Reinforced Concrete Design Formulas 26286b

REINFORCED CONCRETE DESIGN FORMULAS: Working Stress Design Working Stress Design is called Alternate Design Method by NS (National Structural Code of the Philippines) and ACI (American Concrete Institute, ACI).

concrete

Code Reference NS 2010, Section 424: Alternate Design Method ACI 318M-99, Appendix A: Alternate Design Method

Design Principle The design concept of WSD is based on Elastic Theory in which the stress-strain diagram is within the proportional limit and will obey Hooke's Law.

Notation

fc

= allowable compressive stress of concrete

fs

= allowable tensile stress of steel reinforcement

f 'c fy

= specified compressive strength of concrete = specified yield strength of steel reinforcement

Ec = modulus of elasticity of concrete Es

= modulus of elasticity of steel

n = modular ratio M

= design moment

d = distance from extreme concrete fiber to centroid of steel reinforcement

kd

= distance from the neutral axis to the extreme

fiber of concrete

jd = distance between compressive force C and

tensile force T

ρ = ratio of the area of steel to the effective area of

βc

= ratio of long side to short side of concentrated

load or reaction area As = area of steel reinforcement

Actual Stress ≤ Allowable Stress Allowable Stresses (NS 2010, 424.4.1 / ACI 318M, A.3.1) Concrete

1 .

Flexure

2 .

Shear 140 MPa

Steel Reinforcement 1 .

Grade 275 or Grade 350 reinforcement

For nonprestressed reinforcement

Extreme fiber stress in compression

Beams and one-way slabs and footings:

2 .

Grade 420 reinforcement or Shear carried 170 MPaby greater and welded wire fabric concrete, Vc (plain or deformed)

3 .

For flexural reinforcement, 10 Maximum 0.50 shear fy mm or less in diameter, in one-carried by concrete way slabs of not more than 4 mplus shear span but not greater than 200 reinforcement, Vc MPa

Shear carried by concrete, Vc

0.38 √ f 'c

0.9 √ f 'c

Two-way slabs and footings:

3

Shear carried by

1.5

concrete, Vc

Ec =W c 0.043 √ f ' c

but

not greater For normal weight concrete

than

Ec=4700 f ' c Modulus of Elasticity of Steel Reinforcement (NS 408.6.2)

0.9 √ f 'c

Joists:

Modulus of Elasticity Modulus of Elasticity of Concrete (NS 408.6.1) For weight of concrete, Wc between 1500 and 2500 kg/ m

0.45 f ' c

3 .

1 f' 6√ c

Bearing on loaded area

1 2 (1+ ) √ f ' c 2 βc

0.3 f ' c

Es=200000 MPa

Modular Ratio (NS 424.6.4) It shall be permitted to take the modular ratio,

n=

Es Ec

as the nearest

whole number (but not less than 6). Modular Ratio for Beams with Compression Steel (NS 424.6.5) In doubly reinforced flexural , an effective modular ratio of

2n

shall

be used to transform compression reinforcement for stress computations. Assumptions in WSD Plane section remains plane before and after bending. Concrete stress varies from zero at the neutral axis to a maximum at the extreme fiber.

Concrete do not carry tensile stress. All tensile stress will be carried by steel reinforcement. The steel and concrete are perfectly bond so that no slippage between the two will occur.

M =Cy M =T y Bending stresses

I NA =

f c=

bx 3 2 +n A s ( d −x ) 3

Mx I NA

f s M (d−x ) = n I NA Moment of area:

QaboveNa=QbelowNa

1 2 b x =nAs ( d−x ) 2 Tensile and Compressive Forces

T =f s A s

1 C= f c bx 2 C=T Moment in the section

1 y=d− x 3

Working Stress Analysis – Uncracked Stage The beam will behave elastically and remains uncracked. The tensile stress of concrete is below rupture.

Cracking Moment NS 2010, Section 409.6.2.3 Modulus of rupture of concrete,

Cracking moment,

M cr =

f r=0.75 √ f ' c MPa

f r Ig Yt

Where:

Ig

= Moment of inertia of the gross section

neglecting reinforcement

Yt

= distance from centroid of gross section to

extreme tension fiber

For investigation: Factor k:

K=

fc

f c+

fs n

Working Stress Analysis – Cracked Stage General Requirement Actual Stresses ≤ Allowable Stresses

Factor j:

Internal Couple Method Static equilibrium of internal forces

Moment resistance coefficient:

1 J =1− k 3 1 R= f c kj 2

Moment capacity: Use the smallest of the two

1 Mc=Cj d= f c kj bd 2=R bd 2 2 Ms=Tjd= A s f s jd

Transformed Section Method Convert steel area to equivalent concrete area by multiplying As with modular ratio, n.

Doubly reinforced:

I NA =

bx 3 ( 2 + 2 n−1 ) A' s ( x−d ' ) +n A s ( d −x ) 3

Actual stresses (calculate using Flexure Formula) Concrete

f c=

Location of the neutral axis from extreme compression fiber Singly reinforced:

1 2 b x =nAs ( d−x ) 2

Doubly reinforced:

1 2 ' ' b x + ( 2 n−1 ) A s ( x −d )=nA s ( d −x ) 2

Cracked section moment of inertia Singly reinforced: 3

bx I NA = +n A s (d−x )2 3

I NA =I cr

Mx I NA

Tension steel

f s M (d−x ) = n I NA Compression steel for doubly reinforced

f ' s M (x−d ' ) = 2n I NA