slab and beam IHFDS

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CHAPTER I
BEAM AND GIRDER FLOOR
1.1. Data and specifications
A concrete floor slab type beam and girder is considered as a part of a building
(Fig. 1.1), with two transversal spans L = 9. m and longitudinal spans l = !. m.
"he resistant structure is reali#ed of columns and girders, that are part of the floor.
Fig. 1.1
"he floor is reali#ed of o monolith slab that support on beams that transmit the load
to girders (Fig. 1.1).
"he ratio λ =
a
l
is considered bigger than $, the slab being reinforced on one
direction%
$ !& . $
$' . $
. !
> · · λ
.
Concrete (uality is chosen function the e)posure class of the building in
environment conditions. "he design building is a te)tile production hall and there is not
e)cessive humidity. According to "able *.1 and Fig. *.$a, Appendi) *, the construction is
included in class +,1 without corrosion ris- or chemical attac-. For this class the
minimum concrete grade is ,$.$'.
"he floor will have a minimum resistance to fire of ! minutes (/!).
For concrete C20/25 the design strength in compression, using relation $.11 and
"able $.1%
00 . 10
' . 1
$
1 · · ·
c
ck
cc cd
f
f
γ
α
1.mm
$
"ensile strength is obtained with the relation%
1
' . 1
' . 1
1
' .
· · ·
c
ctk
ct ctd
f
f
γ
α
1.mm
$
$ . $ ·
ctm
f
1.mm
$
Steel reinforcement type 2' and welded mesh type 2"13 is used for the slab
and beams.
• For 2'
' ·
yk
f
1.mm
$
40'
1' . 1
'
· ·
yd
f
1.mm
$
• For 2"13 (for d5 &.1 mm)
060
1' . 1
44
· ·
yd
f
1.mm
$
"he finish is reali#ed of mosaic (0 mm depth) on a layer of cement plaster of 0
mm depth.
"he service load is%
n
u
p = ' 1.m
$
.
1.2. Desin of !"a#
1.2.1. $%e"i&ina%' si(in of t)e s"a#
Fle)ural design will ordinarily consist in selecting a slab depth which will permit to
use of an economical low steel ratio and which will not allow unsightly or damaging
deflections.
For slabs elastically fi)ed on perimeter, the rigidity condition is%
h
p
7
0'
1
a, where a is the opening parallel with the short side, a = $.$' m.
h
p
=
$$'
0'
1

= !4.0 mm
And the minimum depth from technological condition to monolith slab, is ' mm,
the depth from the above conditions will be%
h
p
= & mm
From the fire condition the minimum depth is h
p
= 6 mm.
1.2.2. *idt)s fo% #ea&s and i%de%s
Fig. 1.$
From technological conditions the width of beams (b
b
) can be between 1' mm and
$' mm, and the width of girders (b
g
) and edge beams (b
eb
) between $ mm and 0'
mm. From fire resistance condition b> 1$ mm.
8ne chooses%
b
g
= 0 mm
b
eb
= 0 mm
b
b
= $ mm
1.2.+. C"ea% spans
Fig. 1.0
$
$ . 0 .
$' . $
$
1

− ·

− ·
b b
a
l
b eb
c
= $. m
l
c$
= a 9 b
b
= $.$' 9 .$ = $.' m
1.2.,. Loads (according to 2"A2 111.8A:&& ;0< and 111.1:&6 ;4<)
A. =nfactored value
• >ermanent load
: from own:weight of slab (h
p
= & mm)
n
slab
g = .& ⋅ 1. ⋅ 1. ⋅ $' = 1&' 1.m
$
: from plaster weight (h
plaster
= 0 mm)
n
plaster
g
= .0 ⋅ 1. ⋅ 1. ⋅ $1 = !0 1.m
$
: from mosaic weight (h
mosaic
= 0 mm)
n
mosaic
g = 1 1.m
$
(according to 2"A2 ...)
"otal% g
n
= 006 1.m
$
• "emporary load
: service load given in pro?ect data%
n
u
p = ' 1.m
$
"otal load%
n
u
n n
p g q + · = 006 @ ' = 606 1.m
$
3. Aesign Boad
"he design loads are determined by multiplying the unfactored loads by loading
coefficients, n
i
, tht are given in 2"A2 ... function the load type.
• >ermanent load
: from own weight of slab 1&' ⋅ 1.1 = 19$' 1.m
$
: from plaster weight !0 ⋅ 1.0 = 6$ 1.m
$
: from mosaic weight 1 ⋅ 1.0 = 10 1.m
$
"otal% g = 4' 1.m
$
• "emporary load
p
u
= ' ⋅ 1.0 = !' 1.m
$
"otal load% q = g @ p
u
= 4' @ !' = 1'' 1.m
$
Boads on linear meter (for a strip of slab of 1 m width, Fig. 1.$)%
q
n
= 606 1.m
$
⋅ 1. m = 606 1.m
q = 1'' 1.m
$
⋅ 1. m = 1'' 1.m
1.2.5. !tatica" Co&p-tation
For determining the bending moment it considers a strip of 1. m width, that has as
support the beams (points A, 3, ,, A, C, F, D, E, *).
>lastic analysis may be used to determine the bending moments because the
permanent loads are important ;'<.
Fig. 1.4.
: in the first span%
( )
( )
11
. $
!' 4'
11
$
$
1
1
⋅ +
·
⋅ +
·
l
p g
M
c
u
= 06!0 1m
: in the first support%
( )
( )
14
. $
!' 4'
14
$
$
1
⋅ +
·
⋅ +
− ·
l
p g
M
c
u
B
= :014 1m
: in the interior spans and interior supports%
· − ·
M M
C $
...
( ) ( )
1!
' . $
!' 4'
1!
$ $
$
⋅ +
·
⋅ +
·
l
p g
c
u
= $&&1
1m
1.2... !tee" desin
*e"ded &es)
Faterials characteristics
f
yd
= 060 1.mm
$
for 2"13
f
cd
= 10.00 1.mm
$
for , $.$'
2ection data
b = 1 mm
h
p
= & mm Fig. 1.'
G = ' mm
mm d
mm c c c
c h d
tol nom
nom p
' . 4& $ . ' $ &
$ 1 1
$
G
min
s
· − − ·
· + · ∆ + ·
− − ·
"he value c
min
is obtaining from "able '.$ function of e)posure class (+,1) and
structure class. For establishing the structure class the life of the building was considered
' years (class 4) and provisions from "able '.4 that consider a reduced class for
members type flat (4:1=0)
c
min
=1 mm
According to "able !.! the minimum distance to the centroid of resistant steel is a
= $ mm. For a reinforcement with diameter of ' mm, effective distance to the centroid
is%
a
eff
= $@'.$=$$.' mmH$ mm
"he section of tensioned steel will be%
Fi%st span/
:/educed moment is%
M
1
= 060! ⋅ 1
0
1mm
060 . 1$& .
00 . 10 ' . 4& 1
1 60! . 0
lim
$ 0
!
$
1
· < ·
⋅ ⋅

·
⋅ ⋅
· µ µ
cd
f d b
M
*t results singly reinforced for which from "able 0.'a for the value of I=.1$& it
results ω
s
=.10!9
· ⋅ ⋅ ·
yd
cd
s sl
f
f
d b A ω
$ $ 0
$! . $ 1&' . 0 . $$!
060
00 . 10
' . 4& 1 10!9 . cm mm · · ⋅ ⋅
8ne chooses from Anne) +*J% welded mesh 110 DK $4!%L'.!.1:L'.!.1
(A
sleff
= $4! mm
$
).
Fi%st s-ppo%t
M
B
= :014 ⋅ 1
0
1mm
060 . 1 .
00 . 10 ' . 4& 1
1 14 . 0
lim
$ 0
!
$
· < ·
⋅ ⋅

·
⋅ ⋅
· µ µ
cd
B
f d b
M
ω
s
=.1'! and M=.10$
1&4
060
00 . 10
' . 4& 1 1'! .
0
· ⋅ ⋅ ⋅ ·
sl
A
mm
$
= 1.&4 cm
$
8ne chooses% welded mesh 111DK 19! L'.1:L'.1 (A
sleff
= 19! mm
$
).
Inte%io% spans and inte%io% s-ppo%ts
M
$
= :M
C
= M
0
= :M
D
= ... = $&&1 1m
060 . 9$ .
00 . 10 ' . 4& 1
1 &&1 . $
lim
$ 0
!
$
$
· < ·
⋅ ⋅

·
⋅ ⋅
· µ µ
cd
f d b
M
ω
s
=.9&$
& . 1!
060
00 . 10
' . 4& 1 9&$ .
0
· ⋅ ⋅ ⋅ ·
sl
A
mm
$
= 1.!& cm
$
8ne chooses% welded mesh 111DK 19! L'.1:L'.1 (A
sleff
= 19! mm
$
).
"he arrangements of bars for both reinforcing type are presented in schedules 1 and
$.
For constructive provisions for slabs see ,hapter *** (point 1).
"he effective area of tensioned longitudinal reinforcement must be between the
limits%
&' . !1
&' . !1 ' . 4& 1 10 . 10 .
&' . !1 ' . 4& 1
44
$ . $
$! . $! .
ma)
0
0
min
·
¹
¹
¹
)
¹
¹
¹
¹
'
¹
· ⋅ ⋅ · ⋅ ⋅
· ⋅ ⋅ · ⋅
·
d b
d b
f
f
A
yk
ctm
sl
mm
$
19 ' . 4& 1 4 . 4 .
0
ma)
· ⋅ ⋅ · ⋅ ⋅ · d b A
sl
mm
$
19! &' . !1
min , , min ,
· < ·
eff sl sl
A A
mm
$
N
19
ma) ,
·
sl
A
mm
$
"he ma)imum distance between resistant steel is limited to%
$
$
$1 & 0 0
min
ma)
·
¹
)
¹
¹
'
¹
· ⋅ ·
·
mm
h
s
p
mmH
1 ·
eff
s
mm
For the slab reinforced on one direction, there are reinforcements on the other
direction at least%
$ . 49 $4! $ . $ .
,
· ⋅ ·
eff sl
A
mm
$
N
19! ! . 19 1 · ⋅ ·
eff
A
mm
$
(L'.1)
For ta-ing over the local moments on continuous supports from the short direction
and on edge supports in the long direction, constructive welded meshes are provided type
11$ DK 19! L'.1: L'.1, which are e)tended on both sides of support with .$' l
n
1.2.0. Desinin of ties 1cent-%a2
"he ties are reinforced according to standard provisions because the computation is made
only for a current floor, not for the entire structure.
0 · ·
! c
b b
mm H
$'
min ,
·
c
b
mm
$ $
min ,
· · ·
c c
h mm h
mm
! $ 0
,
· × ·
eff c
A
mm
$
H
'
min ,
·
c
A
mm
$
Fig. 1.!
46 ! 1 6 . 1 6 .
$
,
$
· ⋅ ⋅ · ⋅ ⋅ ·
− −
eff c s
A A
mm
$

1 !Φ ⇒ with
4&1
min ,
·
s
A
mm
$
(Fig. 1.!)
"he value of minimum cover with concrete depends on the type of e)ternal finish.
"he value of minimum cover is given function the structure class (4)%
$'
min
· c mm H L
sl
=1 mm
0' 1 $'
min
· + · ∆ + · c c c
nom mm
*f the e)ternal wall is with plaster and lime, the cover layer can be reduced with
'
,
· ∆
add dur
C
mm.
For transversal stirrups (having L
sw
=! mm) the concrete cover is%
$9 ! 0'
,
· − · Φ − ·
s" sl nom #
c c
mm
Finally, the concrete cover will be considered
0' ·
sl
c
mm for longitudinal steel
and
0 ·
s"
c
mm for stirrups.
1.2.3. Anc)o%in of %einfo%ce&ent
1.+. Bea& Desin
1.+.1. $%e"i&ina%' si(in of t)e #ea&
For secondary beam the depth of the cross:section can be considered with the
following value%
• "he rigidity condition%
1$ .. 1
!
1$ .. 1
· ·
l
h
b = '..! mm
*t is adopted h
b
= ' mm
• From the condition%
0 .. $ ·
b
b
b
h
it results that%
0 .. $
'
0 .. $
· ·
b
b
h
b = $..$' mm
*t is adopted b
b
= $ mm.
• From technological condition% the depth must be a multiple of ' mm
• From the condition of fire resistance%
h
b
>1$ mm
For the beam the following si#es are adopted%
h
b
4 500 && and b
b
4 200 &&
1.+.2. !tatica" &ode"
"he beam can be considered as a continuous beam with e(ual opening, that
supports on the girders (Fig. 1.&).
Fig. 1.&.
1.+. !pans
Fig. 1.6.
"he design spans are function of clear spans, (Fig. 1.6) which are determined as it
follows%
l = ! m
$
1
1
b b
l
g eb
c

− ·
l
c$
= '.& m l
c$
= l
!
l
c%
= 1 9 b
g
l
c%
= '.& m l
c%
= l
0
= l
4
= l
'
l
c$
&l
c%
&: clear spans
b
eb
9 width of the edge beam (0 mm)
b
g
9 width of the girder (0 mm)
"he design spans are computed ta-ing into account the provisions given in Anne)
**.
, !
$
4 .
O
$
0 .
min $ & . '
$ 1 1 1
·
1
]
1

¸

⋅ + · + + · a a l l
c eff
m
, !
$
4 .
O
$
0 .
min $ & . '
$ 1 $ $
·
1
]
1

¸

⋅ + · + + · a a l l
c eff
m
1.+.,. Loads
"he loads from the slab are transmitted to the beam from a strip e(ual to a as
uniform load, Fig.1.9.
Fig. 1.9
A. 5nfacto%ed "oad
• >ermanent load (g
n
)
: transmitted by the slab and finish
a = $.$' m
a g g g
n
plaster
n
mosaic
n
slab
⋅ + + ) (
= (1&' @ !0 @ 1) ⋅ $.$' = &!' 1.m
: self weight of the beam
(h
b
9 h
p
) ⋅ b
b
⋅ '
b
= (.' 9 .&) ⋅ .$ ⋅ 1. ⋅ $' = $1' 1.m
"otal permanent load% g
n
= &!' @ $1' = 9&'' 1.m
• "emporary load (p
n
)
p
n
= p
u
n
⋅ a = ' ⋅ $.$' = 11$' 1.m
"otal load% q
n
= g
n
@ p
n
= 9&'' @ 11$' = $1' 1.m.
B. Desin "oad
• >ermanent load (g)
: transmitted by the slab and finish
(g
slab
@ g
mosaic
@ g
plaster
) ⋅ a = (190 @ 6$ @ 10) ⋅ $.$' = 911$.' 1.m
: self weight of the beam
(h
b
9 h
p
) ⋅ b
b
⋅ '
b
⋅ 1.1 = (.' 9 .&) ⋅ .$ ⋅ $' ⋅ 1.1 = $0!' 1.m
"otal permanent load% g = 911$.' @ $0!' = 114&&.' 1.m
• "emporary load
p = p
u
⋅ a = !' ⋅ $.$' = 14!$' 1.m
"otal load% q = g @ p = 114&&.' @ 14!$' = $!$$.' 1.m.
1.+.5. !t%ess Co&p-tation
1.+.5.1. Bendin &o&ents fo% se%6ice stae
For computing to limit state of crac-ing and to limit state of deflection is necessary
to determine the values of bending moments for second stage of wor-ing (service loads).
3ecause of that the bending moments are computed in elastic domain, as for a continuous
beam with constant moment of inertia.
Fig. 1.1.
For determining the bending moments one can use the values given in Appendi) 6.
"he values and the coefficients for the continuous beam with five spans, and the
ma)imum and minimum bending moments are given in "able. 1.1.
"able 1.1.
!ec
tio
n
Coefficient of inf"-ence
!pan
1Teff2
1&2
$e%&anent
"oad
g
17N/&2
8e&po%a%'
"oad
p
17N/&2
M&a91:2 M&a91;2
a # c
1a⋅g:b⋅p2⋅l
⋅l
1a⋅g:c⋅p2⋅l⋅
l
1 ,&$ ,99 :,$! !. 11,4&6 14,!$' &0,691& 14,49!
3 :,1' ,14 :,1$ !. 11,4&6 14,!$' :0$,'40 1&,6!00
$ ,00 ,&9 :,4! !. 11,4&6 14,!$' 49,644' :9,''10
, :,&9 ,0$ :,111 !. 11,4&6 14,!$' :14,$''4 :6$,$4$
0 ,4! ,6' :,09 !. 11,4&6 14,!$' '&,'40' :1,0&&$
For the limit state of crac-ing%
Fig. 1.11.
1.+.5.2. Bendin &o&ents fo% desin "oads
"hese moments are used for designing the ultimate limit state in normal section.
"he bending moments are computed in plastic domain, with the following relations%
According to 2"A2 11&.$:9$, current floors of slab and beams reali#ed of
reinforced concrete and Pprestressed concreteQ when the ratio between long and
temporary loads and total is smaller than .&' is sufficiently the computation of stresses
in plastic domain.
*n the case of beams the ratio is
' . $!1$
14!$'
= .'!
: First span%
11
! ' $!1$
$
11
$
1
1

l q
eff
M

·

·
= 6'4$!.4 1m
: First support%
14
. ! ' . $!1$
$
14
$
1

·

·
l q
eff
M
B
= !&1$.& 1m
: 8ther interior spans and supports%
1!
. ! ' . $!1$
$
1!
$
$
0 $

·

· − · · − ·
l q
eff
M M M M
D C
='6&0 1m
1.+.5.+. !)ea% fo%ces
2hear forces are used in computation at ultimate limit state in inclined sections. "he
values of shear forces are determined in plastic domain with the following relations (Fig.
1.11)%
: in ad?acent sections of edge support%
(
A
= .4' ⋅ q ⋅ l
eff$
= .4' ⋅ $!1$.' ⋅ !. (
A
= &4&& 1
: in first span, ad?acent to first interior support%
(
B
left
= .!' ⋅ q ⋅ l
eff$
= .!' ⋅ $!1$.' ⋅ !. (
B
left
= 11&99 1
: other sections ad?acent to interior supports%
(
B
right
= .'' ⋅ q ⋅ l
eff%
= .'' ⋅ $!1$.' ⋅ !. J
B
right
= 6!106 1
Fig. 1.11.
1.+... Co&p-tation at t)e -"ti&ate "i&it state in no%&a" section
,haracteristics of materials%
f
yd
= 40' 1.mm
$
for 2'
f
cd
= 10.00 1.mm
$
for , $.$'
"he depth of the cross:section is%
mm d
mm c c c
c h d
tol nom
nom g
4! $ . $ 0 '
0 1 $
$
G
min
s
· − − ·
· + · ∆ + ·
− − ·
1.+.0. !tee" Desin
"he beam has flanged section in spans and rectangular section in support.
• Fi%st span
Cffective width of the slab must be established from the conditions given in the
Appendi) **.
Fig. 1.1$.
mm 14$'
1$'
$
$
$
$$'
$ $
b where
mm $$'
mm 14$' $ ' . !1$ $
min
$ 1
$ 1
,
· ⇒
¹
¹
¹
¹
¹
)
¹
¹
¹
¹
¹
¹
'
¹
· − · − · ·
· + +
· + ⋅ · + Σ
·
eff
b
b
b i eff
eff
b
mm
b a
b
b b b
b b
b
Rhere%
mm 1$ ! 6' . $ . mm ' . !1$
! 6' . 1 .
$
1$'
$ . 1 . $ .
$ , 1 $ , 1
· ⋅ ⋅ < ·
· ⋅ ⋅ + ⋅ · ⋅ + · ) b b
eff
"he effective width of the slab is adopted with the smallest value. 2o%
b
eff
= 14$' m
2ection data
b
b
= $ mm h
b
= ' mm
d = 4! mm h
p
= & mm
b
eff
= 14$' mm
F
1
=6'4$!.4 nm
>ositions of the neutral a)is
"he position of neutral a)is is verified%
' 1$' . &
$
14$'
> · ·
b
eff
b
b
919 .
4!
&
1
4!
&
$
14$'
' . 1 1'1 .
00 . 10 4! $
1 4 . 6'4$!
$
0
$
ma)
·
,
_

¸
¸
− ⋅ ·
·


,
_


¸
¸
− ⋅ < ·
⋅ ⋅

·
⋅ ⋅
·
d
h
d
h
b
b
f d b
M
p p
b
eff
cd b
µ
"he neutral a)is is in the flange *+,h
p
-
"he section is computed as a rectangular cross:section having the width beff and the
depth d.
Fig.
$1$ .
00 . 10 4! 14$'
1 4 . 6'4$!
$
0
$
1
·
⋅ ⋅

·
⋅ ⋅
·
cd eff
f d b
M
µ
S=.$1$
4$'
40'
00 . 10
4! 14$' $1$ . · ⋅ ⋅ ⋅ · ⋅ ⋅ ⋅ ·
yd
cd
eff sl
f
f
d b A ω
mm
$
8ne chooses 0G14 A
aeff
= 4!$ mm
$
• Inte%io% !pans
"he width of the flange is%
mm 64 ! & . $ . mm ' . '$$
! & . 1 .
$
1$'
$ .
$ , 1
· ⋅ ⋅ < ·
· ⋅ ⋅ + ⋅ ·
eff
b
mm mm b b b
b eff eff
$$' 1$4' $ ' . '$$ $ $
$ , 1
< · + ⋅ · + ⋅ ·
3ecause
' $$ . !
$
1$4'
> · ·
b
eff
b
b
it results that neutral a)is passes through the
flange and the section is considered as rectangular cross section having the width e(ual to
b
eff

1!& .
00 . 10 4! 1$4'
1 '6&0
$
0
$
$
·
⋅ ⋅

·
⋅ ⋅
·
cd eff
f d b
M
µ
from "able ......it results%
S=.1&'
0&
40'
00 . 10
4! 1$4' 1&' . · ⋅ ⋅ ⋅ · ⋅ ⋅ ⋅ ·
yd
cd
eff sl
f
f
d b A ω
mm
$
8ne chooses $G14 A
aeff
= 06 mm
$
• !tee" desin in fi%st s-ppo%t B
0&$ . 119 .
00 . 10 4! $
1 & . !&1$
lim
$
0
$
· < ·
⋅ ⋅

·
⋅ ⋅
· µ µ
cd b
B
f d b
M
S=.1$&!
& . 0'9
40'
00 . 10
4! $ 1$&! . · ⋅ ⋅ ⋅ · ⋅ ⋅ ⋅ ·
yd
cd
b sl
f
f
d b A ω
mm
$
8ne chooses $G1! A
aeff
= 4$ mm
$
• !tee" desin in inte%io% s-ppo%ts
0&$ . 14 .
00 . 10 4! $
1 '6&0
lim
$
0
$
· < ·
⋅ ⋅

·
⋅ ⋅
· µ µ
cd b
C
f d b
M
S=.1111
$ . 0$0
40'
00 . 10
4! $ 1111 . · ⋅ ⋅ ⋅ · ⋅ ⋅ ⋅ ·
yd
cd
b sl
f
f
d b A ω
mm
$
8ne chooses 1G14@1G1! A
aeff
= 0'4 mm
$
"he reinforcement is use as straight bars in the supports.
*t must verify if the bars places in one row in the cross section of the beam respect
the provisions regarding the minimum distance between bars. *f the concrete is with the
ma)imum si#e of the aggregate d
g
H1! mm, the minimum distance between bars can be
determined according to "able '.'.....
¹
)
¹
¹
'
¹
· + · +
·
·
mm d
mm
s
g
nh
0! ' 01 '
1!
ma)
ma)
φ
"he cover layer is 0 mm, the distance among bars is%
( )
mm s mm s
nh eff nh
0! 49
$
14 0 0 $ $
,
· > ·
⋅ + ⋅ −
·
1.+.3. Desin fo% s)ea%
"he minimum coefficient of transversal reinforcing is determined with the
relation%
6$ .
40'
$
6 . 6 .
min ,
· · ·
yk
ck
"
f
f
ρ
"he ma)imum (uantity of transversal reinforcement is%
!' . 1
40'
00 . 10
$ '4 . 1 ' . ' .
1
ma)
· ⋅ ⋅ ⋅ ⋅ · ⋅ ⋅ ⋅ ⋅ ·
,
_

¸
¸
y"d
cd
c"
s"
f
f
b #
s
A
α
Rhere% .
c"
=1 for reinforced concrete
v
1
=.!
'4 .
$
$
1 ! .
$
1 ·
,
_

¸
¸
− ·
,
_

¸
¸

ck
f
"he ma)imum distance among stirrups on longitudinal and transversal direction
is%
( )
− · > · ⋅ · ·
¹
)
¹
¹
'
¹ · ⋅ · +
·
$ 04' 4! &' . &' .
0
04' 4! &' . 1 &' .
, ma) ,
ma) ,
eff t t
l
/ mm d /
mm
mm ctg d
/
α
• !-ppo%t B "eft
"he design shear force is%
0 d q ( (
left
B
left
red B
69&9$ 4! ' . $!1$ 11&99
,
· ⋅ − · ⋅ − ·
"he capable shear force of the member without specifically shear reinforcement is
computed with relation%
( ) [ ] d b k f C (
cp ck c 1d c 1d
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ · σ ρ η
1
0 . 1
1 1 , ,
1
Rhere
·
cp
σ
a)ial force is neglected
1$ .
' . 1
16 . 16 .
,
· · ·
c
c 1d
C
γ
. 1
1
· η for ordinary concrete
$ 404 . 1
4!
$
1
$
1 < · + · + ·
d
k
$ . 40& ,
4! $
4$
< ·

·

·
d b
A
sl
l
ρ
"he longitudinal steel A
sl
is anchored with the length l
bd
in section *:* is
considered $G1! (A
sl
=4$ mm
$
).
( ) [ ] · · ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ · 0 (
c 1d
0$!1 4! $ $ 40& . 1 404 . 1 1 1$ .
0
1
,
0$.!1 T1
"he minimum value of capable shear force without specific reinforcement can be
determined with relation%
( ) d b k # (
cp c 1d
⋅ ⋅ ⋅ + · σ
1 min min ,
Rhere%
$!& . $ 40 . 1 0' . 0' .
$
0
$
0
min
· ⋅ ⋅ · ⋅ ⋅ ·
ck
f k #
0 (
c 1d
$4'!4 4! $ $!& .
min ,
· ⋅ ⋅ ·
=$4.'!4 T1
20 ( 20 (
1dc c 1d
'!4 . $4 !1 . 0$
min , ,
· > ·
3ecause the design shear is bigger than the capacity of the beam without reinforcement
for shear%
20 ( 20 (
c 1d
left
red B
!1 . 0$ &9$ . 69
, ,
· > ·
⇒ "he shear reinforcement is
necessary.
"he capacity of compressed struts (
1d& ma+
is determined for the ma)imum value of
ctg3=%4
0
tg ctg
f # ! b (
cd c" 1d
0
1 ma) ,
1 ' . $'
' . $
1
' . $
1
00 . 10 '4 . 414 $ 1
1
⋅ ·
+
⋅ ⋅ ⋅ ⋅ ·
+
⋅ ⋅ ⋅ ⋅ ⋅ ·
θ θ
α
where #=.9d
*t can observe that J
3red
left
is between the two values but closer to the inferior
limit, so in this case the shear force is reduced to medium, the minimum (uantity of
transversal steel can be obtained for high value of ctgU.
*t can adopt for ctgU=1.&'
"he distance among stirrups for a diameter of
mm ! φ
and #=.9d=.9V4!=414
is given by%
$
,
! . '! 0 . $6 $ mm A
" s
· ⋅ ·
(two legs)
mm
(
ctg f ! A
s
left
Bred
y"d s"
$ 9 . 196
69&9$
&' . 1 40' 414 !& . '!
≈ ·
⋅ ⋅ ⋅
·
⋅ ⋅ ⋅
·
θ
"he capacity of compressed struts for ctgU=$ is%
0
tg ctg
f # ! b (
cd c" 1d
0
1 ma) ,
1 4 . $06
. $
1
. $
1
00 . 10 '4 . 414 $ 1
1
⋅ ·
+
⋅ ⋅ ⋅ ⋅ ·
+
⋅ ⋅ ⋅ ⋅ ⋅ ·
θ θ
α
left
Bred 1d
( ( >
ma) ,
"he transversal steel percentage is%
6$ . 141' .
$ $
! . '!
min ,
· > ·

·

· ρ ρ
b s
A
s"
eff "
"he basic anchoring length l
b&rqd
for the resisting steel L1! used at superior fiber
is obtained from "able+J*.'.
l
b,rd(
=1$& mm
"he design anchoring length is%
min , , ' 4 0 $ 1 b rdq b bd
l l l > ⋅ ⋅ ⋅ ⋅ ⋅ · α α α α α
Rhere coefficients W are obtained from "able '.9
W
1
=1 for straight bars
( ) 6& .
1!
1! 0
1' . 1 . 1' . 1
$
·
,
_

¸
¸ −
− · − ⋅ − · φ φ α
d
c
Rhere c
d
=min(a.$, c
1
, c)=min('4,0)
a=$:$V0:$V1!=16 mm
W
0
= W
4
=1.
1
'
· α
without transversal pressure
mm l
bd
11' 1$& 1 1 1 6& . 1 · ⋅ ⋅ ⋅ ⋅ ⋅ ·
bd b
brdq
bd
l mm l
mm
mm
mm l
l < · ⇒
¹
¹
¹
)
¹
¹
¹
¹
'
¹
· ⋅ · ⋅
· ⋅ · ⋅
· 061
1
1! 1! 1 1
061 1$& 0 . 0 .
ma)
min , min ,
φ
"he length of bars from the superior part of the beam can be determined using the
Fig. '.40.

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