Reinforced Concrete Beam Design ACI 318 08

Reinforced Concrete Beam Design ACI ACI 318-08 Given: •

f’c = 6000 psi

•

f y = 60 ksi

Required: •

Determine the nominal moment capacity,

•

Determine the nominal shear capacity,

!Mn

!Vn

Assumptions: 1.

Plain Plain sect section ions s remai remain n plain plain (ACI (ACI 318-0 318-08 8 secti section on 10. 10.2.2 2.2))

2.

Maximu Maximum m conc concret rete e stra strain in at extrem extreme e compre compressi ssion on fiber fiber = 0.00 0.003 3 (ACI (ACI section 10.2.3)

3.

Tensil ensile e stren strength gth of conc concret rete e is negl neglect ected ed (10 (10.2. .2.5) 5)

4.

Compre Compressi ssion on steel steel is is negle neglecte cted d in this this calcu calculat lation ion..

Reinforced Concrete Beam Design ACI 318-08

Let’s start by constructing the stress and strain diagrams:

•

Next, we’ll calculate d, the depth from the extreme compression fiber to the center of reinforcement in the tensile zone. – d = h – Clear Spacing – d stirrup – dreinforcement/2 – d = 24” – 1.5” – 0.5” - .75”/2 d = 21.625”


Next, we want to use equilibrium to solve for a, the depth of the Whitney stress block: – As a side note, the Whitney stress block refers to the method ACI provides for approximating the stress in concrete under a given strain. This method was developed by an engineer name Charles Whitney, and was adopted by ACI in 1956 From the rules of equilibrium we know that C must equal T C=T C = 0.85 x f’ c x b x a •

Defined in ACI section 10.2.7.1

•

b = width of compression zone

•

a = depth of Whitney stress block

C = 0.85 x 6000psi x 12” x a = 61,200 lb/in x a T = f s x As •

f s = stress in the steel (we make the assumption that the steel yields, and will later confirm if it does).

•

As = area of tensile steel

T = 60000psi x (3 x 0.44 in 2) = 79,200 lb


Solve for a: •

61,200 lb/in x a = 79,200 lb

Now that we know the depth of the stress block, we can calculate c, the depth to the neutral axis.

•

a = 1.294”

From ACI 318 section 10.2.7.1 – a = "1 x c

"1 is a factor that relates the depth of the Whitney stress block to the depth of the neutral axis based on the concrete strength. It is defined in 10.2.7.3 "1 = 0.65

!

0.85 - ((f’c – 4000psi)/1000)) x 0.05 ! 0.85

"1 = 0.85 – ((6000psi – 4000psi)/1000) x 0.05 = 0.75 c = a / "1 = 1.294”/0.75 c = 1.725”

With c, we can calculate the strain in the steel using similar triangles. With this strain calculated, we can check our assumption that the steel yields, and determine if the section is tension controled c/!c = d/(!c + !t)

!t = (d x !c)/c – !c = (21.625” x 0.003)/1.725” – 0.003 "t =

0.0346


Check our assumption that the steel yields, and determine if the section is tension controlled: •

Determine the strain at which the steel yields: – E = f y/!y • E = Young’s modulus which is generally accepted to be 29,000 ksi for steel • f y = steel yield stress •

–

•

!y = yield strain

!y = 60ksi / 29,000 ksi = 0.00207 < 0.0346 therefore our assumption that the steel yields is correct and the stress in the steel may be taken as 60 ksi at failure

Determine if the section is tension controlled: – Per ACI section 10.3.4 a beam is considered tension controlled if the strain in the extreme tension steel is greater than 0.005. – The calculated steel strain in our section is 0.0346 is greater than 0.005 therefore this beam section is tension controlled.


Next, let’s determine if the beam section satisfies the minimum steel requirements of ACI:

•

Per ACI section 10.5.1, the minimum steel requirement is: – As, min = ((3 x square root(f’ c))/f y) x bw x d " (200/f y) x bw x d • As a side note, the 200/f y minimum controls when f’ c is less than 4,500 psi – As, min = (( 3 x squareroot(6,000 psi))/60,000psi) x 12” x 21.625” – As,min = 1.01 in 2 – As = 3 x 0.44 in 2 = 1.32 in 2 > 1.01 in 2 therefore we satisfy the minimum steel requirements of ACI


Finally, let’s calculate the nominal moment capacity of the section: Using moment equilibrium, we can calculate the moment capacity by taking the moment about the center of the tensile force, or the center of the compressive force.

Calculate the moment about the center of the compressive force:

#Mn = # x T x (d – a/2) •

# = 0.9 for a tension controlled section per ACI 9.3.2.1

• T = Asf y

#Mn = 0.9 x 1.32 in 2 x 60 ksi (21.625in - 1.294in / 2) !Mn =

1,495 kip*in = 124.6 kip*ft


Finally, let’s calculate the nominal moment capacity of the section: Let’s check our solution by calculating the moment capacity about the center of the tensile force.

Calculate the moment about the center of the Tensile force:

#Mn = # x C x (d – a/2) •

# = 0.9 for a tension controlled section per ACI 9.3.2.1

• C = 0.85 x f’c x b x a

#Mn = 0.9 x 0.85 x 6 ksi x 12in x 1.294in (21.625in - 1.294in / 2) !Mn =

1,495 kip*in = 124.6 kip*ft

Our solution checks out!


Now let’s calculate the shear capacity of our section: Per ACI equation 11-2:

#Vn = #(Vc + Vs) • # = 0.75 per ACI section 9.3.2.3 • Vc = shear capacity of the concrete section • Vs = shear capacity of the stirrups Vc = 2 x $ x squareroot(f’ c) x bw x d (ACI eqn 11-4) •

$ = modification factor for lightweight concrete (See ACI section 8.6.1)

Vc = 2 x 1.0 x squareroot(6,000 psi) x 12” x 21.625” = 40,200 lb Vs = (Av x f yt x d)/s (ACI eqn 11-15) • Av = 2 x 0.2 in 2 (there are 2 legs at each stirrup location Vs = (0.4 in2 x 60,000 psi x 21.625 in) / 12 in = 43, 250 lb


Now let’s calculate the shear capacity of our section:

Per ACI equation 11-2:

#Vn = #(Vc + Vs) #Vs = 0.75 x (40,200 lb + 43,250 lb) !Vn =

62,600 lb = 62.6 kip


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Reinforced Concrete Beam Design ACI 318 08

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