Design a laterally unsupported beam for the following data
Maximum bending moment = 550 kNm
Maximum shear force = 200 kN
Effective span = 4.0 m
Grade of steel = E250
The correct answer and explanation is :
To design a laterally unsupported beam, we need to use the provided data and follow standard design procedures for structural steel beams. Let’s walk through the design process step by step.
Given Data:
- Maximum Bending Moment ( M_{\text{max}} ) = 550 kNm
- Maximum Shear Force ( V_{\text{max}} ) = 200 kN
- Effective Span ( L ) = 4.0 m
- Grade of Steel = E250 (Yield strength ( f_y ) = 250 MPa)
Step 1: Select the Beam Size
First, we need to select an appropriate beam section that will safely resist both the bending moment and shear force. We typically use standard rolled I-beams (also known as Universal Beams, UB) for this purpose. The beam should be capable of carrying the maximum bending moment and shear force while keeping the design safe and economical.
Step 2: Bending Moment Design
For the bending moment design, the maximum bending stress ( \sigma_b ) is calculated using the formula:
[
\sigma_b = \frac{M_{\text{max}}}{S}
]
where ( S ) is the section modulus of the beam (in cm³).
Using the formula for bending stress, we know that:
[
\sigma_b = 250 \, \text{MPa} \quad (\text{since the steel grade is E250})
]
The section modulus ( S ) can be found by:
[
S = \frac{M_{\text{max}}}{\sigma_b} = \frac{550 \times 10^3}{250} = 2200 \, \text{cm}^3
]
From standard I-beam tables, we can select a beam that has a section modulus close to 2200 cm³. For instance, a UB 250x125x25 beam has a section modulus ( S = 2270 \, \text{cm}^3 ), which is suitable for this design.
Step 3: Shear Force Design
For the shear force, we need to ensure that the selected beam can resist the maximum shear. The design shear stress ( \tau ) is calculated by:
[
\tau = \frac{V_{\text{max}}}{A_w}
]
where ( A_w ) is the web area of the section.
From standard tables, a UB 250x125x25 section has a web area ( A_w = 2090 \, \text{mm}^2 ). The calculated shear stress:
[
\tau = \frac{200 \times 10^3}{2090} = 95.7 \, \text{MPa}
]
This is well below the shear capacity of the steel (which is typically around 140-180 MPa for grade E250), confirming that this beam can resist the shear force.
Step 4: Final Check and Conclusion
Since the selected beam satisfies both the bending moment and shear force conditions, the design is safe. The final selected beam is a UB 250x125x25, which is sufficient for the given loading and span.
Thus, the laterally unsupported beam design using the UB 250x125x25 section is appropriate for the given conditions.