# Design of Beams and Shafts

Urban and Civil

Nov 15, 2013 (4 years and 6 months ago)

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Chapter Objectives

Design a beam to resist both bending and shear loads

Design a shaft to resist both bending and torsional
moments

1.

Prismatic beam design

2.

Fully stressed beam design

3.

Shaft design

In
-
class Activities

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

See
Chapter 12)

Section strength requirement

allow
d
req
M
S

max
'

PRISMATIC BEAM DESIGN (cont)

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)

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)

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
fasteners.

PRISMATIC BEAM DESIGN (cont)

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)

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
, 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)

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)

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

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

7
a
.

EXAMPLE 1 (cont)

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)

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

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)

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)

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

The wooden T
-
beam shown in Fig. 11

9
a
200mm x 30mm boards. If the allowable bending stress is 12
MPa and the allowable shear stress is 0.8 MPa, determine if 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)

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)

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)

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)

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)

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

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

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)

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)

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

Consider the effects of moments and torsion

SHAFT DESIGN (cont)

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)

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

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)

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)

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