Structural Design Module 2

7/25/2019 Structural Design Module 2

1/94

COURSE OUTLINE

I. Design of Steel Members

A. Beams

B. Columns

Example 1

C. Connections

Example 2

II. Reinforced Concrete

B. WSD

Beam

A. Definition of Terms

Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6

III. Building Frame System

B. Structural System

A. Code Design Criteria

C. Vertical IrregularitiesD. Plan Irregularities

IV. NSCP Provisions

B. Material Specification

A. Design Philosophy

C. Flexural Members

D. Beam-Column

E. Beam-Column Joints

F. Rebar Details

Center for the Designed Environment Profession

# 2 Matulungin Street, House of Architects BuildingTeachers Village, Quezon City 1

STRUCTURAL DESIGN

Center for the Designed Environment

Professions, Inc. (CDEP)

M

odu

le2


2/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



Design of Steel Members

A. BEAMS

Types of Beams According to its Function

Purlin - carries the roof load between trusses or rafters

Rafter - usually a sloping beam carrying the reaction of purlins

Lintel - carries the masonry across the opening made by a door or windowJoist - a closely spaced beams supporting the floor of a building

Stringer - similar to a joist, it carries the flooring of a bridge

Girder - large-sized beams usually carrying the floor beams

Spandrel - spans between columns and support the floors and curtain walls

Grade beam - lowermost spandrel of a building that has no basement.Shaft - circular beam that transmits power to the machinery. Also carries

torsion in addition to shear and flexure


A. Beams


3/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



A. BEAMS

Steel Sections

are classified as compact, non-compact, and slender element.

Allowable flexural stress

Compact sections Fb= 0.66Fy

Non-Compact sections Fb= 0.60FySlender sections Fb0.60Fy

Allowable shear stress Fv= 0.40Fy

Sections

Design of Steel MembersI. Design of Steel Members

A. Beams


4/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 1

A continuos beam is loaded as

shown below. Assuming that the

section is compact, investigate the

adequacy of the beam if Fy= 248

MPa.

Figure

SolutionSolve the reactions

P = 45 KN

A C

w = 8 KN/m

RC

For RA, MC= 0 +

0 = RA(5) - 8 (5)(2.5)

RA= 38 KN

0 = RA + RC- 8 (5)RC= 47 KN

For RC, FV= 0 +

Section

150

300 6

10

280

2m

B

- 45 (2)

- 45

+38

+14

-47

-31

+78

3m

Plot the shear diagram

VA= RA

VA - 8(3)

VA= +38 KN

VBL= +14 KN

VBR= VBL - 45 VBR= -31 KN

VBL=

VBR - 8(2) VC= -47 KNVC=Plot the moment diagram

MA= 0

MB= (3)

Pinned support

MB= 78

MC= MB+ (2) MC= 0

(VA + VBL)MA+

(VBR+ VC)


Example 1


5/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 1 (contd)

150

300 6

10

280

Calculate section capacity

I = INA + Ad2

= (150)112 (10)3 (6)112 (280)

3+

(150) (145)2+ (10)

I = 74.076X106mm4

(2)

(2)

Solving I and c

From flexure formula, f = M/SM = fbS

For compact sections

fb= 0.66Fy

= 0.66(248)

= 163.68 MPaS = section modulus

= I/c

c = 300/2 = 150 mm

c = distance from NA to extreme

fiber in tension/compression

S =74.076X106

150= 493840 mm3

Maximum values

Vmax= VC= 47 KN

Mmax= MB= 78 KNm


Example 1


6/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



Solving the moment capacity

M = (163.68 MPa)

Check shear capacity

fv= 0.40Fy

= 0.40(248)

= 99.20 MPa

Vcap= fvAw

=

(493840 mm3)

M = 80.83x106Nmm

M = 80.83 KNm > 78 KNm

Therefore, OK!

(99.20 MPa) (300x6)

Vcap= 178560 N

Vcap= 178.56 KN > 47 KN

Therefore, OK!

Design of Steel Members

EXAMPLE 1 (contd)


Example 1


7/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



B. COLUMNS


B. Columns

Prevailing design load is axial and failure may be initiated by

overstressing of the material

buckling about the weak axis

For this reason, the equation that determine the allowable stress of

the columns is express in terms of the length and radius of gyration.

For Intermediate Column 2

2EFy

KLr

Cc =

1 -Fa=

KL/rCc

20.50

KL/rCc

KL/rCc

353 +

38

-Fy

For Long Column

Fa=122E23

> 22E

FyKLr

Cc =

18

Where

Fa= allowable axial stress

L = height of column

K = effective length factor

r = radius of gyration

= IAI = moment of inertia

A = cross sectional area

KLr

2


8/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



B. COLUMNS


B. Columns

Values of K

Both ends hinged One end fixed,

other end pinned

Both ends fixed

L

K = 1.0

L

K = 0.7

L

K = 0.5

One end fixed,

other end free

L

K = 2.0

Prevailing design load is axial and failure may be initiated by

overstressing of the material

buckling about the weak axis

For this reason, the equation that determine the allowable stress of

the columns is express in terms of the length and radius of gyration.


9/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 2


Example 2

Calculate the axial capacity of

the column shown if

a) L = 3m

b) L = 6m

Use Fy = 248 MPa, moment if

inertia I = 1.20x106mm4, and

cross-sectional area A = 1550

mm2.

L

Illustration

Solution

22EFy

Cc=

Solve the radius of gyration, r

r =I

A

r =

1.20x106

1550

r = 27.82 mm

22(200000)248

Cc=

Cc= 126.17

Solve Cc


10/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6





IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 2 (contd)


Example 2

Calculate the axial capacity of

the column shown if

a) L = 3m

b) L = 6m

Use Fy = 248 MPa, moment if

inertia I = 1.20x106mm4, and

area A = 1550 mm2.

L

Illustration

Solution

a) If L = 3.0 m, solve KL/r

Int. Column

=0.70(3000)

Cc= 126.17

27.82KLr

= 75.47 1.5bd= 30

> 40 mm

< 150 mm

Therefore OK!

Examples 3, 4, & 5

EXAMPLE 5 (contd)

PDL= 240 KN

PLL= 180 KN

s

COURSE OUTLINE


39/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6




C. Vertical Irregularities

D. Plan Irregularities

IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



C. ULTIMATE STRENGTH DESIGN

Design of Reinforced Concrete

II. Reinforced Concrete Ultimate Strength Design

x Applied Load Section Capacityis greater than 1

Load FactorsU = 1.4D + 1.7LU = 0.75[1.4D + 1.7L + 1.7W]U = 0.9D + 1.3WU = 1.1D + 1.3L + 1.1EU = 0.9D + 1.1EU = 1.4D + 1.7L + 1.7HU = 0.9D + 1.7H (if live/dead load reduces the effect of H)U = 0.75[ 1.4D + 1.4T + 1.7L ]U = 1.4[ D + T ]

Material Strength

fc= strength of concrete at strain of 0.003

fy= is the yield strength of steel

C. USD

COURSE OUTLINE


40/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



BEAMS


II. Reinforced ConcreteFlexure Formula

b

h

s

d

c

d - cT

C

c0.85fc

T

C

0.85fc

aa

d -a

M = T (d -a)or

Solve the internal moment capacity, M

MC= 0 +

M = C (d -a)

MT= 0 +

FH= 0 +

C = T

0.85fc ba = fyAs

0.85fca

b=

fyAs

M = (d -a)fyAs

0.85fcbfyAsM = d -fyAs

0.85fcbfyAsMu= d -fyAs

Where

Mu= is the ultimate moment capacity

As= is the area of reinforcing bar

fy= is the yield strength

d = is the effective depth

b = is the width of the beam

fc= is the compressive strength of concrete

= is the reduction factor equal to 0.90

C. USDBeam

COURSE OUTLINE


41/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




II. Reinforced ConcreteFlexure Formula

b

h

s

d

c

d - cT

C

c0.85fc

T

C

0.85fc

aa

d -a

act=

To ensure yielding,

bd

As max

min

max= 0.75 0.851fcfy600

fy 600+

min=1.4fy

min= 4fyfc

act= is the actual steel ratio

Where

1= 0.85 - 0.05(fc- 30) fcis in MPa

0.65

0.85

C. USDBeam

BEAMS

COURSE OUTLINE


42/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



COLUMNS



C. USD

Column

Formula for axially loaded column

For tied column (= 0.70)

Pult= 0.80[0.85fc(Ag- As) + fyAs]

For spiral column (= 0.75)

Pult= 0.85[0.85fc(Ag- As) + fyAs]

Where

Pultis the ultimate axial capacity

fcis the compressive strength of concrete

fyis the yield strength of steel

Agis the gross cross sectional area of column

Asis the total area of reinforcing bar

COURSE OUTLINE


43/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 5



5m

Calculate the number of 20 mmreinforcing bar needed for the

beam loaded as shown below. Use

fc= 25 MPa, fy= 414 MPa, and

effective depth d = 350mm.

Figure

A

WU= 26 KN/m

RB

B

Section 250

350

Solution

Calculate the ultimate load

PDL= 20 KN

PLL= 16 KN

Pu= 1.4PDL+ 1.7 PLL

= 1.4(20) + 1.7 (16)

Pu= 55.20 KN

Wu= 26 KN/mCalculate the maximum moment

MUW=WuL2

PuL

= 81.25 KNm

81.25 69.0+

Mu= 150.25 KNm

MUP=

Due to uniform load

Due to concentrated load

=(26)(5)2

(55.2)(5)= = 69.0 KNm

Total ultimate load

Mu=

C. USD

Examples 5 and 6

COURSE OUTLINE


44/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 5 (contd)



5m





Figure

A

WU= 26 KN/m

RB

B

Section 250

350

Solution

PDL= 20 KN

PLL= 16 KN

Solve As

0.85fcbfyAsMu= d -fyAs

1.7fcbfy

(As)2Mu -

fyAs(d) =

1.7(25) 250414 (As)2150.25x106 -

(414)350

(0.90)= As

0.0390(As)2403247 -(350)= As

0.039(As)2 403247+(350) = 0As-

As=

a b c

-b b2- 4 a c2 a

=350 3502- 4 (0.039)403247

2 (0.039)As= + 7616.89

As= + 1357.44

C. USD

Examples 5 and 6

COURSE OUTLINE


45/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 5 (contd)



5m





Figure

A

WU= 26 KN/m

RB

B

Section 250

350

Solution

PDL= 20 KN

PLL= 16 KN

Solving for the number of bars

n =AsAb

=1357.44

d24

n = 4.3 Say 5 - 20 mm

1357.44(20)2

4

=

Check ductility limit

act= b dAs

max= 0.75 0.85 1fcfy

600fy 600+

min=1.4fy

min= 4fyfc

=250 (350)

5(314)

act= 0.0179

max= 0.75 0.85 (0.85)25

414

(600)

414 600+max= 0.0194 > actok!1.4414

= = 0.003 < actok!

=4(414)25

= 0.003 < actok!

C. USD

Examples 5 and 6

COURSE OUTLINE

D i f R i f d C


46/94

COURSE OUTLINE


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 6



C. USD

Examples 5 and 6

Design a square column using 20mmreinforcing bar if PDL= 240 KNand PLL= 180 KN. Use fc= 25 MPa, fy= 276 MPa, = 3%, and 10 mmdiameter ties.

Solution

Solve the ultimate load Pult

Pult= 1.4PDL 1.7PLL+

= 1.4(240) 1.7(180)+Pult= 642 KN

From axial load formula

Pult 0.80[0.85fc(Ag- As) + fyAs]

642x103

(0.8)(0.7)[(0.85) (25) (h2

- 0.03h2

)+ (276) (0.03h2)]

642x103(0.56) [ (21.25) (0.97) h2+ 8.28]

h = 199.20 say 200 mm

Figure

b = h

h

PDL= 240 KN

PLL= 180 KN

s

Ag= h2

= As/AgAs= 0.03Ag

COURSE OUTLINE

D i f R i f d C t


47/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details



EXAMPLE 6 (contd)



C. USD

Examples 5 and 6

Design a square column using 20mmreinforcing bar if PDL= 240 KNand PLL= 180 KN. Use fc= 25 MPa, fy= 276 MPa, = 3%, and 10 mmdiameter ties.

Solution

Solve number of bars, Ag= 200x200

Pult 0.80[0.85fc(Ag- As) + fyAs]642x103(0.8)(0.7)[(0.85) (25) (2002- As)

+ (276) As]

642x103(0.56) [ 21.25 +(2002- As) 276 As]

642x10311.9 +(2002- As) 154.56 As

642x103

11.9 +(2002

) - 11.9As 154.56 AsAs= 1163.6 mm2

# of bars =Asd2

=1163.6(20)2

# of bars = 3.7 say 4 pcs

Figure

b = h

h

PDL= 240 KN

PLL= 180 KN

s

Ag= h2

= As/AgAs= 0.03Ag

COURSE OUTLINE

B ildi F S t


48/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details


# 2 Matulungin Street, House of Architects Building

Teachers Village, Quezon City 48

A. CODE DESIGN CRITERIA

Procedure and Limitations for the Design of Structures

Zoning - Indicate the effective peak ground acceleration

0.40g for Zone 4

0.20g for Zone 2

Site Characteristic

A factor greater than or equal to 1.0 introduce to the baseshear formula to account for the variability of soil conditions.

Occupancy

A factor greater than or equal to 1.0 introduce to the base

shear formula to account for the importance of the structure

Configuration

Implies the type of plan and vertical irregularity

Structural System and Height

Implies the response of the building under lateral load

Building Frame Systems



COURSE OUTLINE

B ildi F S t


49/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





Two Major Parameters in the Selection of Design Criteria

Occupancy

Structural Configuration

Four Categories of Occupancy

Essential Facilities

Occupancies having surgery and emergency treatment areasFire and police stations

Garages and shelters for emergency vehicles and emergency

aircraft

Structures and shelters in emergency preparedness centers

Aviation control towers

Structures and equipment in communication centers and other

facilities required for emergency response

Standby power-generating equipment for Category 1 facilities

Tanks and other structures containing housing or supporting

water or fire-suppression material or equipment required for

the protection of category I, II or III structures.




COURSE OUTLINE

B ildi F S t


50/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





Four Categories of OccupancyHazardous Facilities

Occupancies and structures therein housing or supporting

toxic or explosive chemicals or substances

Non-building structures housing, supporting or containing

quantities of toxic or explosive substances.

Special FacilitiesBuildings with an assembly room with an occupant capacity >1000

Educational buildings with a capacity of 300 or more students

Buildings used for college or adult education with a capacity > 500

Institutional buildings with 50 or more incapacitated patients, but

not included in Category I

Mental hospitals, sanitariums, jails, prison and other buildings

where personal liberties of inmates are similarly restrainedAll structures with an occupancy 5,000 or more persons

Structures and equipment in power-generating stations and other

public utility facilities not included in Category I or Category II

above, and required for continued operation.




COURSE OUTLINE

B ildi F S t


51/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





Four Categories of Occupancy

Standard FacilitiesAll structures housing occupancies or having functioned not

listed in Category I, II or III above and Category V below.

Miscellaneous Facilities

Private garages, carports, sheds, agricultural buildings, andfences over 1.8 meters high.




COURSE OUTLINE

B ildi F S t


52/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




Description of Lateral Force Height

Resisting System Limit (Z4)

B. BASIC STRUCTURAL SYSTEM

1. Bearing Wall System

a structural system without a complete vertical load-carrying space

frame. Bearing walls or bracing systems provide support for all or

most gravity loads. Resistance to lateral load is provided by shear

walls or brace frame.

Illustration

1. Light-framed walls with shear panels

Wood structural Panels -------------------------------- 20

All other light-framed walls ---------------------------- 20

2. Shear wall

Concrete --------------------------------------------------- 50

Masonry ---------------------------------------------------- 50

3. Light steel-framed bearing walls tension bracing --- 20

4. Braced frames where bracing carries gravity load

Steel -------------------------------------------------------- 50

Concrete --------------------------------------------------- ***

Heavy Timber -------------------------------------------- 20




COURSE OUTLINE

B ildi F S t


53/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




2. Building Frame System

a structural system with an essentially complete space frame

providing support for gravity loads. Resistance to lateral load is

provided by shear walls or brace frames.

Illustration Description of Lateral Force HeightResisting System Limit (Z4)

1. Steel eccentrically braced frame ------------------------ 75

2. Light-framed walls with shear panels

Wood structural Panels -------------------------------- 20

All other light-framed walls ---------------------------- 20

3. Shear wall

Concrete --------------------------------------------------- 75

Masonry ---------------------------------------------------- 50

4. Ordinary braced frame

Steel -------------------------------------------------------- 50

Concrete --------------------------------------------------- ***

Heavy timber --------------------------------------------- 20

5. Special concentrically steel braced frame ------------ 75





COURSE OUTLINE

B ilding Frame S stems


54/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




3. Moment-Resisting Frame System

a structural system with essentially complete space frame providing

support for gravity loads. Resistance to lateral load is provided

primarily by flexural action of members.

Illustration Description of Lateral Force HeightResisting System Limit (Z4)

1. Special moment-resisting frame

Steel -------------------------------------------------------- NL

Concrete --------------------------------------------------- NL

2. Masonry moment-resisting walls frame --------------- 50

3. Concrete intermediate moment-resisting frame ----- ***

4. Ordinary moment-resisting frame

Steel -------------------------------------------------------- 50Concrete --------------------------------------------------- ***

5. Special truss moment frames of steel ----------------- 75





COURSE OUTLINE



55/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




4. Dual Systemis a combination of moment-resisting frames & shear walls or braced

frames. Moment-resisting frame shall be designed to resist 25 % of the

base shear & 75 % for the

shear walls/braced frame.

Illustration


Resisting System Limit (Z4)1. Shear wall

Concrete with SMRF ------------------------------------ NL

Concrete with steel OMRF or concrete IMRF ---- 50

Masonry with SMRF or steel OMRF ---------------- 50Masonry with concrete IMRF ------------------------- ***

Masonry with masonry MMRWF --------------------- 50

2. Steel eccentrically braced frame

With steel SMRF ----------------------------------------- NL

With steel OMRF ---------------------------------------- 50

3. Ordinary braced frame

Steel with steel SMRF ---------------------------------- NLSteel with steel OMRF ---------------------------------- 50

Concrete w/ concrete SMRF or concrete IMRF -- ***

4. Special concentrically braced frame

Steel with steel SMRF ---------------------------------- NL

Steel with steel OMRF ---------------------------------- 50





COURSE OUTLINE



56/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USDBeam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




5. Cantilevered Column Building System

a structural system relying on cantilevered column elements for

lateral resistance.

Illustration



Cantilevered column elements -------------------------- 10





COURSE OUTLINE



57/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




6. Shear Wall-Frame Interactive System

a combination of shear walls and frames designed to resist lateral

forces in proportion to their relative rigidities, considering interaction

between shear walls and frames on all levels.

Illustration



Concrete ------------------------------------- 50





COURSE OUTLINE



58/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




1. Stiffness Irregularity / Soft Storyis one in which the lateral stiffness is less than 70 percent of that in

the story above or less than 80 percent of the average stiffness of the

three stories above.

Illustration

Soft

story

Soft Story stiffness < 70% of story stiffness above

Soft Story stiffness < 80% of average stiffness 3 stories above

Soft

story

Soft

story

Braced frame Shear wall Stiff column

Note: Need not be considered if the story drift under the lateral force is less than 1.3 times the story drift above

C. VERTICAL STRUCTURAL IRREGULATITIES




COURSE OUTLINE



59/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




2. Weight (mass) Irregularitymass irregularity shall be considered to exist where the effective

mass of any story is more than 150 percent of the effective mass of

an adjacent story. A roof that is lighter than the floor below need not

be considered.

Illustration

Story mass > 150% of the mass of adjacent story

HEAVY

MASS

HEAVY MASS

HEAVY MASS

HEAVY MASS

Note: Need not be considered if the story drift under the lateral force is less than 1.3 times the story drift above





COURSE OUTLINE



60/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




3. Vertical Geometric Irregularity

vertical geometric irregularity shall be considered to exist where the

horizontal dimension of the lateral-force-resisting system in any story is

more than 130 percent of that in an adjacent story. One-story penthouses

need not be considered.

Illustration

Story dimension > 130% of the dimension of adjacent story





COURSE OUTLINE



61/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




4. In-Plane Discontinuity in Vertical Lateral-Force-Resisting Element

an in-plane offset of the lateral-load-resisting elements greater than

the length of those elements.

IllustrationShear wall

Braced frameShear wall





COURSE OUTLINE



62/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




5. Discontinuity in Capacity / Weak Storya weak story is one in which the story strength is less than 80 percent

of that in the story above. The story strength is the total strength of all

seismic-resisting elements sharing the story for the direction under

consideration.

Illustration

weak

story

weak

story

weak

story

Shear wall Braced frame Shear wall

Story strength < 80% of the story strength above





COURSE OUTLINE



63/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




D. PLAN STRUCTURAL IRREGULARITIES

1. Torsional Irregularity (to be considered if diaphragm is not flexible)

torsional irregularly shall be considered to exist when the maximum

story drift, computed including accidental torsion, at one end of the

structure transverse to an axis is more than 1.2 times the average of

the story drifts of the two ends of the structure.

Illustration

1

1

2

2

2> 1.20(1+ 2)/2

P

M




COURSE OUTLINE



64/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




2. Re-Entrant Corners

plan configurations of a structure and its lateral-force-resisting system

contain re-entrant corners, where both projections of the structure

beyond a re-entrant corner are greater than 15 percent of the plan

dimension of the structure in the given direction.

Illustration

L

B

>

0.1

5B

> 0.15L

Re-entrant corner





COURSE OUTLINE



65/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




3. Diaphragm Discontinuity

diaphragm with abrupt discontinuities or variations in stiffness,

including those having cutout or open areas greater than 50 percent of

the gross enclosed area of the diaphragm, or changes in effective

diaphragm stiffness or more than 50 percent from one story to the

next.

Illustration

L

B

Diaphragm discontinuity





COURSE OUTLINE



66/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




4. Out-of-Plane Offsets

discontinuities in a lateral force path, such as out-of-plane offsets of

the vertical elements.

IllustrationLateral-load-resisting

element

Lateral-load-resisting

element





COURSE OUTLINE



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A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




5. Nonparallel System

the vertical lateral-load-resisting elements are not parallel to or symmetric

about the major orthogonal axes of the lateral-force systems.

Illustration


element


element





COURSE OUTLINE

NSCP Provisions for RC Members


68/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





A. DESIGN PHILOSOPHY

The NSCP C101-01 Section 421 contains special requirement

for the design of RC members that are part of the lateral

resisting frame subjected to earthquake motions.

These requirements were established based on the profound

engineering experiences and experiments to ensure good

performance of the structure during earthquakes.

It provides requirements to mitigate earthquake stresses by

increasing the ductility of the structure through the confinement

of concrete with reinforcing steel where plastic hinging mayoccur.

IV. NSCP Provisions


COURSE OUTLINE



69/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





The NSCP C101-01 Section 421 contains special requirement

for the design of RC members that are part of the lateral

resisting frame subjected to earthquake motions.

These requirements were established based on the profound

engineering experiences and experiments to ensure good

performance of the structure during earthquakes.

It provides requirements to mitigate earthquake stresses by

increasing the ductility of the structure through the confinement

of concrete with reinforcing steel where plastic hinging may

occur.


IV. NSCP Provisions


COURSE OUTLINE



70/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





Reinforced concrete structures in high seismic risk must have:

Strength, Ductility, Toughness

The performance criteria of RC members resisting earthquake:

Serviceability Limit State - material remains in the elasticrange and no damage is expected.

Minor - Magnitude 1 - 4 < 10 yrs

Control Limit - some yielding may occur and may have minor

structural damage.

Moderate - Mag. 4 - 6 -10-20 years Survival Limit State - inelastic behavior and may have major

structural damage.

Major - Magnitude 7 and up - 100-500 years


IV. NSCP Provisions


COURSE OUTLINE



71/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




SCOPE

Section 421 contains special requirements for design and

construction of RC members of a structure for which the design

forces, related to earthquake motions, have been determined

based on energy dissipation in the nonlinear range of response.


L

oad,

P

Deformation,

Elastic

Sway Deformation

Force-Displacement Relat ionship Elast ic vs Inelast ic Respo nse

Actual

Code

P

Failure


IV. NSCP Provisions


COURSE OUTLINE



72/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




B. MATERIAL SPECIFICATION

LIMITATIONS ON MATERIAL STRENGTH

Concrete compressive strength

f'c 21 MPa

f'c 17 MPa may be use for footings

Steel reinforcement:

ASTM A706M

Low-alloy steel deformed bars (Grade 60)

welding and bending is important

ASTM A615M Grade 275 and Grade 420 are allowed if fu/fy1.25

Actual fy(specified fy+ 120 MPa) - retests shall notexceed 20 MPa


IV. NSCP Provisions


COURSE OUTLINE



73/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




SCOPE

The following requirements

shall apply to members that:

Frame membersresisting earthquake

induced forces

Factored axial loadproportioned to resist

flexure, Pu0.1f'cAg

C. FLEXURAL MEMBERS

LIMITS ON SECTION

AND OR DIMENSION

Clear span, L 4d (bw/ H) 0.3

bw250 mm

bwB + 1.5H

Side Elevation

Cross-Section

d H

L

B

bw

H


IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE

I D i f St l M bNSCP Provisions for RC Members


74/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




C. FLEXURAL MEMBERS

LONGITUDINAL REINFORCEMENT

Asrequired in the analysis

AsAs,min= (f'c/4fy)bwd or

AsAs,min= (1.4/fy)bwd

As= equivalent of two bars (continuos)

As0.025bwdNote: As,minneed not be satisfied if As supplied is 1/3

greater than Asrequired.

Positive moment strength at joint face shall not be less thanone half of the negative moment strength provided at that

face of the joint. Neither the negative nor the positivemoment strength at any section along member length shall

be less than one fourth the maximum moment strength

provided at face of either joint.


IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE



75/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




LdAstrequired in the analysisequivalent of 2 bars

(1.4/fy)bwd(fc/4fy)bwd0.025bwd

Asbrequired in the analysisequivalent of 2 bars(1.4/fy)bwd(fc/4fy)bwdAst/2

Ldd12dbL/16

At leastAst/3

(To beam centerline)L

C. FLEXURAL MEMBERS

Pointofinflec

tion


IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE



76/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




LAP SPLICES REQUIREMENT

No splices are allowed within joints.

No splices are allowed within 2h from face of joint.

No splices are allowed within 2h from points of flexural yielding

Lap length must be provided with a hoops/spiral with Smin= d/4

or 100 mm.

TRANSVERSE REINFORCEMENT

Hoops shall be provided within:

2h from face of the support

2h from both sides of sections where flexure yielding are

likely to occur.

First hoop shall be located not more than 50 mm from the

face of the supporting element.

C. FLEXURAL MEMBERS

IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE

I Design of Steel MembersNSCP Provisions for RC Members


77/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




C. FLEXURAL MEMBERS


Maximum hoop spacing should be the lesser of;

d/4 8db(longitudinal bars) 24db(hoops) 300 mm

Notes:

Corner and alternate longitudinal bars shall be providedwith lateral support by a tie with included angle not more than

135 degrees. Longitudinal bars shall be no farther than 150

mm from such laterally supported bars.

Where hoops are not required, stirrups with seismic hook atboth ends shall be spaced at a distance not more than d/2

throughout the length of the member.

IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE



78/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




50 mm

2H

H

s d/2

d/48db(longitudinal bars)24db(hoops)300 mm

C. FLEXURAL MEMBERS

IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE



79/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




SHEAR STRENGTH REINFORCEMENT

The design shear forces Veshall be determined fromconsideration of the static forces on the portion of the

member between faces of the joint. It shall be assumed that

moments of opposite sign corresponding to probable

flexural strength Mpract at the joint faces and that the

member is loaded with the tributary gravity load along its

span.

Transverse reinforcement over the confined region shall beproportioned to resist shear assuming Vc= 0 when both of

the following conditions occur:

(MprA+ MprB)/L Ve and

Pu0.05f'cAg

C. FLEXURAL MEMBERS

IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE



80/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




C. FLEXURAL MEMBERS

MPRL MPRR

VL

VRL

VL= (MprA+ MprB)/L - 0.75(1.4DL + 1.7LL)L/2

VR= (MprA+ MprB)/L + 0.75(1.4DL + 1.7LL)L/2

IV. NSCP Provisions

C. Flexural Members

COURSE OUTLINE



81/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




D. BEAM-COLUMN

SCOPE

The following requirements shall apply to members that:

resist earthquake induced forces, and

have a factored axial forces exceeding 0.1Agf'c

LIMITATION ON SECTION DIMENSIONS Least cross-sectional dimension 300mm

Least dimension / dimension 0.4

Limitation on longitudinal reinforcement 0.01 g0.06

H

B

B H 300mmB

H0.40

IV. NSCP Provisions

D. Beam-Column

COURSE OUTLINE



82/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




MINIMUM FLEXURA L STRENGTH

The flexural strength of the columns shall satisfy:

Mc1.2Mg

Where:

Mc= sum of column moments at the center of the joint.

Mg= sum of girder moments at the center of the joint.

MNCT

MNCB

MNGR

MNGL

D. BEAM-COLUMN

IV. NSCP Provisions

D. Beam-Column

COURSE OUTLINE



83/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




RESTRICTION ON LAP SPLICES

Splices are permitted only within the middle half of thecolumn height,

Splice must be designed as tension lap splice, and

Hoop spacing must be the lesser of;

H/4 6db(longitudinal bar)

s = 100 + (350-H)/3, 100 mm s 150 mm


Closed hoops or continuous spirals must be provided to

confine the concrete core, to act as lateral support of thelongitudinal bars, and to resist shear. The amount of

transverse reinforcement must be larger of that required for

confinement or the design shear.

D. BEAM-COLUMN

IV. NSCP Provisions

D. Beam-Column

COURSE OUTLINE



84/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





Confinement reinforcement must be provided within a lengthLofrom each joint face where flexure yielding may occur due

to inelastic lateral displacements.

Where:

Lo

Depth of the member

Clear height / 6

450 mm

whichever is

smaller

For spiral reinforcement (volumetric ratio)

s0.45 (Ag/Ac- 1)f'c/fy

0.12f'c/fyWhere:

s = volume of spiral/volume of confined core

Ac = area of the core out-to-out from transverse bars

D. BEAM-COLUMN

IV. NSCP Provisions

D. Beam-Column

COURSE OUTLINE

I. Design of Steel MembersNSCP Provisions for RC Members


85/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details





Rectangular hoop

Ash = 0.3(shcf'c/fy)(Ag/Ach- 1)

Ash = 0.09shcf'c/fy

Note: For adequate core strength this equations need notbe satisfied.

whichever is

smaller

Hoop spacing shall be the lesser of

H/4

6db(longitudinal bar)

s = 100 + (350-H)/3, 100 mm s 150 mm

Crossties or legs of hoops shall not be spaced farther than350 mm on center in the direction perpendicular to axis of

longitudinal bar.

D. BEAM-COLUMN

IV. NSCP Provisions

D. Beam-Column

COURSE OUTLINE



86/94


A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details




SHEAR REQUIREMENTS

The design shear, Ve, shall be determined from themaximum probable moment strength, Mpr as for beams.

The shear need not exceed the value determined from

joints strengths based on the probable moment strength of

the transverse members framing into the joint.

Vcshall be taken as the larger of the above analysis or asdetermined by analysis of the structure. For confined

region, transverse reinforcement shall be proportioned

assuming Vc= 0.0 if

(MprA+ MprB) / L Ve and

Pu0.05f'cAg

D. BEAM-COLUMN

IV. NSCP Provisions

D. Beam-Column

COURSE OUTLINE



87/94

g

A. Beams

B. Columns

Example 1

C. Connections

Example 2


B. WSD

Beam


Column

Examples 3, 4, & 5

C. USD

Beam

Column

Examples 5 and 6






IV. NSCP Provisions



C. Flexural Members

D. Beam-Column


F. Rebar Details

Structural Design Module 2

Documents

Transcript of Structural Design Module 2