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What will be the minimum stress in a section (20 cm x 25 cm) of a column carrying a load of 30 kN in a plane bisecting 25 cm side, at an eccentricity of 10 cm from the geometric axis of the section?
- a)2.4 MN/m2, tensile
- b)2.4 MN/m2, compressive
- c)1.2 MN/m2, compressive
- d)1.2 MN/m2, tensile
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
What will be the minimum stress in a section (20 cm x 25 cm) of a colu...
Minimum stress in the section = direct stress - bending stress
Direct stress = 30 / (0.2 x 0.25) = 600 kN/m2
M = 30 x 0.1 = 3 kN-m
d = 0.25 m, b = 0.2 m
Z = db2/6 = 0.01/6
Bending stress = 3 x 6/0.01 = 1800 kN/m2
Minimum stress in the section = direct stress - bending stress = 600 - 1800 = -1200 kN/m2 (tensile) = 1.2 MN/m2(tensile)
Direct stress = 30 / (0.2 x 0.25) = 600 kN/m2
M = 30 x 0.1 = 3 kN-m
d = 0.25 m, b = 0.2 m
Z = db2/6 = 0.01/6
Bending stress = 3 x 6/0.01 = 1800 kN/m2
Minimum stress in the section = direct stress - bending stress = 600 - 1800 = -1200 kN/m2 (tensile) = 1.2 MN/m2(tensile)
Most Upvoted Answer
What will be the minimum stress in a section (20 cm x 25 cm) of a colu...
To calculate the minimum stress in the given section of the column, we can use the formula for eccentrically loaded columns:
σ_min = (P/A) - (M*y/I)
Where:
σ_min = minimum stress in the section
P = applied load on the column (30 kN)
A = cross-sectional area of the section (20 cm x 25 cm = 500 cm^2)
M = moment due to eccentricity (P * e)
y = distance from the centroid of the section to the extreme fiber (12.5 cm)
I = moment of inertia of the section
Calculating the moment due to eccentricity:
M = P * e = 30 kN * 10 cm = 300 kNm
Calculating the moment of inertia of the section:
I = (b * h^3) / 12
= (20 cm * (25 cm)^3) / 12
= 104,167 cm^4
Substituting the given values into the formula for minimum stress:
σ_min = (30 kN / 500 cm^2) - (300 kNm * 12.5 cm / 104,167 cm^4)
Simplifying the equation:
σ_min = 0.06 kN/cm^2 - 0.036 kN/cm^2
σ_min = 0.024 kN/cm^2
Converting the stress from kN/cm^2 to MN/m^2:
σ_min = 0.024 MN/m^2
Therefore, the minimum stress in the section of the column is 0.024 MN/m^2, which is equal to 1.2 MN/m^2 when rounded to one decimal place.
Therefore, the correct answer is option D: 1.2 MN/m^2, tensile.
σ_min = (P/A) - (M*y/I)
Where:
σ_min = minimum stress in the section
P = applied load on the column (30 kN)
A = cross-sectional area of the section (20 cm x 25 cm = 500 cm^2)
M = moment due to eccentricity (P * e)
y = distance from the centroid of the section to the extreme fiber (12.5 cm)
I = moment of inertia of the section
Calculating the moment due to eccentricity:
M = P * e = 30 kN * 10 cm = 300 kNm
Calculating the moment of inertia of the section:
I = (b * h^3) / 12
= (20 cm * (25 cm)^3) / 12
= 104,167 cm^4
Substituting the given values into the formula for minimum stress:
σ_min = (30 kN / 500 cm^2) - (300 kNm * 12.5 cm / 104,167 cm^4)
Simplifying the equation:
σ_min = 0.06 kN/cm^2 - 0.036 kN/cm^2
σ_min = 0.024 kN/cm^2
Converting the stress from kN/cm^2 to MN/m^2:
σ_min = 0.024 MN/m^2
Therefore, the minimum stress in the section of the column is 0.024 MN/m^2, which is equal to 1.2 MN/m^2 when rounded to one decimal place.
Therefore, the correct answer is option D: 1.2 MN/m^2, tensile.
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What will be the minimum stress in a section (20 cm x 25 cm) of a column carrying a load of 30 kN in a plane bisecting 25 cm side, at an eccentricity of 10 cm from the geometric axis of the section?a)2.4 MN/m2, tensileb)2.4 MN/m2, compressivec)1.2 MN/m2, compressived)1.2 MN/m2, tensileCorrect answer is option 'D'. Can you explain this answer?
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What will be the minimum stress in a section (20 cm x 25 cm) of a column carrying a load of 30 kN in a plane bisecting 25 cm side, at an eccentricity of 10 cm from the geometric axis of the section?a)2.4 MN/m2, tensileb)2.4 MN/m2, compressivec)1.2 MN/m2, compressived)1.2 MN/m2, tensileCorrect answer is option 'D'. Can you explain this answer? 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 What will be the minimum stress in a section (20 cm x 25 cm) of a column carrying a load of 30 kN in a plane bisecting 25 cm side, at an eccentricity of 10 cm from the geometric axis of the section?a)2.4 MN/m2, tensileb)2.4 MN/m2, compressivec)1.2 MN/m2, compressived)1.2 MN/m2, tensileCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What will be the minimum stress in a section (20 cm x 25 cm) of a column carrying a load of 30 kN in a plane bisecting 25 cm side, at an eccentricity of 10 cm from the geometric axis of the section?a)2.4 MN/m2, tensileb)2.4 MN/m2, compressivec)1.2 MN/m2, compressived)1.2 MN/m2, tensileCorrect answer is option 'D'. Can you explain this answer?.
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