Question

(b) Explain the function, objective and constraints for each case study. (i) A design of cylindrical tie-rod of specified length 1, to carry a tensile force F without failure, it is to be of minimum mass. Think of golf-club shafts. Consider, then a light beam of square section b x b and length 1 loaded in bending which must meet a constraint on its stiffness S, meaning that it must not deflect more than under a load F. (iii) We seek for the cheapest legs of a table (cylindrical column) of a specified height, 1, which will safely support a load F. (9 marks)

Answer 1

(1) Cylindrical tie rod! Function & Tie rods are used to steer a vehicle. A tie rod connects the steering system to the front wheels of a car. Atie rod is car in order to take twins. necessary for Objective & To obtain a tie-rod with minimum mass and adquate strength. Constrains ? Deflection of tie-rod under tensile force ; 8= FL/AE A = Area =Deflection; F = Force , L= Length, A= E= Youngs Moduls. we know, stiffness » K= (2) 8 Now, mass of the rod can be written ase m= A.LB- [p = Density from 7 and we can write E => A- LK - 8 FL AE E Putting ④ in ① for mass calculation. ĽKS (Lk JL. Hence, the minimum mass for our load consideration is shown above, ms

Gi) Function: A golf shaft should have the ability to be flexible in order to throw the ball long distances. distances. There fore, it needs to be minimal in stiffness Now, for the beam with cross section bxb A stiffness s is required so that it should not deflect than our limits. So Minimum load. by? Sa E Smax Objective! deflection under given Constraints & Stiffness 's' that is represented - ,F= Force applied Smax= Maximum deflection under given stifness and load appled. The defection can also be written as. Smax = 2, E2 Young's modulus. Iz Area moment of Thertia. So, and 64 for square 12 be written as; section. F² 3 E 64/12 F13 3 EI also Smax- & which gives; Х 12 3E yu / / E Required dimension based on stiffness s deflection. on 45

As, the length length L is fixed. then i the eqreaction 3 E 45 Can be written as. b4= E L² 45 OR 4 b = WEL3/45 Hence, b is the desired breadth

As, Sa+ Smax and 64 as b4 as and Bat the cross section al area , Hence, larger and shorter the length; the deflection t's lesser. (1) The functions of a leg in table is to prevent buckling of itself and carrying the load of the table. Objective cheapest legs for the table. Constraints & The buckling load fore a column can be written as; F=na² EI/L² Ez Allowed load n= End condition factors. E= Modulus of elasticity L= Length th of column I = Moment Area A table leg can be observed as fixed and Ebase) and a free end (top). T for this condition n=0.25, n=0.25

I for circular cross section is J= 7n4 So, the equation becomes! F= 0,25 x 72 x E x 0.25 x2x44 ² on + E 0.252xx² χ Λ'χE As, the allowable load FRE ; hence a cheap mat erial with high Young's modulus could be Alumenium or wood. Also, the greater the readius, and shorter the length; the greater the load carrying capacity As, Fary and FL t If the dimensions are fixed, so we can also write r / 2= K (constant) So, E K X0.252x73 Requined Youngs modulus.