C along Y axis To be computed

% ---------------------------------------------------------------------------------% Define variables or parameters that are to needed to compute the unknown quantities

syms a b c d F P


50

% Take moments about point 'C' to compute VaVa = (P*(c+d))/d;

% Take moments about point 'A' to compute VcVc = (P*c)/d;

% Take moments about point 'B' to compute FF = (Vc*a)/(b);

% Take moments about point 'C' to compute VbVb = (F*(b+a))/a;

% Display results:fprintf('The reactions at the joints are:\n');fprintf('Va:'); pretty(Va)fprintf('Vb:'); pretty(Vb)fprintf('Vc:'); pretty(Vc)fprintf('The force F = '); pretty(F)


51

Problem 2.5.5 Plow Mechanism

Compute the forces at the connections labeled for the plow mechanism in Figure P6.95, under the assumption that the blade of theplow provides a force of P = 10 kN and W = 1.5 kN. Neglect the weight of the other members of the machine.

MATLAB File

% ---------------------------------------------------------------------------


% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 6-95 on pg. 333 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the forces at the connections% for the plow mechanism, when the blade of plow provides a% force of P = 10kN and W = 1.5kN. Also, note that DE, BF% and GF are two force members.%

% CONCEPT KEYWORDS: Analysis of Structures, Free Body Diagrams%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% P, W Forces that the blade provides Given% Fde,Fbf,Fgf Forces in members DE, BF, GF To be computed% Ax,Ay Reaction at A along X, Y axis To be computed


52

% Fx,Fy Reaction at F along X, Y axis To be computed% Cx,Cy Reaction at C along X, Y axis To be computed% ----------------------------------------------------------------------------% Set default values:

P = 10; W = 1.5;

% From the geometry of figure, the angle CF makes with the horizontal is:Alpha = atan(0.1/0.3);

% The equilibrium equations for the blade portion are:% Summation Fx = Ax + 10 + Fgf*cos(alpha) = 0.% Summation Fy = Ay - 1.5 + Fgf*sin(alpha) = 0.% Summation Ma = W(0.05) - P(0.15) - Fgfcos(alpha)(0.3) = 0.

Fgf = (W*(0.05) - P*(0.15))/(0.3*cos(Alpha)); Ay = W - Fgf*sin(Alpha); Ax = -P - Fgf*cos(Alpha);

% The angle that ED makes with the horizontal is 'Beta':Beta = atan(0.2/0.3);

% Consider the equilibrium of the body 'FDC'. Forces 'Fx', 'Fy' are the pin% forces acting at 'F'. Fde acts at 'D' at an angle 'Beta'. 'Cx', 'Cy' are the% the pin forces at 'C'.% Sum Fx = Cx - Fx + Fdecos(Beta) = 0 ....(A)% Sum Fy = Cy + Fy - Fdesin(Beta) = 0 ....(B)% Md = Cy(0.05) + Cx(0.2) - Fy(0.45) = 0 ....(C)% The angle that FB makes with the horizontal is 'Theta':

Theta = atan(0.3/0.1);% Consider the equilibrium of the bar 'AFE'. Forces 'Fx', 'Fy' are the pin% forces acting at 'F'. Fde acts at 'E' at an angle 'Beta'. Ffb acts at% 'F' at an angle 'Theta'.% Sum Fx = Fx - Fdecos(Beta) - Fgfcos(Alpha) - Fbfcos(Theta) - Ax = 0 ...(D)% Sum Fy = -Fy + Fdesin(Beta) - Fgfsin(Alpha) + Fbfsin(Theta) - Ay = 0 ...(E)% Mf = Fdesin(Beta)(0.2) + Fdecos(Beta)(0.2) + Ay(0.3) - Ax(0.4) = 0% From Mf = 0, we get:

Fde = (Ax*(0.4) - Ay*(0.3))/(sin(Beta)*0.2 + cos(Beta)*0.2);

% Now we solve a set of linear equations 'A' to 'E' described above% to solve for Cx, Cy, Fx, Fy, Fbf.% A = [ Cx Cy Fx Fy Fbf ]

a = [ 1 0 -1 0 0]; b = [ 0 1 0 1 0]; c =[ 0.2 0.05 0 -0.45 0]; d = [ 0 0 1 0 -cos(Theta)]; e = [ 0 0 0 -1 sin(Theta)]; f = Fgf*cos(Alpha)+Fde*cos(Beta)+Ax; g = Fgf*sin(Alpha)-Fde*sin(Beta)+Ay;

A = [ a; b; c; d; e]; B = [ -Fde*cos(Beta); Fde*sin(Beta); 0; f; g]; x = inv(A)*B;

% Display results:fprintf('The reactions at A are Ax = %g kN, Ay = %g kN\n', Ax,Ay);fprintf('The force Fde = %g kN\n', Fde);

fprintf('The force Fgf = %g kN\n', Fgf); fprintf('The reactions at C are Cx = %g kN, Cy = %gkN\n', x(1),x(2)); fprintf('The reactions at F are Fx = %g kN, Fy = %gkN\n', x(3),x(4)); fprintf('The force Fbf = %g kN\n', x(5));


53

2.6 Beam and Frame Analysis

Two problems are solved in this section.



Problem Title

Problem 2.6.1 Problem 7.31 on page 357 Airplane wing acting as a cantileveredbeam

Problem 2.6.2 Problem 7.59 on page 377 Computations of tension in cablesegments


54

Problem 2.6.1 Airplane wing acting a cantilevered beam

A lifting load on an airplane wing is modeled byw x x x lb ft( ) sin( ) /= 1000 , and the wing is modeled as a10-ft-long cantilevered beam. (See Figure P7.31). Computethe reaction forces, as well as the shear force and the bendingmoment as a function of x. Plot the shear and the bendingmoment functions.

MATLAB File: This is the main m-file which calls ‘FV731.m’ and ‘FB731.m’ files to plot the shear and bending functions.

% ---------------------------------------------------------------------------


% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 7-31 on pg. 357 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: To compute the reaction forces and the internal shear force% and bending moment. Plot the SF and BM as a function of 'x'.% The loading is w(x) = 1000*x*sin(x) lb/ft.%% CONCEPT KEYWORDS: Shear Force, Bending Moments, Beams%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% w Distributed load Given% A Area under the load curve To be computed% L Length of the beam Given% AInc Increments to change the length'L' Given% ----------------------------------------------------------------------------

% Set default valuesL = 10;

w = 1000; AInc = 0.1;

AngleToRadians = pi/180;

% To calculate the total force acting on the beam we need to integrate the% distributed load w(x) along the length of the beam. The total load is the% area under the curve of the load w(x). Consider a vertical strip of% infinitesimally small thickness 'dx' at any 'x' having length of 'w(x)'.% Let the area of this strip be dA = w(x)dx. As this strip moves along% the x axis from '0' to 'L', we need to integrate the above equation dA


55

% from '0' to 'L'.x=0:0.1:10;

n = length(x); % Defines the length of the array 'x'

% Use the quad8 function for numerical integrationfor i = 1:n

Vx(i) = quad8 ('fV731',0,x(i),[],[],w,L); Mx(i) = quad8 ('fB731',0,x(i),[],[],w,L); end

% prepare for plotting figure(1); clf;

subplot (2,1,1)plot (x,Vx) % Plots shear force diagramxlabel ('x (ft)')ylabel ('V(x) (lb)')subplot (2,1,2)plot (x,Mx) % Plots bending moment diagramxlabel ('x (ft)')ylabel ('M(x) (lb-ft)')

% ==============================================================================% FURTHER EXERCISES% (1) Solve the same problem by taking a vertical strip moving along the w(x) axis.

MATLAB File fV731.m

function y = fV731 (u,w,L)% define the shear force functiony = w*(sin(u) - u.*sin(u)) - 7850;

MATLAB File fB731.m

function y = fB731(u,w,L)% define the bending moment functiony = 71349 - 7850*u - 2000*cos(u) - 1000*u.*sin(u);


56

Problem 2.6.2 Computation of tension in cable segments

Compute the tension in each cable segment and the angles α β, ,and γ for the traffic light system shown in Figure P7.59. Design

(choose) the hanger length l so that the top of the two lights hangto the same level for ease of viewing. The length of the cable is155 ft. Here, P lb1 125= and P lb2 100= .

MATLAB FILE: This is the main m-file. The equations are defined in m- file ‘sp759_fcn.m’ and the nonlinear equation solver is infile ‘mnewt.m’.

% ---------------------------------------------------------------------------

% AUTHOR: Dhaval Shah% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 7-59 on pg. 377 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: To compute the tension in each cable segment and the angles% alpha, beta, and gamma for the traffic light system. Given that% P1 = 125lb and P2 = 100lb hang from B and C as shown in the% figure.%% CONCEPT KEYWORDS: Internal Forces, Cables.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% L Length of Cable Given% P1, P2 Loads on the cable Given% alpha1 Angle made by cable AB with horz. To be computed% alpha2 Angle made by cable BC with horz. To be computed% alpha3 Angle made by cable CD with horz. To be computed% T1, T2, T3 Tensions in cables AB, BC, CD To be computed% ABx, BCx Horizontal lengths of AB, BC Given% CDx, ADy Lengths of CD, AD as shown in figure Given% ----------------------------------------------------------------------------% Set default values:

L = 155; P1 = 125; P2 = 100; ABx = 100; BCx = 25; CDx = 25; ADy = 0;


57

% put initial guesses into a column vector% T1 T2 T3 alpha1 alpha2 alpha3

IC = [400; 400; 400; 5; 15; 25];

% Use the Newton-Raphson method mnewt to solve the system% on nonlinear equations

soln = mnewt('sp759_fcn', 100, IC, 1.0e-4, 1.0e-4, L, P1, P2,... ABx,BCx,CDx,ADy);

% Display the resultsfprintf('\nFor L = %g ft the solution is\n', L);

fprintf('T1 = %8.3f lb\n', soln(1)); fprintf('T2 = %8.3f lb\n', soln(2));

fprintf('T3 = %8.3f lb\n', soln(3));fprintf('alpha = %8.3f lb\n', soln(4));fprintf('beta = %8.3f lb\n', soln(5));fprintf('gamma = %8.3f lb\n', soln(6));

MATLAB File sp759_fcn.m

function q = sp759_fcn (p, L, P1, P2, ABx, BCx, CDx, ADy)% set up the equilibrium equations

T1 = p(1); T2 = p(2); T3 = p(3);DegreesToRadians = pi/180;alpha1 = p(4)*DegreesToRadians;alpha2 = p(5)*DegreesToRadians;alpha3 = p(6)*DegreesToRadians;

q = zeros(6,1);q(1) = -T1*cos(alpha1) + T2*cos(alpha2);q(2) = -T1*sin(alpha1) + T2*sin(alpha2) - P1;q(3) = -T2*cos(alpha2) + T3*cos(alpha3);q(4) = -T2*sin(alpha2) + T3*sin(alpha3) - P2;q(5) = ABx*tan(alpha1) + BCx*tan(alpha2) + CDx*tan(alpha3) - ADy;

q(6) = sum([ABx BCx CDx]./cos([alpha1 alpha2 alpha3])) - L;


58

2.7 Friction

Two problems are solved in this section.



Problem Title

Problem 2.7.1 Problem 8.29 on page 400 Wedge on an inclined planeProblem 2.7.2 Problem 8.89 on page 412 Leveling of a refrigerator by using a

wedge


59

Problem 2.7.1 Wedge on an inclined plane

A 10-lb force is pushing a 40-lb block, as shown in Figure P8.4.The angle of inclination of the plane is β which varies from 0to 65° in increments of 5° . The coefficient of static frictionbetween the block and the incline is µ s = 0 75. , and the

coefficient of kinetic friction is µ k = 0 65. . Will the block slideon the plane? If it does, will it slide up or down the plane? Whatis the friction force between the block and the plane?

MATLAB File

% ---------------------------------------------------------------------------


% LAST UPDATE: August 4, 2000%% REFERENCE: Problem 8-29 on pg. 400 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with finding out if the block of 40lb,% subjected to a horizontal force of 10lb, starts sliding on% an inclined plane making an angle 'Beta'?%% CONCEPT KEYWORDS: Friction: Coulomb Friction.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% W Weight of the block Given% P Force applied to the block Given% Beta Angle of inclination of the% inclined plane To be varied

% Nus Coefficient of Static Friction Given% Nuk Coefficient of Kinetic Friction Given% Fs Static frictional force To be computed% Fk Kinetic frictional force To be computed% N Normal force perpendicular to% the inclined plane Intermediate qty% AInc Increments to change the angle 'Beta' Given% ----------------------------------------------------------------------------

% Set default valuesW = 40;P = 10;Nus = 0.75; Nuk = 0.65;AInc = 5;

AngletoRadians = pi/180;


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% Now run through the for loop to get the different values of Beta% from 0 to 65 degrees.

J = 0;for I = 0:5:65

J = J + 1; Beta(J) = I; BetaR(J) = I * AngletoRadians;% Define equilbrium equations to obtain the unknowns.% Components perpendicular to the inclined plane are equated.

N = W*cos(BetaR(J)); Fs = Nus * N;% Components along the plane are equated. Fact = (W*sin(BetaR(J)) + P); if Fs < Fact fprintf ('The block will slide for angle %g degrees.\n',... Beta(J)); Fk = Nuk * N; fprintf ('Now that the block slides, the frictional'); fprintf (' force is %g lb\n', Fk); else fprintf (' The block will not slide for angle %g degrees.\n',... Beta(J));

end end

% ==============================================================================% FURTHER EXERCISES% (1) Resolve the above problem if the force P as shown in the figure is at an% inclination of 20 degrees with the inclined plane.% (2) Compute the force P in the above problem such that the block stays at rest% for all the cases of Beta varying from 0 to 65 degrees.


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Problem 2.7.2 Leveling of a refrigerator by using a wedge

It is necessary to level the refrigerator shown in the Figure P8.87

with a 10° wedge system. The weight distribution is such that theweights on all feet of the 800-lb refrigerator are equal. Thecoefficient of friction between the feet and the contacting surfaceis 0.8, the coefficient of static friction between the wedges is 0.2,and the coefficient of static friction between the wedge and thefloor is 0.4. What is the minimum force P required to initiateleveling of the refrigerator?

MATLAB FILE:

% ---------------------------------------------------------------------------



%

% REFERENCE: Problem 8-89 on pg. 400 in Engineering Mechanics Statics by Soutas.%% PROBLEM OVERVIEW: This problem deals with computing the minimum force P that% is applied to the wedge to level the refrigerator. Given that% the wedge system is 10 degrees and the coefficient of friction% between wedges is 0.2, between the feet and wedge is 0.8 and% that between the wedge and the floor is 0.8.%% CONCEPT KEYWORDS: Friction, Coulomb Friction.%% VARIABLE DICTIONARY:% Variable Remarks Status% -------- ------- ------% W Weight of the refrigerator Given% P Force applied to the wedge To be computed% Beta Angle of inclination of the% wedge with horizontal Given% Nus1, Nus2, Nus3 Coefficient of Static Friction Given% Fs1 Frictional force between% feet and wedge To be computed% Fs2 Frictional force between two wedges To be computed% Fs3 Frictional force between% wedge and floor To be computed


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% N Normal force perpendicular to% the inclined plane of the wedge Intermediate qty% ----------------------------------------------------------------------------

% Set absolute values:W = 800;W1 = W/4; % weight on each foot of the refrigeratorNus1 = 0.8; Nus2 = 0.2; Nus3 = 0.4;Beta = 10;AngletoRadians = pi/180;BetaR = Beta * AngletoRadians;

% Consider the foot of the refrigerator in equilibrium and compute the force% normal to its foot

N1 = W1;% Compute the frictional force

Fs1 = N1*Nus1;% Consider the equilibrium of the wedge on which 'P' is applied.% Equating all the forces along the vertical:

N2 = W1/(cos(BetaR) - Nus2*sin(BetaR));

% Equating all the forces along the horizontal:Fs2 = Nus2*N2;P = Fs1 + N2*sin(BetaR) + Fs2*cos(BetaR);

% Display results:fprintf ('The minimum force P required to level the refrigerator');fprintf (' is: %g lb\n', P);

% ==============================================================================% FURTHER EXERCISES% (1) Solve the above problem for Beta = 5 degrees.

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