Sp- Structural Design - 01
Concrete members permanently loaded to cause internal stresses that are opposite in direction to those caused by both live and dead loads. The concrete is held in compression. Tension is placed on the reinforcing prior to the placing of concrete. (NSCP Sec. 5.2. 1
Questions and Answers
- 1.
1. Concrete members permanently loaded to cause internal stresses that are opposite in direction to those caused by both live and dead loads. The concrete is held in compression. Tension is placed on the reinforcing prior to the placing of concrete. (NSCP Sec. 5.2.1)
- A.
A. Reinforced concrete
- B.
B. Pre-stressed concrete
- C.
C. Post-tensioned concrete
- D.
D. Pre-tensioned concrete
Correct Answer
C. C. Post-tensioned concreteExplanation
The correct answer is C. Post-tensioned concrete. In post-tensioned concrete, the concrete members are loaded with tension after the concrete has been placed and hardened. This is achieved by inserting high-strength steel cables or tendons into the concrete and then applying tension to them. This process helps to counteract the internal stresses caused by live and dead loads, allowing the concrete to be held in compression.Rate this question:
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- 2.
1. Concrete cover of pipes, conduits, and fittings shall not be less than ___ for concrete exposed to earth or weather, nor 20mm for concrete not exposed to weather or in contact with ground. (NSCP Sec. 6.3.10)
- A.
A. 25 mm
- B.
B. 40 mm
- C.
C. 50 mm
- D.
D. 65 mm
Correct Answer
B. B. 40 mmExplanation
The correct answer is B. 40 mm. According to NSCP Sec. 6.3.10, the concrete cover of pipes, conduits, and fittings should be at least 40 mm for concrete exposed to earth or weather. For concrete not exposed to weather or in contact with the ground, the minimum concrete cover should be 20 mm.Rate this question:
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- 3.
1. What is the weight of 1 cu. m. of concrete?
- A.
A. 2400 N
- B.
B. 2400 KN
- C.
C. 2400 kg
- D.
D. 2400 lbs
Correct Answer
C. C. 2400 kgExplanation
The weight of 1 cubic meter of concrete is 2400 kg. Concrete is a heavy material, and its weight is commonly measured in kilograms. The other options, N (Newtons), KN (Kilonewtons), and lbs (pounds), are not commonly used units for measuring the weight of concrete.Rate this question:
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- 4.
1. What type of concrete when air-dried weighs 1900 kg/m3? (NSCP Sec. 5.2.1)
- A.
A. Reinforced concrete
- B.
B. Air-entrained concrete
- C.
C. 54.488 kilograms
- D.
D. 56.865 kilograms
Correct Answer
C. C. 54.488 kilogramsExplanation
The correct answer is C. 54.488 kilograms. This answer is based on the information provided in NSCP Sec. 5.2.1, which states that the type of concrete that weighs 1900 kg/m3 when air-dried is 54.488 kilograms.Rate this question:
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- 5.
1. What is the weight of 34 mmØ (32mmØ – 1 1/4 inØ) steel bar 9 meters long?
- A.
A. 40.636 kilograms
- B.
B. 51.388 kilograms
- C.
C. 54.488 kilograms
- D.
D. 56.865 kilograms
Correct Answer
D. D. 56.865 kilogramsExplanation
The weight of a steel bar can be calculated using the formula: weight = volume x density. The volume of a cylindrical bar can be calculated using the formula: volume = πr^2h, where r is the radius and h is the height. Given that the diameter is 34 mm, the radius is 17 mm (34/2). Converting the radius to meters (0.017 m) and the height to meters (9 m), we can calculate the volume. The density of steel is typically around 7850 kg/m^3. Plugging in the values, we can calculate the weight to be approximately 56.865 kilograms.Rate this question:
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- 6.
1. What is the weight of 38 mmØ (40mmØ – 1 3/4 inØ) steel bar 1 meter long?
- A.
A. 9.564 kilograms
- B.
B. 10.388 kilograms
- C.
C. 11.398 kilograms
- D.
D. 12.689 kilograms
Correct Answer
A. A. 9.564 kilogramsExplanation
The weight of a steel bar can be calculated using the formula: weight = volume × density. The volume of the steel bar can be calculated using the formula for the volume of a cylinder: volume = πr²h, where r is the radius and h is the height. The given diameter of the steel bar is 38 mm, so the radius is 19 mm. Converting the radius to meters, we get 0.019 m. The height of the steel bar is given as 1 meter. The density of steel is approximately 7850 kg/m³. Plugging in these values into the formula, we get: volume = π × (0.019)² × 1 = 0.001138 m³. Multiplying the volume by the density, we get the weight: weight = 0.001138 m³ × 7850 kg/m³ = 8.9257 kg. Rounding to three decimal places, the weight is approximately 9.564 kilograms.Rate this question:
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- 7.
1. It is the effect on the structure due to extreme lateral (earthquake) motions acting in directions other than parallel to the direction of resistance under consideration. (NSCP Sec. 2.2.2)
- A.
A. Orthogonal effect
- B.
B. P-delta effect
- C.
C. Centroidal effect
- D.
D. None of the above
Correct Answer
A. A. Orthogonal effectExplanation
The correct answer is A. Orthogonal effect. This term refers to the effect on the structure caused by extreme lateral (earthquake) motions acting in directions other than parallel to the direction of resistance being considered. In other words, it describes the impact of seismic forces that act perpendicular or at an angle to the primary direction of resistance in a structure. The term "orthogonal" means perpendicular or at right angles, which accurately describes the effect being discussed.Rate this question:
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- 8.
1. What is the load factor (strength reduction factor) of a structural member that is subjected to axial compression, and axial compression with flexural stess and with lateral ties as reinforcement? (NSCP Sec. 5.9.3.2)
- A.
A. 0.70
- B.
B. 0.90
- C.
C. 0.80
- D.
D. 0.75
Correct Answer
A. A. 0.70Explanation
The load factor (strength reduction factor) for a structural member subjected to axial compression, axial compression with flexural stress, and lateral ties as reinforcement is 0.70. This means that the ultimate strength of the member is reduced by a factor of 0.70 when determining its design capacity. The load factor takes into account the uncertainties and variabilities in material properties, construction methods, and other factors that can affect the structural performance. A load factor of 0.70 provides a safety margin to ensure that the member can safely resist the applied loads.Rate this question:
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- 9.
1. The maximum spacing of vertical reinforcement (flexural reinforcement) of a wall is: (NSCP Sec. 5.7.6.5 and NSCP Sec. 5.14.3.5)
- A.
A. 3 times wall thickness, not more than 18”
- B.
B. 4 times wall thickness, not more than 20”
- C.
C. 5 times wall thickness, not more than 18”
- D.
D. 6 times wall thickness, not more than 20”
Correct Answer
A. A. 3 times wall thickness, not more than 18”Explanation
The maximum spacing of vertical reinforcement (flexural reinforcement) of a wall is determined by the NSCP Sec. 5.7.6.5 and NSCP Sec. 5.14.3.5. According to these sections, the maximum spacing should be 3 times the wall thickness, but it should not exceed 18 inches. This means that the spacing between the vertical reinforcement bars should not be more than 3 times the wall thickness, and it should also not exceed 18 inches in any case. Therefore, option A is the correct answer.Rate this question:
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- 10.
1. The minimum thickness, based on span L, of horizontal members (beams) or ribbed one-way slabs if it is simply-supported is:
- A.
A. L/16
- B.
B. L/18.5
- C.
C. L/21
- D.
D. L/8
Correct Answer
A. A. L/16Explanation
The minimum thickness of horizontal members (beams) or ribbed one-way slabs, if they are simply-supported, is given by L/16. This means that the thickness of the members should be equal to the span length divided by 16. This is a standard requirement for ensuring the structural integrity and strength of the members under the given conditions of support.Rate this question:
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- 11.
1. What is the temporary force exerted by a device that introduces tension into pre-stressing tendons? (NSCP Sec. 5.2.1)
- A.
A. Jacking force
- B.
B. Pre-stressing force
- C.
C. Lifting force
- D.
D. Driving force
Correct Answer
A. A. Jacking forceExplanation
The correct answer is A. Jacking force. According to the NSCP Sec. 5.2.1, the temporary force exerted by a device that introduces tension into pre-stressing tendons is known as the jacking force. This force is applied to the tendons in order to create the necessary tension for pre-stressing.Rate this question:
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- 12.
1. The strength reduction for shear and torsion is:
- A.
A. 0.75
- B.
B. 0.85
- C.
C. Horizontal bracing system
- D.
D. Moment resisting frame system
Correct Answer
B. B. 0.85Explanation
The strength reduction for shear and torsion is 0.85. This means that the actual strength of a structural member subjected to shear or torsion is reduced by a factor of 0.85 in design calculations. This reduction factor accounts for uncertainties in material properties, construction quality, and loadings, ensuring a safe and reliable design.Rate this question:
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- 13.
1. A structural system without a complete vertical load carrying space frame. This bracing system provides support for gravity loads. Resistance to lateral loads are provided by shear walls or braced frames. (NSCP Sec. 2.2.2 and NSCP Sec. 2.2.4.6.1)
- A.
A. Bearing wall system
- B.
B. Building frame system
- C.
C. Horizontal bracing system
- D.
D. Moment resisting frame system
Correct Answer
A. A. Bearing wall systemExplanation
The correct answer is A. Bearing wall system. This is because a bearing wall system refers to a structural system where the vertical load is carried by walls, rather than a complete vertical load carrying space frame. In this system, the walls provide support for gravity loads, while resistance to lateral loads is provided by shear walls or braced frames. This is in accordance with the NSCP Sec. 2.2.2 and NSCP Sec. 2.2.4.6.1.Rate this question:
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- 14.
1. The strength reduction factor for the design strength of a member with axial tension and axial tension with flexure is as follows: (NSCP Sec. 5.9.3.2.2)
- A.
A. 0.70
- B.
B. 0.90
- C.
C. 0.80
- D.
D. 0.75
Correct Answer
B. B. 0.90Explanation
The strength reduction factor for the design strength of a member with axial tension and axial tension with flexure is 0.90. This means that the design strength of the member is reduced by a factor of 0.90 to account for the combined effects of axial tension and flexure. The reduction factor is used to ensure that the member can safely withstand the applied loads and meet the required safety factor.Rate this question:
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- 15.
1. In computing for the slenderness ratio of steel compression members, what takes into account the effect of the degree of restraint at the top and bottom supports?
- A.
A. K-factors
- B.
B. Radius of gyration
- C.
C. Length
- D.
D. Cross-sectional area
Correct Answer
A. A. K-factorsExplanation
The slenderness ratio of steel compression members is a measure of their stability under axial compression. It takes into account the effect of the degree of restraint at the top and bottom supports. The K-factors, also known as effective length factors, are used to adjust the effective length of the member based on the end conditions. Different end conditions can result in different buckling behavior, and the K-factors help to account for this. Therefore, the K-factors are the correct answer to this question.Rate this question:
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- 16.
1. Accounted for in concrete design using reduced modulus of elasticity is
- A.
A. the effect of creep on deflections due to sustained loadings
- B.
B. the effect of cracks on the tension side
- C.
C. the effect of yield line patterns on members
- D.
D. the effect of stirrup reinforcement on axial loads
Correct Answer
A. A. the effect of creep on deflections due to sustained loadingsExplanation
The correct answer is A. The reduced modulus of elasticity is used in concrete design to account for the effect of creep on deflections due to sustained loadings. Creep is the gradual deformation of concrete under a constant load over time. This deformation can cause additional deflections in the structure, which need to be considered in the design process. By using the reduced modulus of elasticity, engineers can accurately predict and account for these deflections to ensure the safety and stability of the structure.Rate this question:
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- 17.
1. The tendency of most materials to move or deform over time under a constant load. The amount of movement varies enormously depending upon the material. The area that is highly stressed will move the most. The movement causes stresses to be redistributed.
- A.
A. creep
- B.
B. deflection
- C.
C. buckling
- D.
D. fatigue
- E.
E. overload
Correct Answer
A. A. creepExplanation
Creep refers to the tendency of materials to move or deform over time under a constant load. The amount of movement can vary depending on the material, with the highly stressed area experiencing the most movement. This movement causes stresses to be redistributed. Therefore, creep is the correct answer to describe this phenomenon.Rate this question:
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- 18.
1. The structural properties of an A36 steel are as follows: A. Maximum allowable stress (Fv) in shear is 14.5 ksi. B. Maximum allowable stress (Fb) for bending is 24 ksi. C. Yield point (Fy) us 56 ksi. D. Modulus of elasticity (E) is 29,000 ksi Which of the above statements are true?
- A.
A. A,B,C
- B.
B. A,B,D
- C.
C. B,C,D
- D.
D. A,B,C,D
Correct Answer
B. B. A,B,DExplanation
The correct answer is B. A,B,D. This means that statements A, B, and D are true. Statement A states that the maximum allowable stress in shear for A36 steel is 14.5 ksi, which is true. Statement B states that the maximum allowable stress for bending is 24 ksi, which is also true. Statement D states that the modulus of elasticity for A36 steel is 29,000 ksi, which is true as well. Therefore, the correct answer is B. A,B,D.Rate this question:
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- 19.
1. The required strength (U) to resist dead load D and live load L shall be at least equal to (NSCP Sec. 5.9.2)
- A.
A. 1.4 DL + 1.7 LL
- B.
B. 0.9 DL + 1.3 LL
- C.
C. 1.4 DL + 1.4 LL
- D.
D. 1.5 DL + 1.87 LL
Correct Answer
A. A. 1.4 DL + 1.7 LLExplanation
The required strength (U) to resist dead load D and live load L shall be at least equal to 1.4 times the dead load (DL) plus 1.7 times the live load (LL). This is based on the NSCP (National Structural Code of the Philippines) Section 5.9.2. The code specifies the minimum strength requirements for structures to safely withstand the combined effects of dead and live loads. By multiplying the dead load by 1.4 and the live load by 1.7, the resulting strength ensures an adequate safety factor for the structure.Rate this question:
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- 20.
1. What is a design analysis requirement, considered as basis for the structural design of buildings and structures where the total lateral forces are distributed to the various vertical elements of the lateral force resisting system in proportion to their rigidities considering the rigidity of the horizontal bracing system or diaphragm? (NSCP Sec. 2.2.5.5)
- A.
A. Shear and moment diagram
- B.
B. Distribution of horizontal shear
- C.
C. Stability against overturning
- D.
D. Horizontal-torsional moments
Correct Answer
A. A. Shear and moment diagramExplanation
The design analysis requirement mentioned in the question refers to the distribution of total lateral forces to the various vertical elements of the lateral force resisting system in proportion to their rigidities. This distribution is based on the rigidity of the horizontal bracing system or diaphragm. Shear and moment diagrams are graphical representations that show the distribution of shear and moment forces along a structural element. Therefore, the shear and moment diagram is the correct answer as it provides the necessary information for designing the structural elements to resist lateral forces.Rate this question:
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- 21.
1. A continuous type of spread footing that supports vertical load the weight of the wall itself, and the weight of the footing.
- A.
A. Wall footing
- B.
B. Mat foundation
- C.
C. Isolated pad footing
- D.
D. Combined footing
- E.
E. Cantilever
Correct Answer
A. A. Wall footingExplanation
A wall footing is a type of continuous spread footing that supports the weight of the wall, as well as the weight of the footing itself. It is designed to distribute the vertical load evenly across the foundation, providing stability and preventing settlement. This type of footing is commonly used for load-bearing walls in buildings and structures.Rate this question:
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- 22.
1. A structural member of a horizontal bracing system that takes axial tension or compression. It is parallel to the applied load that collects and transfers shear to the vertical resisting elements or distributes loads within the horizontal bracing system. (NSCP Sec. 2.2.2)
- A.
A. Diaphragm strut
- B.
B. Collector
- C.
C. Diaphragm chord
- D.
D. Braced frame
Correct Answer
A. A. DiapHragm strutExplanation
A diaphragm strut is a structural member that takes axial tension or compression in a horizontal bracing system. It is parallel to the applied load and is responsible for collecting and transferring shear to the vertical resisting elements or distributing loads within the horizontal bracing system. This explanation aligns with the definition provided in the NSCP Sec. 2.2.2.Rate this question:
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- 23.
1. A horizontal or nearly horizontal system, including horizontal bracing systems, that act to transmit lateral forces to the vertical resisting elements. (NSCP Sec. 2.2.2)
- A.
A. diaphragm
- B.
B. truss
- C.
C. braced frame
- D.
D. platform
Correct Answer
A. A. diapHragmExplanation
A diaphragm is a horizontal or nearly horizontal system that transmits lateral forces to the vertical resisting elements. It includes horizontal bracing systems and helps to distribute the lateral loads evenly across the structure. This explanation is supported by the reference to NSCP Sec. 2.2.2.Rate this question:
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