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MODULE 2: ANALYSIS AND DESIGN OF REINFORCED CONCRETE TWO-WAY SLABS(WSD)
• Depth Limitation for Two-Way Slabs NSCP
l = clear span in long direction n
β = ratio of long span to short span
α = ave. value of flexural stiffness of beam section on fm
edges of the panel
• Analysis and Design Procedure
• Both directions need to be considered hence the name.
• The Designer must determine the Moments in both the short span and the long span before applying the flexure formulas.
lb
-Ma
la
+Ma
• Determining Moments by the Coefficient Method
-Moment along Short Span
2
-Ma = C
a wtot la
-Moment along Long Span
2
-Mb = C
b wtot lb
+Moment along Short Span +Ma = MDL + MLL
MDL = CaDL WDL la
2 MLL = CaLL WLL la
2 +Moment along Long Span
+Mb = MDL + MLL
MDL = CbDL WDL lb
2 MLL = CbLL WLL lb
2 Where: la = length of the shorter span lb = length of the longer span C = moment coefficient
B-1
B-1
Examples
Determine the required reinforcement for the slab of the structural system shown below.
G-1
f’c=21 Mpa n = 12
Fy = 275 Mpa
Superimposed DL = 3.6 Kpa
LL = 2.4 KPa
5m
G-1
4.0m
Step 1: Determine what the problem provides and requires. The problem requires the amount of reinforcement of the beam (As) Step 2: Determine the slab thickness(t) Step 2: Determine the loads on the slab Given: DL=3.6 Kn/m2 (weight of the slab will be added)
DL=3.6 + 3.0 = 6.6 Kn/m2 LL = 2.4 Kn/m2
WDL = 6.6 ( 1m ) = 6.6 Kn/m WLL = 2.4 ( 1 m] ) = 2.4 Kn/m Then the total load on the joist is simply the sum of these two:
WT = 6.6 + 2.4 = 9.0 Kn/m
When it comes to formulas, beams and slabs share the same
formulas when it comes to flexure. So, the same formula applies
for slabs in terms of As.
M fs j d As =
For two way slabs, the simplest way to determine the required thickness is
Perimeter/180. If calculations will require a thicker slab then bending
provisions will govern/
t = Perimeter / 180
Take note that the uniform load W is in KN/m, while the given date is in
Kn/m2. This simply means you need to multiply the given data to a
“tributary width.” For one way slabs or slabs in general this is taken to be
1m only since we analyse the slabs in 1m strips
Wtslab = unit weight of concrete x thickness = 24 Kn/m3 x 0.125
Wtslab = 3.0 Kn/m2
t = {2 x (5+4)} / 180
t = 0.1m or 100mm use 125mm Since the approximate is exactly 100mm, it
is good practice to use the next
incremental thickness.
Step 3: Compute for the maximum bending moment (M). Negative Moment: Short Span
-Ma = Ca Wt la2
From Tables Ca = 0
-Ma = 0 Kn-m
Long Span -Mb = Cb Wt lb
2
From Tables Cb = 0
-Mb = 0 Kn-m
Positive Moment: Short Span +Ma = MDL + MLL
+MDL = CaDL WDL la2
From Tables CaDL = .056
+MDL = .056(6.6)(4.0)2 = 5.91 Kn-m
+MLL = CaLL WLL la2
From Tables CaLL = .056
+MDL = .056(2.4)(4.0)2 = 2.15 Kn-m
+Ma = 5.91 + 2.15 = 8.06 Kn-m
Long Span +Mb = MDL + MLL
+MDL = CbDL WDL lb2
From Tables CbDL = .023
+MDL = .023(6.6)(5.0)2 = 3.795 Kn-m
+MLL = CbLL WLL lb2
From Tables CbLL = .023
Before we can proceed with this step, there are a few things that must be
established that the tables require. First is the value for “m”
m = la/lb
m = 4.0/5.0
m = .80 this will hold thru for all tables
Next is to determine what “case” the slab in question is. The tables have
nine different cases depending on the edge condition of the slab, The
crosshatch indicates a continuous edge. Otherwise it is deemed
discontinuous. For this case, we have all discountinous edges because
there is no adjacent slab to the one being designed, so we will use Case 1.
Because the edges are discontinuous, the negative
moments are zero.
+MDL = .023(2.4)(5.0)2 = 1.38 Kn-m
+Mb = 3.795 + 1.38 = 5.175 Kn-m Step 4: Compute for As and spacing.
This step is simply filling up the formula
Positive Reinforcement: Short Span
Long Span
M fs j d As =
fs = 0.40 (fy) fy =275 Mpa
fs = 0.40 (275)
fs = 110 MPa
Solve for :
Solve for : d = h - Cc – (φb/2) h = 125mm
Cc = Concrete Cover
= 20 mm
φb = Bar diameter
= 12 mm
d = 125 - 20 – (12/2)
d = 99 mm
Solve for : j = 1 – k/3
k = n n + (fs/fc)
n = 12
fc = .45 fc’
fc = .45 (21)
fc = 9.45 Mpa
k = 12 12 + (110/9.45)
k = 0.508
j = 1 – 0.508/3
j = .8308
8.06 110x1000 (.8308)(.099)
+Asa =
As = .000891 m2 x 10002 = 891 mm2
S = Ab (1000)
As Ab = Area of rebar
Area of 12mm rebar = 113 mm2
S = 113 (1000)
891
s = 126.82 mm say 125 mm
5.175 110x1000 (.8308)(.099)
+Asa =
As = .000572 m2 x 10002 = 572 mm2
Negative Reinforcement:
Short Span -Asdiscont = 891/3 = 297 mm
2 s = 380.5mm
(or simply multiply the spacing by 3)
-sdiscont = 125 x 3 = 375 mm
Long Span -Asdiscont = 572/3 = 191 mm
2 s = 592.7mm
(or simply multiply the spacing by 3)
-sdiscont = 175 x 3 = 525 mm
NSCP states that for discontinuous edges:
-Asdiscont = (+Asmispan)/3
S = Ab (1000)
As Ab = Area of rebar
Area of 12mm rebar = 113 mm2
S = 113 (1000)
572
s = 197.56 mm say 175 mm