Question

Refer to the following figure, stress-strain diagram, for samples of three different materials: A, B, and C. The end of each curve indicates the fracture point.
e) After the conclusion of the tests, if the broken halves of each sample were fitted back together, which sample would be the shortest? Assume that all three samples had the same initial length. f) A circular rod is to be produced, 15 cm long with a diameter of 5.0 mm. This rod must be able to support an applied tensile stress of 230 MPa without plastically deforming. i) List all the materials (A, B, and/or C) that would be appropriate. ii) If a rod(s) was produced from the material (or materials) listed in part (i), which would experience the least amount of total elongation while under the 230 MPa stress? iii) For the material chosen in part ii, determine the amount of total elongation in the rod while under the 230 MPa stress. iv) For the material chosen in part ii, what applied tensile force would cause the rod to break?  

Но тоолуу олиш Ua 350 300 с 250 "В 200 150 A 100 50 2 0 0.1 0.3 0.2 Strain

Answer 1

ANSWER:

E) IF THE TWO PARTS ARE JOINED TOGETHER,THEN THE MATERIAL 'C' WILL BE THE SHORTEST ,SINCE IT DOES NOT UNDERGONE PLASTIC DEFORMATION AND BREAKS BEFORE ENTERING INTO THE PLASTIC REGION. HENCE IT MAY BE TERMED AS BRITTLE MATERIAL.

F)

I) WITHOUT PLASTICALLY DEFORMING THE MATERIAL 'C' SHOULD BE SUITABLE AND IT CAN ALSO SUSTAIN A LOAD OF 230MPa (AS SEEN FROM THE GRAPH THE FRACTURE POINT IS ABOVE 250MPa),WHEREAS 'A' AND 'B' UNDERGOES PLASTIC DEFORMATION BEFORE IT GETS FRACTURE

HENCE MATERIAL 'C' WILL BE SUITABLE

II) THE MATERIAL 'C' WILL EXPERIENCE THE LEAST AMOUNT OF TOTAL ELONGATION ,AS THE LINE OF STRESS VS STRAIN IS MORE STEEP AS COMPARED TO THE OTHER TWO.MEANING THE VALUES OF STRAIN IS LESS (WHERE STRAIN IS THE RATIO OF CHANGE IN LENGTH TO THE ORIGINAL LENGTH),THEREFORE THE CHANGE IN LENGTH WILL BE LESS ,WHICH MEANS THE MATERIAL ELONGATION IS LESS ,HENCE MATERIAL 'C' WILL EXPERIENCE LEAST DEFORMATION.

ALSO FROM THE GRAPH,MATERIAL 'A' AND 'B' UNDERGOES A LARGE PLASTIC REGION WHICH MEANS THE MATERIALS ELONGATES MORE BEFORE FRACTURE.

III) FOR MATERIAL 'C' ,FIRST LET US FIND THE MODULUS OF ELASTICITY (E)

WHICH IS THE SLOPE OF STRESS VS STRAIN IN THE ELASTIC REGION

E = STRESS (POINT(4) -POINT(2)) /RESPECTIVE STRAIN

E = ( 200 - 100 ) / (0.08 - 0.04 ) = 2500 MPa

ALSO ,E = STRESS / STRAIN ,STRAIN = STRESS / E = 230 / 2500 = 0.092

ALSO STRAIN = $\Delta$ L / Lo (CHANGE IN LENGTH / ORIGINAL LENGTH)

STRAIN = Lf - Lo / Lo = 0.092

ALSO PERCENT ELONGATION = (Lf - Lo ) * 100 / Lo = 0.092 * 100 = 9.2 %

IV) FOR MATERIAL 'C' ,ASSUMING THE FRACTURE OCCURS AT 275MPa (FROM GRAPH)

ULTIMATE TENSILE STRENGTH (UTS) = 275 MPa

CONSIDERING THE DIAMTER OF THE MATERIAL = 5 mm

UTS = MAX LOAD / INITIAL AREA , MAX LOAD = UTS * INITIAL AREA = 275 * ($\pi$ * $5^{2}$ / 4 ) = 5399.61 N

HENCE, THE MATERIAL BREAKS AT THE LOAD OF 5399.61 N