Transcript 5.8 SHEAR STRENGTH Where: – V

5.8 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 is
This equation based on the assumption that the stress is constant
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
W shape in the same manner as for the web.
The following Figure shows the shearing stress distribution for a W-shape.
The average stress in the web w is V/ Aw
No big difference between the average stress and the maximum
stress
The web will completely yield long before the flanges begin to
yield.
We can write the equation for the stress in the web at failure as:
Vn
fv 
 0.60Fy
Aw
Where Aw is the area of the web.
The nominal strength corresponding to this limit state is therefore
Vn  0.60FyA w
The relationship between shear strength and the width-thickness
ratio is analogous to that between flexural strength and the width
thickness ratio (for FLB or WLB) and between flexural strength
and unbraced length (for LTB)
This relationship is illustrated in the Figure below :
This relationship given in the AISC as follows:
h
E
For
 2.45
,
tw
Fy
There is no web instabilit y, and
Vn  0.60 Fy Aw
E
h
E
For 2.45
,
 3.07
Fy
tw
Fy
Inelastic web buckling can occur , and

E 
 2.45

Fy 

Vn  0.60 Fy Aw 

h


tw


For 3.07
E
h
, 
 260
Fy
tw
The limit state iselastic web buckling
Vn




 4.52 E 
 Aw 
2 
h  




 tw 




Where, Aw is the area of the web = d*tw
And d is the overall depth of the web.
If h/tw is greater than 260, web stiffeners are required.
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.
Example 5.7
Check the beam in Example 5.6 for shear.
Solution:
From Example 56: A W14X90 with Fy = 50 ksi is used.
Wu = 2.080 kips/ft and L= 40 ft
wuL 2.080 * 40
Vn 

 41.60 kips
2
2
From the Manual
h
 25.9
tw
E
29000
2.45
 2.45
 59.0
Fy
50
h
since
 2.45
tw
E
Fy
The strength is governed by shear yielding of the web
Vn  0.6Fy A w  0.60 * 50 * 14.0 * 0.440  184.8 kips
φVn  0.90 * 184.8  166 kips  41.6 kips
(OK)
The shear design strength is greater than the factored load shear,
so the beam is satisfactory.
BLOCK 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 Deflection
Steel beams are designed for the factored design loads. The moment
capacity, i.e., the factored moment strength (φbMn) should be greater
than the moment (Mu) caused by the factored loads.
A serviceable structure is one that performs satisfactorily, not causing
discomfort or perceptions of unsafety for the occupants or users of
the structure.
For a beam, being serviceable usually means that the deformations,
primarily the vertical sag, or deflection, must be limited.
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 specified values.
Appropriate limits for deflection can be found from the governing
building code.
For the common case of a simply supported, uniformly loaded
beam such as that in the following Figure, the maximum vertical
deflection is:
Deflection limit:
Example 5-9 :
solution
Ponding is one deflection problem that does affect the safety of a
structure.
The AISC specification requires that the roof system have
sufficient stiffness to prevent ponding.
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 moment
Noncompact shapes may fail due to local buckling
Two sets of curves are available, one for W-shapes and one for Cshapes 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 SYSTEMS
5.12 HOLES IN BEAMS
If beam connections are made with bolts, holes will be punched or
drilled in the beam web or flange.
Sometimes electrical conduits and ventilation ducts need large
holes.
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
bending moment.
This is not always be possible, so the effect of the holes must be
accounted for.
The effect of small holes will be small, particularly for flexure.
AISC B10 permits bolt holes in flanges to be ignored when:
0.75 Fu Afn ≥ 0.90 Fy Afg
where, Afg is the gross tension flange area
and, Afn is the net tension flange area.
If this condition is not met, the previous equation is solved for an
effective tension flange area that satisfies that criterion.
A fg  A fe 
0.75Fu A fn
0.9Fy
or
5 Fu
A fe 
A fn
6 Fy
See examples 5.14 and 5.15
5.13 OPEN-WEB STEEL JOISTS
Open-web steel joists are prefabricated trusses of the type shown
in the Figure.
They are used in floor and roof systems.
For a given span, it is lighter in weight and its more easier for
electrical conduits and ventilation ducts than a rolled shape .
More economical than a rolled shape
5.14 Bearing plates and column base plates :
– The function of the plate is to distribute a concentrated load to
the supporting material
– Two types of beam bearing plates are considered

One that transmits the beam reaction to a support

One that transmits a load to the top flange of a beam.
–The design of the bearing plate consists of three steps.

Determine dimension N so that web yielding and web
crippling are prevented

Determine Dimension B so that the area N ×B is sufficient to
prevent the supporting material from being crushed in
bearing.

Determine the thickness t so that the plate has sufficient
bending strength.
Web Yielding
Web yielding is the compressive crushing of a beam web caused by a force
acting on the flange directly above or below the web
When the load is transmitted through a plate, web yielding is assumed to take
place on the nearest section of width tw.
In rolled shape, this section will be at the toe of the fillet, a distance k from the
outside face of the flange
If 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) tw
The nominal strength for web yielding at the support is:
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 is
The design strength is ØRw where Ø = 1.0
Concrete Bearing Strength
Usually, concrete used as the material beam support.
This material must resist the bearing load applied by steel plate
If the plate covers the full area of the support,
'
The nominal strength is:
p
c
P  0.85f A1
If the plate does not cover the full area of the support
Where
Pp  0.85f A 1
'
c
A2
A1
fc is the compressive strength of concrete after 28 days
A1 is the bearing area
A2 is the full area of the support
If area A2 is not concentric with A1, then A2 should be taken as the
largest concentric area that is geometrically similar to A1, as
illustrated in the following Figure
AISC also requires
A2
2
A1
The design bearing strength is ФcPp where Фc=0.60
Plate Thickness
2.222R un 2
t
BNFy
Where
Ru is the support reaction
B is width of the bearing plate
N is length of the bearing plate
Example 5.17
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
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
wall with fc = 3500 psi. the beam made of A992 steel and the plate is
A36.
Solution
The 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 kips
Determine the length of bearing N required to prevent web yielding.
Ru = 2.5k + N) Fy tw = (2.5(1.438) + N) 50*0.430 ≥ 99.73
N ≥ 1.044
Determine the length of bearing N required to prevent web
crippling. Assume N/d > 0.2
1.5 

EFy t f
 4N
 t w 
2 

φ0.40 t w 1  
 0.2 
 Ru


  d

tw
 t f 


1.5 
  4N
29000 * 50 * 0.685
 0.43 
2

075 * 0.40 * 0.43  1  
 0.2 
 99.73

0.43
 0.685 
  21.1

N ≥ 3.0 in
Check the assumption
N/d = 3.0/21.1 = 0.14 < 0.2
(N.G.) so for N/d < 0.2
1.5 



EFy t f
 N  t w 
2 

φ0.40 t w 1  3 



t
tw
 d 


f





 N  0.43
2
075 * 0.40 * 0.43  1  3

21.1

 0.685


 Ru
1.5 






29000 * 50 * 0.685
N ≥ 2.59 in, and N/d = 2.59/21.1 = 0.12< 0.2
0.43
(OK)
 99.73
Try N = 6.0
Assume full area of support is used, the required plate area is
A = Ф * 0.85 * fc * A ≥ Ru
A = 0.60 * 0.85 * 3.5 * A ≥ 99.73
A ≥ 55.87 in2
B = 55.87/6 = 9.31 in
The flange width of a W21x68 is 8.27 < 9.31, use B as 10 in
Compute the required plate thickness, n = (B – 2k)/2
= (10 – 2*1.19)/2 = 3.81 in
t
2.222R un 2

BNFy
Use a PL 1.25 * 6 * 10.
2.222 * 99.73 * 3.81
10 * 6 * 36
2
 1.22 in
Column Base Plates
Major differences between bearing and base plates are:
Bending in bearing plates in one direction, whereas column base
plates are subjected to two-way bending.
Web crippling and web yielding are not factors in column base
plate design.
Design steps:
Determine the allowable strength of the foundation
Determine the required column base plate area
Select column base plate dimension (not less than column
dimension)
Determine the thickness
Where,
L  max (m, n n)
N - 0.95d
m
2
B  0.8b f
n
2
2 X

1
1 1 X
 4db f
X 
 d  b f
 Pu

2
  c Pp
2Pu
t L
0.9BNFy
1
n 
db
f
4
  0.60
c
Pp = nominal bearing strength from
AISC equation
'
p
c 1
P  0.85f A
Pp  0.85f A 1
'
c
A2
A1
Example 5.18
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
A36 base plate for a column dead load of 98 kips and a live load of 145
kips the concrete strength is fc = 3000 psi.
The factored load = 1.2 * 98 + 1.6 * 145 = 349.6
Compute the required bearing area (assume that plat area < pier
area)
A2
 Pu
A1
φ c (0.85)f A 1
'
c
0.6 * 085 * 3 * A 1
A 1  161.1 in
18 * 18
 349.6
A1
2
check
A2
18 * 18

 1.41  2
A1
161.1
(OK)
Also the plate must be at least as large as the column, so
Bfd = 10 * 9.98 = 99.8 < 161.1 in2
(OK)
For B = N = 13 in, A1 provided = 3 * 13 = 169 > 161.1 in2
Check the assumption
Plate area = 169 < pier area = 18 * 18
(OK)
Determine the required thickness
N - 0.95d 13  9.48
m

 1.76 in
2
2
B  0.8b f 13  8
n

 2.5 in
2
2
1
1
n 
db f 
9.98 * 10  2.497 in
4
4
As a conservative simplification, let λ =1, giving
L = max(m, n, λn’) = max (1.76, 2.5, 2.497) = 2.5 in
The required plate thickness is
2Pu
2 * 349.6
t L
 2.5
 0.893 in
0.9BNFy
0.9 * 13 * 13 * 6
Use a PL 1 x 13 x 13
5.15 BIAXIAL BENDING
Please try to understand it alone
May be necessary in some of your projects.
If there is a time, we will discuss it later
Any problem, you can see me in my office within the office hours.
The last section 5.16 canceled.
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