Answer :
To solve this problem, we need to use the formula for the deflection of a cantilever beam subject to a uniformly distributed load (UDL). The deflection [tex]\delta[/tex] at the free end of the cantilever beam can be given by the formula:
[tex]\delta = \frac{{wL^4}}{{8EI}}[/tex]
Where:
- [tex]\delta[/tex] = deflection at the free end (2.7 mm in this case)
- [tex]w[/tex] = uniformly distributed load per unit length, in N/mm (what we need to find)
- [tex]L[/tex] = length of the beam (1500 mm, converting 1.5 m to mm)
- [tex]E[/tex] = modulus of elasticity (given as [tex]2.1 \times 10^5[/tex] N/mm²)
- [tex]I[/tex] = moment of inertia of the beam's cross-section
First, calculate the moment of inertia [tex]I[/tex] for a rectangular section:
[tex]I = \frac{{b \, h^3}}{12}[/tex]
Where:
- [tex]b[/tex] = width of the beam (150 mm)
- [tex]h[/tex] = depth of the beam (200 mm)
Substituting the values:
[tex]I = \frac{{150 \, \times \, (200)^3}}{12} = \frac{{150 \, \times \, 8,000,000}}{12} = 100,000,000 \, \text{mm}^4[/tex]
Now, plug the values into the deflection formula:
[tex]2.7 = \frac{{w \, \times \, (1500)^4}}{{8 \, \times \, 2.1 \times 10^5 \, \times \, 100,000,000}}[/tex]
Calculate [tex]L^4[/tex]:
[tex](1500)^4 = 5,062,500,000,000[/tex]
Substitute back into the equation and solve for [tex]w[/tex]:
[tex]2.7 = \frac{{w \, \times \, 5,062,500,000,000}}{{168,000,000,000,000}}[/tex]
Rearranging to solve for [tex]w[/tex]:
[tex]w = \frac{{2.7 \, \times \, 168,000,000,000,000}}{5,062,500,000,000}[/tex]
[tex]w \approx 0.0897 \, \text{N/mm}[/tex]
Therefore, the uniformly distributed load that the beam should carry to produce a deflection of 2.7 mm at the free end is approximately [tex]0.0897 \, \text{N/mm}[/tex]. This result can be verified through a re-calculation for accuracy.
