Final answer:
For a simply supported beam of 8 m span with a UDL of 50 KN/m, the required modulus of section is at least 3.333 x 10^3 cm^3 to not exceed a bending stress of 120 MPa.
Step-by-step explanation:
To calculate the modulus of section (Z) required for a simply supported beam carrying a uniform distributed load (UDL), we need to use the formula for the maximum bending moment in the beam, which occurs at the center due to symmetry. The formula for the maximum bending moment (M) for a UDL on a simply supported beam is given by:
M = \( \frac{wL^2}{8} \)
where w is the UDL per meter and L is the span of the beam. For a beam of span 8 m and a UDL of 50 KN/m, we have:
M = \( \frac{50 \times 8^2}{8} \) = 400 KN\cdot m = 400 \times 10^3 N\cdot m
Now, knowing the maximum stress (\( \sigma \)) that the material can withstand and the maximum bending moment (M), we can determine the required section modulus (Z) using the formula:
\( \sigma = \frac{M}{Z} \)
Rearranging the formula, we find Z:
Z = \( \frac{M}{\sigma} \) = \( \frac{400 \times 10^3}{120 \times 10^6} \)
Z = 3.333 \times 10^{-3} m^3
To find Z in cm^3 (which is a more common unit in engineering), we multiply by 10^6:
Z = 3.333 \times 10^3 cm^3
Therefore, the required section modulus of the beam should be at least 3.333 \times 10^3 cm^3 to ensure that the bending stress does not exceed 120 MPa.