RCA

Question 1

Steps in WORKING STRESS DESIGN METHOD

Answer 1

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(𝑎𝑙𝑙𝑜𝑤) will govern

Question 2

Modular elasticity/ratio formula

Answer 2

𝑛=𝐸𝑠/𝐸𝑐

Question 3

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

Answer 3

𝑏𝑥 (𝑥/2) =𝑛𝐴𝑠(𝑑−𝑥)

Question 4

Allowable moment corresponding to the stress limit for concrete

Answer 4

𝑀𝑎𝑙𝑙𝑜𝑤 = 𝐶 (𝑑−1/3 𝑥)
where: 𝐶= 1/2 𝑏𝑥(0.45𝑓′𝑐)

Question 5

Allowable moment corresponding to the stress limit for steel

Answer 5

𝑀𝑎𝑙𝑙𝑜𝑤 = 𝑇 (𝑑−1/3 𝑥)
where: 𝑇 = 𝑓𝑠𝐴𝑠

Question 6

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

Answer 6

WSD (Working Stress Design)

Question 7

USD (Ultimate Stress Design) is also called

Answer 7

Strength Design

Question 8

WSD (Working Stress Design)

Answer 8

Alternate Design Method

Question 9

NSCP 2015 emphasizes the use of _____

Answer 9

USD

Question 10

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

Answer 10

inelastic

Question 11

USD Basic assumptions (for concrete)

Answer 11

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.

Question 12

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

Answer 12

0.003

Question 13

USD Basic assumptions (for steel)

Answer 13

Steel stress is 𝝈𝒔=𝑬𝒔𝝐𝒔 when the strain is less than the yield strain.

Steel stress is 𝝈𝒔=𝝈𝒚 when the strain is greater than the yield strain.

Question 14

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

Answer 14

85%

Question 15

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.

Answer 15

-load factors
-strength factors

Question 16

In ASD, the primary design parameter is the ______________.

Answer 16

maximum stress

Question 17

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.

Answer 17

1

Question 18

______ is the newer and more empirically accurate philosophy

Answer 18

LRFD

Question 19

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

Answer 19

• ASD
• LRFD

Question 20

Inputs in structural analysis

Answer 20

E
I
other member properties
loads
constraints

Question 21

Outputs in structural analysis

Answer 21

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

Question 22

Main components of Structural analysis

Answer 22

Modelling
Loading
Calculation

Question 23

loads due to the weight of the permanent components

Answer 23

Dead Loads

Question 24

loads due to the weight of temporary or moving components

Answer 24

Live Loads

Question 25

loads due to pressure and vibration caused by wind

Answer 25

Wind Loads

Question 26

loads induced forces due to ground motion

Answer 26

EQ loads

Question 27

loads due to soil pressure

Answer 27

Soil Lateral Loads

Question 28

loads due to the accumulation of water from rain

Answer 28

Rain Loads

Question 29

hydrostatic and hydrodynamic loads due to surface runoff

Answer 29

Flood Loads

Question 30

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

Answer 30

actual loading conditions

Question 31

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

Answer 31

0.90

Question 32

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

Answer 32

75%

Question 33

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

Answer 33

0.65

Question 34

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

Answer 34

negligible

Question 35

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

Answer 35

significant

Question 36

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

Answer 36

P-delta effect

Question 37

The code usually refers to P-delta effects as _______________

Answer 37

“slenderness effects”

Question 38

effects are caused by lateral forces

Answer 38

P delta (𝑷−∆)

Question 39

effects are caused by geometric imperfections of the member

Answer 39

P sigma (𝑷−𝜹)

Question 40

Examples of RC frames “braced against sidesway”

Answer 40

Diagonal Bracing Shear Wall

Question 41

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 _______________

Answer 41

braced against sidesway

Question 42

formula of r (radius of gyration)

Answer 42

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

Question 43

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

Answer 43

Jackson and Moreland Alignment Charts

Question 44

Effective Length is

Answer 44

k

Question 45

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

Answer 45

Jackson and Moreland Alignment Charts

Question 46

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

Answer 46

a) Moment magnification method after first-order analysis OR b) P-delta analysis (Second order analysis)

Question 47

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

Answer 47

1) Calculate 𝑃𝑐 (6.6.4.4.2) 2) Determine 𝐶𝑚 (6.6.4.5.3) 3) Calculate 𝛿 (6.6.4.5.2) 4) Apply 𝛿 to 𝑀2 (6.6.4.5.1)

Question 48

Steps in Moment Magnification Method (Sway frames):

Answer 48

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)

Question 49

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

Answer 49

iterative geometric updating

Question 50

Steps in Second-order (P-delta) analysis

Answer 50

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