5.8 SHEAR STRENGTH5.8 SHEAR STRENGTH
Where:
–Vu = maximum shear based on the controlling combination of
factored loads
–Ø = resistance factor for shear = 0.90
–Vn = nominal shear strength
Consider a simple beam as shown in
Fig. a. At a distance x from the left
end and at the neutral axis of the
cross section, the state of stress is as
shown in Fig. d. Because this element
is located at the neutral axis, it is not
subjected to flexural stress.
From elementary mechanics of materials, the shearing stress isFrom elementary mechanics of materials, the shearing stress is
This equation based on the assumption that the stress is constant This equation based on the assumption that the stress is constant
across the width b, and it is therefore accurate only for small across the width b, and it is therefore accurate only for small
values of b, so the equation can not be applied to the flange of a values of b, so the equation can not be applied to the flange of a
W shape in the same manner as for the web.W shape in the same manner as for the web.
The average stress in the web w is V/ AThe average stress in the web w is V/ Aww
No big difference between the average stress and the maximum No big difference between the average stress and the maximum
stressstress
The web will completely yield long before the flanges begin to The web will completely yield long before the flanges begin to
yield.yield.
The following Figure shows the shearing stress distribution for a W-shape.The following Figure shows the shearing stress distribution for a W-shape.
We can write the equation for the stress in the web at failure as:We can write the equation for the stress in the web at failure as:
Where AWhere Aww is the area of the web. is the area of the web.
The nominal strength corresponding to this limit state is thereforeThe nominal strength corresponding to this limit state is therefore
The relationship between shear strength and the width-thickness The relationship between shear strength and the width-thickness
ratio is analogous to that between flexural strength and the ratio is analogous to that between flexural strength and the
width thickness ratio (for FLB or WLB) and between flexural width thickness ratio (for FLB or WLB) and between flexural
strength and unbraced length (for LTB) strength and unbraced length (for LTB)
wAy0.60FnV
y0.60FwAnV
vf
This relationship is illustrated in the Figure below :This relationship is illustrated in the Figure below :
This relationship given in the AISC as follows:This relationship given in the AISC as follows:
wyn
yw
AFV
andyinstabilitwebnoisThere
F
E
t
hFor
60.0
,
,45.2
w
y
wyn
ywy
t
h
F
E
AFV
andoccurcanbucklingwebInelastic
F
E
t
h
F
EFor
45.2
60.0
,
07.3,45.2
2
52.4
260,07.3
w
wn
wy
t
h
EAV
t
h
F
EFor
bucklingwebelasticisstatelimitThe
Where, AWhere, Aww is the area of the web = d*t is the area of the web = d*tww
And d is the overall depth of the web.And d is the overall depth of the web.
If h/tIf h/tw w is greater than 260, web stiffeners are required.is greater than 260, web stiffeners are required.
Shear is rarely a problem in rolled steel beams; the usual practice Shear is rarely a problem in rolled steel beams; the usual practice
is to design a beam for flexure and then to check it for shear.is to design a beam for flexure and then to check it for shear.
Example 5.7Example 5.7
Check the beam in Example 5.6 for shear.Check the beam in Example 5.6 for shear.
Solution:Solution:
From Example 56: A W14X90 with FFrom Example 56: A W14X90 with Fyy = 50 ksi is used. = 50 ksi is used.
WWuu = 2.080 kips/ft and L= 40 ft = 2.080 kips/ft and L= 40 ft
kips41.602
40*2.0802
LuwnV
From the ManualFrom the Manual
0.5950
2900045.245.2
9.25
y
w
F
E
t
h
The strength is governed by shear yielding of the webThe strength is governed by shear yielding of the web
yw F
E2.45
t
hsince
(OK)kips41.6kips166184.8*0.90 nφV
kips184.80.440*14.0*50*0.60wAy0.6FnV
The shear design strength is greater than the factored load shear, The shear design strength is greater than the factored load shear,
so the beam is satisfactory.so the beam is satisfactory.
BLOCK SHEARBLOCK SHEAR
To facilitate the connection of beams to other beams so that the
top flanges are at the same elevation, a short length of the top
flange of one of the beams may be cut away, or coped. If a
coped beam is connected with bolts as in Figure, segment ABC
will tend to tear out.
The applied load in this case will be
the vertical beam reaction.
Shear will occur along line AB and
there will be tension along BC. Thus
the block shear strength will be a
limiting value of the reaction.
We covered the computation of block shear strength in Chapter3
Example 5-8
5.9 Deflection5.9 Deflection
Steel beams are designed for the factored design loads. The moment Steel beams are designed for the factored design loads. The moment
capacity, i.e., the factored moment strength (φcapacity, i.e., the factored moment strength (φbbMMnn) should be greater than ) should be greater than
the moment (Mthe moment (Muu) caused by the factored loads. ) caused by the factored loads.
A serviceable structure is one that performs satisfactorily, not causing A serviceable structure is one that performs satisfactorily, not causing
discomfort or perceptions of unsafety for the occupants or users of the discomfort or perceptions of unsafety for the occupants or users of the
structure. structure.
For a beam, being serviceable usually means that the deformations, primarily For a beam, being serviceable usually means that the deformations, primarily
the vertical sag, or deflection, must be limited.the vertical sag, or deflection, must be limited.
The maximum deflection of the designed beam is checked at the service-level The maximum deflection of the designed beam is checked at the service-level
loads. The deflection due to service-level loads must be less than the loads. The deflection due to service-level loads must be less than the
specified values.specified values.
Appropriate limits for deflection can be found from the governing building Appropriate limits for deflection can be found from the governing building
code. code.
For the common case of a simply supported, uniformly loaded For the common case of a simply supported, uniformly loaded beam such as that in the following Figure, the maximum vertical beam such as that in the following Figure, the maximum vertical
deflection is:deflection is:
Deflection limit:Deflection limit:
Example 5-9 :Example 5-9 :
solutionsolution
Ponding is one deflection problem that does affect the safety of a Ponding is one deflection problem that does affect the safety of a
structure.structure.
The AISC specification requires that the roof system have The AISC specification requires that the roof system have
sufficient stiffness to prevent ponding.sufficient stiffness to prevent ponding.
5.10 DESIGN:5.10 DESIGN:
Beam design entails the selection of a cross-sectional shape
that will have enough strength and that will meet
serviceability requirements.
The design process can be outlined as follows:
1. Compute the factored load moment.
2. Select a shape that satisfies this strength requirement.
This can be done in one of two ways:
Assume a shape, compute the design strength, and
compare it with the factored load moment.
Use beam design chart in part 5 at Manual (preferred).
3. Check the shear strength.
4. Check the defection.
Beam Design Charts::
– Many graphs, charts, and tables are available for the
practicing engineer, and these aids can greatly simplify the
design process.
– To determine the efficiency, they are used in design offices.
– You should approach their use with caution and not allow
basic principles to become obscured.
– The curves of design moment versus un braced length given
in Part 5 of the Manual.
– All curves were generated with Fy =50 ksi and Cb = 1.0.
– For other values of Cb simply multiply the design moment
from the chart by Cb.
The following curve described the design moment ФbMn as a function of unbraced length Lb for a particular compact shape.
Remember that design strength can never exceed plastic momentRemember that design strength can never exceed plastic moment
Noncompact shapes may fail due to local buckling
Two sets of curves are available, one for W-shapes and one for C-shapes and MC-
shapes.
Example 5.12
Use A992 steel and select a rolled shape for the beam shown
below. The concentrated load is a service live load, and the
uniform load is 30% dead load and 70% live load. Lateral
bracing is provided at the ends and at mid span. There is no
restriction on deflection.
5.11 FLOOR AND ROOF FRAMING SYSTEMS5.11 FLOOR AND ROOF FRAMING SYSTEMS
5.12 HOLES IN BEAMS5.12 HOLES IN BEAMS
If beam connections are made with bolts, holes will be punched or If beam connections are made with bolts, holes will be punched or
drilled in the beam web or flange.drilled in the beam web or flange.
Sometimes electrical conduits and ventilation ducts need large Sometimes electrical conduits and ventilation ducts need large
holes.holes.
Ideally, holes should be placed in the web only at section of low Ideally, holes should be placed in the web only at section of low
shear, and holes should be made in the flanges at points of low shear, and holes should be made in the flanges at points of low
bending moment.bending moment.
This is not always be possible, so the effect of the holes must be This is not always be possible, so the effect of the holes must be
accounted for.accounted for.
The effect of small holes will be small, particularly for flexure.The effect of small holes will be small, particularly for flexure.
AISC B10 permits bolt holes in flanges to be ignored when: AISC B10 permits bolt holes in flanges to be ignored when:
0.75 F0.75 Fuu A Afnfn ≥ 0.90 F ≥ 0.90 Fyy A Afgfg
where, Awhere, Afg fg is the gross tension flange areais the gross tension flange area
and, Aand, Afn fn is the net tension flange area.is the net tension flange area.
If this condition is not met, the previous equation is solved for an If this condition is not met, the previous equation is solved for an
effective tension flange area that satisfies that criterion. effective tension flange area that satisfies that criterion.
fnAyF
uF
6
5feA
or
y0.9F
fnAu0.75F
feAfgA
See examples 5.14 and 5.15See examples 5.14 and 5.15
5.13 OPEN-WEB STEEL JOISTS5.13 OPEN-WEB STEEL JOISTS
Open-web steel joists are prefabricated trusses of the type shown Open-web steel joists are prefabricated trusses of the type shown
in the Figure.in the Figure.
They are used in floor and roof systems.They are used in floor and roof systems.
For a given span, it is lighter in weight and its more easier for For a given span, it is lighter in weight and its more easier for
electrical conduits and ventilation ducts than a rolled shape .electrical conduits and ventilation ducts than a rolled shape .
More economical than a rolled shape More economical than a rolled shape
5.14 Bearing plates and column base plates :5.14 Bearing plates and column base plates :
– The function of the plate is to distribute a concentrated load to The function of the plate is to distribute a concentrated load to
the supporting materialthe supporting material
– Two types of beam bearing plates are considered Two types of beam bearing plates are considered
One that transmits the beam reaction to a supportOne that transmits the beam reaction to a support
One that transmits a load to the top flange of a beam.One that transmits a load to the top flange of a beam.
–The design of the bearing plate consists of three steps.The design of the bearing plate consists of three steps.
Determine dimension N so that web yielding and web Determine dimension N so that web yielding and web
crippling are prevented crippling are prevented
Determine Dimension B so that the area N Determine Dimension B so that the area N ××B is sufficient to B is sufficient to
prevent the supporting material from being crushed in prevent the supporting material from being crushed in
bearing.bearing.
Determine the thickness t so that the plate has sufficient Determine the thickness t so that the plate has sufficient
bending strength.bending strength.
Web YieldingWeb Yielding
Web yielding is the compressive crushing of a beam web caused by a force Web yielding is the compressive crushing of a beam web caused by a force
acting on the flange directly above or below the webacting on the flange directly above or below the web
When the load is transmitted through a plate, web yielding is assumed to take When the load is transmitted through a plate, web yielding is assumed to take
place on the nearest section of width tplace on the nearest section of width tww..
In rolled shape, this section will be at the toe of the fillet, a distance k from the In rolled shape, this section will be at the toe of the fillet, a distance k from the
outside face of the flangeoutside face of the flange
If the load is assumed to distribute itself at a slop of 1:2.5 as shownIf the load is assumed to distribute itself at a slop of 1:2.5 as shown
The area at the support subject to yielding is (2.5k + N) tThe area at the support subject to yielding is (2.5k + N) tw w
The nominal strength for web yielding at the support is:The nominal strength for web yielding at the support is:
The bearing length N at the support should not be less than k.The bearing length N at the support should not be less than k.
At the interior load, the length of the section subject to yielding isAt the interior load, the length of the section subject to yielding is
The design strength is The design strength is ØRwØRw where where Ø = 1.0Ø = 1.0
Concrete Bearing StrengthConcrete Bearing Strength
Usually, concrete used as the material beam support.Usually, concrete used as the material beam support.
This material must resist the bearing load applied by steel plateThis material must resist the bearing load applied by steel plate
If the plate covers the full area of the support, If the plate covers the full area of the support,
The nominal strength is:The nominal strength is:
If the plate does not cover the full area of the supportIf the plate does not cover the full area of the support
WhereWhere
ffcc is the compressive strength of concrete after 28 days is the compressive strength of concrete after 28 days
AA11 is the bearing area is the bearing area
AA22 is the full area of the support is the full area of the support
1'cp A0.85fP
1
21
'cp A
AA0.85fP
If area AIf area A22 is not concentric with A is not concentric with A11, then A, then A22 should be taken as the should be taken as the
largest concentric area that is geometrically similar to Alargest concentric area that is geometrically similar to A11, as , as
illustrated in the following Figureillustrated in the following Figure
AISC also requiresAISC also requires
The design bearing strength is The design bearing strength is ФФccPPpp where where ФФcc=0.60 =0.60
21
2
A
A
Plate ThicknessPlate Thickness
WhereWhere
RRuu is the support reaction is the support reaction
B is width of the bearing plateB is width of the bearing plate
N is length of the bearing plateN is length of the bearing plate
y
2u
BNF
n2.222Rt
Example 5.17Example 5.17
Design a bearing plate to distribute the reaction of a W 21 x 68 with a Design a bearing plate to distribute the reaction of a W 21 x 68 with a
span length of 15 ft 10 inches center to center of supports. The total span length of 15 ft 10 inches center to center of supports. The total
service load, including the beam weight , is 19 kips with equal parte service load, including the beam weight , is 19 kips with equal parte
dead and live load. The beam is to be supported on reinforced concrete dead and live load. The beam is to be supported on reinforced concrete
wall with fwall with fcc = 3500 psi. the beam made of A992 steel and the plate is = 3500 psi. the beam made of A992 steel and the plate is
A36.A36.
SolutionSolution
The factored load is = 1.2*4.5 + 1.6*4.5 = 12.600 kips/ftThe factored load is = 1.2*4.5 + 1.6*4.5 = 12.600 kips/ft
The reaction is = 12.6*15.83/2 = 99.3 kipsThe reaction is = 12.6*15.83/2 = 99.3 kips
Determine the length of bearing N required to prevent web yielding.Determine the length of bearing N required to prevent web yielding.
RRuu = 2.5k + N) F = 2.5k + N) Fyy t tw w = (2.5(1.438) + N) 50*0.430 ≥ 99.73= (2.5(1.438) + N) 50*0.430 ≥ 99.73
N ≥ 1.044N ≥ 1.044
Determine the length of bearing N required to prevent web Determine the length of bearing N required to prevent web crippling. Assume N/d > 0.2 crippling. Assume N/d > 0.2
N ≥ 3.0 inN ≥ 3.0 inCheck the assumptionCheck the assumptionN/d = 3.0/21.1 = 0.14 < 0.2 (N.G.) so for N/d < 0.2N/d = 3.0/21.1 = 0.14 < 0.2 (N.G.) so for N/d < 0.2
99.730.43
0.685*50*290001.5
0.685
0.430.2
21.1
4N1
20.43*0.40*075
uRwt
ftyEF1.5
ft
wt0.2d
4N12
wt0.40φ
99.730.43
0.685*50*290001.5
0.685
0.43
21.1
N1
20.43*0.40*075
uRwt
ftyEF1.5
ft
wt
d
N12
wt0.40φ
3
3
N ≥ 2.59 in, and N/d = 2.59/21.1 = 0.12< 0.2 (OK)N ≥ 2.59 in, and N/d = 2.59/21.1 = 0.12< 0.2 (OK)
Try N = 6.0Try N = 6.0
Assume full area of support is used, the required plate area isAssume full area of support is used, the required plate area is
A = A = ФФ * 0.85 * f * 0.85 * fcc * A ≥ R * A ≥ Ru u
A = 0.60 * 0.85 * 3.5 * A ≥ 99.73A = 0.60 * 0.85 * 3.5 * A ≥ 99.73
A ≥ 55.87 inA ≥ 55.87 in22
B = 55.87/6 = 9.31 inB = 55.87/6 = 9.31 in
The flange width of a W21x68 is 8.27 < 9.31, use B as 10 inThe flange width of a W21x68 is 8.27 < 9.31, use B as 10 in
Compute the required plate thickness, n = (B – 2k)/2 Compute the required plate thickness, n = (B – 2k)/2
= (10 – 2*1.19)/2 = 3.81 in = (10 – 2*1.19)/2 = 3.81 in
Use a PL 1.25 * 6 * 10.Use a PL 1.25 * 6 * 10.
in1.22
36*6*10
3.81*99.73*2.222
BNF
n2.222R t
2
y
2u
Column Base PlatesColumn Base Plates
Major differences between bearing and base plates are:Major differences between bearing and base plates are:
Bending in bearing plates in one direction, whereas column base Bending in bearing plates in one direction, whereas column base
plates are subjected to two-way bending.plates are subjected to two-way bending.
Web crippling and web yielding are not factors in column base Web crippling and web yielding are not factors in column base
plate design.plate design.
Design steps:Design steps:
Determine the allowable strength of the foundationDetermine the allowable strength of the foundation
Determine the required column base plate area Determine the required column base plate area
Select column base plate dimension (not less than column Select column base plate dimension (not less than column
dimension)dimension)
Determine the thicknessDetermine the thickness
Where, Where, y
u
0.9BNF
2PL t
pc
u
f
f
f
P
P
bd
dbX
X
X
bBn
2
4
111
2
2
8.02
0.95d-Nm
)nn(m,maxL
60.04
1
c
fdbn
Pp = nominal bearing strength from Pp = nominal bearing strength from AISC equation AISC equation
1'cp A0.85fP
1
21
'cp A
AA0.85fP
Example 5.18Example 5.18
A W 10 x 49 is used as a column and is supported by a concrete pier as A W 10 x 49 is used as a column and is supported by a concrete pier as
shown in the Figure. The top surface of pier is 18 in by 18 in. Design an shown in the Figure. The top surface of pier is 18 in by 18 in. Design an
A36 base plate for a column dead load of 98 kips and a live load of 145 A36 base plate for a column dead load of 98 kips and a live load of 145
kips the concrete strength is fkips the concrete strength is fcc = 3000 psi. = 3000 psi.
The factored load = 1.2 * 98 + 1.6 * 145 = 349.6The factored load = 1.2 * 98 + 1.6 * 145 = 349.6
Compute the required bearing area (assume that plat area < pier Compute the required bearing area (assume that plat area < pier
area)area)
(OK)21.41161.1
18*18
A
A
check
in161.1A
349.6A
18*18A*3*085*0.6
PA
AA(0.85)fφ
1
2
21
11
u1
21
'cc
Also the plate must be at least as large as the column, soAlso the plate must be at least as large as the column, so
BBffd = 10 * 9.98 = 99.8 < 161.1 ind = 10 * 9.98 = 99.8 < 161.1 in22 (OK) (OK)
For B = N = 13 in, AFor B = N = 13 in, A11 provided = 3 * 13 = 169 > 161.1 in provided = 3 * 13 = 169 > 161.1 in22
Check the assumptionCheck the assumption
Plate area = 169 < pier area = 18 * 18 (OK)Plate area = 169 < pier area = 18 * 18 (OK)
Determine the required thicknessDetermine the required thickness
in2.49710*9.984
1db
4
1n
in2.52
813
2
0.8bBn
in1.762
9.4813
2
0.95d-Nm
f
f
in0.8936*13*13*0.9
349.6*22.5
0.9BNF
2PL t
y
u
As a conservative simplification, let As a conservative simplification, let λλ =1, giving =1, giving
L = max(m, n, L = max(m, n, λλn’) = max (1.76, 2.5, 2.497) = 2.5 inn’) = max (1.76, 2.5, 2.497) = 2.5 in
The required plate thickness isThe required plate thickness is
Use a PL 1 x 13 x 13Use a PL 1 x 13 x 13
5.15 BIAXIAL BENDING5.15 BIAXIAL BENDING
Please try to understand it alonePlease try to understand it alone
May be necessary in some of your projects.May be necessary in some of your projects.
If there is a time, we will discuss it laterIf there is a time, we will discuss it later
Any problem, you can see me in my office within the office hours.Any problem, you can see me in my office within the office hours.
The last section 5.16 canceled.The last section 5.16 canceled.
لله“ ومماتي ومحياي نسكي و صالتي ان لله“ قل ومماتي ومحياي نسكي و صالتي ان قل
وأنا * أمرت وبذلك له شريك ال العالمين وأنا * رب أمرت وبذلك له شريك ال العالمين رب
" العظيم الله صدق المسلمين " أول العظيم الله صدق المسلمين أول