"The step-cone pulley is to be designed for transmitting a power of at least 0.75 hp. The speed of the input shaft\r\nis 350 rpm and the output speed requirements are 750, 450, 250 and 150 rpm for a fixed center distance of ‘a’\r\nbetween the input and output shafts. The tension on the tight side of the belt is to be kept more than twice that\r\non the slack side. The thickness of the belt is ‘t’ and the coefficient of friction between the belt and the pulley is\r\n.\r\n(1) Formulate the problem of finding the widths and diameters of the steps for minimum weight.\r\n(1hp=745.7W). For some parameters not given, you can assume some realistic values.\r\n(2) Design widths and diameters to minimise the weight (use at least two different optimisation\r\ntechniques/software to find the solution and compare the results)\r\n(3) Discuss the effects of the design variables on the weight of the step-cone pulley.\r\n(4) Prepare a report which provides detailed design information. Design guideline should also be included\r\nin the report (assuming your colleagues will use this guideline in their designs of this type of pulley)\r\n\r\nDesign of a Step-Cone Pulley May 11, 2017 The project is to design a step-cone pulley such that the total weight of the system is minimized. The task involves mechanical and geometric considera- tions, de?ned by the following design equations. 1 Mathematical Model We are given the output angular speeds (750, 450, 250 and 150 rpm), and the constraints that the tension of the tight side should be at least the double that on the slack side, and that the pulley should be able to transmit at least 0.75 hp (560 W). Furthermore, the belt is constrained by the admissible force, which depends on its thickness, and the length of the belt should be constant for all cylinder 1pairs. It is also reasonable to assume that the width of the cylinders are equal to the width of the belt, and that the sum of the radii of the cylinders are less than the distance between the step cones. To formulate this as an optimization problem, we begin by considering a single pair of cylinders, say where the second cylinder is required to have a rotational speed of 150 rpm. Since we want to minimize the weight of the system, we take cylinder diameter, d (and the reciprocal diameter, D), and cylinder width, w, as variables. The system is described by the ratio between the cylinder diameters, k, the rotational frequency n , the angular speed, n rpm and the required minimum amount of transmitted power, P , t k = 350/150= 7/3, n = 350rpm, rpm n = 2pn /60rad/s, rpm P = 0.75hp= 0.75·745.7W, t We also need to de?ne some constants: the belt thickness, t, the allowable tension F , speci?c weight of the belt, ? , friction coe?cient µ, distance be- a belt tween step cones a, and some correction factors: service factor, K , factor of s safety,n ,pulleycorrectionfactor,C ,andvelocitycorrectionfactorC . These fs p v can be found in tables, and the selected values are t = 0.0013(m), F = 6000(N), a 3 ? = 9500(N/m ), belt µ = 0.5, a = 0.5(m), K = 1.5, s n = 1.2, fs C = 0.7, p C = 1.0. v From this data, we can easily compute belt speed, v, torque, T, as v = pdn /60(m/s), rpm T = P ·K ·n /n(Nm), t s fs We have the geometric relations for the wrap angles of the smaller ? and S larger ? , L ? = p-2*arcsin((D-d)/2a)(rad), S ? = p+2*arcsin((D-d)/2a)(rad). L 2De?ning the help factors for ease of notation a = ? ·t, 1 belt a = F ·C ·C , 2 a p v a 1 2 a = (p·d·n /60) 3 rpm g we state the engineering equations for the tension ratio, r, cylinder width, w, 1 maximumallowabletensionF ,hoop(centrifugal)tensionF andinitialtension, c a F . The tight side is denoted by subscript 1f, and the slack side by 2f. i r = exp(µ?), w = (2*T *r/d)/((a -a )(r-1)), 3 2 1 F = a ·w, 3 a F = a ·w, c 2 F =T/d-a ·w, i 2 The ?nal tensions and maximum transmitted power are then F = T/d(r+1)/(r-1), 1f F = F +F ·2/(r+1), 2f c i 1 2 P = (F -F )v. max f f Finally, to compute the weight, we need the unit weight of the belt, W, the length of the belt, L and the volume of the cylinders, V , which are d W = a ·w, 1 p 1 2 2 L = 4a -(D-d) + (D? +d? ), L S 2 \u0012 \u0013 2 2 D d V = w + p/2. cyl 2 2 To calculate the weight, the speci?c weight for the cylinder is assumed to 3 be ? = 80000(N/m ) (stainless steel). To obtain the mass, we simply divide cyl these expressions by the gravitational acceleration g = 9.81. 2 Nonlinear Optimization 2.1 Optimization Method Since this is a nonlinear programming problem, we may use built-in functions in Octave/Matlab. Most such methods are based on solving a sequence of linearized subproblems, such as sequential quadratic programming. The sub- problems can be solved using quadratic programming, which allows a nonlinear objective function with linear constraints. Linearization of the constraints then gives a sequence of subproblems, that can be solved e?ciently. 32.2 Cylinder Optimization Byrelaxingtheconditionthatthebeltlengthshouldbethesameforallcylinder pairs, we decouple the problem into four independent ones. This result is fairly easy to formulate as an optimization problem to get an idea of the dimensions of the optimal system. For the individual cylinder pair optimization, we use the sum of volumes of the two cylinders, times the speci?c weight to obtain the total weight. Dividing by g gives its mass. The objective function is the total mass of the system, so we want to mini- mize m(w,d) =V ? /g+LW/g. cyl cyl The constraints are r = expµ? = 2, F = F , 1f a P = P . max t These are rearranged to give a standard formulation h(z)= 0 (or h(z)= 0, depending on software), where we have introduced the parameter vector z = 1 (w,d,r-2,F -F ,P -P ). a max t f To ?nd a feasible starting solution, we assume d = 0.025 (a tabulated min valuecorrespondingto some ofthe parametersabove), andverifythath(z)= 0. Bysequentialquadraticprogramming(Octave/Matlab)weobtaintheresults in the table below for each cylinder pair. Cylinder 150 rpm 250 rpm 450 rpm 750 rpm w 0.4053 0.4115 0.5148 0.8609 d 0.0250 0.0250 0.0194 0.0117 D 0.0583 0.0350 0.0250 0.0250 L 1.1315 1.0943 1.0698 1.0577 W 5.8057 2.9292 2.3469 3.2446 t P 602 597 985 2744 max 2.3 Pulley Optimization For the optimization of the total transmission system, we require that the belt length L and width w are equal for all cylinder pairs. Firstly, we extend the 1 parameter vector to z = (w,d ,...d ,r - 2,...,r - 2,F - F ,...,F - 1 4 1 4 a a 1f 4 1 4 F ,P -P ,...,P -P ). Secondly, we add a penalty term p(z) to the t t max max 1f objective function, where 4 X 2 p(z) =K (L(1+i%4)-L(i)) , i=1 4with K a large positive number. Afeasiblesolutionisfoundbylettingd = 0.025andverifyingthath(z)= 0. i The result is given in the following table Cylinder 150 rpm 250 rpm 450 rpm 750 rpm w 0.0832 0.0832 0.0832 0.0832 d 0.1685 0.2396 0.2521 0.1796 D 0.3932 0.3354 0.3241 0.3848 L 1.9077 1.9077 1.9077 1.9077 W 48.77 45.27 44.93 48.06 t P 650 611 774 1356 max 3 Discussion 3.1 Widths and Diameters Wecanseetheratherlargedi?erenceinwidth andweightofthetwosystems. It isimportant to notice that the weightincreaseswith the squareofthe diameter, and linearly with the width. In the ?rst case, optimization gives minimum diameter for the smaller cylinder, and compensate the contact area by wider cylinders. In the second case, due to the constraints, the diameters need to be much larger, but also gives narrower cylinders. The constraint that widths and belt length are identical is therefore a rather strong restriction. 3.2 Design Parameters The weight of the system obviously depends on the material choosen. With a stainless steel pulley, the result is approx. 200 kg. We have also assumed a rather modest friction coe?cient µ = 0.5. With increasing µ, the contact area andtherebybothwidthanddiameterscanbereduced. Increasingbeltthickness allows a higher tension F , which also help reducing size and weight. a 4 Matlab/Octave Code 4.1 Cylinder Optimization % Cylinder Optimization matlab = false; % Change ratio here (in three files): k = 7/3; g = 9.81; 5t = 0.0013; dmin = 0.025; Fa = 6000; gammBelt = 9500; gammCyl = 80000; mu = 0.5; a = 0.5; nrpm = 350; nrad0 = 2*pi*nrpm/60; Pt = 560; % = 0.75 hp nfs = 1.2; Ks = 1.5; Cp = 0.7; Cv = 1.0; a1 = gammBelt*t; a3 = Fa*Cp*Cv; v = pi*dmin*nrpm/60; T = Pt*Ks*nfs/nrad0; if (k > 1) dsmall = dmin; dlarge = k*dmin; else dsmall = k*dmin; dlarge = dmin; nrpm = nrpm/k; end a2 = a1/g*(pi*dsmall*nrpm/60)^2; theta = pi-2*asin((dlarge-dsmall)/(2*a)); thetaL = pi+2*asin((dlarge-dsmall)/(2*a)); r0 = exp(mu*theta); w0 = (2*T*r0/dsmall)/((a3-a2)*(r0-1)); L = sqrt(4*a^2-(dlarge-dsmall)^2)+0.5*(dlarge*thetaL+dsmall*theta); W = a1*w0; wt_init = w0*((dlarge/2)^2+(dsmall/2)^2)*pi/2*gammCyl/g; wt_init = wt_init + L*W/g z0 = [w0; dmin] h_init = h(z0) s = input(’Press ENTER to optimize’); 6lb = [0.001; dmin]; ub = [10.0; a]; if (matlab) options = optimoptions(’fmincon’,’Display’,’iter’,’Algorithm’,’sqp’); [z, obj, exitflag, output] = fmincon(@wt, z0, [], [], [], [], lb, ub, @hm, options) else [z, obj, info, iter, nf, lambda] = sqp(z0, @wt, [], @h, lb, ub) end w = z(1); d = z(2); if (k > 1) dsmall = d; dlarge = k*d; else dsmall = k*d; dlarge = d; nrpm = nrpm/k; end a2 = a1/g*(pi*dsmall*nrpm/60)^2; theta = pi-2*asin((dlarge-dsmall)/(2*a)); thetaL = pi+2*asin((dlarge-dsmall)/(2*a)); r = exp(mu*theta); % w = (2*T*r/dsmall)/((a3-a2)*(r-1)) F1a = a3*w; Fc = a2*w; Fi = T/dsmall-a2*w; F1f = T/dsmall*(r+1)/(r-1); F2f = Fc+Fi*2/(r+1); v = pi*dsmall*nrpm/60; Pmax = (F1f-F2f)*v; W = a1*w; wt = w*((dlarge/2)^2+(dsmall/2)^2)*pi/2*gammCyl/g; h_final = h([w d]) w 7dsmall dlarge L = sqrt(4*a^2-(dlarge-dsmall)^2)+0.5*(dlarge*thetaL+dsmall*theta) wt = wt + L*W/g Pmax % Function file h.m function f = h(z) w = z(1); d = z(2); % Change ratio here: k = 7/3; g = 9.81; t = 0.0013; dmin = 0.025; Fa = 6000; gammBelt = 9500; mu = 0.5; a = 0.5; nrpm = 350; nrad0 = 2*pi*nrpm/60; Pt = 560; % = 0.75 hp nfs = 1.2; Ks = 1.5; Cp = 0.7; Cv = 1.0; a1 = gammBelt*t; a3 = Fa*Cp*Cv; v = pi*d*nrpm/60; T = Pt*Ks*nfs/nrad0; if (k > 1) dsmall = d; dlarge = k*d; else dsmall = k*d; dlarge = d; nrpm = nrpm/k; end a2 = a1/g*(pi*dsmall*nrpm/60)^2; 8theta = pi-2*asin((dlarge-dsmall)/(2*a)); thetaL = pi+2*asin((dlarge-dsmall)/(2*a)); r = exp(mu*theta); F1a = a3*w; Fc = a2*w; Fi = T/dsmall-a2*w; F1f = T/dsmall*(r+1)/(r-1); F2f = Fc+Fi*2/(r+1); Pmax = (F1f-F2f)*v; f = [ w; d; r-2.0; F1a-F1f; Pmax-Pt ]; endfunction % Function file wt.m function obj = wt(z) w = z(1); d = z(2); % Change ratio here: k = 7/3; g = 9.81; t = 0.0013; dmin = 0.025; Fa = 6000; gammBelt = 9500; gammCyl = 80000; mu = 0.5; a = 0.5; nrpm = 350; nrad0 = 2*pi*nrpm/60; Pt = 560; % = 0.75 hp nfs = 1.2; Ks = 1.5; Cp = 0.7; 9Cv = 1.0; a1 = gammBelt*t; a3 = Fa*Cp*Cv; v = pi*d*nrpm/60; T = Pt*Ks*nfs/nrad0; if (k > 1) dsmall = d; dlarge = k*d; else dsmall = k*d; dlarge = d; nrpm = nrpm/k; end a2 = a1/g*(pi*dsmall*nrpm/60)^2; theta = pi-2*asin((dlarge-dsmall)/(2*a)); thetaL = pi+2*asin((dlarge-dsmall)/(2*a)); r = exp(mu*theta); L = sqrt(4*a^2-(dlarge-dsmall)^2)+0.5*(dlarge*thetaL+dsmall*theta); W = a1*w; obj = w*((dlarge/2)^2+(dsmall/2)^2)*pi/2*gammCyl/g; obj = obj + L*W/g; endfunction 4.2 Pulley Optimization % Pulley Optimization matlab = false; g = 9.81; t = 0.0013; dmin = 0.025; Fa = 6000; gammBelt = 9500; gammCyl = 80000; mu = 0.5; a = 0.5; nrpm = 350; nrpm0 = nrpm; nrad0 = 2*pi*nrpm/60; Pt = 560; % = 0.75 hp 10"