Chapter Objectives
Design a beam to resist both bending and shear loads
Design a shaft to resist both bending and torsional
moments
Copyright © 2011 Pearson Education South Asia Pte Ltd
1.
Prismatic beam design
2.
Fully stressed beam design
3.
Shaft design
In

class Activities
Copyright © 2011 Pearson Education South Asia Pte Ltd
PRISMATIC BEAM DESIGN
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Basis of beam design
•
Strength concern (i.e. provide safety margin to
normal/shear stress limit)
•
Serviceability concern (i.e. deflection limit
–
See
Chapter 12)
•
Section strength requirement
allow
d
req
M
S
max
'
PRISMATIC BEAM DESIGN (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Choices of section:
•
Steel sections e.g. AISC standard
W 460 X 68
(height = 459 ≈ 460mm
weight = 0.68 kN/m)
PRISMATIC BEAM DESIGN (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Wood sections
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)
Copyright © 2011 Pearson Education South Asia Pte Ltd
Procedures:
•
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)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Normal Stress
–
If the beam is
relatively long
, it is designed by
finding its section modulus using the flexure formula,
Sreq’d = Mmax/
σ
allow.
–
Once Sreq’d is determined, the cross

sectional
dimensions for simple shapes can then be
computed, using Sreq’d = I/c.
–
If rolled

steel sections are to be used, several
possible values of S may be selected from the
tables in Appendix B. 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)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
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
τ
allow
≥ V
max
Q/It.
PRISMATIC BEAM DESIGN (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Shear Stress (cont)
–
If the beam has a
solid rectangular cross section
,
the shear formula becomes
τ
allow ≥ 1.5(V
max
/A),
Eq.7

5, 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
τ
allow
≥ V
max
/A
web
, where A
web
is determined from the product of the beam’s depth
and the web’s thickness.
PRISMATIC BEAM DESIGN (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
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, q
allow
= VQ/I, calculated at
points on the cross section where the fasteners are
located.
EXAMPLE 1
Copyright © 2011 Pearson Education South Asia Pte Ltd
A beam is to be made of steel that has an allowable bending
stress of
σ
allow = 170 MPa and an allowable shear stress of
τ
allow
= 100 MPa. Select an appropriate W shape that will carry
the loading shown in Fig. 11
–
7
a
.
EXAMPLE 1 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
The required section modulus for the beam is determined from the
flexure formula,
Solutions
3
3
3
3
3
3
3
3
3
3
3
3
mm
10
987
100
200
mm
10
984
80
250
mm
10
1060
74
310
mm
10
1030
64
360
mm
10
1200
67
410
mm
10
1120
60
460
S
W
S
W
S
W
S
W
S
W
S
W
EXAMPLE 1 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
The beam having the least weight per foot is chosen, W460 x 60
•
The beam’s weight is
•
From Appendix B, for a W460 x 60,
d =
455 mm and
t
w
=
8 mm.
•
Thus
•
Use a W460 x 60.
Solutions
kN
55
.
3
2
.
3552
6
81
.
9
35
.
60
W
(OK)
MPa
100
MPa
7
.
24
8
455
10
900
3
max
w
avg
A
V
EXAMPLE 2
Copyright © 2011 Pearson Education South Asia Pte Ltd
The laminated wooden beam shown in Fig. 11
–
8
a
supports a
uniform distributed loading of 12kN/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)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Applying the flexure formula,
•
Assuming that the width is
a
,
the height is 1.5a.
Solutions
3
6
3
max
m
00119
.
0
10
9
10
67
.
10
allow
req
M
S
m
147
.
0
m
003160
.
0
75
.
0
5
.
1
00119
.
0
0
3
3
3
12
1
a
a
a
a
a
c
I
S
req
EXAMPLE 2 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
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
MPa
6
.
0
929
.
0
147
.
0
5
.
1
147
.
0
10
20
5
.
1
5
.
1
3
max
max
A
V
(Ans)
mm
183
m
183
.
0
5
.
1
10
20
5
.
1
600
5
.
1
3
max
a
a
a
A
V
allow
EXAMPLE 3
Copyright © 2011 Pearson Education South Asia Pte Ltd
The wooden T

beam shown in Fig. 11
–
9
a
is made from two
200mm x 30mm boards. If the allowable bending stress is 12
MPa and the allowable shear stress is 0.8 MPa, determine if the
beam can safely support the loading shown. Also, specify the
maximum spacing of nails needed to hold the two boards
together if each nail can safely resist 1.50 kN in shear.
EXAMPLE 3 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
The reactions on the beam are shown, and the shear and moment
diagrams are drawn in Fig. 11
–
9
b
.
•
The neutral axis (centroid) will be located from
the bottom of the beam.
Solutions
m
1575
.
0
2
.
0
03
.
0
2
.
0
03
.
0
2
.
0
03
.
0
251
.
0
2
.
0
03
.
0
1
.
0
A
A
y
y
EXAMPLE 3 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Thus
•
Since c = 0.1575 m,
Solutions
4
6
2
3
2
3
m
10
125
.
60
1575
.
0
215
.
0
2
.
0
03
.
0
03
.
0
2
.
0
12
1
1
.
0
1575
.
0
2
.
0
03
.
0
2
.
0
03
.
0
12
1
I
(OK)
Pa
10
24
.
5
10
125
.
60
1575
.
0
10
2
10
12
6
6
3
9
max
I
c
M
allow
EXAMPLE 3 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Maximum shear stress in the beam depends upon the magnitude of
Q
and
t
.
•
We will use the rectangular area below the neutral axis to calculate
Q
,
rather than a two

part composite area above this axis, Fig. 11
–
9
c
•
So
.
Solutions
3
3
m
10
372
.
0
03
.
0
1575
.
0
2
1575
.
0
'
A
y
Q
(OK)
Pa
10
309
03
.
0
10
125
.
60
10
372
.
0
10
5
.
1
10
800
3
6
3
3
3
max
It
Q
V
allow
EXAMPLE 3 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
From the shear diagram it is seen that the shear varies over the entire
span.
•
Since the nails join the flange to the web, Fig. 11
–
9
d
, we have
•
The shear flow for each region is therefore
Solutions
3
3
m
10
0.345
03
.
0
0.2
015
.
0
0725
.
0
'
A
y
Q
kN/m
74
.
5
10
125
.
60
10
345
.
0
10
1
kN/m
61
.
8
10
125
.
60
10
345
.
0
10
5
.
1
6
3
3
6
3
3
I
Q
V
q
I
Q
V
q
CD
CD
BC
BC
EXAMPLE 3 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
One nail can resist 1.50 kN in shear, so the maximum spacing becomes
•
For ease of measuring, use
Solutions
m
261
.
0
74
.
5
5
.
1
m
174
.
0
61
.
8
5
.
1
CD
BC
s
s
(Ans)
mm
250
(Ans)
mm
150
CD
BC
s
s
FULLY STRESSED BEAMS
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
The strength requirements are not the same for every
cross sections of the beam.
•
To optimize the design so as to reduce the beam’s
weight,
non

prismatic beam
is a choice.
•
A beam designed in this manner is called a
fully
stressed beam
; e.g.
EXAMPLE 4
Copyright © 2011 Pearson Education South Asia Pte Ltd
Determine the shape of a fully stressed, simply supported beam
that supports a concentrated force at its center, Fig. 11

11
a
.The
beam has a rectangular cross section of constant width
b
, and
the allowable stress is
σ
allow
.
EXAMPLE 4 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
The internal moment in the beam, Fig. 11
–
11
b
, expressed as a function
of position,
•
The required section modulus is
•
For a cross

sectional area
h
by
b
we have
Solutions
x
P
M
2
x
P
M
S
allow
allow
2
x
b
P
h
x
P
h
bh
c
I
allow
allow
3
2
2
/
2
3
12
1
EXAMPLE 4 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
If
•
By inspection, the depth
h
must therefore vary in a
parabolic
manner with
the distance
x
.
Solutions
L/2,
at x
0
h
h
(Ans)
2
2
3
2
0
2
2
0
x
L
h
h
b
PL
h
allow
SHAFT DESIGN
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Consider the effects of moments and torsion
SHAFT DESIGN (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Resultant moment: (algebraic sum of M
x
and M
y
)
•
Normal stress and shear stress
2
2
y
x
M
M
M
J
Tc
I
Mc
and
SHAFT DESIGN (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
For solid shaft, I =
π
c
4
/4 and J =
π
c
4
/2.
2
2
2
allow
3
/
1
2
2
2
2
3
2
2
T
M
c
or
T
M
c
allow
allow
EXAMPLE 5
Copyright © 2011 Pearson Education South Asia Pte Ltd
The shaft in Fig. 11
–
14
a
is supported by smooth journal
bearings at
A
and
B
. Due to the transmission of power to and
from the shaft, the belts on the pulleys are subjected to the
tensions shown. Determine the smallest diameter of the shaft
using the maximum

shear

stress theory, with
τ
allow
= 50 MPa
EXAMPLE 5 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
The resultant moment is
•
At
B
it is smaller,
Solutions
m
N
5
.
124
5
.
37
75
.
118
2
2
C
M
m
N
75
B
M
EXAMPLE 5 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
•
Since the design is based on the maximum

shear

stress theory, Eq. 11
–
2 applies.
•
Thus, the smallest allowable diameter is
Solutions
m
0117
.
0
5
.
7
5
.
124
10
50
2
2
3
/
1
2
2
6
3
/
1
2
2
T
M
c
allow
(Ans)
mm
3
.
23
01177
.
0
2
d
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