National Structural Code Of The Philippines Trivia
(National Structural Code of the Philippines)
Preaped by:
Arch. /EnP. RSGabitan
Questions and Answers
- 1.
An open building is a structure having all walls at least _____% open. (NSCP 207)
- A.
50%
- B.
60%
- C.
80%
- D.
75%
Correct Answer
C. 80%Explanation
An open building is defined as a structure where at least 80% of the walls are open. This means that the building has a significant amount of open space, allowing for better ventilation and natural lighting. It is important for safety reasons and to comply with building codes that a certain percentage of the walls are open to ensure proper airflow and visibility within the structure. Having 80% of the walls open also promotes a more open and spacious environment within the building.Rate this question:
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- 2.
. Low rise buildings is an enclosed or partially enclosed with mean roof height less than or equal to? (NSCP 207.20)
- A.
70 M
- B.
50 M
- C.
15 M
- D.
18 M
Correct Answer
D. 18 MExplanation
The correct answer is 18 M. According to NSCP 207.20, low rise buildings are defined as enclosed or partially enclosed structures with a mean roof height less than or equal to 18 M.Rate this question:
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- 3.
The wind load importance factor lw for essential facilities is equal to? (NSCP 207.50)
- A.
1.15
- B.
1.0
- C.
2.15
- D.
0.87
Correct Answer
A. 1.15Explanation
The wind load importance factor lw for essential facilities is equal to 1.15. This factor is used to account for the importance of the facility in terms of its function and the potential consequences of failure due to wind loads. A higher importance factor indicates that the facility is more critical and therefore requires a higher level of design consideration and safety measures to withstand wind loads.Rate this question:
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- 4.
The wind load importance factor for hazardous facilities is equal to?
- A.
1.0
- B.
0.87
- C.
1.15
- D.
2.15
Correct Answer
C. 1.15Explanation
The wind load importance factor for hazardous facilities is equal to 1.15. This means that the wind load on hazardous facilities is considered more important and significant compared to other structures. The importance factor takes into account the potential risks and consequences associated with hazardous facilities, such as chemical plants or nuclear power plants, and increases the wind load to ensure the structures are designed to withstand severe weather conditions.Rate this question:
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- 5.
The wind load importance factor for standard occupancy structures is equal to?
- A.
1.0
- B.
1.15
- C.
0.87
- D.
2.0
Correct Answer
A. 1.0Explanation
The wind load importance factor for standard occupancy structures is equal to 1.0. This means that the wind load on these structures is considered to be of normal importance. The importance factor is used to account for the level of consequence or risk associated with a particular structure. In this case, standard occupancy structures are not considered to have any additional importance or risk factors related to wind loads, hence the importance factor is 1.0.Rate this question:
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- 6.
The wind load importance factor for miscellaneous structures is equal to?
- A.
1.0
- B.
1.15
- C.
0.87
- D.
2.0
Correct Answer
C. 0.87Explanation
The wind load importance factor for miscellaneous structures is equal to 0.87. This factor is used to account for the importance or significance of the structure in relation to wind loads. A value of 0.87 suggests that the structure is considered to have a moderate importance in terms of withstanding wind loads.Rate this question:
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- 7.
Large city centers with at least 50% of the buildings having a height greater than 21M. falls on what exposure category for wind loading? (NSCP 207.5.3)
- A.
Exposure A
- B.
Exposure B
- C.
Exposure C
- D.
Exposure D
Correct Answer
A. Exposure AExplanation
Large city centers with at least 50% of the buildings having a height greater than 21M fall under Exposure A for wind loading. This means that these areas are exposed to the highest wind speeds and have the highest potential for damage from wind.Rate this question:
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- 8.
Open terrain with scattered obstructions having heights less than 9M. Falls on what exposure category for wind loading?
- A.
Exposure A
- B.
Exposure B
- C.
Exposure C
- D.
Exposure D
Correct Answer
C. Exposure CExplanation
Exposure C is the correct answer because open terrain with scattered obstructions having heights less than 9M falls under this category for wind loading. This exposure category indicates that the terrain has a moderate amount of obstructions that can slightly reduce wind forces.Rate this question:
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- 9.
Flat unobstructed areas exposed to wind flowing over open water for a distance of at least 2 km falls on what exposure category for wind loading?
- A.
Exposure A
- B.
Exposure B
- C.
Exposure C
- D.
Exposure D
Correct Answer
D. Exposure DExplanation
Flat unobstructed areas exposed to wind flowing over open water for a distance of at least 2 km are categorized as Exposure D for wind loading. This means that the area is highly exposed to wind and experiences the highest wind loads. The absence of any obstructions or barriers allows the wind to flow freely and increases the intensity of the wind. Therefore, structures in this category need to be designed and built to withstand the highest wind forces.Rate this question:
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- 10.
Urban and suburban areas, wooded areas or other terrain with numerous closely spaced obstructions having the size of single family dwelling or larger falls on what exposure category for wind loading?
- A.
Exposure A
- B.
Exposure B
- C.
Exposure C
- D.
Exposure D
Correct Answer
B. Exposure BExplanation
Exposure B is the correct answer because urban and suburban areas, wooded areas, or other terrains with numerous closely spaced obstructions, such as single-family dwellings or larger structures, fall under this category for wind loading. This means that the area is moderately exposed to wind, which can be influenced by nearby obstructions.Rate this question:
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- 11.
Zone 1 of the Philippine map has a wind velocity of ____Kph?
- A.
150 Kph
- B.
250 Kph
- C.
125 Kph
- D.
200 Kph
Correct Answer
B. 250 Kph -
- 12.
Zone 3 of the Philippine map has a wind velocity of ____Kph?
- A.
200 Kph
- B.
125 Kp
- C.
150 Kph
- D.
250 Kph
Correct Answer
B. 125 KpExplanation
Zone 3 of the Philippine map has a wind velocity of 125 Kph. This means that in Zone 3, the average speed of the wind is 125 kilometers per hour.Rate this question:
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- 13.
In testing concrete laboratory cured specimens, no individual strength test (average of 2 cylinders) falls below fc’ by more than _______. (NSCP 405.7.3.3)
- A.
5 Mpa
- B.
4.25 Mpa
- C.
3.50 Mpa
- D.
4.0 Mpa
Correct Answer
C. 3.50 MpaExplanation
In testing concrete laboratory cured specimens, the average strength of 2 cylinders is taken as the individual strength test. According to NSCP 405.7.3.3, no individual strength test should fall below fc’ (the specified compressive strength of concrete) by more than a certain value. The correct answer is 3.50 Mpa, which means that no individual strength test should fall below fc’ by more than 3.50 Mpa.Rate this question:
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- 14.
Spacing for a lateral support for a beam shall not exceed _______times the least width b of compression flange or face. (NSCP 410.5.10)
- A.
40
- B.
50
- C.
60
- D.
30
Correct Answer
B. 50Explanation
According to NSCP 410.5.10, the spacing for a lateral support for a beam should not exceed 50 times the least width b of the compression flange or face. This means that the distance between each lateral support should not be more than 50 times the width of the compression flange or face. This is important to ensure the stability and strength of the beam, as proper lateral support helps prevent buckling and deformation under load.Rate this question:
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- 15.
For a rectangular reinforced concrete compression member, it shall be permitted to take the radius of gyration equal to _______times the overall dimension of the direction of stability is being considered. (NSCP 410.12.20)
- A.
0.30
- B.
0.45
- C.
0.50
- D.
0.75
Correct Answer
A. 0.30Explanation
In accordance with NSCP 410.12.20, for a rectangular reinforced concrete compression member, the radius of gyration can be taken as 0.30 times the overall dimension of the direction of stability being considered.Rate this question:
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- 16.
For members whose design is based on compressive force, the slenderness ratio kL/r preferably should not exceed ________? (NSCP 502.8.1)
- A.
300
- B.
250
- C.
200
- D.
350
Correct Answer
C. 200Explanation
According to NSCP 502.8.1, for members designed based on compressive force, the slenderness ratio kL/r should not exceed 200. This means that the length of the member (L) multiplied by the effective length factor (k) divided by the radius of gyration (r) should not be greater than 200. This is important to prevent buckling and ensure the stability of the member under compressive loads.Rate this question:
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- 17.
members whose design is based on tensile force, the slenderness ratio L/r preferably should not exceed _________.
- A.
200
- B.
300
- C.
250
- D.
350
Correct Answer
B. 300Explanation
The slenderness ratio L/r is a measure of the stability of a member under compressive load. When a member's design is based on tensile force, a higher slenderness ratio can be tolerated compared to when it is based on compressive force. Therefore, a higher value of L/r is preferred. Among the given options, the highest value is 350, which is greater than 300. Hence, the slenderness ratio L/r preferably should not exceed 300.Rate this question:
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- 18.
For pin connected members, the allowable stress on the net area of the pinhole for pin connected members is _________. (NSCP 504.4.1.1)
- A.
0.60 Fy
- B.
0.50 Fy
- C.
0.45 Fy
- D.
0.40 Fy
Correct Answer
C. 0.45 FyExplanation
The allowable stress on the net area of the pinhole for pin connected members is 0.45 Fy.Rate this question:
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- 19.
Other than pin connected members, the allowable tensile stress shall not exceed _______ on the gross area. (NSCP 504.2.1)
- A.
0.60 Fy
- B.
0.60 Fy
- C.
0.33 Fy
- D.
0.66 Fy
Correct Answer
A. 0.60 FyExplanation
The allowable tensile stress on the gross area, other than pin connected members, should not exceed 0.60 times the yield strength (Fy).Rate this question:
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- 20.
For pin connected plates, the minimum net area beyond the pinhole parallel to the axis of the member shall not be less than _______of the net area across the pinhole. (NSCP 504.4.2.1)
- A.
3/4
- B.
2/3
- C.
3/5
- D.
1/3
Correct Answer
B. 2/3Explanation
For pin connected plates, the minimum net area beyond the pinhole parallel to the axis of the member should be at least 2/3 of the net area across the pinhole. This means that the area of the plate beyond the pinhole should be larger than two-thirds of the area across the pinhole. This requirement ensures that there is enough material surrounding the pinhole to provide sufficient strength and stability to the connection.Rate this question:
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- 21.
For pin connected members in which the pin is expected to provide for relative movement between connected parts while under full load, the diameter of the pinhole shall not be more than ______mm greater than the diameter of the pin. (NSCP 504.4.2.3)
- A.
1 mm
- B.
0.5 mm
- C.
0.80 mm
- D.
2 mm
Correct Answer
C. 0.80 mmExplanation
The correct answer is 0.80 mm. According to NSCP 504.4.2.3, for pin connected members where the pin is expected to allow for relative movement between connected parts under full load, the diameter of the pinhole should not exceed 0.80 mm more than the diameter of the pin. This ensures that there is enough clearance for movement while still maintaining structural integrity.Rate this question:
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- 22.
The maximum longitudinal spacing of bolts, nuts and intermittent welds correctly two rolled shapes in contact for a built up section shall not exceed ________. (NSCP 505.5.4)
- A.
700 mm
- B.
500 mm
- C.
600 mm
- D.
400 mm
Correct Answer
C. 600 mmExplanation
The maximum longitudinal spacing of bolts, nuts, and intermittent welds for two rolled shapes in contact for a built-up section should not exceed 600 mm. This means that the distance between each bolt, nut, or intermittent weld should not be more than 600 mm. This requirement ensures that the connection between the two rolled shapes is strong and secure, preventing any potential failure or instability in the built-up section.Rate this question:
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- 23.
The ratio L/r for lacing bars arranged in single system shall not exceed ________. (NSCP 505.5.80)
- A.
140 mm
- B.
200 mm
- C.
250 mm
- D.
100 mm
Correct Answer
A. 140 mmExplanation
According to NSCP 505.5.80, the ratio L/r for lacing bars arranged in a single system should not exceed 140 mm. This means that the length of the lacing bars (L) divided by their radius (r) should not be greater than 140 mm. This limitation ensures that the lacing bars are properly sized and spaced to provide sufficient structural support and prevent failure.Rate this question:
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- 24.
The ratio L/r for lacing bars arranged in double system shall not exceed ________.
- A.
140 mm
- B.
200 mm
- C.
250 mm
- D.
100 mm
Correct Answer
B. 200 mmExplanation
The given question is asking for the maximum ratio L/r for lacing bars arranged in a double system. The correct answer is 200 mm. This means that the length L of the lacing bars should not exceed 200 times the radius r. This ratio is important to ensure the stability and strength of the lacing system. If the ratio exceeds 200 mm, it may compromise the structural integrity of the lacing bars.Rate this question:
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- 25.
For members bent about their strong or weak axes, members with compact sections where the flanges continuously connected to web the allowable bending stress is _________. (NSCP 506.4.1.1)
- A.
0.75 Fy
- B.
0.60 Fy
- C.
0.45 Fy
- D.
0.66 Fy
Correct Answer
D. 0.66 FyExplanation
The correct answer is 0.66 Fy. According to NSCP 506.4.1.1, for members with compact sections where the flanges continuously connected to the web, the allowable bending stress is 0.66 times the yield strength (Fy) of the material.Rate this question:
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- 26.
For box type and tabular textural members that meet the non compact section requirements of section 502.6, the allowable bending stress is ________. (NSCP 506.4.1.1)
- A.
0.75 Fy
- B.
0.60 Fy
- C.
0.45 Fy
- D.
0.66 Fy
Correct Answer
B. 0.60 FyExplanation
For box type and tabular textural members that meet the non-compact section requirements, the allowable bending stress is 0.60 Fy. This means that the maximum stress that can be applied to these members without causing failure is 0.60 times the yield strength (Fy) of the material. This value is determined based on the design standards specified in the NSCP 506.4.1.1.Rate this question:
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- 27.
Bolts and rivets connecting stiffness to the girder web shall be spaced not more than ______mm on centers. (NSCP 507.5.3)
- A.
300 mm
- B.
400 mm
- C.
350 mm
- D.
250 mm
Correct Answer
A. 300 mmExplanation
According to NSCP 507.5.3, the maximum spacing allowed for bolts and rivets connecting stiffness to the girder web is 300 mm. This means that the distance between each bolt or rivet should not exceed 300 mm.Rate this question:
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- 28.
Ira composite beam section, the actual section modulus of the transformed composite section shall be used in calculating the concrete flexural compressed stress and for construction without temporary shores, this stress shall be based upon loading applied after the concrete has reached _____% of its required strength.
- A.
50%
- B.
60%
- C.
80%
- D.
75%
Correct Answer
D. 75%Explanation
In a composite beam section, the actual section modulus of the transformed composite section is used to calculate the concrete flexural compressed stress. When constructing without temporary shores, this stress is based on loading applied after the concrete has reached 75% of its required strength. This means that the concrete needs to have achieved a certain level of strength before it can bear the intended load without the support of temporary shores.Rate this question:
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- 29.
Shear connectors shall have at least ________mm of lateral concrete covering. (NSCP 509.5.8)
- A.
50 mm
- B.
100 mm
- C.
25 mm
- D.
40 mm
Correct Answer
C. 25 mmExplanation
According to NSCP 509.5.8, shear connectors must have a minimum of 25 mm of lateral concrete covering. This means that the connectors should be embedded in the concrete by at least 25 mm to ensure proper structural integrity and load transfer between the concrete and the connectors.Rate this question:
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- 30.
The minimum center to center spacing of stud connectors along the longitudinal axis of supporting composite beam is ____________. (NSCP 509.5.8)
- A.
6 diameter of connector
- B.
10 diameter of connector
- C.
12 diameter of connector
- D.
5 diameter of connector
Correct Answer
A. 6 diameter of connectorExplanation
According to NSCP 509.5.8, the minimum center to center spacing of stud connectors along the longitudinal axis of a supporting composite beam is 6 times the diameter of the connector.Rate this question:
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- 31.
The maximum center to center spacing of stud connectors along the longitudinal axis of supporting composite beam is ____________. (NSCP 509.5.8)
- A.
8 diameter of connector
- B.
6 diameter of connector
- C.
5 diameter of connector
- D.
10 diameter of connector
Correct Answer
A. 8 diameter of connectorExplanation
The maximum center to center spacing of stud connectors along the longitudinal axis of a supporting composite beam is 8 times the diameter of the connector. This means that the distance between each stud connector should be no more than 8 times the diameter of the connector. This requirement ensures proper load transfer and prevents excessive deflection or failure of the composite beam.Rate this question:
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- 32.
Connections carrying calculated stresses, except for lacing, sag bars and girts, shall be designed to support not less than ________Kn of force. (NSCP 510.10.61)
- A.
30
- B.
50
- C.
26.70
- D.
35
Correct Answer
C. 26.70Explanation
Connections carrying calculated stresses, except for lacing, sag bars, and girts, should be designed to support a minimum force of 26.70 kilonewtons (kN). This requirement ensures that the connections can withstand the calculated stresses and maintain the structural integrity of the system. It is important to design connections properly to prevent failure and ensure the safety of the structure.Rate this question:
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- 33.
The connections at ends of tension or compression members in trusses shall develop the force due to the design load, but no less than _______% at the effective strength of the member unless a smaller percentage is justified by engineering analysis that considers other factors including handling, shipping and erection. (NSCP 510.2.5.1)
- A.
50
- B.
70
- C.
65
- D.
100
Correct Answer
A. 50Explanation
The connections at the ends of tension or compression members in trusses need to develop a force that is equal to or greater than a certain percentage of the effective strength of the member. According to NSCP 510.2.5.1, this percentage is not allowed to be less than 50%, unless there is a valid engineering analysis that justifies a smaller percentage. This analysis should take into account factors such as handling, shipping, and erection.Rate this question:
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- 34.
When formed steel decking is a part of the composite beam, the spacing of stud shear connector along the length of the supporting beam or girder shall not exceed _______mm. (NSCP 509.6.1.2)
- A.
800
- B.
750
- C.
900
- D.
1000
Correct Answer
C. 900Explanation
The correct answer is 900. According to NSCP 509.6.1.2, when formed steel decking is a part of the composite beam, the spacing of stud shear connector along the length of the supporting beam or girder should not exceed 900mm.Rate this question:
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- 35.
The minimum sizes of filler weld for plates with thickness greater than 20mm is ________. (NSCP 510.3.3.2)
- A.
10
- B.
8
- C.
15
- D.
20
Correct Answer
B. 8Explanation
According to NSCP 510.3.3.2, the minimum size of filler weld for plates with thickness greater than 20mm is 8.Rate this question:
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- 36.
The minimum sizes of filler weld for plates with thickness 6mm is ________? (NSCP 510.3.3.2)
- A.
3 mm
- B.
6 mm
- C.
8 mm
- D.
10 mm
Correct Answer
A. 3 mmExplanation
According to NSCP 510.3.3.2, the minimum size of filler weld for plates with a thickness of 6mm is 3mm.Rate this question:
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- 37.
The minimum sizes of filler weld for plates with thickness over 12mm to 20mm is ________? (NSCP 510.3.3.2)
- A.
8 mm
- B.
10 mm
- C.
6 mm
- D.
5 mm
Correct Answer
C. 6 mmExplanation
According to NSCP 510.3.3.2, the minimum size of filler weld for plates with thickness over 12mm to 20mm is 6 mm.Rate this question:
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Mar 21, 2023Quiz Edited by
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