Section 15. Deflection under Short-Time Loading

alligatorsavoryUrban and Civil

Nov 26, 2013 (3 years and 6 months ago)

75 views

3

Section 15.


Deflection under Short
-
Time Loading


Deflections of reinforced concrete elements that respond primarily in flexure create design
challenges mainly in relation to six issues:

(1) Damage to partition walls on which they may rest or may come to rest.

(2)
Restraint of door mechanisms in contact with flexural elements

(3) Damage to piping attached near or at mid
-
span of flexural elements

(4) Ponding of water in roof slabs

(5) Unevenness
vibration of floor slabs

(6) Flexibility of floor slabs leading
to perceptible vibrations


Deflection concerns in these contexts are primarily serviceability concerns. As in the case of
flexural cracking, the main driver is the “working stress,” the unit stress in the tensile reinforcement
related to service load. A si
mple and direct procedure to keep deflection low is to keep the steel stress low.
To achieve that result for a given bending moment and span, the basic thinking process is simple. The
product of the cross sectional area of the tensile reinforcement and the

effective depth (As*d) should be
kept high vis
-
à
-
vis the moment demand but within the limits set by

applicable building code
1
.

From the viewpoint of efficient and proper d
esign, it is essential that the deflection problem be
approached as problem in geometry.

In this section we develop a simple procedure to determine the "short
-
time" deflection of
reinforced concrete elements with transverse loads. "Short
-
Time" deflection r
efers to deflection that is not
affected by time
-
dependent volume changes in concrete. Strictly, we are thinking of a load that is applied
in a matter of minutes but not milliseconds or days.

Our goal will be to develop procedures to determine and, more im
portantly, to understand the
deflection response of slender reinforced concrete beams.

To begin with, we need to emphasize that, in design of structures, the object is not to predict
deflection. Rather, the object is to control deflection or to keep it wi
thin bounds to avoid the six problems
listed above. The main factors controlling service
-
load deflection are the magnitude of reinforcement
stress and the effective depth of the section.

To determine the proper depth of the beam should be very simple. All
one has to do is to look at
existing beams and develop a sensitivity to their span
-
to
-
depth ratios. Naturally, the successful span
-
to
-
depth ratio will be different for different boundary conditions and for different load magnitudes and
distributions. Howev
er, for a large number of cases the student will observe that depths of reinforced
concrete beams range from 1/10 to 1/24 of the clear span. Even though it may appear to be an argument
for "know
-
nothingism" the budding designer would be wise to bear in min
d that a calculated deflection is
nothing more than a


guess

especially if time
-
dependent effects are involved
. It is better to put one's faith
in proportions that have led to success in the past rather than believe that a calculated beam deflect
ion is
exact.


Deflection of a Beam Subjected to Constant Moment


Determination of the deflection of a beam is best handled in




1

Typically, building codes will demand that the amount of reinforcement is l
ow enough to ensure that the
reinforcement

will yield before concrete fails if the beam or the slab is subjected to high bending moment. That
requirement is
likely

to keep the reinforcement ratio below 2% unless compressive reinforcement is used.

3

Relation to its strain geometry
. To develop the correct
perspective for treating the deflection problem in geometry,
we start with a very simple case: a
prismatic
and
homogeneous
beam segment
subjected to constant moment
as shown in
Figure
15
-
1
a.


We assume that the beam is weightless (no vertical reactions
needed and no change in moment along the span) and that it
is
responding
in its linear range of response.


To
emphasize the importance of geometry in thinking about deflections
, we break the segment into ten
elements of equal length (
Figure
15
-
1
b)
.


First, we consider the element to the left of mid
-
span (Element 1 in
Figure
15
-
1
c).
We start from
mid
-
span

because we know the slope
of the beam is
zero at
mid
-
span
.
T
he starting rotation is

zero

at the right
boundary of element
1
. W
e

have assumed that the entire beam is subjected to the same moment
magnitude.

The unit curvature, which is proportional to the applied moment,
is summed

over distance
L/10 to obtain
the change in
rotation
,

θ
1
,
over th
e length of element 1.


10
1
L





(
15
-
1
)


φ : Unit curvature in response to applied moment M

L : Length of beam


W
e note that the unit curvature has a unit of
(
1/Leng
t
h
). By itself,
unit
curvature

i
s
a definition of a rate of
change
.
It cannot be sensed at a single location. Two measurements of slope are needed to measure the
value of unit curvature.
Multiplying
the unit curvature
by a
finite
length produces an angle change
.
Equation 15
-
1

also implies that the unit curvature does not change over the length L/10.



Knowing the angle change taking place in Element 1,
w
e rotate the next element on the left (element 2) so
that its right boundary coincides with the left boundary of the first
element. The increase in angle change
from the right to the left boundary of element 2 is the same
, in this case with constant moment,

as that of
element 1. The angle change at the left boundary of element 2 is the sum of the angle change in element 1
plus

that in element 2.



5
L
10
L
10
L
2











(
15
-
2
)


Because the unit curvature in all elements is the same, it follows that the rotation at the support is



L
2
1
10
L
5
right
left











(
15
-
3
)


This derivation considering changes in the geometries of the discrete elements is, in essence, the same as
the definition in the moment
-
area method that specifies, in simple terms, that the change in rotation from
a point
in a beam to another point is the integral of the unit curvature variation between the two points.
Using the moment area method, we would note that the rotation (or slope) at mid
-
span is zero and
conclude that the rotation at the support would be simply th
e area of the unit
-
curvature diagram over one
half of the beam considered


L
2
1
port
sup





(
15
-
4
)


Figure
15
-
1

L

C

3

To determine the deflection at mid
-
span, we refer to
Figure
15
-
1
. We obtain the total deflection in a step
by step procedure by adding the tangential offsets at the left support. The segments are numbered
sequentially from center to the left support.


Deflection increments related to individual segments
are

10
L
5
.
4
10
L
1






10
L
5
.
3
10
L
2






10
L
5
.
2
10
L
3






10
L
5
.
1
10
L
4






10
L
5
.
0
10
L
5







The total deflection

is



5
.
0
5
.
1
5
.
2
5
.
3
5
.
4
100
L
2
5
4
3
2
1
total





















2
total
L
8
1






(
15
-
5
)


We can arrive at the same result directly by using the "Moment
-
Area Method."
For a beam with zero
slope at mid
-
span
the deflection at mid
-
span is
stated
,

in accordance with the moment
-
area method, to be

the deviation at either support of a tangent at mid
-
span.
The cited deviation
is
determined as the first
moment of the area of the unit curvature diagram between the mid
-
span and either support. In this simple
case (with constant unit curvature along the entire beam):


Deflection at mid
-
span = Unit Curvature x Tributary Span x Distance

from support to centroid of the unit
-
curvature diagram



4
L
2
L
total






(
15
-
6
)



2
total
L
8
1






(
15
-
7
)


The moment
-
area method is very simple and straightforward. It becomes a powerful and reliable tool as
long as the engineer does not lose his/her connection with the geometry leading to it. It is to be
understood that

the method is to be use
d
in problems
with small
deformations

[does the student know
what small deflections are?]
.







Deflection of a
n Uncracked

Reinforced Concrete
Beam with
Concentrated Loads

Figure
15
-
2

3


Having discussed the fundamental aspects of the momen
t
-
are
a

method and having emphasized that the deflection
must always be consider
ed in terms of
strain and beam
geometry, it is time to investiga
t
e beam deflections to
check how large or small they are likely to be
.


We start by

considering an uncracked reinforced concrete
beam. In practice an uncracked beam occurs very rarely
but
obt
aining an estimate of its deflection compared with
its span is instructive.


Figure
15
-
3
a shows a simply supported beam
spanning over a distance L with equal loads at third
-
points
of its span. The beam reactions are each (ignoring self
weight)


2
P
R
A


(
15
-
8
)


R
A

: Reaction at left support in response to applied load P

P : Total applied load


resulting in a maximum moment in terms of P and L of



6
L
P
3
L
2
P
max
M





(
15
-
9
)


M
max

: Moment acting over middle third of beam span.




(a) cross section

(b) unit strain

(c) unit stress


Figure
15
-
4


As we did in Section 13, to compute
curvature for an uncracked beam
we
determine the maximum stress
corresponding to the maximum moment


d

b

h



f

h/2

h/2

c

c

y
n

f(y
n
)



Figure
15
-
3

3



M
h
2

I
PL
6
H
2

1
1
12
b

h
3


PL
bh
2

(
15
-
10
)

h : total height of section

b: width of rectangular section

I : Moment of inertia of rectangular section.


From the maximum stress, we obtain the
maximum strain




(
15
-
11
)

ε
max

:strain at extreme fiber

σ
max

: stress at extreme fiber

E : Young's modulus for beam material


The maximum unit curvature is




φ
max
: unit curvature corresponding to M
max



(
15
-
12
)

Using the moment area method, we calculate the deviation at one support of a tangent at mid
-
span to
obtain the deflection at mid
-
span.







































12
L
3
L
6
L
3
L
3
2
2
1
3
L
max
max
max

(
15
-
13
)



2
max
2
max
max
L
h
2
216
23
L
216
23















(
15
-
14
)


The expression refers to a homogeneous beam responding to load within its linear range of response (an
uncracked concrete beam).




3


Figure 5

Next we deal with a cracked reinforced concrete section. We assume that it is reinforced in
tension only and we ignore the tensile strength of concrete. From the strain geometry we obtain
the relationship


c
k
d


s
d
1
k

(
)


3


In section 15, the value of k was deri
ved to be (in terms of the section properties for a
rectangular section reinforced in tension only)

k

n

(
)
2
2


n



n



k = ratio of depth of neutral axis to the effective depth

n = modular ratio, ratio of Young's modulus for steel to that of concrete

ρ = reinforcement
r慴i漬or慴i漠潦=cr潳s
-
獥捴io湡n=慲攠of=t敮eil攠rei湦潲c敭敮e=t漠th攠灲潤o捴=of=
t桥⁷h摴栠慮搠aff散tiv攠摥pt栠潦⁳散ti潮
=
坩t栠k=數灲敳s敤ei渠t敲m猠潦=t桥hpr潰orti敳=of=t桥hs散ti潮Ⱐw攠摥di湥nt桥hm慸amum=
畮ut= 捵rv慴畲攠 f潲= 愠 r散t慮a畬慲= 捯湣ret攠 獥cti潮= rein
f潲c敤e i渠 t敮獩潮= a湤n
r敳灯湤n湧⁩渠it猠si湥慲=r慧e=⁲敳灯p獥⁡s
=

s

s
d
1
k

(
)
=
卵扳pituti湧⁴桥⁣urv慴ur攠e渠n桥h數灲敳si潮of潲⁤敦le捴i潮Ⱐ摥ole捴i潮⁡tid
-
獰s渠n猠
=

midspan
23
216

s

L
2

=
T漠o敶敬潰⁡⁳=n獥f⁰=潰潲ti潮Ⱐo攠r数敡t⁴h攠e扯b攠畳u湧⁲数r敳敮e慴iv攠摩m敮ei潮献o
=
坥⁡獳畭e
=
t桥⁳灡測⁌Ⱐt漠扥′〠bt.
=
e慶i湧=l潯o敤e慴=潴桥h=獩m灬y=獵灰潲t敤e扥bm猬sw攠m慫攠t桥h摥灴栠h漠扥b慰ar潸om慴敬y=
潮o
-
t敮e栠潦=t桥hs灡渠or=㈠ft⸠Ag慩渠l潯oi湧=慴=獥ctio湳nw攠桡h攠獥敮ei渠t數e=bo潫猠慮搠
慲潵湤⁵猬⁷攠獥t⁴h攠睩eth=⁴h攠e散ti潮ot漠o攠ㄠet⸠.
=
3


First we consider the beam in its uncracked state. To simplify the solution we ignore the
influence of the tensile reinforcement on stiffness.

For the section to remain uncracked, the tensile stress of the concrete should not be
exceeded. We consider
a tolerable tensile stress of 400 psi. We assume the Young's
modulus for the concrete to be E
c

= 4,000 ksi. With that stress in the extreme fiber,
the corresponding strain is

f
t
400
psi


h
24
in


L
20
ft


E
c
4000
ksi



t
f
t
E
c



t
1
10
4




The maximum unit curvature


tmax

t
h
2



tmax
8.3
10
6


1
in



Deflection at mid
-
span


tmidspan
23
216

tmax

L
2




tmidspan
0.05
in



In terms of

the span, the deflection amounts to

R

t

tmidspan
L


R

t
0.02
%



The ratio of the span to the calculated deflection is

L

tmidspan
4.7
10
3



3


The ratios above tell us that the deflection is negligibly small. Note that without
comparing the deflection to the span of the beam, it would be difficult t
o judge how
small the calculated deflection is. In the practice of civil engineering, it is always
helpful to compare calculated results as a ratio of a quantity that generalizes the result.

The conclusion does not surprise us. What we have calculated is
an approximation to the
deflection at first cracking of a reinforced concrete girder spanning 20 ft. It should be
very small.


We should also remember that the computed deflection is qualified by assumptions we
have made in arriving at it. We ignored the s
elf
-
weight and the presence of tensile
reinforcement. We ignored any residual stresses (related to temperature and/or shrinkage)
that may exist in the concrete. In fact, the deflection we have computed is not of much
practical significance, but it is usefu
l for us to understand the effect of tensile strength of
concrete on deflections. A question we consider next.

We compute the mid
-
span deflection for the beam under its service load. What could its
service load be? To determine that we assume the beam to b
e reinforced in tension with 4
#8 bars.

Total cross
-
sectional area of tensile reinforcement

A
s
4
0.79

in
2


A
s
3.2
in
2



Yield stress

f
y
60
ksi


Comp. Str. Concrete

f'
c
4000
psi


Young's mod. steel

E
s
29000
ksi


Effective depth

d
h
1.5
in

.5
in



d
22
in



Width of rect. section

b
12
in


We always check the reinforcement ratio

3



A
s
b
d




1.2
%



n
E
s
E
c


n
7


k

n

(
)
2
2


n



n




k
0.34


Nominal flexural strength

M
u
A
s
f
y

d

0.9

1
1
2

f
y

0.85
2
f'
c














M
u
274
kip
ft




Is this a reasonable value? It is always good to check using the simple expression
below.

M
test
A
s
f
y

d

0.9

7
8



M
test
274
kip
ft




The service load may be approximately 2/3 the calculated moment

M
service
2
3
M
u


M
service
183
kip
ft




The steel stress at assumed service load

f
s
M
service
A
s
1
k
3








d



f
s
36
ksi



The steel strain

3



s
f
s
E
s



s
1.2
10
3




Maximum unit curvature at service load


s

s
d
1
k

(
)




s
8.4
10
5


1
in



Deflection at mid
-
span


smidspan
23
216

s

L
2




smidspan
0.5
in



RD
smidspan

smidspan
L


RD
smidspan
0.2
%



Does this result make sense? We look at it in another way.

L

smidspan
464


So the calculated deflection at mid
-
span is approximately 1/450 of the span. It is
approximately ten times the
deflection at cracking.

Is it a reasonable ratio? We need to think about it as we look at beams in buildings and as
we calculate service
-
loaddeflections for other beams.

We also need to recognize two qualifications to our calculated value.

1. We made the

calculation assuming that the section was uniformly cracked throughout
the span. We know that is not correct. Even in the middle third of the beam where the
moment is constant, the number of cracks are bound to be finite. In between cracks, the
section is

likely to be stiffer. That is also true for the two outer spans where the moment
decays toward the supports and cracking is even more limited. The deflection calculated
is likely to be higher than the actual value, assuming that the moduli are correct.


3


2. In an actual construction, the beam is cast in a form. The form is usually
cambered based on the experience of the engineer. The visible deflection, the
difference in elevation between the support and the point of maximum deflection
(usually at mid
-
span
) is likely to be smaller than the calculated value.