Chapter Objectives
Design a beam to resist both bending and shear
loads
A Bridge Deck under Bending Action
Castellated Beams
Post-tensioned Concrete Beam
Lateral Distortion of a Beam Due to Lateral Load
Action
T-Beam Roof on a Office Building
Modern Steel Frame Building
Joint in steel frame building
Beams with Different Boundary/Support
Conditions
Beams with Different Vertical Load Conditions
Triangular distributed load and concentrated load on an over-hang beam.
Beams with Different Vertical Load Conditions
Uniform distributed load a simply supported beam.
Internal Actions of a simply supported beam
loaded with uniform load
Internal Actions of a cantilever beam loaded
with triangular distributed load
Vmax
Mmax
Internal Actions of a simply-supported beam
loaded with an arbitrary loading
Vmax
Mmax
Deformations and Stresses of a Beam Under
Bending Effect
Bending Moment
Undeformed state – No bending
Deformed state – Negative
Bending
Deformation shape of a Cantilever Beam under
Bending Action
“Almost” Undeformed state – No
bending
External Moment
Acting
Deformed state – Positive Bending
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)
Bending Formula
(Maximum Stress at
a arbitrary section)
M
M
c
σx =
y σ max =
Iz
Iz
Bending Formula (Absolute
Maximum Stress at a section
where max. mom. is observed)
σ max
M max
=
c
Iz
Cross Section
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
Cross Section View
S req 'd =
M max
b
σ allow
c
I
S=
c
S: Section modulus
I: Moment of inertia about the
centroidal axis
c: half-height of the section
C
h
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.
Flange Area
Web 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)
m
m3
= 706x103 mm3
W 460 × 60
S = 1120(103 ) mm3
W 410 × 67
S = 1200(103 ) mm3
W 360 × 64
S = 1030(103 ) mm3
W 310 × 74
S = 1060(103 ) mm3
W 250 × 80
S = 984(103 ) mm3
W 200 × 100
S = 987(103 ) mm3
EXAMPLE 1 (cont)
Solutions
• The beam having the least weight per metre is chosen, W460 x 60
• The beam’s weight is
W = (0.60kN / m )(6m ) = 3.60 kN
• From Appendix B (tables), for a W460 x 60, d = 455 mm and tw = 8 mm.
• Thus
τ avg
Vmax 900(103 )
=
=
= 24.7 MPa < 100 MPa (OK)
Aweb (455)(8)
• The beam is designed to be a W460 x 60.
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)
Solutions
• Applying the flexure formula,
S req =
M max
σ allow
( )
( )
10.67 103
3
=
=
0
.
00119
m
9 106
= Vmax
• Assuming that the width is a,
the height is 1.5a.
Sreq
I
= = 0.00119 =
c
1
12
(a )(1.5a )3
(0.75a )
a 3 = 0.003160 m3
a = 0.147 m
Mmax=
EXAMPLE 2 (cont)
Solutions
• Applying the shear formula for rectangular sections,
τ max
Vmax
20(103 )
= 1.5
= (1.5)
= 0.929 > 0.6 MPa
(1.5)(0.147)(0.147)
A
• Since the design fails the shear criterion, the beam must be redesigned
on the basis of shear.
Vmax
τ allow = 1.5
A
20 103
600 = 1.5
a(1.5a )
a = 0.183 m = 183 mm (Ans)
( )
• This larger section will also adequately resist the normal stress.