Flow Chart - One Way Joist Construction

1
Prepared & Compiled by Engr. Qamar uz Zaman, qamaruzzaman@msn.com
LAYOUT—ONE WAY JOIST/RIBBED SLAB
TOP SLAB
JOIST / RIB
Refer to Section 8.13 of ACI 318-08 for
Joist Construction

2
Design of One Way Joist Slab
Design of Top Slab
Assume thickness, t (inches)
Not < S/12 or 2”
ACI Sec 8.13.6.1
Calculate Loads and Load Effects
Wu = 1.2(Wself + WSDL) + 1.6 x Wlive
Mu = WuS2
/12
Where:
Wu = Factored Load (ACI Section 9.2)
Mu = Factored / Ultimate / Design Moment
Wself = t/12 x ɣconcrete (For NSC, ɣconcrete = 150 lbs/ft3
)
S = Clear Spacing between Ribs
WSDL = Superimposed dead load
Wlive = Refer to ASCE-7 2010, Table 4-1
Design of Top Slab
Tensile Strength of Concrete will be considered and checked that the Moment Capacity of Un-cracked
Section is > Mu
Where:
ft = tensile strength of concrete, psi
f’c = Compressive strength of concrete, psi
I = 2nd Moment of Area about axis of bending, in4
Y = Depth of centroid of the section, in
M = Moment capacity of un-cracked section
If M > Mu (No need to calculate Reinforcement, Concrete is providing sufficient resistance.
But provide temperature and shrinkage reinforcement as per ACI 7.12.2.
'
5 ct ff =
y
If
M t
=
Continued on next page…..
Temperature & Shrinkage Reinforcement, As,sh
• Will be provided to meet the requirements of Section 7.12 of ACI Code.
• Work out As,sh
• Find c/c spacing between the bars using Equation S = 12 Ab/As,sh
• Comply to spacing requirements of reinforcing bars as mentioned in Section 7.12.2.2
Where:
Ab = Area of bar, in2
Designed as a continuous beam
supported by the ribs.
A more conservative calculation can be done as:
Where:
If Top Slab thickness is calculated using
above relationship then there is no
requirement of design the slab for flexure.
Provide Temperature and Shrinkage rein-
forcement as per Sec 7.12.2 of the Code
12/
2
2
3
'
tb
t
M
ff
w
u
ct
φ






==
'
3
cw
u
fb
M
t
φ
=

3
Design of Joist/Ribs
Calculate Span Length, Ln
Refer to ACI Section 8.9
Calculate Joist thickness, h
Refer to ACI Table 9.5(a)
Calculate Loads and Load Effects
Wu = Top Slab Load + 1.2 x Wself
Bending Moment:
Mu = WuLn
2
/12 (Single Span, Integral Support) or
Mu = WuLn
2
/8 (Single Span, Simple Support) or
Mu = Use ACI Moment Coefficient for Multi Span
Shear Force:
Vu, at Support = WuLn/2 (Simple Support/Integral Support) or
Vu = Use ACI Shear Coefficient for Multi Span
Where:
Vu = Factored / Ultimate Shear Force
Calculate Reinforcement, As
• Remember that member has been proportioned (b, h), material properties (f’c, fy) are
known.
• Assume Bar # to calculate d
• d = h - concrete cover - 1/2 db
• ɸMn = ɸ x 0.85 fc’ x a x b (d-a/2) Eq (1)
• ɸMn = ɸ x As x fy (d-a/2) Eq (2)
• a = As fy / (0.85 fc’ x a x b) Eq (3)
• Calculate As using Eq (2) and (3), assuming steel is yielding (it will be confirmed later)
• Find out a, using Eq (3)
• a = β1c Eq (4)
Where β1 = For fc′ between 2500 and 4000 psi, β1 shall be taken as 0.85. For fc′ above 4000 psi, β1 shall be reduced line-
arly at a rate of 0.05 for each 1000 psi of strength in excess of 4000 psi, but β1 shall not be taken less than 0.65.
ACI Section 10.2.7.3
• Find c, the depth of Neutral Axis from extreme compression fiber
• Find c/dt , if c/dt < 0.375, Section is Tension Controlled and assumption confirmed
• if c/dt > 0.375, Increase the depth of the Joist to design a Tension Controlled Section
Where:
d = Depth from top compression fiber to centroid to tension reinforcement
db = Dia of bar
Mn = Nominal Moment Capacity of the Section
a = Depth of Whitney Stress block

4
Perform Checks
• Reinforcement Ratios:
⇒ Calculate
ϼmax = 0.75 ϼbal, Where
But not < 200/fy
ϼactual = As / bd
ϼmax > ϼactual > ϼmin
Where:
ϼmax = Maximum reinforcement ratio
ϼmin = Minimum reinforcement ratio
ϼactual = Reinforcement ratio being provided in section
• Reinforcement Spacing:
⇒ Maximum: ACI Section 10.6.4 but not <
⇒ Minimum: ACI Section 7.6 (db but not < 1”)
• Check for Shear:
⇒ Draw Shear Force diagram
⇒ Calculate Vu at critical Section i.e at distance ’d’from face of support
⇒ Calculate
⇒ If ɸVc > Vu (OK), otherwise increase slab thickness or provide Shear reinforcement.








+
=
yy
c
bal
ff
f
87000
87000
85.0
'
ζ
y
c
f
f '
min 3=ζ ACI Section 10.5
dbfV wcc
'
21.1 λφφ = ACI Section 8.13.8 & 11.2.1.1
cC
sf
S 5.2
40000
15 −








= 







=
sf
S
40000
12

Flow Chart - One Way Joist Construction

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