Question

Force (N)	time (s)		Elongation (t*0.05 mm/s)		Engineering stress(Mpa)		Engineering Strain
0	0		0		0		0
1500	5		0.25		53.05978069		0.00625
5000	10		0.5		176.8659356		0.0125
9250	20		1		327.2019809		0.025
11250	25		1.25		397.9483551		0.03125
11750	30		1.5		415.6349487		0.0375
11500	32.5		1.625		406.7916519		0.040625
12000	35		1.75		424.4782455		0.04375
14250	50		2.5		504.0679165		0.0625
15250	60		3		539.4411036		0.075
16000	70		3.5		565.970994		0.0875
16750	85		4.25		592.5008843		0.10625
17250	100		5		610.1874779		0.125
17500	132.5		6.625		619.0307747		0.165625
17250	170		8.5		610.1874779		0.2125
17000	180		9		601.3441811		0.225
16500	190		9.5		583.6575875		0.2375
16250	195		9.75		574.8142908		0.24375
16000	200		10		565.970994		0.25
15000	210		10.5		530.5978069		0.2625
13250	220		11		468.6947294		0.275
12697.1	222.5		11.125		449.1368942		0.278125
Initial Length (mm)		Initial Diameter(mm)		Area = pi*r^2
40		6		28.27

Maximum load capacity: 50,000 N Test rate: 3 mm/min - 0.05 mm/s Steel A Steel B Initial length, Lo 40 mm 40 mm Initial diameter, do 6 mm 6 mm Final length, Lf 48 mm 52 mm Final diameter, df 4.2 mm 3.1 mm  

From the data given in excel file, calculate:

(b) Tensile strength, modulus of elasticity, modulus of resilience, % elongation and % reduction in area.

(c) Plot both engineering stress strain curve and true stress strain curve.

(d) Compare the tensile behaviour of steel A and B in one paragraph

Answer 1

A) ENGINEERING STRESS-STRAIN CURVE

B) TRUE STRAIN = ln ( 1 + ENGINEERING STRAIN )

TRUE STRESS = ENGINEERING STRESS ( 1 + ENGINEERING STRAIN )

C) TRUE STRESS-STRAIN CURVE

D)

TENSILE STRENGTH = MAX LOAD / INITIAL AREA = 17500 / 28.27 = 619.03 N/ = 619.03MPa

MODULUS OF ELASTICITY = SLOPE UNDER PROPORTIONALITY LIMIT

= SLOPE [STRESS (POINT(4) - POINT (3)) / STRAIN {POINT(4) - POINT(3)) ]

= (327.2019809 - 176.8659356 ) / (0.025-0.0125)

= 12026.88 MPa = 12.026 GPa

MODULUS OF RESILIENCE = AREA UNDER PROPORTIONALITY LIMIT

    = AREA OF TRIANGLE = 0.5 * STRESS(POINT(5) - POINT(1)) * STRAIN(POINT(5)-POINT(1)

=0.5 * ( 397.8482551-0) * (0.03125-0) = 6.21794 MPa

E) PERCENT ELONGATION (STEEL A) = (48 -40 / 40 ) *100 = 20 %

PERCENT ELONGATION   (STEEL B) =  (52-40 /40 )*100 = 30%

AREA REDUCTION (STEEL A) = (FINAL AREA -INITIAL AREA) / INITIAL AREA *100

INITIAL AREA (Ai) =pi * 3*3 =28.27 ,FINAL AREA (Af) = pi * 2.1*2.1 =13.85

AREA REDUCTION = (28.27 - 13.85 ) *100 / 28.27 = 51 %

F)PERCENT REDUCTION (STEEL B):

INITIAL AREA (Ai) =pi * 3*3 =28.27 ,FINAL AREA = pi * 1.55*1.55 =7.5476

AREA REDUCTION = (28.27 - 7.5476 )*100 / 28.27 = 73.3%

G) MORE THE ELONGATION MORE WILL BE THE TENSILE STRENGTH OF THE MATERIAL ,HENCE STEEL B HAS MORE TENSILE STRENGTH AS IT'S ELONAGTION IS 30% COMPARED TO STEEL A OF ELONGATION 20%