# Young's Modulus By Four-Point Bending Test The apparatus was set up as above with masses varied at points A and C. The distances from the supports to the ends of the beam were measured and equalised at 145mm. The same was done for the distance between the supports and the midpoint of the beam which was at 125mm. The beam was measured to have a thickness of 3.4mm and width 25mm, the thickness was measured with a calliper for increased precision. The strain gauge and deflection gauge are positioned at point B. By varying the mass at A and C, corresponding results of deflection and strain was obtained.

The free body diagram, along with shear and bending moment diagrams are given below.   From the following diagram, the relationship between δ and the radius, r, of the arc of the beam is derived. The arc formed by the beam has a radius r from point O. Then by Pythagoras

$$FO^2 + FC^2 = CO^2$$

Substituting and rearranging for r

$$r = \frac{n^2}{2\delta}$$

## Results   The load/strain graph gave wildly skewed results when forced through the origin so the SLOPE() function in Excel was used instead. The two slopes were therefore determined as follows

$$\frac{w}{\delta} = 11.928$$
$$\frac{w}{\varepsilon} = 19034.089$$

From the equation

$$\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}$$

and since
$$\sigma = \frac{E}{R}y$$
$$I = \frac{bt^3}{12}$$

The Young's Modulus, E, can be expressed as

$$E = \frac{6m}{bt^2} \frac{W}{\varepsilon}$$
$$E = \frac{6mn^2}{bt^3} \frac{W}{\delta}$$

The two values calculated from the results obtained yield the following

$$E_{\frac{W}{\varepsilon}} = 194819.5 \mathrm{N mm^{-1}}$$
$$E_{\frac{W}{\delta}} = 165017.6 \mathrm{N mm^{-1}}$$