The tensile test was conducted on a mild steel bar. The following data were obtained during the test: Diameter at the steel bar: 16 mm Gauge length at the bar: 80 mm Load at proportionality limit: 72 kN Extension at a load of 60 kN: 0.115 mm Load at failure: 80 kN Final gauge length of bar: 104 mm

Question

The tensile test was conducted on a mild steel bar. The following data were obtained during the test:
Diameter at the steel bar: 16 mm
Gauge length at the bar: 80 mm
Load at proportionality limit: 72 kN
Extension at a load of 60 kN: 0.115 mm
Load at failure: 80 kN
Final gauge length of bar: 104 mm
Diameter of the rod at failure = 12 mm
Determine (a) Young’s modulus, (b) proportionality limit, (c) true breaking stress, (d) percentage elongation and (e) percentage reduction in cross-sectional area.

Accepted Answer

A typical load–deflection plot for a steel bar during tensile test is shown in Figure 1.18.

(a) Young’s modulus (E ) is the ratio of longitudinal stress to longitudinal strain within the proportional limit

[latex] E=\frac{P / A_o}{\Delta L / L_o}=\frac{60 imes 10^3 imes 80}{(\pi / 4) imes 16^2 imes 0.115} \frac{ N }{ mm ^2} [/latex]

therefore E = 207593.4 MPa = 207.59 GPa.

(b) Proportional limit is ratio of load at proportional limit to original cross-sectional area

[latex] ext { Proportional limit }=\frac{72\left(10^3 ight)}{(\pi / 4) imes 16^2} MPa [/latex]

therefore proportional limit is 358.1 MPa.

(c) True breaking stress is the ratio of breaking point load to area at breaking point

[latex] ext { True breaking stress }=\frac{80 imes 10^3}{(\pi / 4) imes 12^2} \frac{ N }{ mm ^2} [/latex]

Thus, the true breaking stress is 707.36 MPa.

(d) Percentage elongation is given by

[latex] \% ext { elongation }=\frac{L_{ ext{f} }-L_{ o }}{L_{ o }} imes 100=\frac{104-80}{80} imes 100 [/latex]

where [latex] L_{ o } ext { and } L_{ ext{f} } [/latex] are initial and final gauge lengths. Therefore the percentage elongation is 30%.

(e) Percentage reduction in area is given by

[latex] \begin{aligned} \% ext { reduction } &=\frac{A_{ o }-A_{ ext{f} }}{A_{ o }} imes 100 \ &=\frac{(\pi / 4) d_{ o }^2-(\pi / 4) d_{ ext{f} }^2}{(\pi / 4) d_{ o }^2} imes 100 \ &=\left\lgroup 1-\frac{d_{ ext{f}}^2}{d_{ o }^2} ight group imes 100 \ &=\left\lgroup 1-\frac{12^2}{16^2} ight group imes 100 \end{aligned} [/latex]

where [latex] d_{ o } ext { and } d_{ ext{f} } [/latex] are initial and final rod diameters. Therefore percentage reduction is 43.75%.

Note: Ductility of steel, quantitatively, is expressed by the parameters calculated in parts (d) and (e) above.

Question 1.3: The tensile test was conducted on a mild steel bar. The foll...

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