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All questions of Torsion for Civil Engineering (CE) Exam
The skew bending theory is based on the __________- a)Plane Elasticity
- b)Plane Deformation
- c)Plane Torque
- d)Plane Strain
Correct answer is option 'B'. Can you explain this answer?
The skew bending theory is based on the __________
a)
Plane Elasticity
b)
Plane Deformation
c)
Plane Torque
d)
Plane Strain
|
Sanvi Kapoor answered |
The skew bending theory is based on the plane deformation approach to plane sections subjected to bending and torsion, the skew bending theory was initially proposed by Lessing with subsequent contributions from Collins, Hsu, Zia, GEsund, Mattock and Elfgreen, of the several researches in this field, Hsu has made the most significant contribution based on the experimental investigations, his work forms the basis of the American, Australian(AS 1481) and Indian code (IS: 1343) provisions.
The failure of prestressed concrete member without additional un tensioned reinforcement, to that of plain concrete is?- a)Zero
- b)Less
- c)1
- d)4
Correct answer is option 'B'. Can you explain this answer?
The failure of prestressed concrete member without additional un tensioned reinforcement, to that of plain concrete is?
a)
Zero
b)
Less
c)
1
d)
4
|
Jay Sharma answered |
The failure of prestressed concrete member without additional un tensioned reinforcement, compared to plain concrete, is less severe.
Prestressed concrete is a type of concrete that is subjected to compressive stresses before it is put into service. This is achieved by applying internal or external forces to the concrete, called prestressing, which counteract the tensile stresses that the member will experience during its service life. This technique improves the overall performance and durability of the concrete member.
On the other hand, plain concrete does not have any additional reinforcement to counteract the tensile stresses. It relies solely on the compressive strength of the concrete to resist these stresses. As a result, plain concrete is more prone to cracking and failure when subjected to excessive tensile forces.
Now, let's discuss why the failure of prestressed concrete without additional un tensioned reinforcement is less severe compared to plain concrete:
1. Reduction in tensile stresses: The main purpose of prestressing is to reduce or eliminate tensile stresses in the concrete member. By applying compressive forces, the concrete is effectively "pre-compressed," which helps to counteract the tensile forces that would otherwise cause cracking and failure in plain concrete.
2. Increased load-carrying capacity: Prestressed concrete can carry higher loads compared to plain concrete. The prestressing forces create a state of initial compression in the member, which increases its load-carrying capacity and allows it to withstand larger external loads without failure.
3. Crack control: The prestressing forces in the concrete help to control and limit the width and propagation of cracks. This is particularly important in structures where crack control is essential for serviceability and durability.
4. Improved structural behavior: Prestressed concrete members exhibit improved structural behavior due to the redistribution of stresses. The prestressing forces help to create a more uniform stress distribution, reducing localized stress concentrations and improving the overall performance of the member.
In conclusion, the failure of prestressed concrete without additional un tensioned reinforcement is less severe compared to plain concrete. The application of prestressing forces helps to reduce tensile stresses, increase load-carrying capacity, control cracks, and improve the structural behavior of the member. These factors contribute to the enhanced performance and durability of prestressed concrete.
Prestressed concrete is a type of concrete that is subjected to compressive stresses before it is put into service. This is achieved by applying internal or external forces to the concrete, called prestressing, which counteract the tensile stresses that the member will experience during its service life. This technique improves the overall performance and durability of the concrete member.
On the other hand, plain concrete does not have any additional reinforcement to counteract the tensile stresses. It relies solely on the compressive strength of the concrete to resist these stresses. As a result, plain concrete is more prone to cracking and failure when subjected to excessive tensile forces.
Now, let's discuss why the failure of prestressed concrete without additional un tensioned reinforcement is less severe compared to plain concrete:
1. Reduction in tensile stresses: The main purpose of prestressing is to reduce or eliminate tensile stresses in the concrete member. By applying compressive forces, the concrete is effectively "pre-compressed," which helps to counteract the tensile forces that would otherwise cause cracking and failure in plain concrete.
2. Increased load-carrying capacity: Prestressed concrete can carry higher loads compared to plain concrete. The prestressing forces create a state of initial compression in the member, which increases its load-carrying capacity and allows it to withstand larger external loads without failure.
3. Crack control: The prestressing forces in the concrete help to control and limit the width and propagation of cracks. This is particularly important in structures where crack control is essential for serviceability and durability.
4. Improved structural behavior: Prestressed concrete members exhibit improved structural behavior due to the redistribution of stresses. The prestressing forces help to create a more uniform stress distribution, reducing localized stress concentrations and improving the overall performance of the member.
In conclusion, the failure of prestressed concrete without additional un tensioned reinforcement is less severe compared to plain concrete. The application of prestressing forces helps to reduce tensile stresses, increase load-carrying capacity, control cracks, and improve the structural behavior of the member. These factors contribute to the enhanced performance and durability of prestressed concrete.
According to IS 456 ∶ 2000, The expression for equivalent shear is given by ________.
WHERE
VU = SHEAR
VE = EQUIVALENT SHEAR
TU = TORSIONAL MOMENT
b = breadth of the beam- a)Ve = Vu + 1.6 Tu/b
- b)Ve = Vu + Tu/b
- c)Ve = Vu − 1.6 Tu/b
- d)Ve = Vu + 1.6 b/Tu
Correct answer is option 'A'. Can you explain this answer?
According to IS 456 ∶ 2000, The expression for equivalent shear is given by ________.
WHERE
VU = SHEAR
VE = EQUIVALENT SHEAR
TU = TORSIONAL MOMENT
b = breadth of the beam
WHERE
VU = SHEAR
VE = EQUIVALENT SHEAR
TU = TORSIONAL MOMENT
b = breadth of the beam
a)
Ve = Vu + 1.6 Tu/b
b)
Ve = Vu + Tu/b
c)
Ve = Vu − 1.6 Tu/b
d)
Ve = Vu + 1.6 b/Tu
|
|
Sanvi Kapoor answered |
As per clause 41.3.1, IS 456:2000, The expression for equivalent shear is given by:
Where Ve = Equivalent Shear, Vu = Nominal Shear, Tu = Torsional Moment, b = Width of beam
Important Points
The equivalent bending moment, Mel given by
, Where Mel = Equivalent bending moment, Mu = Bending moment at the cross section, Tu = Torsional moment, D = Overall depth of beam, b = Width of beam.
Where Ve = Equivalent Shear, Vu = Nominal Shear, Tu = Torsional Moment, b = Width of beamImportant Points
The equivalent bending moment, Mel given by
, Where Mel = Equivalent bending moment, Mu = Bending moment at the cross section, Tu = Torsional moment, D = Overall depth of beam, b = Width of beam.Which type of cables are advantages in reducing the effective shear?- a)Straight
- b)Curved
- c)Trapezoidal
- d)Longitudinal
Correct answer is option 'B'. Can you explain this answer?
Which type of cables are advantages in reducing the effective shear?
a)
Straight
b)
Curved
c)
Trapezoidal
d)
Longitudinal
|
|
Ashish Chakraborty answered |
Straight cables:
- Straight cables do not provide any advantage in reducing the effective shear.
- They have a uniform distribution of forces along their length, which does not contribute to reducing shear.
Trapezoidal cables:
- Trapezoidal cables have varying cross-sectional areas along their length.
- This variation in cross-sectional area can help in reducing the effective shear.
- As the cable tapers from one end to another, the shear force is distributed more evenly along the length, reducing the concentration of shear at any specific point.
Longitudinal cables:
- Longitudinal cables run parallel to the direction of the applied force.
- These cables do not provide any advantage in reducing the effective shear.
- They mainly resist tensile forces and do not contribute significantly to reducing shear.
Curved cables:
- Curved cables have a curved shape along their length.
- The curvature of the cable allows for a redistribution of forces, which helps in reducing the effective shear.
- The curved shape helps in redirecting the shear forces along the cable, resulting in a more uniform distribution of forces and reducing shear concentrations.
Explanation:
In the given options, the only type of cable that provides an advantage in reducing the effective shear is the curved cable.
Curved cables have a curved shape along their length, which helps in redistributing the shear forces. This redistribution results in a more uniform distribution of forces along the cable and reduces shear concentrations. The curved shape of the cable redirects the shear forces, preventing them from accumulating at specific points and reducing the overall shear effect. Therefore, curved cables are advantageous in reducing the effective shear.
- Straight cables do not provide any advantage in reducing the effective shear.
- They have a uniform distribution of forces along their length, which does not contribute to reducing shear.
Trapezoidal cables:
- Trapezoidal cables have varying cross-sectional areas along their length.
- This variation in cross-sectional area can help in reducing the effective shear.
- As the cable tapers from one end to another, the shear force is distributed more evenly along the length, reducing the concentration of shear at any specific point.
Longitudinal cables:
- Longitudinal cables run parallel to the direction of the applied force.
- These cables do not provide any advantage in reducing the effective shear.
- They mainly resist tensile forces and do not contribute significantly to reducing shear.
Curved cables:
- Curved cables have a curved shape along their length.
- The curvature of the cable allows for a redistribution of forces, which helps in reducing the effective shear.
- The curved shape helps in redirecting the shear forces along the cable, resulting in a more uniform distribution of forces and reducing shear concentrations.
Explanation:
In the given options, the only type of cable that provides an advantage in reducing the effective shear is the curved cable.
Curved cables have a curved shape along their length, which helps in redistributing the shear forces. This redistribution results in a more uniform distribution of forces along the cable and reduces shear concentrations. The curved shape of the cable redirects the shear forces, preventing them from accumulating at specific points and reducing the overall shear effect. Therefore, curved cables are advantageous in reducing the effective shear.
Maximum shear stress τcmax for M20 concrete is- a)1.8
- b)2.5
- c)3.1
- d)2.8
Correct answer is option 'D'. Can you explain this answer?
Maximum shear stress τcmax for M20 concrete is
a)
1.8
b)
2.5
c)
3.1
d)
2.8
|
|
Sanvi Kapoor answered |
The shear strength of reinforced concrete with the reinforcement is restricted to some maximum value τcmax depending on the grade of concrete.
Table 20 of IS 456
Stipulates the maximum shear stress of reinforced concrete in beams τcmax as given below in Table. Under no circumstances, the nominal shear stress in beams τv shall exceed τcmax given in the table for different grades of concrete.
Important Points
When the shear stress in concrete exceeds these values it leads to brittle failure due to diagonal compression.
Table 20 of IS 456
Stipulates the maximum shear stress of reinforced concrete in beams τcmax as given below in Table. Under no circumstances, the nominal shear stress in beams τv shall exceed τcmax given in the table for different grades of concrete.
Important Points
When the shear stress in concrete exceeds these values it leads to brittle failure due to diagonal compression.
Shear reinforcement is required to prevent propagation of _______.- a)Flexural crack
- b)Diagonal cracks
- c)Dowel crack
- d)Splitting crack
Correct answer is option 'B'. Can you explain this answer?
Shear reinforcement is required to prevent propagation of _______.
a)
Flexural crack
b)
Diagonal cracks
c)
Dowel crack
d)
Splitting crack
|
Tanvi Shah answered |
Diagonal tension:
Diagonal Tension is the reason for the cracks in shear failure. The diagonal tension crack originates at an angle of 45° and breaks the beam into two pieces. It is present in both the brittle and ductile material.
Shear reinforcement is required to prevent propagation of diagonal cracks.
The diagonal tension cracks develop in the beam, when the strength of the beam in diagonal tension is less than the strength in flexural tension. On this type of failure, the cracks start to develop due to the flexure at mid-span. For the prevention of diagonal tension failure, the shear link is provided.
According to IS 456: 200, the shear reinforcement is must to be provided
Diagonal Tension is the reason for the cracks in shear failure. The diagonal tension crack originates at an angle of 45° and breaks the beam into two pieces. It is present in both the brittle and ductile material.
Shear reinforcement is required to prevent propagation of diagonal cracks.
The diagonal tension cracks develop in the beam, when the strength of the beam in diagonal tension is less than the strength in flexural tension. On this type of failure, the cracks start to develop due to the flexure at mid-span. For the prevention of diagonal tension failure, the shear link is provided.
According to IS 456: 200, the shear reinforcement is must to be provided
- To resist the diagonal tension or principal tension. In the beam diagonal failure occurs when the shear span is higher than three times the value of effective depth.
- To avoid the formation of tension cracks due to any reason given below
- Due to diagonal tension
- Due to the sudden increase in stress
- Due to shrinkage and creep
- To improve the ductility of member
- To improve the rate of main reinforcement
Flexural crack:- In flexural tension failure, under the reinforced condition, the steel yielding first, after that concrete crack and the cracks goes up to a neutral axis from the compressive side of the beam. These types of cracks are avoided by providing the steel and concrete inappropriate manner that neither the section is under reinforced nor over reinforced.
Dowel crack:- It occurs in the road pavement when the concrete pavement is a
A dowel is provided a mechanical connection between concrete slabs.
The bar increases load transfer efficiency and reduces its joint stress, deflection. Hence the dowel crack is avoided by providing an adequate length of dowel bar.
Splitting crack:- This type of crack occurs due to insufficient steel reinforcement and low concrete quality. This type of crack is generally found in concrete columns with non-uniform width.
Dowel crack:- It occurs in the road pavement when the concrete pavement is a
A dowel is provided a mechanical connection between concrete slabs.
The bar increases load transfer efficiency and reduces its joint stress, deflection. Hence the dowel crack is avoided by providing an adequate length of dowel bar.
Splitting crack:- This type of crack occurs due to insufficient steel reinforcement and low concrete quality. This type of crack is generally found in concrete columns with non-uniform width.
As per IS-456: 2000, Side face reinforcemnet shall be provided along the two faces of a reinforced concrete beam, when the depth of the web exceeds- a)500 mm
- b)750 mm
- c)250 mm
- d)400 mm
Correct answer is option 'B'. Can you explain this answer?
As per IS-456: 2000, Side face reinforcemnet shall be provided along the two faces of a reinforced concrete beam, when the depth of the web exceeds
a)
500 mm
b)
750 mm
c)
250 mm
d)
400 mm
|
|
Sanya Agarwal answered |
Side face reinforcement:
Side face reinforcement shall be provided as per CI. 26.5.1.3 and 26.5.17(6) of IS 456:2000
Side face reinforcement shall be provided as per CI. 26.5.1.3 and 26.5.17(6) of IS 456:2000
- Beam depth > 450 mm (if beam subjected to torsion)
- Beam depth > 750 mm (if beam not subjected to torsion)
- Provide @ 0.1% of web area and distribute it equally on both side faces
The members subjected to torque, bending and shear are generally reinforced with __________- a)Parallel and perpendicular reinforcements
- b)Longitudinal and transverse reinforcements
- c)Rectangular and trapezoidal reinforcements
- d)Circular and square reinforcements
Correct answer is option 'B'. Can you explain this answer?
The members subjected to torque, bending and shear are generally reinforced with __________
a)
Parallel and perpendicular reinforcements
b)
Longitudinal and transverse reinforcements
c)
Rectangular and trapezoidal reinforcements
d)
Circular and square reinforcements
|
|
Sanya Agarwal answered |
Members subjected to torque, bending and shear are generally reinforced with longitudinal and transverse reinforcements in order to study the contribution of the longitudinal and transverse reinforcement in resisting flexure, torsion and shear forces, it becomes necessary to analyze the system of forces acting on the warped cross sections of the structural element at the limit state of failure.
The various codes recommend empirical relations to estimate _____- a)Ultimate shear resistance
- b)Ultimate torsional resistance
- c)Ultimate bending resistance
- d)Ultimate load
Correct answer is option 'A'. Can you explain this answer?
The various codes recommend empirical relations to estimate _____
a)
Ultimate shear resistance
b)
Ultimate torsional resistance
c)
Ultimate bending resistance
d)
Ultimate load
|
|
Abhay Kapoor answered |
Empirical Relations for Estimating Ultimate Shear Resistance
In civil engineering, the ultimate shear resistance of various structural elements is a critical parameter that needs to be accurately estimated for the design of safe and efficient structures. Empirical relations are commonly used to estimate the ultimate shear resistance, as they provide a simplified approach based on experimental data and observations. These relations are derived from extensive testing of different structural elements and have been incorporated into various building codes and design guidelines.
Explanation:
Ultimate Shear Resistance:
The ultimate shear resistance refers to the maximum shear force that a structural element can resist before failure occurs. This is a crucial parameter to consider in the design of beams, columns, slabs, and other structural members, as shear forces can significantly affect the overall stability and load-carrying capacity of a structure.
Empirical Relations:
Empirical relations for estimating ultimate shear resistance are derived based on experimental data and observations. These relations provide a simplified approach to estimate the shear resistance without the need for complex mathematical calculations or detailed structural analysis. They are commonly included in building codes and design guidelines to ensure the safe and efficient design of structures.
Various Codes:
Different codes and standards, such as the American Concrete Institute (ACI) Code, Eurocode, and Indian Standards, provide recommendations for estimating the ultimate shear resistance of various structural elements. These codes consider factors such as concrete strength, reinforcement details, geometry, loading conditions, and other relevant parameters to develop empirical relations for different structural elements.
Benefits and Limitations:
The use of empirical relations for estimating ultimate shear resistance offers several benefits, including simplicity, time efficiency, and ease of implementation. They provide a practical approach that can be readily applied in engineering practice. However, it is important to note that these relations are based on experimental data and may have limitations in certain cases. Therefore, they should be used with caution and validated against experimental results or more sophisticated analysis methods when necessary.
Conclusion:
In conclusion, the correct answer to the question is option 'A' - ultimate shear resistance. Empirical relations are commonly recommended in various codes to estimate the ultimate shear resistance of different structural elements. These relations provide a simplified and practical approach for design purposes but should be used judiciously and validated when needed.
In civil engineering, the ultimate shear resistance of various structural elements is a critical parameter that needs to be accurately estimated for the design of safe and efficient structures. Empirical relations are commonly used to estimate the ultimate shear resistance, as they provide a simplified approach based on experimental data and observations. These relations are derived from extensive testing of different structural elements and have been incorporated into various building codes and design guidelines.
Explanation:
Ultimate Shear Resistance:
The ultimate shear resistance refers to the maximum shear force that a structural element can resist before failure occurs. This is a crucial parameter to consider in the design of beams, columns, slabs, and other structural members, as shear forces can significantly affect the overall stability and load-carrying capacity of a structure.
Empirical Relations:
Empirical relations for estimating ultimate shear resistance are derived based on experimental data and observations. These relations provide a simplified approach to estimate the shear resistance without the need for complex mathematical calculations or detailed structural analysis. They are commonly included in building codes and design guidelines to ensure the safe and efficient design of structures.
Various Codes:
Different codes and standards, such as the American Concrete Institute (ACI) Code, Eurocode, and Indian Standards, provide recommendations for estimating the ultimate shear resistance of various structural elements. These codes consider factors such as concrete strength, reinforcement details, geometry, loading conditions, and other relevant parameters to develop empirical relations for different structural elements.
Benefits and Limitations:
The use of empirical relations for estimating ultimate shear resistance offers several benefits, including simplicity, time efficiency, and ease of implementation. They provide a practical approach that can be readily applied in engineering practice. However, it is important to note that these relations are based on experimental data and may have limitations in certain cases. Therefore, they should be used with caution and validated against experimental results or more sophisticated analysis methods when necessary.
Conclusion:
In conclusion, the correct answer to the question is option 'A' - ultimate shear resistance. Empirical relations are commonly recommended in various codes to estimate the ultimate shear resistance of different structural elements. These relations provide a simplified and practical approach for design purposes but should be used judiciously and validated when needed.
The maximum shear stress of circular section is given as __________- a)16T/πD3
- b)20T/πD3
- c)40T/πD3
- d)100T/πD3
Correct answer is option 'A'. Can you explain this answer?
The maximum shear stress of circular section is given as __________
a)
16T/πD3
b)
20T/πD3
c)
40T/πD3
d)
100T/πD3
|
Lavanya Menon answered |
An analysis of principal stresses in prestressed concrete members should include the combined effect of shear stresses due to transverse loads and torsion, together with direct stresses due to flexure and prestress and the maximum stress of circular section is given as:
16T/πD3, D = diameter, T = torsion.
16T/πD3, D = diameter, T = torsion.
In R.C.C. roof, straight bar length of hook taken as (where D is the diameter of the bar)-- a)6 D
- b)8 D
- c)9 D
- d)12 D
Correct answer is option 'C'. Can you explain this answer?
In R.C.C. roof, straight bar length of hook taken as (where D is the diameter of the bar)-
a)
6 D
b)
8 D
c)
9 D
d)
12 D
|
|
Lavanya Menon answered |
Bends and Hooks Forming End Anchorages ( As per IS 2502:1963 )
Here
k = 2 for Mild Steel
k = 3 for Medium Tensile Steel
k = 4 for Cold-worked Steel
As per IS 2502:1963, P-6, Table-II
H = Hook allowance taken as 9d, 11d, 13d, and 17d for k values 2, 3, 4 and 6 respectively and rounded off to the nearest 5 mm, but not less than 75 mm.
B = Bend allowance is taken as 5d, 5.5d, 6d, and 7d for k values 2, 3, 4 and 6 respectively and rounded off to the nearest 5 mm, but not less than 75 mm.
∴ Extra length for one hook = 9d
Here
k = 2 for Mild Steel
k = 3 for Medium Tensile Steel
k = 4 for Cold-worked Steel
As per IS 2502:1963, P-6, Table-II
H = Hook allowance taken as 9d, 11d, 13d, and 17d for k values 2, 3, 4 and 6 respectively and rounded off to the nearest 5 mm, but not less than 75 mm.
B = Bend allowance is taken as 5d, 5.5d, 6d, and 7d for k values 2, 3, 4 and 6 respectively and rounded off to the nearest 5 mm, but not less than 75 mm.
∴ Extra length for one hook = 9d
The distribution of torsion shear stress is uniform in __________
- a)Parallel sections
- b)Rectangular sections
- c)Uniform in Circular sections
- d)Trapezoidal sections
Correct answer is option 'C'. Can you explain this answer?
The distribution of torsion shear stress is uniform in __________
a)
Parallel sections
b)
Rectangular sections
c)
Uniform in Circular sections
d)
Trapezoidal sections
|
|
Sanvi Kapoor answered |
The distribution of torsion shear stress is uniform in circular sections where the magnitude of the shear stress is proportional to the distance from the center and in case of non circular sections involving warping of the cross section, approximate formula have been proposed based on elastic analysis.
In case of structural concrete members subjected to torsion, shear stress develops depending upon the __________- a)Type of bending
- b)Type of tendon
- c)Type of anchorage
- d)Type of cross section
Correct answer is option 'D'. Can you explain this answer?
In case of structural concrete members subjected to torsion, shear stress develops depending upon the __________
a)
Type of bending
b)
Type of tendon
c)
Type of anchorage
d)
Type of cross section
|
Zoya Sharma answered |
In the case of structural concrete members subjected to torsion, shear stresses develop depending upon the type of cross section and magnitude of torque, the shear stresses in association with the flexural stresses may give rise to principal tensile stresses, the value of which when it exceeds tensile strength of the concrete results in the development of cracks on the surface of the member.
Which type of shear reinforcement should be provided for members with thin webs?- a)Maximum shear reinforcement
- b)Minimum shear reinforcement
- c)Nominal shear reinforcement
- d)Tensile reinforcement
Correct answer is option 'C'. Can you explain this answer?
Which type of shear reinforcement should be provided for members with thin webs?
a)
Maximum shear reinforcement
b)
Minimum shear reinforcement
c)
Nominal shear reinforcement
d)
Tensile reinforcement
|
Lekshmi Das answered |
Understanding Shear Reinforcement in Thin Webs
In structural engineering, specifically in the design of reinforced concrete members, shear reinforcement plays a crucial role in ensuring stability and safety. When dealing with members that have thin webs, the appropriate type of shear reinforcement is essential.
What is Shear Reinforcement?
- Shear reinforcement is provided to resist shear forces that occur in structural elements like beams and slabs.
- It helps prevent shear failure, particularly in elements with low shear capacity due to slenderness.
Why Nominal Shear Reinforcement?
- Thin Webs: Members with thin webs (like slender beams) are often more susceptible to shear cracking because of their limited capacity to carry shear loads.
- Nominal Shear Reinforcement: This type is designed to meet the minimum requirements for shear strength. It is not intended to resist all shear forces but provides enough support to prevent premature failure.
- Code Compliance: Building codes often specify the need for nominal shear reinforcement in thin web members to ensure structural integrity.
Benefits of Nominal Shear Reinforcement
- Crack Control: Helps control cracks that may develop due to shear forces, enhancing durability.
- Load Distribution: Aids in distributing shear forces throughout the member, reducing localized stress concentrations.
- Economical Design: Provides a cost-effective solution to enhance the performance of slender or thin-walled members without excessive material use.
In summary, for members with thin webs, providing nominal shear reinforcement is essential to ensure safety, control cracking, and comply with structural design codes.
In structural engineering, specifically in the design of reinforced concrete members, shear reinforcement plays a crucial role in ensuring stability and safety. When dealing with members that have thin webs, the appropriate type of shear reinforcement is essential.
What is Shear Reinforcement?
- Shear reinforcement is provided to resist shear forces that occur in structural elements like beams and slabs.
- It helps prevent shear failure, particularly in elements with low shear capacity due to slenderness.
Why Nominal Shear Reinforcement?
- Thin Webs: Members with thin webs (like slender beams) are often more susceptible to shear cracking because of their limited capacity to carry shear loads.
- Nominal Shear Reinforcement: This type is designed to meet the minimum requirements for shear strength. It is not intended to resist all shear forces but provides enough support to prevent premature failure.
- Code Compliance: Building codes often specify the need for nominal shear reinforcement in thin web members to ensure structural integrity.
Benefits of Nominal Shear Reinforcement
- Crack Control: Helps control cracks that may develop due to shear forces, enhancing durability.
- Load Distribution: Aids in distributing shear forces throughout the member, reducing localized stress concentrations.
- Economical Design: Provides a cost-effective solution to enhance the performance of slender or thin-walled members without excessive material use.
In summary, for members with thin webs, providing nominal shear reinforcement is essential to ensure safety, control cracking, and comply with structural design codes.
Which type of prestressing is advantageous for the members subjected to pure tension?- a)Concentric prestressing
- b)Tangential prestressing
- c)Circular prestressing
- d)Overloaded prestressing
Correct answer is option 'A'. Can you explain this answer?
Which type of prestressing is advantageous for the members subjected to pure tension?
a)
Concentric prestressing
b)
Tangential prestressing
c)
Circular prestressing
d)
Overloaded prestressing
|
|
Lavanya Menon answered |
The research by Humphery and Zia has shown that by suitably adjusting the value of the prestressing force the torsional resistance can be increased by as much as 2.5 times that for the corresponding plain concrete member and for members subjected to pure torsion; concentric prestress is more advantageous than eccentric prestress.
The pre and post tensioned members with bonded tendons bond stress between _______- a)Steel and concrete
- b)Steel and water
- c)Steel and aggregates
- d)Steel and plastic
Correct answer is option 'A'. Can you explain this answer?
The pre and post tensioned members with bonded tendons bond stress between _______
a)
Steel and concrete
b)
Steel and water
c)
Steel and aggregates
d)
Steel and plastic
|
|
Lavanya Menon answered |
Pre tensioned or post tensioned members with bonded tendons develop bond stresses between steel and concrete when the sections are subjected to transverse shear forces due to the rate of change of moment along length of the beam and in the case of type 1 and 2 members, which are uncracked at service loads, the flexural bond stresses developed are computed by considering the complete section.
The space truss, which is composed of longitudinal bars and diagonal concrete truss subjected to twist is known as __________- a)Skew bending theory
- b)Space truss analogy theory
- c)Space truss theory
- d)Compression failed theory
Correct answer is option 'B'. Can you explain this answer?
The space truss, which is composed of longitudinal bars and diagonal concrete truss subjected to twist is known as __________
a)
Skew bending theory
b)
Space truss analogy theory
c)
Space truss theory
d)
Compression failed theory
|
|
Sanya Agarwal answered |
The space truss analogy theory, which is a modification of the planar truss analogy for shear and according to this theory the space truss, which is composed of longitudinal bars and diagonal concrete struts is subjected to twist in which the stirrups and longitudinal bars are considered the tension members and the diagonal concrete struts at an angle θ between the cracks and considered the compression members θ is idealized to 45 degrees.
The behavior of a prestressed concrete member is affected by the relative magnitude of __________- a)Internal actions
- b)External actions
- c)Zero
- d)Constant actions
Correct answer is option 'A'. Can you explain this answer?
The behavior of a prestressed concrete member is affected by the relative magnitude of __________
a)
Internal actions
b)
External actions
c)
Zero
d)
Constant actions
|
|
Sanya Agarwal answered |
The behavior of a prestressed concrete member is affected by the relative magnitude of the internal actions, such as torque, bending moment and shear force, in circular sections and if torsion is small, it has little effect on the overall behavior and the failure are controlled by either flexure or shear.
Lifing of the corners of the slabs is prevented by providing ________ reinforcement.- a)torsion
- b)shear
- c)transverse
- d)longitudinal
Correct answer is option 'A'. Can you explain this answer?
Lifing of the corners of the slabs is prevented by providing ________ reinforcement.
a)
torsion
b)
shear
c)
transverse
d)
longitudinal
|
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Tanvi Shah answered |
- As per IS 456 CI. D-1, Restrained slab is the slab whose corners are prevented from getting lifted up and are provided with suitable reinforcement to resist torsion and are called a restrained slab.
- All four edges of the slab are assumed to be rigidly tied with the beams or walls underneath and the edges may be either continuous (Fixed) or discontinuous.
- A corner restrained in the slab reduces the bending moment and deflection in the middle of the slab just like a fixed beam which reduces the mid-span moment and deflection as compared to simply supported beams.
Important Points
As per IS 456 (B 1.8, B 1.9, B 1.10), Torsion reinforcement is provided as:
B 1.8 Torsion reinforcement is required at the corner where both edges are discontinuous
Reinforcement is provided in 4 layers..
size of mesh = 1/5
Area of steel in each layer = (3/4) x Ast (+ve)
B 1.9: Torsion reinforcement is required where at least one edge is discontinuous
Provided in 4 layers
size of mesh = 1/5
Area of steel in each layer = (3/8) x Ast (+ve)
B 1.10: Where both edges are continuous no need to provide torsional reinforcement.
As per IS 456 (B 1.8, B 1.9, B 1.10), Torsion reinforcement is provided as:
B 1.8 Torsion reinforcement is required at the corner where both edges are discontinuous
Reinforcement is provided in 4 layers..
size of mesh = 1/5
Area of steel in each layer = (3/4) x Ast (+ve)
B 1.9: Torsion reinforcement is required where at least one edge is discontinuous
Provided in 4 layers
size of mesh = 1/5
Area of steel in each layer = (3/8) x Ast (+ve)
B 1.10: Where both edges are continuous no need to provide torsional reinforcement.
When both longitudinal steel and spirals are provided in prestressed members, the ultimate torsional resistance is?- a)Twp+Tws
- b)Ttp+Tts
- c)Tvp+Tvs
- d)Tep+Tes
Correct answer is option 'B'. Can you explain this answer?
When both longitudinal steel and spirals are provided in prestressed members, the ultimate torsional resistance is?
a)
Twp+Tws
b)
Ttp+Tts
c)
Tvp+Tvs
d)
Tep+Tes
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Sanya Agarwal answered |
The use of longitudinal steel or spirals independent of each other does not increase the ultimate torsional resistance but when both longitudinal steel and spirals are provided in prestressed members, the ultimate torsional resistance is enhanced and according can be expressed as Ttp + Tts, where Ttp is the torsional resistance moment of the prestressed concrete section and Tts is the additional torsional resistance moment of the non-prestressed reinforcement, which must consist of spirals and longitudinal steel.
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