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Question 1Structural Systems
Which of the following is classified as a dead load on a building frame?
- Snow accumulated on a roof
- People standing in a corridor
- A permanently hung mechanical unit Correct answer.
- Furniture moved by occupants
Explanation
Dead loads are the permanent, fixed weights of the structure and the components attached to it for the life of the building. A hung mechanical unit is permanently attached and therefore acts as dead load. Occupants, movable furniture, and snow are variable and are treated as live or environmental loads.
Question 2Structural Systems
The section modulus S of a beam cross-section is defined as which ratio?
- Moment of inertia about the centroidal axis
- Moment of inertia over extreme-fiber distance Correct answer.
- Area times the section depth
- Extreme-fiber distance over moment of inertia
Explanation
Bending stress is computed as σ = M/S, and S equals the moment of inertia divided by the distance from the neutral axis to the extreme fiber, S = I/c. It therefore has units of length cubed. Simply using I alone neglects how far the extreme fiber sits from the neutral axis.
Question 3Structural Systems
A simply supported beam of span L carries a uniform load w over its full length. What is the maximum bending moment?
- wL/8
- wL²/8 Correct answer.
- wL²/4
- wL²/2
Explanation
For a simply supported span with a uniform load, the reactions are wL/2 at each end and the maximum moment occurs at midspan. Cutting the beam at midspan gives M = (wL/2)(L/2) − (wL/2)(L/4) = wL²/8.
Question 4Structural Systems
A cantilever beam of length L carries a uniform load w over its full length. Where does the maximum bending moment occur, and what is its magnitude?
- At midspan, wL²/8
- At the free end, zero
- At the fixed support, wL²/2 Correct answer.
- At midspan, wL²/24
Explanation
The bending moment in a uniformly loaded cantilever grows from zero at the free end to a maximum at the support. Summing moments about the cut at the support gives M = wL(L/2) = wL²/2, and the top fibers are in tension.
Question 5Structural Systems
Which foundation system carries load primarily through end bearing or skin friction at depth?
- Spread footing
- Slab on grade
- Driven pile Correct answer.
- Mat foundation
Explanation
Deep foundations such as driven piles bypass weak upper soils and transfer load to deeper strata through end bearing at the tip or friction along the shaft. Spread footings, mats, and slabs on grade are shallow systems that distribute load near the ground surface.
Question 6Structural Systems
For a retaining wall, which earth pressure condition produces the smallest lateral force on the wall?
- Active pressure Correct answer.
- Hydrostatic pressure
- At-rest pressure
- Passive pressure
Explanation
Active pressure develops after the wall yields slightly away from the backfill, allowing the soil mass to mobilise its shear strength and reduce the lateral thrust. At-rest pressure is larger because no movement is permitted, and passive pressure, which requires pushing into the soil, is the largest of the three.
Question 7Structural Systems
Concrete compressive strength f'c used in design is normally established from which test?
- Split-cylinder tension test at 7 days
- Flexural beam test at 14 days
- Cylinder compression test at 28 days Correct answer.
- Rebound hammer test at 28 days
Explanation
Design compressive strength is based on standard molded cylinders loaded in compression at 28 days, when the concrete has essentially reached its specified strength. Test results are averaged and compared against the specified f'c with acceptance criteria.
Question 8Structural Systems
The primary purpose of an expansion joint in a long building is to
- Increase the lateral stiffness of the frame
- Provide a continuous load path for wind uplift
- Allow plastic shrinkage of fresh concrete
- Accommodate movement from thermal expansion and contraction Correct answer.
Explanation
Expansion joints physically separate building segments so that thermally induced lengthening and shortening does not generate large restraint forces or cracking. The joint must remain open and free of rigid elements across its width. Shrinkage cracking is normally handled by contraction joints rather than expansion joints.
Question 9Structural Systems
Under allowable stress design, which load combination is used to check a member for gravity loads only?
- 0.9D + 1.0W
- 1.2D + 1.6L
- D + L Correct answer.
- D + 0.75L + 0.75W
Explanation
Allowable stress design uses service-level loads, and the basic gravity combination is dead plus live load, D + L. The factors 1.2 and 1.6 belong to strength design, while the other combinations include wind.
Question 10Structural Systems
Why does the building code permit a reduction of design live load on members with large tributary areas?
- Concrete gains strength after the design period
- Roof live loads are set by the manufacturer
- Full live load rarely covers the entire tributary area at once Correct answer.
- Live loads are always smaller than dead loads on the member
Explanation
Live load reduction reflects probability: it is unlikely that every square foot of a large tributary area carries its full design live load simultaneously. The permitted reduction increases with tributary area but is capped, and it does not apply to certain high-occupancy or storage areas.
Question 11Structural Systems
A floor beam supporting a plastered ceiling is typically limited to what deflection under live load?
- L/120
- L/240
- L/360 Correct answer.
- L/600
Explanation
Members supporting plaster or other brittle finishes are commonly limited to live-load deflection of L/360 to prevent cracking of the finish. Less sensitive construction may use L/240, and stricter limits such as L/600 are reserved for special cases.
Question 12Structural Systems
For a column that is pinned at both ends, the effective length factor K is
- 0.7
- 1.0 Correct answer.
- 2.0
- 0.5
Explanation
The effective length factor accounts for end restraint, and a column free to rotate but restrained against translation at both ends has K = 1.0. Fixed ends reduce K toward 0.5, while a free-standing cantilever column has K = 2.0.
Question 13Structural Systems
In a rectangular beam subjected to vertical shear, the maximum shear stress occurs at which location?
- At the neutral axis Correct answer.
- At the top fiber
- At one-quarter of the depth
- At the bottom fiber
Explanation
Shear stress varies parabolically over a rectangular section, following τ = VQ/(It). Because Q is largest at the neutral axis, the peak shear stress of 1.5V/A occurs there, while it falls to zero at the extreme fibers.
Question 14Structural Systems
A column carries a total service load of 200 kips and the allowable soil bearing pressure is 4 ksf. What is the minimum required footing area?
- 25 sf
- 80 sf
- 100 sf
- 50 sf Correct answer.
Explanation
Footing area must satisfy service load divided by allowable bearing pressure, A = P/q_allow. Here 200 kips divided by 4 ksf equals 50 square feet. The footing self-weight and any overburden are then added in the final check.
Question 15Structural Systems
At a truss joint with exactly two non-collinear members and no external load applied, the member forces are
- Equal to the larger joint load
- Zero Correct answer.
- Undetermined without member stiffness
- Equal and opposite in tension
Explanation
Summing forces in the direction perpendicular to one member leaves only the component of the second member, which must therefore be zero. With two non-collinear members and no external load, both members are zero-force members in that loading condition.
Question 16Structural Systems
The moment of inertia of an area about an axis parallel to its centroidal axis is obtained by
- Dividing the centroidal moment of inertia by the offset distance
- Adding the area to the centroidal moment of inertia
- Adding Ad² to the centroidal moment of inertia Correct answer.
- Multiplying the area by the offset distance
Explanation
The parallel-axis theorem states I = I_c + Ad², where d is the perpendicular distance between the centroidal axis and the parallel axis. The additional term accounts for the area being shifted away from the centroid and is always positive.
Question 17Structural Systems
Under allowable stress design for wood, the load duration factor CD for a roof subjected to snow load is
- 1.15 Correct answer.
- 1.0
- 1.6
- 1.25
Explanation
Wood exhibits higher strength under short-duration loading, and the duration factor reflects the cumulative time the load acts. Snow load uses CD = 1.15, whereas wind and seismic use 1.6, and permanent loads use 1.0.
Question 18Structural Systems
ASTM A992 steel, commonly used for wide-flange shapes, has a specified minimum yield stress of
- 60 ksi
- 65 ksi
- 36 ksi
- 50 ksi Correct answer.
Explanation
A992 is the standard specification for rolled structural shapes and has a minimum yield stress of 50 ksi with a tensile strength of 65 ksi. A36 steel, at 36 ksi yield, is used less often for new wide-flange work.
Question 19Structural Systems
ASCE 7 Exposure C applies to which terrain?
- Open terrain with scattered low obstructions Correct answer.
- Dense urban centers with tall buildings
- Large bodies of water
- Flat, unobstructed open country
Explanation
Exposure C covers open terrain with scattered obstructions generally under about 30 ft in height, such as grassland or scattered low buildings. Exposure B covers urban and wooded areas, and Exposure D covers flat, unobstructed water surfaces.
Question 20Structural Systems
A flexible diaphragm distributes lateral load to the vertical elements below in which manner?
- It cannot transfer load to shear walls
- In proportion to each element's stiffness
- Only through direct bearing on the foundation
- In proportion to each element's tributary area Correct answer.
Explanation
A flexible diaphragm deflects enough that the vertical resisting elements act largely independently, so load is assigned by tributary area rather than by stiffness. A rigid diaphragm, by contrast, distributes lateral load in proportion to relative stiffness and also resists torsion.
Question 21Structural Systems
In the ASCE 7 equivalent lateral force procedure, Cs is computed as SDS/(R/Ie) and need not exceed which value when T is less than or equal to TL?
- SDS/(R/Ie)
- 0.5 SDS
- 0.044 SDS Ie
- SD1/(T (R/Ie)) Correct answer.
Explanation
The base value of the seismic response coefficient is SDS/(R/Ie), but for structures with period T ≤ TL it need not exceed SD1/(T(R/Ie)), which controls for longer-period buildings. A lower-bound value of 0.044 SDS Ie must still be met.
Question 22Structural Systems
Under Rankine theory for a dry cohesionless backfill, the active earth pressure coefficient Ka equals
- (1 − sin φ)/(1 + sin φ) Correct answer.
- (1 − sin φ)/2
- 1 − sin φ
- tan²(45° + φ/2)
Explanation
Rankine active pressure uses Ka = tan²(45° − φ/2), which is algebraically identical to (1 − sin φ)/(1 + sin φ). The coefficient decreases as the friction angle increases, reflecting greater soil shear strength.
Question 23Structural Systems
For a compact steel wide-flange beam braced at intervals shorter than Lp, the nominal flexural strength Mn is governed by
- Yielding and the plastic moment Mp Correct answer.
- Local flange buckling
- Inelastic lateral-torsional buckling
- Elastic lateral-torsional buckling
Explanation
When the unbraced length is at most Lp and the section is compact, lateral-torsional buckling cannot occur before the full plastic moment Mp = FyZ is reached. Between Lp and Lr the strength falls on an inelastic buckling curve, and beyond Lr it follows the elastic curve.
Question 24Structural Systems
In a singly reinforced concrete beam, if the reinforcement ratio exceeds the balanced ratio, the section
- Develops its full plastic moment
- Becomes preferred for ductile design
- Fails in tension after the steel yields
- Fails in compression before the steel yields Correct answer.
Explanation
Above the balanced ratio the section is compression-controlled: the concrete reaches its crushing strain while the tension steel is still below yield. Such a failure is sudden and non-ductile, so the code limits the reinforcement ratio to a fraction of the balanced value.
Question 25Structural Systems
For a slender reinforced concrete column, the moment magnification factor δns accounts for
- Additional moment from axial load acting on the deflected shape Correct answer.
- Reduction in concrete modulus with age
- Stiffening effect of lateral bracing
- Creep of concrete under sustained axial load alone
Explanation
Slender columns deflect laterally under axial load, and the eccentricity of that load on the displaced shape produces a secondary, or P-Δ, moment. The factor δns = Cm/(1 − Pu/(0.75Pc)) magnifies the factored end moments to include this effect.
Question 26Structural Systems
Which of the following is classified as a dead load in structural design?
- Snow accumulated on a sloped roof
- Weight of a permanently installed mechanical unit suspended from the roof Correct answer.
- Floor live load produced by occupants
- Wind pressure acting on the building envelope
Explanation
Dead loads are the permanent, in-place weights of construction materials and permanently attached equipment or systems. Because the mechanical unit stays in place for the life of the building, its weight belongs in the dead load category even though it is not part of the structural frame.
Question 27Structural Systems
Floor live load in a building code is intended to account for:
- The weight of occupants, furniture, and movable equipment Correct answer.
- The self-weight of the structural frame and floor finishes
- Lateral forces generated by earthquakes
- Volume change caused by temperature fluctuations
Explanation
Live loads represent transient or movable weights that the occupancy imposes on the structure, including people, furnishings, and movable equipment. They are specified as uniformly distributed values tabulated by occupancy type, with concentrated load checks where appropriate.
Question 28Structural Systems
The primary function of a shear wall in a building is to:
- Provide thermal insulation at the building perimeter
- Reduce the dead load transmitted to the foundation
- Carry roof gravity load down to the footings
- Resist lateral loads such as wind and seismic forces Correct answer.
Explanation
A shear wall is a stiff, in-plane vertical element that collects diaphragm forces and transfers them to the foundation, limiting lateral drift. Its defining role is lateral load resistance through in-plane shear and bending, even though it may also carry gravity load.
Question 29Structural Systems
A bearing wall is best described as a wall that:
- Resists only lateral soil pressure from backfill
- Transfers load to the structure only through tension
- Supports vertical loads from the floors or roof above Correct answer.
- Carries no load other than its own self-weight
Explanation
Bearing walls are structural walls that receive and deliver gravity loads from the framing above down to the foundation. Masonry and light-frame walls can serve this role, and openings in them must be framed with headers to maintain the load path.
Question 30Structural Systems
For a simply supported beam carrying a uniformly distributed load over its full span, the maximum bending moment occurs at:
- The free end of the beam
- The supports
- Midspan Correct answer.
- The quarter points of the span
Explanation
On a simply supported span under uniform load, the shear diagram crosses zero at midspan, and bending moment is maximum where shear passes through zero. The moment diagram is parabolic, reaching wL squared divided by 8 at the center.
Question 31Structural Systems
Allowable soil bearing pressure used for footing design is most commonly expressed in:
- Pounds per linear foot
- Kips per square foot Correct answer.
- Foot-pounds of energy
- Kips per inch of settlement
Explanation
Bearing pressure is a stress, defined as force divided by contact area, so it is reported in force per unit area such as kips per square foot. This value is compared against the footing contact stress computed from service loads.
Question 32Structural Systems
A simple (shear) connection in structural steel is designed primarily to transfer:
- Axial tension only
- Torsion between the beam and its support
- Bending moment between the connected members
- Shear force, with negligible moment Correct answer.
Explanation
Simple framing connections such as shear tabs or double angles are detailed to permit rotation and are analyzed as transferring shear reaction only. Moment resistance is deliberately excluded from the design model of a simple connection.
Question 33Structural Systems
Concrete cover over reinforcing steel is provided primarily to:
- Protect the bars from corrosion and provide fire resistance Correct answer.
- Reduce the effective depth of the section
- Increase the dead load of the member for stability
- Eliminate the need for a development length
Explanation
Cover places a durable layer of concrete between the bar surface and the environment, slowing chloride and moisture ingress, and it insulates the steel from fire. Minimum cover values increase with exposure severity and with member size for fire-rating purposes.
Question 34Structural Systems
A two-way slab behaves as one-way when:
- It is cantilevered from walls on all four edges
- It carries only live load and no superimposed dead load
- It is supported on only two opposite sides, or its long-to-short span ratio is 2 or greater Correct answer.
- It is supported on four sides with a long-to-short span ratio less than 2
Explanation
When a panel is supported on two opposite edges only, load can travel in one direction, and when the aspect ratio reaches about 2 the shorter direction dominates so strongly that the panel is designed as one-way. Panels with aspect ratios below 2 deflect and carry load in both directions and are designed as two-way slabs.
Question 35Structural Systems
Shear studs welded to the top flange of a steel beam in a composite floor system are used to:
- Anchor the beam to the column below
- Limit beam deflection under live load
- Increase the fire rating of the steel section
- Develop composite action between the steel beam and the concrete slab Correct answer.
Explanation
Shear connectors transfer horizontal shear between the steel section and the hardened slab so the two act as a single flexural member. This composite action raises the effective moment of inertia and flexural strength compared with the bare steel beam.
Question 36Structural Systems
Gravity loads in a conventional building are delivered to the ground in which sequence?
- Foundation, column, beam, slab
- Slab, beam, column, foundation Correct answer.
- Beam, slab, column, foundation
- Column, foundation, beam, slab
Explanation
The load path begins at the surface where the load is applied, passes into the framing members that support that surface, and then into the vertical elements and finally the foundation. Every element must be capable of receiving the reaction from the element above it.
Question 37Structural Systems
The base plate of a steel column is provided to:
- Isolate the column from thermal expansion of the slab
- Increase the slenderness ratio used in column design
- Spread the column load over a larger area of the concrete footing Correct answer.
- Develop the full moment capacity of the column section
Explanation
The base plate enlarges the bearing area so the compressive stress on the concrete pedestal or footing stays below the allowable bearing value. Plate thickness is then designed for the bending produced by the upward concrete pressure.
Question 38Structural Systems
Weep holes or a drainage layer placed behind a retaining wall serve to:
- Permit the wall to expand under temperature change
- Increase the active earth pressure acting on the wall
- Reduce the bearing capacity of the soil beneath the footing
- Prevent the buildup of water pressure behind the wall Correct answer.
Explanation
Drainage collects and removes seepage so the backfill does not become saturated, which would add hydrostatic pressure on top of the earth pressure. Without drainage, the wall must be designed for a substantially larger total lateral force.
Question 39Structural Systems
Compared with a braced frame, the principal advantage of a moment-resisting frame is that it:
- Carries gravity loads without the use of columns
- Eliminates the need for a foundation system
- Provides greater lateral stiffness for the same weight of steel
- Leaves the bays open without diagonal bracing Correct answer.
Explanation
A moment frame resists lateral load through flexure in its beams and columns, so no diagonal members intrude into the bays. The tradeoff is lower stiffness and larger drift, which usually requires larger members than a braced frame would need.
Question 40Structural Systems
The development length of a reinforcing bar is the length required to:
- Extend the bar beyond the support face for fire protection
- Develop the bar's yield force through bond with the surrounding concrete Correct answer.
- Bend the bar into a standard hook
- Lap two bars of different diameters
Explanation
Development length is the embedment needed for the bar to reach its yield stress without pulling out or splitting the concrete. It depends on bar size, concrete strength, cover, spacing, and confinement by transverse reinforcement.
Question 41Structural Systems
In light-frame wood construction, a header over a window opening is provided to:
- Resist wind suction on the window glass
- Carry the loads from above around the opening into the adjacent wall Correct answer.
- Provide lateral bracing for the wall studs
- Increase the thermal insulation at the opening
Explanation
A header, or lintel, spans the opening and collects the tributary loads from the framing above so they can be delivered to jack studs on each side. Its size is governed by the span of the opening and by the loads it supports.
Question 42Structural Systems
Deep pile foundations are selected instead of spread footings when:
- Sound rock is present at the ground surface
- The upper soils are too weak to carry the loads within a practical footing area Correct answer.
- The water table lies well below the proposed footing elevation
- The building is only one story in height
Explanation
Piles carry load to deeper, competent strata or through friction when shallow soils cannot provide adequate bearing at acceptable settlement. A spread footing depends on the near-surface soil, so poor upper soils are the classic reason to go deep.
Question 43Structural Systems
For a simply supported beam under a uniform load, the maximum deflection is proportional to which power of the span?
- The fourth power of the span Correct answer.
- The square of the span
- The cube of the span
- The span itself
Explanation
The elastic deflection of a uniformly loaded simple beam is five wL to the fourth divided by 384 EI, so span enters to the fourth power. This strong sensitivity is why small increases in span demand disproportionately deeper members.
Question 44Structural Systems
In the equivalent lateral force procedure, the seismic base shear is distributed over the height of a building so that:
- Each floor receives an equal share of the base shear
- The lowest floor receives the largest share
- The share is uniform per unit of floor area
- Upper floors receive a larger share, increasing with height Correct answer.
Explanation
The vertical distribution approximates the first-mode response of the building, in which displacements and accelerations grow toward the top. The resulting story forces are therefore larger at upper levels, producing the greatest story shear at the base.
Question 45Structural Systems
In a braced frame, the diagonal members resist lateral load primarily through:
- Axial force in the braces Correct answer.
- Torsion of the beam-to-column joints
- Friction between the brace and the floor slab
- Bending of the braces about their weak axis
Explanation
A braced frame forms triangulated trusses in which lateral force is carried as tension or compression along the brace axis. Because axial stiffness is much greater than flexural stiffness, braced frames drift far less than moment frames of comparable weight.
Question 46Structural Systems
Second-order (P-delta) effects become most significant in a building that is:
- Loaded laterally only at its base
- Short, stiff, and lightly loaded
- Constructed entirely of short, stocky columns
- Tall, relatively flexible, and carrying large axial loads Correct answer.
Explanation
P-delta amplification grows with the product of axial load and lateral displacement, so tall flexible frames carrying heavy gravity load are most affected. The displaced geometry adds an additional moment that must be included in member design to avoid underestimating demands.
Question 47Structural Systems
A building is classified as having a torsional irregularity when:
- The story drift at one end is significantly larger than at the other end under equivalent lateral loading Correct answer.
- Its plan is symmetric about both principal axes
- It uses a single continuous floor diaphragm
- Its lateral force resisting elements are distributed on all four sides
Explanation
Torsional irregularity exists when the ratio of maximum to average story drift across the plan exceeds the code threshold, revealing that stiffness is unevenly distributed. In such cases the accidental torsion must be amplified and the structure designed for the resulting rotation.
Question 48Structural Systems
For a deep flexural member with a clear span less than about twice its effective depth, design is governed primarily by:
- Axial compression in the member
- Shear, analyzed with a strut-and-tie or nonlinear strain model Correct answer.
- Pure flexure with plane sections remaining plane
- Torsion of the member
Explanation
In a deep beam the span is short relative to the depth, so a large share of the load travels directly to the supports as diagonal compression rather than through flexural action. Conventional beam theory no longer predicts behavior well, so strut-and-tie modeling or nonlinear analysis is used.
Question 49Structural Systems
The Euler buckling load of a pin-ended column is proportional to which expression?
- EI divided by L
- EI divided by L squared Correct answer.
- EA divided by L squared
- EI divided by L to the fourth
Explanation
Euler buckling load equals pi squared times EI divided by the square of the effective length, so stiffness and length enter as shown. The strong inverse dependence on length explains why column capacity drops rapidly as unbraced length increases.
Question 50Structural Systems
In a special moment frame, capacity design intends that inelastic action occur in the:
- Base plates, which are designed to yield first
- Panel zones of the connections, which are left unreinforced
- Columns, which are made weaker than the beams
- Beams, so that plastic hinges form there rather than in the columns Correct answer.
Explanation
The strong-column weak-beam concept requires beam flexural strengths to be exceeded before column strengths, so plastic hinges form in the beams. Beam hinging produces a stable, ductile mechanism, whereas column hinging can create a story collapse mechanism.