Aci 318-11 Example 001.pdf 253p6f

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ACI 318-11 Example 001 SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE EXAMPLE DESCRIPTION The flexural and shear design of a rectangular concrete beam is calculated in this example. A simply ed beam is subjected to an ultimate uniform load of 9.736 k/ft. This example is tested using the ACI 318-11 concrete design code. The flexural and shear reinforcing computed is compared with independent hand calculated results. GEOMETRY, PROPERTIES AND LOADING CL

10"

A

13.5"

2.5" A

Section A-A

10' = 120" Material Properties E= 3600 k/in2 ν= 0.2 G= 1500 k/in2

Section Properties d = 13.5 in b = 10.0 in I = 3,413 in4

Design Properties f’c = 4 k/in2 fy = 60 k/in2

TECHNICAL FEATURES TESTED  Calculation of Flexural reinforcement, As  Enforcement of Minimum tension reinforcement, As,min  Calculation of Shear reinforcement, Av  Enforcement of Minimum shear reinforcing, Av,min

ACI 318-11 Example 001 - 1

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RESULTS COMPARISON Independent results are hand calculated based on the equivalent rectangular stress distribution described in Example 6.1 in Notes on ACI 318-11 Building Code.

Output Parameter

ETABS

Independent

Percent Difference

Design Moment, Mu (k-in)

1460.4

1460.4

0.00%

Tension Reinf, As (in2)

2.37

2.37

0.00%

Design Shear Force, Vu

37.73

37.73

0.00%

Shear Reinf, Av/s (in2/in)

0.041

0.041

0.00%

COMPUTER FILE: ACI 318-11 Ex001 CONCLUSION The computed results show an exact match for the flexural and the shear reinforcing.

ACI 318-11 Example 001 - 2

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HAND CALCULATION

Flexural Design The following quantities are computed for all the load combinations: ϕ = 0.9, Ag = 160 sq-in As,min =

=

200 bw d = 0.450 sq-in (Govern) fy

3 f c' fy

bw d = 0.427 sq-in  f c′ − 4000  0.85 =  1000 

β1 = 0.85 − 0.05 

0.003 = d 5.0625 in 0.003 + 0.005

= cmax

amax = β1cmax = 4.303 in

Combo1 wu = (1.2wd + 1.6wl) = 9.736 k/ft Mu =

∙

wu l 2 = 9.736 102/8 = 121.7 k-ft = 1460.4 k-in 8

The depth of the compression block is given by: a = d − d2 −

2Mu 0.85 f c'ϕb

= 4.183 in (a < amax)

The area of tensile steel reinforcement is given by: Mu 1460.4 = a 0.9 • 60 • (13.5 − 4.183 / 2 )  ϕ fy d −  2 

As

=

As

= 2.37 sq-in

ACI 318-11 Example 001 - 3

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Shear Design The following quantities are computed for all of the load combinations: ϕ

=

0.75 f c′ :

Check the limit of

f c′ = 63.246 psi < 100 psi

The concrete shear capacity is given by: ϕ Vc =

ϕ2

f c′ bd

= 12.807 k

The maximum shear that can be carried by reinforcement is given by: ϕ Vs =

ϕ8

f c′ bd = 51.229 k

The following limits are required in the determination of the reinforcement: (ϕ Vc/2)

= 6.4035 k

(ϕ Vc + ϕ 50 bd)

= 11.466 k

ϕ Vmax = ϕ Vc + ϕ Vs = 64.036 k Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for any load combination is calculated as follows: If Vu ≤ ϕ (Vc/2), Av = 0, s

else if ϕ (Vc/2) < Vu ≤ ϕ Vmax Av (V − φVc )  Av  = u ≥  s φ f ys d  s  min

where:   b  Av  w = max 50    s f  min   yt

  bw ,   f   yt

 3   4 

 f c′  

else if Vu > ϕ Vmax, a failure condition is declared.

ACI 318-11 Example 001 - 4

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Combo1

∙

Vu = 9.736 (5-13.5/12) = 37.727 k = φ (Vc = Vu 37.727 k ≤ φ V / 2 ) 6.4035 k ≤= 64.036 k max   10   10  3   Av   4, 000  ,   = max 50     s  min   60, 000   60, 000  4  in 2  Av  = = max 0.0083, 0.0079 0.0083 { }   in  s  min Av = s

(Vu − φVc ) in 2 in 2 = 0.041 = 0.492 in ft φ f ys d

ACI 318-11 Example 001 - 5