reinforced concrete beam design aci 318 08

8/10/2019 Reinforced Concrete Beam Design ACI 318 08

1/11

Reinforced Concrete Beam Design ACI 318-08

Given:

fc = 6000 psi f y = 60 ksi

Required:

Determine the nominal moment capacity, ! Mn

Determine the nominal shear capacity, ! Vn

Assumptions:

1. Plain sections remain plain (ACI 318-08 section 10.2.2)

2. Maximum concrete strain at extreme compression fiber = 0.00section 10.2.3)

3. Tensile strength of concrete is neglected (10.2.5)

4. Compression steel is neglected in this calculation.

Monday, August 6, 12


2/11

Lets start by constructing the stress and strain diagrams:

Next, well calculate d, the depth from the extreme compression fiber to the center of reinforcement in the tensile zone.

d = h Clear Spacing d stirrup d reinforcement /2 d = 24 1.5 0.5 - .75/2 d = 21.625




3/11

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. Thismethod 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 wiconfirm if it does).

As = area of tensile steel

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




4/11

Solve for a:

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

a = 1.294

With c, we can calculate the strain in the steel using similar triangles. With this straincalculated, we can check our assumption that the steel yields, and determine if thesection 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

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

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 theneutral 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




5/11

Determine the strain at which the steel yields:

E = f y/! y E = Youngs 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.


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



6/11

Next, lets determine if the beam section satisfies the minimum steel requirements of ACI:

Per ACI section 10.5.1, the minimum steel requirement is:

A s, min = ((3 x square root(f c))/f y) x b w x d " (200/f y) x b w x d

As a side note, the 200/f y minimum controls when f c is less than 4,500 psi

A s, min = ((3 x squareroot(6,000 psi))/60,000psi) x 12 x 21.625

A s,min = 1.01 in 2

A s = 3 x 0.44 in 2 = 1.32 in 2 > 1.01 in 2 therefore we satisfy the minimum steel requirements of ACI




7/11

Finally, lets 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 thecompressive 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 = A sf 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




8/11

Finally, lets calculate the nominal moment capacity of the section:

Lets 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 fc 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!




9/11

Now lets calculate the shear capacity of our section:

Per ACI equation 11-2:

# Vn = # (Vc + V s)

# = 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 b w x d (ACI eqn 11-4)

$ = modification factor for lightweight concrete (See AC

Vc = 2 x 1.0 x squareroot(6,000 psi) x 12 x 21.625

= 40,200 lb

Vs = (A v 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 in 2 x 60,000 psi x 21.625 in) / 12 in

= 43, 250 lb




10/11

Now lets calculate the shear capacity of our section:

Per ACI equation 11-2:

# Vn = # (Vc + V s)

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

! Vn = 62,600 lb = 62.6 kip




11/11

For questions or comments, send me an email [email protected]

Check us out on the web at http://www.civilpe.net


mailto:[email protected]:[email protected]

reinforced concrete beam design aci 318 08

Documents