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EQ 1 > RCA > Flashcards

RCA Flashcards

1
Q

Steps in WORKING STRESS DESIGN METHOD

A

STEP 1: Convert the steel area into an equivalent concrete area

STEP 2: Determine 𝒙 by the by moment of area

STEP 3: Compute the allowable moment corresponding to the stress limit for concrete

STEP 4: Compute the allowable moment corresponding to the stress limit for steel

STEP 5: The smaller value of 𝑀(π‘Žπ‘™π‘™π‘œπ‘€) will govern

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2
Q

Modular elasticity/ratio formula

A

𝑛=𝐸𝑠/𝐸𝑐

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3
Q

STEP 2 formula (Determine 𝒙 by the by moment of area)

A

𝑏π‘₯ (π‘₯/2) =𝑛𝐴𝑠(π‘‘βˆ’π‘₯)

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4
Q

Allowable moment corresponding to the stress limit for concrete

A

π‘€π‘Žπ‘™π‘™π‘œπ‘€ = 𝐢 (π‘‘βˆ’1/3 π‘₯)
where: 𝐢= 1/2 𝑏π‘₯(0.45𝑓′𝑐)

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5
Q

Allowable moment corresponding to the stress limit for steel

A

π‘€π‘Žπ‘™π‘™π‘œπ‘€ = 𝑇 (π‘‘βˆ’1/3 π‘₯)
where: 𝑇 = 𝑓𝑠𝐴𝑠

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6
Q

_______ is accurate only for elastic materials and is considered an obsolete design philosophy for RC.

A

WSD (Working Stress Design)

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7
Q

USD (Ultimate Stress Design) is also called

A

Strength Design

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8
Q

WSD (Working Stress Design)

A

Alternate Design Method

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9
Q

NSCP 2015 emphasizes the use of _____

A

USD

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10
Q

Based on the empirical evidence the stress-strain relationship of both concrete and steel is _______________

A

inelastic

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11
Q

USD Basic assumptions (for concrete)

A

Concrete on the tension side is assumed to be cracked at ultimate strength. Strain at the extreme concrete fiber is 0.003 at ultimate strength.

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12
Q

Strain at the extreme concrete fiber is ____ at ultimate strength.

A

0.003

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13
Q

USD Basic assumptions (for steel)

A

Steel stress is πˆπ’”=𝑬𝒔𝝐𝒔 when the strain is less than the yield strain.

Steel stress is πˆπ’”=πˆπ’š when the strain is greater than the yield strain.

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14
Q

Capacity point is reached when the strain in the farthest concrete fiber reaches 0.003. At such point, the stress in the equivalent uniform concrete stress block can be assumed to be ____ of the concrete compressive strength from tests

A

85%

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15
Q

In LRFD, load effects are enlarged by _________; while resistance or capacity is reduced by ___________. The separate set of factors for load effects and capacity recognizes the differing statistical nature of each.

A

-load factors
-strength factors

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16
Q

In ASD, the primary design parameter is the ______________.

A

maximum stress

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17
Q

In ASD, The applied stress is compared to nominal strength divided by a safety factor greater than ____. The safety factor varies depending on the member designed and other conditions.

A

1

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18
Q

______ is the newer and more empirically accurate philosophy

A

LRFD

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19
Q

____ usually results in a more conservative (but less economical) design than _____

A
  • ASD
  • LRFD
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20
Q

Inputs in structural analysis

A

E
I
other member properties
loads
constraints

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21
Q

Outputs in structural analysis

A

Reactions
Member Forces (Mu, Vu, etc.)
Node Displacements
Members deformations

22
Q

Main components of Structural analysis

A

Modelling
Loading
Calculation

23
Q

loads due to the weight of the permanent components

A

Dead Loads

24
Q

loads due to the weight of temporary or moving components

A

Live Loads

25
Q

loads due to pressure and vibration caused by wind

A

Wind Loads

26
Q

loads induced forces due to ground motion

A

EQ loads

27
Q

loads due to soil pressure

A

Soil Lateral Loads

28
Q

loads due to the accumulation of water from rain

A

Rain Loads

29
Q

hydrostatic and hydrodynamic loads due to surface runoff

A

Flood Loads

30
Q

The designer should first consider the ____________ according to the intended use of the area.

A

actual loading conditions

31
Q

In NSCP 2001, the strength reduction factor for axial and moment is simply ____.

A

0.90

32
Q

In NSCP 2001, compression-controlled design is simply disallowed by requiring a steel ratio of not exceeding ____ the balanced steel ratio

A

75%

33
Q

Starting NSCP 2010, compression-controlled design was permitted but with a more conservative strength reduction factor of _____.

A

0.65

34
Q

in first-order analysis, the structural analysis assuming P-delta effects are _________

A

negligible

35
Q

In second-order analysis, the structural analysis assuming P-delta effects are _______

A

significant

36
Q

the phenomenon of force and displacement magnification due to the change of geometry of the member

A

P-delta effect

37
Q

The code usually refers to P-delta effects as _______________

A

β€œslenderness effects”

38
Q

effects are caused by lateral forces

A

P delta (π‘·βˆ’βˆ†)

39
Q

effects are caused by geometric imperfections of the member

A

P sigma (π‘·βˆ’πœΉ)

40
Q

Examples of RC frames β€œbraced against sidesway”

A

Diagonal Bracing
Shear Wall

41
Q

If bracing resisting lateral movement of a story have a total stiffness of at least 12 times the gross lateral stiffness of the columns in the direction considered, it shall be permitted to consider columns within the story to be _______________

A

braced against sidesway

42
Q

formula of r (radius of gyration)

A

r = sqrt (Ig/Ag)

0.30 times the dimension in the direction stability is being considered for rectangular columns

0.25 times the diameter of circular columns

43
Q

The primary design aid to estimate the effective length factor k is the

A

Jackson and Moreland Alignment Charts

44
Q

Effective Length is

A

k

45
Q

provide a graphical determination of k for a column of constant cross section in a multi-bay frame

A

Jackson and Moreland Alignment Charts

46
Q

If slenderness effects are not permitted to be neglected, it is required to perform

A

a) Moment magnification method after first-order analysis OR

b) P-delta analysis (Second order analysis)

47
Q

Steps in Moment Magnification Method (Non-sway frames):

A

1) Calculate 𝑃𝑐 (6.6.4.4.2)
2) Determine πΆπ‘š (6.6.4.5.3)
3) Calculate 𝛿 (6.6.4.5.2)
4) Apply 𝛿 to 𝑀2 (6.6.4.5.1)

48
Q

Steps in Moment Magnification Method (Sway frames):

A

1) Calculate 𝑄 and 𝑃𝑐 (6.6.4.4)
2) Calculate 𝛿𝑠 (6.6.4.6.2)
3) Apply 𝛿𝑠 to 𝑀1 and 𝑀2 (6.6.4.6.1)

49
Q

In Second-order (P-delta) analysis, P-delta analysis requires _____________

A

iterative geometric updating

50
Q

Steps in Second-order (P-delta) analysis

A

Start > Modelling of structure and loads > Perform a first-order analysis > Geometry is altered due to deformation. Remodel the structure

EQ 1 (19 decks)

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