- 10%
- 25%
- 33 1/3 %
- 50%
Correct Answer: B. 25%
Solution:
When the eccentricity of a load on a masonry wall is small (less than 1/24 of the wall thickness), bending stresses are considered negligible. However, when the eccentricity ratio exceeds this, bending stresses must be accounted for. To compensate for these combined stresses, design codes allow for an increase in the permissible compressive stress, typically by a factor of 25%.
- 0.090
- 0.095
- 0.100
- 0.105
Correct Answer: B. 0.095
Solution:
First, calculate the total load: P = 30 kN/m + 20 kN/m = 50 kN/m. Next, find the moment caused by the eccentric load: M = 20 kN/m * 0.0475 m = 0.95 kNm/m. The resultant eccentricity (e) is M/P = 0.95 / 50 = 0.019 m or 19 mm. The eccentricity ratio is e/t = 19 mm / 200 mm = 0.095.
- Less than or equal to 1/24
- Less than or equal to 1/6
- More than 1/24
- Less than or equal to 1/12
Correct Answer: A. Less than or equal to 1/24
Solution:
Design codes for masonry (like IS 1905) state that if the ratio of eccentricity to the thickness of the member is less than or equal to 1/24, the resulting bending stresses are minimal and can be ignored. The member can then be designed for axial compression only.
- High shrinkage
- Less strength
- Both (A) and (B)
- None of above
Correct Answer: A. High shrinkage
Solution:
Rich cement mortars have a higher proportion of cement. Since cement is the component that undergoes drying shrinkage as it cures, a higher cement content leads to high shrinkage. This increased shrinkage induces tensile stresses within the mortar, making it more prone to cracking.
- 1.0H and 1.0H
- 1.5H and 1.5H
- 2.0H and 1.5H
- 2.0H and 2.0H
Correct Answer: D. 2.0H and 2.0H
Solution:
For a free-standing element (a cantilever), which is fixed at the base and free at the top, the effective height (or length for buckling calculation) is taken as twice its actual height (H). This applies to both walls and columns in a free-standing condition. Thus, the effective height is 2.0H and 2.0H.
- Less than or equal to 1/24
- Less than or equal to 1/6
- More than 1/24
- Less than or equal to 1/12
Correct Answer: A. Less than or equal to 1/24
Solution:
This question repeats the concept from Q48. Design codes for masonry (like IS 1905) state that if the ratio of eccentricity to the thickness of the member is less than or equal to 1/24, the resulting bending stresses are minimal and can be ignored. The member can then be designed for axial compression only.
- 0.15 MPa
- 0.125 MPa
- 0.1 MPa
- 0.075 MPa
Correct Answer: A. 0.15 MPa
Solution:
The permissible shear stress in masonry depends on the strength of the mortar. For a 1:1:6 cement-lime-sand mix, which corresponds to a medium-strength mortar, the building codes specify a maximum allowable horizontal shear stress of 0.15 MPa (or 1.5 kg/cm²).
- More
- Less
- The same in both the cases
- 100 % more
Correct Answer: B. Less
Solution:
A free-standing wall lacks lateral support at the top, making it more susceptible to buckling under a vertical load. When an RC slab rests on the wall, it provides this lateral restraint, effectively reducing the wall's slenderness and increasing its stability. Therefore, the direct load capacity of the free-standing wall is less.
- Lateral restraint but not rotational restraint
- Rotational restraint but not lateral restraint
- Both lateral and rotational restraints
- Neither lateral nor rotational restraint
Correct Answer: A. Lateral restraint but not rotational restraint
Solution:
When a floor is anchored to a wall, it prevents the wall from moving horizontally (in or out) at that level, thus providing lateral restraint. However, if the floor joists are not built into or continuous over the wall, the connection is typically considered pinned, meaning it does not prevent the top of the wall from rotating. Therefore, it provides not rotational restraint.
- Less when framed into a system of RC columns.
- Less when supported on brick walls.
- The same in both the cases.
- Equal to the wall thickness or size of column.
Correct Answer: A. Less when framed into a system of RC columns.
Solution:
When a floor system is framed into RC columns, it forms a monolithic structure. This creates a rigid frame action, allowing for more efficient load distribution and negative moment development at the supports. This results in smaller required beam and slab depths compared to a system simply supported on brick walls, which acts as a series of independent, simply supported beams requiring larger depths for the same span and load.
- 19 mm, 0.095
- 19 mm, 0.1
- 22 mm, 0.11
- 24 mm, 0.12
Correct Answer: A. 19 mm, 0.095
Solution:
Total Load (P) = 26 kN/m + 19 kN/m = 45 kN/m. Moment (M) = 19 kN/m * 45 mm = 855 kN.mm/m. Resultant Eccentricity (e) = M/P = 855 / 45 = 19 mm. Eccentricity Ratio = e/t = 19 mm / 200 mm = 0.095.
- 1.00
- 1.25
- 1.50
- 1.75
Correct Answer: B. 1.25
Solution:
For the arching action to fully develop above a lintel, there must be sufficient masonry mass to provide the necessary thrust restraint. Building codes and structural principles dictate that the height of the masonry above the apex of the load distribution triangle must be at least 1.25 times the height of that triangle. This ensures the arch can form and carry the load to the supports without failing.
- 1.2
- 1.4
- 1.6
- 2.0
Correct Answer: B. 1.4
Solution:
The shape of a masonry unit affects its compressive strength. The basic stress is determined for a standard unit shape. For units with a height-to-width ratio of 1.5, codes like IS 1905 allow for an increase in the basic compressive stress. This is because such a shape can exhibit improved performance. The specified multiplication factor for a height/width ratio of 1.5 is 1.4.
