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

The beam deflection at B can be calculated with equation (a); however,
that equation alone will not be sufficient to calculate the deflection at C.
For the beam considered here, the uniform load extends only between
A and B. There are no loads acting on the beam between B and C,
meaning that there will be no bending moment in the beam in that
region. Since there is no moment, the beam will not be bent (i.e.,
curved), so its slope between B and C will be constant. Because the
beam is continuous, its slope between B and C must equal the rotation
angle of the beam at B caused by the uniformly distributed load.
(Note: Since small deflections are assumed, the beam slope dv/dx is
equal to the rotation angle θ and the terms “slope” and “rotation angle”
will be used synonymously.)
From the beam table in Appendix C, the slope at the free end of
this cantilever beam is given as
v
A
80 kN/m
3 m
B
2 m
C
x
3
wL
θ max =− (b)
6EI
The beam deflection at C will be calculated from both Equation (a)
and Equation (b).
Problem-Solving tip: Before beginning the calculation, it is helpful to sketch the
deflected shape of the beam. Next, make a list of the variables that appear in the standard
equations along with the values applicable to the specific beam being analyzed.
Make sure that the units are consistent at this point in the process. In this example, for
instance, all force units will be expressed in terms of kilonewtons (kN) and all length
units will be stated in terms of meters (m). Making a simple list of the variables appearing
in the equations will greatly increase your likelihood of success, and it will
save you a lot of time in checking your work.
Beam deflection at B: Equation (a) will be used to compute the beam deflection at B. For
this beam, we have the following:
w =−80 kN/m
L = 3m
3 2
EI = 130 × 10 kN⋅m
Note: The distributed load w is negative in this instance because the
distributed load on the beam acts opposite to the direction shown in
the beam table. The cantilever span length L is taken as 3 m because
that is the length of the uniformly distributed load.
Substitute the preceding values into Equation (a) to find
v
B
4
=− =− − 4
wL ( 80kN/m)(3 m)
−3
= 6.231 × 10 m = 6.231 mm
3 2
8EI 8(130 × 10 kN⋅m)
v
A
Although the beam is rotated in
this region, it remains straight.
80 kN/m
3 m
B
v B
2 m
θ B
C
v C
x
The positive value indicates an upward deflection, as expected.
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