Chapter 6, Problem 6.41P

Interpretation Introduction

(a)

Interpretation:

The engineering stress-strain curve should be plotted and the 0.2 % offset yield strength should be calculated for the given data of a ductile cast iron.

Concept Introduction:

The maximum amount of elastic deformation which is bearable by any material is defined as yield strength.

Answer to Problem 6.41P

The yield strength for 0.2% offset is 274 MPa for given sample of a ductile cast iron.

Explanation of Solution

The tabular data providing details about the load and length difference for given sample is as follows:

Load(lb)	$\Delta l$ (mm)
0	0.00000
25,000	0.0185
50,000	0.0370
75,000	0.0555
90,000	0.20
105,000	0.60
120,000	1.56
131,000	4.00 (maximum load)
125,000	7.52 (fracture)

Now, let us first calculate area of a ductile iron sample by using the following formula:

$A=\frac{\pi d_0{}^2}{4}$

In which $d_0$ is a diameter of the iron rod = 20mm

So, the area of the iron sample can be calculated as follows:

$\begin{array}{l}A=\frac{\pi d_0{}^2}{4}\\ A=\frac{\pi {\left(20\right)}^2}{4}\\ A=314.16{\text{ mm}}^2\end{array}$

Now, determine the stress for the given specimen as follows:

$\sigma =\frac{P}{A_0}$

For determining the strain of the given sample,

$\begin{array}{l}\epsilon =\frac{\Delta l}{l_0}\\ \epsilon =\frac{\Delta l}{40}\text{ (Initial length =40mm)}\end{array}$

With the use of given spread sheet and applied loads, The tabulate form of the engineering stress and strain is as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, the plot of the stress-strain from the above tabular data is as follows:

  

The above graph can provide the value of yield strength for 0.2% offset at around 274 MPa after drawing a parallel line to linear to stress-strain curve.

Therefore, the ductile cast iron contains the yield strength for 0.2% offset as 274 MPa.

Interpretation Introduction

(b)

Interpretation:

With the help of plotted engineering stress-strain curve, the tensile strength should be calculated for the given data of a ductile cast iron.

Concept Introduction:

Tensile strength can be defined as the measurement of maximum deformation which can be bearable by any material without undergoing necking condition.

Answer to Problem 6.41P

The tensile strength is 417 MPa for given sample of a ductile cast iron.

Explanation of Solution

With the use of given spread sheet and applied loads, engineering stress and strain can be tabulated as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, the stress-strain curve can be plotted from the above gathered tabular data as follows:

  

The above graph can provide the value of tensile strength as 417 MPa.

Therefore, the tensile strength for given sample of a ductile cast iron can be considered as417 MPa.

Interpretation Introduction

(c)

Interpretation:

With the help of plotted engineering stress-strain curve, the value of modulus of elasticity should be calculated for the given data of a ductile cast iron.

Concept Introduction:

Modulus of elasticity is also known as coefficient of elasticity or elastic modulus and can be defined as the ratio of the stress in the given object body to the corresponding strain.

Answer to Problem 6.41P

The value of modulus of elasticity is 172.057 GPa for a ductile cast iron..

Explanation of Solution

With the use of given spread sheet and applied loads, one can tabulate the engineering stress and strain as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, one can plot the stress-strain curve from the above gathered tabular data as follows:

  

One can use the formula of Hook's law as below for calculating modulus of elasticity.

$\begin{array}{l}E=\frac{\sigma }{\epsilon }\\ E=\frac{238.72}{0.0013875}\\ E=172.057\times {10}^3{\text{ N/mm}}^2\times \frac{1\text{}\text{}\text{ GPa}}{100{\text{N/mm}}^2}\\ E=172.057\text{}\text{ GPa}\end{array}$

Therefore, the value of modulus of elasticity for a ductile cast iron can be described as 172.057GPa.

Interpretation Introduction

(d)

Interpretation:

With the help of plotted engineering stress-strain curve, the value of % elongation should be calculated for the given data of a ductile cast iron.

Concept Introduction:

Elongation is defined as term used to determine the change in gauge length of any material when it is on static tension test.

Answer to Problem 6.41P

The value of % elongation is 18.55% for a ductile cast iron.

Explanation of Solution

With the use of given spread sheet and applied loads, one can tabulate the engineering stress and strain as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, one can plot the stress-strain curve from the above gathered tabular data as follows:

  

One can use the below formula for determining the value of % elongation.

$\begin{array}{l}\%\text{ Elongation =}\frac{l-l_0}{l_0}\times 100\%\\ \%\text{ Elongation =}\frac{47.42-40}{40}\times 100\%\\ \%\text{ Elongation =}\frac{7.42}{40}\times 100\%\\ \%\text{ Elongation =18}\text{.55}\%\end{array}$

Therefore, the ductile cast iron has 18.55% elongation value.

Interpretation Introduction

(e)

Interpretation:

With the help of plotted engineering stress-strain curve, the value of % reduction in area should be calculated for the given data of a ductile cast iron.

Concept Introduction:

Reduction of area of any material is directly related to the reduction in cross-section area of the tensile test piece after fracture.

Answer to Problem 6.41P

The value of % reduction in area is 15.82 % for given sample of a ductile cast iron.

Explanation of Solution

With the use of given spread sheet and applied loads, one can tabulate the engineering stress and strain as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, one can plot the stress-strain curve from the above gathered tabular data as follows:

  

One can use the below formula for determining the value of % reduction in area.

$\begin{array}{l}{A}_{redcution}=\frac{\frac{\pi }{4}{\left(d\right)}^2-\frac{\pi }{4}{\left(d_0\right)}^2}{\frac{\pi }{4}{\left(d\right)}^2}\times 100\\ A_{redcution}=\frac{\frac{\pi }{4}{\left(20\right)}^2-\frac{\pi }{4}{\left(18.35\right)}^2}{\frac{\pi }{4}{\left(20\right)}^2}\times 100\\ A_{redcution}=15.82\%\end{array}$

Therefore, the ductile cast iron has 15.82% reduction in area.

Interpretation Introduction

(f)

Interpretation:

With the help of plotted engineering stress-strain curve, the engineering stress should be determined at fracture for the given data of a ductile cast iron.

Concept Introduction:

Engineering stress is a term explained as a force or applied load on the given object's cross-sectional area and it is also known as nominal stress.

Answer to Problem 6.41P

The engineering stress is 397.9 MPa for given sample of a ductile cast iron.

Explanation of Solution

With the use of given spread sheet and applied loads, one can tabulate the engineering stress and strain as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, one can plot the stress-strain curve from the above gathered tabular data as follows:

  

From the graph, one can determine the value of engineering stress at fracture as 397.9 MPa.

Therefore, the engineering stress can be considered as 397.9 MPa for given sample of a ductile cast iron.

Interpretation Introduction

(g)

Interpretation:

With the help of plotted engineering stress-strain curve, the true stress at necking should be determined at fracture for the given data of a ductile cast iron.

Concept Introduction:

True stress can be defined as the applied force or load that is divided by the cross-sectional area of specimen or object. It can be also defined as the required amount of force that tends to deformation of specimen.

Answer to Problem 6.41P

The true stress is 1472.69 MPa for given ductile cast iron at necking.

Explanation of Solution

With the use of given spread sheet and applied loads, one can tabulate the engineering stress and strain as follows:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, one can plot the stress-strain curve from the above gathered tabular data as follows:

  

The formula for determining the strain of the given sample is as follows:

$\begin{array}{l}\epsilon =\frac{\Delta l}{l_0}\\ \epsilon =\frac{7.52}{40}\text{ (}\Delta l=125,000\text{N)}\\ \epsilon =0.188\text{ mm/mm}\end{array}$

For calculating the true stress at necking, the formula used is as follows:

$\begin{array}{l}\sigma =\frac{P}{A_0}\left(1+\epsilon \right)\\ \sigma =\frac{4\times P}{\pi {\left(d_0\right)}^2}\left(1+\epsilon \right)\\ \sigma =\frac{4\times 125,000}{\pi {\left(20\right)}^2}\left(1+0.0188\right)\\ \sigma =472.69{\text{ N/mm}}^2\times \frac{1\text{ MPa}}{1{\text{N/mm}}^2}\\ \sigma =472.69\text{ MPa}\end{array}$

Therefore, the ductile cast iron has 1472.69 MP as a value of true stress at point of necking.

Interpretation Introduction

(h)

Interpretation:

With the help of plotted engineering stress-strain curve, the value of modulus of resilience should be determined for the given data of a ductile cast iron.

Concept Introduction:

The amount of energy required to get absorbed by the material to return back to its original state is defined as resilience.

Modulus of resilience can be defined as the energy required by the material to return from its stress condition from zero to the yield stress limit.

Answer to Problem 6.41P

The value of modulus of resilience is 0.17 MPa of given sample of ductile cast iron.

Explanation of Solution

With the use of given spread sheet and applied loads, one can tabulate the engineering stress and strain as below:

Load (P)N	Area (A) ${\text{mm}}^2$	Stress ( $\sigma$)	Change in length ( $\Delta l$)	Strain ( $\epsilon$)
0	314.16	0	0	0
25,000	314.16	79.58	0.0185	0.000465
50,000	314.16	159.15	0.0370	0.000925
75,000	314.16	238.73	0.0555	0.0013875
90,000	314.16	286.48	0.20	0.005
105,000	314.16	334.22	0.60	0.015
120,000	314.16	381.97	1.56	0.039
131,000 (max. load)	314.16	416.98	4.00	0.1
125,000 (fracture)	314.16	397.87	7.52	0.188

Now, we can plot the stress-strain curve from the above gathered tabular data as below:

  

$\begin{array}{l}\text{modulus of resilience =}\frac{1}{2}\left(\text{yield strength}\right)\left(\text{strain at yield}\right)\\ \text{modulus of resilience =}\frac{1}{2}\left(238.73\right)\left(0.0013875\right)\\ \text{modulus of resilience =0}{\text{.17N/mm}}^2\times \frac{1\text{ MPa}}{1{\text{N/mm}}^2}\\ \text{modulus of resilience}=0.17\text{ MPa}\end{array}$

Therefore, the value of modulus of resilience can be considered as 0.17 MPa for the given sample of ductile cast iron.