A simply supported rectangular beam of span 4 m supports a udl of 40 kN/m. The crosssection of the beam is 200 mm x 400 mm. The max shear stress in the beam is:
1.5 N/mm2
Understanding the Problem: Shear Stress in a Simply Supported Beam
This problem asks us to calculate the maximum shear stress in a simply supported rectangular beam subjected to a uniform distributed load (UDL). We are given the span of the beam, the intensity of the UDL, and the dimensions of the rectangular cross-section.
Steps to Calculate Maximum Shear Stress
To find the maximum shear stress in the beam, we need to follow these steps:
- Determine the maximum shear force acting on the beam.
- Calculate the area of the beam's cross-section.
- Use the formula for maximum shear stress in a rectangular beam.
1. Calculate Maximum Shear Force ($V_{max}$)
For a simply supported beam subjected to a uniform distributed load (UDL) over its entire span, the maximum shear force occurs at the supports. The magnitude of the maximum shear force is given by the formula:
\[V_{max} = \frac{wL}{2}\]
Where:
- \(w\) is the intensity of the uniform distributed load.
- \(L\) is the span of the beam.
Given:
- \(w = 40 \, \text{kN/m} = 40 \times 10^3 \, \text{N/m}\)
- \(L = 4 \, \text{m}\)
Substituting the values:
\[V_{max} = \frac{(40 \times 10^3 \, \text{N/m}) \times (4 \, \text{m})}{2} = \frac{160 \times 10^3 \, \text{N}}{2} = 80 \times 10^3 \, \text{N}\]
So, the maximum shear force is \(80000 \, \text{N}\).
2. Calculate Cross-sectional Area (A)
The beam has a rectangular cross-section with dimensions 200 mm x 400 mm.
Given:
- Width, \(b = 200 \, \text{mm}\)
- Depth, \(d = 400 \, \text{mm}\)
The area of the rectangular cross-section is given by:
\[A = b \times d\]
Substituting the values:
\[A = 200 \, \text{mm} \times 400 \, \text{mm} = 80000 \, \text{mm}^2\]
3. Calculate Maximum Shear Stress ($\tau_{max}$)
For a rectangular cross-section, the maximum shear stress occurs at the neutral axis and is given by the formula:
\[\tau_{max} = 1.5 \times \frac{V_{max}}{A}\]
Where:
- \(V_{max}\) is the maximum shear force.
- \(A\) is the cross-sectional area.
We have \(V_{max} = 80000 \, \text{N}\) and \(A = 80000 \, \text{mm}^2\).
Substituting these values into the formula:
\[\tau_{max} = 1.5 \times \frac{80000 \, \text{N}}{80000 \, \text{mm}^2}\]
\[\tau_{max} = 1.5 \times 1 \, \text{N/mm}^2\]
\[\tau_{max} = 1.5 \, \text{N/mm}^2\]
Conclusion
The maximum shear stress in the simply supported rectangular beam is \(1.5 \, \text{N/mm}^2\). This stress occurs at the neutral axis of the beam cross-section.
Let's compare this result with the given options:
| Option | Value (N/mm2) |
|---|---|
| 1 | 1.9 |
| 2 | 1.75 |
| 3 | 1.5 |
| 4 | 1.0 |
Our calculated maximum shear stress of \(1.5 \, \text{N/mm}^2\) matches Option 3.
Important Questions from Shear Stress and Bending Stress
- For a beam to be classified as a beam of uniform strength, which of the following conditions must be met?
The maximum shear stress in a circular beam is
An increase in load at the free end of a cantilever is likely to cause failure-
The maximum bending stress in a curved beam having symmetrical section always occurs at the
The stresses caused by the bending moment is called -
