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

ML
θ =
6EI
By the values defined previously, the magnitude of the rotation
angle at B is
ML (3,840 kip⋅in.)(192 in.)
θ B = =
= 0.0020177 rad
6 2
6EI 6(60.9 × 10 kip⋅in. )
(b)
v A
A
8 ft
v
B
θB
C
D
320 kip.ft
8 ft 8 ft 8 ft
40 kips
E
x
Beam deflection at A: By inspection, the rotation angle at B must be positive; that is, the
beam slopes upward to the right at the pin support. Since there is no bending moment in
overhang span AB, the beam will not bend between A and B. Its slope will be constant and
equal to θ B . The magnitude of the beam deflection at A is computed from the beam slope:
v = θ L = (0.0020177 rad)(96 in.) = 0.1937 in.
A B AB
By inspection, the overhang will deflect downward at A; therefore, v A = −0.1937 in.
Ans.
Case 3—Downward Deflection of overhang DE
The downward deflection of point E on the overhang span is computed from two considerations.
First, consider a cantilever beam subjected to a concentrated load at its free end.
The deflection at the tip of the cantilever is given by the equation
v
max
3
PL
=− (c)
3EI
v
v
40 kips
From the values
and
P = 40 kips
L = 8ft = 96in.
A
8 ft
B C D
8 ft 8 ft 8 ft
E
v E
x
6 2
EI = 60.9 × 10 kip⋅in.
one component of the beam deflection at E can be computed as
v
E
3 3
PL (40 kips)(96 in.)
=− =−
6 2
3EI 3(60.9 × 10 kip⋅in. )
=−0.1937 in.
(d)
As discussed previously, this cantilever beam case does not account for all of the deflection
at E. Specifically, Equation (c) assumes that the cantilever beam does not rotate at its
support. However, since center span BD is flexible, overhang DE rotates downward as the
center span bends. The magnitude of the rotation angle of the center span caused by the
concentrated moment M can be computed from the following equation:
ML
θ =
3EI
Note: Equation (e) gives the beam rotation angle at the location of
M for a simply supported beam subjected to a concentrated moment
applied at one end. With the values defined for case 2, the rotation
angle of the center span at roller support D can be calculated as
ML (3,840 kip⋅in.)(192 in.)
θ D = =
= 0.0040355 rad
6 2
3EI 3(60.9 × 10 kip⋅in. )
(e)
A
8 ft
v
θ
D
320 kip.ft
B C D
8 ft 8 ft 8 ft
40 kips
E
v E
x
433