6.2 Truss - Two Dimensions

Transcription

1 6.2 Truss - Two Dimensions Trusses make it easy to carry heavy loads. A bridge over the river, roof over your house, the giant crane used in construction or unloading heavy cargo from a ship are some of common examples. Trusses are light too. It is just a framework. You use them to support the forces so that the structure will not come tumbling down. Let us start with a figure of a simple truss in Figure (from Wikipedia) which will handle the load generated by automobiles/trucks on the bridge. On the right I have idealized the bridge in terms of simple elements: (i) the straight members; (ii) the pins; and (iii) the support for the straight elements at the bottom. The bridge is three-dimensional, and I have shown the information for the front plane (that will be two-dimensional). The pins are deliberate to avoid serious static indeterminacy. Example A simple bridge truss -and simplified Example 6.1 In the interest of keeping the calculations modest I have shortened the truss to a simpler geometry and loading as shown in Figure This is an idealized description of the truss. Figure A simple 2D truss There are four joints/pins (n), which we have identified as A, B, C, and D. There are five members (m) in the truss, and we refer to them through the pair of letters AD, AB, BD, BC, and CD. There are two supports for the truss at A and B. The support at A is a pin, and is not allowed to displace. The support at B is termed as a roller support and can move horizontally, but cannot move in the vertical direction. The truss carries two forces/loads in the direction shown. Points C and D are not restricted from moving. You should decide easily that this is a two-dimensional truss. The stability of the truss is given by a simple relation for a two dimensional truss: K = 2*n - 3 If K = m, the truss is structurally stable We seek two types of solution to the problem expressed in Figure First, we would like to know how much load each of the members will carry. This will help us to design the truss so that it will not fail under loading. This is considered as the Method of Joints. Second, we would also like to compute the final geometry of the truss with the loads as some pins will move under the applied loads. We can use this information to ensure that the point C, for example, will not hit against another machine (not shown here). This is the solution to the truss deformation.

2 Alternately, if we are certain that a certain member is critical, then we can seek the force only in that member to design it to avoid failure. This is called the Method of Sections. This could require less calculations for force computation. However, for the displacements all the forces are necessary. To start the calculations we use a free body diagram of the entire truss and apply the equation of statics Support Reactions The FBD of the truss is shown in Figure Here we are replacing the support reactions on the truss and the applied loads. Figure FBD of Example 6.1 The first thing we observe is that there are three unknown reactions. This works exceedingly well since we have three equations of equilibrium to solve them. This is a statically determinate problem. If the roller support at B was changed to a pin support we would have a statically indeterminate problem and would have trouble solving them from the equations of statics alone. We will have to solve for the displacements to use the new extra displacement information at B (the horizontal displacement is now zero). Our first step is to determine the support reactions. Support Reactions: The equilibrium equations: MATLAB Code (The MATLAB Code is broken into continuous segments - next piece of code may depend on previous code. It should be collected in a single file) % Essential Foundations in Mechanics % P. Venkataraman, Dec 2105 % Example 6.1 % Truss 2D %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clc, clear, format compact, format shortg, close all, digits(3) warning off %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fprintf('example 6.1\n') fprintf(' \n') %% Data F1 = 1500; F2 = 1000; % the applied force A = [0 0 0]; B = [0.5,0,0]; C = [1.5,0,0]; D = [0.5,0.5,0];

3 %% Support reaction calculations syms Ax Ay By real % Unknowns rab = B - A; rad = D - A; rac = C - A; FA = [Ax, Ay, 0]; FB = [0,By,0]; FC = [0,-F1,0]; FD = [-F2,0,0]; %% Equilibrium SumF = FA + FB + FC + FD % Sum of Forces SumMA = cross(rad,fd)+ cross(rab,fb)+ cross(rac,fc) % Sum of moments at A sol = solve(sumf(1),sumf(2), SumMA(3)); Ax = double(sol.ax); Ay = double(sol.ay); By = double(sol.by); fprintf('\nreactions:\n') fprintf(' \n') fprintf('ax [N] = '),disp(ax) fprintf('ay [N] = '),disp(ay) fprintf('by [N] = '),disp(by) In the Command Window Example SumF = [ Ax , Ay + By , 0] SumMA = [ 0, 0, By/2-1750] Reactions: Ax [N] = 1000 Ay [N] = By [N] = Member Forces - Method of Joints One important assumption about the force in the members is that the force will be along the member. The member can therefore be in one of two states - tension or compression. The unknowns are just the magnitude of the force in the members. The directions are obtained from the geometry of the problem. The member forces are obtained by considering the equilibrium of each pin or joint. Since we will be developing equilibrium equations for each pin, or a point - there are only two force equilibrium equations available at each pin for a 2D problem. To solve for the forces explicitly we cannot have more than two unknowns. It is useful to cycle through the pins so that the forces are explicitly solved with a numerical value at each pin (true only for statically determinate problems). It is therefore important to start at a pin with not more than two unknown member forces. We have to draw the FBD of the pins. Figure illustrates the FBD of pins A, B, C. Member forces drawn heading away from the pin are in tension. From Figure we should not start our calculation cycle at pin B (Why?).

4 Figure FBD of pins Solving for the member forces is simple but tiresome. The current set of equations are solved using previous known solutions for the reactions. The components of forces can be easily computed through the unit vectors based on the location of the pins. They can be reviewed in the MATLAB code and are not explicitly registered here. For each pin we are applying the equilibrium of the forces, which is shown along the coordinate directions below. A lot of statements in the code are used for printing. They are same as the ones encountered earlier in this book. Pin A: Pin B: Pin C: We solve these equations using MATLAB. The positive values for the member forces indicate tension. The negative values indicate that the member is in compression. MATLAB Code (Remember this piece of code continues the previous code) %% Member force calculations % substitute for known values FA = subs(fa); FB = subs(fb); %% Pin A fprintf('\npin A:\n') fprintf(' \n') syms Fab Fad rab = B-A; eab = rab/norm(rab); rad = D-A; ead = rad/norm(rad); SumA = FA + Fab*eAB + Fad*eAD sola = solve(suma(1),suma(2)); Fab = double(sola.fab); Fad = double(sola.fad); fprintf('fab [N] fprintf('fad [N] = '),disp(fab) = '),disp(fad) %% Pin B fprintf('\npin B:\n')

5 fprintf(' \n') syms Fbc Fbd rbc = C - B; ebc = rbc/norm(rbc); rbd = D - B; ebd = rbd/norm(rbd); SumB = -Fab*eAB + FB + Fbc*eBC + Fbd*eBD solb = solve(sumb(1),sumb(2)); Fbc = double(solb.fbc); Fbd = double(solb.fbd); fprintf('fbc [N] = '),disp(fbc) fprintf('fbd [N] = '),disp(fbd) %% Pin C fprintf('\npin C:\n') fprintf(' \n') syms Fcd rcd = D - C; ecd = rcd/norm(rcd); % only one unknown - need only one equation to solve for it SumC = -Fbc*eBC + FC + Fcd*eCD solc = solve(sumc(1)); Fcd = double(solc); fprintf('fcd [N] = '),disp(fcd) In The Command Window Pin A: SumA = [ Fab + (2^(1/2)*Fad)/ , (2^(1/2)*Fad)/2-2000, 0] Fab [N] = Fad [N] = 2833 Pin B: SumB = [ Fbc , Fbd , 0] Fbc [N] = Fbd [N] = Pin C: SumC = [ (2*5^(1/2)*Fcd)/5, (5^(1/2)*Fcd)/5-1500, 0] Fcd [N] = Displacement of the Truss This is a missing topic in standard engineering curriculum. Calculating the displacement of the truss is not taught in the basic Statics course because it uses the stress and strain of the members to set up the problem. It is not covered in the basic Strength of Materials courses because Trusses are not revisited there. If you enroll in a Finite Element course you are likely to solve for truss displacements numerically. You might solve for truss displacements using energy methods in an advanced course on Strength of Materials. These problems are usually solved using software today. We will solve for the displacements in this section using simple geometry and the relation between the displacement and the area of cross-section of the member, the length of the member, and the modulus of elasticity (the material property). This is based on stress, strain, and Hooke s law. It does get cumbersome with increase in the number of members. Before we venture into the solution let us consider the displacement of the Example truss in Figure

6 Figure Original and displaced (exaggerated) truss The pins of the truss are free to move if they are not constrained by supports. Pin B can move horizontally. Pins C and D can move in any direction. We have shown the displacement of C and D to be positive, a standard procedure for deriving the formula in all of science and engineering. In 2D space we can resolve the general displacement of the pin into two displacement components in the coordinate directions - components in x and y directions. This is illustrated in Figure for the member CD - both as part of the truss and separately as a FBD. Figure Displacement of member CD To solve for the displacements of member CD through its FBD below: Figure FBD of member CD The total change in length of the member CD (δ CD ) along the member CD (ignoring the small changes and based on undeformed original geometry) is the relative difference between the displacement of the end C and the displacement of the end D along the direction of member CD. We will assume that end C displaces more than end D to indicate tension in the member. This is not a limiting assumption. You could have assumed that end D moves more than end C to get the same results. It is advisable to be consistent in the development of the displacement relations. In Figure this is computed as as follows

7 We make use of the definitions of strain and stress in the member CD. We use the Hooke s law to relate the stress and strain in the member to arrive at a relation between the force F CD in the member and the unknown displacements of the ends of the member CD - dc x, dc y, dd x, dd y Since we have invoked the Hooke s law we are limited to small displacements. That is the displacements of the truss will not change the geometry of the truss significantly. This will allow us to identify the angles q based on the pre-deformed geometry. If the displacements are defined as vectors, then the total change of length along the member is the dot product of the unit vector along the member times the displacement vector of the ends. This is easy to incorporate in MATLAB. In Example 6.1, we have to develop one equation relating the force in the member to the displacements of the ends of the member, for each of the members in the example. Each member will have a FBD like Figure to help develop the relations. It is not shown in the subsequent development, but the actual relations should assist you in drawing one, which is suggested before you write the equations. We will make additional assumptions to make the development easy. All members are made up of the same material with the modulus of elasticity value of 11.4x10 6 [Pa]. All of the members have the same cross-sectional area of 50x10-4 [m 2 ]. Member AB: Member AD:

8 Member BC: Member BD: Member DC: We can solve equations (5) - (9) for the unknowns db x, dc x, dc y, dd x, dd y. MATLAB Code: %% Calculating displacements syms dbx dcx dcy ddx ddy % the unknowns % Properties Elas = 11.4e06; Area = 50e-04; % displacement vectors da = [0,0,0]; db = [dbx, 0,0]; dc = [dcx,dcy,0]; dd = [ddx, ddy,0]; % length of members and unit vectors (Copy and Paste) rab = B - A; LAB = norm(rab); eab = rab/lab; rad = D - A; LAD = norm(rad); ead = rad/lad; rbc = C - B; LBC = norm(rbc); ebc = rbc/lbc; rbd = D - B; LBD = norm(rbd); ebd = rbd/lbd; rdc = C - D; LDC = norm(rdc); edc = rdc/ldc; % set up the equations (Copy and Paste) Eq(1) = dot(eab,db)-dot(eab,da) - (Fab*LAB/(Area*Elas)); % Member AB Eq(2) = dot(ead,dd)-dot(ead,da) - (Fad*LAD/(Area*Elas)); % Member AD Eq(3) = dot(ebc,dc)-dot(ebc,db) - (Fbc*LBC/(Area*Elas)); % Member BC Eq(4) = dot(ebd,dd)-dot(ebd,db) - (Fbd*LBD/(Area*Elas)); % Member BD Eq(5) = dot(edc,dc)-dot(edc,dd) - (Fcd*LAB/(Area*Elas)); % Member DC soldis = solve(eq); fprintf('\ndisplacements\n') fprintf(' \n') fprintf('dbx [m] : '),disp(double(soldis.dbx)) fprintf('dcx [m] : '),disp(double(soldis.dcx)) fprintf('dcy [m] : '),disp(double(soldis.dcy)) fprintf('ddx [m] : '),disp(double(soldis.ddx)) fprintf('ddy [m] : '),disp(double(soldis.ddy))

9 In the Command Window Displacements dbx [m] : dcx [m] : dcy [m] : ddx [m] : ddy [m] : We have now obtained (1) the support reaction, (ii) member forces, and (iii) the displacements of the truss. I have used MATLAB for the calculations. They could have just been done by hand Member Forces - Method of Sections The method of sections is a quick way to determine the forces in selected members. It works best if the support reactions are known. Lets start with Figure and have solved for the unknowns A x, A y, and B y. We now seek the value for the force in the member CD. We will cut or section the truss to expose the member forces of interest and apply the equation of statics to the sectioned FBD shown in Figure Figure Applying method of sections We know apply the equilibrium equations to the sectioned truss F bc will not provide a moment about A. Since there are only two unknown we can solve for the forces from the force equation itself MATLAB Code: %% Method of sections - we will use previous definitions where available syms Fcb Fdc % we will use a different variable for forces clear SumF SumMA FCB = Fcb*eBC; FDC = Fdc*eDC; SumF = FA + FB + FD + FCB + FDC solms = solve(sumf(1),sumf(2)); Fcb = double(solms.fcb); Fdc = double(solms.fdc); fprintf('\nmethod of Sections\n') fprintf(' \n') fprintf('fcb [N] = '),disp(fcb) fprintf('fad [N] = '),disp(fdc) % We can check if this solution satisfies the moment equation SumMA = cross(rad,fd) + cross(rad,fdc) + cross(rab,fb); MomA = eval(subs(summa(3))) % should be zero without rounding error

10 In the Command Window Method of Sections Fcb [N] = Fad [N] = 3355 MomA = There are the same values obtained previously. The sum of moments should have been zero but we have chosen to report our results to 3 digits and hence carry round off error. In this example we could have considered the FBD of the other part of the sectioned figure in Figure That would be exactly the same as the FBD of pin C in Figure