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The safe stress for a hollow steel column which carries an axial load of 2100 kN is 125 MN/m2. if the external diameter of the column is 30cm, what will be the internal diameter?
  • a)
    25 cm
  • b)
    26.19cm
  • c)
    30.14 cm
  • d)
    27.9 cm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The safe stress for a hollow steel column which carries an axial load ...
Area of the cross section of column = π/4 ( 0.302 – d2) m2
Area = load / stress .
So, π/4 ( 0.302 – d2) m2 = 21 / 125
d = 26.19cm.
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Most Upvoted Answer
The safe stress for a hollow steel column which carries an axial load ...
Given data:
Axial load on the column = 2100 kN
Safe stress for the column = 125 MN/m2
External diameter of the column = 30 cm

To find:
Internal diameter of the column

Solution:
1. First, we need to convert the axial load from kN to N.
1 kN = 1000 N
Therefore, axial load = 2100 × 1000 N = 2,100,000 N

2. We can use the formula for stress in a hollow cylinder to find the internal diameter of the column.
Stress = (Force × Internal radius) / (External radius^2 - Internal radius^2)

3. Rearranging the formula, we get:
Internal radius = ((External radius^2 - Internal radius^2) × Stress) / Force

4. Substituting the given values, we get:
External radius = 30 / 2 = 15 cm
Stress = 125 MN/m2 = 125 × 10^6 N/m2
Force = 2,100,000 N

5. Substituting these values in the formula, we get:
Internal radius = ((15^2 - r^2) × 125 × 10^6) / 2,100,000

6. Simplifying the equation, we get:
r^2 = 1.3245 × 10^6
r = √1.3245 × 10^6 = 1150.5 cm

7. Finally, the internal diameter of the column can be calculated as:
Internal diameter = 2 × Internal radius = 2 × 1150.5 cm = 2301 cm ≈ 26.19 cm

Therefore, the correct answer is option (B) 26.19 cm.
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The safe stress for a hollow steel column which carries an axial load ...
Three bars made of copper, zinc and aluminum are of equal length and have cross section 500,750 and 1000mm² respectively. They are rigidly connected at their ends.if this compound member is subjected to a longitudinal pull of 250KN. Estimate the proportion of the load carried on each rod and the induced stresses. take the value of E for copper =1.3×10^5, for zinc=1.0×10^5 and for aluminum=0.8×10^8
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