# BEAM BENDING Bending moments and shear stress diagrams

Mechanics

Jul 18, 2012 (5 years and 10 months ago)

1,110 views

Real Life Examples in Mechanics of Solids

24
BEAM BENDING

8. Principle
:
Bending moments and shear stress diagrams

Engage:
Ride a skateboard into class.
Explore:
Discuss the shear forces and bending moments set-up in
the skateboard when you stand on it sideways balanced
on your heels, i.e. approximating a point load. When
you stand of the board more normally, how do the shear
forces and bending moments change? Discuss where
you need to stand to induce a zero bending moment.

You might want to ask students to work in pairs to draw schematics of these loading
schemes.

Explain:
Plot the shear force and bending moment diagrams for the case where you were rocking
on your heels.

Considering the complete beam
Resolve vertically:
PRR
BA
=+

Moments about A:
0=− LRPa
B

Thus:
L
Pb
R
A
=

L
Pa
R
B
=

Considering the cut section (0<x<a)
Resolve vertically:
L
Pb
RF
A
==

Moments:
L
Pbx
xRM
A
==

Considering the cut section (a<x<l)
Resolve vertically:

L
Pa
P
L
Pb
PRF
A
−=−=−=

Moments:

( )
( )
xL
L
Pa
axPxRM
A
−=−−=

A
B
R
A
R
B
L
P
a
b
A
R
A
x
P
a
F
M
A
R
A
x
F
M
F
Pb
L
-Pa
L
x
Pab
L
M
x
A
B
R
A
R
B
L
P
a
b
A
R
A
x
P
a
F
M
A
R
A
x
F
M
A
B
R
A
R
B
L
P
a
b
A
R
A
x
P
a
F
M
A
R
A
x
P
a
F
M
A
R
A
x
F
M
A
R
A
x
F
M
F
Pb
L
-Pa
L
x
F
Pb
L
-Pa
L
x
Pab
L
M
x
Pab
L
M
x
Real Life Examples in Mechanics of Solids

25
Elaborate
When a skateboarder crosses a plank we can determine the position at which the bending
moment is a maximum. The situation can be idealized as shown below:

The shear force and bending moment diagrams can be plotted as previously considering
small sections of beam, i.e.
Taking moments about B gives:
( )
( )
0
22
=−−−−−
xL
P
axL
P
LR
A

Thus,
( )
2
a
xL
L
P
R
A
−−=

For
( )
xs <

Resolving vertically:
A
RF

=

Taking moments:
sRM
A

=

For
( )( )
xasx +<<

Resolving vertically:
A
R
P
F
−=
2

Taking moments:
( )
sRxs
P
M
A
−−=
2

For
( )( )
xas +>

Resolving vertically:
A
RPF

=

Taking moments:
sR
a
xsPM
A

−−=
2

N.B. These diagrams have been plotted
assuming that
(
)
22 Lax >
+
, if this
were not the case then the diagrams
would look slight different. The
symmetry of the situation allows only
this case to be considered. However a
There are two places where M
c
occurs
symmetric about the mid-point of the
beam.
P
A
B
a
x
L
P
A
B
a
x
L
A
R
A
s
F
M
A
R
A
s
F
M
A
R
A
s
P/2
x
F
M
A
R
A
s
P/2
x
F
M
A
R
A
s
P/2
x
F
M
P/2
a
A
R
A
s
P/2
x
F
M
P/2
a
F
s
M
C
M
s
A
B
R
A
R
B
a
P/2
x
L
P/2
s
F
s
M
C
M
s
A
B
R
A
R
B
a
P/2
x
L
P/2
s
F
s
M
C
M
s
A
B
R
A
R
B
a
P/2
x
L
P/2
s
Real Life Examples in Mechanics of Solids

26
For maximum M
C
:
0=

x
M
C
and
( )
axL
L
Px
xRM
AC
−−==
22
2

So,
042 =−−=

axL
x
M
C
and
42
aL
x
−=

Thus,
( )
2
2
164242
22
2
ˆ
aL
L
PaL
a
aL
L
L
P
M
C
−=

−−=

So for a 1.8m plank and typical skate board (a=65cm) carrying a 65kg person,
( )
11665.06.3
316
81.965
ˆ
2
=−
×
×
=
C
M Nm
If the plank is 13cm wide and 1.8cm thick, then the maximum bending stress is
( )
( )
5.16
018.013.0
1166
12
2
ˆˆ
23
=
×
×
===
bh
hM
I
yM
CC
σ MPa
This compares to compressive ultimate strengths for common woods in the range 35 to
55 MPa parallel to the grain and 4 to 10MPa perpendicular to the grain.
Real Life Examples in Mechanics of Solids

27
Evaluate
Example 8.1

Ask students to repeat the analysis above but for unicyclist crossing the plank.
Solution
:

Considering the complete beam
Resolve vertically:
PRR
BA
=+

Moments about A:
0=− LRPs
B

Thus:
(
)
L
sLP
R
A

=

L
Ps
R
B
=

Considering the cut section (0<x<s)
Resolve vertically:
( )
L
sLP
RF
A

==

Moments:
( )
L
xsLP
xRM
A

==

Considering the cut section (s<x<L)
Resolve vertically:

(
)
L
Ps
P
L
sLP
PRF
A
−=−

=−=

Moments:

( )
( )
xL
L
Ps
sxPxRM
A
−=−−=

For maximum bending moment:
0=

s
M
C
and
(
)
(
)
2
sLs
L
P
L
ssLP
xRM
AC
−=

==

So,
02 =−=

sL
s
M
C
and
2
L
s
=

Thus,
287
4
8.181.965
442
ˆ
22
=
××
==

−=
PLLL
L
P
M
C
Nm
A
B
R
A
R
B
L
P
s
A
R
A
x
P
s
F
M
A
R
A
x
F
M
F
P(L-s)
L
-Ps
L
x
M
C
M
x
A
B
R
A
R
B
L
P
s
A
R
A
x
P
s
F
M
A
R
A
x
P
s
F
M
A
R
A
x
F
M
A
R
A
x
F
M
F
P(L-s)
L
-Ps
L
x
F
P(L-s)
L
-Ps
L
x
M
C
M
x
Real Life Examples in Mechanics of Solids

28
( )
( )
41
018.013.0
2876
12
2
ˆˆ
23
max
=
×
×
===
bh
hM
I
yM
CC
σ MPa
Maximum stress of 41MPa induced when unicyclist at the middle. The position could
have been deduced without analysis.

Example 8.2

Ask students to look for two other examples in their everyday life and explain how the
above principles apply to each example.