A beam 300 mm depth and of symmetrical I section has a I = 1 x 108 mm4 and is simply supported over a span of 6 m. Calculate the udl it may carry if the max bending stress is not to exceed 100 N/mm2
14.81 kN/m
Calculating UDL for Beam Bending Stress
This problem involves determining the maximum uniformly distributed load (UDL) a simply supported beam can carry without exceeding a specified maximum bending stress. We are given the beam's dimensions, moment of inertia, span, and the maximum allowable bending stress.
Given Data
| Parameter | Value |
|---|---|
| Beam depth (h) | 300 mm |
| Section | Symmetrical I section |
| Moment of Inertia (I) | 1 x 108 mm4 |
| Span (L) | 6 m |
| Maximum bending stress (σmax) | 100 N/mm2 |
Step-by-Step Calculation of UDL
First, let's ensure all units are consistent. The span is given in meters, while other dimensions and stress are in millimeters and Newtons/mm2. We will convert the span to millimeters:
- Span \(L = 6 \text{ m} = 6 \times 1000 \text{ mm} = 6000 \text{ mm}\)
For a symmetrical I section, the neutral axis is at the mid-depth. The maximum bending stress occurs at the extreme fibers, which are at a distance \(y_{max}\) from the neutral axis.
- Distance to extreme fiber \(y_{max} = \frac{\text{Beam depth}}{2} = \frac{300 \text{ mm}}{2} = 150 \text{ mm}\)
The section modulus (Z) is a geometric property of the beam's cross-section that relates bending stress to bending moment. It is calculated as:
\[Z = \frac{I}{y_{max}}\]Substitute the given values:
\[Z = \frac{1 \times 10^8 \text{ mm}^4}{150 \text{ mm}}\] \[Z = \frac{10^8}{150} \text{ mm}^3\]The relationship between maximum bending stress ($\sigma_{max}$), maximum bending moment ($M_{max}$), and section modulus (Z) is given by the bending formula:
\[\sigma_{max} = \frac{M_{max}}{Z}\]We can rearrange this formula to find the maximum allowable bending moment:
\[M_{max} = \sigma_{max} \times Z\]Substitute the values for $\sigma_{max}$ and Z:
\[M_{max} = 100 \text{ N/mm}^2 \times \frac{10^8}{150} \text{ mm}^3\] \[M_{max} = \frac{100 \times 10^8}{150} \text{ N-mm}\] \[M_{max} = \frac{10^{10}}{150} \text{ N-mm}\] \[M_{max} = \frac{10^9}{15} \text{ N-mm}\]For a simply supported beam subjected to a uniformly distributed load (w) over its entire span (L), the maximum bending moment occurs at the center and is given by the formula:
\[M_{max} = \frac{wL^2}{8}\]We know $M_{max}$ and $L$, and we need to find 'w'. Rearrange the formula to solve for 'w':
\[w = \frac{8 \times M_{max}}{L^2}\]Substitute the calculated value of $M_{max}$ and the span $L$ (in mm):
\[w = \frac{8 \times \left(\frac{10^9}{15}\right) \text{ N-mm}}{(6000 \text{ mm})^2}\] \[w = \frac{\frac{8 \times 10^9}{15} \text{ N-mm}}{36 \times 10^6 \text{ mm}^2}\] \[w = \frac{8 \times 10^9}{15 \times 36 \times 10^6} \text{ N/mm}\] \[w = \frac{8 \times 1000}{15 \times 36} \text{ N/mm}\] \[w = \frac{8000}{540} \text{ N/mm}\] \[w = \frac{800}{54} \text{ N/mm}\] \[w = \frac{400}{27} \text{ N/mm}\] \[w \approx 14.8148 \text{ N/mm}\]The options are given in kN/m or N/mm. Let's convert our result from N/mm to kN/m:
We know that 1 m = 1000 mm and 1 kN = 1000 N.
\[1 \text{ N/mm} = \frac{1 \text{ N}}{1 \text{ mm}} = \frac{1 \times (10^{-3} \text{ kN})}{1 \times (10^{-3} \text{ m})} = \frac{10^{-3} \text{ kN}}{10^{-3} \text{ m}} = 1 \text{ kN/m}\]Therefore, \(w \approx 14.8148 \text{ N/mm}\) is approximately equal to \(14.8148 \text{ kN/m}\).
Comparing Calculated UDL with Options
Our calculated UDL is approximately 14.8148 kN/m.
- Option 1: 14.6 kN/m
- Option 2: 15 N/mm (= 15 kN/m)
- Option 3: 14.81 kN/m
- Option 4: 15.14 kN/m
Comparing the calculated value to the options, 14.81 kN/m is the closest match.
Thus, the maximum uniformly distributed load the beam can carry without exceeding the bending stress limit is approximately 14.81 kN/m.
Important Questions from Beams
For a simply supported beam or slab, the effective span is calculated as:
Which of the following is CORRECT for indeterminate beam condition?
A cantilever beam is one which is -
In case of deep beam or in thin webbed R.C.C members, the first crack formed is-
In case of web crippling, the dispersion of load from bearing plate takes place at:
