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

A
v
v
30 kips
a= 13 ft b=
7 ft
B
C D
4 ft 6 ft 3 ft 7 ft
L=
20 ft
30 kips
a=
13 ft b =7 ft
v C
E
x
x
Case 1—30 kip load on Simple Span
It is essential to note the interval upon which the elastic curve
equation is applicable. Equation (a) gives the beam deflection
at any distance x from the origin up to, but not past, the location
of the concentrated load. That location is denoted by the
term a in the equation. For this beam, a = 13 ft. Since point C
is located at x = 10 ft, the elastic curve equation is applicable
to this case.
The deflected shape of the beam is shown. List the variables
that appear in the elastic curve equation along with their
corresponding values:
P = 30 kips
b = 7ft = 84in.
L = 20 ft = 240 in.
x =10 ft
EI
= 15.022 × 10 kip⋅in.
6 2
L=20 ft
Beam deflection at C: At point C, x = 10 ft = 120 in. Therefore,
the beam deflection at C for this case is
v
C
Pbx
=− − −
6LEI L 2 b 2 x 2
( )
(30 kips)(84 in.)(120 in.)
2 2 2
=−
[(240 in.) − (84 in.) − (120 in.) ]
6 2
6(240 in.)(15.022 × 10 kip⋅in. )
=−0.5053 in.
A
v
20 kips
a = 4 ft b=
16 ft
B
C D
4 ft 6 ft 3 ft 7 ft
L=
20 ft
20 kips
b = 4 ft a = 16 ft
E
v
x
Case 2—20 kip load on Simple Span
Next, consider the simply supported beam with only the
20 kip load. From this sketch, it is apparent that the distance a
from the origin to the point of application of the load is a = 4 ft.
Since C is located at x = 10 ft, the elastic curve equation is not
applicable to this case, because x > a.
However, the elastic curve equation can be used for this
case if we make a simple transformation. The origin of the x–v
coordinate axes will be repositioned at the right end of the
beam, and the positive x direction will be redefined as extending
toward the pin support at the left end of the span. With
this transformation, x < a and the elastic curve equation can
be used.
The variables that appear in the elastic curve equation
and their corresponding values are as follows:
x
A
B
C D
4 ft 6 ft 3 ft 7 ft
L=
20 ft
E
P = 20 kips
b = 4ft = 48in.
L = 20 ft = 240 in.
EI
= 15.022 × 10 kip⋅in.
6 2
430