Flexural Design of

Beam Sections:
Analysis of Continuous
Beams & One-Way Floor
Systems
 Load Paths
Consider the idealized
one-way floor system:

Assume a concentrated
load is applied at p in
the central slab panel.

The one-way slab panel

carries the load to m & n.
Fig. 7-1 Load paths in a one-way floor system.

2
Load Paths

The floor beams carry

the loads to h, i, j &
k on the girders.

Girders transfer the loads

to the columns at W, X,
Y, and Z.

Fig. 7-1 Load paths in a one-way floor system.

3
Tributary Areas

Floor systems in almost all buildings are

designed for uniformly distributed dead qD
& live qL loads.
Total dead load is normally composed of dead
loads superimposed on the floor system, and
the self weight of the floor members.

4
Tributary Areas

Typical live loads of various of various types of

structures are given in the following table:
Table 7-1 Minimum Live Loads.

5
 Tributary Areas

Floor beams are designed to resist area loads

acting within the tributary area.
The area load q (kN/m2)
is multiplied by the
tributary width to get
a line load w (kN/m).

For one-way slab, the

width of analysis strip
is considered 1 m.
Fig. 7-2 Tributary areas for floor beams.

6
Tributary Areas

Fig. 7-3 Width of analysis strip & tributary area for one-way slab strip.

7
Pattern Loadings for Live Load

Largest moments in a continuous beam occurs

when some spans are loaded with LL and
others are not.
Influence lines are used to determine which
spans should be loaded.

An influence line is a graph of the variation of

the moment, shear or other effect at one
particular point in a beam due to a unit load.
8
Pattern Loadings for Live Load

Influence lines can be

used to establish
loading patterns to
maximize the
moments or shears
due to live load.

Fig. 7-4 Influence lines for moments and loading patterns.

9
Pattern Loadings for Live Load

Fig. 7-5 Influence

lines for shears.

10
Pattern Loadings for Live Load

Using this sort of reasoning, ACI Code Section

6.4.2 defines loading patterns to determine
maximum design moments for continuous beams
and one-way slabs:
1. Factored dead load on all spans with factored
live load on 2 adjacent spans and no live load on
any other spans. (–ve Mmax & Vmax at supports
between 2 loaded spans).
2. Factored dead load on all spans with factored
live load on alternate spans. (+ve Mmax at
midspan, –ve Mmax & Vmax at exterior support).
11
Live Load Reduction

ASCE 7-10 allows a reduction on a member having

an influence area AI = KLLAT of 37.16 m2 or
more:
 4.57 
L  L0  0.25 
  SI units 
 K LL AT 

Where L: Reduced design live load per m2

L0 : Unreduced design live load per m2
KLL: Live load element factor
AT : Tributary area in m2

12
Live Load Reduction
Table 7-2 Live Load Element Factor, KLL.

13
Live Load Reduction

L shall not be less than 0.50L0 for members

supporting one floor.

L shall not be less than 0.40L0 for members

supporting two or more floors.

No reduction is allowed for live loads ≥ 4.79 kN/m2

or for structures used for public assembly,
garages or roofs.

14
ACI Moment and Shear Coefficients

Set of approximate moment & shear coeff. for

the analysis and design of non-prestressed
continuous beams and one-way slabs.
The following combination usually governs:
w  1.2w  1.6w
u D L

If w  1.4w happens to govern, then the ACI

u D

moment coeff. cannot be used, and a full

structural analysis is required.
15
ACI Moment and Shear Coefficients

ACI Code Section 6.5.1 gives the requirements

for using moment & shear coefficients:
1. Two or more continuous spans.
2. Approximately Equal Spans, longer of 2
adjacent spans not greater by 1.2 times the
shorter one.
3. Loads are uniformly distributed.
4. LL < 3(DL) (Unfactored)
5. Members are prismatic.

16
 ACI Moment and Shear Coefficients

Typically produce values larger than those

obtained from exact analysis.
If any of these conditions are violated, then a
full structural analysis is required.
ACI coeff. are based
on clear span n

Fig. 7-6 Definitions of clear span and average clear span for use
with ACI moment and shear coefficients.

17
ACI Moment and Shear Coefficients

Maximum +ve and ve moments and shears are

computed from: M  C  w  2
u m u n

w 
V C  
u n

 
u v
2
For all +ve midspan moments, all shears and ve
moments at exterior supports, is the span n

under consideration.
For ve moments at interior supports,  avg  is n

used.
18
Fig. 7-7 ACI moment and shear coefficients. 19
Example 7.1
Consider the continuous floor beam ABCD.
Use the ACI moment
coefficients to find the
design moments at the
critical sections for one
exterior span and the
interior span.

20
Example 7.1
Assume hf = 150 mm, h = 610 mm, bw = 300 mm.

Assume the columns are 460 mm by 460 mm.

Finally assume the floor is to be designed with a

live load of 2.85 kN/m2 and a superimposed dead
load (SDL) of 1 kN/m2.

21
Example 7.1 (Solution)
1. Confirm that ACI moments coeff. can be used.

-There are 2 or more spans.

-The spans are approximately equal: 8.6/7.2 = 1.19
-The loads are uniformly distributed.
-3(DL) = 3(240.15) = 10.8 kN/m2 > LL = 2.85 kN/m2.
-Members are prismatic.

22
 Example 7.1 (Solution)
2. Total factored loads.
qslab + SDL  3.6  1  4.6 kN/m2
wslab + SDL  4.6  3.6   16.56 kN/m

w beam web  24  0.61  0.15 0.3  3.31 kN/m

wD  3.31  16.56  19.87 kN/m
wL  2.85  3.6   10.26 kN/m
wu  1.2 19.87  +1.6 10.26   40.26 kN/m
wu  1.4 19.87   27.82 kN/m  does not govern 
23
Example 7.1 (Solution)
3. Design moments.
a) ve moment at face of support A:
1
Cm  
16
 n  AB  8.6  0.46  8.14 m
1
M u  Cm  wu n     40.26  8.14   166.7 kN.m
2 2

16

24
Example 7.1 (Solution)
3. Design moments.
b) +ve moment at midspan of AB:
1
Cm 
14
 n  AB  8.14 m
1
M u   40.26 8.14   190.5 kN.m
2

14

25
Example 7.1 (Solution)
3. Design moments.
c) +ve moment at midspan of BC:
1
Cm 
16
 n BC  7.2  0.46  6.74 m
1
M u   40.26  6.74   114.3 kN.m
2

16

26
Example 7.1 (Solution)
3. Design moments.
d) ve moment at face of support B:
1
Larger Cm governs: Cm  
10
 n avg  8.14  6.74 / 2  7.44 m
1
M u    40.26  7.44   222.8 kN.m
2

10

27
Example 7.2
Find required As at mid span of beam AB (Mu = 178
kN.m) and at face of support B (Mu = -196 kN.m) of
the continuous floor beam ABCD.
f’c = 28 MPa,
fy = 420 MPa.

28
Example 7.2 (Solution)
1. Determine be.
bw  clear tranverse span  3600 mm

be  bw  16h f  300  16 100   1900 mm

 4  8600 4  2150 mm
be =1900 mm

 be  1900 mm

29 29
Example 7.2 (Solution)
2. Calculate As for +ve moment at mid span of beam AB.
Assume a tension-controlled section: ϕ = 0.9
Assume a < hf (T-beam in +ve bending is treated as a beam of
rectangular section with wide compression zone).
Assume 1 layer of reinforcement: d  600 – 65 = 535 mm
Mu 178(10)6
Rn    0.364 MPa
be d 2
0.9(1900)(535) 2

0.85 f c  2 Rn  0.85(28)  2(0.364) 

 1  1   1  1  
fy  0.85 f c  (420)  0.85(28) 
  0.000864  As   be d  0.000873(1900)(535)  887 mm2
30
 Example 7.2 (Solution)
3. Check if As  As,min. be =1900 mm

0.25 f c 1.4 bw d d = 535 mm

As ,min  bwd 
fy fy

0.25 f c  1.323  1.4

1.4
As ,min   300  535  535 mm2
420

As = 887 mm2 > As,min = 535 mm2.

31
 Example 7.2 (Solution)
4. Check if a < hf . be =1900 mm

As f y (887)(420)
a  d = 535 mm
0.85 f cbe 0.85(28)(1900)
a  8.2 mm  h f  100 mm

5. Check ductility of section.

c  a / 1  8.2 / 0.85  9.6 mm
 d c   535  9.6 
t   s    0.003    0.003  0.16419  0.005
 c   9.6 
Tension controlled section    0.9

32
Example 7.2 (Solution)
6. Calculate As for ve moment at face of support B.

Flange of T beam in Tension Treat section as rectangular.

Assume 1 layer of reinforcement: d  600 – 65 = 535 mm

Mu 196(10)6
Rn    2.536 MPa
bwd 2
0.9(300)(535) 2

0.85 f c  2 Rn  0.85(28)  2(2.536) 

 1  1   1  1  
fy  0.85 f c  (420)  0.85(28) 
  0.006400  As   bwd  0.0064(300)(535)  1027 mm2

33
 Example 7.2 (Solution)
7. Check if As  As,min. be =1900 mm

0.25 f c 1.4 bw d
As ,min  bwd  d = 535 mm
fy fy

0.25 f c  1.323  1.4

1.4
As ,min   300  535  535 mm2
420

As = 1027 mm2 > As,min = 535 mm2.

34
Example 7.2 (Solution)
8. Check if a < 500 mm . be =1900 mm

As f y (1027)(420)
a 
0.85 f cbw 0.85(28)(300) d = 535 mm

a  60.4 mm

9. Check ductility of section.

a  1c  c  a / 1  60.4 / 0.85  71.1 mm
 d c   535  71.1 
t   s    0.003    0.003  0.01957  0.005
 c   71.1 
Tension controlled section    0.9

35