SHEAR AND DIAGONAL TENSION – ULTIMATE STRENGTH
Strength of the PSC Beams without web Reinforcement
The mechanism of diagonal cracking associated with stress and the manner in which
shear reinforcement behaves are no more susceptible to theoretical analysis in prestressed
concrete beams than they are in reinforced concrete. Extensive testings have been done during
the period 1955 – 1965 and these results form the basis of the present AC1 formulas.
Test the results showed that there are basically two (2) types of shear failure in PSC beams:
1. Shear compression failure
2. Web distress failure.
The first is due to diagonal cracking associated with flexural cracks. The second is due to
cracking in regions without flexural cracks
Shear Compression Failure:
diagonal
Flexural
ctacks
Shear diagram
Moment diagram
Consider a section at the shear span:
M
Vcr
Mcr
V
0
d/2 d/2
d
= MO=0 M = Mcr + V
2
2 M Vd M d
Mcr = V
2 V 2
M cr
V= .
M d
V 2
6
5
4
3
2
1
1 2 3 4 5 6
Vci Vd
bw d fc 1
From test results:
Vci Vd M cr
0.60 .
bw d fc 1 M d
bw d fc
1
V 2
M cr
Or Vci = 0.6bwd fc 1 + + Vd (an AC1 formula)
M d
V 2
Where fc1 is in psi:
Where:
Mcr = net flexural cracking moment
M = bending moment due to externally applied load.
Vci = shear due to externally applied load when such cracking is the result of
combined shear and moment.
Vd = shear due to dead load
bw = minimum width of web.
V = shear due to LL. and DL.
The expression for Vci has been modified by dropping the term d/2 in the denomination, thus
M crV
Vci = 0.6 bwd fc 1 + Vd
M max
Or in SI Units
M cr
Vci = 0.05 bwd fc 1 + .V Vd *
Max
Where
Mcr =
I
Y
6 fc 1 f pe f d ; fc1 in psi.
Or
Mcr =
I
Y
0.5 fc 1 f pe f d ; fc1 in MPa *
Where
fpe = stress at extreme tensile fiber due to prestress
fd = stress at extreme tensile fiber due to dead load
Y = distance from the neutral axis to extreme tensile fiber
d = distance from compressive face to central of steel or 0.8h whichever is greater
To fit all test results to the equation, an upper limit has been suggested for V ci, thus
Vci the following:
Vci 1.7bwd fc 1 in psi.
Vci 0.14bwd fc 1 in MPa.
Web – Distress Failure
Consider the principal stress of an element at the neutral axis of a psc beams.
ft ft
From the Mohr Circle of Stress
• r
σ
ft
f1 f2
2
ft
R= v 2
2
ft
f2 = R-
2
2
ft f
v t
2
f2 =
2 2
2 2
f f
f2 t t v2
2 2
2
ft f2
V2 = f 2 2 +
f2
f
V = f2 1 t
f2
Assuming f2 = 3.5 fc 1 in psi.
ft
V = 3.5 fc 1 1
3.5 fc 1
This can be approximated by
V = 3.5 fc 1 +0.3ft
V
The nominal shear stress may also be used as a measure of diagonal tension for web
bw d
distress failure and the stress Vcw is given as
Vcw = bwd 3.5 fc 1 0.3 f pe
Or in SI units,
Vcw = bwd 0.29 fc 1 0.3 f pe *
If the cable is draped,
Vcw = bwd 3.5 fc 1 0.3 f pe V p
or
Vcw=bwd 0.29 fc 1 0.3 f pe V p in
SI units
Where Vcw= shear force at diagonal cracking due to all loads when such cracking is the result
of excessive principal tensile stresses in the web.
VP= vertical component of the effective prestressing force at the section considered..
fpe= longitudinal stress at n.a after losses
P
=n .
A
bw=m width of the web
d= depth of the centroid of steel or 0.80 h
whichever is greater
h= height of the beam
Design of Web Reinforcement
Let Vu= usable ultimate shear force
Vc= diagonal cracking is either Vci or Vcw whichever is smaller.
V1w= shear carried by the web reinforcement
Ø= reduction factor, 0.85 for shear
Vu= Ø (VC + V1w)
The average shear stress is obtained by dividing both sides of the above equation by b wd.
V 1u Av f y
bw d bw S
V1w= Avfy .
d
s
V c + Avfy .
Vu d
Then
s
V S
or Av= u Vc
f yD
The ACI code specifies the minimum Av in square inches as
bwS
AV= 0.34
fy
0r
A ps f pu S d
AV=
80 f y d bw
The ACI code specifies that
S>¾h
Or
S > 24 in. or 610mm.
Where fpu= ultimate tensile strength of prestressing wire
Fy= yield point stress of web reinforcement
S= spacing of web reinforcement
d= distance from extreme compressions fiber to the centroid of the steel at
the section of maximum moment
Av= total area of stirrups, (no. of legs tension area of one leg)
Aps= area of prestressing steel
Example:
For the beam shown, find the shear capacity at a section 1.5m. from the support. Use
the following data:
Aps=1483.5mm2
Fy=275.8 MPa
Fc= 34.5 MPa
fpu=1723 MPa
P= 142KN
=23.56 KN/m3
(Determine the stirrup spacing at 1.5m. from the support. Use 9.375mm. Ø u- stirrup)
w = 20.13 kN/m
125 mm 750 mm
250 mm
3m 6m 3m 500
Solution:
(A) Shear- Compression Failure:
Vi M cr
Vci= 0.05bd f 1c Vd
M max
I
Mcr= 0.5 f 1 c f pc f d
Y
1
I= x500 x750 3 17,578 x10 6 mm 4 .
2
750
Y= 375mm.; e 125mm.; A 500 x750
2
P P Y 1432 x10 3 1432 x10 3 x125 x375
f pe e
A I 500 x750 17.578 x10 6
f pe 3.818 3.818 7.636 MPa
WL2 20.13 x12 2
M max 362.34 KN
8 8
WD= 23.56x0.50 x 0.750 = 8.835KN/m.
Md
WD L
2
WD
x 2 WD
2
2
L 2
MD
8.835
2
12 x1.5 2 29.818 KNm.
M D y 29.818 x375
fd 0.636MPa
I 17.578 x10 6
M cr
17.578 x10 6
375
0.5 34.5 7.636 0.636
M cr 465.78 x10 6 KNmm 465.78 KNm
WD L L 12
Vd W D W D 8.835 1.5 39.75 KN
2 2 2
L 12
Vi W L W D 20.130 835 1.5 130.34 KN
2 2
0.05 x500 x500 130.34 x 465.78
Vci 34.5 39.75
3
10 362.34
Vci 73.42 167.55 39.75 280.72 KN
or
14.500 x500
Vci . 34.5 205.57 KN
103
Controlling Vci = 280.72 KN
(B) Web- distress failure
Vcw bw d 0.29 f 1 c 0.3 f pc VP
d= 375+125= 500mm. 0r d= 0.80x750= 680mm.
3
P 1432 x10
f pc 3.818MPa use d= 600mm.
A 500 x 250
125
V P P. x 1432 x10 3 x 119.333 KN
1500
Vcw
500 x500
103
0.29 34.5 0.3 x3.818 119.333
Vcw 854.4 119.333 973.733KN
Controlling Vc = Vci = 280.72 KN (Shear – compression failure)
Design of Stirrups:
Vu 1.4WD 1.7WL L
x
0.85 2
1.4 x8.835 1.7 x 20.13 12
=
1.5
0.85 2
=
12 . 369 34 . 221
4.5 246.65 KN Vc
0.85
bw s
Provide min AV=0.34 =
fy
AV f y
138 x 275.8
or S = 223mm.
0.34bw 0.34 x500
A f 80d 138 x 275.8 x80 x500
or S = V y 594.9mm.
d 500
A ps f pu 1483.5 x1723
bw 500
Use S = 223mm c/c.
S > 3/4h; S > ¾(750) = 562.5mm
S > ¾ h = S> ¾ (750)= 562.5mm.
Spacing of 223 mm d/c is OK
PSC Beams with Non- Prestressed Reinforcement 1
fc 1
b 0.85fc
cu .003
a/2
Bac=a
C C
dp
d
ds
Apsfps Apsfps
Aps
Asfy Asfy T
As
F H O
0.85 f1cba= Apsfps+ Asfy
A ps f ps d p As f y d s
d=
A ps f ps As f y
M 0
Mn= 0.85f1Cba(d-a/2)
or
Mn= (Apsfps+Asfy)(d-a/2)
or
Mn= Apsfps(dp-a/2)+ Asfy(ds-a/2)
Where fps has been defined before.
For bounded prestressing tendons:
f pu
Fps=fpu(1-0.5Pp )
f 1c
A ps
Where Pp =
bd p
For unbounded tendons
f 1c
fps= fpe+ 69+ in MPa
100 Pp
fps>fpy
fps>(fpe+414)MPa
T- Beams, I- Beams and box- Beams with Non- prestressed Reinforcements.
0.85fc1
b
hf a
bw dp d
d ds
Aps
Apsfps
As Asfy
FOR a> hf
F H 0 Apsfps+Asfy = 0.85fc1(b-bw)hf+0.85fc1bwa
A ps f ps As f y 0.85 fc1 (b bw )h f
a=
0.85 fc1bw
M 0
hf a
Mn = 0.85fc1(b-bw)hf(d- )+0.85fc1bwa(d- )
2 2
