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Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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order to complete this FBD, however, the vertical reaction force A y

and the moment reaction M A must be determined. Perhaps it

might be simpler to consider an FBD of the right portion of the

cantilever, since the reactions at fixed support A do not appear on

that FBD.

An FBD of the right portion of the cantilever beam is shown.

A common mistake at this stage of the analysis is to define the

beam length between section a–a and B as x. Note, however, that

the origin of the x–v coordinate system is located at support A,

with positive x extending to the right. Therefore, to be consistent

with the defined coordinate system, the length of the beam segment

must be denoted L − x. This simple coordinate transformation

is the key to success with this type of problem.

Accordingly, cut through the beam at section a–a, and consider

the beam and its loads between a–a and the free end of the

cantilever at B. Note that a clockwise internal moment M is

shown acting on the beam segment at a–a. Clockwise is the positive

direction for an internal moment acting on the left face of a

bending element, and this direction is consistent with the sign

convention shown in Figure 10.5.

The equilibrium equation for the sum of moments about a–a is

Σ = − −

⎛ L x

M

a−a

w( L x)

⎝ ⎠ − M

2

= 0

Therefore, the bending-moment equation for this beam is

v

M A

w

a

M

A

a

V

x

L – x

B

L

Free-body diagram of the left portion

of the cantilever beam.

v

w (L – x)

w

M

a

x

A y

A

a

V

L – x

B

Free-body diagram of the right portion

of the cantilever beam.

x

M

w

=− L − x

2 ( )2 (a)

Notice that this equation is valid for the interval 0 ≤ x ≤ L. Substitute the expression for

M into Equation (10.1) to obtain

EI d 2

v

2

dx

w

=− L − x

2 ( ) 2

(b)

integration

The first integration of Equation (b) gives

EI dv

dx

w

=+ L − x + C

6 ( )3 1 (c)

where C 1 is a constant of integration. Note the sign change on the first term. Integrating

again gives

w

EIv =− L − x + Cx + C

24 ( )4 1 2 (d)

where C 2 is a second constant of integration.

403

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