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Determine the economical depth of the welded plate girder of span 24 m to carry super
imposed load of 35 kN/m. Avoid stiffeners and use Fe 415 steel :
(A) 1000 mm (B) 800 mm (C) 1400 mm (D) 1200 mm?
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Determine the economical depth of the welded plate girder of span 24 m...
**Problem Statement:**
We are given a welded plate girder with a span of 24 m and a superimposed load of 35 kN/m. We need to determine the economical depth of the girder using Fe 415 steel and without the use of stiffeners.

**Approach:**
To determine the economical depth of the welded plate girder, we will follow the following steps:

1. Calculate the maximum bending moment (Mmax) at the center of the span.
2. Determine the required section modulus (Sreq) based on the maximum bending moment and the allowable stress of Fe 415 steel.
3. Assume a suitable depth for the plate girder and calculate the actual section modulus (Sact) based on the assumed depth and the properties of the girder.
4. Compare the required section modulus (Sreq) with the actual section modulus (Sact) and adjust the depth of the plate girder until they are equal.
5. Once the required section modulus (Sreq) is equal to the actual section modulus (Sact), the corresponding depth of the plate girder will be the economical depth.

**Calculation:**
1. Maximum Bending Moment (Mmax):
- The maximum bending moment occurs at the center of the span.
- Mmax = (w * L^2) / 8, where w is the superimposed load and L is the span of the girder in meters.
- Substituting the given values, we have Mmax = (35 * 24^2) / 8 = 2520 kNm.

2. Required Section Modulus (Sreq):
- The required section modulus is calculated using the formula Sreq = Mmax / σallow, where σallow is the allowable stress of Fe 415 steel.
- For Fe 415 steel, the allowable stress is taken as 165 N/mm^2.
- Substituting the given values, we have Sreq = 2520 / 165 = 15.27 * 10^3 mm^3.

3. Assumed Depth and Actual Section Modulus:
- Assume a suitable depth for the plate girder, let's say "d" mm.
- The actual section modulus is calculated using the formula Sact = bd^2 / 6, where b is the width of the girder.
- The width of the girder can be assumed based on design requirements or standards.
- For this problem, let's assume a width of 300 mm.
- Substituting the given values, we have Sact = (300 * d^2) / 6 = 50 * d^2 mm^3.

4. Comparison and Adjustment:
- Compare the required section modulus (Sreq) with the actual section modulus (Sact) and adjust the depth of the plate girder until they are equal.
- Set Sreq = Sact and solve for "d".
- 15.27 * 10^3 = 50 * d^2
- d^2 = 305.4
- Taking the square root, we have d = 17.48 mm.

5. Economical Depth:
- The calculated depth of 17.48 mm is not a practical depth for a plate girder.
- Therefore, we need to round up the depth to the nearest practical
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Determine the economical depth of the welded plate girder of span 24 m to carry superimposed load of 35 kN/m. Avoid stiffeners and use Fe 415 steel :(A) 1000 mm (B) 800 mm (C) 1400 mm (D) 1200 mm?
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Determine the economical depth of the welded plate girder of span 24 m to carry superimposed load of 35 kN/m. Avoid stiffeners and use Fe 415 steel :(A) 1000 mm (B) 800 mm (C) 1400 mm (D) 1200 mm? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Determine the economical depth of the welded plate girder of span 24 m to carry superimposed load of 35 kN/m. Avoid stiffeners and use Fe 415 steel :(A) 1000 mm (B) 800 mm (C) 1400 mm (D) 1200 mm? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Determine the economical depth of the welded plate girder of span 24 m to carry superimposed load of 35 kN/m. Avoid stiffeners and use Fe 415 steel :(A) 1000 mm (B) 800 mm (C) 1400 mm (D) 1200 mm?.
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