2-3 What is a Distributed Load?

1

2
-
3
What is
a

Distributed Load
?


In winter, a cantilever beam may deflect
because of a

heavy snow

load
. In summer, strong wind
s

may
push

the beam from below. We call
such load
s

'distributed load
s
'.
Log into
GOYA
-
C.

Type
0.5 in the text field of
‘
Amplifica
tion
’

to reduce the deformation. C
lick

'
A
dd load' nine times
and
move each load
to develop the

distribute
d

load as shown in Fig. 2
-
3
-
1
.

Note that the width of each
element

is 10 mm and the magnitude of each
load
is 10 N.
T
his represents
a
distributed load
of 1
N/mm. Though the shear
-
force diagram is
shown in
step
s

at intervals of 10 mm, you will have

a

smoother
diagram

if you distribute 100 loads of 1 N at intervals of 1 mm.




Fig. 2
-
3
-
1 Simulation of distributed load



2

Figure 2
-
3
-
2 shows the forces a
cting on
a
leng
t
h
dx

of

the beam
.
E
quilibrium in the vertical
direction requires
that
dV

=
w.dx

or


dV/dx

=
w





(2.3.1).

If
w

is constant (uniformly distributed load), integration of the above equation leads to
V

=
w.x

+
C
,
where
C

is constant. Because

the shear force is zero at the free end as shown in Fig. 2
-
3
-
1,
the
constant
C

should be zero and we have


V

=
w.x





(2.3.2).

Substituting this into
dM/dx

=
V

and noting that
M

= 0 at the free end (
x

= 0), we have


M

=
w.x
2
/2




(2.3.3).

Note that the b
ending moment diagram in Fig. 2
-
3
-
1 is almost parabolic. If we substitute
w

= 1
N/mm and
x

= 100 mm (the beam length), we have
V

= 100 N and
M

= 5000 N
-
mm

which agree
with the numbers indicated in Fig. 2
-
3
-
1 at the
fixed end
.

a
V
V+dV
dx
w

Fig. 2
-
3
-
2

Equilibrium of

dx


We may also obtain these equations without integration. Cut the beam as shown in Figs. 2
-
3
-
3a
and b, and represent the distributed load by an equivalent concentrated force as shown in Fig.
2
-
3
-
3c
.

T
hen, equilibrium in the horizontal direction leads
to Eq. 2.3.2 and
moment
equilibrium
leads to Eq. 2.3.3.


3


We can approximate the distributed force
by

a few concentrated forces as shown in Fig. 2
-
3
-
3d.
T
h
e broken and solid lines in Fig. 2
-
3
-
2e show the exact and approximate

shear force
diagrams,
respectively. They agree at the free end, mid
-
span, and at the fixed end. The bending moment
diagrams also agree at the same points as shown in Fig. 2
-
3
-
3f. Simulate the concentrated forces
shown in Fig. 2
-
3
-
3d on
GOYA
-
C
. You will find that the d
eflection is also similar to those shown
in Fig. 2
-
3
-
1.


Fig. 2
-
3
-
3 Beam with uniformly distributed load


Example 2
-
3
-
1
:
Assume that you have a beam with the same magnitude of distributed load but
double
the
length.
Determine

the shear force and the bendi
ng moment at
the
fixed end.



4

Solution
: According to Eqs. 2.3.2 and 2.3.3
, the shear force will double and the bending moment
will be four times

that in the previous example.


Example 2
-
3
-
2
:
Draw shear
-
force and bending
-
moment diagrams
caused by

the load
sh
own in

Fig. 2
-
3
-
4.

a
a
50
50
x
0.
2
N/mm

Fig. 2
-
3
-
4 Load on left half


Solution
:
We have
dV/dx

= 0.2 in the left half (0
<

x

<

50) and
V

= 0 at
x

= 0. So,



V

= 0.2
x



in 0
<

x

<

50.

Substituting this into
dM/dx

=
V

and noting
M

= 0 at the free end (
x

= 0)



M

= 0.1
x
2


in
0
<

x

<

50.

For 50
<

x

<

100,
dV/dx

= 0 yields
V

= constant. Considering the continuity of shear force, we
have the shear force diagram shown in Fig. 2
-
3
-
5a.
Because

V

= 10 in 50
<

x

<

100, the slope of
the bending moment should be 10 and we have the bendi
ng moment diagram shown in Fig. 2
-
3
-
5b.

Figure 2
-
3
-
5c shows the concentrated force equivalent to the distributed load. Figures 2
-
3
-
5d and e
show the approximated shear force and bending moment diagrams. Again, they agree with the
exact values at the free e
nd, mid
-
span, and at the fixed end.



5


Fig. 2
-
3
-
5 Solution


Example 2
-
3
-
3
:

Draw shear
-
force and bending
-
moment diagrams
caused by

the
force

in

Fig. 2
-
3
-
6.

Note that
w

is negative because the force

is downward.

a
a
100
x
0.
3
N/mm

Fig. 2
-
3
-
6 Linearly
varying
distributed load


Answer:

Because

the load is
w

=
-
(0.3/100)
x
, we shall integrate
dV/dx

=
-
(0.3/100)
x

noting
V

= 0
at the free end (
x

=

100) and have



15
10
15
4
2





x
V

Substituting this into
dM/dx

=
V

and noting
M

= 0 at
x

= 100


6



1000
15
10
5
4
3






x
x
M

The
se results are shown in Fig
s
. 2
-
3
-
7
a and b
.
Figure 2
-
3
-
7c shows the concentrated force
equivalent to the distributed load. Figures 2
-
3
-
7d and e show the approximate

shear
-
force and
bending
-
moment diagrams. They agree with the exact values at the free end
and at the fixed end.


Fig. 2
-
3
-
7 Solution


Example 2
-
3
-
4
:

Draw
shear
-
force and bending
-
moment diagrams
caused by

the load
s

shown
in

Fig. 2
-
3
-
8. (Hint: you may draw the diagrams caused by distributed load and those caused by
concentrated load separate
ly and add them.)

a
100
1
N/mm
4
0 N

Fig. 2
-
3
-
8
D
istributed load plus concentrated load



7

Solution
:
As we noted in
S
ection

2
-
2
, the shear force and bending moment diagrams caused by the
concentrated load (
-
40 N) are given as shown in Fig. 2
-
3
-
9. Those related to the distrib
uted load
are shown in Fig. 2
-
3
-
10. Adding (superposing) these diagrams, we obtain Figs. 2
-
3
-
11a and b.
Note that the shear force and the slope of the moment are zero at
x

=

40 mm.
Figure 2
-
3
-
5c shows
the concentrated forces equivalent to the distributed l
oad. Figures 2
-
3
-
5d and e show the
approximate

shear force and bending moment diagrams. They agree with the exact values at the
free end, mid
-
span, and at the fixed end.
Figure 2
-
3
-
12 shows the results obtained by
GOYA
-
C
.
Note
that

the deflection at the
beam end is only 3 mm, which is less than 1/10 of the deflection
caused by the distributed load shown in Fig. 2
-
3
-
1 (44 mm). The clue
to explain this result may be
found
in the bending
-
moment diagram, which shows that the beam deflects up near the fixed

end
and down near the middle. The upward deflection tends to reduce the downward deflection.





Fig. 2
-
3
-
9

Contribution of concentrated load

Fig. 2
-
3
-
10 Contribution of distributed load


8


Fig. 2
-
3
-
11 Solution


Fig. 2
-
3
-
12 Simulation using
GOYA
-
C


Exa
mple 2
-
3
-
5
:

Plot

shear
-
force and bending
-
moment diagrams caused by the load
in

Fig.
2
-
3
-
13
.

(Do not use
GOYA
-
C

before making your own calculations.)

a
40
60
1
N/mm
4
0 N


9

Fig. 2
-
3
-
13


Solution
:
T
he shear force and bending moment diagrams caused by the concentrated load (
-
40 N
)
are given as shown in Fig. 2
-
3
-
14. Adding this diagram to the one in Fig. 2
-
3
-
10, we obtain Fig.
2
-
3
-
15. Note that the shear force is discontinuous at
x

=

40 mm. Figure 2
-
3
-
16 shows the results
simulated by
GOYA
-
C
. Note
that

the deflection at the beam en
d is 24 mm, which is eight times
that obtained for the loading condition in Fig. 2
-
3
-
12 (3 mm). The clue lies again in the bending
moment diagram, which shows that the beam deflects upward throughout its length.
Approximate
the bending moment diagram using

concentrated forces equivalent to the distributed load and
compare with the exact diagram.





Fig. 2
-
3
-
14
Contribution of concentrated load



Fig. 2
-
3
-
15 Solution



10


Fig. 2
-
3
-
16 Deformation


We have distinguished the effects of a concentrated load from

those of a distributed load. However,
if we look at the experiment shown in Fig. 2
-
3
-
17 closely, we may notice that the finger applying
the load has a finite length in the direction of the span. The applied load should be
modeled

as a
distribut
ed load as
shown in Fig. 2
-
3
-
18a
. Such modeling leads to the shear
-
force diagram shown
in Fig. 2
-
3
-
18b
. In other words, the equation
dV/dx

=
w

also applies to concentrated loads.

a
Pencase
Desk
Push



a
a
(a
)
Distribute
d
load
(b
)
Shea
r
force
dx
dx
V

Fig. 2
-
3
-
17 Push a cantilever beam



Fig. 2
-
3
-
18 Distributed load


Example 2
-
3
-
6
:

Draw shear
-
force and bending
-
moment
diagrams
caused by

the
distributed
load
shown in

Fig. 2
-
3
-
19
.

a
a
2 m
[N/m]
2 m
x
w


10
sin

2
x






Fig. 2
-
3
-
19


11


Solution
: Integrating
the
distributed
load

leads to the shear force.

1
2
cos
20
C
x
dx
w
V














where
C
1

is an integral constant. Because
V

= 0 at the free e
nd (
x

= 0), we have



20
1


C

Figure 2
-
3
-
20a

shows the shear force diagram. Integrating the shear force leads to the bending
moment.

2
2
20
2
sin
40
C
x
x
dx
V
M
















where
C
2

is an integral constant. Because
M

= 0 at the free end (
x

= 0), we have
C
2

= 0.
Figure
2
-
3
-
20b

shows the bending
-
moment diagram. Because the diagram indicates that the bottom of the
beam is compressed, the beam deflects d
ownward as shown in Fig. 2
-
3
-
20c
.


Let us represent the distributed load by a pair of concentrate
d loads as s
hown in Fig. 2
-
3
-
21a
,
noting that




40
2
sin
2
0








dx
x

The shear force and bending moment diagram caused by these

loads are shown in Fig. 2
-
3
-
21b and
c
. They agree with the exact values at the free end,
at
mid
-
span and at the fixed end.


12






Fig. 2
-
3
-
20





Fig. 2
-
3
-
21

Example 2
-
3
-
7
:

Construct
the shear
-
force diagrams
and
evaluate

the

loads corresponding

to the
bending moment

in

Fig.
2
-
3
-
22
, where

]
N.m
[
2
cos
20







x
M



Solution
: Differentiating
the
bending moment leads to the shear
force.










x
dx
dM
V
2
sin
10



This is shown in Fig. 2
-
3
-
23a
. Differentiating
the
shear force
leads to the distributed load.










x
dx
dV
w
2
cos
5
2




Fig.
2
-
3
-
22


a
a
2 m
2 m
x
2
0
[N.m]
T
o
p
compression
Botto
m
compression
(a
)
Shea
r
force
(b
)
Load
(c
)
Deformation
10


[N]

Fig.
2
-
3
-
23


13

Because the bending moment at the free end is 20 N
-
m, there should be a couple at the end
,

and
the load should be as show
n in Fig.
2
-
3
-
23b
. The bending moment diagram indicates that the beam
deflects as shown in Fig.
2
-
3
-
23c
.


Design your own beam (Part3)
: We want to design a beam that can carry a mini elephant,
whose weight is (the last two figures of your ID#)/100 plus 1
kg. Assume that the gravity
acceleration is 10 m/s
2

and each leg carries the same amount of gravity force. The density of the
beam is 5 x 10
-
5

kg/mm
3
. The tensile force on the wall (
T

in the figure) must be smaller than 100 N.
What is the required beam dep
th if the beam
is prismatic (
its
section
remains the same
along its
span)

(Fig. 2
-
3
-
24a) or if it has varying depth
as specified in
Fig. 2
-
3
-
24b? Check your results
using GOYA
-
C.

Hint: If the depth is constant, the distributed load ca
used by the self
-
weight is

g
h
b
w





,

where


: density (
5 x 10
-
5

kg/mm
3
),

b
: beam width (10 mm),



h
: beam depth (unknown), and
g
: acceleration of gravity (
10 m/s
2
)

as shown in Fig.
2
-
3
-
24c
. The load
w

will vary if the beam depth varies. I
n both cases, you will
derive equations in terms of
h

and solve them. The depth required for case b (varying) is smaller
than that required for case a (uniform). You can find beams with varying depths in actual
structures such as
those supporting
balconies

or elevated highways.
GOYA
-
C

cannot simulate a
beam with varying depth. You can check your result by using a varying load as shown in
Fig.
2
-
3
-
24e

with the depth at the fixed end,

h
.


14



(e) Simulation by
GOYA
-
C

Fig.
2
-
3
-
24

Design your beam