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.

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