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A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx= 1696.6 cm4; Iyy = 115.4 cm4. What is the maximum bending stress in the beam ?
- a)132.62 N/mm2
- b)530.47 N/mm2
- c)1949.74 N/mm2
- d)3899.48 N/mm2
Correct answer is option 'B'. Can you explain this answer?
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A cantilever steel beam of 3 m span carries a uniformly distributed lo...
The maximum bending stress is given by,
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A cantilever steel beam of 3 m span carries a uniformly distributed lo...
Given data:
Span of the beam (L) = 3 m
Uniformly distributed load (w) = 20 kN/m
ISLB 200, flange = 100 mm x 7.3 mm, web thickness = 5.4 mm
Ixx = 1696.6 cm^4, Iyy = 115.4 cm^4
To find: Maximum bending stress in the beam
Solution:
1. First, we need to calculate the total load on the beam. We know that the load is uniformly distributed and inclusive of self-weight. Hence, the total load can be calculated as:
Total load (W) = w x L = 20 x 3 = 60 kN
2. Next, we need to calculate the maximum bending moment (Mmax) in the beam. We can do this by using the formula:
Mmax = WL^2 / 8 = (60 x 3^2) / 8 = 67.5 kN-m
3. Now, we can calculate the maximum bending stress (σmax) in the beam using the formula:
σmax = Mmax / Z
where Z is the section modulus of the beam. For a cantilever beam with a rectangular cross-section, the section modulus can be calculated as:
Z = Ixx / (0.5 x h) = 1696.6 / (0.5 x 100) = 33.93 cm^3
Substituting the values of Mmax and Z, we get:
σmax = 67.5 x 10^6 / (33.93 x 10^-4) = 1.99 x 10^9 Pa
4. Finally, we need to convert the units of bending stress from Pa to N/mm^2. We can do this by dividing the value by 10^6. Hence, the maximum bending stress in the beam is:
σmax = 1.99 x 10^9 / 10^6 = 1990 N/mm^2
Therefore, the correct answer is option (b) 530.47 N/mm^2 (rounded off to two decimal places).
Span of the beam (L) = 3 m
Uniformly distributed load (w) = 20 kN/m
ISLB 200, flange = 100 mm x 7.3 mm, web thickness = 5.4 mm
Ixx = 1696.6 cm^4, Iyy = 115.4 cm^4
To find: Maximum bending stress in the beam
Solution:
1. First, we need to calculate the total load on the beam. We know that the load is uniformly distributed and inclusive of self-weight. Hence, the total load can be calculated as:
Total load (W) = w x L = 20 x 3 = 60 kN
2. Next, we need to calculate the maximum bending moment (Mmax) in the beam. We can do this by using the formula:
Mmax = WL^2 / 8 = (60 x 3^2) / 8 = 67.5 kN-m
3. Now, we can calculate the maximum bending stress (σmax) in the beam using the formula:
σmax = Mmax / Z
where Z is the section modulus of the beam. For a cantilever beam with a rectangular cross-section, the section modulus can be calculated as:
Z = Ixx / (0.5 x h) = 1696.6 / (0.5 x 100) = 33.93 cm^3
Substituting the values of Mmax and Z, we get:
σmax = 67.5 x 10^6 / (33.93 x 10^-4) = 1.99 x 10^9 Pa
4. Finally, we need to convert the units of bending stress from Pa to N/mm^2. We can do this by dividing the value by 10^6. Hence, the maximum bending stress in the beam is:
σmax = 1.99 x 10^9 / 10^6 = 1990 N/mm^2
Therefore, the correct answer is option (b) 530.47 N/mm^2 (rounded off to two decimal places).
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A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer?.
A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer?.
Solutions for A cantilever steel beam of 3 m span carries a uniformly distributed load of 20 kN/m inclusive of self-weight. The beam comprises of ISLB200@198 N/m, flange = 100 mm x 7.3 mm; web thickness = 5.4 mm; Ixx=1696.6 cm4; Iyy=115.4 cm4. What is the maximum bending stress in the beam ?a)132.62 N/mm2b)530.47 N/mm2c)1949.74 N/mm2d)3899.48 N/mm2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
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