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OBC Building Structural practice questions

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Question 1Structural Design Basis, Loads, and Limit States

Which load type is classified as a dead load in a building's structural design basis?

  1. The weight of movable furniture and occupants
  2. The weight of temporary construction equipment
  3. The lateral pressure from wind gusts
  4. The self-weight of the floor slabs and fixed partitions Correct answer.

Explanation

Dead loads are the permanent, immovable weights of the structure itself and attached fixed elements such as slabs, beams, finishes, and fixed partitions. Their magnitude stays essentially constant throughout the structure's service life. Movable, temporary, and environmental effects are classified under other load categories.

Question 2Structural Design Basis, Loads, and Limit States

In limit states design, what does the strength (ultimate) limit state primarily ensure?

  1. That vibration stays within comfort thresholds
  2. That deflection remains visually acceptable
  3. That materials never crack under service loads
  4. That the structure resists the factored design loads without collapse Correct answer.

Explanation

The strength or ultimate limit state addresses safety by requiring that the factored resistance equals or exceeds the factored load effects, preventing collapse or structural failure. Serviceability concerns such as deflection, vibration, and cracking are handled by a separate limit state. Confusing the two is a common design-basis error.

Question 3Structural Design Basis, Loads, and Limit States

What is the main purpose of applying load factors in a limit states design format?

  1. To convert service loads into permanent dead loads
  2. To eliminate the need for resistance factors
  3. To reduce material quantities in the design
  4. To account for uncertainty and variability in load magnitudes Correct answer.

Explanation

Load factors magnify nominal loads to reflect the statistical variability and uncertainty in load estimation, so that the probability of exceeding the design load is acceptably low. They are paired with resistance factors that account for material and dimensional uncertainty. Together they calibrate the design to a target reliability.

Question 4Structural Design Basis, Loads, and Limit States

Which statement correctly describes a live load in accordance with typical building codes?

  1. It is a permanent load fixed for the life of the structure
  2. It varies in magnitude and location during normal occupancy Correct answer.
  3. It always acts vertically upward on the roof
  4. It excludes the weight of occupants

Explanation

Live loads are occupancy-related loads that change in magnitude and position, such as people, furniture, and stored goods. Because they are variable, codes specify design values by occupancy type and often require pattern loading for continuous members. They are distinct from permanent dead loads.

Question 5Structural Design Basis, Loads, and Limit States

A structure is designed so that a local failure does not propagate into a widespread collapse. Which design concept does this describe?

  1. Fatigue resistance
  2. Ductility demand
  3. Serviceability control
  4. Robustness against disproportionate collapse Correct answer.

Explanation

Robustness, or resistance to disproportionate collapse, ensures that damage from an accidental event remains localized rather than triggering progressive failure of the whole structure. It is often achieved through tying, continuity, and alternate load paths. Ductility supports this but is not the same concept.

Question 6Structural Design Basis, Loads, and Limit States

What does the term 'characteristic value' of a load represent in structural design?

  1. The absolute maximum load ever recorded
  2. A statistical value with a defined probability of not being exceeded Correct answer.
  3. The load at which first cracking occurs
  4. The average load over the design life

Explanation

A characteristic load is a statistical fractile chosen so that there is only a small defined probability that the actual load exceeds it. Design values are then derived by applying partial factors to this characteristic value. It is not simply the mean or an absolute maximum.

Question 7Structural Design Basis, Loads, and Limit States

Which combination best represents the serviceability limit state check for a floor beam?

  1. Checking column buckling under factored axial load
  2. Checking dead plus live load deflection against a span limit Correct answer.
  3. Checking connection weld strength
  4. Checking factored bending capacity

Explanation

Serviceability checks verify that the structure performs acceptably under normal service conditions, such as limiting deflection under service loads to a fraction of the span. Strength checks under factored loads belong to the ultimate limit state. Deflection control protects finishes and occupant comfort.

Question 8Structural Design Basis, Loads, and Limit States

Why do design codes require consideration of load combinations rather than a single load acting alone?

  1. Because single loads are always underestimated
  2. Because combinations reduce the required resistance factors
  3. Because dead loads are ignored in isolation
  4. Because loads rarely act simultaneously at their individual maxima Correct answer.

Explanation

Structures are subjected to several load types that vary independently, and it is improbable that all reach their peak at the same instant. Load combinations with accompanying factors capture realistic co-occurring effects. Designing for each load alone would either be unsafe or wasteful.

Question 9Structural Design Basis, Loads, and Limit States

A simply supported floor beam spans 8 m and carries a uniformly distributed service load of 20 kN/m. What is the maximum bending moment at midspan?

  1. 160 kN·m Correct answer.
  2. 640 kN·m
  3. 320 kN·m
  4. 80 kN·m

Explanation

For a simply supported beam with a uniformly distributed load, the maximum moment is wL²/8. Substituting gives 20 × 8² / 8 = 20 × 64 / 8 = 160 kN·m. This midspan moment is the basis for the flexural design of the section.

Question 10Structural Design Basis, Loads, and Limit States

A tributary area carries a uniformly distributed load of 5 kN/m². What total load acts on a column supporting a tributary area of 24 m²?

  1. 60 kN
  2. 120 kN Correct answer.
  3. 240 kN
  4. 29 kN

Explanation

The total load equals the distributed load multiplied by the tributary area, that is 5 kN/m² × 24 m² = 120 kN. The tributary area represents the floor region whose load is delivered to that column. This total is then used in the column design with appropriate load factors.

Question 11Structural Design Basis, Loads, and Limit States

A beam of span 6 m supports a point load of 30 kN at midspan. What is the maximum bending moment?

  1. 360 kN·m
  2. 90 kN·m
  3. 180 kN·m
  4. 45 kN·m Correct answer.

Explanation

For a simply supported beam with a central point load, the maximum moment is PL/4. Substituting gives 30 × 6 / 4 = 45 kN·m. The moment diagram is triangular, peaking at midspan under the load.

Question 12Structural Design Basis, Loads, and Limit States

A factored load combination yields a design bending moment of 200 kN·m. If the section's design moment resistance is 250 kN·m, what is the utilization ratio?

  1. 1.25
  2. 0.40
  3. 0.60
  4. 0.80 Correct answer.

Explanation

The utilization ratio is the design action divided by the design resistance, that is 200 / 250 = 0.80. A value below 1.0 indicates the section is adequate for the applied moment. Values above 1.0 would signal an overstressed section.

Question 13Structural Design Basis, Loads, and Limit States

Which factor primarily governs the choice of a partial safety factor for a permanent (dead) load?

  1. The wind speed at the site
  2. The number of floors in the building
  3. The uncertainty in its magnitude and the consequence of unfavorable deviation Correct answer.
  4. The color and finish of the structural material

Explanation

Partial factors for permanent loads reflect the statistical uncertainty in estimating self-weight and the seriousness of an unfavorable deviation. Because dead loads are relatively well defined, their factors are usually smaller than those for variable loads. Wind speed and building height affect other load factors.

Question 14Structural Design Basis, Loads, and Limit States

A designer must verify a cantilever slab against overturning. Which limit state is being checked?

  1. Fatigue limit state
  2. Ultimate limit state of static equilibrium Correct answer.
  3. Serviceability deflection
  4. Fire resistance limit state

Explanation

Overturning is a stability check belonging to the ultimate limit state of equilibrium, where destabilizing actions must not exceed stabilizing resistances by a defined margin. It is treated separately from strength checks because it concerns rigid-body stability. Deflection and fatigue are different concerns.

Question 15Structural Design Basis, Loads, and Limit States

What is the purpose of a load pattern (checkerboard) analysis for continuous beams?

  1. To compute the natural frequency of the floor
  2. To determine the maximum support reactions only
  3. To find the most adverse design actions by placing live load on selected spans Correct answer.
  4. To reduce the dead load applied to the beam

Explanation

Because live load may act on some spans and not others, pattern loading is used to find the worst-case bending moments and shears in continuous members. Placing live load on alternate spans can produce moments larger than a fully loaded arrangement. This governs the reinforcement design at critical sections.

Question 16Structural Design Basis, Loads, and Limit States

A building is located in a region with high seismicity. Which design basis concept determines the seismic design category?

  1. The color of the building's facade
  2. The combination of seismic hazard and the structure's occupancy and risk Correct answer.
  3. The number of elevators in the building
  4. The type of roof covering

Explanation

Seismic design categories are assigned based on the site's seismic hazard together with the structure's risk or occupancy importance. Higher categories impose stricter detailing and analysis requirements. Occupancy and hazard together drive the required level of seismic protection.

Question 17Structural Design Basis, Loads, and Limit States

A structure is required to remain operational immediately after a design earthquake. Which performance objective does this correspond to?

  1. Life safety
  2. Collapse prevention
  3. Immediate occupancy Correct answer.
  4. No performance objective

Explanation

Immediate occupancy is a performance objective requiring the building to remain usable with limited structural and non-structural damage after the design event. Life safety permits more damage but protects occupants, while collapse prevention is the least demanding. These objectives form a hierarchy of seismic performance.

Question 18Structural Design Basis, Loads, and Limit States

What does the term 'load duration factor' account for in timber design?

  1. That timber shrinks under sustained load
  2. That wood strength decreases over the building's life
  3. That wood strength increases for short-duration loads Correct answer.
  4. That dead loads dominate timber design

Explanation

Wood exhibits a duration-of-load effect in which its strength is higher under short-duration loads such as wind or impact than under permanent loads. Design codes apply a load duration factor greater than unity for brief events. This reflects the viscoelastic nature of wood.

Question 19Structural Design Basis, Loads, and Limit States

Why is the self-weight of a structure often estimated before member sizes are finalized?

  1. Because dead loads are excluded from load combinations
  2. Because self-weight is negligible in all buildings
  3. Because dead load is needed to size members, yet it depends on those sizes Correct answer.
  4. Because self-weight is treated as a live load

Explanation

Dead load depends on member dimensions, but those dimensions cannot be chosen until the total load is known, creating a circular dependency. Designers therefore assume preliminary member sizes, estimate self-weight, and iterate until convergence. This iterative process is standard in structural design.

Question 20Structural Design Basis, Loads, and Limit States

A wind load on a tall building produces both a base shear and an overturning moment. Which action is most critical for the foundation design?

  1. The base shear only
  2. The overturning moment and its associated uplift Correct answer.
  3. The building's natural frequency
  4. The roof deflection only

Explanation

Overturning moment from lateral wind creates varying axial forces in the foundations, including uplift on the windward side and increased compression on the leeward side. Foundation design must resist these axial effects in addition to shear. Base shear alone does not capture the uplift demand.

Question 21Structural Design Basis, Loads, and Limit States

In a load combination, why is the accompanying variable action reduced by a combination factor rather than taken at its full characteristic value?

  1. Because combination factors increase the dead load
  2. Because variable actions are ignored in ultimate limit states
  3. Because simultaneous occurrence of all variable actions at their peaks is improbable Correct answer.
  4. Because variable actions are always smaller than dead loads

Explanation

Combination factors reduce accompanying variable actions to reflect the low probability that all variable loads reach their characteristic values simultaneously. This produces a realistic and economical design envelope. The leading variable action is still taken at its full characteristic value.

Question 22Structural Design Basis, Loads, and Limit States

A designer is evaluating a structure for progressive collapse. Which strategy best provides an alternate load path if a column is removed?

  1. Reducing the reinforcement in the affected beams
  2. Providing continuity and tying between members Correct answer.
  3. Removing all lateral bracing near the column
  4. Increasing the dead load on the affected bay

Explanation

Alternate load path design ensures that if a column is lost, the loads can be redistributed through continuous and well-tied members. Continuity and tying allow the structure to bridge over the removed element. This is a core robustness strategy in progressive collapse design.

Question 23Structural Design Basis, Loads, and Limit States

Why must the design of a structure consider both ultimate and serviceability limit states even though they use different load levels?

  1. Because serviceability checks govern all member sizing
  2. Because the two limit states use identical load factors
  3. Because safety against collapse and acceptable in-service performance are separate requirements Correct answer.
  4. Because ultimate checks alone guarantee deflection control

Explanation

Ultimate limit state checks ensure safety against collapse under factored loads, while serviceability checks ensure the structure performs acceptably under normal service loads. A structure can satisfy one and fail the other, so both must be verified. They serve distinct and complementary purposes.

Question 24Structural Design Basis, Loads, and Limit States

A structure's resistance factor is reduced. What is the effect on the design, assuming the factored load effects are unchanged?

  1. The applied loads are reduced
  2. The dead load is reclassified as live load
  3. The required section capacity must increase Correct answer.
  4. The required section capacity may decrease

Explanation

A reduced resistance factor lowers the design resistance for a given section, so a larger or stronger section is needed to satisfy the inequality that design resistance exceeds design action. The factored loads are unaffected by this change. This illustrates the balance between load and resistance factors.

Question 25Structural Design Basis, Loads, and Limit States

Which term describes the magnitude of a load that has a specified probability of not being exceeded during a defined reference period?

  1. Design load
  2. Service load
  3. Characteristic load Correct answer.
  4. Nominal load

Explanation

The characteristic load is a statistical value chosen so that it has a small, defined probability of being exceeded within the reference period. Design values are then obtained by applying partial factors to this characteristic value. It is the basic reference magnitude from which factored loads are derived.

Question 26Structural Design Basis, Loads, and Limit States

Which limit state is concerned with deflection, cracking, and vibration that impair normal use of a structure?

  1. Serviceability limit state Correct answer.
  2. Accidental limit state
  3. Ultimate limit state
  4. Fatigue limit state

Explanation

The serviceability limit state addresses conditions that affect the function, appearance, or comfort of the structure under normal service conditions. Deflection, cracking, and vibration are typical serviceability criteria. Ultimate limit states instead concern safety against collapse or structural failure.

Question 27Structural Design Basis, Loads, and Limit States

A permanent action has a characteristic value of 40 kN and a partial safety factor of 1.35. What is its design value?

  1. 29.6 kN
  2. 54.0 kN Correct answer.
  3. 40.0 kN
  4. 45.4 kN

Explanation

The design value of an action is found by multiplying the characteristic value by the relevant partial safety factor. Here 40 kN multiplied by 1.35 gives 54.0 kN. This factored value is then used in the ultimate limit state load combination.

Question 28Structural Design Basis, Loads, and Limit States

In limit state design, what does the term 'design resistance' represent?

  1. The resistance computed from characteristic material strengths reduced by a partial factor Correct answer.
  2. The average resistance measured from laboratory tests on the material
  3. The resistance of the structure under its self-weight alone
  4. The resistance required to satisfy serviceability deflection limits

Explanation

Design resistance is obtained by dividing the characteristic resistance by a material partial safety factor, giving a conservative lower-bound strength. It accounts for variability in material properties and member geometry. The design check compares this resistance against the factored load effect.

Question 29Structural Design Basis, Loads, and Limit States

A floor carries a permanent action of 6 kN/m² and an imposed action of 4 kN/m². Using factors of 1.35 for permanent and 1.5 for imposed actions, what is the total design load per square metre?

  1. 10.0 kN/m²
  2. 13.0 kN/m²
  3. 14.1 kN/m² Correct answer.
  4. 12.6 kN/m²

Explanation

Each action is multiplied by its own partial factor before summing: 6 times 1.35 gives 8.1, and 4 times 1.5 gives 6.0. The total design load is therefore 14.1 kN/m². Factoring separately reflects the different variability of permanent and variable actions.

Question 30Structural Design Basis, Loads, and Limit States

Which of the following best describes the role of a combination factor applied to an accompanying variable action?

  1. It converts a variable action into a permanent action for the whole design life
  2. It increases the characteristic value to cover construction tolerances
  3. It ensures the structure remains serviceable after an earthquake
  4. It accounts for the reduced probability that several variable actions act simultaneously at their full values Correct answer.

Explanation

When several variable actions are combined, it is unlikely that all reach their characteristic values at the same moment. The combination factor reduces accompanying actions to reflect this lower joint probability. The leading variable action is generally kept at its full characteristic value.

Question 31Structural Design Basis, Loads, and Limit States

A cantilever beam of span 2 m carries a point load of 15 kN at its free end. What is the maximum bending moment at the support?

  1. 60 kN·m
  2. 7.5 kN·m
  3. 15 kN·m
  4. 30 kN·m Correct answer.

Explanation

For a cantilever with a point load at the free end, the maximum moment at the fixed support equals the load multiplied by the cantilever length. Thus 15 kN times 2 m gives 30 kN·m. The moment varies linearly from zero at the free end to this maximum at the support.

Question 32Structural Design Basis, Loads, and Limit States

A column supports a tributary area of 30 m². The roof carries a permanent load of 5 kN/m² and a snow load of 1.2 kN/m². Using factors of 1.35 and 1.5, what is the total design axial load on the column?

  1. 150.0 kN
  2. 202.5 kN
  3. 186.0 kN
  4. 256.5 kN Correct answer.

Explanation

The permanent load over the tributary area is 5 times 30 equals 150 kN, factored to 202.5 kN. The snow load is 1.2 times 30 equals 36 kN, factored to 54 kN. Summing the factored effects gives 256.5 kN total design axial load.

Question 33Structural Design Basis, Loads, and Limit States

Why is a reduced accompanying value used for wind when it is combined with a dominant imposed floor load in a gravity combination?

  1. Because wind loads never act on occupied buildings
  2. Because the likelihood of peak wind and peak imposed load occurring together is lower than either alone Correct answer.
  3. Because wind loads are treated as permanent actions in the combination
  4. Because wind loads are always smaller than imposed loads

Explanation

Combining full characteristic values of two independent variable actions would be overly conservative. The probability that both reach their peak simultaneously is lower than for either alone, so the accompanying action is reduced. This keeps the design economical while maintaining an acceptable reliability level.

Question 34Structural Design Basis, Loads, and Limit States

A structure is designed so that if one column is removed, the remaining members can redistribute the loads and prevent collapse. Which design concept does this illustrate?

  1. Alternate load path provision Correct answer.
  2. Base isolation of the structure
  3. Ductility-based energy dissipation
  4. Moment redistribution in continuous beams

Explanation

Providing an alternate load path allows loads to be carried by other members when a primary element is lost, preventing disproportionate collapse. This is a key strategy for robustness and progressive-collapse resistance. It is distinct from plastic redistribution of moments within a single member.

Question 35Structural Design Basis, Loads, and Limit States

A beam has a design moment resistance of 180 kN·m and is subjected to a factored moment of 150 kN·m. By what percentage is the utilization ratio relative to capacity?

  1. 120 percent
  2. 17 percent
  3. 83 percent Correct answer.
  4. 50 percent

Explanation

The utilization ratio is the applied design effect divided by the design resistance, here 150 divided by 180, which equals 0.83 or 83 percent. A value below 100 percent indicates the section is adequate. The remaining margin is about 17 percent of capacity.

Question 36Structural Design Basis, Loads, and Limit States

A designer reduces the material partial safety factor for a steel member while keeping the factored load effects unchanged. What is the effect on the design?

  1. The factored load effect increases proportionally
  2. The design resistance decreases and the member becomes less adequate
  3. The serviceability deflection limit becomes more stringent
  4. The design resistance increases and the member becomes more adequate Correct answer.

Explanation

Design resistance is the characteristic resistance divided by the material partial factor. Reducing that factor therefore raises the design resistance for the same characteristic strength. With unchanged load effects, the member gains capacity margin and becomes more adequate.

Question 37Structural Analysis and Design Methods

Which structural system resists lateral loads primarily through the bending stiffness of rigidly connected beams and columns?

  1. Shear wall system
  2. Flat plate slab system
  3. Moment-resisting frame Correct answer.
  4. Braced frame

Explanation

A moment-resisting frame relies on rigid beam-to-column joints so that lateral displacement produces bending moments and shears in the members, with stiffness governed by member sizes rather than diagonal bracing. Braced frames and shear walls resist lateral load through axial and in-plane shear action, which is a different mechanism.

Question 38Structural Analysis and Design Methods

In a simply supported beam carrying a uniformly distributed load over its full span, where does the maximum bending moment occur?

  1. At the point of contraflexure
  2. At midspan Correct answer.
  3. At the supports
  4. At quarter points

Explanation

For a uniformly loaded simply supported beam the moment diagram is parabolic, zero at both ends and reaching its peak at midspan where shear passes through zero. Moment is maximum where the shear force changes sign, which for symmetric loading is the center of the span.

Question 39Structural Analysis and Design Methods

What does the dead load of a structure represent in design calculations?

  1. The force from wind pressure on cladding
  2. The effect of temperature change on members
  3. The weight of occupants and movable furnishings
  4. The self-weight of permanent construction Correct answer.

Explanation

Dead load is the self-weight of all permanent materials, including structural members, finishes, partitions, and fixed services. Live, wind, and thermal effects are separate load categories with their own factors and combinations.

Question 40Structural Analysis and Design Methods

A cantilever beam of length L carries a concentrated load P at its free end. What is the maximum bending moment in the beam?

  1. Zero at the fixed end
  2. P times L Correct answer.
  3. P times L divided by 2
  4. P times L squared divided by 8

Explanation

For a cantilever with an end point load, the moment at the fixed support equals the load multiplied by its lever arm to that support, giving P times L, and it decreases linearly to zero at the free end. The other expressions correspond to different support or loading conditions.

Question 41Structural Analysis and Design Methods

Which statement correctly describes the purpose of load combinations in limit-states design?

  1. They eliminate loads that act only briefly
  2. They average all loads to obtain a single service condition
  3. They convert dynamic loads into static equivalents only
  4. They apply factored loads and compare effects with factored resistance Correct answer.

Explanation

Load combinations pair factored load effects with factored member resistance so that the probability of failure stays acceptably low across credible loading scenarios. Each combination represents a distinct governing condition, such as gravity alone or gravity plus wind, and all must be checked.

Question 42Structural Analysis and Design Methods

What is the primary function of a shear wall in a building?

  1. To reduce foundation settlement
  2. To provide in-plane lateral stiffness and strength Correct answer.
  3. To carry gravity loads from the roof only
  4. To support the floor slab in bending

Explanation

A shear wall is a vertical plate acting in its own plane to resist lateral forces, providing substantial stiffness and strength against wind and seismic demand. It may also carry gravity load, but that is not its distinguishing function.

Question 43Structural Analysis and Design Methods

In structural analysis, what does static indeterminacy of a structure mean?

  1. The members have identical stiffness
  2. The structure is unstable under lateral load
  3. The support reactions cannot be found from equilibrium alone Correct answer.
  4. The structure cannot carry any load

Explanation

A statically indeterminate structure has more unknown reactions or internal forces than the available equilibrium equations, so compatibility of deformations and member stiffness must also be used. Determinate structures, by contrast, can be solved from equilibrium alone.

Question 44Structural Analysis and Design Methods

A simply supported beam spans 6 m and carries a uniform load of 10 kN per meter, including self-weight. What is the maximum bending moment?

  1. 45 kN-m Correct answer.
  2. 90 kN-m
  3. 60 kN-m
  4. 30 kN-m

Explanation

The midspan moment of a simply supported beam under uniform load is w times L squared divided by eight. With w equal to 10 and L equal to 6, the result is 10 times 36 divided by 8, which is 45 kN-m.

Question 45Structural Analysis and Design Methods

Why is a rigid diaphragm assumption commonly used in building analysis?

  1. It removes the need to check column axial forces
  2. It lets floor inertial forces be distributed to vertical elements by relative stiffness Correct answer.
  3. It allows torsional effects to be ignored entirely
  4. It guarantees the building remains elastic during earthquakes

Explanation

When a floor slab is idealized as rigid in its own plane, it has only three degrees of freedom, and lateral forces are shared among vertical elements in proportion to their stiffness and position. This simplifies the model while capturing the dominant distribution of forces, including torsion from eccentricity.

Question 46Structural Analysis and Design Methods

A reinforced concrete beam is designed as under-reinforced. What does this mean for its failure mode?

  1. The beam fails in shear before flexure
  2. The beam remains elastic up to failure
  3. Concrete crushes before the steel yields
  4. Steel yields before the concrete crushes Correct answer.

Explanation

Under-reinforced sections contain less steel than the balanced ratio, so the tension steel reaches yield while the concrete is still below its crushing strain. This produces a ductile, gradual failure with warning deflection, which is the preferred design outcome.

Question 47Structural Analysis and Design Methods

In a steel moment frame, what is the purpose of the strong-column weak-beam design philosophy?

  1. To prevent any yielding in the structure
  2. To ensure columns yield before beams
  3. To force plastic hinges into beams rather than columns Correct answer.
  4. To minimize the weight of the beam sections

Explanation

By making column flexural capacity exceed that of the framing beams, plastic hinges form in the beams during severe seismic response, preserving overall frame stability. Column hinging can create a soft story and risk collapse, so it is deliberately discouraged.

Question 48Structural Analysis and Design Methods

A two-story building has a lateral stiffness of 20 kN per mm at the first story and 40 kN per mm at the second story. Where does the largest story drift tend to occur under a uniform lateral load pattern?

  1. At the first story Correct answer.
  2. Drift is zero at the first story
  3. At the second story
  4. Drift is equal at both stories

Explanation

Story drift equals the story shear divided by the story stiffness. Lower stories carry the cumulative shear from all stories above, and here the first story is also the less stiff, so its drift is larger than the second story's.

Question 49Structural Analysis and Design Methods

What is the main reason for providing a redundancy factor in seismic design provisions?

  1. To convert seismic demand to an equivalent static force
  2. To increase the building mass used in analysis
  3. To reduce the design base shear
  4. To account for the number of elements that must fail to cause instability Correct answer.

Explanation

Redundancy provisions recognize that a structural system with many load-resisting elements has reserve capacity if some yield or fail, while a system with few such elements is more vulnerable. The factor penalizes configurations lacking such reserve.

Question 50Structural Analysis and Design Methods

A simply supported beam has a point load of 20 kN at midspan and spans 4 m. What is the maximum shear force in the beam?

  1. 40 kN
  2. 20 kN
  3. 10 kN Correct answer.
  4. 5 kN

Explanation

By symmetry each support carries half the applied point load, so the reaction and the maximum shear are 10 kN. The shear diagram is constant at that value from each support to midspan, where it changes sign.