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

Note: Throughout the previous chapters in this book, the symbol v has been used to denote
deflection perpendicular to the longitudinal axis of a beam. In this chapter, velocity
has been introduced as an additional consideration, and the symbol v is also used to denote
velocity. In this example, both beam deflections and the velocity of a block are being
considered. The context of the problem and the subscripts used with the symbol v clearly
indicate whether a deflection or a velocity is intended; however, the reader is cautioned to
examine the context in which the symbol v is used.
By the conservation of energy, the work that is performed on the post must equal the
kinetic energy of the block:
1 1
P v = mv
2 2
max max 0 2
Note: On the left-hand side of this equation, v max is a displacement. On the right-hand
side, v 0 is a velocity.
Therefore, the maximum velocity of the block must not exceed
v
0
Pmaxvmax
(435.2 N)(0.013865 m)
= = = 0.448 m/s Ans.
m
30 kg
This problem can also be solved with the impact factor given in Equation (17.26).
The weight of the block is (30 kg)(9.807 m/s 2 ) = 294.2 N. If this force were gradually
applied horizontally to the post at B, the static deflection would be
v
st
3 3
PL st
(294.2 N)(850 mm)
= = = 9.373 mm
2 4
3EI 3(200,000 N/mm )(32,127.7 mm )
The impact factor can be calculated from the static and dynamic deflections at B:
vmax
13.866 mm
n = = =
v 9.373 mm 1.479
st
From Equation (17.26), the impact factor for a weight moving horizontally is
n =
v
gv
0 2 st
This equation can be solved for the maximum velocity v 0 :
v
0
=
2
n gv
st
2 2
= (1.479) (9.807 m/s )(0.009373 m)
= 0.448 m/s
Ans.
742