A bar of metal 25 mm in diameter is tested on a length of 250 mm. In tension the following results were recorded: A torsion test gave the following results: Represent these results in graphical form and hence determine Young’s modulus, E, the modulus of rigidity, G, Poisson’s ratio, v, and the bulk

Question

A bar of metal 25 mm in diameter is tested on a length of 250 mm. In tension the following results were recorded:

Table P.8.2(a)
Load (kN)	10.4	31.2	51.0	72.8
Extension (mm)	0.036	0.089	0.140	0.191

A torsion test gave the following results:

Table P.8.2(b)
Torque (kNm)	0.051	0.052	0.253	0.354
Angle of twist (degrees)	0.24	0.71	1.175	1.642

Represent these results in graphical form and hence determine Young’s modulus, E, the modulus of rigidity, G, Poisson’s ratio, v, and the bulk modulus, K, for the metal. (Note: see Chapter 11 for torque–angle of twist relationship).

Accepted Answer

From Eq. (7.8) Young’s modulus E is equal to the slope of the stress–strain curve. Then, since stress=load/area and strain=extension/original length, E=slope of the load-extension curve multiplied by (original length/area of cross section).

[latex] \frac{\sigma}{\varepsilon}=E [/latex] (7.8)

From the results given the slope of the load-extension curve [latex] \simeq 402.6 kN / mm [/latex]. Then

[latex] E \simeq \frac{402.6 imes 10^{3} imes 250}{\left(\frac{\pi imes 25^{2}}{4} ight)} \simeq 205000 N / mm ^{2} [/latex]

From Eq. (11.4) the modulus of rigidity is given by

[latex] \frac{T}{J}=\frac{ au}{r}=G \frac{ heta}{L} [/latex] (11.4)

[latex] G=\frac{T L}{ heta J} [/latex]

Therefore the slope of the torque-angle of twist (in radians) graph multiplied by (L/J ) is equal to G. From the results given the slope of the torque-angle of twist graph is [latex] \simeq 12.38 kNm / rad [/latex]. Therefore

[latex] G \simeq \frac{12.38 imes 10^{6} imes 250}{\left(\frac{\pi imes 25^{4}}{32} ight)} \simeq 80700 N / mm ^{2} [/latex]

Having obtained E and G the value of Poisson’s ratio may be found from Eq. (7.21), i.e.

[latex] G=\frac{E}{2(1+v)} ext { or } E=2 G(1+v) [/latex] (7.21)

[latex] v=\left(\frac{E}{2 G} ight)-1 \simeq 0.27 [/latex]

Finally, the bulk modulus K may be found using either of Eqs (7.22) or (7.23). From Eq. (7.22)

[latex] K=\frac{E}{3(1-2 v)} [/latex] (7.22)

[latex] K=\frac{2 G(1+v)}{3(1-2 v)} [/latex] (7.23)

[latex] K \simeq \frac{E}{3(1-2 v)} \simeq 148500 N / mm ^{2} [/latex]

Question 8.2: A bar of metal 25 mm in diameter is tested on a length of 25...

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