Answer :
To determine how wide the beam should be, we need to use the concepts of stress in beams under a uniformly distributed load. Let's break down the problem:
Understanding the problem:
- The beam has a span of 4 meters (m).
- It carries a uniformly distributed load of 0.25 tons per meter (t/m).
- The beam is of rectangular cross-section with a depth (height) of 25 centimeters (cm).
- The stress in the material must not exceed 0.2 tons per square centimeter (t/cm²).
Converting units:
- Since the problem is given in metric units, ensure consistency in the units used. Here, 1 ton = 1,000 kilograms, and we need square centimeters for stress.
Calculate the total load on the beam:
[
\text{Total load} = \text{Uniformly distributed load} \times \text{span length}
]
[
= 0.25 ; \text{t/m} \times 4 ; \text{m} = 1 ; \text{ton}
]
Calculate the maximum bending moment (M):
- For a simple beam with a uniformly distributed load, the maximum bending moment can be calculated by:
[
M = \frac{wL^2}{8}
]
Where:
- [tex]w[/tex] = load per unit length = 0.25 t/m
- [tex]L[/tex] = length of the span in meters = 4 m
[
M = \frac{0.25 \times 4^2}{8} = 0.5 ; \text{t}\cdot\text{m} = 500 ; \text{t}\cdot\text{cm}
]
Using the bending stress formula:
The formula for maximum stress ([tex]\sigma[/tex]) in a beam is given by:
[
\sigma = \frac{M}{Z}
]
Where [tex]Z[/tex] is the section modulus for a rectangle, [tex]Z = \frac{bd^2}{6}[/tex]; [tex]b[/tex] is the width, and [tex]d[/tex] the depth of the beam.
Calculating the width (b):
Rearrange the stress formula to solve for [tex]b[/tex]:
[
0.2 = \frac{500}{\frac{b \times 25^2}{6}}
]
Simplify:
[
0.2 = \frac{500 \times 6}{b \times 625}
]
[
0.2 = \frac{3000}{625b}
]
[
0.2b = 4.8
]
[
b = \frac{4.8}{0.2} = 24; \text{cm}
]
Therefore, the beam should be 24 centimeters wide to ensure the stress does not exceed 0.2 t/cm².
