CONSTRUCTION | Structural Design | 10 Questions (EASY)

STRUCTURAL DESIGN & CONCEPTUALIZATION

(10 Questions, Difficulty Level: EASY)

by Raison John J. Bassig

Q#1: The formula for stress.
a. Deformation / Length
b. Area / Force
c. Length / Deformation
d. Force / Area
e. Overtime Work + Deadline + Delayed Salary + Nagging Boss

In simple terms, stress (σ) is the internal force (F) of particles of a continuous material exerting on each other over an area (A). It is expressed in Pascal (Pa) using SI Units or Pound-force per square inch (psi) using English Units. So,

σ = F / A

Therefore, the correct answer is d. Force / Area.

Q#2: Loads consisting of wind pressure and suctions, earthquake, rainwater on flat roof, flood, snow and forces caused by temperature differentials:
a. Live Loads
b. Impact Loads
c. Dead Loads
d. Environmental Loads

According to the 2015 (7th Edition) of the National Structural Code of the Philippines (NSCP),

"LIVE LOADS are those loads produced by the use and occupancy of the building or other structure and do NOT included dead load, construction load, or environmental loads."
(Sec. 202, Definitions of the 2015 NSCP)

"LOAD, LIVE is a load that is NOT permanently applied to a structure, but is likely to occur during the service life of the structure (EXCLUDING environmental loads); or loads meeting specific criteria found in the general building code; without load factors."
(Sec. 402.3, Terminology of the 2015 NSCP)

"IMPACT LOADS...xxx...involve unusual VIBRATION and IMPACT forces...xxx...for cranes...xxx...for heliport and helistop landing areas." (Sec. 206.3 of the 2015 NSCP)

"DEAD LOADS consist of the weight of all materials and fixed equipment incorporated into the building or other structure."
(Sec. 202, Definitions of the 2015 NSCP)

"LOAD, DEAD is the weight of the members, supported structure, and permanent attachments or accessories that are likely to be present on a structure in service; or loads meeting specific criteria found in the general building code; without load factors."
(Sec. 402.3, Terminology of the 2015 NSCP)

Note that Impact Loads are INCLUDED in Live Loads, while Environmental Loads are EXCLUDED in Live Loads.

"ENVIRONMENTAL LOAD consist of WIND pressure and suctions, EARTHQUAKE loads, RAINWATER in flat roof, SNOW and forces caused by TEMPERATURE differentials."
(Simplified Methods of Building Construction by Max Fajardo)

(Sec. 202, 206, and 402 of the 2015 NSCP, and Simplified Methods of Building Construction by Max Fajardo)

Therefore, the correct answer is d. Environmental Loads.

Q#3: Ratio of normal stress to corresponding strain for tensile and compressive stresses below proportional limit of material.
a. Modulus of Elasticity
b. Yield Strength
c. Hooke's Law
d. Poisson's Ratio

Hooke's Law, in physics, is a law which states that the Force (F) needed to compress or extend a spring by a certain distance (x) scales linearly with respect to such distance.

F = -kx
where k is the stiffness constant factor

Poisson's Ratio (ν) is the negative of the ratio of transverse strain (ε_trans) to axial strain (ε_strain).

ν = - [ε_trans / ε_strain]
where:
ε_trans is negative for axial tension, positive for axial compression
ε_strain is positive for axial tension, negative for axial compression

Yield Strength (f_y) is a specified minimum yield strength or yield point of reinforcement.

Modulus of Elasticity (E) is the ratio of normal stress (σ) to corresponding strain (ε).

E = σ / ε

Since Stress (σ) = Force (F) / Area (A), and Strain (ε) = Elongation (e) / Original Length (L_o), then,

E = [F/A] / [e/L_o]
E = FL_o / Ae

Therefore, the correct answer is a. Modulus of Elasticity.

Q#4: It is an instrument which measures the strong motions of an earthquake in the ground.
a. Accelerograph
b. Odometer
c. Tachometer
d. Tomograph

An accelerograph (or also known as a seismograph) is an instrument used to record the acceleration of the ground with respect to time due to strong ground motion. It is used to monitor the response of the structure to seismic movements (earthquakes).

An odometer is an instrument used to measure the distance travelled by a vehicle, usually expressed in kilometers (km).

A tachometer is an instrument used to measure the rotational speed of a shaft or disc, usually expressed in revolutions per minute (RPM).

A tomograph is an instrument used to sectioning (creating imagery by sections) by means of penetrating waves.

Therefore, the correct answer is a. Accelerograph.

Q#5: The square root of the quotient of the moment of inertia and the cross-sectional area will give you what value?
a. Section Modulus
b. Polar Moment of Inertia
c. Slenderness Ratio
d. Radius of Gyration

Section Modulus is a geometric property, dependent on the shape or section, used in designing flexural members for shear and tension (considering area), for compression (considering radius of gyration), and for stiffness (considering moment of inertia and polar moment of inertia). Generally, the Elastic Section Modulus (S) is used where,

S = I / y
where:
I is the Area Moment of Inertia (or Second Moment of Inertia)
y is the distance of the neutral axis from any given fiber

Polar Moment of Inertia (I_z) is the resistance to torsional deformation of a body in relation to its shape, commonly round or circular shapes.

Slenderness Ratio is the ratio of a column's effective length (L) to its least radius of gyration (R_g), which is a measure of a column's tendency to buckle.

Slenderness Ratio = L / R_g

Radius of Gyration (R_g) is used in estimating the stiffness of a column and is described as the distribution of cross-sectional area (A) in a column around its centroidal axis with the mass of the body (Area Moment of Inertia, I).

R_g² = I / A
R_g = √(I / A)

Therefore, the correct answer is d. Radius of Gyration.

Q#6: What is the Area Moment of Inertia of a square column with a cross-sectional area of 0.2025 sq.m.?
a. 8,437.50 m⁴
b. 3,417,187,500 mm⁴
c. 875,787,780,761,718,750 mm⁴
d. 41,006,250,000 mm⁴

Recall the formula for the Area Moment of Inertia (I) for a SQUARE section,

I = b⁴ / 12 (for a square section)

Since area (A) of the square cross-section is given at 0.2025 sq.m., then, dimension of its side (b) is,

A = b²
A = 0.2025 sq.m. = 202,500 sq.mm.
b² = 202,500 sq.mm.
b = √(202,500 sq.mm.)
b = 450 mm.

Solving for Area Moment of Intertia (I),

I = (450 mm)⁴ / 12
I = 3,417,187,500 mm⁴

Therefore, the correct answer is b. 3,417,187,500 mm⁴.

Q#7: The length of frame element over which flexural yielding is intended to occur due to earthquake design displacements, extending not less than a distance (h) from the critical section where flexural yielding initiates.
a. Transfer Length
b. Plastic Hinge Region
c. Effective Depth of Section
d. Tension-Controlled Section

According to the 2010 (6th Edition) and 2015 (7th Edition) of the National Structural Code of the Philippines (NSCP),

"TRANSFER LENGTH is the length of embedded pretensioned strand/reinforcement required to transfer the effective prestress to the concrete."
(Chapter 4, Sec. 402, Definitions of the 2010 NSCP, and Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)

"PLASTIC HINGE REGION is the length of frame element over which flexural yielding is intended to occur due to earthquake design displacements, extending not less than a distance h from the critical section where flexural yielding initiates."
(Chapter 4, Sec. 402, Definitions of the 2010 NSCP, and Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)

"EFFECTIVE DEPTH OF SECTION (d) is the distance measured from extreme compression fiber to centroid of tension reinforcement."
(Chapter 4, Sec. 402, Definitions of the 2010 NSCP, and Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)

"TENSION-CONTROLLED SECTION is a cross-section in which the net tensile strain in the extreme tension steel at nominal strength is greater than or equal to 0.005."
(Chapter 4, Sec. 402, Definitions of the 2010 NSCP, and Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)

Therefore, the correct answer is b. Plastic Hinge Region (or Plastic Hinge Zone, PHZ).

Q#8: Loop of reinforcing bar or wire enclosing longitudinal reinforcement in compressive members.
a. Stirrup
b. Strap
c. Axle
d. Tie

According to the 2010 (6th Edition) and 2015 (7th Edition) of the National Structural Code of the Philippines (NSCP),

"STIRRUP is reinforcement used to resist shear and torsion stresses in a structural member; typically bars, wires, or welded wire fabric (plain or deformed) bent into L, U or rectangular shapes and located perpendicular to or at an angle to longitudinal reinforcement.  The term "stirrups" is usually applied to lateral reinforcement in flexural members and the term "ties" to those in compression members."
(Chapter 4, Sec. 402, Definitions of the 2010 NSCP)

"STIRRUP is an reinforcement used to resist shear and torsion stresses in a structural member; typically deformed bars, derformed wires, or welded wire reinforcement either single leg or bent into L, U or rectangular shapes and located perpendicular to or at an angle to longitudinal reinforcement."
(Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)

"TIE is a loop of reinforcing bar or wire enclosing longitudinal reinforcement; a continuously wound bar or wire in the form of a circle, rectangle, or other polygon shape without re-entrant corners is acceptable."
(Chapter 4, Sec. 402, Definitions of the 2010 NSCP, and Chapter 4, Sec. 402.3 Terminology of the 2015 NSCP)

Note that the keyword in my question is "COMPRESSIVE" members - NOT flexural members. A column is an example of a compressive member; while a beam is an example of a flexural member.

So, Ties are for Columns (Compression Members); Stirrups are for Beams (Flexural Members)

Therefore, the correct answer is d. Tie.

Q#9: Given a simply-supported beam with a point load at its center. What is the approximate shape of its moment diagram?
a. Triangle
b. Quadrilateral
c. 2 Square or Rectangle Shapes
d. Semi-Ellipse

A SEMI-ELLIPSE-shaped MOMENT diagram applies to SIMPLY-SUPPORTED beams with a UNIFORMLY-DISTRIBUTED LOAD. The SHEAR diagram will be triangular (sloped) and the MOMENT diagram will be curved (forming a semi-ellipse).

The shape of the SHEAR DIAGRAM of a SIMPLY-SUPPORTED beam with a CONCENTRATED LOAD is rectangular (vertical & horizontal) due to the reactions at the support (vertical), the lack of other loads along the beam (horizontal), up to the midpoint where the concentrated load is located (vertical).

Thus, the shape of the MOMENT DIAGRAM of the same beam and load is triangular (sloped) from zero moment at the supports, and proceeds linearly at a slope to the maximum moment at the midpoint where the concentrated load is located.

Refer to my sketch of shear and moment diagrams of different typical beam types:

Therefore, the correct answer is a. Triangle.

Q#10: In a fixed beam with uniformly-distributed load, where is/are the maximum moment/s located?
a. At the supports
b. At the midspan
c. At both the midspan and the supports
d. At point of contraflexure

If the beam is a FIXED BEAM with a CONCENTRATED LOAD (or "point load") located at its midspan, then, the maximum moment/s for that beam will be found at both its midspan and its fixed ends (its supports).

If the beam is onlya SIMPLY-SUPPORTED BEAM (i.e., NOT fixed), whether the load is uniformly-distributed along its entire length or the load is concentrated at its midspan, then, the maximum moment for such beam is onnly found at its midspan (there will be zero moment at the supports).

Since my question calls for a FIXED BEAM with UNIFORM LOAD, the moment at midspan for such beam is calculated as [wL² / 24] while the moment at its fixed ends (supports) is calculated as [wL² / 12] (greater than the moment at midspan). Hence, the maximum moment for fixed beams with uniform load is at its supports (or its fixed ends).

Refer to my sketch of shear and moment diagrams of different typical beam types:

Therefore, the correct answer is a. At the supports (or fixed ends).