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All questions of Reinforced Cement Concrete for Civil Engineering (CE) Exam
For a simply supported beam of span 15m, the minimum effective depth to satisfy the vertical deflection limits should be- a)600 mm
- b)750 mm
- c)900 mm
- d)more than 1 m
Correct answer is option 'B'. Can you explain this answer?
For a simply supported beam of span 15m, the minimum effective depth to satisfy the vertical deflection limits should be
a)
600 mm
b)
750 mm
c)
900 mm
d)
more than 1 m
 | Sanya Agarwal answered |
For s.s slab span/effective depth=20
eff. Depth=span/20
=15000/20
=750mm
If the size of panel in a flat slab is 6m x 6m, then as per Indian Standard Code, the widths of column strip and middle strip are- a)3.0 m and 1.5 m
- b)1.5 m and 3.0 m
- c)3.0 m and 3.0 m
- d)1.5 m and 1.5 m
Correct answer is option 'C'. Can you explain this answer?
If the size of panel in a flat slab is 6m x 6m, then as per Indian Standard Code, the widths of column strip and middle strip are
a)
3.0 m and 1.5 m
b)
1.5 m and 3.0 m
c)
3.0 m and 3.0 m
d)
1.5 m and 1.5 m
 | Poulomi Patel answered |
Explanation:
Flat slab is a reinforced concrete slab supported by columns without any form of beam. The column strip is a strip of the slab that is adjacent to the column and the middle strip is the remaining area of the slab.
As per Indian Standard Code IS 456:2000, the width of the column strip and middle strip is determined by the following formula:
Column strip width = Effective length of panel / 5
Middle strip width = Effective length of panel / 10
For a panel size of 6m x 6m, the effective length of the panel is the same as the span of the slab, which is 6m.
Using the formula above, we can calculate the widths of the column strip and middle strip as follows:
Column strip width = 6m / 5 = 1.2m
Middle strip width = 6m / 10 = 0.6m
However, as per IS 456:2000, the minimum width of the column strip and middle strip should not be less than half the width of the panel. Therefore, the final widths of the column strip and middle strip are:
Column strip width = Maximum of (1.2m, 3m/2) = 3m
Middle strip width = Maximum of (0.6m, 3m/2) = 1.5m
Hence, option C (3.0m and 3.0m) is incorrect and the correct answer is option C (3.0m and 1.5m).
Flat slab is a reinforced concrete slab supported by columns without any form of beam. The column strip is a strip of the slab that is adjacent to the column and the middle strip is the remaining area of the slab.
As per Indian Standard Code IS 456:2000, the width of the column strip and middle strip is determined by the following formula:
Column strip width = Effective length of panel / 5
Middle strip width = Effective length of panel / 10
For a panel size of 6m x 6m, the effective length of the panel is the same as the span of the slab, which is 6m.
Using the formula above, we can calculate the widths of the column strip and middle strip as follows:
Column strip width = 6m / 5 = 1.2m
Middle strip width = 6m / 10 = 0.6m
However, as per IS 456:2000, the minimum width of the column strip and middle strip should not be less than half the width of the panel. Therefore, the final widths of the column strip and middle strip are:
Column strip width = Maximum of (1.2m, 3m/2) = 3m
Middle strip width = Maximum of (0.6m, 3m/2) = 1.5m
Hence, option C (3.0m and 3.0m) is incorrect and the correct answer is option C (3.0m and 1.5m).
Minimum clear cover (in mm) to the main steel bars in slab, beam, column and footing respectively are?a)10, 15, 20, 25b)15, 25, 40, 50c) 20, 25, 30, 40d)20, 35, 40, 75Correct answer is option 'B'. Can you explain this answer?
 | Rutuja Pillai answered |
Option (b) is correct:
In IS : 456 – 1978‚ the clear covers are specified as :
(i) For longitudinal reinforcement bar in a column, not less than 40 mm‚ or less than the diameter of such bar.
(ii) For longitudinal reinforcing bar in a beam, not less than 25mm, or less than the diameter of such bar.
(iii) For tensile, compressive, shear or other reinforcement in a slab, not less than 15 mm, or less than the diameter of such bar.
(iv) in other cases the clear cover to the reinforcement bars shall not be less than 40 min of diameter of such bar whichever is more. The clear cover should not exceed 75 mm in any case. According to IS 456 : 2000 nominal cover shall not be less than the diameter of bar. The nominal cover is specified to meet durability requirement in Table 16 on page 47 of the IS Code. To meet specified period of fire resistance the clear covers are given in Table 16A on page 47. For longitudinal reinforcing bars the nominal cover shall not be less than 40 mm. For footings minimum cover shall be 50 mm.
As per the provisions of IS : 456–2000, in the limit state method for design of beams, the limiting value of the depth of neutral axis in a reinforced concrete beam of effective depth 'd' is given as- a)0.53 d
- b)0.48
- c)0.46 d
- d)any of the above depending on the different grades of steel
Correct answer is option 'D'. Can you explain this answer?
As per the provisions of IS : 456–2000, in the limit state method for design of beams, the limiting value of the depth of neutral axis in a reinforced concrete beam of effective depth 'd' is given as
a)
0.53 d
b)
0.48
c)
0.46 d
d)
any of the above depending on the different grades of steel
 | Zoya Sharma answered |
Limiting value of depth of neutral axis depends on grade of concrete.
The minimum reinforced using mild steel in slab should not be less than.- a)0.10%
- b)0.12%
- c)0.15%
- d)12 cm
Correct answer is option 'C'. Can you explain this answer?
The minimum reinforced using mild steel in slab should not be less than.
a)
0.10%
b)
0.12%
c)
0.15%
d)
12 cm
 | Aditya Deshmukh answered |
As per IS 456, the minimum reinforcement in either direction of slabs shall not be less than 0.15 percent of total cross-sectional area in case of mild steel and 0.12 percent of total cross section in case of high strength deformed bars.
In RCC beams, as the percentage area of tensile steel increase- a)depth of neutral axis increase
- b)depth of neutral axis decrease
- c)depth of neutral axis does not change
- d)lever arm increase
Correct answer is option 'A'. Can you explain this answer?
In RCC beams, as the percentage area of tensile steel increase
a)
depth of neutral axis increase
b)
depth of neutral axis decrease
c)
depth of neutral axis does not change
d)
lever arm increase
 | Shivam Sharma answered |
As the percentage area of steel increases, the depth of neutral axis will increases.
In a balanced section, increase in area of steel will result in higher tensile stress carrying capacity. To counteract or generate equivalent compressive stress/force, neutral axis has to come down.
In the limit state method, balanced design of a reinforced concrete beam gives- a)smallest concrete section and maximum area of reinforcement
- b)largest concrete section and maximum area of reinforcement
- c)smallest concrete section and minimum area of reinforcement
- d)largest concrete section and minimum area of reinforcement
Correct answer is option 'A'. Can you explain this answer?
In the limit state method, balanced design of a reinforced concrete beam gives
a)
smallest concrete section and maximum area of reinforcement
b)
largest concrete section and maximum area of reinforcement
c)
smallest concrete section and minimum area of reinforcement
d)
largest concrete section and minimum area of reinforcement
 | Lavanya Menon answered |
Balanced Design means both concrete and steel will fail at the same time. Thus, it gives optimum smallest concrete section for a given reinforcement and maximum reinforcement for a given concrete section.
A reinforced concrete beam of 10 mEffective span and 1 m effective depth is supported on 500 mm x 500 mm columns. If the total uniformly distributed load on the beam is 10 MN/m, the design shear force for the beam is- a)50 MN
- b)47.5 MN
- c)37.5 MN
- d)43 MN
Correct answer is option 'C'. Can you explain this answer?
A reinforced concrete beam of 10 mEffective span and 1 m effective depth is supported on 500 mm x 500 mm columns. If the total uniformly distributed load on the beam is 10 MN/m, the design shear force for the beam is
a)
50 MN
b)
47.5 MN
c)
37.5 MN
d)
43 MN
| | Arka Choudhary answered |
The shear force should be calculated at critical section i.e. 1 m away from the face of the column. Effective span of beam is to be taken as the clear span plus effective depth of beam or centre to centre spacing of supports which ever is less. The location for shear calculation will be,
In case of simply supported beam , the clear distance between lateral restraints shall not exceed- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In case of simply supported beam , the clear distance between lateral restraints shall not exceed
a)
b)
c)
d)
| | Lavanya Menon answered |
As per IS 456:2000 for lateral stability, the clear span of
Simply supported beam should be less than 60b or 250(b2)/d
Cantilever Beam should be less than 25b or 100(b2)/d
Simply supported beam should be less than 60b or 250(b2)/d
Cantilever Beam should be less than 25b or 100(b2)/d
Half of the main steel in a simply supported slab is bent up near the support at a distance of x from the centre of slab bearing where x is equal to- a)1/3
- b)1/5
- c)1/7
- d)1/10
Correct answer is option 'C'. Can you explain this answer?
Half of the main steel in a simply supported slab is bent up near the support at a distance of x from the centre of slab bearing where x is equal to
a)
1/3
b)
1/5
c)
1/7
d)
1/10
 | Anand Kumar answered |
Given:
Half of the main steel is bent up near the support at a distance of x from the centre of slab bearing.
To find:
Value of x.
Solution:
Let's assume the total width of the slab be 2L, and the distance of x from the centre of the slab be 'a'.
- The reinforcement steel near the support is bent up to resist the negative bending moment.
- At the centre of the slab, there is no bending moment, so there is no need to provide any reinforcement.
- As we move towards the support, the bending moment increases, so we need to provide more reinforcement.
- Therefore, the reinforcement steel is bent up at a distance of a from the centre of the slab bearing.
Now, we can use the formula for the moment of inertia of a rectangular section to find the value of x.
Moment of inertia of a rectangular section:
Ixx = (b * h^3) / 12
where,
b = width of the slab
h = depth of the slab
- The maximum bending moment occurs at the support, which is equal to wL^2/8.
- Since half of the main steel is bent up, the remaining half resists the bending moment.
- Therefore, the effective depth of the slab is h/2.
- We can assume that the reinforcement steel is placed at a distance of a from the centre of the slab bearing, and the remaining distance is (L-a).
- The moment of inertia of the slab can be calculated as follows:
Ixx = (b * h^3) / 12 + (0.5 * Ast * (a - h/2)^2) + (0.5 * Ast * (L - a - h/2)^2)
where,
Ast = area of steel reinforcement
- Equating the bending moment to the moment of resistance, we get:
wL^2/8 = (f_y / y) * Ast * (d - a/2)
where,
f_y = yield strength of reinforcement steel
y = distance from the extreme compression fibre to the centroid of the reinforcement steel
d = effective depth of the slab
- Substituting for 'd' and 'y', we get:
wL^2/8 = (f_y / (h/2 + a/2)) * Ast * (h/2 - a/2)
- Simplifying the equation, we get:
Ast = (wL^2 * (h/2 + a/2)) / (8f_y * (h/2 - a/2))
- Substituting the value of Ast in the equation for Ixx, we get:
Ixx = (b * h^3) / 12 + (0.5 * (wL^2 * (h/2 + a/2)) / (8f_y * (h/2 - a/2))) * (a - h/2)^2 + (0.5 * (wL^2 * (h/2 + a/2)) / (8f_y * (h/2 - a/2))) * (L - a - h/2)^2
- Simplifying the equation, we get:
Ixx = (b * h^3) / 12 + (wL^4 * (h/2 + a/2)) / (128f_y * (h/
Half of the main steel is bent up near the support at a distance of x from the centre of slab bearing.
To find:
Value of x.
Solution:
Let's assume the total width of the slab be 2L, and the distance of x from the centre of the slab be 'a'.
- The reinforcement steel near the support is bent up to resist the negative bending moment.
- At the centre of the slab, there is no bending moment, so there is no need to provide any reinforcement.
- As we move towards the support, the bending moment increases, so we need to provide more reinforcement.
- Therefore, the reinforcement steel is bent up at a distance of a from the centre of the slab bearing.
Now, we can use the formula for the moment of inertia of a rectangular section to find the value of x.
Moment of inertia of a rectangular section:
Ixx = (b * h^3) / 12
where,
b = width of the slab
h = depth of the slab
- The maximum bending moment occurs at the support, which is equal to wL^2/8.
- Since half of the main steel is bent up, the remaining half resists the bending moment.
- Therefore, the effective depth of the slab is h/2.
- We can assume that the reinforcement steel is placed at a distance of a from the centre of the slab bearing, and the remaining distance is (L-a).
- The moment of inertia of the slab can be calculated as follows:
Ixx = (b * h^3) / 12 + (0.5 * Ast * (a - h/2)^2) + (0.5 * Ast * (L - a - h/2)^2)
where,
Ast = area of steel reinforcement
- Equating the bending moment to the moment of resistance, we get:
wL^2/8 = (f_y / y) * Ast * (d - a/2)
where,
f_y = yield strength of reinforcement steel
y = distance from the extreme compression fibre to the centroid of the reinforcement steel
d = effective depth of the slab
- Substituting for 'd' and 'y', we get:
wL^2/8 = (f_y / (h/2 + a/2)) * Ast * (h/2 - a/2)
- Simplifying the equation, we get:
Ast = (wL^2 * (h/2 + a/2)) / (8f_y * (h/2 - a/2))
- Substituting the value of Ast in the equation for Ixx, we get:
Ixx = (b * h^3) / 12 + (0.5 * (wL^2 * (h/2 + a/2)) / (8f_y * (h/2 - a/2))) * (a - h/2)^2 + (0.5 * (wL^2 * (h/2 + a/2)) / (8f_y * (h/2 - a/2))) * (L - a - h/2)^2
- Simplifying the equation, we get:
Ixx = (b * h^3) / 12 + (wL^4 * (h/2 + a/2)) / (128f_y * (h/
Flexural collapse in over-reinforced beam is due toa)Compression failureb)Tension failurec)Yielding of steeld)None of theseCorrect answer is option 'A'. Can you explain this answer?
 | Neha Choudhury answered |
RCC beams have two stages of faliure (the beam is considered to be failed at first stage itself).
The primary faliure (occurs first) in an over reinforced beam is compression faliure, as the crushing of concrete occurs before yielding of thae tension reinforcement.In an under reinforced beam the secondary faliure is compression faliure as crushing of concrete occurs after yielding of tension steel.The collapse of an reinforced beam occurs at the secondary stage. For an over reinforced beam second faliure is tension faliure (yielding of tension reinforcement when the stress in it reaches yield stress on further loading after the primary faliure,crushing of concrete).
The main reason for providing number of reinforcing bars at a support in a simply supported beam is to resist in that zone- a)compressive stress
- b)shear stress
- c)bond stress
- d)tensile stress
Correct answer is option 'C'. Can you explain this answer?
The main reason for providing number of reinforcing bars at a support in a simply supported beam is to resist in that zone
a)
compressive stress
b)
shear stress
c)
bond stress
d)
tensile stress
| | Simran Mukherjee answered |
Reinforcing bars are provided at the support to develop adequate bond between concrete and steel, so that adequate bond stress is developed between them within development length.
In the reinforced concrete slab, the spacing between main reinforcement should nto exceed- a)Three times its effective depth
- b)Four times its effective depth
- c)Five times its effective depth
- d)Six times its effective depth
Correct answer is option 'A'. Can you explain this answer?
In the reinforced concrete slab, the spacing between main reinforcement should nto exceed
a)
Three times its effective depth
b)
Four times its effective depth
c)
Five times its effective depth
d)
Six times its effective depth
 | Swara Dasgupta answered |
The maximum diameter of bar used in slab should not exceed 1/8 of the total thickness of slab. Maximum spacing of main bar is restricted to 3 times effective depth or 300 mm whichever is less. For distribution bars the maximum spacing is specified as 5 times the effective depth or 450 mm whichever is less.
Permissible shear stress in concrete is a function of- a)Grade of concrete
- b)Grade of steel
- c)Percentage of steel reinforcement
- d)Percentage of steel reinforcement grade of concrete
Correct answer is option 'D'. Can you explain this answer?
Permissible shear stress in concrete is a function of
a)
Grade of concrete
b)
Grade of steel
c)
Percentage of steel reinforcement
d)
Percentage of steel reinforcement grade of concrete
 | Mrinalini Sen answered |
Whenever the value of actual shear stress exceeds the permissible shear stress of the concrete used, the shear reinforcement must be provided. The purpose of shear reinforcement is to prevent failure in shear, and to increase beam ductility and subsequently the likelihood of sudden failure will be reduced.
For footing minimum cover should be- a)25 mm
- b)35 mm
- c)50 mm
- d)60 mm
Correct answer is option 'C'. Can you explain this answer?
For footing minimum cover should be
a)
25 mm
b)
35 mm
c)
50 mm
d)
60 mm
| | Rithika Kaur answered |
Understanding Minimum Cover in Footings
The minimum cover for reinforcement in footings is a critical consideration in civil engineering. It ensures the durability and structural integrity of the concrete structure.
Why is Minimum Cover Important?
- Protection Against Corrosion: Adequate cover protects the reinforcing steel from environmental effects, such as moisture and chemicals, which can lead to corrosion.
- Fire Resistance: Sufficient cover enhances the fire resistance of the structure, as it provides a barrier to heat, preventing the steel from reaching critical temperatures.
- Structural Integrity: Proper cover helps in distributing loads effectively and maintaining the overall stability of the footing.
Recommended Minimum Cover for Footings
- The commonly accepted minimum cover for footings is 50 mm. This is essential for various reasons:
- Standards Compliance: Building codes and standards, such as those from the American Concrete Institute (ACI) and other regulatory bodies, often specify this minimum to ensure safety and performance.
- Concrete Protection: A cover of 50 mm ensures that the concrete adequately encases the reinforcement, providing sufficient shielding from external elements.
Conclusion
In conclusion, the correct answer to the question regarding the minimum cover for footings is option 'C' (50 mm). Adhering to this requirement is essential for ensuring the longevity and safety of concrete structures.
The minimum cover for reinforcement in footings is a critical consideration in civil engineering. It ensures the durability and structural integrity of the concrete structure.
Why is Minimum Cover Important?
- Protection Against Corrosion: Adequate cover protects the reinforcing steel from environmental effects, such as moisture and chemicals, which can lead to corrosion.
- Fire Resistance: Sufficient cover enhances the fire resistance of the structure, as it provides a barrier to heat, preventing the steel from reaching critical temperatures.
- Structural Integrity: Proper cover helps in distributing loads effectively and maintaining the overall stability of the footing.
Recommended Minimum Cover for Footings
- The commonly accepted minimum cover for footings is 50 mm. This is essential for various reasons:
- Standards Compliance: Building codes and standards, such as those from the American Concrete Institute (ACI) and other regulatory bodies, often specify this minimum to ensure safety and performance.
- Concrete Protection: A cover of 50 mm ensures that the concrete adequately encases the reinforcement, providing sufficient shielding from external elements.
Conclusion
In conclusion, the correct answer to the question regarding the minimum cover for footings is option 'C' (50 mm). Adhering to this requirement is essential for ensuring the longevity and safety of concrete structures.
When shear reinforcement is not provided in flat slab, the calculated shear stress at the critical section shall not exceed Ks ζc, where Ks is Where βc is ratio of short side to long side of column capital.- a)(1 + βc)
- b)(0 .6 + βc)
- c)(0.5 + βc)
- d)(1.5 + βc)
Correct answer is option 'C'. Can you explain this answer?
When shear reinforcement is not provided in flat slab, the calculated shear stress at the critical section shall not exceed Ks ζc, where Ks is
Where βc is ratio of short side to long side of column capital.
a)
(1 + βc)
b)
(0 .6 + βc)
c)
(0.5 + βc)
d)
(1.5 + βc)
| | Anjana Mukherjee answered |
The ratio of the maximum and minimum percentage.of longitudinal reinforcement that can be provided in a column is- a)6
- b)7.5
- c)5
- d)8
Correct answer is option 'B'. Can you explain this answer?
The ratio of the maximum and minimum percentage.of longitudinal reinforcement that can be provided in a column is
a)
6
b)
7.5
c)
5
d)
8
| | Asha Menon answered |
Maximum reinforcement = 6%
Minimum reinforcement = 0.8%
Minimum reinforcement = 0.8%
A strap footing is a special type of- a)Strip footing
- b)Combined footing
- c)Raft footing
- d)Spread footing
Correct answer is option 'B'. Can you explain this answer?
A strap footing is a special type of
a)
Strip footing
b)
Combined footing
c)
Raft footing
d)
Spread footing
| | Mehul Choudhury answered |
A strap footing consists of a spread footings of two columns connected by a strap beam. This type of footing is useful when the external column is very near to the property line so that its footing cannot be spread beyond the property line.
If the nominal shear stress (ζv) at a section does not exceed the permissible shear stress (τc)- a)minimum shear reinforcement is still provided
- b)shear reinforcement is provided to resist the nominal shear stress
- c)no shear reinforcement is provided
- d)shear reinforcement is provided for the difference of the two
Correct answer is option 'A'. Can you explain this answer?
If the nominal shear stress (ζv) at a section does not exceed the permissible shear stress (τc)
a)
minimum shear reinforcement is still provided
b)
shear reinforcement is provided to resist the nominal shear stress
c)
no shear reinforcement is provided
d)
shear reinforcement is provided for the difference of the two
| | Anshul Chakraborty answered |
Τ) on a material is given in pounds per square inch (psi) and the material's thickness (t) is given in inches, then the formula to calculate the shear force (F) required to shear the material is:
F = τ * A
where A is the area of the material being sheared, given by:
A = t * w
where w is the width of the material. Therefore, the general formula for calculating the shear force required to shear a rectangular material is:
F = τ * t * w
For example, if a material has a nominal shear stress of 10,000 psi and a thickness of 0.5 inches, and the width of the material is 2 inches, then the shear force required to shear the material would be:
F = 10,000 psi * 0.5 in * 2 in
F = 5,000 pounds
Therefore, a force of 5,000 pounds would be required to shear this material under these conditions.
F = τ * A
where A is the area of the material being sheared, given by:
A = t * w
where w is the width of the material. Therefore, the general formula for calculating the shear force required to shear a rectangular material is:
F = τ * t * w
For example, if a material has a nominal shear stress of 10,000 psi and a thickness of 0.5 inches, and the width of the material is 2 inches, then the shear force required to shear the material would be:
F = 10,000 psi * 0.5 in * 2 in
F = 5,000 pounds
Therefore, a force of 5,000 pounds would be required to shear this material under these conditions.
Consider the following statements:
1. Development length includes anchorage value of hooks in tension reinforcement.
2. The development length of each bar of bundled bars shall be that for the individual bar, increased by, 33% for three bars in contact.
3. Deformed bars may be used without end anchorage provided development length requirement is satisfied.
Which of these statements is/are correct?- a)Only 2
- b)Both 1 and 3
- c)Both 1 and 2
- d)1,2 and 3
Correct answer is option 'B'. Can you explain this answer?
Consider the following statements:
1. Development length includes anchorage value of hooks in tension reinforcement.
2. The development length of each bar of bundled bars shall be that for the individual bar, increased by, 33% for three bars in contact.
3. Deformed bars may be used without end anchorage provided development length requirement is satisfied.
Which of these statements is/are correct?
1. Development length includes anchorage value of hooks in tension reinforcement.
2. The development length of each bar of bundled bars shall be that for the individual bar, increased by, 33% for three bars in contact.
3. Deformed bars may be used without end anchorage provided development length requirement is satisfied.
Which of these statements is/are correct?
a)
Only 2
b)
Both 1 and 3
c)
Both 1 and 2
d)
1,2 and 3
| | Rajat Sen answered |
In case of bundled bars in contact the development length is increased than that for individual bar by,
1. 10 % for two bars in contact
2. 20% for three bars in contact
3. 33% for four bars in content
1. 10 % for two bars in contact
2. 20% for three bars in contact
3. 33% for four bars in content
The load carrying capacity of helically reinforced column as compared to that, of a tied column is about- a)5% less
- b)10% less
- c)5% more
- d)10% more
Correct answer is option 'C'. Can you explain this answer?
The load carrying capacity of helically reinforced column as compared to that, of a tied column is about
a)
5% less
b)
10% less
c)
5% more
d)
10% more
| | Lekshmi Rane answered |
Load carrying capacity of helically reinforced column as compared to a tied column:
Introduction:
The load carrying capacity of a column is directly related to its strength and stiffness. Columns can be reinforced in various ways to increase their strength and stiffness. Helical reinforcement is one such method that is used to increase the load carrying capacity of columns.
Helical reinforcement in columns:
Helical reinforcement involves wrapping a helical reinforcement around the column. This reinforcement is made up of a continuous steel bar that is bent into a helix shape. The helix is then wrapped around the column at a certain pitch. The pitch of the helix determines the spacing of the reinforcement.
Tied columns:
A tied column is a type of column that is reinforced with vertical ties. These ties are made up of steel bars that are placed at regular intervals along the length of the column. The ties are tied together at the top and bottom of the column to form a cage-like structure.
Load carrying capacity:
The load carrying capacity of a column depends on its strength and stiffness. Helical reinforcement increases the strength and stiffness of a column by providing a continuous reinforcement around the column. This reinforcement helps to distribute the load evenly across the column and prevent it from buckling.
The load carrying capacity of a helically reinforced column is about 5% more than that of a tied column. This is because helical reinforcement provides a continuous reinforcement around the column, which increases its strength and stiffness. Tied columns, on the other hand, have discrete reinforcement in the form of ties, which may not distribute the load evenly across the column.
Conclusion:
Helical reinforcement is an effective method of increasing the load carrying capacity of columns. It provides a continuous reinforcement around the column, which increases its strength and stiffness. The load carrying capacity of a helically reinforced column is about 5% more than that of a tied column.
Introduction:
The load carrying capacity of a column is directly related to its strength and stiffness. Columns can be reinforced in various ways to increase their strength and stiffness. Helical reinforcement is one such method that is used to increase the load carrying capacity of columns.
Helical reinforcement in columns:
Helical reinforcement involves wrapping a helical reinforcement around the column. This reinforcement is made up of a continuous steel bar that is bent into a helix shape. The helix is then wrapped around the column at a certain pitch. The pitch of the helix determines the spacing of the reinforcement.
Tied columns:
A tied column is a type of column that is reinforced with vertical ties. These ties are made up of steel bars that are placed at regular intervals along the length of the column. The ties are tied together at the top and bottom of the column to form a cage-like structure.
Load carrying capacity:
The load carrying capacity of a column depends on its strength and stiffness. Helical reinforcement increases the strength and stiffness of a column by providing a continuous reinforcement around the column. This reinforcement helps to distribute the load evenly across the column and prevent it from buckling.
The load carrying capacity of a helically reinforced column is about 5% more than that of a tied column. This is because helical reinforcement provides a continuous reinforcement around the column, which increases its strength and stiffness. Tied columns, on the other hand, have discrete reinforcement in the form of ties, which may not distribute the load evenly across the column.
Conclusion:
Helical reinforcement is an effective method of increasing the load carrying capacity of columns. It provides a continuous reinforcement around the column, which increases its strength and stiffness. The load carrying capacity of a helically reinforced column is about 5% more than that of a tied column.
What should be the minimum grade of reinforced concrete in and around sea coast construction?- a)M 35
- b)M 30
- c)M 25
- d)M 20
Correct answer is option 'B'. Can you explain this answer?
What should be the minimum grade of reinforced concrete in and around sea coast construction?
a)
M 35
b)
M 30
c)
M 25
d)
M 20
| | Madhurima Banerjee answered |
Minimum Grade of Reinforced Concrete for Sea Coast Construction
Reinforced concrete is commonly used in sea coast construction due to its durability, strength, and resistance to corrosion. However, the concrete used in such construction should be of a high grade to withstand the harsh marine environment. The minimum grade of reinforced concrete for sea coast construction is M30.
Explanation:
M30 grade concrete has a compressive strength of 30 N/mm2 after 28 days of curing. This grade is suitable for moderate exposure conditions, including marine environments with low to moderate chloride content. The use of M30 grade concrete ensures that the structure can withstand the effects of sea water, such as corrosion, erosion, and salt attack.
Factors Affecting the Grade of Reinforced Concrete:
The choice of the correct grade of reinforced concrete for sea coast construction depends on several factors such as:
1. Exposure Conditions: The level of exposure to marine environment determines the minimum grade of reinforced concrete required.
2. Design Requirements: The structural requirements and design loadings also influence the grade of concrete used.
3. Durability Requirements: The durability requirements of the structure influence the choice of concrete grade.
4. Construction Method: The construction method, including formwork, curing, and placement, can affect the concrete strength.
Conclusion:
In conclusion, the minimum grade of reinforced concrete for sea coast construction is M30, which provides sufficient strength and durability to withstand the harsh marine environment. However, the grade of concrete used should be selected based on the specific requirements of the project, including exposure conditions, design requirements, durability requirements, and construction method.
Reinforced concrete is commonly used in sea coast construction due to its durability, strength, and resistance to corrosion. However, the concrete used in such construction should be of a high grade to withstand the harsh marine environment. The minimum grade of reinforced concrete for sea coast construction is M30.
Explanation:
M30 grade concrete has a compressive strength of 30 N/mm2 after 28 days of curing. This grade is suitable for moderate exposure conditions, including marine environments with low to moderate chloride content. The use of M30 grade concrete ensures that the structure can withstand the effects of sea water, such as corrosion, erosion, and salt attack.
Factors Affecting the Grade of Reinforced Concrete:
The choice of the correct grade of reinforced concrete for sea coast construction depends on several factors such as:
1. Exposure Conditions: The level of exposure to marine environment determines the minimum grade of reinforced concrete required.
2. Design Requirements: The structural requirements and design loadings also influence the grade of concrete used.
3. Durability Requirements: The durability requirements of the structure influence the choice of concrete grade.
4. Construction Method: The construction method, including formwork, curing, and placement, can affect the concrete strength.
Conclusion:
In conclusion, the minimum grade of reinforced concrete for sea coast construction is M30, which provides sufficient strength and durability to withstand the harsh marine environment. However, the grade of concrete used should be selected based on the specific requirements of the project, including exposure conditions, design requirements, durability requirements, and construction method.
What is the minimum nominal percentage longitudinal reinforcement to be provided in a concrete pedestal as per relevant IS code?- a)0.4
- b)0.2
- c)0.15
- d)0.1
Correct answer is option 'C'. Can you explain this answer?
What is the minimum nominal percentage longitudinal reinforcement to be provided in a concrete pedestal as per relevant IS code?
a)
0.4
b)
0.2
c)
0.15
d)
0.1
| | Avinash Mehta answered |
The minimum nominal percentage longitudinal reinforcement to be provided in a concrete pedestal as per relevant IS code (IS 456-2000) is 0.15%. This means that at least 0.15% of the cross-sectional area of the pedestal should be made up of longitudinal reinforcement, such as steel bars or welded wire fabric. This minimum requirement is in place to ensure that the pedestal has adequate strength and ductility to resist the forces to which it will be subjected.
It is important to mention that this answer should be considered as general and it might change depending on the specific code or standard of the country and the type of structure that the pedestal is going to be part of, it is always important to check the specific code that applies to the project.
The maximum percentage of moment redistribution allowed in RCC beams is- a)10%
- b)20%
- c)30%
- d)40%
Correct answer is option 'C'. Can you explain this answer?
The maximum percentage of moment redistribution allowed in RCC beams is
a)
10%
b)
20%
c)
30%
d)
40%
| | Gowri Singh answered |
As per clause 37.1.1 of IS: 456-2000 the ultimate moment of resistance at any section should not be less than 70% of the factored moment at that section as obtained from the elastic moment envelope (considering all loading combinations).
The span to depth ratio limit is specified in IS : 456-1978 for the reinforced concrete beams, in order to ensure that the- a)tensile crack width is below a limit
- b)shear failure is avoided
- c)stress in the tension reinforcement is less than the allowable value
- d)deflection of the beam is below a limiting value
Correct answer is option 'D'. Can you explain this answer?
The span to depth ratio limit is specified in IS : 456-1978 for the reinforced concrete beams, in order to ensure that the
a)
tensile crack width is below a limit
b)
shear failure is avoided
c)
stress in the tension reinforcement is less than the allowable value
d)
deflection of the beam is below a limiting value
| | Nikhil Majumdar answered |
Span to Depth Ratio Limit and Its Significance in Reinforced Concrete Beams
The span to depth ratio limit is an important aspect of the design of reinforced concrete beams. It is specified in IS : 456-1978, which is the Indian Standard Code of Practice for Reinforced Concrete Structures. The limit is designed to ensure that the deflection of the beam is below a limiting value.
What is Span to Depth Ratio?
The span to depth ratio is the ratio of the clear span of the beam to its depth. In other words, it is the distance between the supports divided by the depth of the beam.
Why is Span to Depth Ratio Important?
The span to depth ratio is an important factor in determining the strength and deflection of the beam. A beam with a high span to depth ratio will have a greater deflection and will be weaker than a beam with a low span to depth ratio.
What is the Limit for Span to Depth Ratio?
The limit for the span to depth ratio is specified in IS : 456-1978. According to the code, the span to depth ratio should not exceed the following values:
- Simply supported beams: 20
- Continuous beams: 26
Why is Deflection Important?
Deflection is the amount of deformation that a beam experiences when it is subjected to a load. Excessive deflection can lead to cracking and failure of the beam. Therefore, it is important to limit the deflection of the beam to a safe value. The span to depth ratio limit ensures that the deflection of the beam is below a limiting value.
Conclusion
The span to depth ratio limit is an important aspect of the design of reinforced concrete beams. It ensures that the deflection of the beam is below a limiting value, which is important for the safety and performance of the structure. Designers and engineers must follow the specified limit to ensure the quality and durability of the reinforced concrete structure.
The span to depth ratio limit is an important aspect of the design of reinforced concrete beams. It is specified in IS : 456-1978, which is the Indian Standard Code of Practice for Reinforced Concrete Structures. The limit is designed to ensure that the deflection of the beam is below a limiting value.
What is Span to Depth Ratio?
The span to depth ratio is the ratio of the clear span of the beam to its depth. In other words, it is the distance between the supports divided by the depth of the beam.
Why is Span to Depth Ratio Important?
The span to depth ratio is an important factor in determining the strength and deflection of the beam. A beam with a high span to depth ratio will have a greater deflection and will be weaker than a beam with a low span to depth ratio.
What is the Limit for Span to Depth Ratio?
The limit for the span to depth ratio is specified in IS : 456-1978. According to the code, the span to depth ratio should not exceed the following values:
- Simply supported beams: 20
- Continuous beams: 26
Why is Deflection Important?
Deflection is the amount of deformation that a beam experiences when it is subjected to a load. Excessive deflection can lead to cracking and failure of the beam. Therefore, it is important to limit the deflection of the beam to a safe value. The span to depth ratio limit ensures that the deflection of the beam is below a limiting value.
Conclusion
The span to depth ratio limit is an important aspect of the design of reinforced concrete beams. It ensures that the deflection of the beam is below a limiting value, which is important for the safety and performance of the structure. Designers and engineers must follow the specified limit to ensure the quality and durability of the reinforced concrete structure.
At the time of initial tensioning, the maximum tensile stress immediately behind the anchorage should not exceed which one of the following?- a)0.50 x ultimate tensile stress
- b)0.60 x ultimate tensile stress
- c)0.70 x ultimate tensile stress
- d)0.80 x ultimate tensile stress
Correct answer is option 'D'. Can you explain this answer?
At the time of initial tensioning, the maximum tensile stress immediately behind the anchorage should not exceed which one of the following?
a)
0.50 x ultimate tensile stress
b)
0.60 x ultimate tensile stress
c)
0.70 x ultimate tensile stress
d)
0.80 x ultimate tensile stress
| | Megha Choudhury answered |
As per clause 18.5.1 of IS: 1343-1980, at the time of initial tensioning, the maximum tensile stress immediately behind the anchorages shall not exceed 80% of the ultimate tensile strength of the wire or bar or strand.
A reinforced concrete member is subjected to combined action of combined action of compressive axial force and bending moment. If ec is the best least compressive strain in the member, fy, the yield stress of steel and, Es, the modulus of elasticity of steel, the maximum permissible compressive strain in concrete member will be- a)0.002
- b)0.002 fy(1.15 Es)
- c)0.0035 – 0.75 ec
- d)0.0035
Correct answer is option 'C'. Can you explain this answer?
A reinforced concrete member is subjected to combined action of combined action of compressive axial force and bending moment. If ec is the best least compressive strain in the member, fy, the yield stress of steel and, Es, the modulus of elasticity of steel, the maximum permissible compressive strain in concrete member will be
a)
0.002
b)
0.002 fy(1.15 Es)
c)
0.0035 – 0.75 ec
d)
0.0035
| | Milan Ghosh answered |
Ans.
Option (c)
The bond strength between steel and concrete is due to- a)friction
- b)adhesion
- c)both friction and adhesion
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The bond strength between steel and concrete is due to
a)
friction
b)
adhesion
c)
both friction and adhesion
d)
None of these
 | Rajeev Sharma answered |
the correct answer is option "c) both friction and adhesion."
The bond strength between steel and concrete is the force that holds the two materials together. This bond strength is due to both friction and adhesion between the steel and concrete.
Friction is the force that opposes the relative motion of two surfaces in contact. In the case of steel and concrete, the friction between the two materials helps to hold them together.
Adhesion is the attraction between the atoms or molecules of two different materials. In the case of steel and concrete, the adhesion between the two materials helps to hold them together.
Together, both friction and adhesion contribute to the overall bond strength between steel and concrete. It is important to consider the bond strength between these two materials when designing structures that use both steel and concrete, as it can significantly affect the structural performance of the structure.
If the depth of neutral axis for a singly reinforced balanced rectangular section is represented by kd in working stress design, then the value of k for a balanced section, - a)depends on σst only
- b)depends on σcbc only
- c)depends on both σst and σcbc
- d)is independent of both σst and σcbc
Correct answer is option 'A'. Can you explain this answer?
If the depth of neutral axis for a singly reinforced balanced rectangular section is represented by kd in working stress design, then the value of k for a balanced section,
a)
depends on σst only
b)
depends on σcbc only
c)
depends on both σst and σcbc
d)
is independent of both σst and σcbc
| | Nikhil Majumdar answered |
The value of k is given by,
For a balanced section,
For a balanced section,
Consider the following statements:
1. In a helical reinforced concrete column, the concrete core is subjected to tri-axial state of stress.
2. Helically reinforced columns are very much suitable for earthquake resistant structures.
Which of these statements is/are correct? ,- a)1 only
- b)2 only
- c)Both 1 and 2
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Consider the following statements:
1. In a helical reinforced concrete column, the concrete core is subjected to tri-axial state of stress.
2. Helically reinforced columns are very much suitable for earthquake resistant structures.
Which of these statements is/are correct? ,
1. In a helical reinforced concrete column, the concrete core is subjected to tri-axial state of stress.
2. Helically reinforced columns are very much suitable for earthquake resistant structures.
Which of these statements is/are correct? ,
a)
1 only
b)
2 only
c)
Both 1 and 2
d)
None of these
| | Harshad Iyer answered |
Confinement of concrete by providing transverse reinforcement in the form of steel hoops and spirals increases the strength and ductility of concrete. This reduces the tendency for internal cracking and volume increase just prior to failure.
A beam of rectangular cross-section (b x d) is subjected to a torque T. What is the maximum torsional stress induced in the beam (b < d and α is a constant)?- a)T/αb2d
- b)T/αbd2
- c)T/αbd
- d)T/bd
Correct answer is option 'A'. Can you explain this answer?
A beam of rectangular cross-section (b x d) is subjected to a torque T. What is the maximum torsional stress induced in the beam (b < d and α is a constant)?
a)
T/αb2d
b)
T/αbd2
c)
T/αbd
d)
T/bd
| | Shraddha Datta answered |
Torsion constant of a rectangular section of width b and depth d (b < d) may be expressed as,
J = b3d
J = b3d
For T, L and / sections torsion constant,
where bi and di are the dimensions of each of the component rectangles into which the section may be divided.
Torsional shear stress for rectangular section
Torsional shear stress for rectangular section
For T, L and / sections torsional shear stress may be calculated for each component rectangle by considering them subjected to torsional moment,
The critical section of finding maximum bending moment for footing under masonry wall is located- a)at the middle of the wall
- b)at the edge of the wail
- c)halfway between the middle and edge of the wall
- d)at a distance equal to effective depth of footing from the edge of the wall
Correct answer is option 'C'. Can you explain this answer?
The critical section of finding maximum bending moment for footing under masonry wall is located
a)
at the middle of the wall
b)
at the edge of the wail
c)
halfway between the middle and edge of the wall
d)
at a distance equal to effective depth of footing from the edge of the wall
| | Avinash Mehta answered |
The critical section of finding maximum bending moment for footing under masonry wall is located halfway between the middle and edge of the wall. The maximum bending moment occurs where the load is transferred to the footing and the load is largest, and this typically occurs at the location of maximum leverage, which is halfway between the middle and the edge of the wall.
This is due to the fact that the load is transferred to the footing through the wall and it tends to be more concentrated at the corners, which creates more leverage than in the middle of the wall. By considering this section the designer can ensure that the footing and the foundation is properly designed to resist the expected forces and moments.
The Load carrying capacity of a column designed by working stress method is 500 kN. The ultimate collapse load of the column is- a)500 kN
- b)662.5 kN
- c)750 kN
- d)1100 kN
Correct answer is option 'C'. Can you explain this answer?
The Load carrying capacity of a column designed by working stress method is 500 kN. The ultimate collapse load of the column is
a)
500 kN
b)
662.5 kN
c)
750 kN
d)
1100 kN
| | Rahul Chauhan answered |
The ultimate collapse load on the column is given by,
Wu = Ws x 1.5 = 500 x 1.5 = 750 kN
Wu = Ws x 1.5 = 500 x 1.5 = 750 kN
A prestressed concrete beam has a cross- section with the following properties:
A = 46,400 mm2, I = 75.8 x 107 mm4,
ybottom = 244 mm, ytop = 156 mm.
It is subjected to a prestressing force at an eccentricity ‘e’ so as to have a zero stress at the top fibre. The value of ‘e’ is given by- a)66.66 mm
- b)66.95 mm
- c)104.72 mm
- d)133.33 mm
Correct answer is option 'C'. Can you explain this answer?
A prestressed concrete beam has a cross- section with the following properties:
A = 46,400 mm2, I = 75.8 x 107 mm4,
ybottom = 244 mm, ytop = 156 mm.
It is subjected to a prestressing force at an eccentricity ‘e’ so as to have a zero stress at the top fibre. The value of ‘e’ is given by
A = 46,400 mm2, I = 75.8 x 107 mm4,
ybottom = 244 mm, ytop = 156 mm.
It is subjected to a prestressing force at an eccentricity ‘e’ so as to have a zero stress at the top fibre. The value of ‘e’ is given by
a)
66.66 mm
b)
66.95 mm
c)
104.72 mm
d)
133.33 mm
| | Sankar Rane answered |
For zero stress at top fibre,
P/A = Pe/Z
⇒
⇒
P/A = Pe/Z
⇒
⇒
Which one of the following statements about the percentage of tensile steel required to produce a balanced reinforced concrete section is correct- a)Reduces as the yield strength of steel increase
- b)Remains unchanged irrespective of the yield strength of steel
- c)Is the same for a given quality of steel irrespective of whether working stress method is followed or ultimates load method
- d)Is only a function of the modulus of elasticity of steel
Correct answer is option 'C'. Can you explain this answer?
Which one of the following statements about the percentage of tensile steel required to produce a balanced reinforced concrete section is correct
a)
Reduces as the yield strength of steel increase
b)
Remains unchanged irrespective of the yield strength of steel
c)
Is the same for a given quality of steel irrespective of whether working stress method is followed or ultimates load method
d)
Is only a function of the modulus of elasticity of steel
| | Anmol Choudhary answered |
Percentage of Tensile Steel in a Balanced Reinforced Concrete Section
In reinforced concrete, the steel reinforcement and concrete work together to resist applied loads. The amount of steel reinforcement required in a reinforced concrete section varies depending on various factors such as the type of loading, the yield strength of steel, and the properties of concrete.
The percentage of tensile steel required to produce a balanced reinforced concrete section is the amount of steel needed to develop the full strength of concrete in tension and compression. This percentage is important to determine the maximum load that a reinforced concrete section can withstand.
Correct Answer: Option C
Explanation:
The percentage of tensile steel required to produce a balanced reinforced concrete section is the same for a given quality of steel irrespective of whether working stress method or ultimate load method is followed.
This is because the percentage of tensile steel required to produce a balanced reinforced concrete section depends on the properties of concrete, such as its compressive strength, and the quality of steel used, such as its yield strength. These factors remain the same whether the working stress method or ultimate load method is used.
Therefore, option C is the correct answer.
In reinforced concrete, the steel reinforcement and concrete work together to resist applied loads. The amount of steel reinforcement required in a reinforced concrete section varies depending on various factors such as the type of loading, the yield strength of steel, and the properties of concrete.
The percentage of tensile steel required to produce a balanced reinforced concrete section is the amount of steel needed to develop the full strength of concrete in tension and compression. This percentage is important to determine the maximum load that a reinforced concrete section can withstand.
Correct Answer: Option C
Explanation:
The percentage of tensile steel required to produce a balanced reinforced concrete section is the same for a given quality of steel irrespective of whether working stress method or ultimate load method is followed.
This is because the percentage of tensile steel required to produce a balanced reinforced concrete section depends on the properties of concrete, such as its compressive strength, and the quality of steel used, such as its yield strength. These factors remain the same whether the working stress method or ultimate load method is used.
Therefore, option C is the correct answer.
The main reinforcement of a RC slab consists of 10 mm bars at 10 cm spacing. If it is desired to replace 10 mm bars by 12 mm bars, then the spacing of 12 mm bars should be- a)12cm
- b)14cm
- c)14.40 cm
- d)16 cm
Correct answer is option 'C'. Can you explain this answer?
The main reinforcement of a RC slab consists of 10 mm bars at 10 cm spacing. If it is desired to replace 10 mm bars by 12 mm bars, then the spacing of 12 mm bars should be
a)
12cm
b)
14cm
c)
14.40 cm
d)
16 cm
| | Anuj Chakraborty answered |
The horizontal distance between parallel main reinforcement bars shall not be more than three times the effective depth of solid slab or 300 mm whichever is smaller. The total reinforcement in the slab should remain same. By replacing 10 mm bars by 12 mm bars, the spacing will increase as
What is the anchorage value of a standard hook of a reinforcement bar of diameter D?- a)4D
- b)8D
- c)12D
- d)16D
Correct answer is option 'D'. Can you explain this answer?
What is the anchorage value of a standard hook of a reinforcement bar of diameter D?
a)
4D
b)
8D
c)
12D
d)
16D
| | Prasad Desai answered |
As per clause 26.2.2.1 of IS : 456-2000, the anchorage value of a standard U-type hook shall be equal to 16 times the diameter of the bar
In a Pedestal, the factor by which the effective length should not exceed the least lateral dimension is- a)2
- b)3
- c)4
- d)5
Correct answer is option 'B'. Can you explain this answer?
In a Pedestal, the factor by which the effective length should not exceed the least lateral dimension is
a)
2
b)
3
c)
4
d)
5
| | Yash Verma answered |
Very short columns with effective length less than three times least lateral dimension are called pedestal columns.
Spacing of shear stirrups in a rectangular RC beam is- a)Kept constant througout the span
- b)Decreased towards the centre of the beam
- c)Increased at ends
- d)Increased towards the centre of the span
Correct answer is option 'D'. Can you explain this answer?
Spacing of shear stirrups in a rectangular RC beam is
a)
Kept constant througout the span
b)
Decreased towards the centre of the beam
c)
Increased at ends
d)
Increased towards the centre of the span
 | Moumita Dasgupta answered |
The correct answer is option 'D': Increased towards the centre of the span.
Explanation:
Shear stirrups are used in reinforced concrete (RC) beams to enhance the shear resistance of the beam. They provide confinement to the concrete in the shear zone and help in preventing the shear failure of the beam.
When designing shear stirrups in an RC beam, it is important to consider the variation of the shear force along the span of the beam. The shear force is usually maximum at the supports and decreases towards the center of the span. Therefore, the spacing of shear stirrups is increased towards the center of the span to provide adequate shear resistance.
Here is a detailed explanation of why the spacing of shear stirrups is increased towards the center of the span:
1. Shear force distribution:
In a simply supported RC beam, the shear force distribution is triangular, with the maximum shear force occurring at the supports and decreasing linearly towards the center of the span. This distribution is due to the bending moment distribution in the beam.
2. Shear failure mechanism:
Shear failure in RC beams typically occurs in a diagonal tension zone, known as the shear zone, which extends from the support towards the center of the span. The diagonal cracks develop in this zone when the shear force exceeds the shear resistance of the concrete.
3. Need for shear stirrups:
To prevent shear failure, shear stirrups are provided in the shear zone of the beam. They resist the diagonal tension forces and help in transferring the shear force from the beam to the stirrups.
4. Stirrup spacing:
The spacing of shear stirrups is determined based on the maximum shear force and the shear resistance of the stirrups. As the shear force is maximum at the supports and decreases towards the center of the span, the spacing of shear stirrups is increased towards the center of the span.
- At the supports, where the shear force is maximum, the spacing between the stirrups is kept smaller to provide sufficient resistance against the shear force.
- As we move towards the center of the span, where the shear force is lower, the spacing between the stirrups can be increased as the shear force is less and the demand for shear resistance is lower.
This variation in the spacing of shear stirrups along the beam ensures that there is adequate shear resistance throughout the span to prevent shear failure.
Explanation:
Shear stirrups are used in reinforced concrete (RC) beams to enhance the shear resistance of the beam. They provide confinement to the concrete in the shear zone and help in preventing the shear failure of the beam.
When designing shear stirrups in an RC beam, it is important to consider the variation of the shear force along the span of the beam. The shear force is usually maximum at the supports and decreases towards the center of the span. Therefore, the spacing of shear stirrups is increased towards the center of the span to provide adequate shear resistance.
Here is a detailed explanation of why the spacing of shear stirrups is increased towards the center of the span:
1. Shear force distribution:
In a simply supported RC beam, the shear force distribution is triangular, with the maximum shear force occurring at the supports and decreasing linearly towards the center of the span. This distribution is due to the bending moment distribution in the beam.
2. Shear failure mechanism:
Shear failure in RC beams typically occurs in a diagonal tension zone, known as the shear zone, which extends from the support towards the center of the span. The diagonal cracks develop in this zone when the shear force exceeds the shear resistance of the concrete.
3. Need for shear stirrups:
To prevent shear failure, shear stirrups are provided in the shear zone of the beam. They resist the diagonal tension forces and help in transferring the shear force from the beam to the stirrups.
4. Stirrup spacing:
The spacing of shear stirrups is determined based on the maximum shear force and the shear resistance of the stirrups. As the shear force is maximum at the supports and decreases towards the center of the span, the spacing of shear stirrups is increased towards the center of the span.
- At the supports, where the shear force is maximum, the spacing between the stirrups is kept smaller to provide sufficient resistance against the shear force.
- As we move towards the center of the span, where the shear force is lower, the spacing between the stirrups can be increased as the shear force is less and the demand for shear resistance is lower.
This variation in the spacing of shear stirrups along the beam ensures that there is adequate shear resistance throughout the span to prevent shear failure.
In case of 2-way slab, the limiting deflection of the slab is- a)primarily a function of the long span
- b)primarily a function of the short span
- c)independent of long or short spans
- d)dependent on both long and short spans
Correct answer is option 'B'. Can you explain this answer?
In case of 2-way slab, the limiting deflection of the slab is
a)
primarily a function of the long span
b)
primarily a function of the short span
c)
independent of long or short spans
d)
dependent on both long and short spans
| | Garima Kulkarni answered |
The strip of a two way slab may be checked against shorter span to effective depth ratios.
In the design of two-way slab restrained at all edges, torsional reinforcement required is- a)0.375 times the area of steel provided at midspan in the same direction
- b)0.75 times the area of steel provided at midspan in the same direction
- c)0.375 times the area of steel provided in the shorter span
- d)nil
Correct answer is option 'B'. Can you explain this answer?
In the design of two-way slab restrained at all edges, torsional reinforcement required is
a)
0.375 times the area of steel provided at midspan in the same direction
b)
0.75 times the area of steel provided at midspan in the same direction
c)
0.375 times the area of steel provided in the shorter span
d)
nil
| | Gitanjali Chauhan answered |
The design of two-way slab restrained at all edges requires torsional reinforcement. The correct option is (B) 0.75 times the area of steel provided at midspan in the same direction.
Explanation:
Torsional reinforcement is provided to resist the twisting or torsional forces that may develop in a slab due to the restraint provided at the edges. In a two-way slab, the torsional reinforcement is provided in both directions perpendicular to each other.
The amount of torsional reinforcement required is dependent on the amount of steel provided at midspan in the same direction. As per the IS code, the torsional reinforcement required is 0.75 times the area of steel provided at midspan in the same direction.
This means that if the area of steel provided at midspan in a particular direction is As, then the area of torsional reinforcement required in that direction would be 0.75As.
Option (A) 0.375 times the area of steel provided at midspan in the same direction is incorrect as it is only half of the correct value.
Option (C) 0.375 times the area of steel provided in the shorter span is also incorrect as it is not related to the amount of steel provided at midspan in the same direction.
Option (D) nil is incorrect as torsional reinforcement is always required in a two-way slab restrained at all edges.
Therefore, the correct option is (B) 0.75 times the area of steel provided at midspan in the same direction.
Explanation:
Torsional reinforcement is provided to resist the twisting or torsional forces that may develop in a slab due to the restraint provided at the edges. In a two-way slab, the torsional reinforcement is provided in both directions perpendicular to each other.
The amount of torsional reinforcement required is dependent on the amount of steel provided at midspan in the same direction. As per the IS code, the torsional reinforcement required is 0.75 times the area of steel provided at midspan in the same direction.
This means that if the area of steel provided at midspan in a particular direction is As, then the area of torsional reinforcement required in that direction would be 0.75As.
Option (A) 0.375 times the area of steel provided at midspan in the same direction is incorrect as it is only half of the correct value.
Option (C) 0.375 times the area of steel provided in the shorter span is also incorrect as it is not related to the amount of steel provided at midspan in the same direction.
Option (D) nil is incorrect as torsional reinforcement is always required in a two-way slab restrained at all edges.
Therefore, the correct option is (B) 0.75 times the area of steel provided at midspan in the same direction.
A reduction factor Cr to load carrying capacity of a long column is given by- a)Cr = 1.25 - (Le/24b)
- b)Cr = 1 - (Le/48b)
- c)Cr = 1.25 - (Le/48b)
- d)Cr = 1.5 - (Le/48b)
Correct answer is option 'C'. Can you explain this answer?
A reduction factor Cr to load carrying capacity of a long column is given by
a)
Cr = 1.25 - (Le/24b)
b)
Cr = 1 - (Le/48b)
c)
Cr = 1.25 - (Le/48b)
d)
Cr = 1.5 - (Le/48b)
| | Avinash Mehta answered |
The correct answer is option 'C', A reduction factor, Cr, to the load carrying capacity of a long column is given by Cr = 1.25 - (Le/48b) where Le is the effective length of the column and b is the least lateral dimension of the column cross-section.
This equation is known as the effective length factor, and it is used to account for the buckling of long columns. Column buckling occurs when a column is subject to a compressive load and begins to deform laterally, or buckle, before it reaches the yield point of the material. The buckling load is much lower than the material's yield strength and as the column gets longer, it becomes more susceptible to buckling. Therefore, the effective length factor is used to reduce the load capacity of a long column to account for the likelihood of buckling.
It is important to note that the above equation is valid only for specific case where column has one end fixed and other end hinged, the equation might not be appropriate for other types of end conditions.
According to IS 456, the slenderness ratio for a short, column should not exceed - a)12
- b)18
- c)24
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
According to IS 456, the slenderness ratio for a short, column should not exceed
a)
12
b)
18
c)
24
d)
None of these
| | Divya Kulkarni answered |
Definition of Slenderness Ratio:
The slenderness ratio of a column is the ratio of its effective length (l) to its least radius of gyration (r). It is a measure of how slender or stocky a column is.
Importance of Slenderness Ratio:
The slenderness ratio is an important factor in the design of columns because it determines the column's ability to resist buckling. Buckling is a failure mode in which the column bends or buckles out of its straight shape under compressive loads.
Short Column:
A short column is one that fails primarily due to crushing of the material rather than buckling. In other words, the failure of a short column is governed by its compressive strength rather than its slenderness ratio.
IS 456 Guidelines:
According to IS 456:2000, the code of practice for plain and reinforced concrete, the slenderness ratio for a short column should not exceed 12.
Reasoning:
The specific value of 12 is chosen based on the assumption that a column with a slenderness ratio less than 12 can be considered short and will predominantly fail due to crushing rather than buckling. This assumption is based on empirical evidence and experience from past structures.
Limitations:
It is important to note that the slenderness ratio is not the only factor to consider in column design. Other factors such as the material strength, column dimensions, and load conditions also play a significant role. Additionally, the slenderness ratio limit of 12 is applicable for normal conditions and may vary for special cases or specific design requirements.
Conclusion:
In conclusion, according to IS 456:2000, the slenderness ratio for a short column should not exceed 12. This limit is based on the assumption that columns with a slenderness ratio less than 12 will predominantly fail due to crushing rather than buckling. However, it is important to consider other factors and design requirements in column design.
The slenderness ratio of a column is the ratio of its effective length (l) to its least radius of gyration (r). It is a measure of how slender or stocky a column is.
Importance of Slenderness Ratio:
The slenderness ratio is an important factor in the design of columns because it determines the column's ability to resist buckling. Buckling is a failure mode in which the column bends or buckles out of its straight shape under compressive loads.
Short Column:
A short column is one that fails primarily due to crushing of the material rather than buckling. In other words, the failure of a short column is governed by its compressive strength rather than its slenderness ratio.
IS 456 Guidelines:
According to IS 456:2000, the code of practice for plain and reinforced concrete, the slenderness ratio for a short column should not exceed 12.
Reasoning:
The specific value of 12 is chosen based on the assumption that a column with a slenderness ratio less than 12 can be considered short and will predominantly fail due to crushing rather than buckling. This assumption is based on empirical evidence and experience from past structures.
Limitations:
It is important to note that the slenderness ratio is not the only factor to consider in column design. Other factors such as the material strength, column dimensions, and load conditions also play a significant role. Additionally, the slenderness ratio limit of 12 is applicable for normal conditions and may vary for special cases or specific design requirements.
Conclusion:
In conclusion, according to IS 456:2000, the slenderness ratio for a short column should not exceed 12. This limit is based on the assumption that columns with a slenderness ratio less than 12 will predominantly fail due to crushing rather than buckling. However, it is important to consider other factors and design requirements in column design.
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