+M

b

-Mb

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