Lec09 Shear in RC Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

15-Mar-13
CE 370: Prof. A. Charif 1
CE 370
REINFORCED CONCRETE-I
Prof. A. Charif
Shear in RC Beams
Shear in Beams
2
• Transverse loads on
beams cause
bending moment M
and shear force V
• Shear force V is the
first derivative of
the moment M
dx
dM
V 

15-Mar-13
Shear in Beams
3
• Bending moment causes normal stresses with compression
resisted by concrete and tension by longitudinal steel bars
• Bending design delivers required longitudinal steel
• Shear force causes shear stress and shear design is performed
independently
• Shear failure is brittle and
dangerous
• Shear design must deliver a
shear strength equal to or
greater than flexural strength
• Shear is resisted by concrete
and steel stirrups
Shear stress in an uncracked section
4
Ib
VQ
v
I
My
f 
• Flexural stress is horizontal (compression in top, tension in bot.)
• Shear stress has equal vertical and horizontal components
• This leads to inclined principal stresses (normal stress, no shear)
beamofWidth
fiberaboveareaofmomentFirst
inertiaofMomentforceShear



b
Q
IV

15-Mar-13
5
f
v
v
ff
fp
2
tan2
22
2
2








principal stresses
• The two normal principal stresses (one is tension and other is
compression) are inclined and this angle is 450 at the neutral
axis level (f = 0)
• The inclination angle changes in top (compression side) and
bottom (tension side) regions
6
Shear stress and inclined cracking
• Principal stress
trajectories are flat at
top (compression zone)
and deep at bottom
(tension zone)
• Inclined cracks are due
to combined flexure
and shear
• Diagonal tension in
tension side must be
resisted by adequate
web reinforcement

15-Mar-13
Average Shear Stress between Cracks
7
jdb
V
v
xb
T
v
jd
xV
jd
M
T
jd
M
T
ww







 isstressshearAverage
db
V
v
w

:ACI/SBC
Average
shear stress
must not
exceed
certain limit
Modes of Shear Failure
in Normal Beams
• Shear behavior and analysis of RC beams is quite complex.
• Several experimental studies have been conducted to identify
the various modes of failure due to combined shear and
bending
• Shear behavior and failure is closely related to shear span to
depth ratio a /d
• For normal beams, shear failure modes are :
 Diagonal tension failure
 Flexural shear failure
 Diagonal compression failure
8

15-Mar-13
Diagonal tension Failure
(Web-shear cracks)
• Web shear cracks occur around neutral axis under large shear force
and small bending moment. These cracks are normally at 450 with
the horizontal and form near the mid-depth of sections and move
on a diagonal path to the tension surface.
• Occur at ends of beams at simple supports and at inflection points
in continuous beams
9
Flexural-shear Failure
(Flexure-shear cracks)
• For flexure-shear cracks to occur, the moment must be larger
than the cracking moment and the shear must be rather large.
• The cracks run at angles of about 450 with the beam axis and
often start at the top of the flexural cracks.
10

15-Mar-13
Diagonal compression failure
• Diagonal compression failure occurs under large
shear force. It is characterized by the crushing of
concrete. Normally it occurs in beams which are
reinforced against heavy shear.
11
12
Modes of Shear Failure
in Short and Deep Beams

15-Mar-13
Forces in a cracked beam
without stirrups
• Shear strength mechanism of RC members includes :
– Concrete compression force in uncracked region
– Aggregate interlocking in cracked zone
– Shear across longitudinal steel bars known as dowel force
– Shear reinforcement, if present, will also resist the shear force
13
14
Forces in a beam with vertical stirrups

15-Mar-13
15
Types of Shear Reinforcement (Stirrups)
• Steel stirrups may be inclined along diagonal tension but are
usually vertical
• Part of longitudinal bars may also be bent up from bottom to top
to play role of shear reinforcement and then top steel
• Nowadays,
stirrups are
often vertical
16
Types of Shear Reinforcement
• Inclined stirrups and bent up bars are not very practical
because of the high labor costs for positioning them.
• Nowadays, stirrups are often vertical
Bent up bars

15-Mar-13
17
Types of Shear Reinforcement (Stirrups)
Behavior of RC beams with web
reinforcement (Truss Analogy)
 Concrete in compression is top chord
 Longitudinal tension steel is bottom chord
 Stirrups form truss verticals
 Concrete between diagonal cracks form the truss diagonals
18

15-Mar-13
Shear Stress Transfer in RC Beams
• Diagonal cracks will occur in beams with shear reinforcement
at almost the same loads that they occur in beams of the same
size without shear reinforcement.
• The shear reinforcement makes its presence known only after
the cracks begin to form. At that time, beams must have
sufficient shear reinforcing to resist the shear force not resisted
by the concrete.
• After a shear crack has developed in a beam, only a little shear
can be transferred across the crack unless web reinforcing is
used to bridge the gap. When such reinforcing is present, it
keeps the pieces of concrete on the two sides of the crack from
separating.
19
Benefits of Stirrups
 Stirrups carry shear across the crack directly
 Promote aggregate interlock
 Confine the core of the concrete in the beam thereby
increasing strength and ductility
 Confine the longitudinal bars and prevent concrete
cover from splitting off the beam
 Hold the pieces of concrete on either side of the
crack together and prevent the crack from
propagating into the compression region
20

15-Mar-13
CE 370
REINFORCED CONCRETE-I
Prof. A. Charif
Shear analysis and design of RC beams
according to SBC / ACI Codes
• SBC, ACI and other codes neglect aggregate interlock and
dowel action
• Shear force is resisted by compression concrete and by steel
stirrups only
• Nominal shear strength is therefore:
22
strengthshearStirrups:
strengthshearConcrete:
s
c
scn
V
V
VVV 
Steel stirrups may be inclined along diagonal tension but are
usually vertical
Shear Analysis and Design

15-Mar-13
23
shearin0.75With
strengthshearStirrups:
strengthshearConcrete:
with




s
c
scnun
V
V
VVVVV
Nominal shear strength is provided by concrete and stirrups only
Design shear strength must be equal to or greater than ultimate shear
• Concrete may provide enough strength to resist ultimate shear
but SBC / ACI require stirrups if:
222
c
u
cuc
n
V
V
VVV
V



24
scnun VVVVV  with
• SBC / ACI nominal shear
strength of concrete in beams is : db
f
V w
'
c
c
6

• For axially loaded members (columns) the nominal shear
strength of concrete is :
db
f
A
N
VN
db
f
A
N
VN
w
'
c
g
u
cu
w
'
c
g
u
cu
6
3.0
1:0)(Tension
614
1:0)(nCompressio



















15-Mar-13
25
Shear Analysis and Design per SBC / ACI
 
areasection-crossStirrup:
cossin:stirrupsInclined
:stirrupsVertical
v
yvs
yv
yvss
A
s
d
fAV
s
dfA
fAnV
 

• Assuming a 450 inclined crack,
the number of vertical stirrups
crossed by the crack is ns = d/s
where s is the stirrup spacing
• Assuming that they have yielded
the stirrups shear strength is :
Stirrup Section Area
• A stirrup has at least one leg but usually two legs or more.
26
diameterStirrup:
4
:legswithStirrup
2
s
s
v
yv
s
d
d
nA
n
s
dfA
V



15-Mar-13
Section Adequacy for Shear
• Before performing design, the section must first be checked
whether it is sufficient to resist shear according to SBC / ACI.
• The section is insufficient to resist shear and must be increased if :
27
6
5:ifshearforsectionntInsufficie
:isconditionthestress,shearaverageofIn terms
5:ifshearforsectionntInsufficie
6
but
3
2
3
2 ''
'
c
w
u
u
cu
w
'
c
c
wcc
u
wcs
f
db
V
v
VV
db
f
V
dbfV
V
dbfV







Beam Shear Design
Required Vertical Stirrup Spacing
28
 
trequiremensteelminimumUse0if:Note
:spacingstirrupRequired
designOptimalwith











c
u
s
c
u
yv
c
uyv
c
u
susc
unscnun
V
V
V
V
V
dfA
s
V
V
s
dfA
V
V
VVVV
VVVVVVV





22
c
u
c
n
V
V
V
V


• SBC / ACI specify that stirrups are required in beams if :

15-Mar-13
Minimum Web Reinforcement
and Maximum Stirrup Spacing
SBC / ACI minimum web steel area :
29
y
wc
v
f
sbf
A









3
1
,
16
Max
'
min,
 
 
 
 










(b)Case300,25.0Min3If
(a)Case600,5.0Min3If
(b)Case300,25.0Min2If
(a)Case600,5.0Min2If
max
max
max
max
mmdsVV
mmdsVV
mmdsVV
mmdsVV
cu
cu
cs
cs


SBC / ACI maximum stirrup spacing (geometry considerations) :
Critical Shear Sections
• Load transfer between beams and supports is performed through
conic shaped zones with 450 angles.
• Shear failures occur at critical sections located at a distance d from
the face of the support, where d is the depth of tension steel
• Ultimate shear force considered for design is computed at the
critical section
30

15-Mar-13
31
Critical Shear Sections
Critical Shear Section for Design
• Ultimate shear used for design is computed at the critical section
at a distance d from the support face (d = Tension steel depth)
• It is obtained from support and mid-span values using shear force
envelope diagram.
• Most modern structural analysis methods use clear length Ln (clear
distance between support faces)
32
nL
Stirrup reinforcement in beams
is usually symmetric about
mid-point except in special
cases such as cantileversLn/2d
VuL/2
Vud
Vu0

15-Mar-13
• For a simply supported beam, the ultimate shear force values at
the support and mid-span are :
33
82
2/0
nLu
uL
nu
u
Lw
V
Lw
V 
wu : Total factored (ultimate) uniform load (= 1.4wD + 1.7wL)
wLu : Factored live load (= 1.7wL)
Shear force at support is obtained with ultimate uniform load
applied on all the span
The mid-span value is obtained by applying factored live load on
half the span only (along with factored dead load on all span).
Ln/2d
VuL/2
Vud
Vu0
34
Ln/2d
VuL/2
Vud
Vu0
From similar triangles we obtain the
critical shear value at distance d :
 
8
,
2
with
2
2/0
2/00
nLu
uL
nu
u
uLu
n
uud
Lw
V
Lw
V
VV
L
d
VV


L
1w
2w
SFD82
21
L
w
L
w 
8
2
L
w







8
3
2
21
L
w
L
w
Mid-span shear value

15-Mar-13
Shear Design - Summary
1. Determine ultimate shear force at distance d from support face,
to be used for design
2. Compute concrete shear strength Vc
3. Check for section adequacy (whether it is sufficient for shear).
If insufficient, the section must be increased.
4. Check whether stirrups are required
5. Compute maximum stirrup spacing
6. Compute minimum shear steel area
7. Compute required stirrup spacing
8. Select appropriate stirrup diameter and spacing to satisfy all
conditions (maximum spacing, minimum shear steel, required
spacing)
35
36
Shear Design - Summary
 
 
 
y
wc
v
cu
cu
c
u
cu
w
'
c
c
nLu
uL
nu
uuLu
n
uudu
f
sbf
A
mmdsVV
mmdsVV
V
V
VV
db
f
V
Lw
V
Lw
VVV
L
d
VVV


















3
1
,
16
MaxareasteelshearMinimum/6
(b)300,25.0Min3If
(a)600,5.0Min3If
:spacingstirrupMaximum/5
2
ifrequiredareStirrups/4
5ifshearforsectionntInsufficie/3
6
/2
8
,
2
with
2
/1
'
min,
max
max
2/02/00





15-Mar-13
37
Shear Design – Summary Cont.
7step-spacingstirrupRequiredc/
6step-areasteelshearMinimumb/
5step-spacinggeometryMaximuma/
:conditionsthreeesatisfy thorder toinspacingstirrup
andlegsofnumberdiameter,stirrupeappropriatSelect/8
discardedisspacingrequiredthen,0if:Note
spacingstirrupRequired/7



c
u
s
c
u
yv
V
V
V
V
V
dfA
s


• In practice the stirrup diameter ds is usually fixed and the
designer must determine the required number of legs and spacing.
• ds depends on the diameter of the main bars
• In buildings, ds is 10 mm for columns, 8 to10 mm for beams
• In bridges ds is higher (12 to 16 mm)
38
Shear Design – Summary Cont.
The three SBC / ACI requirements (steps 5 to 7) can be reformulated
using the stirrup spacing as the main design variable.
The designer usually selects first the stirrup diameter ds, the number of
legs n, and then determines the required stirrup spacing s, while
satisfying all code requirements. Given its diameter ds and the number
of legs n, the stirrup area is:
4
2
s
v
dn
A


The minimum shear steel area requirement, expressed in terms of
spacing s, can thus be transformed to another spacing requirement:
w
yv
c
y
wc
v
b
fA
f
s
f
sbf
A


















0.3,
0.16
MinspacingsteelMaximum
3
1
,
16
MaxareasteelshearMinimum
'
2
max
'
min,

15-Mar-13
39
Shear Design - Stirrup Spacing Summary
 
 
 
(b)byorbycontrolledisitifincreasedbeonlycanSpacing
(b)casebycontrolledisitunlessunchangedremains
decreases.whenincreases.shearultimateondependsOnly
notdoesbut)and(onondependand:Notes
,,Min:spacingAdopted
0.3,
0.16
Min:spacingsteelMinimum
(b)3If300,25.0Min
(a)3If600,5.0Min
:spacingGeometry
1
max
3
max
1
max
3
max
3
max
1
max
3
max
2
max
3
max
2
max
1
max
3
max
'
2
max
1
max
ss
s
VsVs
sdnAss
ssss
V
V
dfA
s
b
fA
f
s
VVmmd
VVmmd
s
uu
sv
c
u
yv
w
yv
c
cu
cu






















40
Shear design
and spacing
c
u
yv
w
yv
c
cu
cu
V
V
dfA
s
b
fA
f
s
mm
d
sVV
mm
d
sVV



































3
max
'
2
max
1
max
1
max
0.3,
0.16
Min
300,
4
Min:3
600,
2
Min:3
(b)
(a)
Vu
cV5.0
cV3
cV
cV5
requiredstirrupsNo
3
maxs
(a)Case1
max s
2
(a)Case
(b)Case1
max s
sectionntInsufficie

15-Mar-13
41
?5 cu VV  dimensionsIncrease
sectionntInsufficie
?5.0 cu VV  requirednotStirrups
?3 cu VV  (b)300,
4
Min1
max 





 mm
d
s(a)600,
2
Min1
max 





 mm
d
s
w
yv
c
b
fA
f
s








 0.3,
0.16
Min
'
2
max
?cu VV 
 2
max
1
max ,Min sss 
c
u
yv
V
V
dfA
s



3
max
 3
max
2
max
1
max ,,Min ssss 
Yes
No
Yes
Yes
Yes
No
No
No
db
f
V
V
w
'
c
c
u
6
75.0,known


Shear design Flowchart
Shear Design - Observations
• The designed stirrup spacing may turn out inadequate in many
ways. It could be either too large or too small.
• If the spacing is too large, then the number of legs or (and) stirrup
diameter may be decreased.
• If the spacing is too small, then either the number of legs or
stirrup diameter must be increased.
• A complete and adequate design strategy is to start using the
minimum stirrup diameter and minimum number of legs.
• If the spacing is too small (less than 100 mm), increase the
number of legs, but if the number of legs is excessive then
increase the stirrup diameter.
42

15-Mar-13
Shear Design - Full Automatic Algorithm
Data includes minimum stirrup diameter dsmin as well as minimum
and maximum numbers of legs nmin and nmax
Leg increment = 1 , Stirrup diameter increment = 2 mm
• Compute s1
max
• Start with dsmin
• A - Start with nmin
• B - Compute stirrup area
• Compute s2
max , s3
max and final spacing value
• If spacing value is ok stop
• If spacing too small then :
 If n < nmax : n = n + 1 and goto B
 If n = nmax : ds = ds + 2.0 and goto A
43
Stirrup Design Algorithm
(Stirrup Diameter, Number of Legs and Spacing)
1. Compute maximum geometry spacing s1
max
2. Start with minimum stirrup diameter dsmin
3. Start with minimum leg number nmin
4. Compute stirrup area
5. Compute minimum steel spacing s2
max
6. If Vu > Vc : s = Min (s1
max , s2
max)
7. If Vu ≤ Vc : Compute required steel spacing s3
max and
s = Min (s1
max , s2
max , s3
max)
8. If s ≥ 100 mm then stop
9. If n < nmax then n = n + 1 and goto 4
10. If n = nmax then ds = ds + 2 and goto
44

15-Mar-13
Shear Design Problem 1
• A simply supported beam is subjected to uniform
loading composed of dead load (including self weight)
of 27.0 kN/m and live load of 17.5 kN/m.
• The beam clear span length is 9.6 m and the section
dimensions are 300 x 600 mm.
• Steel depth is d = 540 mm
• Design the beam for shear using 10 mm stirrups and the
following material data :
45
MPafMPaf yc 42025'

Solution 1
The ultimate load is :
wu = 1.4 x 27.0 + 1.7 x 17.5 = 37.8 + 29.75 = 67.55 kN/m
Factored live load is : wLu = 1.7 x 17.5 = 29.75 kN/m
Ultimate shear force at support and mid-span as well as value at the
critical section (at a distance d from the support) are:
46
 
  kNV
VV
L
d
VV
kN
L
wV
kN
L
wV
ud
uLu
n
uud
n
LuuL
n
uu
78.2917.3524.324
6.9
54.02
24.324
2
7.35
8
6.9
75.29
8
24.324
2
6.9
55.67
2
2/00
2/
0







15-Mar-13
47
Solution 1 – Cont.
designforusedis
78.291
7.35
24.324
62.505.0
2/
0
ud
ud
uL
u
c
V
kNV
kNV
kNV
kNV




Ln/2 = 4.8 m0.54m
VuL/2
Vud
Vu0
cV5.0
Concrete nominal shear strength is :
kNNdb
f
V w
c
c 0.135135000540300
6
25
6
'

2
22
6.235
4
103
4
3:stirrupsleg3Use
requiredareStirrups625.5050.:trequiremenStirrup
OKSection78.29125.50613575.055
:checkadequacySection
mm
dn
An
VkNV
kNVVkNV
s
v
udc
uduc






48
  (a)Case0.270600,5.0Min
75.303378.291:spacinggeometryMaximum
1
max mmmmds
kNVV cud

 
mms
b
fA
f
s
w
yv
c
5.989
300
4206.235
0.3,
25
0.16
Min
0.3,
0.16
2
max
'
2
max


















mms
V
V
dfA
s
c
ud
yv
3.210
10000.135
75.0
78.291
5404206.235
3
max
3
max














15-Mar-13
Question 1: We used 3 legs. What happens if we use 2 or 4 legs?
49
 
spacingmm200auseWe
210,,Min:spacingAdopted
3.210:spacingstirrupRequired
5.989:spacingsteelMinimum
0.270:spacingmaximumGeometry
:summarytrequiremenspacingMaximum
3
max
2
max
1
max
3
max
2
max
1
max





mmsssss
mms
mms
mms
• A stirrup is assumed to resist shear over a distance
extending s/2 on each side of the stirrup
• The first stirrup must thus be located at a distance s/2 from
the support face
• First stirrup at 100 mm, then constant 200 mm spacing
• Using more than three legs is wasteful as spacing becomes
controlled by geometry maximum spacing
• Question 2 : Can we change spacing when shear force is reduced?
50
 
mms
ssss
mms
mms
140
,,Min
2.140
7.659
:spacingsteelMinimum
:legsTwo
3
max
2
max
1
max
3
max
2
max




)(0.270:unchangedisspacingmaximumGeometry 1
max amms 
 
mmss
ssss
mms
mms
270
,,Min
4.280
4.1319
:spacingsteelMinimum
:legsFour
1
max
3
max
2
max
1
max
3
max
2
max





15-Mar-13
Stirrup Requirement
and Spacing Variation
51
• When shear force is reduced, we can not only change
(increase) stirrup spacing, but we may be able to stop
stirrups completely when they are no longer required
• Remember : Stirrups are required in beams in all
zones where:
cu VV 5.0
cu VV 5.0
• Stirrups may therefore be required near the beam
support and not required in other zones (around the
mid-span) where shear force is smaller :
52
Stirrups are no longer required when : cu VV 5.0
Ln/2
VuL/2
Vu0
x0
cV5.0
Distance x0 over which stirrups are required is obtained using
similar triangles :









2/0
0
0
5.0
2 uLu
cun
VV
VVL
x

• x0 may be greater or less than half-span
• In half-span of the beam, stirrups are
provided over a distance :







2
,Min 0
n
st
L
xL
Stirrup Requirement
and Spacing Variation

15-Mar-13
53
• Geometry maximum spacing s1
max is independent of shear steel
area Av and thus on number of legs n and stirrup diameter ds,
• If the final stirrup spacing is controlled by the geometry limit
s1
max case (a), it cannot be increased. The number of legs, or
stirrup diameter, may be decreased (if they are not minimal).
• Minimum leg number is usually 2 and minimum stirrup diameter
is 10 mm for columns and 8 to 10 mm for beams
• If the final spacing is controlled by any of the other two limits
s2
max or s3
max , then the design is adopted unless the spacing is
too small (less than 75 to 100 mm), in which case the number of
legs must be increased
• If the final spacing is controlled by the required steel limit s3
max
this means that it can be varied (increased) at a reduced value of
shear force located at a certain distance from the support.
Stirrup Requirement and Spacing Variation
Stirrup Design Algorithm
(Variation of Stirrup Spacing)
• If the final stirrup spacing is controlled by required steel limit s3
max
then it can be varied (increased) at a reduced value of shear force
located at a certain distance from the support
• Spacing variation is performed if the required number of stirrups
over distance Lst is important (say greater than 10)
• The new spacing value is usually 50 to 100 mm greater than the
previous one
54
2
max
1
max
1
12
andlimitstwotheexceednotmustspacingnewThe
100to50:generalIn
100to50
ss
mmss
mmss
jj 



15-Mar-13
Stirrup Design Algorithm – Cont.
• The value of shear force corresponding to the new spacing is
deduced from optimal design condition:
55







2
2
s
dfA
VVV
V
s
dfA
V
yv
cuc
uyv
s 

• Location x2 of this new shear force value is obtained using
similar triangles:









2/0
20
2
2 uLu
uun
VV
VVL
x
Ln/2
VuL/2
Vu0
x2
Vu2
56
 


1
1
1
1
1
1
2
:lengthstirrupCumulated
:spacingsNext5.0:spacingFirst
:distancegivenoverstirrupsofnumberRequired
i
ii
i
si
j
sj
j
s
s
snL
s
D
n
s
D
n
D
Spacing Variation

15-Mar-13
• The number n1 of stirrups with the first spacing value is such :
57
• Spacing variation (increase) may be performed more than once
• The distance left for the second spacing is :
• The approximate number of stirrups with spacing s2 is :
• If this number exceeds ten to twelve, then a further spacing
variation may be considered provided the limits are not already
reached.
5.0
2 1
2
12
1
1121 
s
x
nx
s
snxLs
2
1
1112
s
snLLLR stsst 
2
2
2
s
R
n 
• The next shear force value and its location are given by :
58
• The exact distance used by stirrups with preceding spacing and
their number are:
3
2 2/0
0

















 j
VV
VVL
x
s
dfA
VV
uLu
ujun
j
j
yv
cuj 
1
*
1
1
1
2
1
2
1
*
1
2 






 





 
j
j
j
j
i
iij
j
i
sijj
s
x
n
s
snxLxx
• Remaining distance and approximate number of stirrups with
new spacing sj are:
j
j
j
j
iist
j
i
sistj
s
R
n
s
snLLLR  

 2
1
1
1
1
1

15-Mar-13
Spacing variation algorithm
• Compute spacing s1 at critical section
• If s1 is controlled by s2
max or by s1
max case (a) then stop
• j = 2
A - Choose new spacing sj = sj-1 + 50 to 100 mm
• Deduce Vuj and xj
• Deduce number of stirrups with previous spacing nj-1
• Remaining distance and number of stirrups Rj and nj
• If number nj is large enough and spacing variation is still possible
then : j = j+1 and gotoA
Note : If s1 is controlled by by s1
max case (b), then the second
spacing s2 must be located inside case (a) zone, i.e. with Vu ≤ 3 Vc
59
(Same as Problem 1)
• A simply supported beam is subjected to uniform
loading composed of dead load (including self weight)
of 27.0 kN/m and live load of 17.5 kN/m.
• Steel depth is d = 540 mm
60
MPafMPaf yc 42025'


15-Mar-13
Solution 2
• The ultimate load, shear force values and concrete shear strength
are the same as before.
61
kNdb
f
V
kNV
kNV
kNV
kNV
w
c
c
ud
uL
u
c
0.135
6
78.291
7.35
24.324
62.505.0
'
2/
0





Ln/2 = 4.8 m0.54m
VuL/2
Vud
Vu0
cV5.0
requiredareStirrups62.50
2
:trequiremenStirrup
OKSection78.29125.50613575.055


ud
c
uduc
VkN
V
kNVVkNV


62
• Distance x0 beyond which stirrups are not required is :
mmm
VV
VVL
x
uLu
cun
4552552.4
7.3524.324
13575.05.024.324
2
6.95.0
2 2/0
0
0 

















  (a)0.270600,5.0Min
75.303378.291
:spacinggeometryMaximum
1
max mmmmds
kNVV cud

 
• This distance x0 is smaller than half-span. Stirrups are thus
required over a distance : mmx
L
xL n
st 4552
2
,Min 00 







15-Mar-13
63
mms
b
fA
f
s
mm
dn
An =
w
yv
c
s
v
7.659
300
42008.157
0.3,
25
0.16
Min
0.3,
0.16
08.157
4
100
2
4
2:legstwoStart with
2
max
'
2
max
2
2


















 

mm
V
V
dfA
s
c
ud
yv
2.140
10000.135
75.0
78.291
54042008.157
:spacingstirrupRequired 3
max 












64
 
spacingmm100auseWe
mm50ofmultiplesasvaluesspacingselectusuallyWe
)byd(controlle140
,,Mins:spacingAdopted
3
max
3
max
2
max
1
max
3
max
2
max
1
max






smms
sss
mms
mms
mms

15-Mar-13
Spacing Variation
65
mmm
VV
VVL
x
kN
s
dfA
VV
mms
ss
uLu
uun
yv
cu
746746.0
7.3524.324
38.27924.324
2
6.9
2
:isvalueforceshearthisoflocationThe
38.279
1000
1
540
150
42008.157
13575.0
:isforceshearingCorrespond
150spacingsecondachooseWe
,limits
twotheofanyexceednotdoesitprovidedincreasedbemaySpacing
2/0
20
2
2
2
2
2
max
1
max


































Spacing Variation
66
The total number of stirrups with first spacing is :
896.7
2
1
100
746
2
1
2
1
1
2
12
1
1121  n
s
x
nx
s
snxLs
mmLLR
mm
s
snLLLR
sst
ssst
38027504552
750
2
100
1008
2
12
1
11112


The approximate number of stirrups with spacing s2 is :
3.25
150
3802
2
2
2 
s
R
n
The first stirrup is at a distance s1/2 = 50 mm. Seven more
stirrups are needed to cover this distance x2 (= 746 mm)
The remaining distance for spacing s2 is :

15-Mar-13
Spacing Variation
67
This large number allows for further spacing variation (which is
still possible).
We chose a new (third) value of 250 mm (which corresponds in
fact to geometry maximum spacing) : s3 = 250 mm
The corresponding shear force value and its location are :
3.25
150
3802
2
2
2 
s
R
n
mmm
VV
VVL
x
kN
s
dfA
VV
uLu
uun
yv
cu
1932932.1
7.3524.324
13.20824.324
2
6.9
2
13.208
1000
1
250
54042008.157
13575.0
2/0
30
3
3
3
































Spacing Variation
68
mmxkNVu 193213.208 33 
The exact distance used by stirrups with spacing s2 and their
number are:
mmsnL
n
s
x
n
mmLxx
s
s
12001508
888.7
150
1182
11827501932
222
2
2
*
2
2
13
*
2



Remaining distance and approximate number of stirrups with
spacing s3 are :
mmsnL
n
s
R
n
mmLLLR
s
ssst
275025011
1141.10
250
2602
260212007504552
333
3
3
3
3
213




15-Mar-13
Solution 2- Summary
Stirrups required over a distance Lst = 4552 mm (less than half-span)
Use of two-leg 10 mm stirrups as follows:
1. Eight stirrups with spacing s1 = 100 mm. First stirrup located at
s1/2 = 50 mm, and then seven stirrups with spacing s1 = 100 mm
(Ls1 = 50 + 7 x 100 = 750 mm)
2. Eight stirrups with spacing s2 = 150 mm (Ls2 = 8 x 150 = 1200 mm
and Ls1 + Ls2 = 1950 mm)
3. Eleven stirrups with spacing s3 = 250 mm (Ls3 = 2750 mm , and
Ls1 + Ls2 + Ls3 = 4700 mm)
69
70
Figure produced by
RC-TOOL software
implementing all
previous theory
Same results with
shear force diagrams
Solution 2 - Summary
 
DemandOfferSafety
:(offer)capacityDesign
:(demand)shearUltimate
:diagramsforceShear

 scn
n
u
VVV
V
V



15-Mar-13
(Same as Problem 2)
• Same beam and same data except that we use 8-mm stirrups.
• For a two-leg 8-mm stirrup, shear steel area is
71
mm
V
V
dfA
s
mm
b
fA
f
s
mms
mm
dn
A
c
ud
yv
w
yv
c
s
v
7.89
10000.135
75.0
78.291
54042053.100
2.422420
300
53.100
0.3,
25
16
Min0.3,
0.16
Min
:arespacingrequiredandspacingsteelMinimum
(a)0.270unchnagedisspacingmaximumGeometry
53.10032
4
64
2
4
3
max
'
2
max
1
max
2
2

































72
Solution 3
• The final spacing value controlled by the required spacing limit
(89.7 mm) is small and less than the minimum limit of 100 mm.
• To increase the required spacing, we must increase the number of
legs (or the stirrup diameter)
• We keep using the same 8-mm stirrup diameter and increase the
number of legs to three. The maximum geometry spacing remains
unchanged but the other two will increase:
mm
V
V
dfA
s
mm
b
fA
f
s
mm
dn
Amms
c
ud
yv
w
yv
c
s
v
6.134
10000.135
75.0
78.291
5404208.150
3.633420
300
8.150
0.3,
25
16
Min0.3,
0.16
Min
8.15048
4
64
3
4
0.270
3
max
'
2
max
2
2
1
max

































15-Mar-13
73
 
spacingmm100auseWe
)byd(controlle134
:stirrupsmm8leg-for threesummarytrequiremenspacingMaximum
3
max
3
max
2
max
1
max
3
max
2
max
1
max






smms
ssss
mms
mms
mms
Spacing Variation
74
mmm
VV
VVL
x
kN
s
dfA
VV
mms
ss
uLu
uun
yv
cu
865865.0
7.3524.324
26.27224.324
2
6.9
2
26.272
1000
1
540
150
4208.150
13575.0
,limitstwothe
ofanyexceednotdoesitprovidedincreasedbemaySpacing
2/0
20
2
2
2
2
2
max
1
max



































15-Mar-13
Spacing Variation
75
mmL
n
s
x
nx
s
snxL
s
s
950100950
1015.9
2
1
100
865
2
1
2
1
1
1
2
12
1
1121


mmLLR sst 3602950455212 
01.24
150
3602
2
2
2 
s
R
n
The first stirrup is at a distance s1/2 = 50 mm. Nine more
stirrups are needed to cover this distance x2 (= 865 mm)
Spacing Variation
76
This large number allows for further spacing variation.
01.24
150
3602
2
2
2 
s
R
n
mmm
VV
VVL
x
kN
s
dfA
VV
uLu
uun
yv
cu
2003003.2
7.3524.324
85.20324.324
2
6.9
2
85.203
1000
1
250
5404208.150
13575.0
2/0
30
3
3
3

































15-Mar-13
Spacing Variation
77
mmxkNVu 200385.203 33 
number are:
mmsnL
n
s
x
n
mmLxx
s
s
12001508
802.7
150
1053
10539502003
222
2
2
*
2
2
13
*
2



Remaining distance and number of stirrups with spacing s3 are :
mmsnL
n
s
R
n
mmLLLR
s
ssst
250025010
107.9
250
2402
240212009504552
333
3
3
3
3
213



Use of three-leg 8 mm stirrups as follows:
1. Ten stirrups with spacing s1 = 100 mm and the first stirrup is at
s1/2 = 50 mm, (Ls1 = 50 + 9 x 100 = 950 mm)
2. Eight stirrups with s2 = 150 mm (Ls2 = 8 x 150 = 1200 mm and
Ls1 + Ls2 = 2150 mm)
3. Ten stirrups with s3 = 250 mm (Ls3 = 10 x 250 = 2500 mm and
Ls1 + Ls2 + Ls3 = 4650 mm)
• Total : 28 three-leg stirrups
• The same results are delivered by RC-TOOL software as shown
in the figure
78

15-Mar-13
79
RC-TOOL Solution of Problem 3
Comments about Problem 3
• It may happen that the required number of legs at the critical
section is greater than the minimum limit. This value can be used
throughout the beam but it can also be reduced for lower shear
force values.
• When the number of legs necessary at the critical section is greater
than the minimum limit (as in example 3), it can be reduced at a
certain distance when the spacing is controlled by the maximum
geometry spacing s1
max
• The last ten stirrups with 250 mm spacing can thus be replaced
with two legged stirrups with an adequate spacing to be calculated.
• Using two legs for the third spacing will also affect the number of
stirrups using the second spacing (this number n2 will increase as
the stopping of spacing 2 will be delayed).
80

15-Mar-13
Reducing number of legs
• Reducing the number of legs to two, we have for
8-mm stirrups:
81
mms
mm
b
fA
f
s
mms
mm
dn
A
w
yv
c
s
v
250spacingchosenthetoequalbeshouldIt
laterestimatedbelimit willspacingstirrupRequired
2.422420
300
53.100
0.3,
25
16
Min0.3,
0.16
Min
:isspacingsteelMinimum
0.270unchnagedisspacingmaximumGeometry
53.10032
4
64
2
4
3
'
2
max
1
max
2
2


















 

82
Using two legs and a value of 250 mm (which corresponds in
mms
mm
V
V
dfA
s
mmm
VV
VVL
x
kN
s
dfA
VV
c
ud
yv
uLu
uun
yv
cu
250spacingofchoiceourconfirmsThis
0.250
10000.135
75.0
65.169
54042053.100
:limitspacingrequiredingcorrespondCheck
2572572.2
7.3524.324
65.16924.324
2
6.9
2
65.169
1000
1
250
54042053.100
13575.0
3
3
max
2/0
30
3
3
3
















































15-Mar-13
83
mmxkNVu 257265.169 33 
The exact distance used by stirrups with spacing s2 (with three
legs) and their number are:
mmsnL
n
s
x
n
mmLxx
s
s
165015011
1181.10
150
1622
16229502572
222
2
2
*
2
2
13
*
2



Remaining distance and number of stirrups with spacing s3 are :
mmsnL
n
s
R
n
mmLLLR
s
ssst
20002508
88.7
250
1952
195216509504552
333
3
3
3
3
213



Summary
First, use of three-leg 8 mm stirrups as follows:
1. Ten stirrups with spacing s1 = 100 mm. First stirrup located at
s1/2 = 50 mm, (Ls1 = 50 + 9 x 100 = 950 mm)
2. Eleven stirrups with s2 = 150 mm (Ls2 = 11 x 150 = 1650 mm and
Ls1 + Ls2 = 2600 mm)
3. Use two leg stirrups : Eight stirrups with spacing s3 = 250 mm
(Ls3 = 8 x 250 = 2000 mm and Ls1 + Ls2 + Ls3 = 4600 mm)
• Total : 29 stirrups of 8-mm diameter composed of :
 21 three-leg stirrups and 8 two-leg stirrups
84

15-Mar-13
85
(High shear force value)
• We study the same previous beam but with higher
loading.
• Dead load (including self weight) = 40.0 kN/m
• Live load = 20.0 kN/m.
• Tension steel depth is d = 540 mm
86
MPafMPaf yc 42025'


15-Mar-13
Solution 4
wu = 1.4 x 40.0 + 1.7 x 20.0 = 56.0 + 34.0 = 90.0 kN/m
Ultimate shear force at support and mid-span as well as value at
distance d are:
87
 
  kNV
VV
L
d
VV
kN
L
wV
kN
L
wV
ud
uLu
n
uud
n
LuuL
n
uu
99.3878.400.432
6.9
54.02
0.432
2
8.40
8
6.9
0.34
8
0.432
2
6.9
0.90
2
2/00
2/
0






88
Solution 4
designforusedis
99.387
8.40
0.432
75.3033
62.505.0
2/
0
ud
ud
uL
u
c
c
V
kNV
kNV
kNV
kNV
kNV







Ln/2 = 4.8 m0.54m
VuL/2
Vud
Vu0
cV5.0
cV3
requiredareStirrups62.50
2
:trequiremenStirrup
OKSection99.38725.50613575.055
0.135135000540300
6
25
6
:shearConcrete
'



ud
c
uduc
w
c
c
VkN
V
kNVVkNV
kNNdb
f
V



15-Mar-13
89
 
mms
b
fA
f
s
mm
dn
An =
mmmmds
kNVV
w
yv
c
s
v
cud
7.659
300
42008.157
0.3,
25
0.16
Min
0.3,
0.16
08.157
4
100
2
4
2:legstwoStart with
(b)casebycontrolled(b)0.135300,25.0Min
2
max
'
2
max
2
2
1
max
























Distance x0 over which stirrups are required is :
mmx
L
xL
L
x
mmm
VV
VVL
x
n
st
n
uLu
cun
4679
2
,Min:distancestirrupRequired.
2
ansmaller thjust
4679679.4
8.400.432
13575.05.00.432
2
6.95.0
2
000
2/0
0
0

























90
)(unchanged0.135
77.139
10000.135
75.0
99.387
54042062.235
and
6.989
300
42062.235
0.3,
25
0.16
Min0.3,
0.16
Min
62.235
4
100
3
4
3:threetolegsofnumber
theincreasethereforeWemm).100than(lesssmallisvaluespacingThis
18.93
10000.135
75.0
99.387
54042008.157
1
max
3
max
'
2
max
2
2
3
max
mms
mm
V
V
dfA
s
mm
b
fA
f
s
mm
dn
An
mm
V
V
dfA
s
c
ud
yv
w
yv
c
s
v
c
ud
yv

















































15-Mar-13
91
 
spacingmm100auseWe
)byd(controlle135
,,Mins:spacingAdopted
1
max
3
max
2
max
1
max
3
max
2
max
1
max






smms
sss
mms
mms
mms
92
mmsn
s
L
n
s
x
nx
s
sn
mmm
VV
VVL
x
s
s
NkV
s
s
s
h
h
uLu
cun
h
c
16501001650)1(
2
1724.16
2
1
100
1574
2
1
2
:spacingmm100withdistancecover thistorequiredstirrupsofNumber
1574574.1
8.400.432
13575.030.432
2
6.93
2
:doubledbewillwhichbeyondDistance
doubled.valueitsand(a)casebycontrolledbecomewill
),75.3033than(lessesshear valureducedFor
force.shearhighaofbecause(b)casebycontrolledis
problem,in thisHoweverchanged.bereforecannot theIt
limitgeometrymaximumby thecontrolledisspacingFinal
11
1
1
1
1
1
1
11
2/0
0
1
max
1
max
1
max
1
max























15-Mar-13
93
 
   
part.secondin thespacingmm200aadoptWe
20.204
10000.135
75.0
525.297
54042062.235
)(unchanged6.9890.270:arespacingsThe
value.forceshearfor thisspacingstirrupnewdesignnowcanWe
525.2978.400.432
9600
16502
0.432
2
:islocationat thisforceshearThe
0.270600,5.0Min
:doubledisspacingmaximumgeometrythe
span),-halftheofpart(seconddistancethisBeyond
16501001650
3
max
2
max
1
max
2/0
1
01
1
max
1



















mm
V
V
dfA
s
mmsmms
kNVV
L
L
VV
mmmmds
mmL
c
ud
yv
uLu
n
s
us
s

Spacing Variation
94
mmm
VV
VVL
x
kN
s
dfA
VV
mms
s
R
n
mmLLR
uLu
uun
yv
cu
sst
2091091.2
8.400.432
57.2610.432
2
6.9
2
57.261
1000
1
540
250
42062.235
13575.0
250spacingthirdachooseWe
possibleisincreasespacingFurther15.15
200
3029
:spacingsecondwithstirrupsofnumberRequired
302916504679distancestirrupRemaining
2/0
30
3
3
3
3
2
2
2
12





































15-Mar-13
Spacing Variation
95
mmxkNVu 209157.261 33 
number are:
mmsnL
n
s
x
n
mmLxx
s
s
6002003
3205.2
200
441
44116502091
222
2
2
*
2
2
13
*
2



spacing s3 are :
mmsnL
n
s
R
n
mmLLLR
s
ssst
250025010
10716.9
250
2429
242960016504679
333
3
3
3
3
213



Use of three-leg 10 mm stirrups as follows:
1. Seventeen stirrups with spacing s1 = 100 mm. First stirrup located
at s1/2 = 50 mm (Ls1 = 50 + 16 x 100 =1650 mm)
2. Three stirrups with s2 = 200 mm (Ls2 = 3 x 200 = 600 mm and
Ls1 + Ls2 = 2250 mm)
3. Ten stirrups with s3 = 250 mm (Ls3 = 10 x 250 = 2500 mm and
Ls1 + Ls2 + Ls3 = 4750 mm)
96

15-Mar-13
97
RC-TOOL output
98
RC-TOOL output – Reduction of leg number

15-Mar-13
Problem 5
• A simply supported beam is subjected to
uniform loading composed of dead load
(including self weight) of 60.0 kN/m and
live load of 90.0 kN/m.
• Span length and section data are shown
• Design the beam for shear using 10 mm
stirrups
99
MPafMPaf yc 42030'

550d
625
375
304
75
300 300
4200
mmLLmmL n 42003004500 
Solution 5
wu = 1.4 x 60.0 + 1.7 x 90.0 = 237.0 kN/m
Ultimate shear force at support and mid-span as well as value at
distance d are:
100
 
  kNV
VV
L
d
VV
kN
L
wV
kN
L
wV
ud
uLu
n
uud
n
LuuL
n
uu
4.388325.807.497
2.4
55.02
7.497
2
325.80
8
2.4
0.153
8
7.497
2
2.4
0.237
2
2/00
2/
0







15-Mar-13
101
designforused4.388
325.807.497
6.705.0
2/0
kNV
kNVkNV
kNV
ud
uLu
c



Ln/2 = 2.1 m0.55m
VuL/2
Vud
Vu0
cV5.0
span-halffulloverrequiredStirrups
2
325.80
requiredareStirrups6.7050.:trequiremenStirrup
OKSection4.388125.7063.18875.055
3.188188279550375
6
30
6
:shearConcrete
2/
'




c
uL
udc
uduc
w
c
c
V
V
VkNV
kNVVkNV
kNNdb
f
V



102
  (a)Case0.275600,5.0Min
1
max mmmmds
kNVV cud

 
mms
b
fA
f
s
mm
dn
An =
w
yv
c
s
v
9.513
375
42008.157
0.3,
30
0.16
Min
0.3,
0.16
08.157
4
100
2
4
2:legstwoStart with
2
max
'
2
max
2
2


















 

mm
LL
xL
mmm
VV
VVL
x
nn
st
uLu
cun
2100
22
,Min
2149149.2
325.807.497
3.18875.05.07.497
2
2.45.0
2
0
2/0
0
0


























15-Mar-13
103
mm
V
V
dfA
s
c
ud
yv
1.110
10003.188
75.0
4.388
55042008.157
:spacingstirrupRequired 3
max 












 
spacingmm100auseWe
)byd(controlle110
3
max
3
max
2
max
1
max
3
max
2
max
1
max






smms
ssss
mms
mms
mms
Spacing Variation
104
mmm
VV
VVL
x
kN
s
dfA
VV
mms
ss
uLu
uun
yv
cu
881881.0
325.807.497
65.3227.497
2
2.4
2
65.322
1000
1
150
55042008.157
3.18875.0
andlimitstwothe
ofanyexceednotdoesitprovidedincreasedbemaySpacing
2/0
20
2
2
2
2
2
max
1
max



































15-Mar-13
Spacing Variation
105
mm
s
snL
n
s
x
n
s 950
2
100
15010
2
1031.9
2
1
100
881
2
1
1
111
1
1
2
1


mmLLR sst 1150950210012 
67.7
150
1150
2
2
2 
s
R
n
Spacing Variation
106
• We can thus use 8 stirrups with spacing 150 mm or perform a
new spacing variation
• The number of stirrups with spacing s2 is not very large
• We may decide that there is no need for a new spacing
variation
• RC-TOOL software performs spacing variation if the stirrup
number exceeds five
67.7
150
1150
2
2
2 
s
R
n

15-Mar-13
Solution 5a - Summary
Stirrups required over a distance Lst = 2100 mm (Half-span)
s1/2 = 50 mm (Ls1 = 950 mm)
2. Eight stirrups with spacing s2 = 150 mm (Ls2 = 1200 mm and
Ls1 + Ls2 = 2150 mm)
3. The extra 50 mm can be easily corrected by many ways, such as
adding one more stirrup at a spacing s1 :
4. Eleven stirrups with spacing s1 = 100 mm. First stirrup located
at s1/2 = 50 mm (Ls1 = 1050 mm)
5. Seven stirrups with spacing s2 = 150 mm (Ls2 = 1050 mm and
Ls1 + Ls2 = 2100 mm)
 Total of 18 stirrups in half-span and 35 in the beam.
107
RC-TOOL Solution 5b
More Spacing Variation
108
mmm
VV
VVL
x
kN
s
dfA
VV
uLu
uun
yv
cu
1246246.1
325.807.497
08.2507.497
2
2.4
2
08.250
1000
1
250
55042008.157
3.18875.0
2/0
30
3
3
3
































67.7
150
1150
2
2
2 
s
R
n

15-Mar-13
109
mmxkNVu 124608.250 33 
number are:
mmLn
s
x
n
mm
s
snxx
s 3001502297.1
150
296
296
2
100
100101246
2
22
2
*
2
2
1
113
*
2


spacing s3 are :
34.3
250
850
8503009502100
3
3
3
3
213


n
s
R
n
mmLLLR ssst
RC-TOOL Solution 5b
Spacing Variation
110
span-halfsecondtheof
symmetrybycoveredbewillandspacinghalfthanlessisspan-midto
distanceremainingthat themeans)4.3(because),3(takeWe
34.3
250
850
8503009502100
33
3
3
3
3
213



nn
n
s
R
n
mmLLLR ssst
RC-TOOL Solution 5b
Spacing Variation

15-Mar-13
RC-TOOL Solution 5b - Summary
Stirrups required over a distance Lst = 2100 mm (Half-span)
s1/2 = 50 mm (Ls1 = 950 mm)
2. Two stirrups with spacing s2 = 150 mm (Ls2 = 300 mm and
Ls1 + Ls2 = 1250 mm)
3. Three stirrups with spacing s3 = 250 mm (Ls23 = 750 mm and
Ls1 + Ls2 + Ls23 = 2000 mm)
4. Remaining distance to mid-span is 100 mm and is less than
half of spacing s3
111
112
RC-TOOL Solution
Total of 15 stirrups
in half-span and
30 in beam
(five less than in
solution 1)

Lec09 Shear in RC Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

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