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A 200 mm thick brick masonary wall made of modular brick carries an axial load of 30 kN/m from wall above and an eccentric load of 20 kN/m from RCC floor acting at a distance of47.5 mm from the centre line of the wall. The resultant eccentricity will be
- a)19
- b)20
- c)21
- d)22
Correct answer is option 'A'. Can you explain this answer?
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A 200 mm thick brick masonary wall made of modular brick carries an ax...
Axial and Eccentric Loads on a Brick Masonry Wall
Given data:
Thickness of wall (t) = 200 mm
Axial load (P) = 30 kN/m
Eccentric load (W) = 20 kN/m
Distance of eccentric load from center line (e) = 47.5 mm
To find: Resultant eccentricity
Solution:
1. Calculation of Area:
The cross-sectional area of the wall can be calculated as:
A = t x 1 (assuming 1 m length of the wall)
A = 0.2 m x 1 m
A = 0.2 m²
2. Calculation of Axial Stress:
The axial stress on the wall can be calculated as:
σa = P/A
σa = 30 kN/m / 0.2 m²
σa = 150 kN/m²
3. Calculation of Eccentric Stress:
The eccentric stress on the wall can be calculated as:
σe = We/I + Pe/A
where,
I = moment of inertia of the wall cross-section
I = (t x (1)³) / 12
I = (0.2 m x 1³) / 12
I = 0.00167 m⁴
σe = (20 kN/m x 0.0475 m) / 0.00167 m⁴ + (30 kN/m / 0.2 m²)
σe = 570.06 kN/m²
4. Calculation of Resultant Eccentricity:
The resultant eccentricity can be calculated as:
e' = e x (σa / (σe - σa))
e' = 0.0475 m x (150 kN/m² / (570.06 kN/m² - 150 kN/m²))
e' = 0.019 m
e' = 19 mm
Therefore, the resultant eccentricity will be 19 mm.
Answer: Option A (19)
Given data:
Thickness of wall (t) = 200 mm
Axial load (P) = 30 kN/m
Eccentric load (W) = 20 kN/m
Distance of eccentric load from center line (e) = 47.5 mm
To find: Resultant eccentricity
Solution:
1. Calculation of Area:
The cross-sectional area of the wall can be calculated as:
A = t x 1 (assuming 1 m length of the wall)
A = 0.2 m x 1 m
A = 0.2 m²
2. Calculation of Axial Stress:
The axial stress on the wall can be calculated as:
σa = P/A
σa = 30 kN/m / 0.2 m²
σa = 150 kN/m²
3. Calculation of Eccentric Stress:
The eccentric stress on the wall can be calculated as:
σe = We/I + Pe/A
where,
I = moment of inertia of the wall cross-section
I = (t x (1)³) / 12
I = (0.2 m x 1³) / 12
I = 0.00167 m⁴
σe = (20 kN/m x 0.0475 m) / 0.00167 m⁴ + (30 kN/m / 0.2 m²)
σe = 570.06 kN/m²
4. Calculation of Resultant Eccentricity:
The resultant eccentricity can be calculated as:
e' = e x (σa / (σe - σa))
e' = 0.0475 m x (150 kN/m² / (570.06 kN/m² - 150 kN/m²))
e' = 0.019 m
e' = 19 mm
Therefore, the resultant eccentricity will be 19 mm.
Answer: Option A (19)
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A 200 mm thick brick masonary wall made of modular brick carries an axial load of 30 kN/m from wall above and an eccentric load of 20 kN/m from RCC floor acting at a distance of47.5 mm from the centre line of the wall. The resultant eccentricity will bea)19b)20c)21d)22Correct answer is option 'A'. Can you explain this answer?
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