CONSTRUCTION | Structural Design | 14 Questions (EASY-MODERATE)
STRUCTURAL DESIGN & CONCEPTUALIZATION
(14 Questions, Difficulty Level:EASY to MODERATE)
by Raison John J. Bassig
Q#1: It is an essentially vertical truss system that provides resistance to lateral loads and provides stability for the structural system.
a. Moment Frame
b. Space Frame
c. Gravity Frame
d. Braced Frame
According to the 2015 (7th Edition) of the National Structural Code of the Philippines (NSCP),
Moment Frame:
"MOMENT FRAME is a frame in which beams, slabs, columns, and joints resist forces predominantly shear, and axial force; Beams or slabs are predominantly horizontal or nearly horizontal; Columns are predominantly vertical or nearly vertical."
(Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)
"MOMENT FRAME refers to framing system that provides resistance to lateral loads and provides stability to the structural system, primarily by shear and flexure of the framing members and their connections."
(Chapter 5, Part 1, Definitions of the 2015 NSCP)
Space Frame:
"SPACE FRAME is a three-dimensional structural system, without bearing walls, composed of members interconnected so as to function as a complete self-contained unit with or without the aid of horizontal diaphragms or floor-bracing systems."
(Chapter 2, Definitions of the 2015 NSCP)
Gravity Frame:
"GRAVITY FRAME is a portion of the framing system not included in the lateral load resisting system."
(Chapter 5, Part 1, Definitions of the 2015 NSCP)
Braced Frame:
"BRACED FRAME is essentially a vertical truss system of the concentric or eccentric type that is provided to resist lateral forces."
(Chapter 2, Definitions of the 2015 NSCP)
"BRACED FRAME is an essentially vertical truss system that provides resistance to lateral loads and provides stability for the structural system."
(Chapter 5, Part 1, Definitions of the 2015 NSCP)
Therefore, the correct answer is d. Braced Frame.
Q#2: Concrete covers shall be provided for reinforced concrete for protection of reinforcement bars from rusting. For prestressed concrete cast-in-place slabs and walls in contact with the ground, what is the minimum specified concrete cover?
a. 40 mm
b. 75 mm
c. 25 mm
d. 1-1/2"
According to the 2015 (7th Edition) and 2010 (6th Edition) of the National Structural Code of the Philippines (NSCP),
(Chapter 4, Table 420.6.1.3.2 of the 2015 NSCP and Chapter 4, Sec. 407.8.3 of the 2010 NSCP)
Therefore, the correct answer is c. 25 mm.
Q#3: What is the level of indeterminacy of a beam with a fixed support at one end and a roller or hinge at the other end?
a. First degree
b. Second degree
c. Third degree
d. Fourth degree
The Level of Indeterminacy is simply given as:
Level of indeterminacy = Unknown Reactions - Known Equations
As beams are assumed to only have shear and flexural loads (with NO axial loads), the known equilibrium equations in beams would be 2, that is:
ΣV=0; and
ΣM=0
Recall that beams with Fixed Ends/Restrained Ends have 3 reactions (Horizontal, Vertical, and Moment), while beams with Simply-Supported Ends (such as Pinned/Hinged Ends or Roller Ends) have NO Moment reactions (where Pinned/Hinged supports have 2 reactions - Horizontal and Vertical; Roller supports have 1 reaction - Vertical). Since typical beams have NO axial load considerations (unless special forces are considered), horizontal reactions for such beams connections is 0.
So, in my given question, a beam with one fixed and and one hinged end is actually a PROPPED BEAM, where the fixed end would have 2 reactions (Vertical and Moment) and the hinged end would have 1 reaction (Vertical) - their horizontal reactions cancel out.
Since we have 2 known equations, then,
Level of indeterminacy = 3 Unknowns - 2 Equations = 1 (i.e., First degree)
Therefore, the level of indeterminacy is a. First degree.
Q#4: For normal weight of concrete, the modulus of elasticity of concrete (Ec) may be taken as:
a. 4,700√f'c MPa
b. 3,700√f'c MPa
c. 2,700√f'c MPa
d. 5,700√f'c psi
According to the National Structural Code of the Philippines (NSCP),
"For normal weight concrete Ec shall be permitted to be taken as 4700√f'c"
(Sec. 419.2.2.1.b of the 2015 NSCP and Sec. 408.6.1 of the 2010 NSCP)
Note that the value given in the NSCP of 4700√f'c is in MPa (metric units). The value used for Ec in psi (English units) is 57000√f'c (see ACI 318 Code).
Therefore, the correct answer is a. 4,700√f'c MPa (or 57,000√f'c psi).
Q#5: What is the minimum live load of the first floor for Group A (Residential) Dwellings?
a. 200 kg/m2
b. 150 kg/m2
c. 250 kg/m2
d. 120 kg/m2
According to the 1977 National Building Code of the Philippines (PD1096) and its 2004 Implementing Rules and Regulations (IRR) under Minimum Requirements for GROUP A Dwellings:
"The LIVE LOAD of the FIRST FLOOR shall be at least 200 KILOGRAMS PER SQ. METER and for the second floor, at least 150 kilograms per sq. meter."
(Sec. 708.f of PD1096 and Sec. 708.6 of the 2004 IRR of PD1096)
"The wind load for roofs shall be at least 120 kilograms per sq. meter for vertical projection."
(Sec. 708.g of PD1096 and Sec. 708.7 of the 2004 IRR of PD1096)
So, in my given choices above,
a. 200 kilograms per square meter ---> FIRST FLOOR minimum live load
b. 150 kilograms per square meter ---> SECOND FLOOR minimum live load
c. 250 kilograms per square meter ---> (no such provision in the code)
d. 120 kilograms per square meter ---> ROOF minimum wind load
Therefore, the correct answer is a. 200 kg/m2.
Q#6: The formula for strain.
a. Force / Area
b. Length / Deformation
c. Area / Force
d. Deformation / Length
Strain (ε) is a measure of deformation representing the displacement between particles in the body relative to a reference length. It is expressed as the ratio of total deformation or elongation (e) to the initial dimension or length (Lo) of the material body in which the forces are being applied.
ε = e / Lo where e = Change in Length = ΔL = Final Length (l) - Original Length (Lo)
Therefore, the correct answer is d. Deformation / Length.
Q#7: What is the Modulus of Elasticity of non-prestressed steel reinforcement (Es)?
a. 200,000 kN/mm2
b. 200,000,000,000 N/m2
c. 200 kN/m2
d. 200,000,000 N/cm2
According to the National Structural Code of the Philippines (NSCP),
"Modulus of elasticity, Es, for nonprestressed bars and wires shall be permitted to be taken as 200,000 MPa."
(Sec. 420.2.2.2 of the 2015 NSCP and Sec. 408.6.2 of the 2010 NSCP)
Recall that Pascal (Pa) is a unit of Stress (σ), which is descriptive of the Force acting over a unit Area (F/A). The SI unit for force is Newton (N), while the SI unit for area is the Square Meter (m2). So,
1 N / m2 = 1 Pa
Now, since MPa (Mega Pascals) is equal to 1,000,000 Pa, converting the given Es = 200,000 MPa to Pascals is simply,
Es = 200,000 MPa x (1,000,000 Pa / 1 MPa) = 200,000,000,000 Pa
And, since 1 Pa = 1 N/m2, we have
Es = 200,000,000,000 Pa = 200,000,000,000 N/m2 (Choice b.)
For additional information, the equivalent Es = 200,000 MPa converted into other units can be any of the following:
= 200,000 N/mm^2
= 20,000,000 N/cm^2
= 200,000,000,000 N/m^2 or 200,000,000,000 Pa
= 200,000,000 kN/m^2 or 200,000,000 kPa
= 200 GPa
= 2,000,000 bar (non-SI unit)
= 29,007.547546 ksi
= 29,007,547.546 psi
Therefore, the correct answer is b. 200,000,000,000 N/m2.
Q#8: Maximum moment of concentrated load on a simply supported beam is:
a. PL4 / EI
b. PL / 2
c. PL / 4
d. PL / 8
Refer to my attached drawings of typical beams and their maximum shear/moment locations -- depending on the load and type of beam.
Note that the question pertains to SIMPLY-SUPPORTED (and NOT fixed ends) and the load is a CONCENTRATED (POINT) LOAD (and NOT uniformly-distributed).
PL / 8 --> Max. Moment for Fixed-End Beams with Point Load
PL / 4 --> Max. Moment for Simply-Supported Beams with Point Load
Therefore, the correct answer is c. PL / 4.
Q#9: What is the minimum/nominal yield strength (fy) of "Structural Grade" reinforcement bars?
a. 33,000 psi
b. 40,000 psi
c. 50,000 psi
d. 60,000 psi
Grades of reinforcement bars are often based on ASTM standards as used in the United States. The most common grades of rebars are ASTM Grade 40 (which has an fy=40ksi) and ASTM Grade 60 (fy=60ksi). Higher grades are also available such as ASTM Grade 75 (fy=75ksi) and ASTM Grade 80 (fy=80ksi). Older/historical grades such as ASTM Grade 33 (fy=33ksi) are also available but not common.
In the Philippines, which uses a PNS standard that is also based on ASTM's metric unit equivalent, there are three common types of grades of rebars that are widely in use (though other special grades may also be available on special orders):
Q#10: According to the National Structural Code of the Philippines (NSCP), if the strain in concrete reaches _____, it begins to crack.
a. 0.0021
b. 0.003
c. 0.005
d. 0.021
According to the National Structural Code of the Philippines (NSCP),
"Maximum strain at the extreme concrete compression fiber shall be assumed equal to 0.003."
(Sec. 422.2.2.1 of the 2015 NSCP and Sec. 410.3.3 of the 2010 NSCP)
(Sec. 422.2.2.1 of the 2015 NSCP and Sec. 410.3.3 of the 2010 NSCP)
Q#11: You specified a concrete mix of 3,000psi. As per ACI and NSCP, what then can the Modulus of Elasticity of this concrete be taken at?
a. Approx. 16,830 MPa
b. Approx. 18,160 MPa
c. Approx. 23,090 MPa
d. Approx. 21,380 MPa
As we have determined in Q#4, the Modulus of Elasticity of Concrete, Ec, is given as [4,700√f'c MPa] based on the National Structural Code of the Philippines (NSCP) provisions (or [57,000√f'c psi] based on ACI provisions),
(Sec. 419.2.2.1.b of the 2015 NSCP and Sec. 408.6.1 of the 2010 NSCP)
In my question, f'c (i.e., the specified compressive strength of concrete), is given as 3,000 psi (english unit). Recall conversion of pressure units,
f'c = 3,000 psi x 1MPa / 145 psi = 20.7 MPa or
f'c = 3,000 psi x 0.006895 MPa = 20.7 MPa
Now solving Ec,
Ec = 4700 x √(20.7 MPa) = Approx. 21,383.708752225... MPa
Therefore, the correct answer is d. Approx. 21,380 MPa.
Q#12: Given a fixed end beam of length (L) and a uniformly-distributed load (w) across its entire length. Determine the moment at its support.
a. wL2 / 24
b. wL2 / 12
c. wL2 / 8
d. wL2 / 4
[wL2 / 24] (choice a.) pertains to the moment at MIDPOINT of a FIXED END beam with a UNIFORMLY-DISTRIBUTED load.
[wL2 / 8] (choice c.) pertains to the moment at MIDPOINT of a SIMPLY-SUPPORTED beam with a UNIFORMLY-DISTRIBUTED load. *(It also pertains (but not shown in my drawing below) to the moment at FIXED SUPPORT of a beam FIXED AT ONE END AND SIMPLY-SUPPORTED AT OTHER END with a UNIFORMLY-DISTRIBUTED load.)
[wL2 / 4] (choice d.) pertains to just HALF of the maximum moment at SUPPORT of a CANTILEVER beam with UNIFORMLY-DISTRIBUTED load *(Note that max moment of this beam is wL^2 / 2 at the support).
Since my question calls for a FIXED BEAM with UNIFORM LOAD, the moment at midspan for such beam is calculated as [wL2 / 24] (choice a.) while the moment at its fixed ends (supports) is calculated as [wL2 / 12] (choice d.) (note that the moment is greater in the supports than at midspan).
Refer to my sketch of shear and moment diagrams of different typical beam types:
Therefore, the correct answer is b. wL2 / 12.
Q#13: If a column's longitudinal reinforcement bars are to be enclosed by circular ties, what would be the minimum number of such bars?
a. 4
b. 6
c. 8
d. 10
Take note that circular ties are NOT the same as spirals. According to the National Structural Code of the Philippines (NSCP),
"For non-prestressed columns...xxx..., the minimum number of longitudinal bars shall be...xxx... FOUR WITHIN RECTANGULAR OR CIRCULAR TIES;"
(Sec. 410.7.3.1.b of the 2015 NSCP, 7th Ed.)
"Minimum number of longitudinal bars in compression members shall be 4 FOR BARS WITHIN RECTANGULAR OR CIRCULAR TIES, 3 for bars within triangular ties, and 6 for bars enclosed by sprials...xxx..."
(Sec. 410.10.2 of the 2010 NSCP, 6th Ed.)
Since my question calls for main bars with CIRCULAR TIES (i.e., NOT spirals), then, the minimum quantity of such bars may be four (4) - NOT six (6).
Therefore, the correct answer is a. 4.
Q#14: Choose the correct provision for the maximum size of fillet weld if thickness of material is 6mm and larger:
a. Thickness of material minus 10%
b. Thickness of material minus 20%
c. Thickness of material minus 5%
d. Thickness of material minus 2.0mm
According to the National Structural Code of the Philippines (NSCP),
"The maximum size of fillet welds of connected parts shall be:
1. Along edges of material less than 6mm thick, not greater than the thickness of the material.
2. Along edges of material 6mm or more in thickness, not greater than the thickness of the material minus 2mm,...xxx..."
(Sec. 510.2.2.2 of the 2015 NSCP, 7th Ed.)
Therefore, the correct answer is d. Thickness of material minus 2.0mm.
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