Question

ViewHelp Review .E. A A. Search ROD AaBbc Aabe ODE ABAC AaBbc O CH RODARICA Emphasing Thing? S S T The Spar Sutte. Des u A E - QUESTION 2 A. The table below lists the Young's modulus €, the yield strength (Oy), ultimate tensile strength (.) and ductile (Er) for there materials. (% Material Brass Copper Steel E (GPa) 100 110 207 Oy (MPa) 345 250 450 (MPa) 420 290 550 Calculate the modulus of resilience and the toughness of the brass alloy. An application requires that rods of square cross section, 10.0 mm x 10.0 mm, and 100 mm in length must experience no permanent deformation and must not elongate more than 0.20 mm when subjected to an axial load of 30 kN. For each of the available materials, determine whether it is suitable fir this application Please justify your answers. B. Is it possible for a material to have a resilience greater than its toughness? Why or why not? QUESTION 3 . .. .. . . .

Answer 1

Among the materials provided only steel is able to resist a deformation less than 0.2mm under given loading. This can be justified by finding the elongation of each sample under given conditions. The modulus of toughness cannot be determined directly from the stress-strain graph. The reason is, linear relationship between stress and strain ceases to exist after the elastic limit. Therefore the only way to find the entire area of the stress-strain graph up to failure point is to use the curve fitting method. A triangular and rectangular portion can be used to approximate this area.

eno - Modulus of Resilience (Uh) is the strain energy stored up to elastic limit / yielding point, per unit volunne. Es Youngs Modulus 1. UM = GEL OEL (brass) = Oy Cbrass) = 345MPa. Ebrass = 100 Pa. UM = (345) = '1-19025 MPa 100x103. Modulus of Toughness (Tm) is the strain energy stored per unit volume till fracture point.

Fracture i pont ----11- (stress) ___ The area under this stress-strain graph till fracture point gives the Modulus of #toughness. touch 0 . Let us take a small strip of width . 'de and ordinate si I Area of this small strip S dA= 6 de i Total area upto fracture point TM= 6 de

But since the relationship b/w G&E exists only upto elastic linnt we won't be able to find the area after elastic limit. Therefore some other method like curve fitting approximation need to be adopted - Guil - - Bfailure Би юу point- (D) / L Ey Eu Efr 1. The whole area of G-E graph up to failure point can be approximated as consisting of triangular portion of

and a rectangular portion ABOC. 1 2 .: Tom = Area of triangle & Area of Rectayle The point. À is the average ordinate limit of of by and ou. TM = 1x (ou tou) Ey + (-Ey) (ou 100) Ey= strain at yield point Ega e = 3 = 0.00345 Cursos es = (20432,5) =3 82.51 (f) brass = 22% = 0.22 -. TM X 382.5x0.00345+ © 22-0-00345)382.S TM = 83.49 MPa

100mm (11) . for not experiencing permanent de formation, the stress should not exceed yield points Elongation where AL=PL de EA : P= load applied E= Youngs Modules A= Cross-sectional area L= Lenth of rod, •Brass bras Pa 30kn, 22100 mm, A= 100 mm ER 100 G Pa. . DL= 3oxoline = 0.3 mm I looxcotx Too . Ara 0.2mm , herce brass cannot be used

Copper Al coprzer = 30x108 109 110x10x1QO = 0.272 mm Al copper > 0.2mm - Copper cannot be used Steel is can Alsteel = 30x108xive 207x10x100 1 c0.144 mm - Alsteel < 0.2mm .. Only steel can be used above application, for the

Resilience for materials cannot - be greater than modulus of - Toughness Because, modulus of resilience is the strain energy per unit volume Calculated upto elastic limit, whereas modulus of toughness is strain energy per unit volume upto fracture point- Elastic limit and fracture limit are two different values ; it cannot be greater than fracture limit value