desing of beams [uyumluluk modu]kisi.deu.edu.tr/ozgur.ozcelik/ekonomi/arch...
TRANSCRIPT
Beams with Different Vertical Load Conditions
Triangular distributed load and concentrated load on an over-hang beam.
Deformation shape of a Cantilever Beam under
Bending Action
Deformed state – Positive Bending
“Almost” Undeformed state – No
bending
External Moment
Acting
Stress and Strain Profile of a Section under
Bending
Observations:
• Maximum strain and stress occur at the outer
most layers/fibers,
• Strain and stress profiles vary linearly with the
distance y from the centroid,
• There is no strain and stress at the neutral axis
(neutral axis is the centroid of the section),
• The value c defines the distance from the
centroid to the outer most layer.
Stress and Strain Profile of a Section under
Bending
Bending Formula
(General)
yI
M
z
x =σ
Bending Formula
(Maximum Stress at
a arbitrary section)
cI
M
z
=maxσ
Bending Formula (Absolute
Maximum Stress at a section
where max. mom. is observed)
cI
M
z
maxmax =σ
PRISMATIC BEAM DESIGN
• Basis of beam design
• Strength concern (i.e., provide safety margin to
normal/shear stress limit)
• Serviceability concern (i.e., deflection limit)
• Section strength requirement
allow
dreq
MS
σmax
' =
c
IS =
S: Section modulus
I: Moment of inertia about the
centroidal axis
c: half-height of the section
Ch
b
c
Cross Section View
PRISMATIC BEAM DESIGN (cont)
• Choices of section (refer to standard tables):
• Section modulus are given for example for
standard steel sections in codes such as AISC
standard
W 460 X 68
Height = 459 ≈ 460 mm
Weight = 0.68 kN/m
PRISMATIC BEAM DESIGN (cont)
• For other sections made of other materials such as
wood, you need to calculate it yourself,
Nominal dimensions (in multiple of 25mm) e.g. 50 (mm)
x 100 (mm) actual or “dressed” dimensions are smaller,
e.g. 50 x 100 is 38 x 89.
• Built-up sections
PRISMATIC BEAM DESIGN (cont)
Procedure:
• Shear and Moment Diagram
• Determine the maximum shear and moment in
the beam. Often this is done by constructing the
beam’s shear and moment diagrams.
• For built-up beams, shear and moment diagrams
are useful for identifying regions where the shear
and moment are excessively large and may
require additional structural reinforcement or
fasteners.
PRISMATIC BEAM DESIGN (cont)
• Normal Stress
– If the beam is relatively long, it is designed by
finding its section modulus and using the flexure
formula, Sreq’d = Mmax/σallow.
– Once Sreq’d is determined, the cross-sectional
dimensions for simple shapes can then be
computed, using Sreq’d = I/c.
– If standard steel sections are to be used, several
possible values of S may be selected from the
tables. Of these, choose the one having the smallest
cross-sectional area, since this beam has the least
weight and is therefore the most economical.
PRISMATIC BEAM DESIGN (cont)
• Normal Stress (cont.)
– Make sure that the selected section modulus, S, is
slightly greater than Sreq’d, so that the additional
moment created by the beam’s weight is
considered.
• Shear Stress
– Normally beams that are short and carry large
loads, especially those made of wood, are first
designed to resist shear and then later checked
against the allowable-bending-stress requirements.
– Using the shear formula, check to see that the
allowable shear stress is not exceeded; that is, use
τallow ≥ (Vmax /As), As is the shear area and is a
function of the cross sectional shape.
PRISMATIC BEAM DESIGN (cont)
• Shear Stress (cont.)
– If the beam has a solid rectangular cross section,
the shear formula becomes τallow ≥ 1.5(Vmax/A), and
if the cross section is a wide flange, it is generally
appropriate to assume that the shear stress is
constant over the cross-sectional area of the beam’s
web so that τallow ≥ Vmax/Aweb, where Aweb is
determined from the product of the beam’s depth
and the web’s thickness.
Web Area
Flange Area
Flange Area
PRISMATIC BEAM DESIGN (cont)
• Adequacy of Fasteners
– The adequacy of fasteners used on built-up beams
depends upon the shear stress the fasteners can
resist. Specifically, the required spacing of nails or
bolts of a particular size is determined from the
allowable shear flow, qallow = VQ/I, calculated at
points on the cross section where the fasteners are
located.
EXAMPLE 1
A beam is to be made of steel that has an allowable bending
stress of σallow = 170 MPa and an allowable shear stress of τallow
= 100 MPa. Select an appropriate W shape that will carry the
loading shown in Fig. a.
EXAMPLE 1 (cont)
( )( )( )( )( )( ) 33
33
33
33
33
33
mm 10987 100200
mm 10984 80250
mm 101060 74310
mm 101030 64360
mm 101200 67410
mm 101120 60460
=×
=×
=×
=×
=×
=×
SW
SW
SW
SW
SW
SW
m3m
= 706x103 mm3
EXAMPLE 1 (cont)
• The beam having the least weight per metre is chosen, W460 x 60
• The beam’s weight is
• From Appendix B (tables), for a W460 x 60, d = 455 mm and tw = 8 mm.
• Thus
• The beam is designed to be a W460 x 60.
Solutions
( )( ) kN 60.36/60.0 == mmkNW
( )( )( )
(OK) MPa 100 MPa 7.248455
10900 3
max <===web
avgA
Vτ
EXAMPLE 2
The laminated wooden beam shown in Fig. 11–8a supports a
uniform distributed loading of 12 kN/m. If the beam is to have a
height-to-width ratio of 1.5, determine its smallest width.
The allowable bending stress is 9 MPa and the allowable shear
stress is 0.6 MPa. Neglect the weight of the beam.
EXAMPLE 2 (cont)
• Applying the flexure formula,
• Assuming that the width is a,
the height is 1.5a.
Solutions
( )( )
3
6
3
max m 00119.0109
1067.10===
allow
req
MS
σ
( )( )( )
m 147.0
m 003160.0
75.0
5.100119.0
33
3
121
=
=
===
a
a
a
aa
c
ISreq
Mmax=
= Vmax
EXAMPLE 2 (cont)
• Applying the shear formula for rectangular sections,
• Since the design fails the shear criterion, the beam must be redesigned
on the basis of shear.
• This larger section will also adequately resist the normal stress.
Solutions
( )( )
( )( )( ) MPa6.0929.0
147.0147.05.1
10205.15.1
3
maxmax >===
A
Vτ
( )( )
(Ans) mm 183 m 183.0
5.1
10205.1600
5.1
3
max
==
=
=
a
aa
A
Vallowτ