SHEAR DESIGN OF A BEAM WITH OVERHANG
CIVL4390 Structural Design 2 [November 15th 2010]
QUESTION:
Design the beam shown below against shear. The component is subjected to a live load of 29.7kN/m
and a dead load, including self weight, of 15.7kN/m. Use a compressive strength of 20MPa and
reinforcement yield strength of 400MPa. The beam will not be exposed to earth or weather during its
service life expectancy. The beam contains reinforcement at an effective depth of 434mm and was
designed to a height of 500mm and a width of 300mm.
wf
A
B
a=300mm
l=10,000mm
SOLUTION:
STEP 1:
Calculate factored load:
wf = 1.25wDL + 1.5wLL = 1.25(15.7kN/m) + 1.5(29.7kN/m)
wf = 64.2kN/m ◄
STEP 2:
Draw the shear force envelope:
The first step in deriving the shear force envelope is to draw the diagram for a simply supported beam,
with overhang, subjected to the full distributed loading wf. The sign convention in shear design is not a
primary concern since positive and negative shear forces require the same shear reinforcing spacing as
well as a similar vertical reinforcement location and orientation.
Vf
Shear Envelope for Design
(Influence Lines)
Zero Shear @ Midspan
(Full Loading Condition)
wf (l2-a2)/(2l)
wf a
x
5.45m
4.55m
3m
-wf (l2+a2)/(2l)
Instructor : H.M.Vogel
Page | 1 of 6
SHEAR DESIGN OF A BEAM WITH OVERHANG
CIVL4390 Structural Design 2 [November 15th 2010]
Moving the live load along the span of the beam will change the zero shear force condition at 4.55 m
to that of a non-zero shear force condition. An influence line for the shear at 4.55m should therefore be
drawn to determine the maximum shear force that can occur at this location. Placing a roller and
applying a vertical shear condition draw the influence line. The deflected shape that results is a visual
representation of the shear force that arises at 4.55m from moving a unit load along the length of the
beam:
1
0.55
4.55m
3m
3m
10m
5.45 m
-0.30
0.3
-0.45
Influence Line @ 4.55 mm
Influence Line for Left Reaction
According to the influence line, the maximum negative shear at 4.55m occurs when the live load is
placed on the left portion (4.55m) of the beam span in addition to the overhang (3m). The maximum
positive shear at 4.55m occurs when the live load is placed on the right half (5.45m) of the beam span
between the supports. The actual magnitude of these shear forces is obtained by multiplying areas
under the influence line that correspond to these regions by the factored live load wfLL.
Vfms,neg = -0.45(4.55m)(1/2)wfLL -0.30(3m)(1/2)wfLL
Vfms,neg = -0.45(4.55m)(1/2)(44.6kN/m) -0.30(3m)(1/2) (44.6kN/m)
Vfms,neg = -65.7kN ◄
Vfms,pos = 0.55(5.45m)(1/2)wfLL
Vfms,pos = 0.55(5.45m)(1/2)(44.6kN/m)
Vfms,pos = 66.8kN ◄
The shear force values at the edge of the supports can be conservatively connected to those at midspan
such that a shear envelope such as the one shown on the previous page can be obtained. The shear
force at the support is given by:
Vfsupport = 1.0(10m)(1/2)wfLL + wf DL(l2-a2)/(2l)
Vfsupport = 1.0(10m)(1/2)(44.6kN/m) + (19.6kN/m)((10m)2-(3m)2)/(2(10m))
Vfsupport = 312.2kN ◄
STEP 3:
Determine whether the size of the section is sufficient as well as the factored shear resistance of
concrete:
VRmax = 0.25φcλf'cbwdv/103
VRmax = 0.25(0.65)(1.0)(20MPa)(300mm)(360mm)/103
VRmax = 351kN > Vfd @ all sections so the section is adequate to resist shear
Instructor : H.M.Vogel
Page | 2 of 6
SHEAR DESIGN OF A BEAM WITH OVERHANG
CIVL4390 Structural Design 2 [November 15th 2010]
Since we are going to provide transverse reinforcement (stirrups) to resist shear forces within the
section, the value for β can be taken as 0.18 when calculating the contribution of concrete to shear
resistance Vc:
Smallest of:
0.9d = 0.9(434) = 391 mm
0.72h = 0.72(500) = 360 mm
Vc = φcλβ√f'cbwdv/103
Vc = (0.65)(1.0)(0.18)(√20MPa)(300mm)(360mm)/103
Vc = 56.5kN
The simplified method for shear design presented in the CSA A23.3-04 code requires stirrups to be
provided whenever the factored shear Vf occurring at a distance equal to dv from the support exceeds
half the concrete contribution to shear resistance 0.5Vc. The factored shear at dv from support A can be
determined from the shear envelope diagram on the basis of similar triangles:
Factored Shear Force at dv:
[Vf - Vfms,pos]/4.55m = [Vfd - Vfms,pos]/[4.55m-dv]
Vfd = Vfms,pos + [4.55m-dv][Vf - Vfms,pos]/4.55m
Vfd = 66.8kN + [4.55-0.36m][312.2kN-66.8kN]/4.55m
Vfd = 292.8kN ◄
Vf
Vf = 312.2kN
Vfd = 292.8kN (Critical Section)
Vms = 66.8 kN
x
dv = 360mm
4.55m-dv
Since shear force values at midspan and at the critical section are beyond 0.5Vc (28.3kN), stirrups will
be required along the entire length of the beam. The next step will consist of determining the
maximum spacing of stirrups for the design process.
STEP 4:
Determine maximum spacing requirements:
Based on Section Depth: Evaluate:
Vfs = 0.125φcλf'cbwdv/103
Vfs = 0.125(0.65)(1.0)(20MPa)(300mm)(360mm)/103 = 175.5kN
Since Vfd>Vfs smax is the Smallest of:
[1] 0.35dv = 0.35(360) = 126mm ◄ governs – 125mm say
[2] 300mm
Based on Minimum Area: Av > Av,min = 0.06√f'cbws/fy
s < fyAv / [0.06√f'cbw] = (400MPa)(200mm2) / [0.06√20MPa(300mm)]
s < 993.8mm
Instructor : H.M.Vogel
Page | 3 of 6
SHEAR DESIGN OF A BEAM WITH OVERHANG
CIVL4390 Structural Design 2 [November 15th 2010]
Based on values reported from consideration of section depth and minimum shear reinforcing area
requirements, the maximum center-to-center spacing should be taken as 125mm whenever Vf > Vfs
along the beam. It should be noted that the maximum spacing requirements based on beam depth are
doubled (~0.7dv or 600mm) whenever Vf < Vfs. We therefore need to determine where this happens
along the beam before the actual spacing is evaluated:
Distance where Vf = Vfs:
STEP 5:
[Vf - Vfms,pos]/4.55m = [Vfs - Vfms,pos]/[x]
x = 4.55m[Vfs - Vfms,pos]/[Vf - Vfms,pos]
x = 4.55m[175.5kN - 66.8kN]/[312.2kN - 66.8kN]
x = 2.02m from ~midspan towards the left support ◄
Calculate required spacing:
The design for spacing starts at the critical section where shear is equal to that obtained at dv from the
support (292.3kN). We require the shear resistance VR = Vc + Vs to exceed the factored shear at this
section:
VR > Vfd
Vc + Vs > Vfd
Vc + φsfyAvdvcotθ/s > Vfd
s < φsfyAvdvcotθ/[Vfd - Vc] = (0.85)(400MPa)(200mm2)(360mm)(cot35o)/[292.3kN-56.5kN]
s < 148.3mm ◄
Since the spacing requirements of 148.3mm are larger than those for maximum spacing based on beam
depth, we can adopt a spacing of 120mm until a distance of x=2.02m from ~midspan is reached, at
which point a spacing of 250mm (~0.7dv) can be used. We have the following zones:
Vf
Vfd = 292.8kN (Critical Section)
Vf = 175.5kN
Maximum Spacing Commencement
Vf = 312.2kN
Vms = 66.8kN
dv = 360mm
Instructor : H.M.Vogel
ZONE I
ZONE II
2175mm
2015mm
x
Page | 4 of 6
SHEAR DESIGN OF A BEAM WITH OVERHANG
CIVL4390 Structural Design 2 [November 15th 2010]
ZONE II:
l2 = 2015mm – 250mm/2 = 1890mm
no.stirrups = l2/s +1 = 1890mm/250mm +1 = 8.56 stirrups say 8 stirrups (rounding down)
l2,new = s/2 + (no.stirrups -1)s = 250mm/2 + (7)(250mm) = 1875mm
8-10M Stirrups @ 250mm o.c.
ZONE I:
l1 = 4550mm – dv/2 – l2 = 4550mm – 360mm/2 – 1875mm = 2495mm
no.stirrups = l1/s +1 = 2495mm/120mm +1 = 21.8 stirrups say 22 stirrups (rounding up)
l1,new = (no.stirrups -1)s = (21)(120mm) = 2520mm
22-10M Stirrups @ 120mm o.c.
95mm
STIRRUP LAYOUT:
22-10M Stirrups
@ 120mm o.c.
60mm
30-10M Stirrups
@ 120mm o.c.
16-10M Stirrups
@ 250mm o.c.
11-10M Stirrups
@ 250mm o.c.
155mm
440mm
a=3000mm
l=10,000mm
It should be noted that the stirrup layout shown above also includes the design of the second region
(5.45m before point B) as well as the third region (overhang). Steps 3 through to 6 have been repeated
for these regions with the following shear forces and distances at the critical as well as maximum
spacing sections:
Vf
312.2kN
dv = 360mm
(No Stirrups)
292.8kN 175.5kN
192.6kN
66.8kN
ZONE I
dv = 360mm
2175mm
Instructor : H.M.Vogel
ZONE II
2015mm
-65.7kN
2106mm
2984mm
ZONE II
ZONE I
175.5kN
0.5Vc=28.3kN
x
440mm (No Stirrups)
2294mm (ZONE II)
266mm (<dv - No Stirrups)
349.9kN
175.5kN
331.1kN
Page | 5 of 6
SHEAR DESIGN OF A BEAM WITH OVERHANG
CIVL4390 Structural Design 2 [November 15th 2010]
STEP 6:
Check anchorage of tensile reinforcement at the supports (s=120mm and dv=360mm
everywhere): Note that the tension reinforcement at support B is placed at the top.
dv
Sum of moments about Cc:
Cc
dv/2
Vf
Inclined
Crack
Ts
Vs
dv
► Ts ≥ Vf -0.5Vs
Ts ≥ Vf - 0.5(0.85)(200)(400)(360)/(120(103)tan35o)
Ts≥ 166.5kN (support A)
204.2kN (support B - left)
46.9kN (support B - right)
► Ts≤ φsAsfy
As ≥ Ts / φsfy
As ≥ 489.7mm2 (support A)
600.6mm2 (support B - left)
137.9mm2 (support B - right)
You must ensure that tensile steel reinforcement in
excess of these values is provided at all these sections to
prevent anchorage failure.
Instructor : H.M.Vogel
Page | 6 of 6