Question

The following figures were obtained in a standard tensile test on a specimen of low carbon steel with a circular sectional area: Diameter of specimen, 50 mm; Gauge length (the length of a specimen used for testing purposes), 100 mm; Minimum diameter after fracture, 10 mm. Load (kN) 2.2 4.5 6.8 9.2 11.8 14.4 17 19.6 22.2 24.7 Extension (μm) 5.6 11.9 18.2 24.5 31.5 38.5 45.5 52.5 59.5 66.5 Load (kN) 27.13 28.6 29.1 29.6 30.2 31 31.7 32 32.9 34.8 Extension (μm) 73.5 81.2 89.6 112 224 448 672 840 1120 1680 Load (kN) 35.8 37 38.7 39.5 40 39.6 35.7 28 Extension (μm) 1960 2520 3640 5600 7840 11200 13440 14560 (a) Generate engineering stress vs engineering strain plot (use a computer tool for accurate plotting. Hand sketch is Not allowed). (b) Determine Young’s modulus of elasticity. (c) Determine the ultimate tensile stress. (d) Determine the percentage reduction of area. (e) Determine the percentage elongation. (f) Determine true stress at fracture.

Answer 1

Answer:

See Explaination

Explanation:

1)here for given stress strain curve graph is given as follows

where for getting stress,S=F/A=4F/(pi*(50*10^-3)^2)

for strain=e=dl/l=dl*10^-3/100 mm/mm or m/m

2)so graph is as follows

3)for getting youngs modulus of elasticity we must know slope of graph stress verses strain and for straight line in elastic region upto 12 point we have elastic region and from that we get E as

E=slope of graph for first 12 points=S/e=14.5665*10^9/.812=17.9390*10^9 N/m2

4)for getting ultimate tensile stress at which specimen bears maximum load without failure so we get UTS as

UTS=maximum load/area=40*10^6/1.9634=20.3728*10^6 N/m2

5)percentage reduction in area is given by

percentage reduction in area=[original area-final area/original area]*100

Percent reduction=[5062-10^2]*100/50^2=96%

6)percentage elongation is given by

percent elongation=[final length-original length/original length]*100

final length at fractureis=14.56+100=114.56 mm

so we get percent elongation as=[114.56-100/100]*100=14.56%

7)true fracture stress is given by load at fracture devided by true area at fracture

Sf=load/(true area)=4*28*10^3/(pi*(10*10^-3)^2)=356.5070*10^6 N/m2