Bond, Anchorage and Development Length MCQ Quiz - Objective Question with Answer for Bond, Anchorage and Development Length - Download Free PDF
Last updated on May 21, 2025
Latest Bond, Anchorage and Development Length MCQ Objective Questions
Bond, Anchorage and Development Length Question 1:
In rebaring work, the minimum lap splice length for tension in a column bar as per IS 456:2000 should be:
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 1 Detailed Solution
Explanation:
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According to IS 456:2000 (Clause 26.2), the minimum lap splice length for tension in a column bar is determined based on the diameter of the bar.
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The lap length for tension bars in columns is given by 40 times the diameter of the bar. This ensures that there is sufficient bond strength and that the rebar is effectively transferred from one piece of reinforcement to the next.
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This length is designed to prevent slippage or failure at the splice joint under tensile loads. The lap splice length ensures continuity of reinforcement across the joint, maintaining the structural integrity of the column.
Additional InformationFactors Affecting Lap Splice Length
The minimum lap splice length is influenced by several factors, including:
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Bar Diameter: Larger bars require longer lap lengths to ensure adequate bonding between bars.
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Concrete Grade: Higher-strength concrete may allow for shorter lap lengths because of the better bonding characteristics between concrete and rebar.
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Reinforcement Type: For higher-strength reinforcement, lap lengths can sometimes be reduced, but only with careful design considerations and following relevant code provisions.
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Type of Force (Tension vs. Compression): The lap length is generally longer for tension bars because tension forces demand more effective transfer than compression, where direct bearing between bars is more significant.
Bond, Anchorage and Development Length Question 2:
Which of the following conditions necessitates the use of rebar couplers instead of lapping?
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 2 Detailed Solution
Explanation:
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Rebar couplers are used when lapping (overlapping two rebars to achieve continuity) is not feasible due to insufficient space between bars or other constraints like structural elements or layout requirements.
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When there isn't enough space to safely overlap rebars, rebar couplers are a better alternative as they provide a mechanical connection between two bars, ensuring continuity without needing the overlap.
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This method is especially useful in areas where space is limited or in structures requiring high-strength connections.
Additional Information
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IS 13620:2002 (Indian Standard for Rebar Couplers) provides guidelines for the use of couplers in reinforced concrete structures. This standard specifies the design, testing, and installation of couplers and ensures that the coupling method meets the required strength and durability criteria.
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IS 456:2000 also addresses rebar connections, although it primarily focuses on lap lengths, which might not be feasible in every case.
Bond, Anchorage and Development Length Question 3:
Which test is most appropriate to assess the corrosion potential of rebars in an existing RCC structure?
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 3 Detailed Solution
Explanation:
Half-Cell Potential Test
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The Half-Cell Potential Test is the most widely used method for evaluating the corrosion potential of reinforcing steel (rebars) in reinforced concrete structures.
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This test measures the electrical potential between the reinforcement and a reference electrode (usually a copper/copper sulfate electrode), and the results provide an indication of whether the rebar is prone to corrosion.
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The test gives insight into the corrosion activity without causing any damage to the structure. It’s effective for identifying areas that are at risk of corrosion, allowing for targeted remediation.
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IS 456 and other standards recommend this method as a reliable indicator of the corrosion state of reinforcement in concrete structures.
Additional InformationRebound Hammer Test
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The Rebound Hammer Test is primarily used to measure the surface hardness of concrete, which gives an indirect indication of concrete strength.
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While it provides useful data for assessing the quality of the concrete, it is not specifically designed for evaluating the corrosion potential of the rebars.
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Rebound Hammer cannot detect the corrosion status of reinforcing bars, as it focuses on surface hardness, not the internal condition of the reinforcement.
Ultrasonic Pulse Velocity Test
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The Ultrasonic Pulse Velocity (UPV) test is used to assess the quality and uniformity of concrete by measuring the speed at which an ultrasonic pulse travels through the concrete.
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It can detect cracks, voids, and honeycombing within the concrete, but it does not provide direct information on the corrosion status of the reinforcement.
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UPV is generally used to assess the concrete condition but not the condition of the rebars or their corrosion potential.
Core Cutting Test
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Core Cutting involves extracting a sample of concrete from the structure for laboratory testing, which provides detailed information on the concrete quality, strength, and aggregate properties.
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While it may help in assessing concrete composition or strength, it does not directly assess the corrosion condition of reinforcing steel within the concrete. It’s more destructive and expensive compared to the Half-Cell Potential Test, which can assess corrosion in place.
Bond, Anchorage and Development Length Question 4:
In case of differential settlement, which crack pattern is most commonly observed in brick masonry walls?
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 4 Detailed Solution
Explanation:
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Differential settlement occurs when different parts of a structure settle at different rates, leading to uneven stresses in the walls.
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This uneven settlement can cause a distortion of the masonry, resulting in cracks.
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In brick masonry walls, diagonal cracks typically form at the corners of openings (such as windows and doors) due to concentrated stresses in these regions.
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These cracks usually occur because the settlement causes torsional or shearing forces, which are most prominent near the edges of openings where the structural continuity is disrupted.
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his is the most common crack pattern associated with differential settlement, as the structural load paths are disturbed, leading to stress concentrations at these points.
Additional InformationTypes of Cracks:
1. Shrinkage Cracks
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Cause: Shrinkage in concrete or masonry due to moisture loss, hydration, or temperature changes.
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Appearance: Fine and shallow cracks, usually seen in newly laid concrete or masonry. They often appear randomly over surfaces.
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Location: Common in plaster and exposed concrete surfaces, especially during the drying phase after construction.
2. Settlement Cracks
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Cause: Foundation settlement due to soil compaction, load redistribution, or structural changes. The foundation may settle unevenly, leading to stress in the walls.
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Appearance: Vertical or diagonal cracks, usually at the corners of openings (windows, doors) or along walls.
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Location: Most commonly seen in load-bearing walls and masonry structures.
3. Thermal Cracks
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Cause: Temperature fluctuations in concrete and masonry. Expansion and contraction due to heating or cooling cycles can create internal stresses.
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Appearance: Fine hairline cracks, typically circular or radial.
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Location: Seen on exposed surfaces or concrete floors exposed to sunlight.
4. Differential Settlement Cracks
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Cause: Uneven settlement of foundation, which can occur if the foundation is built on non-homogeneous soil or if the foundation fails to adequately support one part of the structure.
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Appearance: Diagonal cracks often originating from corners of openings such as windows or doors.
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Location: Typically found near openings, or along load-bearing walls.
Bond, Anchorage and Development Length Question 5:
Which of the following is NOT a common technique for rebar anchoring during repairs?
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 5 Detailed Solution
Explanation:
Laser welding
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Laser welding is not commonly used for rebar anchoring during repairs.
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While laser welding is an advanced technique for joining metals, it is generally not suitable for bonding steel reinforcement to concrete or for repairing concrete structures.
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Laser welding is highly specialized and involves precise and high-intensity heat, making it unfit for reinforcing steel embedding into concrete.
Additional InformationChemical anchoring
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Chemical anchoring involves the use of epoxies, resins, or other bonding agents to anchor rebars into drilled holes in concrete.
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The chemical resin helps to bond the rebar securely to the concrete, making it an effective and widely used method for post-installation anchoring.
Mechanical anchoring
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Mechanical anchoring uses devices like mechanical anchors, bolts, or screws to attach rebars to concrete.
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This technique provides reliable load transfer and is often used for anchoring rebars during repairs, especially when drilling and embedding isn't feasible.
Epoxy bonding
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Epoxy bonding is similar to chemical anchoring but specifically uses epoxy-based resins to bond the reinforcement bars into the concrete.
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It is a widely used method for structural repairs where deep embedding or physical mechanical anchors may be difficult.
Top Bond, Anchorage and Development Length MCQ Objective Questions
If design bond stress = 1.5 N/mm2 is assumed, then the development length of an Fe 500 HYSD bar of nominal diameter 12 mm - which is fully stressed in tension - will be:
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 6 Detailed Solution
Download Solution PDFConcept:
Development length:
(i) The cal culated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or a combination thereof.
(ii) Development length can be calculated as:
\({L_d} = \frac{{ϕ × 0.87{f_y}}}{{4{τ _{bd}}}}\)
Where, ϕ = Diameter of bar
τbd = Design bond stress = Permissible value of average bond stress
The value of bond stress is increased by 60% for a deformed bar in tension and a further increase of 25% is made for bars in compression.
Calculation:
Given,
ϕ = 12 mm
τbd = 1.5 N/mm2
So, τbd = 1.5 × 1.6 N/mm2
Development length, \({L_d} = \frac{{ϕ × 0.87{f_y}}}{{4{τ _{bd}}}}\)
\({L_d} = \frac{{12 × 0.87 × 500}}{{4×1.6 × 1.5}}\) = 543.75 mm ≈ 544 mm
The designed bond stress of M20 grade concrete is:
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 7 Detailed Solution
Download Solution PDFAs per clause No. 26.2.1.1, Design bond stress in limit state method for plain bars in tension for different grades of concrete are:
S. No |
Concrete Grade |
Design Bond Stress (MPa) |
1. |
M-20 |
1.2 |
2. |
M-25 |
1.4 |
3. |
M-30 |
1.5 |
4. |
M-35 |
1.7 |
5. |
M 40 and above |
1.9 |
In limit state design, the design bond stress for deformed bars (conforming to IS 1786) for M20 concrete mentioned in IS 456 ∶ 2000 is (in N/mm2 units):
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 8 Detailed Solution
Download Solution PDFConcept:
Design bond stress:
- Bond stress is the result of the bonding between the concrete surface and the reinforcement steel.
- It varies depending upon the type of concrete and type of reinforcement used.
- If plane rounded steel is used as reinforcement then the bond stress will be less, if the same concrete is used with HYSD steel as reinforcement then the bond stress is higher.
- The bond between steel and concrete is mainly due to pure adhesive resistance, frictional resistance, and mechanical resistance.
According to IS 456 ∶ 2000, the design bond stress in limit state method for plain bars in tension shall be as below:
Grade of concrete | M20 | M25 | M30 | M35 | M40 and above |
Design bond stress(N/mm2) | 1.2 | 1.4 | 1.5 | 1.7 | 1.9 |
Note:
- For deformed bars conforming to IS 1786, these values shall be increased by 60 percent.
- For bars in compression. the values of bond stress for bars in tension shall be increased by 25 percent.
Calculation:
Design bond stress for deformed bars (conforming to IS 1786) for M20 concrete
= 1.2 \(\times\) 1.60 = 1.92 N/mm2
The development length in compression for a 20 mm diameter deformed bar of grade Fe415 embedded in concrete of grade M25 whose design bond stress is 1.40 N/mm2 is -
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 9 Detailed Solution
Download Solution PDFConcept:
Development length:
(i) The calculated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or a combination thereof.
(ii) Development length can be calculated as:
\({L_d} = \frac{{ϕ × 0.87{f_y}}}{{4{τ _{bd}}}}\)
Where, ϕ = Diameter of bar
τbd = Design bond stress = Permissible value of average bond stress
The value of bond stress is increased by 60% for a deformed bar in tension and a further increase of 25% is made for bars in compression.
Calculation:
Given,
ϕ = 20 mm, τbd = 1.4 N/mm2 (In tension), fy = 415 MPa
∵ Bar in compression, so Increased the value of τbd by 25% and for deformed bar it is increased by 60 %
So, τbd = 1.4 × 1.25 × 1.6 = 2.8 N/mm2
Development length, \({L_d} = \frac{{ϕ × 0.87{f_y}}}{{4{τ _{bd}}}}\)
\({L_d} = \frac{{20 × 0.87 × 415}}{{4 × 2.8}}\) = 645 mm
In the steel reinforcement calculation for RCC, the additional length for two 45° bent – ups in reinforcing bars is:
Answer (Detailed Solution Below)
0.84 times the total depth of beam or slab minus bottom and top cover
Bond, Anchorage and Development Length Question 10 Detailed Solution
Download Solution PDFConcept:
Bar Type | Length of Hooks | Total length of bar |
Straight bar | 18d | l + 18d |
Bent - up at one end only | 18d | l + 18d + 0.42D |
Double bent - up bar | 18d | l + 18d + 2 × 0.42D |
Overlap of bars | 18d | Overall length at joint = [ (40 - 45) d + 18d ] |
Details of bar shape:
Calculate the development length of a 10 mm ϕ bar for M 25 grade concrete provide if the stress in bar at the section at design load is 30 MPa.
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 11 Detailed Solution
Download Solution PDFConcept:
Development Length:
A development length is the amount of rebar length that is needed to be embedded or projected into concrete to create desired bond strength between the two materials and also to develop required stress in steel at that section.
As per IS 456: 2000, clause 26.2.1,
The development length is given by:
\({L_d} = \frac{{\phi {\sigma _s}}}{{4{τ _{bd}}}}\)
where
ϕ = nominal diameter of the bar
σs = stress in the bar at the section considered at design load
τbd = design bond stress
Calculation:
Given data,
Grade of concrete = M25
ϕ = 10 mm
σs = 30 MPa
τbd = 1.4 MPa for M25
The development length is given by:
\({L_d} = \frac{{\phi {\sigma _s}}}{{4{τ _{bd}}}}\)
\({L_d} = \frac{{10{\times 30}}}{{4{\times 1.4}}}\) = 53.5 mm
Important Points
Grade of Concrete | M20 | M25 | M30 | M35 | M40 & Above |
Design bond stress(τbd) |
1.2 | 1.4 | 1.5 | 1.7 | 1.9 |
For deformed bars conforming to IS 1786, the design bond stress in limit state method shall be:
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 12 Detailed Solution
Download Solution PDFConcept:
The value of bond stress is increased by 60% when deformed bars are used as having better interlocking bond formation in concrete.
The value of bond stress is increased by 25% when the bar is in compression.
Important Points
The bond stress for different grades of concrete as per different established provisions are as follows:
Grade |
M15 |
M20 |
M25 |
M30 |
M35 |
M40 |
WSM |
0.6 |
0.8 |
0.9 |
1.0 |
1.1 |
1.2 |
LSM |
|
1.2 |
1.4 |
1.5 |
1.7 |
1.9 |
The bond stresses value tabulated above is for plain bar in tension.
According to IS 456 ∶ 2000, the design bond stress (in N/mm2 units) in limit state method for HYSD bars (Fe415 grade) for M30 grade concrete is:
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 13 Detailed Solution
Download Solution PDFConcept:
Design bond stress:
- Bond stress is the result of the bonding between the concrete surface and the reinforcement steel.
- It varies depending upon the type of concrete and type of reinforcement used.
- If plane rounded steel is used as reinforcement then the bond stress will be less, if the same concrete is used with HYSD steel as reinforcement then the bond stress is higher.
- The bond between steel and concrete is mainly due to pure adhesive resistance, frictional resistance, and mechanical resistance.
According to IS 456 ∶ 2000, the design bond stress in the limit state method for plain bars in tension shall be as below:
Grade of concrete | M20 | M25 | M30 | M35 | M40 and above |
Design bond stress(N/mm2) | 1.2 | 1.4 | 1.5 | 1.7 | 1.9 |
Note:
- For deformed bars conforming to IS 1786, these values shall be increased by 60 percent.
- For bars in compression. the values of bond stress for bars in tension shall be increased by 25 percent.
Calculation:
Design bond stress for deformed bars (conforming to IS 1786) for M30 concrete
= 1.5 × 1.60 = 2.4 N/mm2
According to IS 456, the anchorage value of a standard U-type hook shall be equal to X times the diameter of the bar, where X is:
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 14 Detailed Solution
Download Solution PDFConcept:
Anchorage Length:
(i) Anchorage length is the equivalent length of the reinforcement bar which is considered to be available when a straight bar is bent through some angle. It is provided only at the support. Anchorage length is provided if sufficient development length cannot be provided inside the support.
As per IS 456: 2000:
Anchorage Value of Bends & Hooks |
||
Bends |
45° |
4ϕ |
90° |
8ϕ |
|
135° |
12ϕ |
|
Hooks |
U-type Hook (180°) |
16ϕ |
Where, ϕ = nominal diameter of the bar
Calculate the anchorage value of 45° and 90° bend for a 25 mm diameter bar.
Answer (Detailed Solution Below)
Bond, Anchorage and Development Length Question 15 Detailed Solution
Download Solution PDFExplanation:
As per IS 456: 2000,
- A standard hook has an anchorage value equivalent to a straight length of 16ϕ.
- The anchorage value of the standard U-type hook shall be 16 times the diameter of the bar.
- The anchorage value of the standard bend shall be considered as 4 times the diameter of the bar for each 45o bend subject to a maximum value of 16 times the diameter of the bar.
Calculation:
Given:
Diameter = ϕ = 25 mm
Anchorage value for 45° = 25 × 4 = 100 mm
Anchorage value for 90° = 25 × 8 = 200 mm