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1. NOMENCLATURE

Notation to be used in the discussion complies with NSCP 2001

Ag = Gross area of section (Ac + As)

Ac = Area of concrete; gross area – area of reinforcing

As = Area of reinforcing

A’s = Area of compressive reinforcement in a doubly reinforced section

Av = Area of shear reinforcement

A 1 = Loaded area in bearing

A 2 = Gross area of bearing support member

D = Dead loads, or related internal moments and forces

E = Load effects of earthquake, or related internal moments and forces

Ec = Modulus of Elasticity of concrete

Es = Modulus of Elasticity of steel

L = Live loads, or related internal moments and forces

M = Design moment

V = Design shear force

a = Depth of equivalent rectangular stress block (USD)

b = Width of compression face of member

bw = Width of stem in a T-beam

c = Distance from extreme compression fiber to the neutral axis

d = Effective depth, from extreme compression fiber to centroid of tensile reinforcement

e = Eccentricity of a non-axial load, form the centroid of the section to the point of

application of the load

fc = Unit compressive stress in concrete

f’c = Specified compressive strength of concrete

fr = Modulus of rupture of concrete

fs = Stress in reinforcement

fy = Specified yield stress of steel

h = Overall thickness of member; unbraced height of a wall

jd = Length fo internal moment arm (WSD)

kd = Distance from extreme compression fiber ot the neutral axis (WSD)

n = Modular ratio of elasticity: E s/Ec

ρ = Percent of reinforcing with working strength design expressed as a ratio: A s/Ag

s = Spacing of stirrups

t = Thickness of solid slab

φ = Strength reduction factor (USD)

3. DESIGN METHODS

3.1 Working Stress Design (WSD)

SERVICE LOAD CONDITIONS – no load factors applied

ALLOWABLE STRESSES: f c = 0.45 f’c

f (^) s = 0.5 f (^) y → if not given

3.2 Ultimate Strength Design (USD)

REQUIRED STRENGTH – combination of factored forces

Load Combinations *extracted from NSCP 2001, p. 4-

Dead Load and Live load: U = 1.4D + 1.7L

Dead Load, Live Load & Wind Load U = 0.75(1.4D + 1.7L +1.7W)

Dead Load and Wind Load U = 0.9D + 1.3W

Dead Load, Live Load & Earthquake U = 1.32D + 1.1(f 1 )L + 1.1E

f 1 = 1.0 for floors of public

assembly, garage live load

and for L > 4.8 kPa

= 0.5 for other live loads

Dead Load & Earthquake U = 0.9D + 1.1E

DESIGN STRENGTH – nominal strength multiplied by strength reduction factor φ

Reduction Factors *extracted from NSCP 2001, p. 4-

Flexure, without axial (Moment in beams) ……………….. 0.

Axial compression w/ or w/o Flexure

(Axial load and Moment in Tied columns) ………... 0.

(Axial load and Moment in Spiral columns)……….. 0.

Shear and Torsion ………………………………………….. 0.

Bearing on concrete ………………………………………... 0.

4. LOAD ANALYSIS

4.1 SLABS

4.1.1 ONE WAY SLAB – when 0. 5 L

S

m

n

n = <

Sn

L (^) n 1 m

L

w

L

B

w w w

1 m

AREA LOAD ON SLAB:

• DEAD LOAD – slab weight and floor fixed loads

c SDL

slab SDL

D t W

D W W

=γ ⋅ +

• LIVE LOAD – depends on type of occupancy
• REQUIRED FLOOR LOAD

o WSD: W =D+L

o USD: Wu = 1. 4 D+ 1. 7 L

• UNIFORM TRIBUTARY LOAD

o WSD: w =W× 1

o USD: w^ u =Wu×^1

4.3 COLUMNS

• REQUIRED LOAD

o WSD: W =D+L

o USD: Wu = 1. 4 D+ 1. 7 L

• TRIBUTARY COLUMN LOAD PER FLOOR

o WSD: P =W×( S×L) +ωb ×(S +L)

o USD: Pu =Wu×( S×L) +ωub×(S +L)

5. GENERAL REQUIREMENTS OF REINFORCED CONCRETE STRUCTURES

5.1 MINIMUM CONCRETE COVER, IN SITU

*based from NSCP 2001, p. 4-

Exposed to earth (ex. Foundation slab, tie beams)……….………. 75 mm

Exposed to weather (ex .exterior beams and columns)…..………. 50 mm

Not exposed to weather (ex .interior beams and columns)………. 40 mm

Slabs and walls……………………………………………………...… 20 mm

5.2 SPACING OF REINFORCEMENT

5.2.1 SLABS

 MAXIMUM SPACING = 3 tor 450 mm

S S

20 mm clear cover

S 1

L 1

S 2

A B (^) C

1

2

3

D S 3

L 2

5.2.2 BEAMS

5.2.3 COLUMNS

5.3 HOOKS AND BENDS

*NSCP, 407

Bend diameter:

D =

b b

b b

b b

6d d is 10 - 25mm

8d d is 28 - 32mm

10d d is 32mm

→

→

→

12d b b stirrup / tie

 6d     

Bend diameter:

D =

b b

b b

b b

6d d is 10 - 25mm

8d d is 28 - 32mm

10d d is 32mm

→

→

→ b

4d

( min )

65mm

b

db

8d

150 mm

≥

l ≥

b

db

8d

150 mm

≥

l ≥

36 mm

36 mm

1.5db but not less than 40mm

Stirrup or tie

vcs ≥ 25 mm

cs ≥dband 25 mm

Bars in double layer

b b

Bundled bars

cs ≥de = nd b

Clear concrete cover

cc ≥ 40 mm

6.2 FLEXURE FORMULA, Working Stress Design (WSD)

S ingly R einforced R ectangular B eam (SRRB)

 Straight-line stress distribution (triangular compressive stress block)

 Maximum allowable compressive stress of 0.45f’c

 Maximum allowable tensile stress on steel of 0.5f (^) y (if not given)

T R A N S F O R M E D S E C T I O N M E T H O D

 Modular Ratio, c

s

E

E

n =

 Location of Neutral Axis - ΣM (^) NA = 0 and solve for x

nA (d x) 2

x b (^) s

2

= − 

 Moment of Inertia of transformed section – inertia about the N.A.

( ) ( )

2 s

3

NA nA d x 3

b x I = + −

 Compressive force, f(^ xb)^ volumeofstressblock 2

C = c →

 Tensile force, T =As fs

 Resisting Moment,  

x Td 3

x M Cd

 Extreme concrete stress,

( )

NA

c

Mx f I

 Steel bar stress,

( )

NA

s

Md x f n I

b

h

d jd

x/

C

T

x

fc

n

f (^) s

BEAM SECTION

As

TRANSFORMED SECTION STRESS DIAGRAM

NA

b

d - x

M

GENERAL EQUATIONS FOR SRRB.

 Modular ratio,

c

s

E

E

n =

 Steel ratio, bd

A (^) s ρ =

 k^ (^ n)^2 n n

2 = ρ + ρ −ρ

k j = 1 −

 Compressive force, fkdb 2

C = (^) c Resisting Moment, M =Cjd

 Actual concrete stress, (^ )^ bdkj

2 M

fkdbjd f 2

M

c c 2

 Tensile force, T = As fs Resisting Moment, M =Tjd

 Actual Steel Stress, ( ) Ajd

M

M Af jd f s

= s s → s=

DESIGN FOR BALANCED SECTION

 When there is simultaneous limiting stresses in the concrete and steel

c

s

b

nf

f 1

k

 Balanced steel ratio

s

c bal 2 f

kf ρ =

b

h

d jd

kd/

C

T

kd

fc

BEAM SECTION

As

ELASTIC STRESS DIAGRAM

NA

M

7. Estimate the axial load carried by column E at the fourth floor from total factored floor load using

tributary area method.

8. Estimate the axial load carried by column E at the ground floor from total factored floor load

using tributary area method.

9. Calculate the factored uniform load on 1-m strip of slab ABED.
10. Calculate the ultimate bending moments for the design of slab ABED by ACI Coefficients

Method.

11. Calculate the factored uniform load on girders DE and EF.
12. Calculate the ultimate bending moments for the design of girder DE and EF by ACI Coefficients

Method.

13. Calculate the shear force at faces of supports for members DE and EF by ACI Coefficients

Method.

NSCP PROVISION : ACI MOMENT COEFFICIENTS

POSITIVE MOMENT

2 Mu =Cwuln

End spans

Discontinuous end unrestrained……………………..

Discontinuous end integral with support………………

Interior spans………………………………………….

1/

1/

1/

NEGATIVE MOMENT

2 Mu =Cwuln

At exterior face of first interior support

Two spans……………………………………………...

More than two spans…………………………………. At other faces of interior supports……………………..

At face of all supports for slabs with spans not exceeding 3 meters; and

beams where ratio of sum of column stiffness to beam stiffness exceeds

eight at each end of the span…………………………………….

At interior face of exterior support for members built integrally with supports:

Where support is a spandrel beam…………………

Where support is a column………………………….

1/

1/

1/

1/

1/

1/

SHEAR

At face of first interior support …………………… wun 2

1. 15 l

At face of all other supports ……………………….. wun 2

1 l

Where C = coefficient

wu = factored floor load

ln = CLEAR span for +M and V;

= AVERAGE adjacent CLEAR span for -M

B E A M D E S I G N A N D A N A L Y S I S B Y W S D :

SIT. A: A hollow beam with cross-section given below is simply supported over a span of 4 m. The cracking

moment of the beam is 78 kN·m.

1. Find the maximum uniform load in kN/m the beam can carry without cracking.
2. Calculate the modulus of rupture in MPa of concrete used for the beam.
3. If the hollow part of the beam is replaced with a square of side 200 mm, what is the new cracking moment in kN·m?

SIT. B: A rectangular beam 300 mm x 500 mm reinforced with 3-φ28 mm bars having its centroid 75 mm

from the bottom edge of the beam. Use f’c = 21 MPa and fy = 275 MPa.

1. Calculate the elastic bending stresses if is to sustain a dead load moment of 30 kN∙m and live load

moment of 40 kN∙m.

2. Calculate the safe bending moment the beam can sustain without exceeding allowable stresses

under working load conditions.

SIT. C: The dimensions of the reinforced concrete T-beam in the figure below are b 1 = 500 mm, h 1 = 150

mm, b = 250 mm and h = 500 mm. If n = 8 and As = 3300 mm^2 , determine the following:

1

h 1

h

1. The maximum stress produced in concrete by a positive bending moment of 120 kN-m
2. The maximum stress produced in steel.
3. The maximum bending moment applied without exceeding fc = 12 MPa and fs = 140 MPa.

SIT. D: A rectangular concrete beam of concrete with f’c = 21 MPa and steel reinforcing with f (^) s = 138 MPa

must sustain a total service uniform load of 50 kN/m over a simple span of 5 m. Select the beam

dimensions and the reinforcing for a section with tension reinforcing only. Assume b = 0.56d, d (^) b = 20

mm, stirrup diameter = 10 mm.

600 mm

300 mm

150 mm

6.3.1 Singly Reinforced Rectangular Beams (SRRB)

SRRB BASIC EQUATIONS:

1. STATIC EQUILIBRIUM,

[ ]

c

y

s

c

y

c

sy

f'

f

1. 85

d a

bd

A

1. 85 f'

fd a

1. 85 f'b

Af C T a

ρ ω=

ω

ρ=

ρ

= → =

where

where

2. DESIGN MOMENT CAPACITY,

 

  

 φ = ⋅ ⋅ −

 

  

 φ = ⋅ ⋅ −

2

a M 0. 9 T d

2

a M 0. 9 C d

n

n

FLEXURE EQUATIONS: 

 ρ φ = ⋅ρ − c

2 y n y f'

f M 0. 9 fbd 1 0. 59

φ M = 0. 9 ⋅bd ⋅f'cω⋅ ( 1 − 0. 59 ω)

2 n

NSCP 2001 PROVISIONS

 MINIMUM TENSION STEEL RATIO

y y

c min f

butnotlessthan 4 f

f' ρ =

 MAXIMUM TENSION STEEL RATIO

 ρmax = 0. 75 ρbal

d – a/

a c

0.85f’c

b

h

d

As

NA

εs =fsE s

εc = 0. 003

T =As f y

C = 0. 85 f'c ab

STRESS

STRAIN

a/

BALANCED CONDITION

 When there is simultaneous yielding of concrete and steel

y

b 600 f

600 d c

 BALANCED STEEL RATIO

= ⋅β × y y

c bal 1 600 f

f

f' ρ 0. 85

D E T E R M I N A T I O N O F S T E E L A R E A

(GIVEN Mu and beam dimensions)

METHOD 1:





















 

  

 φ =φ −

= +

=φ

2

a M 0. 85 f' abd

M 1. 4 M 1. 7 M

M M

n c

u DL LL

u n

CALCULATE a

y

c s f

1. 85 f' ab A =

→ redesignas DRB

< → = =ρ

s s min

s smin s smin min

A A

A A A A bd

METHOD 2 :

( )

 





 



φ =φ ω⋅ − ω

= +

=φ

M bdf' 1 0. 59

M 1. 4 M 1. 7 M

M M

c

2 n

u DL LL

u n

CALCULATE ω

y

c

f

ωf' ρ=

ρ>ρ → redesignas DRB

ρ≤ρ → =ρ

ρ<ρ → =ρ

max

max s

s min

A bd

A bd min

D E T E R M I N A T I O N O F M O M E N T C A P A C I T Y

(GIVEN beam dimensions and tension steel bars)

METHOD 1 :

[ ]

( )

( )  

  

 − = β = =

=

c

600 d c

1. 85 f' a c b A f

C T

c 1 s s

CALCULATE c, fs and check

fs < f y NON YIELDING STEEL, T = As fs , a =cβ 1

φM (^) n =φAsfs⋅ (d −a/ 2 )

fs ≥ f y STEEL YIELDS, max T = As fy ,

1. 85 f'b

Af a c

sy

φM (^) n =φAsfy⋅ ( d−a/ 2 )

METHOD 2:

bd

A (^) s ρ =

ρ <ρ bal STEEL YIELDS, c

y

f'

ρf ω=

ρ >ρ bal NON YIELING STEEL, c

s f'

ρf ω=

( )

( )  

  

 − = β = =

=

c

600 d c

1. 85 f' a c b A f

C T

c 1 s s

φM =φbdf'cω⋅ ( 1 − 0. 59 ω)

2 n

6.3.3 T-BEAMS

EFFECTIVE FLANGE WIDTH, b (^) f

NSCP Provision:

5.8.10.1 In T-beam construction, the flange and web shall be built integrally or otherwise effectively bonded

together.

5.8.10.2 Width of slab effective as a T-beam flange shall not exceed one-quarter of the span length of the

beam, and the effective overhanging flange width on each side of the web shall not exceed:

a) eight times the slab thickness, and

b) one-half the clear distance to the next web.

5.8.10.3 For beams with a slab on one side only, the effective overhanging flange width shall not exceed:

a) one-twelfth of the span of the beam,

b) six times the slab thickness, and

c) one-half the clear distance to the next web.

5.8.10.4 Isolated beams, in which the T-shape is used to provide a flange for additional compression area,

shall have a flange thickness not less than one-half the width of web and an effective flange width

not more than four times the width of the web.

t 1

b w S^2 S^3

t 2 t 3

S (^1) b w

Interior Beam b f b f

f t 2 w

1 2 w

f

b 8t 8t b

L / 4

S S b 2

b smallest

= + +

=

= +

=

Exterior Beam

f 3 w

w

3 w

f

b 6t b

L /12 b

S b 2

b smallest

= +

= +

= +

=

L

6.3.3.1 WITHOUT T-ACTION, Wide Rectangular Beam

 Compression area is within the flange only

 The beam behaves as a wide RECTANGULAR beam

 a < t (^) f

DESIGN

(Given Mu, required As)

ANALYSIS

(Given As, required φM^ n)

f

f f c f

f u

f u

a t

t M 0.85f ' t b d

M M

M M

φ = φ (^)  −   

φ ≥ →

φ < →

Assume :

design as w/o T - action

design w/ T - action

s y

c f

f

f

A f a 0.85f ' b

a t

a t

Assume as wide rectangular beam :

calculate

analyze w/o T - action

analyze w T - action

u s y

s y

c f

a M A f d 2

A f a 0.85f ' b

= φ (^)  −   

Calculate As by quadratic formula

Check minimum requirements:

s

w

min

min s min w

A

b d

ok

A b d

ρ =

ρ > ρ →

ρ ≤ ρ → = ρ

n s s

s s

c f

s s 1

s y y

s y s

a M A f d 2

from C T

A f a 0.85f' b

from strain diagram:

600d a c calculate f 600 f

check : if f f , use f

if f f , use f

φ = φ (^)  −   

=

• β

bf

h

d

A s

tf

C (^) f = 0.85f’cab (^) f

d – a/

0.85f’c

T = A sfy

a

STRESS BLOCK for the Flange

T-BEAM W/o T-action

φM n