Part (a)
For 1 plank
$I_1 = \dfrac{100(20^3)}{12}$
$I_1 = \frac{200\,000}{3} \, \text{mm}^4$
For the whole beam
$I = 6I_1 = 6(200\,000/3)$
$I = 400\,000 \, \text{mm}^4$
$\dfrac{M}{EI} = \dfrac{M_1}{E_1 I_1}$ where E1 = E
$\dfrac{M}{E(400\,000)} = \dfrac{M_1}{E(\frac{200\,000}{3})}$
$\dfrac{M_1}{M} = \dfrac{1}{6}$ answer
Part (b)
For 1 plank
$I_1 = \dfrac{100(10^3)}{12}$
$I_1 = \frac{25\,000}{3} \, \text{mm}^4$
For the whole beam
$I = 12I_1 = 12(25\,000/3)$
$I = 100\,000 \, \text{mm}^4$
$\dfrac{M}{EI} = \dfrac{M_1}{E_1 I_1}$ where E1 = E
$\dfrac{M}{E(100\,000)} = \dfrac{M_1}{E(\frac{200\,000}{3})}$
$\dfrac{M_1}{M} = \dfrac{1}{12}$ ) answer
I think what is ask is the…
I think what is ask is the comparison of the strength of the built-up beam consisting of 6 planks - 100 mm x 20 mm dimension, resulting in an overall dimensions of 100 mm X 120 mm to the strength of a solid beam with a dimension of 100mm x 120mm.. we compare the section modulus of the beams. The beam cross section with higher section modulus is stronger since it experiences less flexural stress (assuming built up beam and solid beam are subjected to the same loading (same bending moment)). I'm kind of like confuse with the solution above, comparing the strength of 1 plank to the strength of 6 planks combine...
Yes you are absolutely right…
In reply to I think what is ask is the… by Aerisk16
Yes you are absolutely right. We flagged this page for revision. Note that after revision is made, our comments will be put to archive. Thank you for calling our attention, highly appreciated.